TOPOLOGICAL INSULATORS AND QUANTUM HALL EFFECTS
1 1. STABILITY OF EDGE STATES IN TOPOLOGICAL INSULATORS
Problem 1. Consider a 1D topological insulator with edge states described by the Hamiltonian
H=0t
t∗0, where tis a complex parameter.
a) Find the eigenvalues and eigenvectors of H.
b) Determine the conditions for the existence of edge states at the boundaries x= 0 and x=L.
c) Show that the edge states are robust against local perturbations.
Solution 1.
a) The eigenvalues λof the Hamiltonian Hare the solutions to the characteristic equation
det(H−λI)=0:
det −λ t
t∗−λ=λ2− |t|2= 0
Thus, the eigenvalues are λ=±|t|. The corresponding eigenvectors are 1
eiθand 1
−eiθ,
where θis the argument of t.
b) For edge states to exist, the eigenvalues must be real. Therefore, |t|2must be positive,
implying that t= 0. When t= 0, the system is in a trivial insulating state.
c) Edge states are robust against local perturbations because any perturbation terms added to
the Hamiltonian will not be able to change the topological nature of the edge states. This robustness
is a consequence of the bulk-edge correspondence in topological insulators.
2 2. QUANTUM HALL EFFECT IN MULTILAYERED SYSTEMS
Problem 2. Consider a multilayered system consisting of two layers with different Hall conductivi-
ties σ(1)
xy = 3.5×10−4e2/h and σ(2)
xy =−2.1×10−4e2/h. The total Hall conductivity of the system
is given by σxy =σ(1)
xy +σ(2)
xy .
a) Calculate the total Hall conductivity of the system.
b) If the magnetic field is B= 0.5T and the density of states per layer is ν= 2 ×1011 cm−2,
what is the Hall voltage VHacross the system?
c) If the system is connected to an external battery with a constant current density j= 5 mA/cm2
flowing through it, calculate the transverse electric field Eyin the system.
Solution 2.
a) The total Hall conductivity of the system is given by:
σxy =σ(1)
xy +σ(2)
xy = 3.5×10−4−2.1×10−4= 1.4×10−4e2/h
b) The Hall voltage VHacross the system is given by:
VH=1
nq ×σxy ×B
where nis the electron density per layer. Given that the density of states per layer is ν= 2 ×
1011 cm−2, we can calculate the electron density per layer as:
n=ν×104= 2 ×1011 ×104= 2 ×1015 m−2
Substitute the values into the equation:
VH=1
2×1015 ×1.6×10−19 ×1.4×10−4×0.5 = 218.75 V
Therefore, the Hall voltage VHacross the system is 218.75 V.
c) The transverse electric field Eyin the system is given by:
Ey=VH
d
where dis the distance between the layers. Given the current density j= 5 mA/cm2and the
electron density per layer n, we can calculate the drift velocity vd:
vd=j
nq =5×10−3
2×1015 ×1.6×10−19 = 1.56 ×10−3m/s
Assuming steady-state conditions, the drift velocity is related to the transverse electric field by
vd=µEy, where µis the electron mobility. Rearranging for Eywe have Ey=vd
µ.
Hence, we need to know the electron mobility to calculate Ey.
3 3. TOPOLOGICAL INSULATORS IN THE PRESENCE OF DISORDER
Problem 3. Consider a 1D topological insulator system described by the Hamiltonian:
H=−t
N−1
X
n=1
(c†
n+1cn+c†
ncn+1) + V
N
X
n=1
nc†
ncn
where c†
nand cnare creation and annihilation operators at site n,t= 1 is the hopping parameter,
Vis the strength of the disorder potential, and N= 6 is the total number of sites. Assume periodic
boundary conditions.
a) Find the energy eigenvalues and eigenfunctions of this Hamiltonian.
b) Calculate the Chern number of this system.
c) Determine the topological phase of the system based on the Chern number.
Solution 3.
a) To find the energy eigenvalues and eigenfunctions of the Hamiltonian, we first write it in
momentum space. The Hamiltonian in momentum space is:
H=X
k
ψ†
kH(k)ψk
where ψk= [ck, c−k]Tis the two-component wavefunction in momentum space and
H(k) = −2tcos(k)σx+V nσz
is the Hamiltonian in momentum space, σxand σzare Pauli matrices.
The energy eigenvalues are given by the diagonalization of H(k). Solving for the eigenvalues,
we get:
E=±p4t2cos2(k) + V2n2
The corresponding eigenvectors can be obtained by solving the eigenvector equations.
b) To calculate the Chern number of this system, we need to find the Berry curvature which is
given by:
F(k) = i⟨∂ku|∂k′u⟩−⟨∂k′u|∂ku⟩
where uis the wavefunction of the system.
After calculating the Berry curvature, the Chern number is given by integrating the Berry cur-
vature over the 1st Brillouin zone:
C=1
2πZdkdk′F(k)
c) Based on the Chern number, we can determine the topological phase of the system. If the
Chern number is non-zero, the system is in a topologically non-trivial phase.
4 4. CHIRAL EDGE STATES IN QUANTUM HALL SYSTEMS
Problem 4. Consider an integer Quantum Hall system with a chiral edge state. The chiral
edge state is described by a wave function of the form ψ(x) = Aeikx, where Ais the normalization
constant, kis the wave vector, and xis the position along the edge.
a) If the Fermi level of the system is EF= 2eV, and the edge state has a linear energy dispersion
relation E(k)=¯hvFkwith Fermi velocity vF= 105m/s, what is the wave vector kof the edge state?
b) Calculate the group velocity of the edge state.
c) Determine the direction of propagation (clockwise or counterclockwise) of the edge state.
Solution 4.
a) Given that the energy of the edge state is given by E(k)=¯hvFkand the Fermi level is
EF= 2eV, we set E(k) = EFand solve for k:
¯hvFk=EF
¯hvFk= 2eV
k=2eV
¯hvF
k=2×1.6×10−19C×105m/s
6.63 ×10−34m2kg/s×105m/s
k≈4.8×106m−1
Therefore, the wave vector of the edge state is k≈4.8×106m−1.
b) The group velocity of the edge state is given by the derivative of the energy dispersion relation
with respect to wave vector:
vg=dE
dk
vg=d(¯hvFk)
dk
vg= ¯hvF
Therefore, the group velocity of the edge state is vg= ¯hvF= 6.63 ×10−34m2kg/s×105m/s≈
6.63 ×10−29m/s.
c) Since the edge state is described by a wave function of the form ψ(x) = Aeikx, which has a
positive wave vector k, the edge state propagates in the clockwise direction along the edge.
I’m glad to help! Here is a numerical problem question on Topological Insulators and Quantum
Hall Effects:
5 5. TOPOLOGICAL PHASE TRANSITIONS IN QUANTUM HALL EFFECTS
Problem 5. Consider a 2D electron gas in a square lattice with a magnetic field applied per-
pendicular to the plane. The Hamiltonian for this system is given by:
H=X
r tX
i
c†
r+aicr+vX
i
eiθi
rc†
r+aicr+H.c.!
where crare the annihilation operators at lattice sites r,tis the nearest-neighbor hopping pa-
rameter, vis the strength of Rashba spin-orbit coupling, θi
ris the angle of the magnetic field at site
rwith respect to direction i, and aiare the lattice vectors.
Given that the strength of the Rashba spin-orbit coupling varies smoothly across the system with
domain walls separating regions with different coupling strengths, calculate the conditions that lead
to a topological phase transition in this system.
Solution 5. To determine the conditions for a topological phase transition, we need to look at
the Chern number of the system. The Chern number is given by:
C=1
2πZd2kF(k)
where F(k) = ∇k×A(k)is the Berry curvature and A(k) = −i⟨uk|∇k|uk⟩is the Berry con-
nection.
At the topological phase transition point, the energy gap at the Dirac points closes. This occurs
when v= 0 which happens at the domain walls where the coupling strength changes sign.
Therefore, the condition for a topological phase transition in this system is when the Rashba
spin-orbit coupling strength vchanges sign, leading to the closing of the energy gap at the Dirac
points.
I can provide a sample of a problem for you.
6 6. SPIN HALL EFFECT IN TOPOLOGICAL INSULATORS
Problem 6. Consider an electron moving in a two-dimensional topological insulator with the
following Hamiltonian:
H=3vkx−iλky
vkx+iλky−3
where v= 2 meV ·nm, λ= 1 meV ·nm, and kxand kyare the components of the wave vector.
Calculate the eigenvalues of this Hamiltonian.
Solution 6. To find the eigenvalues, we need to solve the characteristic equation given by
det(H−εI)=0, where εis the eigenvalue.
Substitute Hinto the characteristic equation:
det 3vkx−iλky
vkx+iλky−3−ε1 0
0 1= 0
Simplify this equation and solve for ε:
det 3−ε vkx−iλky
vkx+iλky−3−ε= 0
Expanding the determinant gives:
(3 −ε)(−3−ε)−(vkx−iλky)(vkx+iλky)=0
Solving this equation gives the two eigenvalues ε1and ε2.
Thus, the eigenvalues of the given Hamiltonian are:
ε1= 3 −q9 + v2k2
x+λ2k2
y
ε2= 3 + q9 + v2k2
x+λ2k2
y
7 7. QUANTUM SPIN HALL EFFECT IN TWO-DIMENSIONAL SYSTEMS
Problem 7. Consider a two-dimensional system with spin-orbit coupling described by the Hamil-
tonian
H=E αk−
αk+−E,
where k±=kx±iky,αis the strength of the spin-orbit coupling, and Eis the energy.
a) Find the energy eigenvalues of the system.
b) Determine the corresponding eigenvectors.
c) Show that this system exhibits the quantum spin Hall effect.
Solution 7.
a) To find the energy eigenvalues of the system, we need to solve the equation det(H−EI)=0,
where Iis the identity matrix.
Expanding the determinant, we have:
det(H−EI) = det E−E αk−
αk+−E−E
= (E+E)(E+E)−α2k−k+
= 4E2−α2kxky.
Setting this equal to zero gives us the energy eigenvalues E=±α
2pkxky.
b) To find the eigenvectors, let’s consider the eigenvalue E=α
2pkxky:
For E=α
2pkxky, the eigenvector u
vmust satisfy (H−EI)u
v= 0. Solving this system of
equations, we find the eigenvector corresponds to −ky
αkx.
Similarly, for E=−α
2pkxky, the eigenvector corresponds to ky
αkx.
c) The system exhibits the quantum spin Hall effect since it is characterized by a non-trivial Z2
topological invariant, which indicates the presence of helical edge states that are protected against
backscattering.
8 8. INTERACTION EFFECTS IN TOPOLOGICAL INSULATORS
Problem 8. Consider a 2D topological insulator described by the Hamiltonian H=−3v(kx−iky)
v(kx+iky) 3 ,
where vis a constant with units of velocity.
a) Calculate the energy spectrum of this system.
b) Determine the Chern number of this topological insulator.
c) Suppose there is an additional term in the Hamiltonian given by Hint =λσz, where λis a real
constant. How does this interaction affect the energy spectrum of the system?
Solution 8.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian H. The
eigenvalues of Hare given by solving the characteristic equation |H−EI|= 0, where Iis the
identity matrix.
We have: Det −3−E v(kx−iky)
v(kx+iky) 3 −E= (E+ 3)(E−3) −v2(kx+iky)(kx−iky) = E2−
9−v2(k2
x+k2
y).
This gives us the energy spectrum: E=±qv2(k2
x+k2
y)+9.
b) To calculate the Chern number, we first need to find the Berry curvature Ω(k). The Berry
curvature is given by Ω(k) = ∇ × A(k), where A(k) = −i⟨uk|∇k|uk⟩is the Berry connection.
We can calculate the Berry curvature and integrate it over the Brillouin zone to find the Chern
number.
c) The additional interaction term Hint =λσzintroduces a Zeeman-like splitting in the energy
levels. This term shifts the energies by ±λ, depending on the spin orientation.
9 9. TOPOLOGICAL INSULATORS IN MAGNETIC FIELDS
Problem 9. Consider a 2D topological insulator with a lattice constant a= 1 nm and a magnetic
field B=Bˆzapplied perpendicular to the material. The Fermi energy is EF= 100 meV and the
electron charge is e= 1.6×10−19 C.
Given that the magnetic field strength is B= 2 T and the electron velocity in the material is
v= 106m/s, calculate:
a) The cyclotron frequency of the electrons in the magnetic field.
b) The magnetic length.
c) The Landau level index of the first excited state.
Solution 9.
a) The cyclotron frequency ωcof the electrons in a magnetic field is given by
ωc=eB
m
where eis the electron charge, Bis the magnetic field strength, and mis the electron mass.
Given e= 1.6×10−19 C, B= 2 T, and m= 9.1×10−31 kg, we have
ωc=1.6×10−19 ×2
9.1×10−31 = 0.351 ×1012 rad/s
Therefore, the cyclotron frequency of the electrons in the magnetic field is 0.351 ×1012 rad/s.
b) The magnetic length lBis given by
lB=r¯h
eB
where ¯his the reduced Planck constant.
Given ¯h= 1.05 ×10−34 Js and e= 1.6×10−19 C, we have
lB=r1.05 ×10−34
1.6×10−19 ×2= 2.59 ×10−9m
Therefore, the magnetic length is 2.59 ×10−9m.
c) The Landau level index nof the first excited state can be calculated using the equation
En= ¯hωcn+1
2
where Enis the energy of the n-th Landau level.
Given ¯h= 1.05 ×10−34 Js, ωc= 0.351 ×1012 rad/s, and EF= 100 meV, we can rearrange the
equation to solve for n:
n=EF
¯hωc−1
2=100 ×10−3
1.05 ×10−34 ×0.351 ×1012 −1
2= 4.48
Therefore, the Landau level index of the first excited state is approximately 4.48.
10 10. FRACTIONAL QUANTUM HALL EFFECT IN TOPOLOGICAL SYSTEMS
Problem 10. Consider a 2D electron gas in a strong magnetic field with filling fraction ν=4
3.
The system has an effective magnetic length of lB= 10 nm and an electron charge e= 1.6×10−19
C. Calculate:
a) The magnetic field Bin Tesla.
b) The Hall conductance σxy in units of e2/h where h= 6.63 ×10−34 J s is the Planck constant.
Solution 10.
a) The magnetic field Bcan be related to the magnetic length lBas B=2π
l2
B
. Substituting
lB= 10 nm = 10 ×10−9m, we have:
B=2π
(10 ×10−9)2=2π
100 ×10−18 =2π
10−16 ≈6.28 ×1015 T
Therefore, the magnetic field B≈6.28 ×1015 T.
b) The Hall conductance is given by σxy =νe2
h. Substituting ν=4
3,e= 1.6×10−19 C, and
h= 6.63 ×10−34 J s, we get:
σxy =4
3×(1.6×10−19)2
6.63 ×10−34 =4
3×2.56 ×10−38
6.63 ×10−34 =10.24 ×10−38
6.63 ×10−34 =10.24
6.63 ×10−4S
σxy ≈1.54 ×10−4e2/h
Therefore, the Hall conductance σxy ≈1.54 ×10−4e2/h.
11 11. DISORDER-INDUCED LOCALIZATION IN QUANTUM HALL SYSTEMS
Problem 11. Consider a 2D square lattice with a magnetic field applied perpendicular to the
plane, giving rise to a Quantum Hall effect. At zero disorder, the system has a Hall conductivity of
σxy = 2e2/h.
a) If a weak disorder is introduced into the system, how does the value of the Hall conductivity
change?
b) Calculate the localization length ξof the system with weak disorder, given that the mean free
path lis 10 lattice spacings and the Fermi wavelength λFis 5 lattice spacings.
Solution 11.
a) Introduction of weak disorder into the system does not change the value of the Hall conduc-
tivity. This is because the Hall conductivity is a topological invariant and is robust against weak
disorder.
b) The localization length ξof the system can be calculated using the relation ξ=lλF
l2.
Substituting l= 10 and λF= 5 into the formula, we get:
ξ= 10 5
10 2= 10(0.25) = 2.5lattice spacings.
Therefore, the localization length of the system with weak disorder is ξ= 2.5lattice spacings.
12 12. TOPOLOGICAL INSULATORS WITH TIME-REVERSAL SYMMETRY
Problem 12. Consider a two-dimensional topological insulator described by the Bernevig-
Hughes-Zhang (BHZ) model Hamiltonian given by:
H(k) = (M−Bk2)σz+Akxσx−Akyσy
where σiare the Pauli matrices, M=−1,A= 1,B= 1, and ¯h= 1.
a) Determine the energy eigenvalues E(k)for this Hamiltonian.
b) Identify the topological invariants present in the BHZ model.
c) Find the topological phase diagram for the BHZ model in the M−Bparameter space.
Solution 12. a) To find the energy eigenvalues E(k), we diagonalize the Hamiltonian H(k):
H(k) = (M−Bk2)σz+Akxσx−Akyσy
The eigenvalues are given by E(k) = ±p(M−Bk2)2+A2k2.
b) The topological invariants for the BHZ model are the Chern number and the Z2invariant.
The Chern number is given by ν=1
2πRR dkxdkyFxy(k), where Fxy(k)is the Berry curvature.
It can be calculated using the formula Fxy(k) = ˆ
d(k)·(∂kxˆ
d(k)×∂kyˆ
d(k)).
The Z2invariant is determined using the parity of the number of edge states on a finite system.
c) The topological phase diagram for the BHZ model in the M−Bparameter space can be
determined by analyzing the topological invariants at different values of Mand B. By calculating the
Chern number and checking the presence of edge states, we can identify the different topological
phases in the phase diagram.
13 13. EDGE TRANSPORT IN QUANTUM HALL EFFECTS
Problem 13. Consider a quantum Hall system with a Hall conductance of σxy = 2e2/h and
a longitudinal conductance of σxx = 0. The system has N= 6 chiral edge modes moving in the
positive xdirection and N= 4 chiral edge modes moving in the negative xdirection. Assume that
the charge of an electron is −e.
a) Calculate the Hall current in the positive xdirection.
b) Calculate the Hall voltage across the system.
c) Determine the Hall resistance of the system.
Solution 13. a) The Hall current in the positive xdirection is given by the formula:
IHall =σxyVHall
where VHall is the Hall voltage. Since σxy = 2e2/h and the charge of an electron is −e, we have:
IHall = (2e2/h)·(−e)·(6) = −12e2/h
So, the Hall current in the positive xdirection is −12e2/h.
b) The Hall voltage across the system can be found by rearranging the formula for Hall current:
VHall =IHall
σxy
=−12e2/h
2e2/h =−6
Thus, the Hall voltage across the system is −6.
c) The Hall resistance of the system is given by:
RHall =VHall
IHall
=−6
−12e2/h =1
2h/e2
Therefore, the Hall resistance of the system is 1/2h/e2.
14 14. PROXIMITY EFFECTS IN TOPOLOGICAL INSULATOR HETEROSTRUCTURES
Problem 14. Consider a heterostructure composed of a normal insulator (NI) and a topological
insulator (TI) with a proximity-induced superconducting pairing in the TI region. The Hamiltonian
for this system is given by:
H=HNI +HT I +Hint
where HNI is the Hamiltonian of the normal insulator, HT I is the Hamiltonian of the topological
insulator, and Hint represents the interaction between the two regions. The system is described
by the Bogoliubov-de Gennes Hamiltonian:
HBdG =HNI i∆
−i∆HT I
where ∆is the pairing potential in the TI region.
Given that HNI =ϵN∆N
∆N−ϵN,HT I =ϵT∆T
∆T−ϵT, where ϵN= 2,∆N= 1,ϵT= 3,∆T= 2,
and ∆=1. Calculate the energy spectrum of the BdG Hamiltonian.
Solution 14. The Bogoliubov-de Gennes Hamiltonian is given by:
HBdG =
2 1 i0
1−2 0 i
−i032
0−i2−3
Expanding the determinant of the matrix HBdG −λI = 0, where λis the eigenvalue, we get:
det
2−λ1i0
1−2−λ0i
−i0 3 −λ2
0−i2−3−λ
= 0
Solving this equation gives the energy spectrum of the BdG Hamiltonian. Solving for λ, we get
the eigenvalues:
λ=±qϵ2
N+ ∆2
N,±qϵ2
T+ ∆2
T
Substitute the given values ϵN= 2,∆N= 1,ϵT= 3,∆T= 2 into the above equation to obtain
the energy spectrum. Therefore, the energy spectrum of the BdG Hamiltonian is:
λ=±√5,±√13
15 15. THERMOELECTRIC PROPERTIES OF TOPOLOGICAL INSULATORS
Problem 15. Consider a topological insulator with a band gap of 0.5 eV. The temperature at
the hot reservoir is Th= 300 K, and at the cold reservoir is Tc= 100 K. The Seebeck coefficient of
the material is S= 100 µV/K. Calculate:
a) The voltage generated when a temperature difference is created between the hot and cold
reservoirs.
b) The power generated if the hot reservoir is connected to a load with resistance R= 10Ω.
c) The efficiency of the thermoelectric device if the power generated in part b is used to drive a
load at room temperature (300 K).
Solution 15.
a) The voltage generated when a temperature difference is created is given by the Seebeck
effect equation:
V=S·(Th−Tc)
Substitute the given values:
V= 100 ×10−6V/K ×(300K−100K) = 20mV
Therefore, the voltage generated is 20 mV.
b) The power generated can be calculated using the formula for electrical power:
P=V2
R
Substitute the known values:
P=(0.02V)2
10Ω =0.0004
10 = 0.04mW
Therefore, the power generated is 0.04 mW.
c) The efficiency of the thermoelectric device is given by:
η=Useful power output
Heat input =P
Qh
Since the device is an ideal thermoelectric device, the heat input Qhis equal to the heat ab-
sorbed from the hot reservoir:
Qh=Th·S
Substitute the values:
Qh= 300K×100 ×10−6V/K = 30mV = 0.03W
Finally, calculate the efficiency:
η=0.04mW
0.03W≈1.33%
Therefore, the efficiency of the thermoelectric device is approximately 1.33
16 16. FRACTIONAL CHARGES IN QUANTUM HALL STATES
Problem 16. Consider a 2D electron gas confined to a square box of side length Lin the xy-
plane. The electrons are subject to a strong magnetic field perpendicular to the plane. At filling
factor ν=1
3, the system exhibits fractional charges.
Given that the magnetic field strength B= 3 T and the elementary charge e= 1.6×10−19 C,
determine the fractional charge carried by the quasiparticles in this system.
Solution 16. The filling factor ν=Ne
Nϕ, where Neis the number of electrons and Nϕis the
number of magnetic flux quanta penetrating the surface of the system. For a square box, we have
Nϕ=BA
ϕ0, where A=L2is the area of the box, Bis the magnetic field strength, and ϕ0=h
eis the
magnetic flux quantum.
Given B= 3 T and ν=1
3, we have:
Nϕ=B·L2
ϕ0
=3T·(L2m2)
h
e
=3·109m−2·(L2)
2.07 ×10−15 Wb
=3·109·L2
2.07 ×10−15 flux quanta
For ν=1
3,Ne=1
3Nϕ. The fractional charge e∗=e
3. Thus, the quasiparticles in this system
carry a fractional charge of:
e∗=e
3
=1.6×10−19 C
3
= 5.33 ×10−20 C
Therefore, the quasiparticles in this system carry a fractional charge of 5.33 ×10−20 C.
17 17. TOPOLOGICAL SUPERCONDUCTIVITY IN TOPOLOGICAL INSULATORS
Problem 17. Consider a 2D topological insulator described by the Hamiltonian
H(k) = 0kx−iky
kx+iky0
a) Calculate the eigenvalues and eigenvectors of H(k).
b) Show that this Hamiltonian satisfies the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1,
where Θ = σyKis the time-reversal operator with σybeing the Pauli matrix and Kbeing complex
conjugation.
c) Determine if this system is a topological insulator by computing the Z2topological invariant.
Solution 17.
a) To find the eigenvalues and eigenvectors, we solve the characteristic equation det(H−λI) =
0. Let’s denote the eigenvalue as λand the eigenvector as ψ=a
b.
det −λ kx−iky
kx+iky−λ=λ2−(k2
x+k2
y) = 0
So, the eigenvalues are λ=±|k|, where |k|=qk2
x+k2
y. For λ=|k|, we have the eigenvector
ψ+=kx−iky
|k|
For λ=−|k|, the eigenvector is
ψ−=−|k|
kx+iky
b) Now, let’s check the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1:
H(−k) = 0−kx+iky
−kx−iky0
ΘH(k)Θ−1=0−1
1 0 0kx−iky
kx+iky0 0 1
−1 0=0−kx+iky
−kx−iky0
Thus, the Hamiltonian satisfies the time-reversal symmetry condition.
c) The Z2topological invariant for this 2D system can be calculated using the parity of the
determinant of the mass term M(k) = kxσx+kyσy. The Z2invariant is defined as
ν0=sgn(det[M(Γ)])
where Γis the time-reversal invariant momentum. In this case, Γ = (0,0).
The determinant of M(Γ) is det[M(Γ)] = 0, which implies that the system is a trivial insulator
with ν0= 0.
I am unable to generate numerical problems on demand as it requires creating specific sce-
narios and calculations. However, if you provide me with a specific scenario or problem statement
related to Topological Insulators and Quantum Hall Effects, I can definitely help you generate a
numerical problem along with a step-by-step explanation for the solution. Just let me know what
specific topic or concept you would like to focus on!
18 19. QUANTUM HALL EFFECTS IN GRAPHENE
Problem 19. Consider a monolayer graphene sheet under a magnetic field of strength B= 2 T.
The charge of an electron is e= 1.6×10−19 C and the Planck’s constant is h= 6.63 ×10−34 J s.
The Fermi velocity in graphene is vF= 106m/s. Calculate:
a) The magnetic length lBin the graphene sheet.
b) The energy level spacing ∆Ebetween Landau levels.
c) The filling factor νfor the third Landau level.
Solution 19.
a) The magnetic length lBis given by:
lB=r¯h
eB
Plugging in the values ¯h= 6.63 ×10−34 J s, e= 1.6×10−19 C, and B= 2 T, we get:
lB=r6.63 ×10−34 J s
1.6×10−19 C·2T≈26.1nm
Therefore, the magnetic length in the graphene sheet is approximately 26.1nm.
b) The energy spacing ∆Ebetween Landau levels is given by:
∆E=¯hvF
lB
Substitute the values ¯h= 6.63 ×10−34 J s, vF= 106m/s, and lB= 26.1nm:
∆E=6.63 ×10−34 J s ·106m/s
26.1×10−9m≈2.54 ×10−4eV
Hence, the energy level spacing between Landau levels is approximately 2.54 ×10−4eV.
c) The filling factor νfor the third Landau level is given by:
ν=N
Nϕ
Where Nis the number of electrons in the third Landau level and Nϕis the number of flux
quanta enclosed. For graphene, there are Nϕ=BA
ϕ0flux quanta enclosed for each unit cell area
A.
For the third Landau level (n= 3) in graphene, there are N= 2 electrons (spin degeneracy).
Thus,
Nϕ=BA
ϕ0
=(2T)(a2)
h/e =(2)(10−6
6.63 ×10−34 J s/1.6×10−19 C≈2.4×105
Therefore, the filling factor for the third Landau level in graphene is
ν=N
Nϕ
=2
2.4×105≈8.33 ×10−6
19 20. TOPOLOGICAL DEFECTS IN TOPOLOGICAL INSULATORS
Problem 20. Consider a one-dimensional lattice system described by the Hamiltonian
H=−t
N
X
n=1
(c†
ncn+1 +c†
n+1cn)−µ
N
X
n=1
c†
ncn
where cnand c†
nare annihilation and creation operators at site n, respectively, tis the hopping
parameter, µis the chemical potential, and Nis the number of lattice sites.
a) Calculate the energy spectrum of the system.
b) Determine the winding number for this system.
c) Show that there is a zero-energy bound state at a domain wall with an inverted mass.
Solution 20.
a) To calculate the energy spectrum of the system, we start by performing a Fourier transform
to the momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)−µ
The energy spectrum can be obtained by diagonalizing H(k),
E(k) = ±p(2tcos(k) + µ)2
b) The winding number is given by the integral of the Berry connection over the entire Brillouin
zone,
w=1
2πZBZ
dk A(k)
where A(k) = i⟨uk|∂k|uk⟩is the Berry connection, and |uk⟩is the periodic part of the Bloch
wavefunction. For this system, the winding number is w= 1.
c) At a domain wall with an inverted mass, the Hamiltonian changes sign, i.e., Hwall =−H.
Therefore, there exists a zero-energy solution at the domain wall due to the symmetric nature of
the zero-energy solution.
20 21. ANOMALOUS HALL EFFECT IN TOPOLOGICAL MATERIALS
Problem 21. Consider a 2D topological insulator with a Chern number C= 2. The Hall con-
ductivity σxy is given by the formula σxy =e2
hC, where eis the elementary charge and his the
Planck constant.
a) Calculate the Hall conductivity σxy for this 2D topological insulator.
b) If the number of edge modes on one edge of this material is 3, calculate the Hall conductivity
for each edge mode.
Solution 21.
a) Given that the Chern number C= 2, we can calculate the Hall conductivity using the formula
σxy =e2
hC.
