SCHRODINGER EQUATION SOLUTIONS FOR
VARIOUS POTENTIALS
1 INTRODUCTION
The time-independent Schrödinger equation is given by:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝑉(𝑥)𝜓=𝐸𝜓
where 𝜓 is the wavefunction, 𝑉(𝑥) is the potential energy, 𝐸 is the total energy, 𝑚 is the mass
of the particle, and ℏ is the reduced Planck constant.
2 PROBLEMS AND SOLUTIONS
2.1 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
1. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
2. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
3. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
4. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
5. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
6. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
7. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.2 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
1. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
2. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
3. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
4. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
5. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
6. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
7. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.3 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
1. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
2. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
3. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
4. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
5. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
6. These transcendental equations must be solved numerically to find the energy levels
7. The number of bound states depends on the well depth and width
2.4 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
1. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
2. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
3. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
4. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
5. This leads to the condition:
ℏ2𝜅
𝑚=𝛼
6. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
7. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.5 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
1. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
2. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
3. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
4. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
5. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
6. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
7. This equation must be solved numerically to find the energy levels
8. The wavefunctions are linear combinations of Airy functions
2.6 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
9. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
10. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
11. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
12. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
13. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
14. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
15. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.7 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
16. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
17. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
18. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
19. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
20. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
21. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
22. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.8 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
23. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
24. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
25. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
26. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
27. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
28. These transcendental equations must be solved numerically to find the energy levels
29. The number of bound states depends on the well depth and width
2.9 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
30. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
31. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
32. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
33. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
34. This leads to the condition: ℏ2𝜅
𝑚=𝛼
35. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
36. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.10 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
37. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
38. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
39. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
40. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
41. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
42. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
43. This equation must be solved numerically to find the energy levels
44. The wavefunctions are linear combinations of Airy functions
2.11 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
45. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
46. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
47. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
48. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
49. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
50. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
51. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.12 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
52. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
53. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
54. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
55. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
56. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
57. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
58. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.13 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
59. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
60. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
61. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
62. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
63. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
64. These transcendental equations must be solved numerically to find the energy levels
65. The number of bound states depends on the well depth and width
2.14 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
66. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
67. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
68. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
69. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
70. This leads to the condition: ℏ2𝜅
𝑚=𝛼
71. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
72. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.15 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
73. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
74. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
75. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
76. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
77. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
78. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
79. This equation must be solved numerically to find the energy levels
80. The wavefunctions are linear combinations of Airy functions
2.16 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
81. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
82. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
83. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
84. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
85. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
86. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
87. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.17 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
88. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
89. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
90. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
91. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
92. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
93. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
94. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.18 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
95. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
96. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
97. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
98. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
99. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
100. These transcendental equations must be solved numerically to find the energy levels
101. The number of bound states depends on the well depth and width
2.19 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
102. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
103. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
104. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
105. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
106. This leads to the condition: ℏ2𝜅
𝑚=𝛼
107. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
108. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.20 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
109. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
110. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
111. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
112. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
113. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
114. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
115. This equation must be solved numerically to find the energy levels
116. The wavefunctions are linear combinations of Airy functions
2.21 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
117. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
118. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
119. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
120. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
121. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
122. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
123. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.22 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
124. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
125. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
126. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
127. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
128. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
129. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
130. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.23 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
131. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
132. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
133. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
134. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
135. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
136. These transcendental equations must be solved numerically to find the energy levels
137. The number of bound states depends on the well depth and width
2.24 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
138. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
139. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
140. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
141. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
142. This leads to the condition: ℏ2𝜅
𝑚=𝛼
143. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
144. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.25 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
145. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
146. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
147. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
148. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
149. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
150. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
151. This equation must be solved numerically to find the energy levels
152. The wavefunctions are linear combinations of Airy functions
2.26 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
153. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
154. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
155. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
156. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
157. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
158. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
159. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.27 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
160. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
161. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
162. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
163. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
164. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
165. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
166. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.28 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
167. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
168. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
169. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
170. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
171. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
172. These transcendental equations must be solved numerically to find the energy levels
173. The number of bound states depends on the well depth and width
2.29 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
174. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
175. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
176. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
177. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
178. This leads to the condition: ℏ2𝜅
𝑚=𝛼
179. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
180. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.30 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
181. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
182. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
183. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
184. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
185. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
186. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
187. This equation must be solved numerically to find the energy levels
188. The wavefunctions are linear combinations of Airy functions
2.31 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
189. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
190. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
191. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
192. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
193. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
194. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
195. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.32 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
196. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
197. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
198. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
199. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
200. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
201. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
202. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.33 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
203. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
204. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
205. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
206. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
207. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
208. These transcendental equations must be solved numerically to find the energy levels
209. The number of bound states depends on the well depth and width
2.34 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
210. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
211. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
212. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
213. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
214. This leads to the condition: ℏ2𝜅
𝑚=𝛼
215. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
216. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.35 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
217. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
218. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
219. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
220. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
221. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
222. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
223. This equation must be solved numerically to find the energy levels
224. The wavefunctions are linear combinations of Airy functions
2.36 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
225. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
226. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
227. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
228. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
229. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
230. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
231. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.37 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
232. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
233. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
234. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
235. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
236. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
237. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
238. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.38 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
239. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
240. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
241. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
242. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
243. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
244. These transcendental equations must be solved numerically to find the energy levels
245. The number of bound states depends on the well depth and width
2.39 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
246. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
247. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
248. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
249. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
250. This leads to the condition:
ℏ2𝜅
𝑚=𝛼
251. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
252. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.40 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
253. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
254. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
255. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
256. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
257. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
258. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
259. This equation must be solved numerically to find the energy levels
260. The wavefunctions are linear combinations of Airy functions
2.41 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
261. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
262. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
263. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
264. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
265. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
266. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
267. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.42 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
268. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
269. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
270. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
271. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
272. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
273. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
274. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.43 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
275. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
276. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
277. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
278. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
279. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
280. These transcendental equations must be solved numerically to find the energy levels
281. The number of bound states depends on the well depth and width
2.44 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
282. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
283. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
284. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
285. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
286. This leads to the condition: ℏ2𝜅
𝑚=𝛼
287. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
288. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.45 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
289. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
290. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
291. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
292. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
293. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
294. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
295. This equation must be solved numerically to find the energy levels
296. The wavefunctions are linear combinations of Airy functions
2.46 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
297. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
298. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
299. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
300. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
301. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
302. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
303. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.47 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
304. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
305. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
306. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
307. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
308. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
309. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
310. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.48 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
311. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
312. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
313. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
314. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
315. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
316. These transcendental equations must be solved numerically to find the energy levels
317. The number of bound states depends on the well depth and width
2.49 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
318. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
319. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
320. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
321. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
322. This leads to the condition: ℏ2𝜅
𝑚=𝛼
323. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
324. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.50 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
325. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
326. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
327. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
328. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
329. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
330. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
331. This equation must be solved numerically to find the energy levels
332. The wavefunctions are linear combinations of Airy functions
2.51 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
333. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
334. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
335. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
336. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
337. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
338. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
339. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.52 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
340. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
341. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
342. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
343. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
344. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
345. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
346. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.53 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
347. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
348. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
349. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
350. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
351. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
352. These transcendental equations must be solved numerically to find the energy levels
353. The number of bound states depends on the well depth and width
2.54 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
354. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
355. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
356. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
357. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
358. This leads to the condition: ℏ2𝜅
𝑚=𝛼
359. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
360. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.55 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
361. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
362. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
363. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
364. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
365. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
366. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
367. This equation must be solved numerically to find the energy levels
368. The wavefunctions are linear combinations of Airy functions
2.56 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
369. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
370. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
371. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
372. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
373. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
374. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
375. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.57 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
376. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
377. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
378. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
379. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
380. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
381. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
382. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.58 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
383. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
384. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
385. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
386. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
387. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
388. These transcendental equations must be solved numerically to find the energy levels
389. The number of bound states depends on the well depth and width
2.59 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
390. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
391. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
392. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
393. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
394. This leads to the condition: ℏ2𝜅
𝑚=𝛼
395. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
396. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.60 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
397. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
398. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
399. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
400. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
401. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
402. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
403. This equation must be solved numerically to find the energy levels
404. The wavefunctions are linear combinations of Airy functions
2.61 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
405. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
406. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
407. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
408. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
409. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
410. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
411. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.62 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
412. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
413. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
414. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
415. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
416. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
417. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
418. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.63 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
419. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
420. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
421. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
422. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
423. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
424. These transcendental equations must be solved numerically to find the energy levels
425. The number of bound states depends on the well depth and width
2.64 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
426. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
427. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
428. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
429. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
430. This leads to the condition: ℏ2𝜅
𝑚=𝛼
431. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
432. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.65 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
433. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
434. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
435. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
436. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
437. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
438. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
439. This equation must be solved numerically to find the energy levels
440. The wavefunctions are linear combinations of Airy functions
2.66 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
441. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
442. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
443. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
444. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
445. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
446. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
447. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.67 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
448. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
449. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
450. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
451. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
452. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
453. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
454. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.68 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
455. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
456. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
457. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
458. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
459. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
460. These transcendental equations must be solved numerically to find the energy levels
461. The number of bound states depends on the well depth and width
2.69 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
462. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
463. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
464. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
465. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
466. This leads to the condition: ℏ2𝜅
𝑚=𝛼
467. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
468. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.70 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
469. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
470. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
471. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
472. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
473. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
474. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
475. This equation must be solved numerically to find the energy levels
476. The wavefunctions are linear combinations of Airy functions
2.71 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
477. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
478. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
479. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
480. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
481. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
482. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
483. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.72 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
484. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
485. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
486. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
487. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
488. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
489. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
490. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.73 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
491. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
492. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
493. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
494. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
495. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
496. These transcendental equations must be solved numerically to find the energy levels
497. The number of bound states depends on the well depth and width
2.74 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
498. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
499. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
500. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
501. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
502. This leads to the condition:
ℏ2𝜅
𝑚=𝛼
503. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
504. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.75 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
505. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
506. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
507. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
508. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
509. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
510. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
511. This equation must be solved numerically to find the energy levels
512. The wavefunctions are linear combinations of Airy functions
2.76 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
513. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
514. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
515. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
516. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
517. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
518. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
519. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.77 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
520. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
521. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
522. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
523. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
524. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
525. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
526. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.78 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
527. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
528. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
529. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
530. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
531. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
532. These transcendental equations must be solved numerically to find the energy levels
533. The number of bound states depends on the well depth and width
2.79 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
534. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
535. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
536. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
537. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
538. This leads to the condition: ℏ2𝜅
𝑚=𝛼
539. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
540. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.80 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
541. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
542. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
543. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
544. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
545. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
546. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
547. This equation must be solved numerically to find the energy levels
548. The wavefunctions are linear combinations of Airy functions
2.81 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
549. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
550. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
551. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
552. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
553. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
554. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
555. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.82 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
556. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
557. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
558. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
559. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
560. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
561. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
562. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.83 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
563. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
564. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
565. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
566. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
567. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
568. These transcendental equations must be solved numerically to find the energy levels
569. The number of bound states depends on the well depth and width
2.84 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
570. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
571. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
572. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
573. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
574. This leads to the condition: ℏ2𝜅
𝑚=𝛼
575. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
576. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.85 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
577. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
578. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
579. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
580. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
581. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
582. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
583. This equation must be solved numerically to find the energy levels
584. The wavefunctions are linear combinations of Airy functions
2.86 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
585. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
586. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
587. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
588. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
589. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
590. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
591. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.87 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
592. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
593. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
594. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
595. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
596. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
597. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
598. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.88 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
599. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
600. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
601. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
602. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
603. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
604. These transcendental equations must be solved numerically to find the energy levels
605. The number of bound states depends on the well depth and width
2.89 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
606. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
607. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
608. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
609. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
610. This leads to the condition: ℏ2𝜅
𝑚=𝛼
611. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
612. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.90 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
613. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
614. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
615. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
616. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
617. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
618. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
619. This equation must be solved numerically to find the energy levels
620. The wavefunctions are linear combinations of Airy functions
2.91 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
621. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
622. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
623. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
624. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
625. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
626. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
627. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.92 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
628. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
629. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
630. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
631. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
632. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
633. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
634. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.93 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
635. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
636. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
637. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
638. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
639. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
640. These transcendental equations must be solved numerically to find the energy levels
641. The number of bound states depends on the well depth and width
2.94 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
642. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
643. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
644. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
645. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
646. This leads to the condition: ℏ2𝜅
𝑚=𝛼
647. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
648. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.95 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
649. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
650. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
651. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
652. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
653. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
654. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
655. This equation must be solved numerically to find the energy levels
656. The wavefunctions are linear combinations of Airy functions
2.96 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
657. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
658. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
659. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
660. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
661. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
662. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
663. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.97 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
664. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
665. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
666. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
667. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
668. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
669. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
670. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.98 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
671. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
672. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
673. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
674. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
675. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
676. These transcendental equations must be solved numerically to find the energy levels
677. The number of bound states depends on the well depth and width
2.99 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
678. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
679. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
680. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
681. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
682. This leads to the condition: ℏ2𝜅
𝑚=𝛼
683. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
684. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.100 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
685. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
686. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
687. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
688. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
689. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
690. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
691. This equation must be solved numerically to find the energy levels
692. The wavefunctions are linear combinations of Airy functions
2.101 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
693. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
694. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
695. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
696. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
697. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
698. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
699. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.102 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
700. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
701. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
702. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
703. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
704. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
705. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
706. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.103 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
707. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
708. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
709. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
710. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
711. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
712. These transcendental equations must be solved numerically to find the energy levels
713. The number of bound states depends on the well depth and width
2.104 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
714. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
715. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
716. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
717. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
718. This leads to the condition: ℏ2𝜅
𝑚=𝛼
719. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
720. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.105 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
721. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
722. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
723. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
724. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
725. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
726. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
727. This equation must be solved numerically to find the energy levels
728. The wavefunctions are linear combinations of Airy functions
2.106 PROBLEM 1: INFINITE SQUARE WELL
Solve the Schrödinger equation for a particle in an infinite square well of width 𝐿.
Solution:
729. The potential is defined as:
𝑉(𝑥)={0for 0<𝑥<𝐿
∞otherwise
730. Inside the well, the Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2=𝐸𝜓
731. The general solution is:
𝜓(𝑥)=𝐴sin(𝑘𝑥)+𝐵cos(𝑘𝑥), where 𝑘=√2𝑚𝐸
ℏ2
732. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
733. This gives 𝐵=0 and 𝑘𝐿=𝑛𝜋, where 𝑛 is a positive integer
734. The energy levels are:
𝐸𝑛=𝑛2𝜋2ℏ2
2𝑚𝐿2
735. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√2
𝐿sin(𝑛𝜋𝑥
𝐿)
2.107 PROBLEM 2: QUANTUM HARMONIC OSCILLATOR
Find the energy levels and wavefunctions for a quantum harmonic oscillator.
Solution:
736. The potential is 𝑉(𝑥)=1
2𝑘𝑥2=1
2𝑚𝜔2𝑥2
737. The Schrödinger equation becomes:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+1
2𝑚𝜔2𝑥2𝜓=𝐸𝜓
738. Introduce dimensionless variables:
𝜉=√𝑚𝜔
ℏ𝑥, 𝜖=2𝐸
ℏ𝜔
739. The equation transforms to: 𝑑2𝜓
𝑑𝜉2+(𝜖−𝜉2)𝜓=0
740. The solution involves Hermite polynomials:
𝜓𝑛(𝜉)=𝑁𝑛𝐻𝑛(𝜉)𝑒−𝜉2/2
741. The energy levels are:
𝐸𝑛=ℏ𝜔(𝑛+1
2), 𝑛=0,1,2,...
