Physics and Electronics

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Chapter 12

Canonical Transformations

12.1 Transforming Hamilton’s Equations

Lagrange’s equations are left invariant under any change of coordinates. The same cannot be said for Hamilton’s equations, which change form under a general transformation (q,p,t) 7→ (Q,P,t).

Are there any transformations that leave Hamilton’s equations unchanged? Yes. They are called canonical transformations. We can recognise them in two ways: directly, from Hamilton’s equations, and indirectly, from the Principle of Least Action.

Let qi 7→ Qi(q1, ...,qn,p1, ...,pn, t) and pi 7→ Pi(q1, ...,qn,p1, ...,pn, t) be a canonical trans- formation (i = 1, ...,n). Then, the new coordinates Qi and new momenta Pi satisfy Hamilton’s equations

dQi dt

= ∂K

∂Pi , (12.1)

dPi dt

= − ∂K

∂Qi , (12.2)

where K(Q,P) = H ( q(Q,P),p(Q,P)

) is the same Hamiltonian H(q,p) as in the original

(q,p) system, but with the transformation inverted to write the q’s and p’s in terms of the Q’s and P ’s.1

Let us express both sides of (12.1) in terms of the q’s and the p’s. (Ignore the explicit t dependence for simplicity). We get

dQi dt

= ∂Qi ∂qj

dqj dt

+ ∂Qi ∂pj

dpj dt

(12.3)

= ∂Qi ∂qj

∂H

∂pj − ∂Qi ∂pj

∂H

∂qj (12.4)

and

∂K

∂Pi = ∂H

∂qj

∂qj ∂Pi

+ ∂H

∂pj

∂pj ∂Pi

. (12.5)

Since the q’s and p’s are independent, so too are ∂H/∂qj and ∂H/∂pj. Upon equating coefficients, equations (12.1), (12.4), and (12.5) imply

1K(Q, P) is shorthand for K(Q1, ..., Qn, P1, ..., Pn) here and below.

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− ∂Qi ∂pj

= ∂qj ∂Pi

(12.6)

and ∂Qi ∂qj

= ∂pj ∂Pi

. (12.7)

The LHSs of (12.6) and (12.7) are calculated in terms of the q’s and p’s; the RHSs are calculated in terms of the Q’s and P ’s; and the equality is demonstrated by writing one set of coordinates and momenta in terms of the other. Two analogous conditions follow from (12.2).

An alternative, indirect approach to deriving a canonical transformation is via a generating function. According to the Principle of Least Action, the motion extremises the action

S =

∫ t2 t1

dt L(t,q, q̇) (12.8)

=

∫ t2 t1

dt [pjq̇j −H(t,q,p)] . (12.9)

A canonical transformation leaves Hamilton’s equations unchanged, so equivalently the motion extremises

S =

∫ t2 t1

dt [ PjQ̇j −K(t,Q,P )

] . (12.10)

Equations (12.9) and (12.10) imply

pjq̇j −H(t,q,p) = PjQ̇j −K(t,Q,P ) + dG

dt . (12.11)

We can also multiply pjq̇j−H(t,q,p) by some arbitrary constant λ, but that is equivalent to rescaling the dummy variable t in the integral in (12.9). In (12.11), the arbitrary function G is called a generating function.

The function G can take several forms, commonly termed types 1, 2, 3, and 4. For example, consider G = G(q1, ...,qn,Q1, ...,Qn, t). Substituting in (12.11) we obtain

pjq̇j −H(t,q,p) = PjQ̇j −K(t,Q,P ) + ∂G

∂t + ∂G

∂qj

dqj dt

+ ∂G

∂Qj

dQj dt

. (12.12)

Since coordinates and momenta are independent, (12.12) implies

pj = ∂G

∂qj , (12.13)

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Pj = − ∂G

∂Qj , (12.14)

K = H + ∂G

∂t . (12.15)

Practically speaking, (12.13) comprises n equations relating the p’s, q’s, and Q’s, which can be inverted to obtain Qi(q1, ...,qn,p1, ...,pn, t) for i = 1, ...,n. We substitute these re- sults into (12.14) to get Pi(q1, ...,qn,p1, ...,pn, t) for i = 1, ...,n. Finally, putting everything into (12.15), we get the new Hamiltonian K in terms of the Qi’s and Pi’s.