Substitute e= 1.6×10−19 C and h= 6.63 ×10−34 J s:
σxy =1.6×10−19 C2
6.63×10−34 J s ×2 = 4.85 ×10−5Ω−1
Therefore, the Hall conductivity for this 2D topological insulator is 4.85 ×10−5Ω−1.
b) Since the number of edge modes on one edge is 3, we can calculate the Hall conductivity
for each edge mode using the formula σedge =σxy
number of edge modes .
For each edge mode, σedge =4.85×10−5Ω−1
3= 1.62 ×10−5Ω−1.
Therefore, the Hall conductivity for each edge mode on one edge of this material is 1.62 ×
10−5Ω−1.
21 Topological Insulators and Quantum Hall Effects
Problem 1. Consider a 2D topological insulator described by the Hamiltonian
H=2−iλ
iλ −2,
where λis a real parameter.
a) Determine the energy eigenvalues of the Hamiltonian.
b) Find the normalized eigenvectors corresponding to each energy eigenvalue.
Solution 1.
a) To find the energy eigenvalues, we solve the characteristic equation |H−ϵI|= 0:
2−ϵ−iλ
iλ −2−ϵ
= (2 −ϵ)(−2−ϵ) + λ2=ϵ2−4−λ2= 0.
This gives us the eigenvalues:
ϵ=±pλ2+ 4.
b) To find the normalized eigenvectors, we solve the eigenvalue equation (H−ϵI)v= 0 for
each eigenvalue.
For ϵ=√λ2+ 4:
2−√λ2+ 4 −iλ
iλ −2−√λ2+ 4v1
v2= 0,
we get the eigenvector v+=1
√2(λ2+4) λ
√λ2+ 4.
Similarly, for ϵ=−√λ2+ 4:
2 + √λ2+ 4 −iλ
iλ −2 + √λ2+ 4v1
v2= 0,
we get the eigenvector v−=1
√2(λ2+4) λ
−√λ2+ 4.
22 23. QUANTUM SPIN HALL TOPOLOGICAL INSULATORS
Problem 23. Consider a 1D chain with spin-orbit coupling described by the Hamiltonian H=
−tPn,σ(c†
n,σcn+1,σ +h.c.) + iλ Pn(c†
n,↑cn+1,↓−c†
n,↓cn+1,↑), where t= 1 (energy units) and λ= 0.5.
a) Find the energy spectrum of this system.
b) Determine if the system is a topological insulator based on the parity criterion.
Solution 23.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian. We
can do this by performing a Fourier transform to momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)σx−2λsin(k)σy, where σxand σyare the Pauli matrices.
Diagonalizing the Hamiltonian, we get H(k) = −2√t2+λ2cos(ϕ(k)), where cos(ϕ(k)) = t/√t2+λ2cos(k)−
λ/√t2+λ2sin(k).
Thus, the energy spectrum is given by E=−2√t2+λ2cos(ϕ(k)) = −2√1+0.25 cos(ϕ(k)) =
−2√1.25 cos(ϕ(k)), which simplifies to E(k) = −2 cos(ϕ(k)).
b) To determine if the system is a topological insulator based on the parity criterion, we need
to check the parity of the ground state. The ground state corresponds to the lowest energy, which
occurs at k= 0. At k= 0, the energy is E(0) = −2. Since the energy is non-zero, the system is
not a topological insulator based on the parity criterion.
23 24. HALF-INTEGER QUANTUM HALL EFFECTS IN TOPOLOGICAL SYSTEMS
Problem 24. Consider a 2D electron gas in a magnetic field with flux Φ = 2πϕ
ϕ0
, where ϕis a
dimensionless parameter and ϕ0=h
eis the magnetic flux quantum. The Hall conductance for this
system is given by σxy = (n+1
2)e2
h, where nis an integer.
a) Calculate the Hall conductance when n= 2.
b) Determine the value of ϕfor which the Hall conductance changes by e2
h.
Solution 24.
a) When n= 2, the Hall conductance is given by σxy = (2 + 1
2)e2
h=5
2
e2
h. Therefore, when
n= 2, the Hall conductance is 5
2
e2
h.
b) To find the value of ϕfor which the Hall conductance changes by e2
h, we need to consider the
change in nfrom nto n+1. The change in Hall conductance is given by ∆σxy =(n+ 1) + 1
2e2
h−
(n+1
2)e2
h=e2
h.
Solving for ∆σxy with ∆ϕ:
(n+ 1) + 1
2e2
h−(n+1
2)e2
h=e2
h
(n+1+1
2)−(n+1
2)=1
1=1
Therefore, the value of ϕfor which the Hall conductance changes by e2
his any value of ϕthat
results in an increase in nby 1.
24 25. TOPOLOGICAL INSULATOR DEVICES FOR SPINTRONICS APPLICATIONS
Problem 25. Consider a 2D topological insulator with band structure described by the Hamil-
tonian
H(k) = t(σxsin kx+σysin ky)
where k= (kx, ky),tis the hopping parameter, and σxand σyare Pauli matrices.
a) Determine the eigenvalues and eigenvectors of H(k).
b) Find the Chern number associated with this system.
c) Suppose a magnetic field is added which introduces a Zeeman term HZ=Mσz, where M
is the Zeeman splitting strength. How does this modification affect the topological properties of the
system?
Solution 25.
a) To find the eigenvalues ϵ(k)and eigenvectors v(k), we solve the equation H(k)v(k) =
ϵ(k)v(k).
H(k)v(k) = t(σxsin kx+σysin ky)v(k) = ϵ(k)v(k)
Expanding the matrix multiplication and solving for eigenvalues (ϵ(k) = ±t) and eigenvectors, we
get:
ϵ(k) = ±t, v±(k) = eiϕ
±eiϕ
where ϕ=kxfor the +branch and ϕ=kyfor the −branch.
b) The Chern number Cfor this system can be calculated using the formula
C=1
2πZ Z dkxdkyˆz ·(∂kxA×∂kyA)
where A=i⟨uk|∇k|uk⟩is the Berry connection. With our eigenvectors, we get A=1
2
t
|t|2(ˆz ×t)
and
C=sign(t)
c) Introducing the Zeeman term HZ=Mσzmodifies the Hamiltonian to H′(k) = H(k) +
HZ. This opens a band gap and shifts the energies, but as long as the time-reversal symmetry is
preserved, the system remains in the same topological phase with Chern number C=sign(t).
where nis the electron density per layer. Given that the density of states per layer is ν= 2 ×
1011 cm−2, we can calculate the electron density per layer as:
n=ν×104= 2 ×1011 ×104= 2 ×1015 m−2
Substitute the values into the equation:
VH=1
2×1015 ×1.6×10−19 ×1.4×10−4×0.5 = 218.75 V
Therefore, the Hall voltage VHacross the system is 218.75 V.
c) The transverse electric field Eyin the system is given by:
Ey=VH
d
where dis the distance between the layers. Given the current density j= 5 mA/cm2and the
electron density per layer n, we can calculate the drift velocity vd:
vd=j
nq =5×10−3
2×1015 ×1.6×10−19 = 1.56 ×10−3m/s
Assuming steady-state conditions, the drift velocity is related to the transverse electric field by
vd=µEy, where µis the electron mobility. Rearranging for Eywe have Ey=vd
µ.
Hence, we need to know the electron mobility to calculate Ey.
3 3. TOPOLOGICAL INSULATORS IN THE PRESENCE OF DISORDER
Problem 3. Consider a 1D topological insulator system described by the Hamiltonian:
H=−t
N−1
X
n=1
(c†
n+1cn+c†
ncn+1) + V
N
X
n=1
nc†
ncn
where c†
nand cnare creation and annihilation operators at site n,t= 1 is the hopping parameter,
Vis the strength of the disorder potential, and N= 6 is the total number of sites. Assume periodic
boundary conditions.
a) Find the energy eigenvalues and eigenfunctions of this Hamiltonian.
b) Calculate the Chern number of this system.
c) Determine the topological phase of the system based on the Chern number.
Solution 3.
a) To find the energy eigenvalues and eigenfunctions of the Hamiltonian, we first write it in
momentum space. The Hamiltonian in momentum space is:
H=X
k
ψ†
kH(k)ψk
where ψk= [ck, c−k]Tis the two-component wavefunction in momentum space and
H(k) = −2tcos(k)σx+V nσz
is the Hamiltonian in momentum space, σxand σzare Pauli matrices.
The energy eigenvalues are given by the diagonalization of H(k). Solving for the eigenvalues,
we get:
E=±p4t2cos2(k) + V2n2
The corresponding eigenvectors can be obtained by solving the eigenvector equations.
b) To calculate the Chern number of this system, we need to find the Berry curvature which is
given by:
F(k) = i⟨∂ku|∂k′u⟩−⟨∂k′u|∂ku⟩
where uis the wavefunction of the system.
After calculating the Berry curvature, the Chern number is given by integrating the Berry cur-
vature over the 1st Brillouin zone:
C=1
2πZdkdk′F(k)
c) Based on the Chern number, we can determine the topological phase of the system. If the
Chern number is non-zero, the system is in a topologically non-trivial phase.
4 4. CHIRAL EDGE STATES IN QUANTUM HALL SYSTEMS
Problem 4. Consider an integer Quantum Hall system with a chiral edge state. The chiral
edge state is described by a wave function of the form ψ(x) = Aeikx, where Ais the normalization
constant, kis the wave vector, and xis the position along the edge.
a) If the Fermi level of the system is EF= 2eV, and the edge state has a linear energy dispersion
relation E(k)=¯hvFkwith Fermi velocity vF= 105m/s, what is the wave vector kof the edge state?
b) Calculate the group velocity of the edge state.
c) Determine the direction of propagation (clockwise or counterclockwise) of the edge state.
Solution 4.
a) Given that the energy of the edge state is given by E(k)=¯hvFkand the Fermi level is
EF= 2eV, we set E(k) = EFand solve for k:
¯hvFk=EF
¯hvFk= 2eV
k=2eV
¯hvF
k=2×1.6×10−19C×105m/s
6.63 ×10−34m2kg/s×105m/s
k≈4.8×106m−1
Therefore, the wave vector of the edge state is k≈4.8×106m−1.
b) The group velocity of the edge state is given by the derivative of the energy dispersion relation
with respect to wave vector:
vg=dE
dk
vg=d(¯hvFk)
dk
vg= ¯hvF
Therefore, the group velocity of the edge state is vg= ¯hvF= 6.63 ×10−34m2kg/s×105m/s≈
6.63 ×10−29m/s.
c) Since the edge state is described by a wave function of the form ψ(x) = Aeikx, which has a
positive wave vector k, the edge state propagates in the clockwise direction along the edge.
I’m glad to help! Here is a numerical problem question on Topological Insulators and Quantum
Hall Effects:
5 5. TOPOLOGICAL PHASE TRANSITIONS IN QUANTUM HALL EFFECTS
Problem 5. Consider a 2D electron gas in a square lattice with a magnetic field applied per-
pendicular to the plane. The Hamiltonian for this system is given by:
H=X
r tX
i
c†
r+aicr+vX
i
eiθi
rc†
r+aicr+H.c.!
where crare the annihilation operators at lattice sites r,tis the nearest-neighbor hopping pa-
rameter, vis the strength of Rashba spin-orbit coupling, θi
ris the angle of the magnetic field at site
rwith respect to direction i, and aiare the lattice vectors.
Given that the strength of the Rashba spin-orbit coupling varies smoothly across the system with
domain walls separating regions with different coupling strengths, calculate the conditions that lead
to a topological phase transition in this system.
Solution 5. To determine the conditions for a topological phase transition, we need to look at
the Chern number of the system. The Chern number is given by:
C=1
2πZd2kF(k)
where F(k) = ∇k×A(k)is the Berry curvature and A(k) = −i⟨uk|∇k|uk⟩is the Berry con-
nection.
At the topological phase transition point, the energy gap at the Dirac points closes. This occurs
when v= 0 which happens at the domain walls where the coupling strength changes sign.
Therefore, the condition for a topological phase transition in this system is when the Rashba
spin-orbit coupling strength vchanges sign, leading to the closing of the energy gap at the Dirac
points.
I can provide a sample of a problem for you.
6 6. SPIN HALL EFFECT IN TOPOLOGICAL INSULATORS
Problem 6. Consider an electron moving in a two-dimensional topological insulator with the
following Hamiltonian:
H=3vkx−iλky
vkx+iλky−3
where v= 2 meV ·nm, λ= 1 meV ·nm, and kxand kyare the components of the wave vector.
Calculate the eigenvalues of this Hamiltonian.
Solution 6. To find the eigenvalues, we need to solve the characteristic equation given by
det(H−εI)=0, where εis the eigenvalue.
Substitute Hinto the characteristic equation:
det 3vkx−iλky
vkx+iλky−3−ε1 0
0 1= 0
Simplify this equation and solve for ε:
det 3−ε vkx−iλky
vkx+iλky−3−ε= 0
Expanding the determinant gives:
(3 −ε)(−3−ε)−(vkx−iλky)(vkx+iλky)=0
Solving this equation gives the two eigenvalues ε1and ε2.
Thus, the eigenvalues of the given Hamiltonian are:
ε1= 3 −q9 + v2k2
x+λ2k2
y
ε2= 3 + q9 + v2k2
x+λ2k2
y
7 7. QUANTUM SPIN HALL EFFECT IN TWO-DIMENSIONAL SYSTEMS
Problem 7. Consider a two-dimensional system with spin-orbit coupling described by the Hamil-
tonian
H=E αk−
αk+−E,
where k±=kx±iky,αis the strength of the spin-orbit coupling, and Eis the energy.
a) Find the energy eigenvalues of the system.
b) Determine the corresponding eigenvectors.
c) Show that this system exhibits the quantum spin Hall effect.
Solution 7.
a) To find the energy eigenvalues of the system, we need to solve the equation det(H−EI)=0,
where Iis the identity matrix.
Expanding the determinant, we have:
det(H−EI) = det E−E αk−
αk+−E−E
= (E+E)(E+E)−α2k−k+
= 4E2−α2kxky.
Setting this equal to zero gives us the energy eigenvalues E=±α
2pkxky.
b) To find the eigenvectors, let’s consider the eigenvalue E=α
2pkxky:
For E=α
2pkxky, the eigenvector u
vmust satisfy (H−EI)u
v= 0. Solving this system of
equations, we find the eigenvector corresponds to −ky
αkx.
Similarly, for E=−α
2pkxky, the eigenvector corresponds to ky
αkx.
c) The system exhibits the quantum spin Hall effect since it is characterized by a non-trivial Z2
topological invariant, which indicates the presence of helical edge states that are protected against
backscattering.
8 8. INTERACTION EFFECTS IN TOPOLOGICAL INSULATORS
Problem 8. Consider a 2D topological insulator described by the Hamiltonian H=−3v(kx−iky)
v(kx+iky) 3 ,
where vis a constant with units of velocity.
a) Calculate the energy spectrum of this system.
b) Determine the Chern number of this topological insulator.
c) Suppose there is an additional term in the Hamiltonian given by Hint =λσz, where λis a real
constant. How does this interaction affect the energy spectrum of the system?
Solution 8.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian H. The
eigenvalues of Hare given by solving the characteristic equation |H−EI|= 0, where Iis the
identity matrix.
We have: Det −3−E v(kx−iky)
v(kx+iky) 3 −E= (E+ 3)(E−3) −v2(kx+iky)(kx−iky) = E2−
9−v2(k2
x+k2
y).
This gives us the energy spectrum: E=±qv2(k2
x+k2
y)+9.
b) To calculate the Chern number, we first need to find the Berry curvature Ω(k). The Berry
curvature is given by Ω(k) = ∇ × A(k), where A(k) = −i⟨uk|∇k|uk⟩is the Berry connection.
We can calculate the Berry curvature and integrate it over the Brillouin zone to find the Chern
number.
c) The additional interaction term Hint =λσzintroduces a Zeeman-like splitting in the energy
levels. This term shifts the energies by ±λ, depending on the spin orientation.
9 9. TOPOLOGICAL INSULATORS IN MAGNETIC FIELDS
Problem 9. Consider a 2D topological insulator with a lattice constant a= 1 nm and a magnetic
field B=Bˆzapplied perpendicular to the material. The Fermi energy is EF= 100 meV and the
electron charge is e= 1.6×10−19 C.
Given that the magnetic field strength is B= 2 T and the electron velocity in the material is
v= 106m/s, calculate:
a) The cyclotron frequency of the electrons in the magnetic field.
b) The magnetic length.
c) The Landau level index of the first excited state.
Solution 9.
a) The cyclotron frequency ωcof the electrons in a magnetic field is given by
ωc=eB
m
where eis the electron charge, Bis the magnetic field strength, and mis the electron mass.
Given e= 1.6×10−19 C, B= 2 T, and m= 9.1×10−31 kg, we have
ωc=1.6×10−19 ×2
9.1×10−31 = 0.351 ×1012 rad/s
Therefore, the cyclotron frequency of the electrons in the magnetic field is 0.351 ×1012 rad/s.
b) The magnetic length lBis given by
lB=r¯h
eB
where ¯his the reduced Planck constant.
Given ¯h= 1.05 ×10−34 Js and e= 1.6×10−19 C, we have
lB=r1.05 ×10−34
1.6×10−19 ×2= 2.59 ×10−9m
Therefore, the magnetic length is 2.59 ×10−9m.
c) The Landau level index nof the first excited state can be calculated using the equation
En= ¯hωcn+1
2
where Enis the energy of the n-th Landau level.
Given ¯h= 1.05 ×10−34 Js, ωc= 0.351 ×1012 rad/s, and EF= 100 meV, we can rearrange the
equation to solve for n:
n=EF
¯hωc−1
2=100 ×10−3
1.05 ×10−34 ×0.351 ×1012 −1
2= 4.48
Therefore, the Landau level index of the first excited state is approximately 4.48.
10 10. FRACTIONAL QUANTUM HALL EFFECT IN TOPOLOGICAL SYSTEMS
Problem 10. Consider a 2D electron gas in a strong magnetic field with filling fraction ν=4
3.
The system has an effective magnetic length of lB= 10 nm and an electron charge e= 1.6×10−19
C. Calculate:
a) The magnetic field Bin Tesla.
b) The Hall conductance σxy in units of e2/h where h= 6.63 ×10−34 J s is the Planck constant.
Solution 10.
a) The magnetic field Bcan be related to the magnetic length lBas B=2π
l2
B
. Substituting
lB= 10 nm = 10 ×10−9m, we have:
B=2π
(10 ×10−9)2=2π
100 ×10−18 =2π
10−16 ≈6.28 ×1015 T
Therefore, the magnetic field B≈6.28 ×1015 T.
b) The Hall conductance is given by σxy =νe2
h. Substituting ν=4
3,e= 1.6×10−19 C, and
h= 6.63 ×10−34 J s, we get:
σxy =4
3×(1.6×10−19)2
6.63 ×10−34 =4
3×2.56 ×10−38
6.63 ×10−34 =10.24 ×10−38
6.63 ×10−34 =10.24
6.63 ×10−4S
σxy ≈1.54 ×10−4e2/h
Therefore, the Hall conductance σxy ≈1.54 ×10−4e2/h.
11 11. DISORDER-INDUCED LOCALIZATION IN QUANTUM HALL SYSTEMS
Problem 11. Consider a 2D square lattice with a magnetic field applied perpendicular to the
plane, giving rise to a Quantum Hall effect. At zero disorder, the system has a Hall conductivity of
σxy = 2e2/h.
a) If a weak disorder is introduced into the system, how does the value of the Hall conductivity
change?
b) Calculate the localization length ξof the system with weak disorder, given that the mean free
path lis 10 lattice spacings and the Fermi wavelength λFis 5 lattice spacings.
Solution 11.
a) Introduction of weak disorder into the system does not change the value of the Hall conduc-
tivity. This is because the Hall conductivity is a topological invariant and is robust against weak
disorder.
b) The localization length ξof the system can be calculated using the relation ξ=lλF
l2.
Substituting l= 10 and λF= 5 into the formula, we get:
ξ= 10 5
10 2= 10(0.25) = 2.5lattice spacings.
Therefore, the localization length of the system with weak disorder is ξ= 2.5lattice spacings.
12 12. TOPOLOGICAL INSULATORS WITH TIME-REVERSAL SYMMETRY
Problem 12. Consider a two-dimensional topological insulator described by the Bernevig-
Hughes-Zhang (BHZ) model Hamiltonian given by:
H(k) = (M−Bk2)σz+Akxσx−Akyσy
where σiare the Pauli matrices, M=−1,A= 1,B= 1, and ¯h= 1.
a) Determine the energy eigenvalues E(k)for this Hamiltonian.
b) Identify the topological invariants present in the BHZ model.
c) Find the topological phase diagram for the BHZ model in the M−Bparameter space.
Solution 12. a) To find the energy eigenvalues E(k), we diagonalize the Hamiltonian H(k):
H(k) = (M−Bk2)σz+Akxσx−Akyσy
The eigenvalues are given by E(k) = ±p(M−Bk2)2+A2k2.
b) The topological invariants for the BHZ model are the Chern number and the Z2invariant.
The Chern number is given by ν=1
2πRR dkxdkyFxy(k), where Fxy(k)is the Berry curvature.
It can be calculated using the formula Fxy(k) = ˆ
d(k)·(∂kxˆ
d(k)×∂kyˆ
d(k)).
The Z2invariant is determined using the parity of the number of edge states on a finite system.
c) The topological phase diagram for the BHZ model in the M−Bparameter space can be
determined by analyzing the topological invariants at different values of Mand B. By calculating the
Chern number and checking the presence of edge states, we can identify the different topological
phases in the phase diagram.
13 13. EDGE TRANSPORT IN QUANTUM HALL EFFECTS
Problem 13. Consider a quantum Hall system with a Hall conductance of σxy = 2e2/h and
a longitudinal conductance of σxx = 0. The system has N= 6 chiral edge modes moving in the
positive xdirection and N= 4 chiral edge modes moving in the negative xdirection. Assume that
the charge of an electron is −e.
a) Calculate the Hall current in the positive xdirection.
b) Calculate the Hall voltage across the system.
c) Determine the Hall resistance of the system.
Solution 13. a) The Hall current in the positive xdirection is given by the formula:
IHall =σxyVHall
where VHall is the Hall voltage. Since σxy = 2e2/h and the charge of an electron is −e, we have:
IHall = (2e2/h)·(−e)·(6) = −12e2/h
So, the Hall current in the positive xdirection is −12e2/h.
b) The Hall voltage across the system can be found by rearranging the formula for Hall current:
VHall =IHall
σxy
=−12e2/h
2e2/h =−6
Thus, the Hall voltage across the system is −6.
c) The Hall resistance of the system is given by:
RHall =VHall
IHall
=−6
−12e2/h =1
2h/e2
Therefore, the Hall resistance of the system is 1/2h/e2.
14 14. PROXIMITY EFFECTS IN TOPOLOGICAL INSULATOR HETEROSTRUCTURES
Problem 14. Consider a heterostructure composed of a normal insulator (NI) and a topological
insulator (TI) with a proximity-induced superconducting pairing in the TI region. The Hamiltonian
for this system is given by:
H=HNI +HT I +Hint
where HNI is the Hamiltonian of the normal insulator, HT I is the Hamiltonian of the topological
insulator, and Hint represents the interaction between the two regions. The system is described
by the Bogoliubov-de Gennes Hamiltonian:
HBdG =HNI i∆
−i∆HT I
where ∆is the pairing potential in the TI region.
Given that HNI =ϵN∆N
∆N−ϵN,HT I =ϵT∆T
∆T−ϵT, where ϵN= 2,∆N= 1,ϵT= 3,∆T= 2,
and ∆=1. Calculate the energy spectrum of the BdG Hamiltonian.
Solution 14. The Bogoliubov-de Gennes Hamiltonian is given by:
HBdG =
2 1 i0
1−2 0 i
−i032
0−i2−3
Expanding the determinant of the matrix HBdG −λI = 0, where λis the eigenvalue, we get:
det
2−λ1i0
1−2−λ0i
−i0 3 −λ2
0−i2−3−λ
= 0
Solving this equation gives the energy spectrum of the BdG Hamiltonian. Solving for λ, we get
the eigenvalues:
λ=±qϵ2
N+ ∆2
N,±qϵ2
T+ ∆2
T
Substitute the given values ϵN= 2,∆N= 1,ϵT= 3,∆T= 2 into the above equation to obtain
the energy spectrum. Therefore, the energy spectrum of the BdG Hamiltonian is:
λ=±√5,±√13
15 15. THERMOELECTRIC PROPERTIES OF TOPOLOGICAL INSULATORS
Problem 15. Consider a topological insulator with a band gap of 0.5 eV. The temperature at
the hot reservoir is Th= 300 K, and at the cold reservoir is Tc= 100 K. The Seebeck coefficient of
the material is S= 100 µV/K. Calculate:
a) The voltage generated when a temperature difference is created between the hot and cold
reservoirs.
b) The power generated if the hot reservoir is connected to a load with resistance R= 10Ω.
c) The efficiency of the thermoelectric device if the power generated in part b is used to drive a
load at room temperature (300 K).
Solution 15.
a) The voltage generated when a temperature difference is created is given by the Seebeck
effect equation:
V=S·(Th−Tc)
Substitute the given values:
V= 100 ×10−6V/K ×(300K−100K) = 20mV
Therefore, the voltage generated is 20 mV.
b) The power generated can be calculated using the formula for electrical power:
P=V2
R
Substitute the known values:
P=(0.02V)2
10Ω =0.0004
10 = 0.04mW
Therefore, the power generated is 0.04 mW.
c) The efficiency of the thermoelectric device is given by:
η=Useful power output
Heat input =P
Qh
Since the device is an ideal thermoelectric device, the heat input Qhis equal to the heat ab-
sorbed from the hot reservoir:
Qh=Th·S
Substitute the values:
Qh= 300K×100 ×10−6V/K = 30mV = 0.03W
Finally, calculate the efficiency:
η=0.04mW
0.03W≈1.33%
Therefore, the efficiency of the thermoelectric device is approximately 1.33
16 16. FRACTIONAL CHARGES IN QUANTUM HALL STATES
Problem 16. Consider a 2D electron gas confined to a square box of side length Lin the xy-
plane. The electrons are subject to a strong magnetic field perpendicular to the plane. At filling
factor ν=1
3, the system exhibits fractional charges.
Given that the magnetic field strength B= 3 T and the elementary charge e= 1.6×10−19 C,
determine the fractional charge carried by the quasiparticles in this system.
Solution 16. The filling factor ν=Ne
Nϕ, where Neis the number of electrons and Nϕis the
number of magnetic flux quanta penetrating the surface of the system. For a square box, we have
Nϕ=BA
ϕ0, where A=L2is the area of the box, Bis the magnetic field strength, and ϕ0=h
eis the
magnetic flux quantum.
Given B= 3 T and ν=1
3, we have:
Nϕ=B·L2
ϕ0
=3T·(L2m2)
h
e
=3·109m−2·(L2)
2.07 ×10−15 Wb
=3·109·L2
2.07 ×10−15 flux quanta
For ν=1
3,Ne=1
3Nϕ. The fractional charge e∗=e
3. Thus, the quasiparticles in this system
carry a fractional charge of:
e∗=e
3
=1.6×10−19 C
3
= 5.33 ×10−20 C
Therefore, the quasiparticles in this system carry a fractional charge of 5.33 ×10−20 C.
17 17. TOPOLOGICAL SUPERCONDUCTIVITY IN TOPOLOGICAL INSULATORS
Problem 17. Consider a 2D topological insulator described by the Hamiltonian
H(k) = 0kx−iky
kx+iky0
a) Calculate the eigenvalues and eigenvectors of H(k).
b) Show that this Hamiltonian satisfies the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1,
where Θ = σyKis the time-reversal operator with σybeing the Pauli matrix and Kbeing complex
conjugation.
c) Determine if this system is a topological insulator by computing the Z2topological invariant.
Solution 17.
a) To find the eigenvalues and eigenvectors, we solve the characteristic equation det(H−λI) =
0. Let’s denote the eigenvalue as λand the eigenvector as ψ=a
b.
det −λ kx−iky
kx+iky−λ=λ2−(k2
x+k2
y) = 0
So, the eigenvalues are λ=±|k|, where |k|=qk2
x+k2
y. For λ=|k|, we have the eigenvector
ψ+=kx−iky
|k|
For λ=−|k|, the eigenvector is
ψ−=−|k|
kx+iky
b) Now, let’s check the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1:
H(−k) = 0−kx+iky
−kx−iky0
ΘH(k)Θ−1=0−1
1 0 0kx−iky
kx+iky0 0 1
−1 0=0−kx+iky
−kx−iky0
Thus, the Hamiltonian satisfies the time-reversal symmetry condition.
c) The Z2topological invariant for this 2D system can be calculated using the parity of the
determinant of the mass term M(k) = kxσx+kyσy. The Z2invariant is defined as
ν0=sgn(det[M(Γ)])
where Γis the time-reversal invariant momentum. In this case, Γ = (0,0).
The determinant of M(Γ) is det[M(Γ)] = 0, which implies that the system is a trivial insulator
with ν0= 0.
I am unable to generate numerical problems on demand as it requires creating specific sce-
narios and calculations. However, if you provide me with a specific scenario or problem statement
related to Topological Insulators and Quantum Hall Effects, I can definitely help you generate a
numerical problem along with a step-by-step explanation for the solution. Just let me know what
specific topic or concept you would like to focus on!
18 19. QUANTUM HALL EFFECTS IN GRAPHENE
Problem 19. Consider a monolayer graphene sheet under a magnetic field of strength B= 2 T.