742. The normalized wavefunctions are:
𝜓𝑛(𝑥)=√1
2𝑛𝑛!(𝑚𝜔
𝜋ℏ)1/4𝐻𝑛(√𝑚𝜔
ℏ𝑥)𝑒−𝑚𝜔𝑥2/(2ℏ)
2.108 PROBLEM 3: FINITE SQUARE WELL
Determine the condition for bound states in a finite square well of depth 𝑉0 and width 2𝑎.
Solution:
743. The potential is:
𝑉(𝑥)={−𝑉0for |𝑥|<𝑎
0otherwise
744. Inside the well (|𝑥|<𝑎), the solution is:
𝜓𝑖𝑛(𝑥)=𝐴cos(𝑘𝑥) (even) or 𝐵sin(𝑘𝑥) (odd)
745. where 𝑘=√2𝑚(𝐸+𝑉0)
ℏ2
746. Outside the well (|𝑥|>𝑎), the solution is:
𝜓𝑜𝑢𝑡(𝑥)=𝐶𝑒−𝛼|𝑥| where 𝛼=√2𝑚|𝐸|
ℏ2
747. Matching the wavefunctions and their derivatives at 𝑥=±𝑎 leads to:
𝑘tan(𝑘𝑎)=𝛼 (even states)
𝑘cot(𝑘𝑎)=−𝛼 (odd states)
748. These transcendental equations must be solved numerically to find the energy levels
749. The number of bound states depends on the well depth and width
2.109 PROBLEM 4: DELTA FUNCTION POTENTIAL
Find the bound state energy and wavefunction for a delta function potential 𝑉(𝑥)=−𝛼𝛿(𝑥).
Solution:
750. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2−𝛼𝛿(𝑥)𝜓=𝐸𝜓
751. For 𝑥≠0, the equation reduces to:
𝑑2𝜓
𝑑𝑥2=2𝑚|𝐸|
ℏ2𝜓
752. The bound state solution (for 𝐸<0) is:
𝜓(𝑥)=𝐴𝑒−𝜅|𝑥|, where 𝜅=√2𝑚|𝐸|
ℏ2
753. Integrating the Schrödinger equation across 𝑥=0 gives:
−ℏ2
2𝑚[𝜓′(0+)−𝜓′(0−)]−𝛼𝜓(0)=0
754. This leads to the condition:
ℏ2𝜅
𝑚=𝛼
755. The bound state energy is:
𝐸=−𝑚𝛼2
2ℏ2
756. The normalized wavefunction is:
𝜓(𝑥)=√𝑚𝛼
ℏ2𝑒−𝑚𝛼
ℏ2|𝑥|
2.110 PROBLEM 5: PARTICLE IN A BOX WITH LINEAR POTENTIAL
Solve the Schrödinger equation for a particle in a box of width 𝐿 with a linear potential 𝑉(𝑥)=
𝐹𝑥.
Solution:
757. The Schrödinger equation is:
−ℏ2
2𝑚𝑑2𝜓
𝑑𝑥2+𝐹𝑥𝜓=𝐸𝜓
758. Introduce dimensionless variables:
𝑦=(2𝑚𝐹
ℏ2)1/3(𝑥−𝐸
𝐹), 𝜖=(2𝑚
ℏ2𝐹2)1/3𝐸
759. The equation transforms to Airy’s equation:
𝑑2𝜓
𝑑𝑦2−𝑦𝜓=0
760. The general solution is: 𝜓(𝑦)=𝑐1𝐴𝑖(𝑦)+𝑐2𝐵𝑖(𝑦)
761. Boundary conditions: 𝜓(0)=𝜓(𝐿)=0
762. This leads to a transcendental equation for the energy levels:
𝐴𝑖(−𝜖)𝐵𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)=𝐴𝑖(−𝜖+(2𝑚𝐹
ℏ2)1/3𝐿)𝐵𝑖(−𝜖)
763. This equation must be solved numerically to find the energy levels
764. The wavefunctions are linear combinations of Airy functions