Exercise. Let G = qkQk (Einstein summation convention understood). Prove that this generates the canonical transformation qi 7→ Qi,pi 7→ Pi with Pi = −qi,Qi = pi,K = H.

Other types of generating functions work in a similar fashion. They are discussed in Goldstein’s Classical Mechanics.

12.2 Action-angle Variables

Action-angle variables are constructed via a type-2 canonical transformation. We don’t worry about the explicit form of the transformation here. (For anybody who is curi- ous, the generating function equals Hamilton’s characteristic function, S + Ht, when H does not depend on t explicitly.) All we need to know is that the transformation exists, that its associated generalised coordinates are called “angles,” θi, and that its associated generalised momenta are called “actions,” Ji, defined by

J1 =

∮ p1dq1, (12.16)

J2 =

∮ p2dq2, (12.17)

and so on. The symbol ∮

denotes an integral over a closed orbit in phase space of constant energy, e.g., E1 = E1(q1,p1) = constant. Physically, Ji equals the area of phase space enclosed by the orbit.

With the above definitions, it can be shown (exercise: prove it!) that the transformed Hamiltonian K is a function of the actions only, not the angles, i.e., K = K(J1, ...,Jn). Hamilton’s equations then imply

dJi dt

= − ∂K

∂θi (12.18)

= 0, (12.19)

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i.e., all the actions J1, ...,Jn are constants. Hamilton’s equations also imply

dθi dt

= ∂K

∂Ji (12.20)

= constant, (12.21)

where (12.21) follows from (12.20), because K is a function of Ji, ...,Jn only, and J1, ...,Jn are all constant. In other words, each angle θi increases linearly with time as if in uniform circular motion with angular velocity ∂K/∂Ji.

As a worked example, consider a one-dimensional simple harmonic oscillator with energy E. The trajectory of the oscillator satisfies

1 = p(t)2

2mE + kq(t)2

2E . (12.22)

Equation (12.22) describes an ellipse in phase space (the q-p plane) with semimajor and semiminor axes (2mE)1/2 and (2E/k)1/2 respectively (or vice versa). The action is defined as the area of the ellipse, J =

∮ pdq =

∮ pdq dt dt = π(2mE)1/2(2E/k)1/2. Hence we arrive

at

J = 2πE/ω, (12.23)

where ω = (k/m)1/2 is the natural frequency.

One important application of action-angle variables is to systems that are nearly Hamilto- nian. Examples include (i) systems with a small amount of dissipation, where H decreases slowly with t (compared to the natural oscillation frequency, say); and (ii) systems whose control parameters change slowly, even when there is zero dissipation. In both cases, one can prove that the actions are approximately conserved on the slow time-scale of the relevant change. In this context, the actions are called adiabatic invariants.

Let us return to the example of the simple harmonic oscillator. Consider a pendulum whose length changes slowly, e.g., a bob on a string where you run your finger slowly along the string to bring the suspension point closer to the bob. As the pendulum short- ens, ω increases. But what happens to the amplitude? Does it stay constant, because the amplitude and period of a linear pendulum are independent? No! H depends explic- itly (albeit slowly) on t, so things are not that simple. On the other hand, the action J = 2πE/ω is approximately conserved. Hence E increases in proportion to ω, i.e., the amplitude increases. Does this agree with your own experience? (Exercise: try it with a bob on a string!)

Action-angle variables are practically useful in a wide variety of classical problems, from the orbits of stars and dark matter particles in a galaxy to the motion of charged particles in nonuniform but slowly-varying electromagnetic fields, e.g., in the Earth’s radiation belts or in a tokamak (thermonuclear fusion reactor). The conservation of action variables also

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connects deeply to the quantization of quantities like angular momentum in quantum mechanics, as recognised first by Bohr in his early pioneering model of the hydrogen atom.