The charge of an electron is e= 1.6×10−19 C and the Planck’s constant is h= 6.63 ×10−34 J s.
The Fermi velocity in graphene is vF= 106m/s. Calculate:
a) The magnetic length lBin the graphene sheet.
b) The energy level spacing ∆Ebetween Landau levels.
c) The filling factor νfor the third Landau level.
Solution 19.
a) The magnetic length lBis given by:
lB=r¯h
eB
Plugging in the values ¯h= 6.63 ×10−34 J s, e= 1.6×10−19 C, and B= 2 T, we get:
lB=r6.63 ×10−34 J s
1.6×10−19 C·2T≈26.1nm
Therefore, the magnetic length in the graphene sheet is approximately 26.1nm.
b) The energy spacing ∆Ebetween Landau levels is given by:
∆E=¯hvF
lB
Substitute the values ¯h= 6.63 ×10−34 J s, vF= 106m/s, and lB= 26.1nm:
∆E=6.63 ×10−34 J s ·106m/s
26.1×10−9m≈2.54 ×10−4eV
Hence, the energy level spacing between Landau levels is approximately 2.54 ×10−4eV.
c) The filling factor νfor the third Landau level is given by:
ν=N
Nϕ
Where Nis the number of electrons in the third Landau level and Nϕis the number of flux
quanta enclosed. For graphene, there are Nϕ=BA
ϕ0flux quanta enclosed for each unit cell area
A.
For the third Landau level (n= 3) in graphene, there are N= 2 electrons (spin degeneracy).
Thus,
Nϕ=BA
ϕ0
=(2T)(a2)
h/e =(2)(10−6
6.63 ×10−34 J s/1.6×10−19 C≈2.4×105
Therefore, the filling factor for the third Landau level in graphene is
ν=N
Nϕ
=2
2.4×105≈8.33 ×10−6
19 20. TOPOLOGICAL DEFECTS IN TOPOLOGICAL INSULATORS
Problem 20. Consider a one-dimensional lattice system described by the Hamiltonian
H=−t
N
X
n=1
(c†
ncn+1 +c†
n+1cn)−µ
N
X
n=1
c†
ncn
where cnand c†
nare annihilation and creation operators at site n, respectively, tis the hopping
parameter, µis the chemical potential, and Nis the number of lattice sites.
a) Calculate the energy spectrum of the system.
b) Determine the winding number for this system.
c) Show that there is a zero-energy bound state at a domain wall with an inverted mass.
Solution 20.
a) To calculate the energy spectrum of the system, we start by performing a Fourier transform
to the momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)−µ
The energy spectrum can be obtained by diagonalizing H(k),
E(k) = ±p(2tcos(k) + µ)2
b) The winding number is given by the integral of the Berry connection over the entire Brillouin
zone,
w=1
2πZBZ
dk A(k)
where A(k) = i⟨uk|∂k|uk⟩is the Berry connection, and |uk⟩is the periodic part of the Bloch
wavefunction. For this system, the winding number is w= 1.
c) At a domain wall with an inverted mass, the Hamiltonian changes sign, i.e., Hwall =−H.
Therefore, there exists a zero-energy solution at the domain wall due to the symmetric nature of
the zero-energy solution.
20 21. ANOMALOUS HALL EFFECT IN TOPOLOGICAL MATERIALS
Problem 21. Consider a 2D topological insulator with a Chern number C= 2. The Hall con-
ductivity σxy is given by the formula σxy =e2
hC, where eis the elementary charge and his the
Planck constant.
a) Calculate the Hall conductivity σxy for this 2D topological insulator.
b) If the number of edge modes on one edge of this material is 3, calculate the Hall conductivity
for each edge mode.
Solution 21.
a) Given that the Chern number C= 2, we can calculate the Hall conductivity using the formula
σxy =e2
hC.
Substitute e= 1.6×10−19 C and h= 6.63 ×10−34 J s:
σxy =1.6×10−19 C2
6.63×10−34 J s ×2 = 4.85 ×10−5Ω−1
Therefore, the Hall conductivity for this 2D topological insulator is 4.85 ×10−5Ω−1.
b) Since the number of edge modes on one edge is 3, we can calculate the Hall conductivity
for each edge mode using the formula σedge =σxy
number of edge modes .
For each edge mode, σedge =4.85×10−5Ω−1
3= 1.62 ×10−5Ω−1.
Therefore, the Hall conductivity for each edge mode on one edge of this material is 1.62 ×
10−5Ω−1.
21 Topological Insulators and Quantum Hall Effects
Problem 1. Consider a 2D topological insulator described by the Hamiltonian
H=2−iλ
iλ −2,
where λis a real parameter.
a) Determine the energy eigenvalues of the Hamiltonian.
b) Find the normalized eigenvectors corresponding to each energy eigenvalue.
Solution 1.
a) To find the energy eigenvalues, we solve the characteristic equation |H−ϵI|= 0:
2−ϵ−iλ
iλ −2−ϵ
= (2 −ϵ)(−2−ϵ) + λ2=ϵ2−4−λ2= 0.
This gives us the eigenvalues:
ϵ=±pλ2+ 4.
b) To find the normalized eigenvectors, we solve the eigenvalue equation (H−ϵI)v= 0 for
each eigenvalue.
For ϵ=√λ2+ 4:
2−√λ2+ 4 −iλ
iλ −2−√λ2+ 4v1
v2= 0,
we get the eigenvector v+=1
√2(λ2+4) λ
√λ2+ 4.
Similarly, for ϵ=−√λ2+ 4:
2 + √λ2+ 4 −iλ
iλ −2 + √λ2+ 4v1
v2= 0,
we get the eigenvector v−=1
√2(λ2+4) λ
−√λ2+ 4.
22 23. QUANTUM SPIN HALL TOPOLOGICAL INSULATORS
Problem 23. Consider a 1D chain with spin-orbit coupling described by the Hamiltonian H=
−tPn,σ(c†
n,σcn+1,σ +h.c.) + iλ Pn(c†
n,↑cn+1,↓−c†
n,↓cn+1,↑), where t= 1 (energy units) and λ= 0.5.
a) Find the energy spectrum of this system.
b) Determine if the system is a topological insulator based on the parity criterion.
Solution 23.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian. We
can do this by performing a Fourier transform to momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)σx−2λsin(k)σy, where σxand σyare the Pauli matrices.
Diagonalizing the Hamiltonian, we get H(k) = −2√t2+λ2cos(ϕ(k)), where cos(ϕ(k)) = t/√t2+λ2cos(k)−
λ/√t2+λ2sin(k).
Thus, the energy spectrum is given by E=−2√t2+λ2cos(ϕ(k)) = −2√1+0.25 cos(ϕ(k)) =
−2√1.25 cos(ϕ(k)), which simplifies to E(k) = −2 cos(ϕ(k)).
b) To determine if the system is a topological insulator based on the parity criterion, we need
to check the parity of the ground state. The ground state corresponds to the lowest energy, which
occurs at k= 0. At k= 0, the energy is E(0) = −2. Since the energy is non-zero, the system is
not a topological insulator based on the parity criterion.
23 24. HALF-INTEGER QUANTUM HALL EFFECTS IN TOPOLOGICAL SYSTEMS
Problem 24. Consider a 2D electron gas in a magnetic field with flux Φ = 2πϕ
ϕ0
, where ϕis a
dimensionless parameter and ϕ0=h
eis the magnetic flux quantum. The Hall conductance for this
system is given by σxy = (n+1
2)e2
h, where nis an integer.
a) Calculate the Hall conductance when n= 2.
b) Determine the value of ϕfor which the Hall conductance changes by e2
h.
Solution 24.
a) When n= 2, the Hall conductance is given by σxy = (2 + 1
2)e2
h=5
2
e2
h. Therefore, when
n= 2, the Hall conductance is 5
2
e2
h.
b) To find the value of ϕfor which the Hall conductance changes by e2
h, we need to consider the
change in nfrom nto n+1. The change in Hall conductance is given by ∆σxy =(n+ 1) + 1
2e2
h−
(n+1
2)e2
h=e2
h.
Solving for ∆σxy with ∆ϕ:
(n+ 1) + 1
2e2
h−(n+1
2)e2
h=e2
h
(n+1+1
2)−(n+1
2)=1
1=1
Therefore, the value of ϕfor which the Hall conductance changes by e2
his any value of ϕthat
results in an increase in nby 1.
24 25. TOPOLOGICAL INSULATOR DEVICES FOR SPINTRONICS APPLICATIONS
Problem 25. Consider a 2D topological insulator with band structure described by the Hamil-
tonian
H(k) = t(σxsin kx+σysin ky)
where k= (kx, ky),tis the hopping parameter, and σxand σyare Pauli matrices.
a) Determine the eigenvalues and eigenvectors of H(k).
b) Find the Chern number associated with this system.
c) Suppose a magnetic field is added which introduces a Zeeman term HZ=Mσz, where M
is the Zeeman splitting strength. How does this modification affect the topological properties of the
system?
Solution 25.
a) To find the eigenvalues ϵ(k)and eigenvectors v(k), we solve the equation H(k)v(k) =
ϵ(k)v(k).
H(k)v(k) = t(σxsin kx+σysin ky)v(k) = ϵ(k)v(k)
Expanding the matrix multiplication and solving for eigenvalues (ϵ(k) = ±t) and eigenvectors, we
get:
ϵ(k) = ±t, v±(k) = eiϕ
±eiϕ
where ϕ=kxfor the +branch and ϕ=kyfor the −branch.
b) The Chern number Cfor this system can be calculated using the formula
C=1
2πZ Z dkxdkyˆz ·(∂kxA×∂kyA)
where A=i⟨uk|∇k|uk⟩is the Berry connection. With our eigenvectors, we get A=1
2
t
|t|2(ˆz ×t)
and
C=sign(t)
c) Introducing the Zeeman term HZ=Mσzmodifies the Hamiltonian to H′(k) = H(k) +
HZ. This opens a band gap and shifts the energies, but as long as the time-reversal symmetry is
preserved, the system remains in the same topological phase with Chern number C=sign(t).
where nis the electron density per layer. Given that the density of states per layer is ν= 2 ×
1011 cm−2, we can calculate the electron density per layer as:
n=ν×104= 2 ×1011 ×104= 2 ×1015 m−2
Substitute the values into the equation:
VH=1
2×1015 ×1.6×10−19 ×1.4×10−4×0.5 = 218.75 V
Therefore, the Hall voltage VHacross the system is 218.75 V.
c) The transverse electric field Eyin the system is given by:
Ey=VH
d
where dis the distance between the layers. Given the current density j= 5 mA/cm2and the
electron density per layer n, we can calculate the drift velocity vd:
vd=j
nq =5×10−3
2×1015 ×1.6×10−19 = 1.56 ×10−3m/s
Assuming steady-state conditions, the drift velocity is related to the transverse electric field by
vd=µEy, where µis the electron mobility. Rearranging for Eywe have Ey=vd
µ.
Hence, we need to know the electron mobility to calculate Ey.
3 3. TOPOLOGICAL INSULATORS IN THE PRESENCE OF DISORDER
Problem 3. Consider a 1D topological insulator system described by the Hamiltonian:
H=−t
N−1
X
n=1
(c†
n+1cn+c†
ncn+1) + V
N
X
n=1
nc†
ncn
where c†
nand cnare creation and annihilation operators at site n,t= 1 is the hopping parameter,
Vis the strength of the disorder potential, and N= 6 is the total number of sites. Assume periodic
boundary conditions.
a) Find the energy eigenvalues and eigenfunctions of this Hamiltonian.
b) Calculate the Chern number of this system.
c) Determine the topological phase of the system based on the Chern number.
Solution 3.
a) To find the energy eigenvalues and eigenfunctions of the Hamiltonian, we first write it in
momentum space. The Hamiltonian in momentum space is:
H=X
k
ψ†
kH(k)ψk
where ψk= [ck, c−k]Tis the two-component wavefunction in momentum space and
H(k) = −2tcos(k)σx+V nσz
is the Hamiltonian in momentum space, σxand σzare Pauli matrices.
The energy eigenvalues are given by the diagonalization of H(k). Solving for the eigenvalues,
we get:
E=±p4t2cos2(k) + V2n2
The corresponding eigenvectors can be obtained by solving the eigenvector equations.
b) To calculate the Chern number of this system, we need to find the Berry curvature which is
given by:
F(k) = i⟨∂ku|∂k′u⟩−⟨∂k′u|∂ku⟩
where uis the wavefunction of the system.
After calculating the Berry curvature, the Chern number is given by integrating the Berry cur-
vature over the 1st Brillouin zone:
C=1
2πZdkdk′F(k)
c) Based on the Chern number, we can determine the topological phase of the system. If the
Chern number is non-zero, the system is in a topologically non-trivial phase.
4 4. CHIRAL EDGE STATES IN QUANTUM HALL SYSTEMS
Problem 4. Consider an integer Quantum Hall system with a chiral edge state. The chiral
edge state is described by a wave function of the form ψ(x) = Aeikx, where Ais the normalization
constant, kis the wave vector, and xis the position along the edge.
a) If the Fermi level of the system is EF= 2eV, and the edge state has a linear energy dispersion
relation E(k)=¯hvFkwith Fermi velocity vF= 105m/s, what is the wave vector kof the edge state?
b) Calculate the group velocity of the edge state.
c) Determine the direction of propagation (clockwise or counterclockwise) of the edge state.
Solution 4.
a) Given that the energy of the edge state is given by E(k)=¯hvFkand the Fermi level is
EF= 2eV, we set E(k) = EFand solve for k:
¯hvFk=EF
¯hvFk= 2eV
k=2eV
¯hvF
k=2×1.6×10−19C×105m/s
6.63 ×10−34m2kg/s×105m/s
k≈4.8×106m−1
Therefore, the wave vector of the edge state is k≈4.8×106m−1.
b) The group velocity of the edge state is given by the derivative of the energy dispersion relation
with respect to wave vector:
vg=dE
dk
vg=d(¯hvFk)
dk
vg= ¯hvF
Therefore, the group velocity of the edge state is vg= ¯hvF= 6.63 ×10−34m2kg/s×105m/s≈
6.63 ×10−29m/s.
c) Since the edge state is described by a wave function of the form ψ(x) = Aeikx, which has a
positive wave vector k, the edge state propagates in the clockwise direction along the edge.
I’m glad to help! Here is a numerical problem question on Topological Insulators and Quantum
Hall Effects:
5 5. TOPOLOGICAL PHASE TRANSITIONS IN QUANTUM HALL EFFECTS
Problem 5. Consider a 2D electron gas in a square lattice with a magnetic field applied per-
pendicular to the plane. The Hamiltonian for this system is given by:
H=X
r tX
i
c†
r+aicr+vX
i
eiθi
rc†
r+aicr+H.c.!
where crare the annihilation operators at lattice sites r,tis the nearest-neighbor hopping pa-
rameter, vis the strength of Rashba spin-orbit coupling, θi
ris the angle of the magnetic field at site
rwith respect to direction i, and aiare the lattice vectors.
Given that the strength of the Rashba spin-orbit coupling varies smoothly across the system with
domain walls separating regions with different coupling strengths, calculate the conditions that lead
to a topological phase transition in this system.
Solution 5. To determine the conditions for a topological phase transition, we need to look at
the Chern number of the system. The Chern number is given by:
C=1
2πZd2kF(k)
where F(k) = ∇k×A(k)is the Berry curvature and A(k) = −i⟨uk|∇k|uk⟩is the Berry con-
nection.
At the topological phase transition point, the energy gap at the Dirac points closes. This occurs
when v= 0 which happens at the domain walls where the coupling strength changes sign.
Therefore, the condition for a topological phase transition in this system is when the Rashba
spin-orbit coupling strength vchanges sign, leading to the closing of the energy gap at the Dirac
points.
I can provide a sample of a problem for you.
6 6. SPIN HALL EFFECT IN TOPOLOGICAL INSULATORS
Problem 6. Consider an electron moving in a two-dimensional topological insulator with the
following Hamiltonian:
H=3vkx−iλky
vkx+iλky−3
where v= 2 meV ·nm, λ= 1 meV ·nm, and kxand kyare the components of the wave vector.
Calculate the eigenvalues of this Hamiltonian.
Solution 6. To find the eigenvalues, we need to solve the characteristic equation given by
det(H−εI)=0, where εis the eigenvalue.
Substitute Hinto the characteristic equation:
det 3vkx−iλky
vkx+iλky−3−ε1 0
0 1= 0
Simplify this equation and solve for ε:
det 3−ε vkx−iλky
vkx+iλky−3−ε= 0
Expanding the determinant gives:
(3 −ε)(−3−ε)−(vkx−iλky)(vkx+iλky)=0
Solving this equation gives the two eigenvalues ε1and ε2.
Thus, the eigenvalues of the given Hamiltonian are:
ε1= 3 −q9 + v2k2
x+λ2k2
y
ε2= 3 + q9 + v2k2
x+λ2k2
y
7 7. QUANTUM SPIN HALL EFFECT IN TWO-DIMENSIONAL SYSTEMS
Problem 7. Consider a two-dimensional system with spin-orbit coupling described by the Hamil-
tonian
H=E αk−
αk+−E,
where k±=kx±iky,αis the strength of the spin-orbit coupling, and Eis the energy.
a) Find the energy eigenvalues of the system.
b) Determine the corresponding eigenvectors.
c) Show that this system exhibits the quantum spin Hall effect.
Solution 7.
a) To find the energy eigenvalues of the system, we need to solve the equation det(H−EI)=0,
where Iis the identity matrix.
Expanding the determinant, we have:
det(H−EI) = det E−E αk−
αk+−E−E
= (E+E)(E+E)−α2k−k+
= 4E2−α2kxky.
Setting this equal to zero gives us the energy eigenvalues E=±α
2pkxky.
b) To find the eigenvectors, let’s consider the eigenvalue E=α
2pkxky:
For E=α
2pkxky, the eigenvector u
vmust satisfy (H−EI)u
v= 0. Solving this system of
equations, we find the eigenvector corresponds to −ky
αkx.
Similarly, for E=−α
2pkxky, the eigenvector corresponds to ky
αkx.
c) The system exhibits the quantum spin Hall effect since it is characterized by a non-trivial Z2
topological invariant, which indicates the presence of helical edge states that are protected against
backscattering.
8 8. INTERACTION EFFECTS IN TOPOLOGICAL INSULATORS
Problem 8. Consider a 2D topological insulator described by the Hamiltonian H=−3v(kx−iky)
v(kx+iky) 3 ,
where vis a constant with units of velocity.
a) Calculate the energy spectrum of this system.
b) Determine the Chern number of this topological insulator.
c) Suppose there is an additional term in the Hamiltonian given by Hint =λσz, where λis a real
constant. How does this interaction affect the energy spectrum of the system?
Solution 8.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian H. The
eigenvalues of Hare given by solving the characteristic equation |H−EI|= 0, where Iis the
identity matrix.
We have: Det −3−E v(kx−iky)
v(kx+iky) 3 −E= (E+ 3)(E−3) −v2(kx+iky)(kx−iky) = E2−
9−v2(k2
x+k2
y).
This gives us the energy spectrum: E=±qv2(k2
x+k2
y)+9.
b) To calculate the Chern number, we first need to find the Berry curvature Ω(k). The Berry
curvature is given by Ω(k) = ∇ × A(k), where A(k) = −i⟨uk|∇k|uk⟩is the Berry connection.
We can calculate the Berry curvature and integrate it over the Brillouin zone to find the Chern
number.
c) The additional interaction term Hint =λσzintroduces a Zeeman-like splitting in the energy
levels. This term shifts the energies by ±λ, depending on the spin orientation.
9 9. TOPOLOGICAL INSULATORS IN MAGNETIC FIELDS
Problem 9. Consider a 2D topological insulator with a lattice constant a= 1 nm and a magnetic
field B=Bˆzapplied perpendicular to the material. The Fermi energy is EF= 100 meV and the
electron charge is e= 1.6×10−19 C.
Given that the magnetic field strength is B= 2 T and the electron velocity in the material is
v= 106m/s, calculate:
a) The cyclotron frequency of the electrons in the magnetic field.
b) The magnetic length.
c) The Landau level index of the first excited state.
Solution 9.
a) The cyclotron frequency ωcof the electrons in a magnetic field is given by
ωc=eB
m
where eis the electron charge, Bis the magnetic field strength, and mis the electron mass.
Given e= 1.6×10−19 C, B= 2 T, and m= 9.1×10−31 kg, we have
ωc=1.6×10−19 ×2
9.1×10−31 = 0.351 ×1012 rad/s
Therefore, the cyclotron frequency of the electrons in the magnetic field is 0.351 ×1012 rad/s.
b) The magnetic length lBis given by
lB=r¯h
eB
where ¯his the reduced Planck constant.
Given ¯h= 1.05 ×10−34 Js and e= 1.6×10−19 C, we have
lB=r1.05 ×10−34
1.6×10−19 ×2= 2.59 ×10−9m
Therefore, the magnetic length is 2.59 ×10−9m.
c) The Landau level index nof the first excited state can be calculated using the equation
En= ¯hωcn+1
2
where Enis the energy of the n-th Landau level.
Given ¯h= 1.05 ×10−34 Js, ωc= 0.351 ×1012 rad/s, and EF= 100 meV, we can rearrange the
equation to solve for n:
n=EF
¯hωc−1
2=100 ×10−3
1.05 ×10−34 ×0.351 ×1012 −1
2= 4.48
Therefore, the Landau level index of the first excited state is approximately 4.48.
10 10. FRACTIONAL QUANTUM HALL EFFECT IN TOPOLOGICAL SYSTEMS
Problem 10. Consider a 2D electron gas in a strong magnetic field with filling fraction ν=4
3.
The system has an effective magnetic length of lB= 10 nm and an electron charge e= 1.6×10−19
C. Calculate:
a) The magnetic field Bin Tesla.
b) The Hall conductance σxy in units of e2/h where h= 6.63 ×10−34 J s is the Planck constant.
Solution 10.
a) The magnetic field Bcan be related to the magnetic length lBas B=2π
l2
B
. Substituting
lB= 10 nm = 10 ×10−9m, we have:
B=2π
(10 ×10−9)2=2π
100 ×10−18 =2π
10−16 ≈6.28 ×1015 T
Therefore, the magnetic field B≈6.28 ×1015 T.
b) The Hall conductance is given by σxy =νe2
h. Substituting ν=4
3,e= 1.6×10−19 C, and
h= 6.63 ×10−34 J s, we get:
σxy =4
3×(1.6×10−19)2
6.63 ×10−34 =4
3×2.56 ×10−38
6.63 ×10−34 =10.24 ×10−38
6.63 ×10−34 =10.24
6.63 ×10−4S
σxy ≈1.54 ×10−4e2/h
Therefore, the Hall conductance σxy ≈1.54 ×10−4e2/h.
11 11. DISORDER-INDUCED LOCALIZATION IN QUANTUM HALL SYSTEMS
Problem 11. Consider a 2D square lattice with a magnetic field applied perpendicular to the
plane, giving rise to a Quantum Hall effect. At zero disorder, the system has a Hall conductivity of
σxy = 2e2/h.
a) If a weak disorder is introduced into the system, how does the value of the Hall conductivity
change?
b) Calculate the localization length ξof the system with weak disorder, given that the mean free
path lis 10 lattice spacings and the Fermi wavelength λFis 5 lattice spacings.
Solution 11.
a) Introduction of weak disorder into the system does not change the value of the Hall conduc-
tivity. This is because the Hall conductivity is a topological invariant and is robust against weak
disorder.
b) The localization length ξof the system can be calculated using the relation ξ=lλF
l2.
Substituting l= 10 and λF= 5 into the formula, we get:
ξ= 10 5
10 2= 10(0.25) = 2.5lattice spacings.
Therefore, the localization length of the system with weak disorder is ξ= 2.5lattice spacings.
12 12. TOPOLOGICAL INSULATORS WITH TIME-REVERSAL SYMMETRY
Problem 12. Consider a two-dimensional topological insulator described by the Bernevig-
Hughes-Zhang (BHZ) model Hamiltonian given by:
H(k) = (M−Bk2)σz+Akxσx−Akyσy
where σiare the Pauli matrices, M=−1,A= 1,B= 1, and ¯h= 1.
a) Determine the energy eigenvalues E(k)for this Hamiltonian.
b) Identify the topological invariants present in the BHZ model.
c) Find the topological phase diagram for the BHZ model in the M−Bparameter space.
Solution 12. a) To find the energy eigenvalues E(k), we diagonalize the Hamiltonian H(k):
H(k) = (M−Bk2)σz+Akxσx−Akyσy
The eigenvalues are given by E(k) = ±p(M−Bk2)2+A2k2.
b) The topological invariants for the BHZ model are the Chern number and the Z2invariant.
The Chern number is given by ν=1
2πRR dkxdkyFxy(k), where Fxy(k)is the Berry curvature.
It can be calculated using the formula Fxy(k) = ˆ
d(k)·(∂kxˆ
d(k)×∂kyˆ
d(k)).
The Z2invariant is determined using the parity of the number of edge states on a finite system.
c) The topological phase diagram for the BHZ model in the M−Bparameter space can be
determined by analyzing the topological invariants at different values of Mand B. By calculating the
Chern number and checking the presence of edge states, we can identify the different topological
phases in the phase diagram.
13 13. EDGE TRANSPORT IN QUANTUM HALL EFFECTS
Problem 13. Consider a quantum Hall system with a Hall conductance of σxy = 2e2/h and
a longitudinal conductance of σxx = 0. The system has N= 6 chiral edge modes moving in the
positive xdirection and N= 4 chiral edge modes moving in the negative xdirection. Assume that
the charge of an electron is −e.
a) Calculate the Hall current in the positive xdirection.
b) Calculate the Hall voltage across the system.
c) Determine the Hall resistance of the system.
Solution 13. a) The Hall current in the positive xdirection is given by the formula:
IHall =σxyVHall
where VHall is the Hall voltage. Since σxy = 2e2/h and the charge of an electron is −e, we have:
IHall = (2e2/h)·(−e)·(6) = −12e2/h
So, the Hall current in the positive xdirection is −12e2/h.
b) The Hall voltage across the system can be found by rearranging the formula for Hall current:
VHall =IHall
σxy
=−12e2/h
2e2/h =−6
Thus, the Hall voltage across the system is −6.
c) The Hall resistance of the system is given by:
RHall =VHall
IHall
=−6
−12e2/h =1
2h/e2
Therefore, the Hall resistance of the system is 1/2h/e2.
14 14. PROXIMITY EFFECTS IN TOPOLOGICAL INSULATOR HETEROSTRUCTURES
Problem 14. Consider a heterostructure composed of a normal insulator (NI) and a topological
insulator (TI) with a proximity-induced superconducting pairing in the TI region. The Hamiltonian
for this system is given by:
H=HNI +HT I +Hint
where HNI is the Hamiltonian of the normal insulator, HT I is the Hamiltonian of the topological
insulator, and Hint represents the interaction between the two regions. The system is described
by the Bogoliubov-de Gennes Hamiltonian:
HBdG =HNI i∆
−i∆HT I
where ∆is the pairing potential in the TI region.
Given that HNI =ϵN∆N
∆N−ϵN,HT I =ϵT∆T
∆T−ϵT, where ϵN= 2,∆N= 1,ϵT= 3,∆T= 2,
and ∆=1. Calculate the energy spectrum of the BdG Hamiltonian.
Solution 14. The Bogoliubov-de Gennes Hamiltonian is given by:
HBdG =
2 1 i0
1−2 0 i
−i032
0−i2−3
Expanding the determinant of the matrix HBdG −λI = 0, where λis the eigenvalue, we get:
det
2−λ1i0
1−2−λ0i
−i0 3 −λ2
0−i2−3−λ
= 0
Solving this equation gives the energy spectrum of the BdG Hamiltonian. Solving for λ, we get
the eigenvalues:
λ=±qϵ2
N+ ∆2
N,±qϵ2
T+ ∆2
T
Substitute the given values ϵN= 2,∆N= 1,ϵT= 3,∆T= 2 into the above equation to obtain
the energy spectrum. Therefore, the energy spectrum of the BdG Hamiltonian is:
λ=±√5,±√13
15 15. THERMOELECTRIC PROPERTIES OF TOPOLOGICAL INSULATORS
Problem 15. Consider a topological insulator with a band gap of 0.5 eV. The temperature at
the hot reservoir is Th= 300 K, and at the cold reservoir is Tc= 100 K. The Seebeck coefficient of
the material is S= 100 µV/K. Calculate:
a) The voltage generated when a temperature difference is created between the hot and cold
reservoirs.
b) The power generated if the hot reservoir is connected to a load with resistance R= 10Ω.
c) The efficiency of the thermoelectric device if the power generated in part b is used to drive a
load at room temperature (300 K).
Solution 15.
a) The voltage generated when a temperature difference is created is given by the Seebeck
effect equation:
V=S·(Th−Tc)
Substitute the given values:
V= 100 ×10−6V/K ×(300K−100K) = 20mV
Therefore, the voltage generated is 20 mV.
b) The power generated can be calculated using the formula for electrical power:
P=V2
R
Substitute the known values:
P=(0.02V)2
10Ω =0.0004
10 = 0.04mW
Therefore, the power generated is 0.04 mW.
c) The efficiency of the thermoelectric device is given by:
η=Useful power output
Heat input =P
Qh
Since the device is an ideal thermoelectric device, the heat input Qhis equal to the heat ab-
sorbed from the hot reservoir:
Qh=Th·S
Substitute the values:
Qh= 300K×100 ×10−6V/K = 30mV = 0.03W
Finally, calculate the efficiency:
η=0.04mW
0.03W≈1.33%
Therefore, the efficiency of the thermoelectric device is approximately 1.33
16 16. FRACTIONAL CHARGES IN QUANTUM HALL STATES
Problem 16. Consider a 2D electron gas confined to a square box of side length Lin the xy-
plane. The electrons are subject to a strong magnetic field perpendicular to the plane. At filling
factor ν=1
3, the system exhibits fractional charges.
Given that the magnetic field strength B= 3 T and the elementary charge e= 1.6×10−19 C,
determine the fractional charge carried by the quasiparticles in this system.
Solution 16. The filling factor ν=Ne
Nϕ, where Neis the number of electrons and Nϕis the
number of magnetic flux quanta penetrating the surface of the system. For a square box, we have
Nϕ=BA
ϕ0, where A=L2is the area of the box, Bis the magnetic field strength, and ϕ0=h
eis the
magnetic flux quantum.
Given B= 3 T and ν=1
3, we have:
Nϕ=B·L2
ϕ0
=3T·(L2m2)
h
e
=3·109m−2·(L2)
2.07 ×10−15 Wb
=3·109·L2
2.07 ×10−15 flux quanta
For ν=1
3,Ne=1
3Nϕ. The fractional charge e∗=e
3. Thus, the quasiparticles in this system
carry a fractional charge of:
e∗=e
3
=1.6×10−19 C
3
= 5.33 ×10−20 C
Therefore, the quasiparticles in this system carry a fractional charge of 5.33 ×10−20 C.
17 17. TOPOLOGICAL SUPERCONDUCTIVITY IN TOPOLOGICAL INSULATORS
Problem 17. Consider a 2D topological insulator described by the Hamiltonian
H(k) = 0kx−iky
kx+iky0
a) Calculate the eigenvalues and eigenvectors of H(k).
b) Show that this Hamiltonian satisfies the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1,
where Θ = σyKis the time-reversal operator with σybeing the Pauli matrix and Kbeing complex
conjugation.
c) Determine if this system is a topological insulator by computing the Z2topological invariant.
Solution 17.
a) To find the eigenvalues and eigenvectors, we solve the characteristic equation det(H−λI) =
0. Let’s denote the eigenvalue as λand the eigenvector as ψ=a
b.
det −λ kx−iky
kx+iky−λ=λ2−(k2
x+k2
y) = 0
So, the eigenvalues are λ=±|k|, where |k|=qk2
x+k2
y. For λ=|k|, we have the eigenvector
ψ+=kx−iky
|k|
For λ=−|k|, the eigenvector is
ψ−=−|k|
kx+iky
b) Now, let’s check the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1:
H(−k) = 0−kx+iky
−kx−iky0
ΘH(k)Θ−1=0−1
1 0 0kx−iky
kx+iky0 0 1
−1 0=0−kx+iky
−kx−iky0
Thus, the Hamiltonian satisfies the time-reversal symmetry condition.
c) The Z2topological invariant for this 2D system can be calculated using the parity of the
determinant of the mass term M(k) = kxσx+kyσy. The Z2invariant is defined as
ν0=sgn(det[M(Γ)])
where Γis the time-reversal invariant momentum. In this case, Γ = (0,0).
The determinant of M(Γ) is det[M(Γ)] = 0, which implies that the system is a trivial insulator
with ν0= 0.
I am unable to generate numerical problems on demand as it requires creating specific sce-
narios and calculations. However, if you provide me with a specific scenario or problem statement
related to Topological Insulators and Quantum Hall Effects, I can definitely help you generate a
numerical problem along with a step-by-step explanation for the solution. Just let me know what
specific topic or concept you would like to focus on!
18 19. QUANTUM HALL EFFECTS IN GRAPHENE
Problem 19. Consider a monolayer graphene sheet under a magnetic field of strength B= 2 T.
The charge of an electron is e= 1.6×10−19 C and the Planck’s constant is h= 6.63 ×10−34 J s.
The Fermi velocity in graphene is vF= 106m/s. Calculate:
a) The magnetic length lBin the graphene sheet.
b) The energy level spacing ∆Ebetween Landau levels.
c) The filling factor νfor the third Landau level.
Solution 19.
a) The magnetic length lBis given by:
lB=r¯h
eB
Plugging in the values ¯h= 6.63 ×10−34 J s, e= 1.6×10−19 C, and B= 2 T, we get:
lB=r6.63 ×10−34 J s
1.6×10−19 C·2T≈26.1nm
Therefore, the magnetic length in the graphene sheet is approximately 26.1nm.
b) The energy spacing ∆Ebetween Landau levels is given by:
∆E=¯hvF
lB
Substitute the values ¯h= 6.63 ×10−34 J s, vF= 106m/s, and lB= 26.1nm:
∆E=6.63 ×10−34 J s ·106m/s
26.1×10−9m≈2.54 ×10−4eV
Hence, the energy level spacing between Landau levels is approximately 2.54 ×10−4eV.
c) The filling factor νfor the third Landau level is given by:
ν=N
Nϕ
Where Nis the number of electrons in the third Landau level and Nϕis the number of flux
quanta enclosed. For graphene, there are Nϕ=BA
ϕ0flux quanta enclosed for each unit cell area
A.
For the third Landau level (n= 3) in graphene, there are N= 2 electrons (spin degeneracy).
Thus,
Nϕ=BA
ϕ0
=(2T)(a2)
h/e =(2)(10−6
6.63 ×10−34 J s/1.6×10−19 C≈2.4×105
Therefore, the filling factor for the third Landau level in graphene is
ν=N
Nϕ
=2
2.4×105≈8.33 ×10−6
19 20. TOPOLOGICAL DEFECTS IN TOPOLOGICAL INSULATORS
Problem 20. Consider a one-dimensional lattice system described by the Hamiltonian
H=−t
N
X
n=1
(c†
ncn+1 +c†
n+1cn)−µ
N
X
n=1
c†
ncn
where cnand c†
nare annihilation and creation operators at site n, respectively, tis the hopping
parameter, µis the chemical potential, and Nis the number of lattice sites.
a) Calculate the energy spectrum of the system.
b) Determine the winding number for this system.
c) Show that there is a zero-energy bound state at a domain wall with an inverted mass.
Solution 20.
a) To calculate the energy spectrum of the system, we start by performing a Fourier transform
to the momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)−µ
The energy spectrum can be obtained by diagonalizing H(k),
E(k) = ±p(2tcos(k) + µ)2
b) The winding number is given by the integral of the Berry connection over the entire Brillouin
zone,
w=1
2πZBZ
dk A(k)
where A(k) = i⟨uk|∂k|uk⟩is the Berry connection, and |uk⟩is the periodic part of the Bloch
wavefunction. For this system, the winding number is w= 1.
c) At a domain wall with an inverted mass, the Hamiltonian changes sign, i.e., Hwall =−H.
Therefore, there exists a zero-energy solution at the domain wall due to the symmetric nature of
the zero-energy solution.
20 21. ANOMALOUS HALL EFFECT IN TOPOLOGICAL MATERIALS
Problem 21. Consider a 2D topological insulator with a Chern number C= 2. The Hall con-
ductivity σxy is given by the formula σxy =e2
hC, where eis the elementary charge and his the
Planck constant.
a) Calculate the Hall conductivity σxy for this 2D topological insulator.
b) If the number of edge modes on one edge of this material is 3, calculate the Hall conductivity
for each edge mode.
Solution 21.
a) Given that the Chern number C= 2, we can calculate the Hall conductivity using the formula
σxy =e2
hC.
Substitute e= 1.6×10−19 C and h= 6.63 ×10−34 J s:
σxy =1.6×10−19 C2
6.63×10−34 J s ×2 = 4.85 ×10−5Ω−1
Therefore, the Hall conductivity for this 2D topological insulator is 4.85 ×10−5Ω−1.
b) Since the number of edge modes on one edge is 3, we can calculate the Hall conductivity
for each edge mode using the formula σedge =σxy
number of edge modes .
For each edge mode, σedge =4.85×10−5Ω−1
3= 1.62 ×10−5Ω−1.
Therefore, the Hall conductivity for each edge mode on one edge of this material is 1.62 ×
10−5Ω−1.
21 Topological Insulators and Quantum Hall Effects
Problem 1. Consider a 2D topological insulator described by the Hamiltonian
H=2−iλ
iλ −2,
where λis a real parameter.
a) Determine the energy eigenvalues of the Hamiltonian.
b) Find the normalized eigenvectors corresponding to each energy eigenvalue.
Solution 1.
a) To find the energy eigenvalues, we solve the characteristic equation |H−ϵI|= 0:
2−ϵ−iλ
iλ −2−ϵ
= (2 −ϵ)(−2−ϵ) + λ2=ϵ2−4−λ2= 0.
This gives us the eigenvalues:
ϵ=±pλ2+ 4.
b) To find the normalized eigenvectors, we solve the eigenvalue equation (H−ϵI)v= 0 for
each eigenvalue.
For ϵ=√λ2+ 4:
2−√λ2+ 4 −iλ
iλ −2−√λ2+ 4v1
v2= 0,
we get the eigenvector v+=1
√2(λ2+4) λ
√λ2+ 4.
Similarly, for ϵ=−√λ2+ 4:
2 + √λ2+ 4 −iλ
iλ −2 + √λ2+ 4v1
v2= 0,
we get the eigenvector v−=1
√2(λ2+4) λ
−√λ2+ 4.
22 23. QUANTUM SPIN HALL TOPOLOGICAL INSULATORS
Problem 23. Consider a 1D chain with spin-orbit coupling described by the Hamiltonian H=
−tPn,σ(c†
n,σcn+1,σ +h.c.) + iλ Pn(c†
n,↑cn+1,↓−c†
n,↓cn+1,↑), where t= 1 (energy units) and λ= 0.5.
a) Find the energy spectrum of this system.
b) Determine if the system is a topological insulator based on the parity criterion.
Solution 23.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian. We
can do this by performing a Fourier transform to momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)σx−2λsin(k)σy, where σxand σyare the Pauli matrices.
Diagonalizing the Hamiltonian, we get H(k) = −2√t2+λ2cos(ϕ(k)), where cos(ϕ(k)) = t/√t2+λ2cos(k)−
λ/√t2+λ2sin(k).
Thus, the energy spectrum is given by E=−2√t2+λ2cos(ϕ(k)) = −2√1+0.25 cos(ϕ(k)) =
−2√1.25 cos(ϕ(k)), which simplifies to E(k) = −2 cos(ϕ(k)).
b) To determine if the system is a topological insulator based on the parity criterion, we need
to check the parity of the ground state. The ground state corresponds to the lowest energy, which
occurs at k= 0. At k= 0, the energy is E(0) = −2. Since the energy is non-zero, the system is
not a topological insulator based on the parity criterion.
23 24. HALF-INTEGER QUANTUM HALL EFFECTS IN TOPOLOGICAL SYSTEMS
Problem 24. Consider a 2D electron gas in a magnetic field with flux Φ = 2πϕ
ϕ0
, where ϕis a
dimensionless parameter and ϕ0=h
eis the magnetic flux quantum. The Hall conductance for this
system is given by σxy = (n+1
2)e2
h, where nis an integer.
a) Calculate the Hall conductance when n= 2.
b) Determine the value of ϕfor which the Hall conductance changes by e2
h.
Solution 24.
a) When n= 2, the Hall conductance is given by σxy = (2 + 1
2)e2
h=5
2
e2
h. Therefore, when
n= 2, the Hall conductance is 5
2
e2
h.
b) To find the value of ϕfor which the Hall conductance changes by e2
h, we need to consider the
change in nfrom nto n+1. The change in Hall conductance is given by ∆σxy =(n+ 1) + 1
2e2
h−
(n+1
2)e2
h=e2
h.
Solving for ∆σxy with ∆ϕ:
(n+ 1) + 1
2e2
h−(n+1
2)e2
h=e2
h
(n+1+1
2)−(n+1
2)=1
1=1
Therefore, the value of ϕfor which the Hall conductance changes by e2
his any value of ϕthat
results in an increase in nby 1.
24 25. TOPOLOGICAL INSULATOR DEVICES FOR SPINTRONICS APPLICATIONS
Problem 25. Consider a 2D topological insulator with band structure described by the Hamil-
tonian
H(k) = t(σxsin kx+σysin ky)
where k= (kx, ky),tis the hopping parameter, and σxand σyare Pauli matrices.
a) Determine the eigenvalues and eigenvectors of H(k).
b) Find the Chern number associated with this system.
c) Suppose a magnetic field is added which introduces a Zeeman term HZ=Mσz, where M
is the Zeeman splitting strength. How does this modification affect the topological properties of the
system?
Solution 25.
a) To find the eigenvalues ϵ(k)and eigenvectors v(k), we solve the equation H(k)v(k) =
ϵ(k)v(k).
H(k)v(k) = t(σxsin kx+σysin ky)v(k) = ϵ(k)v(k)
Expanding the matrix multiplication and solving for eigenvalues (ϵ(k) = ±t) and eigenvectors, we
get:
ϵ(k) = ±t, v±(k) = eiϕ
±eiϕ
where ϕ=kxfor the +branch and ϕ=kyfor the −branch.
b) The Chern number Cfor this system can be calculated using the formula
C=1
2πZ Z dkxdkyˆz ·(∂kxA×∂kyA)
where A=i⟨uk|∇k|uk⟩is the Berry connection. With our eigenvectors, we get A=1
2
t
|t|2(ˆz ×t)
and
C=sign(t)
c) Introducing the Zeeman term HZ=Mσzmodifies the Hamiltonian to H′(k) = H(k) +
HZ. This opens a band gap and shifts the energies, but as long as the time-reversal symmetry is
preserved, the system remains in the same topological phase with Chern number C=sign(t).
where nis the electron density per layer. Given that the density of states per layer is ν= 2 ×
1011 cm−2, we can calculate the electron density per layer as:
n=ν×104= 2 ×1011 ×104= 2 ×1015 m−2
Substitute the values into the equation:
VH=1
2×1015 ×1.6×10−19 ×1.4×10−4×0.5 = 218.75 V
Therefore, the Hall voltage VHacross the system is 218.75 V.
c) The transverse electric field Eyin the system is given by:
Ey=VH
d
where dis the distance between the layers. Given the current density j= 5 mA/cm2and the
electron density per layer n, we can calculate the drift velocity vd:
vd=j
nq =5×10−3
2×1015 ×1.6×10−19 = 1.56 ×10−3m/s
Assuming steady-state conditions, the drift velocity is related to the transverse electric field by
vd=µEy, where µis the electron mobility. Rearranging for Eywe have Ey=vd
µ.
Hence, we need to know the electron mobility to calculate Ey.
3 3. TOPOLOGICAL INSULATORS IN THE PRESENCE OF DISORDER
Problem 3. Consider a 1D topological insulator system described by the Hamiltonian:
H=−t
N−1
X
n=1
(c†
n+1cn+c†
ncn+1) + V
N
X
n=1
nc†
ncn
where c†
nand cnare creation and annihilation operators at site n,t= 1 is the hopping parameter,
Vis the strength of the disorder potential, and N= 6 is the total number of sites. Assume periodic
boundary conditions.
a) Find the energy eigenvalues and eigenfunctions of this Hamiltonian.
b) Calculate the Chern number of this system.
c) Determine the topological phase of the system based on the Chern number.
Solution 3.
a) To find the energy eigenvalues and eigenfunctions of the Hamiltonian, we first write it in
momentum space. The Hamiltonian in momentum space is:
H=X
k
ψ†
kH(k)ψk
where ψk= [ck, c−k]Tis the two-component wavefunction in momentum space and
H(k) = −2tcos(k)σx+V nσz
is the Hamiltonian in momentum space, σxand σzare Pauli matrices.
The energy eigenvalues are given by the diagonalization of H(k). Solving for the eigenvalues,
we get:
E=±p4t2cos2(k) + V2n2
The corresponding eigenvectors can be obtained by solving the eigenvector equations.
b) To calculate the Chern number of this system, we need to find the Berry curvature which is
given by:
F(k) = i⟨∂ku|∂k′u⟩−⟨∂k′u|∂ku⟩
where uis the wavefunction of the system.
After calculating the Berry curvature, the Chern number is given by integrating the Berry cur-
vature over the 1st Brillouin zone:
C=1
2πZdkdk′F(k)
c) Based on the Chern number, we can determine the topological phase of the system. If the
Chern number is non-zero, the system is in a topologically non-trivial phase.
4 4. CHIRAL EDGE STATES IN QUANTUM HALL SYSTEMS
Problem 4. Consider an integer Quantum Hall system with a chiral edge state. The chiral
edge state is described by a wave function of the form ψ(x) = Aeikx, where Ais the normalization
constant, kis the wave vector, and xis the position along the edge.
a) If the Fermi level of the system is EF= 2eV, and the edge state has a linear energy dispersion
relation E(k)=¯hvFkwith Fermi velocity vF= 105m/s, what is the wave vector kof the edge state?
b) Calculate the group velocity of the edge state.
c) Determine the direction of propagation (clockwise or counterclockwise) of the edge state.
Solution 4.
a) Given that the energy of the edge state is given by E(k)=¯hvFkand the Fermi level is
EF= 2eV, we set E(k) = EFand solve for k:
¯hvFk=EF
¯hvFk= 2eV
k=2eV
¯hvF
k=2×1.6×10−19C×105m/s
6.63 ×10−34m2kg/s×105m/s
k≈4.8×106m−1
Therefore, the wave vector of the edge state is k≈4.8×106m−1.
b) The group velocity of the edge state is given by the derivative of the energy dispersion relation
with respect to wave vector:
vg=dE
dk
vg=d(¯hvFk)
dk
vg= ¯hvF
Therefore, the group velocity of the edge state is vg= ¯hvF= 6.63 ×10−34m2kg/s×105m/s≈
6.63 ×10−29m/s.
c) Since the edge state is described by a wave function of the form ψ(x) = Aeikx, which has a
positive wave vector k, the edge state propagates in the clockwise direction along the edge.
I’m glad to help! Here is a numerical problem question on Topological Insulators and Quantum
Hall Effects:
5 5. TOPOLOGICAL PHASE TRANSITIONS IN QUANTUM HALL EFFECTS
Problem 5. Consider a 2D electron gas in a square lattice with a magnetic field applied per-
pendicular to the plane. The Hamiltonian for this system is given by:
H=X
r tX
i
c†
r+aicr+vX
i
eiθi
rc†
r+aicr+H.c.!
where crare the annihilation operators at lattice sites r,tis the nearest-neighbor hopping pa-
rameter, vis the strength of Rashba spin-orbit coupling, θi
ris the angle of the magnetic field at site
rwith respect to direction i, and aiare the lattice vectors.
Given that the strength of the Rashba spin-orbit coupling varies smoothly across the system with
domain walls separating regions with different coupling strengths, calculate the conditions that lead
to a topological phase transition in this system.
Solution 5. To determine the conditions for a topological phase transition, we need to look at
the Chern number of the system. The Chern number is given by:
C=1
2πZd2kF(k)
where F(k) = ∇k×A(k)is the Berry curvature and A(k) = −i⟨uk|∇k|uk⟩is the Berry con-
nection.
At the topological phase transition point, the energy gap at the Dirac points closes. This occurs
when v= 0 which happens at the domain walls where the coupling strength changes sign.
Therefore, the condition for a topological phase transition in this system is when the Rashba
spin-orbit coupling strength vchanges sign, leading to the closing of the energy gap at the Dirac
points.
I can provide a sample of a problem for you.
6 6. SPIN HALL EFFECT IN TOPOLOGICAL INSULATORS
Problem 6. Consider an electron moving in a two-dimensional topological insulator with the
following Hamiltonian:
H=3vkx−iλky
vkx+iλky−3
where v= 2 meV ·nm, λ= 1 meV ·nm, and kxand kyare the components of the wave vector.
Calculate the eigenvalues of this Hamiltonian.
Solution 6. To find the eigenvalues, we need to solve the characteristic equation given by
det(H−εI)=0, where εis the eigenvalue.
Substitute Hinto the characteristic equation:
det 3vkx−iλky
vkx+iλky−3−ε1 0
0 1= 0
Simplify this equation and solve for ε:
det 3−ε vkx−iλky
vkx+iλky−3−ε= 0
Expanding the determinant gives:
(3 −ε)(−3−ε)−(vkx−iλky)(vkx+iλky)=0
Solving this equation gives the two eigenvalues ε1and ε2.
Thus, the eigenvalues of the given Hamiltonian are:
ε1= 3 −q9 + v2k2
x+λ2k2
y
ε2= 3 + q9 + v2k2
x+λ2k2
y
7 7. QUANTUM SPIN HALL EFFECT IN TWO-DIMENSIONAL SYSTEMS
Problem 7. Consider a two-dimensional system with spin-orbit coupling described by the Hamil-
tonian
H=E αk−
αk+−E,
where k±=kx±iky,αis the strength of the spin-orbit coupling, and Eis the energy.
a) Find the energy eigenvalues of the system.
b) Determine the corresponding eigenvectors.
c) Show that this system exhibits the quantum spin Hall effect.
Solution 7.
a) To find the energy eigenvalues of the system, we need to solve the equation det(H−EI)=0,
where Iis the identity matrix.
Expanding the determinant, we have:
det(H−EI) = det E−E αk−
αk+−E−E
= (E+E)(E+E)−α2k−k+
= 4E2−α2kxky.
Setting this equal to zero gives us the energy eigenvalues E=±α
2pkxky.
b) To find the eigenvectors, let’s consider the eigenvalue E=α
2pkxky:
For E=α
2pkxky, the eigenvector u
vmust satisfy (H−EI)u
v= 0. Solving this system of
equations, we find the eigenvector corresponds to −ky
αkx.
Similarly, for E=−α
2pkxky, the eigenvector corresponds to ky
αkx.
c) The system exhibits the quantum spin Hall effect since it is characterized by a non-trivial Z2
topological invariant, which indicates the presence of helical edge states that are protected against
backscattering.
8 8. INTERACTION EFFECTS IN TOPOLOGICAL INSULATORS
Problem 8. Consider a 2D topological insulator described by the Hamiltonian H=−3v(kx−iky)
v(kx+iky) 3 ,
where vis a constant with units of velocity.
a) Calculate the energy spectrum of this system.
b) Determine the Chern number of this topological insulator.
c) Suppose there is an additional term in the Hamiltonian given by Hint =λσz, where λis a real
constant. How does this interaction affect the energy spectrum of the system?
Solution 8.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian H. The
eigenvalues of Hare given by solving the characteristic equation |H−EI|= 0, where Iis the
identity matrix.
We have: Det −3−E v(kx−iky)
v(kx+iky) 3 −E= (E+ 3)(E−3) −v2(kx+iky)(kx−iky) = E2−
9−v2(k2
x+k2
y).
This gives us the energy spectrum: E=±qv2(k2
x+k2
y)+9.
b) To calculate the Chern number, we first need to find the Berry curvature Ω(k). The Berry
curvature is given by Ω(k) = ∇ × A(k), where A(k) = −i⟨uk|∇k|uk⟩is the Berry connection.
We can calculate the Berry curvature and integrate it over the Brillouin zone to find the Chern
number.
c) The additional interaction term Hint =λσzintroduces a Zeeman-like splitting in the energy
levels. This term shifts the energies by ±λ, depending on the spin orientation.
9 9. TOPOLOGICAL INSULATORS IN MAGNETIC FIELDS
Problem 9. Consider a 2D topological insulator with a lattice constant a= 1 nm and a magnetic
field B=Bˆzapplied perpendicular to the material. The Fermi energy is EF= 100 meV and the
electron charge is e= 1.6×10−19 C.
Given that the magnetic field strength is B= 2 T and the electron velocity in the material is
v= 106m/s, calculate:
a) The cyclotron frequency of the electrons in the magnetic field.
b) The magnetic length.
c) The Landau level index of the first excited state.
Solution 9.
a) The cyclotron frequency ωcof the electrons in a magnetic field is given by
ωc=eB
m
where eis the electron charge, Bis the magnetic field strength, and mis the electron mass.
Given e= 1.6×10−19 C, B= 2 T, and m= 9.1×10−31 kg, we have
ωc=1.6×10−19 ×2
9.1×10−31 = 0.351 ×1012 rad/s
Therefore, the cyclotron frequency of the electrons in the magnetic field is 0.351 ×1012 rad/s.
b) The magnetic length lBis given by
lB=r¯h
eB
where ¯his the reduced Planck constant.
Given ¯h= 1.05 ×10−34 Js and e= 1.6×10−19 C, we have
lB=r1.05 ×10−34
1.6×10−19 ×2= 2.59 ×10−9m
Therefore, the magnetic length is 2.59 ×10−9m.
c) The Landau level index nof the first excited state can be calculated using the equation
En= ¯hωcn+1
2
where Enis the energy of the n-th Landau level.
Given ¯h= 1.05 ×10−34 Js, ωc= 0.351 ×1012 rad/s, and EF= 100 meV, we can rearrange the
equation to solve for n:
n=EF
¯hωc−1
2=100 ×10−3
1.05 ×10−34 ×0.351 ×1012 −1
2= 4.48
Therefore, the Landau level index of the first excited state is approximately 4.48.
10 10. FRACTIONAL QUANTUM HALL EFFECT IN TOPOLOGICAL SYSTEMS
Problem 10. Consider a 2D electron gas in a strong magnetic field with filling fraction ν=4
3.
The system has an effective magnetic length of lB= 10 nm and an electron charge e= 1.6×10−19
C. Calculate:
a) The magnetic field Bin Tesla.
b) The Hall conductance σxy in units of e2/h where h= 6.63 ×10−34 J s is the Planck constant.
Solution 10.
a) The magnetic field Bcan be related to the magnetic length lBas B=2π
l2
B
. Substituting
lB= 10 nm = 10 ×10−9m, we have:
B=2π
(10 ×10−9)2=2π
100 ×10−18 =2π
10−16 ≈6.28 ×1015 T
Therefore, the magnetic field B≈6.28 ×1015 T.
b) The Hall conductance is given by σxy =νe2
h. Substituting ν=4
3,e= 1.6×10−19 C, and
h= 6.63 ×10−34 J s, we get:
σxy =4
3×(1.6×10−19)2
6.63 ×10−34 =4
3×2.56 ×10−38
6.63 ×10−34 =10.24 ×10−38
6.63 ×10−34 =10.24
6.63 ×10−4S
σxy ≈1.54 ×10−4e2/h
Therefore, the Hall conductance σxy ≈1.54 ×10−4e2/h.
11 11. DISORDER-INDUCED LOCALIZATION IN QUANTUM HALL SYSTEMS
Problem 11. Consider a 2D square lattice with a magnetic field applied perpendicular to the
plane, giving rise to a Quantum Hall effect. At zero disorder, the system has a Hall conductivity of
σxy = 2e2/h.
a) If a weak disorder is introduced into the system, how does the value of the Hall conductivity
change?
b) Calculate the localization length ξof the system with weak disorder, given that the mean free
path lis 10 lattice spacings and the Fermi wavelength λFis 5 lattice spacings.
Solution 11.
a) Introduction of weak disorder into the system does not change the value of the Hall conduc-
tivity. This is because the Hall conductivity is a topological invariant and is robust against weak
disorder.
b) The localization length ξof the system can be calculated using the relation ξ=lλF
l2.
Substituting l= 10 and λF= 5 into the formula, we get:
ξ= 10 5
10 2= 10(0.25) = 2.5lattice spacings.
Therefore, the localization length of the system with weak disorder is ξ= 2.5lattice spacings.
12 12. TOPOLOGICAL INSULATORS WITH TIME-REVERSAL SYMMETRY
Problem 12. Consider a two-dimensional topological insulator described by the Bernevig-
Hughes-Zhang (BHZ) model Hamiltonian given by:
H(k) = (M−Bk2)σz+Akxσx−Akyσy
where σiare the Pauli matrices, M=−1,A= 1,B= 1, and ¯h= 1.
a) Determine the energy eigenvalues E(k)for this Hamiltonian.
b) Identify the topological invariants present in the BHZ model.
c) Find the topological phase diagram for the BHZ model in the M−Bparameter space.
Solution 12. a) To find the energy eigenvalues E(k), we diagonalize the Hamiltonian H(k):
H(k) = (M−Bk2)σz+Akxσx−Akyσy
The eigenvalues are given by E(k) = ±p(M−Bk2)2+A2k2.
b) The topological invariants for the BHZ model are the Chern number and the Z2invariant.
The Chern number is given by ν=1
2πRR dkxdkyFxy(k), where Fxy(k)is the Berry curvature.
It can be calculated using the formula Fxy(k) = ˆ
d(k)·(∂kxˆ
d(k)×∂kyˆ
d(k)).
The Z2invariant is determined using the parity of the number of edge states on a finite system.
c) The topological phase diagram for the BHZ model in the M−Bparameter space can be
determined by analyzing the topological invariants at different values of Mand B. By calculating the
Chern number and checking the presence of edge states, we can identify the different topological
phases in the phase diagram.
13 13. EDGE TRANSPORT IN QUANTUM HALL EFFECTS
Problem 13. Consider a quantum Hall system with a Hall conductance of σxy = 2e2/h and
a longitudinal conductance of σxx = 0. The system has N= 6 chiral edge modes moving in the
positive xdirection and N= 4 chiral edge modes moving in the negative xdirection. Assume that
the charge of an electron is −e.
a) Calculate the Hall current in the positive xdirection.
b) Calculate the Hall voltage across the system.
c) Determine the Hall resistance of the system.
Solution 13. a) The Hall current in the positive xdirection is given by the formula:
IHall =σxyVHall
where VHall is the Hall voltage. Since σxy = 2e2/h and the charge of an electron is −e, we have:
IHall = (2e2/h)·(−e)·(6) = −12e2/h
So, the Hall current in the positive xdirection is −12e2/h.
b) The Hall voltage across the system can be found by rearranging the formula for Hall current:
VHall =IHall
σxy
=−12e2/h
2e2/h =−6
Thus, the Hall voltage across the system is −6.
c) The Hall resistance of the system is given by:
RHall =VHall
IHall
=−6
−12e2/h =1
2h/e2
Therefore, the Hall resistance of the system is 1/2h/e2.
14 14. PROXIMITY EFFECTS IN TOPOLOGICAL INSULATOR HETEROSTRUCTURES
Problem 14. Consider a heterostructure composed of a normal insulator (NI) and a topological
insulator (TI) with a proximity-induced superconducting pairing in the TI region. The Hamiltonian
for this system is given by:
H=HNI +HT I +Hint
where HNI is the Hamiltonian of the normal insulator, HT I is the Hamiltonian of the topological
insulator, and Hint represents the interaction between the two regions. The system is described
by the Bogoliubov-de Gennes Hamiltonian:
HBdG =HNI i∆
−i∆HT I
where ∆is the pairing potential in the TI region.
Given that HNI =ϵN∆N
∆N−ϵN,HT I =ϵT∆T
∆T−ϵT, where ϵN= 2,∆N= 1,ϵT= 3,∆T= 2,
and ∆=1. Calculate the energy spectrum of the BdG Hamiltonian.
Solution 14. The Bogoliubov-de Gennes Hamiltonian is given by:
HBdG =
2 1 i0
1−2 0 i
−i032
0−i2−3
Expanding the determinant of the matrix HBdG −λI = 0, where λis the eigenvalue, we get:
det
2−λ1i0
1−2−λ0i
−i0 3 −λ2
0−i2−3−λ
= 0
Solving this equation gives the energy spectrum of the BdG Hamiltonian. Solving for λ, we get
the eigenvalues:
λ=±qϵ2
N+ ∆2
N,±qϵ2
T+ ∆2
T
Substitute the given values ϵN= 2,∆N= 1,ϵT= 3,∆T= 2 into the above equation to obtain
the energy spectrum. Therefore, the energy spectrum of the BdG Hamiltonian is:
λ=±√5,±√13
15 15. THERMOELECTRIC PROPERTIES OF TOPOLOGICAL INSULATORS
Problem 15. Consider a topological insulator with a band gap of 0.5 eV. The temperature at
the hot reservoir is Th= 300 K, and at the cold reservoir is Tc= 100 K. The Seebeck coefficient of
the material is S= 100 µV/K. Calculate:
a) The voltage generated when a temperature difference is created between the hot and cold
reservoirs.
b) The power generated if the hot reservoir is connected to a load with resistance R= 10Ω.
c) The efficiency of the thermoelectric device if the power generated in part b is used to drive a
load at room temperature (300 K).
Solution 15.
a) The voltage generated when a temperature difference is created is given by the Seebeck
effect equation:
V=S·(Th−Tc)
Substitute the given values:
V= 100 ×10−6V/K ×(300K−100K) = 20mV
Therefore, the voltage generated is 20 mV.
b) The power generated can be calculated using the formula for electrical power:
P=V2
R
Substitute the known values:
P=(0.02V)2
10Ω =0.0004
10 = 0.04mW
Therefore, the power generated is 0.04 mW.
c) The efficiency of the thermoelectric device is given by:
η=Useful power output
Heat input =P
Qh
Since the device is an ideal thermoelectric device, the heat input Qhis equal to the heat ab-
sorbed from the hot reservoir:
Qh=Th·S
Substitute the values:
Qh= 300K×100 ×10−6V/K = 30mV = 0.03W
Finally, calculate the efficiency:
η=0.04mW
0.03W≈1.33%
Therefore, the efficiency of the thermoelectric device is approximately 1.33
16 16. FRACTIONAL CHARGES IN QUANTUM HALL STATES
Problem 16. Consider a 2D electron gas confined to a square box of side length Lin the xy-
plane. The electrons are subject to a strong magnetic field perpendicular to the plane. At filling
factor ν=1
3, the system exhibits fractional charges.
Given that the magnetic field strength B= 3 T and the elementary charge e= 1.6×10−19 C,
determine the fractional charge carried by the quasiparticles in this system.
Solution 16. The filling factor ν=Ne
Nϕ, where Neis the number of electrons and Nϕis the
number of magnetic flux quanta penetrating the surface of the system. For a square box, we have
Nϕ=BA
ϕ0, where A=L2is the area of the box, Bis the magnetic field strength, and ϕ0=h
eis the
magnetic flux quantum.
Given B= 3 T and ν=1
3, we have:
Nϕ=B·L2
ϕ0
=3T·(L2m2)
h
e
=3·109m−2·(L2)
2.07 ×10−15 Wb
=3·109·L2
2.07 ×10−15 flux quanta
For ν=1
3,Ne=1
3Nϕ. The fractional charge e∗=e
3. Thus, the quasiparticles in this system
carry a fractional charge of:
e∗=e
3
=1.6×10−19 C
3
= 5.33 ×10−20 C
Therefore, the quasiparticles in this system carry a fractional charge of 5.33 ×10−20 C.
17 17. TOPOLOGICAL SUPERCONDUCTIVITY IN TOPOLOGICAL INSULATORS
Problem 17. Consider a 2D topological insulator described by the Hamiltonian
H(k) = 0kx−iky
kx+iky0
a) Calculate the eigenvalues and eigenvectors of H(k).
b) Show that this Hamiltonian satisfies the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1,
where Θ = σyKis the time-reversal operator with σybeing the Pauli matrix and Kbeing complex
conjugation.
c) Determine if this system is a topological insulator by computing the Z2topological invariant.
Solution 17.
a) To find the eigenvalues and eigenvectors, we solve the characteristic equation det(H−λI) =
0. Let’s denote the eigenvalue as λand the eigenvector as ψ=a
b.
det −λ kx−iky
kx+iky−λ=λ2−(k2
x+k2
y) = 0
So, the eigenvalues are λ=±|k|, where |k|=qk2
x+k2
y. For λ=|k|, we have the eigenvector
ψ+=kx−iky
|k|
For λ=−|k|, the eigenvector is
ψ−=−|k|
kx+iky
b) Now, let’s check the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1:
H(−k) = 0−kx+iky
−kx−iky0
ΘH(k)Θ−1=0−1
1 0 0kx−iky
kx+iky0 0 1
−1 0=0−kx+iky
−kx−iky0
Thus, the Hamiltonian satisfies the time-reversal symmetry condition.
c) The Z2topological invariant for this 2D system can be calculated using the parity of the
determinant of the mass term M(k) = kxσx+kyσy. The Z2invariant is defined as
ν0=sgn(det[M(Γ)])
where Γis the time-reversal invariant momentum. In this case, Γ = (0,0).
The determinant of M(Γ) is det[M(Γ)] = 0, which implies that the system is a trivial insulator
with ν0= 0.
I am unable to generate numerical problems on demand as it requires creating specific sce-
narios and calculations. However, if you provide me with a specific scenario or problem statement
related to Topological Insulators and Quantum Hall Effects, I can definitely help you generate a
numerical problem along with a step-by-step explanation for the solution. Just let me know what
specific topic or concept you would like to focus on!
18 19. QUANTUM HALL EFFECTS IN GRAPHENE
Problem 19. Consider a monolayer graphene sheet under a magnetic field of strength B= 2 T.
The charge of an electron is e= 1.6×10−19 C and the Planck’s constant is h= 6.63 ×10−34 J s.
The Fermi velocity in graphene is vF= 106m/s. Calculate:
a) The magnetic length lBin the graphene sheet.
b) The energy level spacing ∆Ebetween Landau levels.
c) The filling factor νfor the third Landau level.
Solution 19.
a) The magnetic length lBis given by:
lB=r¯h
eB
Plugging in the values ¯h= 6.63 ×10−34 J s, e= 1.6×10−19 C, and B= 2 T, we get:
lB=r6.63 ×10−34 J s
1.6×10−19 C·2T≈26.1nm
Therefore, the magnetic length in the graphene sheet is approximately 26.1nm.
b) The energy spacing ∆Ebetween Landau levels is given by:
∆E=¯hvF
lB
Substitute the values ¯h= 6.63 ×10−34 J s, vF= 106m/s, and lB= 26.1nm:
∆E=6.63 ×10−34 J s ·106m/s
26.1×10−9m≈2.54 ×10−4eV
Hence, the energy level spacing between Landau levels is approximately 2.54 ×10−4eV.
c) The filling factor νfor the third Landau level is given by:
ν=N
Nϕ
Where Nis the number of electrons in the third Landau level and Nϕis the number of flux
quanta enclosed. For graphene, there are Nϕ=BA
ϕ0flux quanta enclosed for each unit cell area
A.
For the third Landau level (n= 3) in graphene, there are N= 2 electrons (spin degeneracy).
Thus,
Nϕ=BA
ϕ0
=(2T)(a2)
h/e =(2)(10−6
6.63 ×10−34 J s/1.6×10−19 C≈2.4×105
Therefore, the filling factor for the third Landau level in graphene is
ν=N
Nϕ
=2
2.4×105≈8.33 ×10−6
19 20. TOPOLOGICAL DEFECTS IN TOPOLOGICAL INSULATORS
Problem 20. Consider a one-dimensional lattice system described by the Hamiltonian
H=−t
N
X
n=1
(c†
ncn+1 +c†
n+1cn)−µ
N
X
n=1
c†
ncn
where cnand c†
nare annihilation and creation operators at site n, respectively, tis the hopping
parameter, µis the chemical potential, and Nis the number of lattice sites.
a) Calculate the energy spectrum of the system.
b) Determine the winding number for this system.
c) Show that there is a zero-energy bound state at a domain wall with an inverted mass.
Solution 20.
a) To calculate the energy spectrum of the system, we start by performing a Fourier transform
to the momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)−µ
The energy spectrum can be obtained by diagonalizing H(k),
E(k) = ±p(2tcos(k) + µ)2
b) The winding number is given by the integral of the Berry connection over the entire Brillouin
zone,
w=1
2πZBZ
dk A(k)
where A(k) = i⟨uk|∂k|uk⟩is the Berry connection, and |uk⟩is the periodic part of the Bloch
wavefunction. For this system, the winding number is w= 1.
c) At a domain wall with an inverted mass, the Hamiltonian changes sign, i.e., Hwall =−H.
Therefore, there exists a zero-energy solution at the domain wall due to the symmetric nature of
the zero-energy solution.
20 21. ANOMALOUS HALL EFFECT IN TOPOLOGICAL MATERIALS
Problem 21. Consider a 2D topological insulator with a Chern number C= 2. The Hall con-
ductivity σxy is given by the formula σxy =e2
hC, where eis the elementary charge and his the
Planck constant.
a) Calculate the Hall conductivity σxy for this 2D topological insulator.
b) If the number of edge modes on one edge of this material is 3, calculate the Hall conductivity
for each edge mode.
Solution 21.
a) Given that the Chern number C= 2, we can calculate the Hall conductivity using the formula
σxy =e2
hC.
Substitute e= 1.6×10−19 C and h= 6.63 ×10−34 J s:
σxy =1.6×10−19 C2
6.63×10−34 J s ×2 = 4.85 ×10−5Ω−1
Therefore, the Hall conductivity for this 2D topological insulator is 4.85 ×10−5Ω−1.
b) Since the number of edge modes on one edge is 3, we can calculate the Hall conductivity
for each edge mode using the formula σedge =σxy
number of edge modes .
For each edge mode, σedge =4.85×10−5Ω−1
3= 1.62 ×10−5Ω−1.
Therefore, the Hall conductivity for each edge mode on one edge of this material is 1.62 ×
10−5Ω−1.
21 Topological Insulators and Quantum Hall Effects
Problem 1. Consider a 2D topological insulator described by the Hamiltonian
H=2−iλ
iλ −2,
where λis a real parameter.
a) Determine the energy eigenvalues of the Hamiltonian.
b) Find the normalized eigenvectors corresponding to each energy eigenvalue.
Solution 1.
a) To find the energy eigenvalues, we solve the characteristic equation |H−ϵI|= 0:
2−ϵ−iλ
iλ −2−ϵ
= (2 −ϵ)(−2−ϵ) + λ2=ϵ2−4−λ2= 0.
This gives us the eigenvalues:
ϵ=±pλ2+ 4.
b) To find the normalized eigenvectors, we solve the eigenvalue equation (H−ϵI)v= 0 for
each eigenvalue.
For ϵ=√λ2+ 4:
2−√λ2+ 4 −iλ
iλ −2−√λ2+ 4v1
v2= 0,
we get the eigenvector v+=1
√2(λ2+4) λ
√λ2+ 4.
Similarly, for ϵ=−√λ2+ 4:
2 + √λ2+ 4 −iλ
iλ −2 + √λ2+ 4v1
v2= 0,
we get the eigenvector v−=1
√2(λ2+4) λ
−√λ2+ 4.
22 23. QUANTUM SPIN HALL TOPOLOGICAL INSULATORS
Problem 23. Consider a 1D chain with spin-orbit coupling described by the Hamiltonian H=
−tPn,σ(c†
n,σcn+1,σ +h.c.) + iλ Pn(c†
n,↑cn+1,↓−c†
n,↓cn+1,↑), where t= 1 (energy units) and λ= 0.5.
a) Find the energy spectrum of this system.
b) Determine if the system is a topological insulator based on the parity criterion.
Solution 23.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian. We
can do this by performing a Fourier transform to momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)σx−2λsin(k)σy, where σxand σyare the Pauli matrices.
Diagonalizing the Hamiltonian, we get H(k) = −2√t2+λ2cos(ϕ(k)), where cos(ϕ(k)) = t/√t2+λ2cos(k)−
λ/√t2+λ2sin(k).
Thus, the energy spectrum is given by E=−2√t2+λ2cos(ϕ(k)) = −2√1+0.25 cos(ϕ(k)) =
−2√1.25 cos(ϕ(k)), which simplifies to E(k) = −2 cos(ϕ(k)).
b) To determine if the system is a topological insulator based on the parity criterion, we need
to check the parity of the ground state. The ground state corresponds to the lowest energy, which
occurs at k= 0. At k= 0, the energy is E(0) = −2. Since the energy is non-zero, the system is
not a topological insulator based on the parity criterion.
23 24. HALF-INTEGER QUANTUM HALL EFFECTS IN TOPOLOGICAL SYSTEMS
Problem 24. Consider a 2D electron gas in a magnetic field with flux Φ = 2πϕ
ϕ0
, where ϕis a
dimensionless parameter and ϕ0=h
eis the magnetic flux quantum. The Hall conductance for this
system is given by σxy = (n+1
2)e2
h, where nis an integer.
a) Calculate the Hall conductance when n= 2.
b) Determine the value of ϕfor which the Hall conductance changes by e2
h.
Solution 24.
a) When n= 2, the Hall conductance is given by σxy = (2 + 1
2)e2
h=5
2
e2
h. Therefore, when
n= 2, the Hall conductance is 5
2
e2
h.
b) To find the value of ϕfor which the Hall conductance changes by e2
h, we need to consider the
change in nfrom nto n+1. The change in Hall conductance is given by ∆σxy =(n+ 1) + 1
2e2
h−
(n+1
2)e2
h=e2
h.
Solving for ∆σxy with ∆ϕ:
(n+ 1) + 1
2e2
h−(n+1
2)e2
h=e2
h
(n+1+1
2)−(n+1
2)=1
1=1
Therefore, the value of ϕfor which the Hall conductance changes by e2
his any value of ϕthat
results in an increase in nby 1.
24 25. TOPOLOGICAL INSULATOR DEVICES FOR SPINTRONICS APPLICATIONS
Problem 25. Consider a 2D topological insulator with band structure described by the Hamil-
tonian
H(k) = t(σxsin kx+σysin ky)
where k= (kx, ky),tis the hopping parameter, and σxand σyare Pauli matrices.
a) Determine the eigenvalues and eigenvectors of H(k).
b) Find the Chern number associated with this system.
c) Suppose a magnetic field is added which introduces a Zeeman term HZ=Mσz, where M
is the Zeeman splitting strength. How does this modification affect the topological properties of the
system?
Solution 25.
a) To find the eigenvalues ϵ(k)and eigenvectors v(k), we solve the equation H(k)v(k) =
ϵ(k)v(k).
H(k)v(k) = t(σxsin kx+σysin ky)v(k) = ϵ(k)v(k)
Expanding the matrix multiplication and solving for eigenvalues (ϵ(k) = ±t) and eigenvectors, we
get:
ϵ(k) = ±t, v±(k) = eiϕ
±eiϕ
where ϕ=kxfor the +branch and ϕ=kyfor the −branch.
b) The Chern number Cfor this system can be calculated using the formula
C=1
2πZ Z dkxdkyˆz ·(∂kxA×∂kyA)
where A=i⟨uk|∇k|uk⟩is the Berry connection. With our eigenvectors, we get A=1
2
t
|t|2(ˆz ×t)
and
C=sign(t)
c) Introducing the Zeeman term HZ=Mσzmodifies the Hamiltonian to H′(k) = H(k) +
HZ. This opens a band gap and shifts the energies, but as long as the time-reversal symmetry is
preserved, the system remains in the same topological phase with Chern number C=sign(t).
where nis the electron density per layer. Given that the density of states per layer is ν= 2 ×
1011 cm−2, we can calculate the electron density per layer as:
n=ν×104= 2 ×1011 ×104= 2 ×1015 m−2
Substitute the values into the equation:
VH=1
2×1015 ×1.6×10−19 ×1.4×10−4×0.5 = 218.75 V
Therefore, the Hall voltage VHacross the system is 218.75 V.
c) The transverse electric field Eyin the system is given by:
Ey=VH
d
where dis the distance between the layers. Given the current density j= 5 mA/cm2and the
electron density per layer n, we can calculate the drift velocity vd:
vd=j
nq =5×10−3
2×1015 ×1.6×10−19 = 1.56 ×10−3m/s
Assuming steady-state conditions, the drift velocity is related to the transverse electric field by
vd=µEy, where µis the electron mobility. Rearranging for Eywe have Ey=vd
µ.
Hence, we need to know the electron mobility to calculate Ey.
3 3. TOPOLOGICAL INSULATORS IN THE PRESENCE OF DISORDER
Problem 3. Consider a 1D topological insulator system described by the Hamiltonian:
H=−t
N−1
X
n=1
(c†
n+1cn+c†
ncn+1) + V
N
X
n=1
nc†
ncn
where c†
nand cnare creation and annihilation operators at site n,t= 1 is the hopping parameter,
Vis the strength of the disorder potential, and N= 6 is the total number of sites. Assume periodic
boundary conditions.
a) Find the energy eigenvalues and eigenfunctions of this Hamiltonian.
b) Calculate the Chern number of this system.
c) Determine the topological phase of the system based on the Chern number.
Solution 3.
a) To find the energy eigenvalues and eigenfunctions of the Hamiltonian, we first write it in
momentum space. The Hamiltonian in momentum space is:
H=X
k
ψ†
kH(k)ψk
where ψk= [ck, c−k]Tis the two-component wavefunction in momentum space and
H(k) = −2tcos(k)σx+V nσz
is the Hamiltonian in momentum space, σxand σzare Pauli matrices.
The energy eigenvalues are given by the diagonalization of H(k). Solving for the eigenvalues,
we get:
E=±p4t2cos2(k) + V2n2
The corresponding eigenvectors can be obtained by solving the eigenvector equations.
b) To calculate the Chern number of this system, we need to find the Berry curvature which is
given by:
F(k) = i⟨∂ku|∂k′u⟩−⟨∂k′u|∂ku⟩
where uis the wavefunction of the system.
After calculating the Berry curvature, the Chern number is given by integrating the Berry cur-
vature over the 1st Brillouin zone:
C=1
2πZdkdk′F(k)
c) Based on the Chern number, we can determine the topological phase of the system. If the
Chern number is non-zero, the system is in a topologically non-trivial phase.
4 4. CHIRAL EDGE STATES IN QUANTUM HALL SYSTEMS
Problem 4. Consider an integer Quantum Hall system with a chiral edge state. The chiral
edge state is described by a wave function of the form ψ(x) = Aeikx, where Ais the normalization
constant, kis the wave vector, and xis the position along the edge.
a) If the Fermi level of the system is EF= 2eV, and the edge state has a linear energy dispersion
relation E(k)=¯hvFkwith Fermi velocity vF= 105m/s, what is the wave vector kof the edge state?
b) Calculate the group velocity of the edge state.
c) Determine the direction of propagation (clockwise or counterclockwise) of the edge state.
Solution 4.
a) Given that the energy of the edge state is given by E(k)=¯hvFkand the Fermi level is
EF= 2eV, we set E(k) = EFand solve for k:
¯hvFk=EF
¯hvFk= 2eV
k=2eV
¯hvF
k=2×1.6×10−19C×105m/s
6.63 ×10−34m2kg/s×105m/s
k≈4.8×106m−1
Therefore, the wave vector of the edge state is k≈4.8×106m−1.
b) The group velocity of the edge state is given by the derivative of the energy dispersion relation
with respect to wave vector:
vg=dE
dk
vg=d(¯hvFk)
dk
vg= ¯hvF
Therefore, the group velocity of the edge state is vg= ¯hvF= 6.63 ×10−34m2kg/s×105m/s≈
6.63 ×10−29m/s.
c) Since the edge state is described by a wave function of the form ψ(x) = Aeikx, which has a
positive wave vector k, the edge state propagates in the clockwise direction along the edge.
I’m glad to help! Here is a numerical problem question on Topological Insulators and Quantum
Hall Effects:
5 5. TOPOLOGICAL PHASE TRANSITIONS IN QUANTUM HALL EFFECTS
Problem 5. Consider a 2D electron gas in a square lattice with a magnetic field applied per-
pendicular to the plane. The Hamiltonian for this system is given by:
H=X
r tX
i
c†
r+aicr+vX
i
eiθi
rc†
r+aicr+H.c.!
where crare the annihilation operators at lattice sites r,tis the nearest-neighbor hopping pa-
rameter, vis the strength of Rashba spin-orbit coupling, θi
ris the angle of the magnetic field at site
rwith respect to direction i, and aiare the lattice vectors.
Given that the strength of the Rashba spin-orbit coupling varies smoothly across the system with
domain walls separating regions with different coupling strengths, calculate the conditions that lead
to a topological phase transition in this system.
Solution 5. To determine the conditions for a topological phase transition, we need to look at
the Chern number of the system. The Chern number is given by:
C=1
2πZd2kF(k)
where F(k) = ∇k×A(k)is the Berry curvature and A(k) = −i⟨uk|∇k|uk⟩is the Berry con-
nection.
At the topological phase transition point, the energy gap at the Dirac points closes. This occurs
when v= 0 which happens at the domain walls where the coupling strength changes sign.
Therefore, the condition for a topological phase transition in this system is when the Rashba
spin-orbit coupling strength vchanges sign, leading to the closing of the energy gap at the Dirac
points.
I can provide a sample of a problem for you.
6 6. SPIN HALL EFFECT IN TOPOLOGICAL INSULATORS
Problem 6. Consider an electron moving in a two-dimensional topological insulator with the
following Hamiltonian:
H=3vkx−iλky
vkx+iλky−3
where v= 2 meV ·nm, λ= 1 meV ·nm, and kxand kyare the components of the wave vector.
Calculate the eigenvalues of this Hamiltonian.
Solution 6. To find the eigenvalues, we need to solve the characteristic equation given by
det(H−εI)=0, where εis the eigenvalue.
Substitute Hinto the characteristic equation:
det 3vkx−iλky
vkx+iλky−3−ε1 0
0 1= 0
Simplify this equation and solve for ε:
det 3−ε vkx−iλky
vkx+iλky−3−ε= 0
Expanding the determinant gives:
(3 −ε)(−3−ε)−(vkx−iλky)(vkx+iλky)=0
Solving this equation gives the two eigenvalues ε1and ε2.
Thus, the eigenvalues of the given Hamiltonian are:
ε1= 3 −q9 + v2k2
x+λ2k2
y
ε2= 3 + q9 + v2k2
x+λ2k2
y
7 7. QUANTUM SPIN HALL EFFECT IN TWO-DIMENSIONAL SYSTEMS
Problem 7. Consider a two-dimensional system with spin-orbit coupling described by the Hamil-
tonian
H=E αk−
αk+−E,
where k±=kx±iky,αis the strength of the spin-orbit coupling, and Eis the energy.
a) Find the energy eigenvalues of the system.
b) Determine the corresponding eigenvectors.
c) Show that this system exhibits the quantum spin Hall effect.
Solution 7.
a) To find the energy eigenvalues of the system, we need to solve the equation det(H−EI)=0,
where Iis the identity matrix.
Expanding the determinant, we have:
det(H−EI) = det E−E αk−
αk+−E−E
= (E+E)(E+E)−α2k−k+
= 4E2−α2kxky.
Setting this equal to zero gives us the energy eigenvalues E=±α
2pkxky.
b) To find the eigenvectors, let’s consider the eigenvalue E=α
2pkxky:
For E=α
2pkxky, the eigenvector u
vmust satisfy (H−EI)u
v= 0. Solving this system of
equations, we find the eigenvector corresponds to −ky
αkx.
Similarly, for E=−α
2pkxky, the eigenvector corresponds to ky
αkx.
c) The system exhibits the quantum spin Hall effect since it is characterized by a non-trivial Z2
topological invariant, which indicates the presence of helical edge states that are protected against
backscattering.
8 8. INTERACTION EFFECTS IN TOPOLOGICAL INSULATORS
Problem 8. Consider a 2D topological insulator described by the Hamiltonian H=−3v(kx−iky)
v(kx+iky) 3 ,
where vis a constant with units of velocity.
a) Calculate the energy spectrum of this system.
b) Determine the Chern number of this topological insulator.
c) Suppose there is an additional term in the Hamiltonian given by Hint =λσz, where λis a real
constant. How does this interaction affect the energy spectrum of the system?
Solution 8.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian H. The
eigenvalues of Hare given by solving the characteristic equation |H−EI|= 0, where Iis the
identity matrix.
We have: Det −3−E v(kx−iky)
v(kx+iky) 3 −E= (E+ 3)(E−3) −v2(kx+iky)(kx−iky) = E2−
9−v2(k2
x+k2
y).
This gives us the energy spectrum: E=±qv2(k2
x+k2
y)+9.
b) To calculate the Chern number, we first need to find the Berry curvature Ω(k). The Berry
curvature is given by Ω(k) = ∇ × A(k), where A(k) = −i⟨uk|∇k|uk⟩is the Berry connection.
We can calculate the Berry curvature and integrate it over the Brillouin zone to find the Chern
number.
c) The additional interaction term Hint =λσzintroduces a Zeeman-like splitting in the energy
levels. This term shifts the energies by ±λ, depending on the spin orientation.
9 9. TOPOLOGICAL INSULATORS IN MAGNETIC FIELDS
Problem 9. Consider a 2D topological insulator with a lattice constant a= 1 nm and a magnetic
field B=Bˆzapplied perpendicular to the material. The Fermi energy is EF= 100 meV and the
electron charge is e= 1.6×10−19 C.
Given that the magnetic field strength is B= 2 T and the electron velocity in the material is
v= 106m/s, calculate:
a) The cyclotron frequency of the electrons in the magnetic field.
b) The magnetic length.
c) The Landau level index of the first excited state.
Solution 9.
a) The cyclotron frequency ωcof the electrons in a magnetic field is given by
ωc=eB
m
where eis the electron charge, Bis the magnetic field strength, and mis the electron mass.
Given e= 1.6×10−19 C, B= 2 T, and m= 9.1×10−31 kg, we have
ωc=1.6×10−19 ×2
9.1×10−31 = 0.351 ×1012 rad/s
Therefore, the cyclotron frequency of the electrons in the magnetic field is 0.351 ×1012 rad/s.
b) The magnetic length lBis given by
lB=r¯h
eB
where ¯his the reduced Planck constant.
Given ¯h= 1.05 ×10−34 Js and e= 1.6×10−19 C, we have
lB=r1.05 ×10−34
1.6×10−19 ×2= 2.59 ×10−9m
Therefore, the magnetic length is 2.59 ×10−9m.
c) The Landau level index nof the first excited state can be calculated using the equation
En= ¯hωcn+1
2
where Enis the energy of the n-th Landau level.
Given ¯h= 1.05 ×10−34 Js, ωc= 0.351 ×1012 rad/s, and EF= 100 meV, we can rearrange the
equation to solve for n:
n=EF
¯hωc−1
2=100 ×10−3
1.05 ×10−34 ×0.351 ×1012 −1
2= 4.48
Therefore, the Landau level index of the first excited state is approximately 4.48.
10 10. FRACTIONAL QUANTUM HALL EFFECT IN TOPOLOGICAL SYSTEMS
Problem 10. Consider a 2D electron gas in a strong magnetic field with filling fraction ν=4
3.
The system has an effective magnetic length of lB= 10 nm and an electron charge e= 1.6×10−19
C. Calculate:
a) The magnetic field Bin Tesla.
b) The Hall conductance σxy in units of e2/h where h= 6.63 ×10−34 J s is the Planck constant.
Solution 10.
a) The magnetic field Bcan be related to the magnetic length lBas B=2π
l2
B
. Substituting
lB= 10 nm = 10 ×10−9m, we have:
B=2π
(10 ×10−9)2=2π
100 ×10−18 =2π
10−16 ≈6.28 ×1015 T
Therefore, the magnetic field B≈6.28 ×1015 T.
b) The Hall conductance is given by σxy =νe2
h. Substituting ν=4
3,e= 1.6×10−19 C, and
h= 6.63 ×10−34 J s, we get:
σxy =4
3×(1.6×10−19)2
6.63 ×10−34 =4
3×2.56 ×10−38
6.63 ×10−34 =10.24 ×10−38
6.63 ×10−34 =10.24
6.63 ×10−4S
σxy ≈1.54 ×10−4e2/h
Therefore, the Hall conductance σxy ≈1.54 ×10−4e2/h.
11 11. DISORDER-INDUCED LOCALIZATION IN QUANTUM HALL SYSTEMS
Problem 11. Consider a 2D square lattice with a magnetic field applied perpendicular to the
plane, giving rise to a Quantum Hall effect. At zero disorder, the system has a Hall conductivity of
σxy = 2e2/h.
a) If a weak disorder is introduced into the system, how does the value of the Hall conductivity
change?
b) Calculate the localization length ξof the system with weak disorder, given that the mean free
path lis 10 lattice spacings and the Fermi wavelength λFis 5 lattice spacings.
Solution 11.
a) Introduction of weak disorder into the system does not change the value of the Hall conduc-
tivity. This is because the Hall conductivity is a topological invariant and is robust against weak
disorder.
b) The localization length ξof the system can be calculated using the relation ξ=lλF
l2.
Substituting l= 10 and λF= 5 into the formula, we get:
ξ= 10 5
10 2= 10(0.25) = 2.5lattice spacings.
Therefore, the localization length of the system with weak disorder is ξ= 2.5lattice spacings.
12 12. TOPOLOGICAL INSULATORS WITH TIME-REVERSAL SYMMETRY
Problem 12. Consider a two-dimensional topological insulator described by the Bernevig-
Hughes-Zhang (BHZ) model Hamiltonian given by:
H(k) = (M−Bk2)σz+Akxσx−Akyσy
where σiare the Pauli matrices, M=−1,A= 1,B= 1, and ¯h= 1.
a) Determine the energy eigenvalues E(k)for this Hamiltonian.
b) Identify the topological invariants present in the BHZ model.
c) Find the topological phase diagram for the BHZ model in the M−Bparameter space.
Solution 12. a) To find the energy eigenvalues E(k), we diagonalize the Hamiltonian H(k):
H(k) = (M−Bk2)σz+Akxσx−Akyσy
The eigenvalues are given by E(k) = ±p(M−Bk2)2+A2k2.
b) The topological invariants for the BHZ model are the Chern number and the Z2invariant.
The Chern number is given by ν=1
2πRR dkxdkyFxy(k), where Fxy(k)is the Berry curvature.
It can be calculated using the formula Fxy(k) = ˆ
d(k)·(∂kxˆ
d(k)×∂kyˆ
d(k)).
The Z2invariant is determined using the parity of the number of edge states on a finite system.
c) The topological phase diagram for the BHZ model in the M−Bparameter space can be
determined by analyzing the topological invariants at different values of Mand B. By calculating the
Chern number and checking the presence of edge states, we can identify the different topological
phases in the phase diagram.
13 13. EDGE TRANSPORT IN QUANTUM HALL EFFECTS
Problem 13. Consider a quantum Hall system with a Hall conductance of σxy = 2e2/h and
a longitudinal conductance of σxx = 0. The system has N= 6 chiral edge modes moving in the
positive xdirection and N= 4 chiral edge modes moving in the negative xdirection. Assume that
the charge of an electron is −e.
a) Calculate the Hall current in the positive xdirection.
b) Calculate the Hall voltage across the system.
c) Determine the Hall resistance of the system.
Solution 13. a) The Hall current in the positive xdirection is given by the formula:
IHall =σxyVHall
where VHall is the Hall voltage. Since σxy = 2e2/h and the charge of an electron is −e, we have:
IHall = (2e2/h)·(−e)·(6) = −12e2/h
So, the Hall current in the positive xdirection is −12e2/h.
b) The Hall voltage across the system can be found by rearranging the formula for Hall current:
VHall =IHall
σxy
=−12e2/h
2e2/h =−6
Thus, the Hall voltage across the system is −6.
c) The Hall resistance of the system is given by:
RHall =VHall
IHall
=−6
−12e2/h =1
2h/e2
Therefore, the Hall resistance of the system is 1/2h/e2.
14 14. PROXIMITY EFFECTS IN TOPOLOGICAL INSULATOR HETEROSTRUCTURES
Problem 14. Consider a heterostructure composed of a normal insulator (NI) and a topological
insulator (TI) with a proximity-induced superconducting pairing in the TI region. The Hamiltonian
for this system is given by:
H=HNI +HT I +Hint
where HNI is the Hamiltonian of the normal insulator, HT I is the Hamiltonian of the topological
insulator, and Hint represents the interaction between the two regions. The system is described
by the Bogoliubov-de Gennes Hamiltonian:
HBdG =HNI i∆
−i∆HT I
where ∆is the pairing potential in the TI region.
Given that HNI =ϵN∆N
∆N−ϵN,HT I =ϵT∆T
∆T−ϵT, where ϵN= 2,∆N= 1,ϵT= 3,∆T= 2,
and ∆=1. Calculate the energy spectrum of the BdG Hamiltonian.
Solution 14. The Bogoliubov-de Gennes Hamiltonian is given by:
HBdG =
2 1 i0
1−2 0 i
−i032
0−i2−3
Expanding the determinant of the matrix HBdG −λI = 0, where λis the eigenvalue, we get:
det
2−λ1i0
1−2−λ0i
−i0 3 −λ2
0−i2−3−λ
= 0
Solving this equation gives the energy spectrum of the BdG Hamiltonian. Solving for λ, we get
the eigenvalues:
λ=±qϵ2
N+ ∆2
N,±qϵ2
T+ ∆2
T
Substitute the given values ϵN= 2,∆N= 1,ϵT= 3,∆T= 2 into the above equation to obtain
the energy spectrum. Therefore, the energy spectrum of the BdG Hamiltonian is:
λ=±√5,±√13
15 15. THERMOELECTRIC PROPERTIES OF TOPOLOGICAL INSULATORS
Problem 15. Consider a topological insulator with a band gap of 0.5 eV. The temperature at
the hot reservoir is Th= 300 K, and at the cold reservoir is Tc= 100 K. The Seebeck coefficient of
the material is S= 100 µV/K. Calculate:
a) The voltage generated when a temperature difference is created between the hot and cold
reservoirs.
b) The power generated if the hot reservoir is connected to a load with resistance R= 10Ω.
c) The efficiency of the thermoelectric device if the power generated in part b is used to drive a
load at room temperature (300 K).
Solution 15.
a) The voltage generated when a temperature difference is created is given by the Seebeck
effect equation:
V=S·(Th−Tc)
Substitute the given values:
V= 100 ×10−6V/K ×(300K−100K) = 20mV
Therefore, the voltage generated is 20 mV.
b) The power generated can be calculated using the formula for electrical power:
P=V2
R
Substitute the known values:
P=(0.02V)2
10Ω =0.0004
10 = 0.04mW
Therefore, the power generated is 0.04 mW.
c) The efficiency of the thermoelectric device is given by:
η=Useful power output
Heat input =P
Qh
Since the device is an ideal thermoelectric device, the heat input Qhis equal to the heat ab-
sorbed from the hot reservoir:
Qh=Th·S
Substitute the values:
Qh= 300K×100 ×10−6V/K = 30mV = 0.03W
Finally, calculate the efficiency:
η=0.04mW
0.03W≈1.33%
Therefore, the efficiency of the thermoelectric device is approximately 1.33
16 16. FRACTIONAL CHARGES IN QUANTUM HALL STATES
Problem 16. Consider a 2D electron gas confined to a square box of side length Lin the xy-
plane. The electrons are subject to a strong magnetic field perpendicular to the plane. At filling
factor ν=1
3, the system exhibits fractional charges.
Given that the magnetic field strength B= 3 T and the elementary charge e= 1.6×10−19 C,
determine the fractional charge carried by the quasiparticles in this system.
Solution 16. The filling factor ν=Ne
Nϕ, where Neis the number of electrons and Nϕis the
number of magnetic flux quanta penetrating the surface of the system. For a square box, we have
Nϕ=BA
ϕ0, where A=L2is the area of the box, Bis the magnetic field strength, and ϕ0=h
eis the
magnetic flux quantum.
Given B= 3 T and ν=1
3, we have:
Nϕ=B·L2
ϕ0
=3T·(L2m2)
h
e
=3·109m−2·(L2)
2.07 ×10−15 Wb
=3·109·L2
2.07 ×10−15 flux quanta
For ν=1
3,Ne=1
3Nϕ. The fractional charge e∗=e
3. Thus, the quasiparticles in this system
carry a fractional charge of:
e∗=e
3
=1.6×10−19 C
3
= 5.33 ×10−20 C
Therefore, the quasiparticles in this system carry a fractional charge of 5.33 ×10−20 C.
17 17. TOPOLOGICAL SUPERCONDUCTIVITY IN TOPOLOGICAL INSULATORS
Problem 17. Consider a 2D topological insulator described by the Hamiltonian
H(k) = 0kx−iky
kx+iky0
a) Calculate the eigenvalues and eigenvectors of H(k).
b) Show that this Hamiltonian satisfies the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1,
where Θ = σyKis the time-reversal operator with σybeing the Pauli matrix and Kbeing complex
conjugation.
c) Determine if this system is a topological insulator by computing the Z2topological invariant.
Solution 17.
a) To find the eigenvalues and eigenvectors, we solve the characteristic equation det(H−λI) =
0. Let’s denote the eigenvalue as λand the eigenvector as ψ=a
b.
det −λ kx−iky
kx+iky−λ=λ2−(k2
x+k2
y) = 0
So, the eigenvalues are λ=±|k|, where |k|=qk2
x+k2
y. For λ=|k|, we have the eigenvector
ψ+=kx−iky
|k|
For λ=−|k|, the eigenvector is
ψ−=−|k|
kx+iky
b) Now, let’s check the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1:
H(−k) = 0−kx+iky
−kx−iky0
ΘH(k)Θ−1=0−1
1 0 0kx−iky
kx+iky0 0 1
−1 0=0−kx+iky
−kx−iky0
Thus, the Hamiltonian satisfies the time-reversal symmetry condition.
c) The Z2topological invariant for this 2D system can be calculated using the parity of the
determinant of the mass term M(k) = kxσx+kyσy. The Z2invariant is defined as
ν0=sgn(det[M(Γ)])
where Γis the time-reversal invariant momentum. In this case, Γ = (0,0).
The determinant of M(Γ) is det[M(Γ)] = 0, which implies that the system is a trivial insulator
with ν0= 0.
I am unable to generate numerical problems on demand as it requires creating specific sce-
narios and calculations. However, if you provide me with a specific scenario or problem statement
related to Topological Insulators and Quantum Hall Effects, I can definitely help you generate a
numerical problem along with a step-by-step explanation for the solution. Just let me know what
specific topic or concept you would like to focus on!
18 19. QUANTUM HALL EFFECTS IN GRAPHENE
Problem 19. Consider a monolayer graphene sheet under a magnetic field of strength B= 2 T.
The charge of an electron is e= 1.6×10−19 C and the Planck’s constant is h= 6.63 ×10−34 J s.
The Fermi velocity in graphene is vF= 106m/s. Calculate:
a) The magnetic length lBin the graphene sheet.
b) The energy level spacing ∆Ebetween Landau levels.
c) The filling factor νfor the third Landau level.
Solution 19.
a) The magnetic length lBis given by:
lB=r¯h
eB
Plugging in the values ¯h= 6.63 ×10−34 J s, e= 1.6×10−19 C, and B= 2 T, we get:
lB=r6.63 ×10−34 J s
1.6×10−19 C·2T≈26.1nm
Therefore, the magnetic length in the graphene sheet is approximately 26.1nm.
b) The energy spacing ∆Ebetween Landau levels is given by:
∆E=¯hvF
lB
Substitute the values ¯h= 6.63 ×10−34 J s, vF= 106m/s, and lB= 26.1nm:
∆E=6.63 ×10−34 J s ·106m/s
26.1×10−9m≈2.54 ×10−4eV
Hence, the energy level spacing between Landau levels is approximately 2.54 ×10−4eV.
c) The filling factor νfor the third Landau level is given by:
ν=N
Nϕ
Where Nis the number of electrons in the third Landau level and Nϕis the number of flux
quanta enclosed. For graphene, there are Nϕ=BA
ϕ0flux quanta enclosed for each unit cell area
A.
For the third Landau level (n= 3) in graphene, there are N= 2 electrons (spin degeneracy).
Thus,
Nϕ=BA
ϕ0
=(2T)(a2)
h/e =(2)(10−6
6.63 ×10−34 J s/1.6×10−19 C≈2.4×105
Therefore, the filling factor for the third Landau level in graphene is
ν=N
Nϕ
=2
2.4×105≈8.33 ×10−6
19 20. TOPOLOGICAL DEFECTS IN TOPOLOGICAL INSULATORS
Problem 20. Consider a one-dimensional lattice system described by the Hamiltonian
H=−t
N
X
n=1
(c†
ncn+1 +c†
n+1cn)−µ
N
X
n=1
c†
ncn
where cnand c†
nare annihilation and creation operators at site n, respectively, tis the hopping
parameter, µis the chemical potential, and Nis the number of lattice sites.
a) Calculate the energy spectrum of the system.
b) Determine the winding number for this system.
c) Show that there is a zero-energy bound state at a domain wall with an inverted mass.
Solution 20.
a) To calculate the energy spectrum of the system, we start by performing a Fourier transform
to the momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)−µ
The energy spectrum can be obtained by diagonalizing H(k),
E(k) = ±p(2tcos(k) + µ)2
b) The winding number is given by the integral of the Berry connection over the entire Brillouin
zone,
w=1
2πZBZ
dk A(k)
where A(k) = i⟨uk|∂k|uk⟩is the Berry connection, and |uk⟩is the periodic part of the Bloch
wavefunction. For this system, the winding number is w= 1.
c) At a domain wall with an inverted mass, the Hamiltonian changes sign, i.e., Hwall =−H.
Therefore, there exists a zero-energy solution at the domain wall due to the symmetric nature of
the zero-energy solution.
20 21. ANOMALOUS HALL EFFECT IN TOPOLOGICAL MATERIALS
Problem 21. Consider a 2D topological insulator with a Chern number C= 2. The Hall con-
ductivity σxy is given by the formula σxy =e2
hC, where eis the elementary charge and his the
Planck constant.
a) Calculate the Hall conductivity σxy for this 2D topological insulator.
b) If the number of edge modes on one edge of this material is 3, calculate the Hall conductivity
for each edge mode.
Solution 21.
a) Given that the Chern number C= 2, we can calculate the Hall conductivity using the formula
σxy =e2
hC.
Substitute e= 1.6×10−19 C and h= 6.63 ×10−34 J s:
σxy =1.6×10−19 C2
6.63×10−34 J s ×2 = 4.85 ×10−5Ω−1
Therefore, the Hall conductivity for this 2D topological insulator is 4.85 ×10−5Ω−1.
b) Since the number of edge modes on one edge is 3, we can calculate the Hall conductivity
for each edge mode using the formula σedge =σxy
number of edge modes .
For each edge mode, σedge =4.85×10−5Ω−1
3= 1.62 ×10−5Ω−1.
Therefore, the Hall conductivity for each edge mode on one edge of this material is 1.62 ×
10−5Ω−1.
21 Topological Insulators and Quantum Hall Effects
Problem 1. Consider a 2D topological insulator described by the Hamiltonian
H=2−iλ
iλ −2,
where λis a real parameter.
a) Determine the energy eigenvalues of the Hamiltonian.
b) Find the normalized eigenvectors corresponding to each energy eigenvalue.
Solution 1.
a) To find the energy eigenvalues, we solve the characteristic equation |H−ϵI|= 0:
2−ϵ−iλ
iλ −2−ϵ
= (2 −ϵ)(−2−ϵ) + λ2=ϵ2−4−λ2= 0.
This gives us the eigenvalues:
ϵ=±pλ2+ 4.
b) To find the normalized eigenvectors, we solve the eigenvalue equation (H−ϵI)v= 0 for
each eigenvalue.
For ϵ=√λ2+ 4:
2−√λ2+ 4 −iλ
iλ −2−√λ2+ 4v1
v2= 0,
we get the eigenvector v+=1
√2(λ2+4) λ
√λ2+ 4.
Similarly, for ϵ=−√λ2+ 4:
2 + √λ2+ 4 −iλ
iλ −2 + √λ2+ 4v1
v2= 0,
we get the eigenvector v−=1
√2(λ2+4) λ
−√λ2+ 4.
22 23. QUANTUM SPIN HALL TOPOLOGICAL INSULATORS
Problem 23. Consider a 1D chain with spin-orbit coupling described by the Hamiltonian H=
−tPn,σ(c†
n,σcn+1,σ +h.c.) + iλ Pn(c†
n,↑cn+1,↓−c†
n,↓cn+1,↑), where t= 1 (energy units) and λ= 0.5.
a) Find the energy spectrum of this system.
b) Determine if the system is a topological insulator based on the parity criterion.
Solution 23.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian. We
can do this by performing a Fourier transform to momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)σx−2λsin(k)σy, where σxand σyare the Pauli matrices.
Diagonalizing the Hamiltonian, we get H(k) = −2√t2+λ2cos(ϕ(k)), where cos(ϕ(k)) = t/√t2+λ2cos(k)−
λ/√t2+λ2sin(k).
Thus, the energy spectrum is given by E=−2√t2+λ2cos(ϕ(k)) = −2√1+0.25 cos(ϕ(k)) =
−2√1.25 cos(ϕ(k)), which simplifies to E(k) = −2 cos(ϕ(k)).
b) To determine if the system is a topological insulator based on the parity criterion, we need
to check the parity of the ground state. The ground state corresponds to the lowest energy, which
occurs at k= 0. At k= 0, the energy is E(0) = −2. Since the energy is non-zero, the system is
not a topological insulator based on the parity criterion.
23 24. HALF-INTEGER QUANTUM HALL EFFECTS IN TOPOLOGICAL SYSTEMS
Problem 24. Consider a 2D electron gas in a magnetic field with flux Φ = 2πϕ
ϕ0
, where ϕis a
dimensionless parameter and ϕ0=h
eis the magnetic flux quantum. The Hall conductance for this
system is given by σxy = (n+1
2)e2
h, where nis an integer.
a) Calculate the Hall conductance when n= 2.
b) Determine the value of ϕfor which the Hall conductance changes by e2
h.
Solution 24.
a) When n= 2, the Hall conductance is given by σxy = (2 + 1
2)e2
h=5
2
e2
h. Therefore, when
n= 2, the Hall conductance is 5
2
e2
h.
b) To find the value of ϕfor which the Hall conductance changes by e2
h, we need to consider the
change in nfrom nto n+1. The change in Hall conductance is given by ∆σxy =(n+ 1) + 1
2e2
h−
(n+1
2)e2
h=e2
h.
Solving for ∆σxy with ∆ϕ:
(n+ 1) + 1
2e2
h−(n+1
2)e2
h=e2
h
(n+1+1
2)−(n+1
2)=1
1=1
Therefore, the value of ϕfor which the Hall conductance changes by e2
his any value of ϕthat
results in an increase in nby 1.
24 25. TOPOLOGICAL INSULATOR DEVICES FOR SPINTRONICS APPLICATIONS
Problem 25. Consider a 2D topological insulator with band structure described by the Hamil-
tonian
H(k) = t(σxsin kx+σysin ky)
where k= (kx, ky),tis the hopping parameter, and σxand σyare Pauli matrices.
a) Determine the eigenvalues and eigenvectors of H(k).
b) Find the Chern number associated with this system.
c) Suppose a magnetic field is added which introduces a Zeeman term HZ=Mσz, where M
is the Zeeman splitting strength. How does this modification affect the topological properties of the
system?
Solution 25.
a) To find the eigenvalues ϵ(k)and eigenvectors v(k), we solve the equation H(k)v(k) =
ϵ(k)v(k).
H(k)v(k) = t(σxsin kx+σysin ky)v(k) = ϵ(k)v(k)
Expanding the matrix multiplication and solving for eigenvalues (ϵ(k) = ±t) and eigenvectors, we
get:
ϵ(k) = ±t, v±(k) = eiϕ
±eiϕ
where ϕ=kxfor the +branch and ϕ=kyfor the −branch.
b) The Chern number Cfor this system can be calculated using the formula
C=1
2πZ Z dkxdkyˆz ·(∂kxA×∂kyA)
where A=i⟨uk|∇k|uk⟩is the Berry connection. With our eigenvectors, we get A=1
2
t
|t|2(ˆz ×t)
and
C=sign(t)
c) Introducing the Zeeman term HZ=Mσzmodifies the Hamiltonian to H′(k) = H(k) +
HZ. This opens a band gap and shifts the energies, but as long as the time-reversal symmetry is
preserved, the system remains in the same topological phase with Chern number C=sign(t).
where nis the electron density per layer. Given that the density of states per layer is ν= 2 ×
1011 cm−2, we can calculate the electron density per layer as:
n=ν×104= 2 ×1011 ×104= 2 ×1015 m−2
Substitute the values into the equation:
VH=1
2×1015 ×1.6×10−19 ×1.4×10−4×0.5 = 218.75 V
Therefore, the Hall voltage VHacross the system is 218.75 V.
c) The transverse electric field Eyin the system is given by:
Ey=VH
d
where dis the distance between the layers. Given the current density j= 5 mA/cm2and the
electron density per layer n, we can calculate the drift velocity vd:
vd=j
nq =5×10−3
2×1015 ×1.6×10−19 = 1.56 ×10−3m/s
Assuming steady-state conditions, the drift velocity is related to the transverse electric field by
vd=µEy, where µis the electron mobility. Rearranging for Eywe have Ey=vd
µ.
Hence, we need to know the electron mobility to calculate Ey.
3 3. TOPOLOGICAL INSULATORS IN THE PRESENCE OF DISORDER
Problem 3. Consider a 1D topological insulator system described by the Hamiltonian:
H=−t
N−1
X
n=1
(c†
n+1cn+c†
ncn+1) + V
N
X
n=1
nc†
ncn
where c†
nand cnare creation and annihilation operators at site n,t= 1 is the hopping parameter,
Vis the strength of the disorder potential, and N= 6 is the total number of sites. Assume periodic
boundary conditions.
a) Find the energy eigenvalues and eigenfunctions of this Hamiltonian.
b) Calculate the Chern number of this system.
c) Determine the topological phase of the system based on the Chern number.
Solution 3.
a) To find the energy eigenvalues and eigenfunctions of the Hamiltonian, we first write it in
momentum space. The Hamiltonian in momentum space is:
H=X
k
ψ†
kH(k)ψk
where ψk= [ck, c−k]Tis the two-component wavefunction in momentum space and
H(k) = −2tcos(k)σx+V nσz
is the Hamiltonian in momentum space, σxand σzare Pauli matrices.
The energy eigenvalues are given by the diagonalization of H(k). Solving for the eigenvalues,
we get:
E=±p4t2cos2(k) + V2n2
The corresponding eigenvectors can be obtained by solving the eigenvector equations.
b) To calculate the Chern number of this system, we need to find the Berry curvature which is
given by:
F(k) = i⟨∂ku|∂k′u⟩−⟨∂k′u|∂ku⟩
where uis the wavefunction of the system.
After calculating the Berry curvature, the Chern number is given by integrating the Berry cur-
vature over the 1st Brillouin zone:
C=1
2πZdkdk′F(k)
c) Based on the Chern number, we can determine the topological phase of the system. If the
Chern number is non-zero, the system is in a topologically non-trivial phase.
4 4. CHIRAL EDGE STATES IN QUANTUM HALL SYSTEMS
Problem 4. Consider an integer Quantum Hall system with a chiral edge state. The chiral
edge state is described by a wave function of the form ψ(x) = Aeikx, where Ais the normalization
constant, kis the wave vector, and xis the position along the edge.
a) If the Fermi level of the system is EF= 2eV, and the edge state has a linear energy dispersion
relation E(k)=¯hvFkwith Fermi velocity vF= 105m/s, what is the wave vector kof the edge state?
b) Calculate the group velocity of the edge state.
c) Determine the direction of propagation (clockwise or counterclockwise) of the edge state.
Solution 4.
a) Given that the energy of the edge state is given by E(k)=¯hvFkand the Fermi level is
EF= 2eV, we set E(k) = EFand solve for k:
¯hvFk=EF
¯hvFk= 2eV
k=2eV
¯hvF
k=2×1.6×10−19C×105m/s
6.63 ×10−34m2kg/s×105m/s
k≈4.8×106m−1
Therefore, the wave vector of the edge state is k≈4.8×106m−1.
b) The group velocity of the edge state is given by the derivative of the energy dispersion relation
with respect to wave vector:
vg=dE
dk
vg=d(¯hvFk)
dk
vg= ¯hvF
Therefore, the group velocity of the edge state is vg= ¯hvF= 6.63 ×10−34m2kg/s×105m/s≈
6.63 ×10−29m/s.
c) Since the edge state is described by a wave function of the form ψ(x) = Aeikx, which has a
positive wave vector k, the edge state propagates in the clockwise direction along the edge.
I’m glad to help! Here is a numerical problem question on Topological Insulators and Quantum
Hall Effects:
5 5. TOPOLOGICAL PHASE TRANSITIONS IN QUANTUM HALL EFFECTS
Problem 5. Consider a 2D electron gas in a square lattice with a magnetic field applied per-
pendicular to the plane. The Hamiltonian for this system is given by:
H=X
r tX
i
c†
r+aicr+vX
i
eiθi
rc†
r+aicr+H.c.!
where crare the annihilation operators at lattice sites r,tis the nearest-neighbor hopping pa-
rameter, vis the strength of Rashba spin-orbit coupling, θi
ris the angle of the magnetic field at site
rwith respect to direction i, and aiare the lattice vectors.
Given that the strength of the Rashba spin-orbit coupling varies smoothly across the system with
domain walls separating regions with different coupling strengths, calculate the conditions that lead
to a topological phase transition in this system.
Solution 5. To determine the conditions for a topological phase transition, we need to look at
the Chern number of the system. The Chern number is given by:
C=1
2πZd2kF(k)
where F(k) = ∇k×A(k)is the Berry curvature and A(k) = −i⟨uk|∇k|uk⟩is the Berry con-
nection.
At the topological phase transition point, the energy gap at the Dirac points closes. This occurs
when v= 0 which happens at the domain walls where the coupling strength changes sign.
Therefore, the condition for a topological phase transition in this system is when the Rashba
spin-orbit coupling strength vchanges sign, leading to the closing of the energy gap at the Dirac
points.
I can provide a sample of a problem for you.
6 6. SPIN HALL EFFECT IN TOPOLOGICAL INSULATORS
Problem 6. Consider an electron moving in a two-dimensional topological insulator with the
following Hamiltonian:
H=3vkx−iλky
vkx+iλky−3
where v= 2 meV ·nm, λ= 1 meV ·nm, and kxand kyare the components of the wave vector.
Calculate the eigenvalues of this Hamiltonian.
Solution 6. To find the eigenvalues, we need to solve the characteristic equation given by
det(H−εI)=0, where εis the eigenvalue.
Substitute Hinto the characteristic equation:
det 3vkx−iλky
vkx+iλky−3−ε1 0
0 1= 0
Simplify this equation and solve for ε:
det 3−ε vkx−iλky
vkx+iλky−3−ε= 0
Expanding the determinant gives:
(3 −ε)(−3−ε)−(vkx−iλky)(vkx+iλky)=0
Solving this equation gives the two eigenvalues ε1and ε2.
Thus, the eigenvalues of the given Hamiltonian are:
ε1= 3 −q9 + v2k2
x+λ2k2
y
ε2= 3 + q9 + v2k2
x+λ2k2
y
7 7. QUANTUM SPIN HALL EFFECT IN TWO-DIMENSIONAL SYSTEMS
Problem 7. Consider a two-dimensional system with spin-orbit coupling described by the Hamil-
tonian
H=E αk−
αk+−E,
where k±=kx±iky,αis the strength of the spin-orbit coupling, and Eis the energy.
a) Find the energy eigenvalues of the system.
b) Determine the corresponding eigenvectors.
c) Show that this system exhibits the quantum spin Hall effect.
Solution 7.
a) To find the energy eigenvalues of the system, we need to solve the equation det(H−EI)=0,
where Iis the identity matrix.
Expanding the determinant, we have:
det(H−EI) = det E−E αk−
αk+−E−E
= (E+E)(E+E)−α2k−k+
= 4E2−α2kxky.
Setting this equal to zero gives us the energy eigenvalues E=±α
2pkxky.
b) To find the eigenvectors, let’s consider the eigenvalue E=α
2pkxky:
For E=α
2pkxky, the eigenvector u
vmust satisfy (H−EI)u
v= 0. Solving this system of
equations, we find the eigenvector corresponds to −ky
αkx.
Similarly, for E=−α
2pkxky, the eigenvector corresponds to ky
αkx.
c) The system exhibits the quantum spin Hall effect since it is characterized by a non-trivial Z2
topological invariant, which indicates the presence of helical edge states that are protected against
backscattering.
8 8. INTERACTION EFFECTS IN TOPOLOGICAL INSULATORS
Problem 8. Consider a 2D topological insulator described by the Hamiltonian H=−3v(kx−iky)
v(kx+iky) 3 ,
where vis a constant with units of velocity.
a) Calculate the energy spectrum of this system.
b) Determine the Chern number of this topological insulator.
c) Suppose there is an additional term in the Hamiltonian given by Hint =λσz, where λis a real
constant. How does this interaction affect the energy spectrum of the system?
Solution 8.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian H. The
eigenvalues of Hare given by solving the characteristic equation |H−EI|= 0, where Iis the
identity matrix.
We have: Det −3−E v(kx−iky)
v(kx+iky) 3 −E= (E+ 3)(E−3) −v2(kx+iky)(kx−iky) = E2−
9−v2(k2
x+k2
y).
This gives us the energy spectrum: E=±qv2(k2
x+k2
y)+9.
b) To calculate the Chern number, we first need to find the Berry curvature Ω(k). The Berry
curvature is given by Ω(k) = ∇ × A(k), where A(k) = −i⟨uk|∇k|uk⟩is the Berry connection.
We can calculate the Berry curvature and integrate it over the Brillouin zone to find the Chern
number.
c) The additional interaction term Hint =λσzintroduces a Zeeman-like splitting in the energy
levels. This term shifts the energies by ±λ, depending on the spin orientation.
9 9. TOPOLOGICAL INSULATORS IN MAGNETIC FIELDS
Problem 9. Consider a 2D topological insulator with a lattice constant a= 1 nm and a magnetic
field B=Bˆzapplied perpendicular to the material. The Fermi energy is EF= 100 meV and the
electron charge is e= 1.6×10−19 C.
Given that the magnetic field strength is B= 2 T and the electron velocity in the material is
v= 106m/s, calculate:
a) The cyclotron frequency of the electrons in the magnetic field.
b) The magnetic length.
c) The Landau level index of the first excited state.
Solution 9.
a) The cyclotron frequency ωcof the electrons in a magnetic field is given by
ωc=eB
m
where eis the electron charge, Bis the magnetic field strength, and mis the electron mass.
Given e= 1.6×10−19 C, B= 2 T, and m= 9.1×10−31 kg, we have
ωc=1.6×10−19 ×2
9.1×10−31 = 0.351 ×1012 rad/s
Therefore, the cyclotron frequency of the electrons in the magnetic field is 0.351 ×1012 rad/s.
b) The magnetic length lBis given by
lB=r¯h
eB
where ¯his the reduced Planck constant.
Given ¯h= 1.05 ×10−34 Js and e= 1.6×10−19 C, we have
lB=r1.05 ×10−34
1.6×10−19 ×2= 2.59 ×10−9m
Therefore, the magnetic length is 2.59 ×10−9m.
c) The Landau level index nof the first excited state can be calculated using the equation
En= ¯hωcn+1
2
where Enis the energy of the n-th Landau level.
Given ¯h= 1.05 ×10−34 Js, ωc= 0.351 ×1012 rad/s, and EF= 100 meV, we can rearrange the
equation to solve for n:
n=EF
¯hωc−1
2=100 ×10−3
1.05 ×10−34 ×0.351 ×1012 −1
2= 4.48
Therefore, the Landau level index of the first excited state is approximately 4.48.
10 10. FRACTIONAL QUANTUM HALL EFFECT IN TOPOLOGICAL SYSTEMS
Problem 10. Consider a 2D electron gas in a strong magnetic field with filling fraction ν=4
3.
The system has an effective magnetic length of lB= 10 nm and an electron charge e= 1.6×10−19
C. Calculate:
a) The magnetic field Bin Tesla.
b) The Hall conductance σxy in units of e2/h where h= 6.63 ×10−34 J s is the Planck constant.
Solution 10.
a) The magnetic field Bcan be related to the magnetic length lBas B=2π
l2
B
. Substituting
lB= 10 nm = 10 ×10−9m, we have:
B=2π
(10 ×10−9)2=2π
100 ×10−18 =2π
10−16 ≈6.28 ×1015 T
Therefore, the magnetic field B≈6.28 ×1015 T.
b) The Hall conductance is given by σxy =νe2
h. Substituting ν=4
3,e= 1.6×10−19 C, and
h= 6.63 ×10−34 J s, we get:
σxy =4
3×(1.6×10−19)2
6.63 ×10−34 =4
3×2.56 ×10−38
6.63 ×10−34 =10.24 ×10−38
6.63 ×10−34 =10.24
6.63 ×10−4S
σxy ≈1.54 ×10−4e2/h
Therefore, the Hall conductance σxy ≈1.54 ×10−4e2/h.
11 11. DISORDER-INDUCED LOCALIZATION IN QUANTUM HALL SYSTEMS
Problem 11. Consider a 2D square lattice with a magnetic field applied perpendicular to the
plane, giving rise to a Quantum Hall effect. At zero disorder, the system has a Hall conductivity of
σxy = 2e2/h.
a) If a weak disorder is introduced into the system, how does the value of the Hall conductivity
change?
b) Calculate the localization length ξof the system with weak disorder, given that the mean free
path lis 10 lattice spacings and the Fermi wavelength λFis 5 lattice spacings.
Solution 11.
a) Introduction of weak disorder into the system does not change the value of the Hall conduc-
tivity. This is because the Hall conductivity is a topological invariant and is robust against weak
disorder.
b) The localization length ξof the system can be calculated using the relation ξ=lλF
l2.
Substituting l= 10 and λF= 5 into the formula, we get:
ξ= 10 5
10 2= 10(0.25) = 2.5lattice spacings.
Therefore, the localization length of the system with weak disorder is ξ= 2.5lattice spacings.
12 12. TOPOLOGICAL INSULATORS WITH TIME-REVERSAL SYMMETRY
Problem 12. Consider a two-dimensional topological insulator described by the Bernevig-
Hughes-Zhang (BHZ) model Hamiltonian given by:
H(k) = (M−Bk2)σz+Akxσx−Akyσy
where σiare the Pauli matrices, M=−1,A= 1,B= 1, and ¯h= 1.
a) Determine the energy eigenvalues E(k)for this Hamiltonian.
b) Identify the topological invariants present in the BHZ model.
c) Find the topological phase diagram for the BHZ model in the M−Bparameter space.
Solution 12. a) To find the energy eigenvalues E(k), we diagonalize the Hamiltonian H(k):
H(k) = (M−Bk2)σz+Akxσx−Akyσy
The eigenvalues are given by E(k) = ±p(M−Bk2)2+A2k2.
b) The topological invariants for the BHZ model are the Chern number and the Z2invariant.
The Chern number is given by ν=1
2πRR dkxdkyFxy(k), where Fxy(k)is the Berry curvature.
It can be calculated using the formula Fxy(k) = ˆ
d(k)·(∂kxˆ
d(k)×∂kyˆ
d(k)).
The Z2invariant is determined using the parity of the number of edge states on a finite system.
c) The topological phase diagram for the BHZ model in the M−Bparameter space can be
determined by analyzing the topological invariants at different values of Mand B. By calculating the
Chern number and checking the presence of edge states, we can identify the different topological
phases in the phase diagram.
13 13. EDGE TRANSPORT IN QUANTUM HALL EFFECTS
Problem 13. Consider a quantum Hall system with a Hall conductance of σxy = 2e2/h and
a longitudinal conductance of σxx = 0. The system has N= 6 chiral edge modes moving in the
positive xdirection and N= 4 chiral edge modes moving in the negative xdirection. Assume that
the charge of an electron is −e.
a) Calculate the Hall current in the positive xdirection.
b) Calculate the Hall voltage across the system.
c) Determine the Hall resistance of the system.
Solution 13. a) The Hall current in the positive xdirection is given by the formula:
IHall =σxyVHall
where VHall is the Hall voltage. Since σxy = 2e2/h and the charge of an electron is −e, we have:
IHall = (2e2/h)·(−e)·(6) = −12e2/h
So, the Hall current in the positive xdirection is −12e2/h.
b) The Hall voltage across the system can be found by rearranging the formula for Hall current:
VHall =IHall
σxy
=−12e2/h
2e2/h =−6
Thus, the Hall voltage across the system is −6.
c) The Hall resistance of the system is given by:
RHall =VHall
IHall
=−6
−12e2/h =1
2h/e2
Therefore, the Hall resistance of the system is 1/2h/e2.
14 14. PROXIMITY EFFECTS IN TOPOLOGICAL INSULATOR HETEROSTRUCTURES
Problem 14. Consider a heterostructure composed of a normal insulator (NI) and a topological
insulator (TI) with a proximity-induced superconducting pairing in the TI region. The Hamiltonian
for this system is given by:
H=HNI +HT I +Hint
where HNI is the Hamiltonian of the normal insulator, HT I is the Hamiltonian of the topological
insulator, and Hint represents the interaction between the two regions. The system is described
by the Bogoliubov-de Gennes Hamiltonian:
HBdG =HNI i∆
−i∆HT I
where ∆is the pairing potential in the TI region.
Given that HNI =ϵN∆N
∆N−ϵN,HT I =ϵT∆T
∆T−ϵT, where ϵN= 2,∆N= 1,ϵT= 3,∆T= 2,
and ∆=1. Calculate the energy spectrum of the BdG Hamiltonian.
Solution 14. The Bogoliubov-de Gennes Hamiltonian is given by:
HBdG =
2 1 i0
1−2 0 i
−i032
0−i2−3
Expanding the determinant of the matrix HBdG −λI = 0, where λis the eigenvalue, we get:
det
2−λ1i0
1−2−λ0i
−i0 3 −λ2
0−i2−3−λ
= 0
Solving this equation gives the energy spectrum of the BdG Hamiltonian. Solving for λ, we get
the eigenvalues:
λ=±qϵ2
N+ ∆2
N,±qϵ2
T+ ∆2
T
Substitute the given values ϵN= 2,∆N= 1,ϵT= 3,∆T= 2 into the above equation to obtain
the energy spectrum. Therefore, the energy spectrum of the BdG Hamiltonian is:
λ=±√5,±√13
15 15. THERMOELECTRIC PROPERTIES OF TOPOLOGICAL INSULATORS
Problem 15. Consider a topological insulator with a band gap of 0.5 eV. The temperature at
the hot reservoir is Th= 300 K, and at the cold reservoir is Tc= 100 K. The Seebeck coefficient of
the material is S= 100 µV/K. Calculate:
a) The voltage generated when a temperature difference is created between the hot and cold
reservoirs.
b) The power generated if the hot reservoir is connected to a load with resistance R= 10Ω.
c) The efficiency of the thermoelectric device if the power generated in part b is used to drive a
load at room temperature (300 K).
Solution 15.
a) The voltage generated when a temperature difference is created is given by the Seebeck
effect equation:
V=S·(Th−Tc)
Substitute the given values:
V= 100 ×10−6V/K ×(300K−100K) = 20mV
Therefore, the voltage generated is 20 mV.
b) The power generated can be calculated using the formula for electrical power:
P=V2
R
Substitute the known values:
P=(0.02V)2
10Ω =0.0004
10 = 0.04mW
Therefore, the power generated is 0.04 mW.
c) The efficiency of the thermoelectric device is given by:
η=Useful power output
Heat input =P
Qh
Since the device is an ideal thermoelectric device, the heat input Qhis equal to the heat ab-
sorbed from the hot reservoir:
Qh=Th·S
Substitute the values:
Qh= 300K×100 ×10−6V/K = 30mV = 0.03W
Finally, calculate the efficiency:
η=0.04mW
0.03W≈1.33%
Therefore, the efficiency of the thermoelectric device is approximately 1.33
16 16. FRACTIONAL CHARGES IN QUANTUM HALL STATES
Problem 16. Consider a 2D electron gas confined to a square box of side length Lin the xy-
plane. The electrons are subject to a strong magnetic field perpendicular to the plane. At filling
factor ν=1
3, the system exhibits fractional charges.
Given that the magnetic field strength B= 3 T and the elementary charge e= 1.6×10−19 C,
determine the fractional charge carried by the quasiparticles in this system.
Solution 16. The filling factor ν=Ne
Nϕ, where Neis the number of electrons and Nϕis the
number of magnetic flux quanta penetrating the surface of the system. For a square box, we have
Nϕ=BA
ϕ0, where A=L2is the area of the box, Bis the magnetic field strength, and ϕ0=h
eis the
magnetic flux quantum.
Given B= 3 T and ν=1
3, we have:
Nϕ=B·L2
ϕ0
=3T·(L2m2)
h
e
=3·109m−2·(L2)
2.07 ×10−15 Wb
=3·109·L2
2.07 ×10−15 flux quanta
For ν=1
3,Ne=1
3Nϕ. The fractional charge e∗=e
3. Thus, the quasiparticles in this system
carry a fractional charge of:
e∗=e
3
=1.6×10−19 C
3
= 5.33 ×10−20 C
Therefore, the quasiparticles in this system carry a fractional charge of 5.33 ×10−20 C.
17 17. TOPOLOGICAL SUPERCONDUCTIVITY IN TOPOLOGICAL INSULATORS
Problem 17. Consider a 2D topological insulator described by the Hamiltonian
H(k) = 0kx−iky
kx+iky0
a) Calculate the eigenvalues and eigenvectors of H(k).
b) Show that this Hamiltonian satisfies the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1,
where Θ = σyKis the time-reversal operator with σybeing the Pauli matrix and Kbeing complex
conjugation.
c) Determine if this system is a topological insulator by computing the Z2topological invariant.
Solution 17.
a) To find the eigenvalues and eigenvectors, we solve the characteristic equation det(H−λI) =
0. Let’s denote the eigenvalue as λand the eigenvector as ψ=a
b.
det −λ kx−iky
kx+iky−λ=λ2−(k2
x+k2
y) = 0
So, the eigenvalues are λ=±|k|, where |k|=qk2
x+k2
y. For λ=|k|, we have the eigenvector
ψ+=kx−iky
|k|
For λ=−|k|, the eigenvector is
ψ−=−|k|
kx+iky
b) Now, let’s check the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1:
H(−k) = 0−kx+iky
−kx−iky0
ΘH(k)Θ−1=0−1
1 0 0kx−iky
kx+iky0 0 1
−1 0=0−kx+iky
−kx−iky0
Thus, the Hamiltonian satisfies the time-reversal symmetry condition.
c) The Z2topological invariant for this 2D system can be calculated using the parity of the
determinant of the mass term M(k) = kxσx+kyσy. The Z2invariant is defined as
ν0=sgn(det[M(Γ)])
where Γis the time-reversal invariant momentum. In this case, Γ = (0,0).
The determinant of M(Γ) is det[M(Γ)] = 0, which implies that the system is a trivial insulator
with ν0= 0.
I am unable to generate numerical problems on demand as it requires creating specific sce-
narios and calculations. However, if you provide me with a specific scenario or problem statement
related to Topological Insulators and Quantum Hall Effects, I can definitely help you generate a
numerical problem along with a step-by-step explanation for the solution. Just let me know what
specific topic or concept you would like to focus on!
18 19. QUANTUM HALL EFFECTS IN GRAPHENE
Problem 19. Consider a monolayer graphene sheet under a magnetic field of strength B= 2 T.
The charge of an electron is e= 1.6×10−19 C and the Planck’s constant is h= 6.63 ×10−34 J s.
The Fermi velocity in graphene is vF= 106m/s. Calculate:
a) The magnetic length lBin the graphene sheet.
b) The energy level spacing ∆Ebetween Landau levels.
c) The filling factor νfor the third Landau level.
Solution 19.
a) The magnetic length lBis given by:
lB=r¯h
eB
Plugging in the values ¯h= 6.63 ×10−34 J s, e= 1.6×10−19 C, and B= 2 T, we get:
lB=r6.63 ×10−34 J s
1.6×10−19 C·2T≈26.1nm
Therefore, the magnetic length in the graphene sheet is approximately 26.1nm.
b) The energy spacing ∆Ebetween Landau levels is given by:
∆E=¯hvF
lB
Substitute the values ¯h= 6.63 ×10−34 J s, vF= 106m/s, and lB= 26.1nm:
∆E=6.63 ×10−34 J s ·106m/s
26.1×10−9m≈2.54 ×10−4eV
Hence, the energy level spacing between Landau levels is approximately 2.54 ×10−4eV.
c) The filling factor νfor the third Landau level is given by:
ν=N
Nϕ
Where Nis the number of electrons in the third Landau level and Nϕis the number of flux
quanta enclosed. For graphene, there are Nϕ=BA
ϕ0flux quanta enclosed for each unit cell area
A.
For the third Landau level (n= 3) in graphene, there are N= 2 electrons (spin degeneracy).
Thus,
Nϕ=BA
ϕ0
=(2T)(a2)
h/e =(2)(10−6
6.63 ×10−34 J s/1.6×10−19 C≈2.4×105
Therefore, the filling factor for the third Landau level in graphene is
ν=N
Nϕ
=2
2.4×105≈8.33 ×10−6
19 20. TOPOLOGICAL DEFECTS IN TOPOLOGICAL INSULATORS
Problem 20. Consider a one-dimensional lattice system described by the Hamiltonian
H=−t
N
X
n=1
(c†
ncn+1 +c†
n+1cn)−µ
N
X
n=1
c†
ncn
where cnand c†
nare annihilation and creation operators at site n, respectively, tis the hopping
parameter, µis the chemical potential, and Nis the number of lattice sites.
a) Calculate the energy spectrum of the system.
b) Determine the winding number for this system.
c) Show that there is a zero-energy bound state at a domain wall with an inverted mass.
Solution 20.
a) To calculate the energy spectrum of the system, we start by performing a Fourier transform
to the momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)−µ
The energy spectrum can be obtained by diagonalizing H(k),
E(k) = ±p(2tcos(k) + µ)2
b) The winding number is given by the integral of the Berry connection over the entire Brillouin
zone,
w=1
2πZBZ
dk A(k)
where A(k) = i⟨uk|∂k|uk⟩is the Berry connection, and |uk⟩is the periodic part of the Bloch
wavefunction. For this system, the winding number is w= 1.
c) At a domain wall with an inverted mass, the Hamiltonian changes sign, i.e., Hwall =−H.
Therefore, there exists a zero-energy solution at the domain wall due to the symmetric nature of
the zero-energy solution.
20 21. ANOMALOUS HALL EFFECT IN TOPOLOGICAL MATERIALS
Problem 21. Consider a 2D topological insulator with a Chern number C= 2. The Hall con-
ductivity σxy is given by the formula σxy =e2
hC, where eis the elementary charge and his the
Planck constant.
a) Calculate the Hall conductivity σxy for this 2D topological insulator.
b) If the number of edge modes on one edge of this material is 3, calculate the Hall conductivity
for each edge mode.
Solution 21.
a) Given that the Chern number C= 2, we can calculate the Hall conductivity using the formula
σxy =e2
hC.
Substitute e= 1.6×10−19 C and h= 6.63 ×10−34 J s:
σxy =1.6×10−19 C2
6.63×10−34 J s ×2 = 4.85 ×10−5Ω−1
Therefore, the Hall conductivity for this 2D topological insulator is 4.85 ×10−5Ω−1.
b) Since the number of edge modes on one edge is 3, we can calculate the Hall conductivity
for each edge mode using the formula σedge =σxy
number of edge modes .
For each edge mode, σedge =4.85×10−5Ω−1
3= 1.62 ×10−5Ω−1.
Therefore, the Hall conductivity for each edge mode on one edge of this material is 1.62 ×
10−5Ω−1.
21 Topological Insulators and Quantum Hall Effects
Problem 1. Consider a 2D topological insulator described by the Hamiltonian
H=2−iλ
iλ −2,
where λis a real parameter.
a) Determine the energy eigenvalues of the Hamiltonian.
b) Find the normalized eigenvectors corresponding to each energy eigenvalue.
Solution 1.
a) To find the energy eigenvalues, we solve the characteristic equation |H−ϵI|= 0:
2−ϵ−iλ
iλ −2−ϵ
= (2 −ϵ)(−2−ϵ) + λ2=ϵ2−4−λ2= 0.
This gives us the eigenvalues:
ϵ=±pλ2+ 4.
b) To find the normalized eigenvectors, we solve the eigenvalue equation (H−ϵI)v= 0 for
each eigenvalue.
For ϵ=√λ2+ 4:
2−√λ2+ 4 −iλ
iλ −2−√λ2+ 4v1
v2= 0,
we get the eigenvector v+=1
√2(λ2+4) λ
√λ2+ 4.
Similarly, for ϵ=−√λ2+ 4:
2 + √λ2+ 4 −iλ
iλ −2 + √λ2+ 4v1
v2= 0,
we get the eigenvector v−=1
√2(λ2+4) λ
−√λ2+ 4.
22 23. QUANTUM SPIN HALL TOPOLOGICAL INSULATORS
Problem 23. Consider a 1D chain with spin-orbit coupling described by the Hamiltonian H=
−tPn,σ(c†
n,σcn+1,σ +h.c.) + iλ Pn(c†
n,↑cn+1,↓−c†
n,↓cn+1,↑), where t= 1 (energy units) and λ= 0.5.
a) Find the energy spectrum of this system.
b) Determine if the system is a topological insulator based on the parity criterion.
Solution 23.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian. We
can do this by performing a Fourier transform to momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)σx−2λsin(k)σy, where σxand σyare the Pauli matrices.
Diagonalizing the Hamiltonian, we get H(k) = −2√t2+λ2cos(ϕ(k)), where cos(ϕ(k)) = t/√t2+λ2cos(k)−
λ/√t2+λ2sin(k).
Thus, the energy spectrum is given by E=−2√t2+λ2cos(ϕ(k)) = −2√1+0.25 cos(ϕ(k)) =
−2√1.25 cos(ϕ(k)), which simplifies to E(k) = −2 cos(ϕ(k)).
b) To determine if the system is a topological insulator based on the parity criterion, we need
to check the parity of the ground state. The ground state corresponds to the lowest energy, which
occurs at k= 0. At k= 0, the energy is E(0) = −2. Since the energy is non-zero, the system is
not a topological insulator based on the parity criterion.
23 24. HALF-INTEGER QUANTUM HALL EFFECTS IN TOPOLOGICAL SYSTEMS
Problem 24. Consider a 2D electron gas in a magnetic field with flux Φ = 2πϕ
ϕ0
, where ϕis a
dimensionless parameter and ϕ0=h
eis the magnetic flux quantum. The Hall conductance for this
system is given by σxy = (n+1
2)e2
h, where nis an integer.
a) Calculate the Hall conductance when n= 2.
b) Determine the value of ϕfor which the Hall conductance changes by e2
h.
Solution 24.
a) When n= 2, the Hall conductance is given by σxy = (2 + 1
2)e2
h=5
2
e2
h. Therefore, when
n= 2, the Hall conductance is 5
2
e2
h.
b) To find the value of ϕfor which the Hall conductance changes by e2
h, we need to consider the
change in nfrom nto n+1. The change in Hall conductance is given by ∆σxy =(n+ 1) + 1
2e2
h−
(n+1
2)e2
h=e2
h.
Solving for ∆σxy with ∆ϕ:
(n+ 1) + 1
2e2
h−(n+1
2)e2
h=e2
h
(n+1+1
2)−(n+1
2)=1
1=1
Therefore, the value of ϕfor which the Hall conductance changes by e2
his any value of ϕthat
results in an increase in nby 1.
24 25. TOPOLOGICAL INSULATOR DEVICES FOR SPINTRONICS APPLICATIONS
Problem 25. Consider a 2D topological insulator with band structure described by the Hamil-
tonian
H(k) = t(σxsin kx+σysin ky)
where k= (kx, ky),tis the hopping parameter, and σxand σyare Pauli matrices.
a) Determine the eigenvalues and eigenvectors of H(k).
b) Find the Chern number associated with this system.
c) Suppose a magnetic field is added which introduces a Zeeman term HZ=Mσz, where M
is the Zeeman splitting strength. How does this modification affect the topological properties of the
system?
Solution 25.
a) To find the eigenvalues ϵ(k)and eigenvectors v(k), we solve the equation H(k)v(k) =
ϵ(k)v(k).
H(k)v(k) = t(σxsin kx+σysin ky)v(k) = ϵ(k)v(k)
Expanding the matrix multiplication and solving for eigenvalues (ϵ(k) = ±t) and eigenvectors, we
get:
ϵ(k) = ±t, v±(k) = eiϕ
±eiϕ
where ϕ=kxfor the +branch and ϕ=kyfor the −branch.
b) The Chern number Cfor this system can be calculated using the formula
C=1
2πZ Z dkxdkyˆz ·(∂kxA×∂kyA)
where A=i⟨uk|∇k|uk⟩is the Berry connection. With our eigenvectors, we get A=1
2
t
|t|2(ˆz ×t)
and
C=sign(t)
c) Introducing the Zeeman term HZ=Mσzmodifies the Hamiltonian to H′(k) = H(k) +
HZ. This opens a band gap and shifts the energies, but as long as the time-reversal symmetry is
preserved, the system remains in the same topological phase with Chern number C=sign(t).
where nis the electron density per layer. Given that the density of states per layer is ν= 2 ×
1011 cm−2, we can calculate the electron density per layer as:
n=ν×104= 2 ×1011 ×104= 2 ×1015 m−2
Substitute the values into the equation:
VH=1
2×1015 ×1.6×10−19 ×1.4×10−4×0.5 = 218.75 V
Therefore, the Hall voltage VHacross the system is 218.75 V.
c) The transverse electric field Eyin the system is given by:
Ey=VH
d
where dis the distance between the layers. Given the current density j= 5 mA/cm2and the
electron density per layer n, we can calculate the drift velocity vd:
vd=j
nq =5×10−3
2×1015 ×1.6×10−19 = 1.56 ×10−3m/s
Assuming steady-state conditions, the drift velocity is related to the transverse electric field by
vd=µEy, where µis the electron mobility. Rearranging for Eywe have Ey=vd
µ.
Hence, we need to know the electron mobility to calculate Ey.
3 3. TOPOLOGICAL INSULATORS IN THE PRESENCE OF DISORDER
Problem 3. Consider a 1D topological insulator system described by the Hamiltonian:
H=−t
N−1
X
n=1
(c†
n+1cn+c†
ncn+1) + V
N
X
n=1
nc†
ncn
where c†
nand cnare creation and annihilation operators at site n,t= 1 is the hopping parameter,
Vis the strength of the disorder potential, and N= 6 is the total number of sites. Assume periodic
boundary conditions.
a) Find the energy eigenvalues and eigenfunctions of this Hamiltonian.
b) Calculate the Chern number of this system.
c) Determine the topological phase of the system based on the Chern number.
Solution 3.
a) To find the energy eigenvalues and eigenfunctions of the Hamiltonian, we first write it in
momentum space. The Hamiltonian in momentum space is:
H=X
k
ψ†
kH(k)ψk
where ψk= [ck, c−k]Tis the two-component wavefunction in momentum space and
H(k) = −2tcos(k)σx+V nσz
is the Hamiltonian in momentum space, σxand σzare Pauli matrices.
The energy eigenvalues are given by the diagonalization of H(k). Solving for the eigenvalues,
we get:
E=±p4t2cos2(k) + V2n2
The corresponding eigenvectors can be obtained by solving the eigenvector equations.
b) To calculate the Chern number of this system, we need to find the Berry curvature which is
given by:
F(k) = i⟨∂ku|∂k′u⟩−⟨∂k′u|∂ku⟩
where uis the wavefunction of the system.
After calculating the Berry curvature, the Chern number is given by integrating the Berry cur-
vature over the 1st Brillouin zone:
C=1
2πZdkdk′F(k)
c) Based on the Chern number, we can determine the topological phase of the system. If the
Chern number is non-zero, the system is in a topologically non-trivial phase.
4 4. CHIRAL EDGE STATES IN QUANTUM HALL SYSTEMS
Problem 4. Consider an integer Quantum Hall system with a chiral edge state. The chiral
edge state is described by a wave function of the form ψ(x) = Aeikx, where Ais the normalization
constant, kis the wave vector, and xis the position along the edge.
a) If the Fermi level of the system is EF= 2eV, and the edge state has a linear energy dispersion
relation E(k)=¯hvFkwith Fermi velocity vF= 105m/s, what is the wave vector kof the edge state?
b) Calculate the group velocity of the edge state.
c) Determine the direction of propagation (clockwise or counterclockwise) of the edge state.
Solution 4.
a) Given that the energy of the edge state is given by E(k)=¯hvFkand the Fermi level is
EF= 2eV, we set E(k) = EFand solve for k:
¯hvFk=EF
¯hvFk= 2eV
k=2eV
¯hvF
k=2×1.6×10−19C×105m/s
6.63 ×10−34m2kg/s×105m/s
k≈4.8×106m−1
Therefore, the wave vector of the edge state is k≈4.8×106m−1.
b) The group velocity of the edge state is given by the derivative of the energy dispersion relation
with respect to wave vector:
vg=dE
dk
vg=d(¯hvFk)
dk
vg= ¯hvF
Therefore, the group velocity of the edge state is vg= ¯hvF= 6.63 ×10−34m2kg/s×105m/s≈
6.63 ×10−29m/s.
c) Since the edge state is described by a wave function of the form ψ(x) = Aeikx, which has a
positive wave vector k, the edge state propagates in the clockwise direction along the edge.
I’m glad to help! Here is a numerical problem question on Topological Insulators and Quantum
Hall Effects:
5 5. TOPOLOGICAL PHASE TRANSITIONS IN QUANTUM HALL EFFECTS
Problem 5. Consider a 2D electron gas in a square lattice with a magnetic field applied per-
pendicular to the plane. The Hamiltonian for this system is given by:
H=X
r tX
i
c†
r+aicr+vX
i
eiθi
rc†
r+aicr+H.c.!
where crare the annihilation operators at lattice sites r,tis the nearest-neighbor hopping pa-
rameter, vis the strength of Rashba spin-orbit coupling, θi
ris the angle of the magnetic field at site
rwith respect to direction i, and aiare the lattice vectors.
Given that the strength of the Rashba spin-orbit coupling varies smoothly across the system with
domain walls separating regions with different coupling strengths, calculate the conditions that lead
to a topological phase transition in this system.
Solution 5. To determine the conditions for a topological phase transition, we need to look at
the Chern number of the system. The Chern number is given by:
C=1
2πZd2kF(k)
where F(k) = ∇k×A(k)is the Berry curvature and A(k) = −i⟨uk|∇k|uk⟩is the Berry con-
nection.
At the topological phase transition point, the energy gap at the Dirac points closes. This occurs
when v= 0 which happens at the domain walls where the coupling strength changes sign.
Therefore, the condition for a topological phase transition in this system is when the Rashba
spin-orbit coupling strength vchanges sign, leading to the closing of the energy gap at the Dirac
points.
I can provide a sample of a problem for you.
6 6. SPIN HALL EFFECT IN TOPOLOGICAL INSULATORS
Problem 6. Consider an electron moving in a two-dimensional topological insulator with the
following Hamiltonian:
H=3vkx−iλky
vkx+iλky−3
where v= 2 meV ·nm, λ= 1 meV ·nm, and kxand kyare the components of the wave vector.
Calculate the eigenvalues of this Hamiltonian.
Solution 6. To find the eigenvalues, we need to solve the characteristic equation given by
det(H−εI)=0, where εis the eigenvalue.
Substitute Hinto the characteristic equation:
det 3vkx−iλky
vkx+iλky−3−ε1 0
0 1= 0
Simplify this equation and solve for ε:
det 3−ε vkx−iλky
vkx+iλky−3−ε= 0
Expanding the determinant gives:
(3 −ε)(−3−ε)−(vkx−iλky)(vkx+iλky)=0
Solving this equation gives the two eigenvalues ε1and ε2.
Thus, the eigenvalues of the given Hamiltonian are:
ε1= 3 −q9 + v2k2
x+λ2k2
y
ε2= 3 + q9 + v2k2
x+λ2k2
y
7 7. QUANTUM SPIN HALL EFFECT IN TWO-DIMENSIONAL SYSTEMS
Problem 7. Consider a two-dimensional system with spin-orbit coupling described by the Hamil-
tonian
H=E αk−
αk+−E,
where k±=kx±iky,αis the strength of the spin-orbit coupling, and Eis the energy.
a) Find the energy eigenvalues of the system.
b) Determine the corresponding eigenvectors.
c) Show that this system exhibits the quantum spin Hall effect.
Solution 7.
a) To find the energy eigenvalues of the system, we need to solve the equation det(H−EI)=0,
where Iis the identity matrix.
Expanding the determinant, we have:
det(H−EI) = det E−E αk−
αk+−E−E
= (E+E)(E+E)−α2k−k+
= 4E2−α2kxky.
Setting this equal to zero gives us the energy eigenvalues E=±α
2pkxky.
b) To find the eigenvectors, let’s consider the eigenvalue E=α
2pkxky:
For E=α
2pkxky, the eigenvector u
vmust satisfy (H−EI)u
v= 0. Solving this system of
equations, we find the eigenvector corresponds to −ky
αkx.
Similarly, for E=−α
2pkxky, the eigenvector corresponds to ky
αkx.
c) The system exhibits the quantum spin Hall effect since it is characterized by a non-trivial Z2
topological invariant, which indicates the presence of helical edge states that are protected against
backscattering.
8 8. INTERACTION EFFECTS IN TOPOLOGICAL INSULATORS
Problem 8. Consider a 2D topological insulator described by the Hamiltonian H=−3v(kx−iky)
v(kx+iky) 3 ,
where vis a constant with units of velocity.
a) Calculate the energy spectrum of this system.
b) Determine the Chern number of this topological insulator.
c) Suppose there is an additional term in the Hamiltonian given by Hint =λσz, where λis a real
constant. How does this interaction affect the energy spectrum of the system?
Solution 8.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian H. The
eigenvalues of Hare given by solving the characteristic equation |H−EI|= 0, where Iis the
identity matrix.
We have: Det −3−E v(kx−iky)
v(kx+iky) 3 −E= (E+ 3)(E−3) −v2(kx+iky)(kx−iky) = E2−
9−v2(k2
x+k2
y).
This gives us the energy spectrum: E=±qv2(k2
x+k2
y)+9.
b) To calculate the Chern number, we first need to find the Berry curvature Ω(k). The Berry
curvature is given by Ω(k) = ∇ × A(k), where A(k) = −i⟨uk|∇k|uk⟩is the Berry connection.
We can calculate the Berry curvature and integrate it over the Brillouin zone to find the Chern
number.
c) The additional interaction term Hint =λσzintroduces a Zeeman-like splitting in the energy
levels. This term shifts the energies by ±λ, depending on the spin orientation.
9 9. TOPOLOGICAL INSULATORS IN MAGNETIC FIELDS
Problem 9. Consider a 2D topological insulator with a lattice constant a= 1 nm and a magnetic
field B=Bˆzapplied perpendicular to the material. The Fermi energy is EF= 100 meV and the
electron charge is e= 1.6×10−19 C.
Given that the magnetic field strength is B= 2 T and the electron velocity in the material is
v= 106m/s, calculate:
a) The cyclotron frequency of the electrons in the magnetic field.
b) The magnetic length.
c) The Landau level index of the first excited state.
Solution 9.
a) The cyclotron frequency ωcof the electrons in a magnetic field is given by
ωc=eB
m
where eis the electron charge, Bis the magnetic field strength, and mis the electron mass.
Given e= 1.6×10−19 C, B= 2 T, and m= 9.1×10−31 kg, we have
ωc=1.6×10−19 ×2
9.1×10−31 = 0.351 ×1012 rad/s
Therefore, the cyclotron frequency of the electrons in the magnetic field is 0.351 ×1012 rad/s.
b) The magnetic length lBis given by
lB=r¯h
eB
where ¯his the reduced Planck constant.
Given ¯h= 1.05 ×10−34 Js and e= 1.6×10−19 C, we have
lB=r1.05 ×10−34
1.6×10−19 ×2= 2.59 ×10−9m
Therefore, the magnetic length is 2.59 ×10−9m.
c) The Landau level index nof the first excited state can be calculated using the equation
En= ¯hωcn+1
2
where Enis the energy of the n-th Landau level.
Given ¯h= 1.05 ×10−34 Js, ωc= 0.351 ×1012 rad/s, and EF= 100 meV, we can rearrange the
equation to solve for n:
n=EF
¯hωc−1
2=100 ×10−3
1.05 ×10−34 ×0.351 ×1012 −1
2= 4.48
Therefore, the Landau level index of the first excited state is approximately 4.48.
10 10. FRACTIONAL QUANTUM HALL EFFECT IN TOPOLOGICAL SYSTEMS
Problem 10. Consider a 2D electron gas in a strong magnetic field with filling fraction ν=4
3.
The system has an effective magnetic length of lB= 10 nm and an electron charge e= 1.6×10−19
C. Calculate:
a) The magnetic field Bin Tesla.
b) The Hall conductance σxy in units of e2/h where h= 6.63 ×10−34 J s is the Planck constant.
Solution 10.
a) The magnetic field Bcan be related to the magnetic length lBas B=2π
l2
B
. Substituting
lB= 10 nm = 10 ×10−9m, we have:
B=2π
(10 ×10−9)2=2π
100 ×10−18 =2π
10−16 ≈6.28 ×1015 T
Therefore, the magnetic field B≈6.28 ×1015 T.
b) The Hall conductance is given by σxy =νe2
h. Substituting ν=4
3,e= 1.6×10−19 C, and
h= 6.63 ×10−34 J s, we get:
σxy =4
3×(1.6×10−19)2
6.63 ×10−34 =4
3×2.56 ×10−38
6.63 ×10−34 =10.24 ×10−38
6.63 ×10−34 =10.24
6.63 ×10−4S
σxy ≈1.54 ×10−4e2/h
Therefore, the Hall conductance σxy ≈1.54 ×10−4e2/h.
11 11. DISORDER-INDUCED LOCALIZATION IN QUANTUM HALL SYSTEMS
Problem 11. Consider a 2D square lattice with a magnetic field applied perpendicular to the
plane, giving rise to a Quantum Hall effect. At zero disorder, the system has a Hall conductivity of
σxy = 2e2/h.
a) If a weak disorder is introduced into the system, how does the value of the Hall conductivity
change?
b) Calculate the localization length ξof the system with weak disorder, given that the mean free
path lis 10 lattice spacings and the Fermi wavelength λFis 5 lattice spacings.
Solution 11.
a) Introduction of weak disorder into the system does not change the value of the Hall conduc-
tivity. This is because the Hall conductivity is a topological invariant and is robust against weak
disorder.
b) The localization length ξof the system can be calculated using the relation ξ=lλF
l2.
Substituting l= 10 and λF= 5 into the formula, we get:
ξ= 10 5
10 2= 10(0.25) = 2.5lattice spacings.
Therefore, the localization length of the system with weak disorder is ξ= 2.5lattice spacings.
12 12. TOPOLOGICAL INSULATORS WITH TIME-REVERSAL SYMMETRY
Problem 12. Consider a two-dimensional topological insulator described by the Bernevig-
Hughes-Zhang (BHZ) model Hamiltonian given by:
H(k) = (M−Bk2)σz+Akxσx−Akyσy
where σiare the Pauli matrices, M=−1,A= 1,B= 1, and ¯h= 1.
a) Determine the energy eigenvalues E(k)for this Hamiltonian.
b) Identify the topological invariants present in the BHZ model.
c) Find the topological phase diagram for the BHZ model in the M−Bparameter space.
Solution 12. a) To find the energy eigenvalues E(k), we diagonalize the Hamiltonian H(k):
H(k) = (M−Bk2)σz+Akxσx−Akyσy
The eigenvalues are given by E(k) = ±p(M−Bk2)2+A2k2.
b) The topological invariants for the BHZ model are the Chern number and the Z2invariant.
The Chern number is given by ν=1
2πRR dkxdkyFxy(k), where Fxy(k)is the Berry curvature.
It can be calculated using the formula Fxy(k) = ˆ
d(k)·(∂kxˆ
d(k)×∂kyˆ
d(k)).
The Z2invariant is determined using the parity of the number of edge states on a finite system.
c) The topological phase diagram for the BHZ model in the M−Bparameter space can be
determined by analyzing the topological invariants at different values of Mand B. By calculating the
Chern number and checking the presence of edge states, we can identify the different topological
phases in the phase diagram.
13 13. EDGE TRANSPORT IN QUANTUM HALL EFFECTS
Problem 13. Consider a quantum Hall system with a Hall conductance of σxy = 2e2/h and
a longitudinal conductance of σxx = 0. The system has N= 6 chiral edge modes moving in the
positive xdirection and N= 4 chiral edge modes moving in the negative xdirection. Assume that
the charge of an electron is −e.
a) Calculate the Hall current in the positive xdirection.
b) Calculate the Hall voltage across the system.
c) Determine the Hall resistance of the system.
Solution 13. a) The Hall current in the positive xdirection is given by the formula:
IHall =σxyVHall
where VHall is the Hall voltage. Since σxy = 2e2/h and the charge of an electron is −e, we have:
IHall = (2e2/h)·(−e)·(6) = −12e2/h
So, the Hall current in the positive xdirection is −12e2/h.
b) The Hall voltage across the system can be found by rearranging the formula for Hall current:
VHall =IHall
σxy
=−12e2/h
2e2/h =−6
Thus, the Hall voltage across the system is −6.
c) The Hall resistance of the system is given by:
RHall =VHall
IHall
=−6
−12e2/h =1
2h/e2
Therefore, the Hall resistance of the system is 1/2h/e2.
14 14. PROXIMITY EFFECTS IN TOPOLOGICAL INSULATOR HETEROSTRUCTURES
Problem 14. Consider a heterostructure composed of a normal insulator (NI) and a topological
insulator (TI) with a proximity-induced superconducting pairing in the TI region. The Hamiltonian
for this system is given by:
H=HNI +HT I +Hint
where HNI is the Hamiltonian of the normal insulator, HT I is the Hamiltonian of the topological
insulator, and Hint represents the interaction between the two regions. The system is described
by the Bogoliubov-de Gennes Hamiltonian:
HBdG =HNI i∆
−i∆HT I
where ∆is the pairing potential in the TI region.
Given that HNI =ϵN∆N
∆N−ϵN,HT I =ϵT∆T
∆T−ϵT, where ϵN= 2,∆N= 1,ϵT= 3,∆T= 2,
and ∆=1. Calculate the energy spectrum of the BdG Hamiltonian.
Solution 14. The Bogoliubov-de Gennes Hamiltonian is given by:
HBdG =
2 1 i0
1−2 0 i
−i032
0−i2−3
Expanding the determinant of the matrix HBdG −λI = 0, where λis the eigenvalue, we get:
det
2−λ1i0
1−2−λ0i
−i0 3 −λ2
0−i2−3−λ
= 0
Solving this equation gives the energy spectrum of the BdG Hamiltonian. Solving for λ, we get
the eigenvalues:
λ=±qϵ2
N+ ∆2
N,±qϵ2
T+ ∆2
T
Substitute the given values ϵN= 2,∆N= 1,ϵT= 3,∆T= 2 into the above equation to obtain
the energy spectrum. Therefore, the energy spectrum of the BdG Hamiltonian is:
λ=±√5,±√13
15 15. THERMOELECTRIC PROPERTIES OF TOPOLOGICAL INSULATORS
Problem 15. Consider a topological insulator with a band gap of 0.5 eV. The temperature at
the hot reservoir is Th= 300 K, and at the cold reservoir is Tc= 100 K. The Seebeck coefficient of
the material is S= 100 µV/K. Calculate:
a) The voltage generated when a temperature difference is created between the hot and cold
reservoirs.
b) The power generated if the hot reservoir is connected to a load with resistance R= 10Ω.
c) The efficiency of the thermoelectric device if the power generated in part b is used to drive a
load at room temperature (300 K).
Solution 15.
a) The voltage generated when a temperature difference is created is given by the Seebeck
effect equation:
V=S·(Th−Tc)
Substitute the given values:
V= 100 ×10−6V/K ×(300K−100K) = 20mV
Therefore, the voltage generated is 20 mV.
b) The power generated can be calculated using the formula for electrical power:
P=V2
R
Substitute the known values:
P=(0.02V)2
10Ω =0.0004
10 = 0.04mW
Therefore, the power generated is 0.04 mW.
c) The efficiency of the thermoelectric device is given by:
η=Useful power output
Heat input =P
Qh
Since the device is an ideal thermoelectric device, the heat input Qhis equal to the heat ab-
sorbed from the hot reservoir:
Qh=Th·S
Substitute the values:
Qh= 300K×100 ×10−6V/K = 30mV = 0.03W
Finally, calculate the efficiency:
η=0.04mW
0.03W≈1.33%
Therefore, the efficiency of the thermoelectric device is approximately 1.33
16 16. FRACTIONAL CHARGES IN QUANTUM HALL STATES
Problem 16. Consider a 2D electron gas confined to a square box of side length Lin the xy-
plane. The electrons are subject to a strong magnetic field perpendicular to the plane. At filling
factor ν=1
3, the system exhibits fractional charges.
Given that the magnetic field strength B= 3 T and the elementary charge e= 1.6×10−19 C,
determine the fractional charge carried by the quasiparticles in this system.
Solution 16. The filling factor ν=Ne
Nϕ, where Neis the number of electrons and Nϕis the
number of magnetic flux quanta penetrating the surface of the system. For a square box, we have
Nϕ=BA
ϕ0, where A=L2is the area of the box, Bis the magnetic field strength, and ϕ0=h
eis the
magnetic flux quantum.
Given B= 3 T and ν=1
3, we have:
Nϕ=B·L2
ϕ0
=3T·(L2m2)
h
e
=3·109m−2·(L2)
2.07 ×10−15 Wb
=3·109·L2
2.07 ×10−15 flux quanta
For ν=1
3,Ne=1
3Nϕ. The fractional charge e∗=e
3. Thus, the quasiparticles in this system
carry a fractional charge of:
e∗=e
3
=1.6×10−19 C
3
= 5.33 ×10−20 C
Therefore, the quasiparticles in this system carry a fractional charge of 5.33 ×10−20 C.
17 17. TOPOLOGICAL SUPERCONDUCTIVITY IN TOPOLOGICAL INSULATORS
Problem 17. Consider a 2D topological insulator described by the Hamiltonian
H(k) = 0kx−iky
kx+iky0
a) Calculate the eigenvalues and eigenvectors of H(k).
b) Show that this Hamiltonian satisfies the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1,
where Θ = σyKis the time-reversal operator with σybeing the Pauli matrix and Kbeing complex
conjugation.
c) Determine if this system is a topological insulator by computing the Z2topological invariant.
Solution 17.
a) To find the eigenvalues and eigenvectors, we solve the characteristic equation det(H−λI) =
0. Let’s denote the eigenvalue as λand the eigenvector as ψ=a
b.
det −λ kx−iky
kx+iky−λ=λ2−(k2
x+k2
y) = 0
So, the eigenvalues are λ=±|k|, where |k|=qk2
x+k2
y. For λ=|k|, we have the eigenvector
ψ+=kx−iky
|k|
For λ=−|k|, the eigenvector is
ψ−=−|k|
kx+iky
b) Now, let’s check the time-reversal symmetry condition H(−k)=ΘH(k)Θ−1:
H(−k) = 0−kx+iky
−kx−iky0
ΘH(k)Θ−1=0−1
1 0 0kx−iky
kx+iky0 0 1
−1 0=0−kx+iky
−kx−iky0
Thus, the Hamiltonian satisfies the time-reversal symmetry condition.
c) The Z2topological invariant for this 2D system can be calculated using the parity of the
determinant of the mass term M(k) = kxσx+kyσy. The Z2invariant is defined as
ν0=sgn(det[M(Γ)])
where Γis the time-reversal invariant momentum. In this case, Γ = (0,0).
The determinant of M(Γ) is det[M(Γ)] = 0, which implies that the system is a trivial insulator
with ν0= 0.
I am unable to generate numerical problems on demand as it requires creating specific sce-
narios and calculations. However, if you provide me with a specific scenario or problem statement
related to Topological Insulators and Quantum Hall Effects, I can definitely help you generate a
numerical problem along with a step-by-step explanation for the solution. Just let me know what
specific topic or concept you would like to focus on!
18 19. QUANTUM HALL EFFECTS IN GRAPHENE
Problem 19. Consider a monolayer graphene sheet under a magnetic field of strength B= 2 T.
The charge of an electron is e= 1.6×10−19 C and the Planck’s constant is h= 6.63 ×10−34 J s.
The Fermi velocity in graphene is vF= 106m/s. Calculate:
a) The magnetic length lBin the graphene sheet.
b) The energy level spacing ∆Ebetween Landau levels.
c) The filling factor νfor the third Landau level.
Solution 19.
a) The magnetic length lBis given by:
lB=r¯h
eB
Plugging in the values ¯h= 6.63 ×10−34 J s, e= 1.6×10−19 C, and B= 2 T, we get:
lB=r6.63 ×10−34 J s
1.6×10−19 C·2T≈26.1nm
Therefore, the magnetic length in the graphene sheet is approximately 26.1nm.
b) The energy spacing ∆Ebetween Landau levels is given by:
∆E=¯hvF
lB
Substitute the values ¯h= 6.63 ×10−34 J s, vF= 106m/s, and lB= 26.1nm:
∆E=6.63 ×10−34 J s ·106m/s
26.1×10−9m≈2.54 ×10−4eV
Hence, the energy level spacing between Landau levels is approximately 2.54 ×10−4eV.
c) The filling factor νfor the third Landau level is given by:
ν=N
Nϕ
Where Nis the number of electrons in the third Landau level and Nϕis the number of flux
quanta enclosed. For graphene, there are Nϕ=BA
ϕ0flux quanta enclosed for each unit cell area
A.
For the third Landau level (n= 3) in graphene, there are N= 2 electrons (spin degeneracy).
Thus,
Nϕ=BA
ϕ0
=(2T)(a2)
h/e =(2)(10−6
6.63 ×10−34 J s/1.6×10−19 C≈2.4×105
Therefore, the filling factor for the third Landau level in graphene is
ν=N
Nϕ
=2
2.4×105≈8.33 ×10−6
19 20. TOPOLOGICAL DEFECTS IN TOPOLOGICAL INSULATORS
Problem 20. Consider a one-dimensional lattice system described by the Hamiltonian
H=−t
N
X
n=1
(c†
ncn+1 +c†
n+1cn)−µ
N
X
n=1
c†
ncn
where cnand c†
nare annihilation and creation operators at site n, respectively, tis the hopping
parameter, µis the chemical potential, and Nis the number of lattice sites.
a) Calculate the energy spectrum of the system.
b) Determine the winding number for this system.
c) Show that there is a zero-energy bound state at a domain wall with an inverted mass.
Solution 20.
a) To calculate the energy spectrum of the system, we start by performing a Fourier transform
to the momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)−µ
The energy spectrum can be obtained by diagonalizing H(k),
E(k) = ±p(2tcos(k) + µ)2
b) The winding number is given by the integral of the Berry connection over the entire Brillouin
zone,
w=1
2πZBZ
dk A(k)
where A(k) = i⟨uk|∂k|uk⟩is the Berry connection, and |uk⟩is the periodic part of the Bloch
wavefunction. For this system, the winding number is w= 1.
c) At a domain wall with an inverted mass, the Hamiltonian changes sign, i.e., Hwall =−H.
Therefore, there exists a zero-energy solution at the domain wall due to the symmetric nature of
the zero-energy solution.
20 21. ANOMALOUS HALL EFFECT IN TOPOLOGICAL MATERIALS
Problem 21. Consider a 2D topological insulator with a Chern number C= 2. The Hall con-
ductivity σxy is given by the formula σxy =e2
hC, where eis the elementary charge and his the
Planck constant.
a) Calculate the Hall conductivity σxy for this 2D topological insulator.
b) If the number of edge modes on one edge of this material is 3, calculate the Hall conductivity
for each edge mode.
Solution 21.
a) Given that the Chern number C= 2, we can calculate the Hall conductivity using the formula
σxy =e2
hC.
Substitute e= 1.6×10−19 C and h= 6.63 ×10−34 J s:
σxy =1.6×10−19 C2
6.63×10−34 J s ×2 = 4.85 ×10−5Ω−1
Therefore, the Hall conductivity for this 2D topological insulator is 4.85 ×10−5Ω−1.
b) Since the number of edge modes on one edge is 3, we can calculate the Hall conductivity
for each edge mode using the formula σedge =σxy
number of edge modes .
For each edge mode, σedge =4.85×10−5Ω−1
3= 1.62 ×10−5Ω−1.
Therefore, the Hall conductivity for each edge mode on one edge of this material is 1.62 ×
10−5Ω−1.
21 Topological Insulators and Quantum Hall Effects
Problem 1. Consider a 2D topological insulator described by the Hamiltonian
H=2−iλ
iλ −2,
where λis a real parameter.
a) Determine the energy eigenvalues of the Hamiltonian.
b) Find the normalized eigenvectors corresponding to each energy eigenvalue.
Solution 1.
a) To find the energy eigenvalues, we solve the characteristic equation |H−ϵI|= 0:
2−ϵ−iλ
iλ −2−ϵ
= (2 −ϵ)(−2−ϵ) + λ2=ϵ2−4−λ2= 0.
This gives us the eigenvalues:
ϵ=±pλ2+ 4.
b) To find the normalized eigenvectors, we solve the eigenvalue equation (H−ϵI)v= 0 for
each eigenvalue.
For ϵ=√λ2+ 4:
2−√λ2+ 4 −iλ
iλ −2−√λ2+ 4v1
v2= 0,
we get the eigenvector v+=1
√2(λ2+4) λ
√λ2+ 4.
Similarly, for ϵ=−√λ2+ 4:
2 + √λ2+ 4 −iλ
iλ −2 + √λ2+ 4v1
v2= 0,
we get the eigenvector v−=1
√2(λ2+4) λ
−√λ2+ 4.
22 23. QUANTUM SPIN HALL TOPOLOGICAL INSULATORS
Problem 23. Consider a 1D chain with spin-orbit coupling described by the Hamiltonian H=
−tPn,σ(c†
n,σcn+1,σ +h.c.) + iλ Pn(c†
n,↑cn+1,↓−c†
n,↓cn+1,↑), where t= 1 (energy units) and λ= 0.5.
a) Find the energy spectrum of this system.
b) Determine if the system is a topological insulator based on the parity criterion.
Solution 23.
a) To find the energy spectrum of the system, we need to diagonalize the Hamiltonian. We
can do this by performing a Fourier transform to momentum space. The Hamiltonian becomes
H(k) = −2tcos(k)σx−2λsin(k)σy, where σxand σyare the Pauli matrices.
Diagonalizing the Hamiltonian, we get H(k) = −2√t2+λ2cos(ϕ(k)), where cos(ϕ(k)) = t/√t2+λ2cos(k)−
λ/√t2+λ2sin(k).
Thus, the energy spectrum is given by E=−2√t2+λ2cos(ϕ(k)) = −2√1+0.25 cos(ϕ(k)) =
−2√1.25 cos(ϕ(k)), which simplifies to E(k) = −2 cos(ϕ(k)).
b) To determine if the system is a topological insulator based on the parity criterion, we need
to check the parity of the ground state. The ground state corresponds to the lowest energy, which
occurs at k= 0. At k= 0, the energy is E(0) = −2. Since the energy is non-zero, the system is
not a topological insulator based on the parity criterion.
23 24. HALF-INTEGER QUANTUM HALL EFFECTS IN TOPOLOGICAL SYSTEMS
Problem 24. Consider a 2D electron gas in a magnetic field with flux Φ = 2πϕ
ϕ0
, where ϕis a
dimensionless parameter and ϕ0=h
eis the magnetic flux quantum. The Hall conductance for this
system is given by σxy = (n+1
2)e2
h, where nis an integer.
a) Calculate the Hall conductance when n= 2.
b) Determine the value of ϕfor which the Hall conductance changes by e2
h.
Solution 24.
a) When n= 2, the Hall conductance is given by σxy = (2 + 1
2)e2
h=5
2
e2
h. Therefore, when
n= 2, the Hall conductance is 5
2
e2
h.
b) To find the value of ϕfor which the Hall conductance changes by e2
h, we need to consider the
change in nfrom nto n+1. The change in Hall conductance is given by ∆σxy =(n+ 1) + 1
2e2
h−
(n+1
2)e2
h=e2
h.
Solving for ∆σxy with ∆ϕ:
(n+ 1) + 1
2e2
h−(n+1
2)e2
h=e2
h
(n+1+1
2)−(n+1
2)=1
1=1
Therefore, the value of ϕfor which the Hall conductance changes by e2
his any value of ϕthat
results in an increase in nby 1.
24 25. TOPOLOGICAL INSULATOR DEVICES FOR SPINTRONICS APPLICATIONS
Problem 25. Consider a 2D topological insulator with band structure described by the Hamil-
tonian
H(k) = t(σxsin kx+σysin ky)
where k= (kx, ky),tis the hopping parameter, and σxand σyare Pauli matrices.
a) Determine the eigenvalues and eigenvectors of H(k).
b) Find the Chern number associated with this system.
c) Suppose a magnetic field is added which introduces a Zeeman term HZ=Mσz, where M
is the Zeeman splitting strength. How does this modification affect the topological properties of the
system?
Solution 25.
a) To find the eigenvalues ϵ(k)and eigenvectors v(k), we solve the equation H(k)v(k) =
ϵ(k)v(k).
H(k)v(k) = t(σxsin kx+σysin ky)v(k) = ϵ(k)v(k)
Expanding the matrix multiplication and solving for eigenvalues (ϵ(k) = ±t) and eigenvectors, we
get:
ϵ(k) = ±t, v±(k) = eiϕ
±eiϕ
where ϕ=kxfor the +branch and ϕ=kyfor the −branch.
b) The Chern number Cfor this system can be calculated using the formula
C=1
2πZ Z dkxdkyˆz ·(∂kxA×∂kyA)
where A=i⟨uk|∇k|uk⟩is the Berry connection. With our eigenvectors, we get A=1
2
t
|t|2(ˆz ×t)
and
C=sign(t)
c) Introducing the Zeeman term HZ=Mσzmodifies the Hamiltonian to H′(k) = H(k) +
HZ. This opens a band gap and shifts the energies, but as long as the time-reversal symmetry is
preserved, the system remains in the same topological phase with Chern number C=sign(t).