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Chapter 13

Cycles from Delays

Reference: J. D. Murray, Mathematical Biology (3rd ed.), Springer

13.1 Nicholson’s blowflies

In 1957, Nicholson performed a famous series of controlled experiments on the popula- tion growth of the Australian sheep-blowfly (Lucilia cuprina). When the food level and temperature are held constant, the blowfly population varies cyclically with a period of 35-40 days. The period is independent of the food level.

The cyclic behaviour is surprising. Let x(t) be the blowfly population (number of adult blowflies per unit area of the food source) as a function of time, k be the net intrinsic growth rate (births minus deaths), and X be the carrying capacity of the food source. Then, in the popular logistic model of population growth,

dx(t)

dt = kx(t) [X −x(t)] , (13.1)

one expects x(t) to increase monotonically up to the carrying capacity X and stay there. The curve exhibits a point of inflection at x = X/2 if x(0) < X/2.

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In 1975, the renowned Australian physicist-turned-mathematical-biologist Robert May showed that the cycle arises, because a delay of T ≈ 11 days elapses, before a blowfly larva matures into an adult. Under these circumstances, the logistic model generalises to

dx(t)

dt = kx(t) [X −x(t−T)] . (13.2)

Equation (13.2) is an example of a delay differential equation. It is not local in time. Delays are common — and vitally important — in ecological applications (e.g., gestation period, time to reach maturity), in economics (e.g., repayment of a loan), and in numerous fascinating systems in physics.

13.2 Cycles

The behaviour of a nonlocal equation like (13.2) is completely different to its local coun- terpart with T = 0. To see why, consider a situation where the population is per- turbed slightly away from its steady-state value (which equals the carrying capacity), i.e., x(t) = X + x(1)(t), with

∣∣x(1)(t)∣∣ � X. To linear order, equation (13.2) becomes dx(1)(t)

dt = −kXx(1)(t−T). (13.3)

Even though (13.3) contains a delay, it is still linear in x(1)(t), so its solutions are expo- nentials, x(1) ∝ exp (λt). Substituting into (13.3) and cancelling the common exponential factor on both sides, we get

λ = −kXe−λT . (13.4)

If λ is purely real, (13.4) has no valid solutions. But our earlier study of linear oscillations shows that λ can be complex. Writing λ = λr + iλi, we find that the real and imaginary parts of equation (13.4) evaluate to

λr = −kXe−λrT cos (λiT), (13.5)

λi = kXe −λrT sin (λiT). (13.6)

You can show graphically on the complex plane that (13.5) and (13.6) admit multiple solutions with λr 6 0 and λi 6= 0, which correspond to decaying oscillations. For example:

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For the special case λr = 0, corresponding to a persistent oscillation which does not decay, (13.5) and (13.6) imply cos (λiT) = 0 and λi = kX respectively, i.e.,

kXT = π

2 . (13.7)

Equation (13.7) is an important result physically. It says that a system that exhibits exponential growth or decay for T = 0 changes behaviour dramatically and exhibits long- lived cycles, when T is of order the reciprocal of the natural growth rate, i.e., T ∼ (kX)−1. The period of the cycles is 2π/λi = 4T from (13.6). This agrees with the behaviour of Nicholson’s blowflies, where we observe periods of 35-40 days and T ≈ 11 days (and therefore infer kX ≈ 0.14 days−1). Note that 4T, the period, is independent of X and hence the food level, as observed.

You can convince yourself graphically that (13.5) and (13.6) admit a countably infinite number of solutions λ1,λ2, .... Equation (13.3) is linear, so the Principle of Superposition applies: the general solution to (13.3) is given by x(1)(t) =

∑∞ j=1 Aj exp [λ

(j)t], where A1,A2, ... are complex-valued constants determined by the initial conditions. But how is this possible? Superficially, it looks like there is only one initial condition, namely the value of x(1)(0), which is definitely not enough to pin down the values of an infinite number of constants A1,A2, .... The resolution of this “paradox” is the following: a delay equation like (13.3) is well posed, only once we specify initial data on the interval 0 6 t 6 T , not just at t = 0. [Imagine you are a computer stepping forward the solution one increment at a time, from t = T to T + ∆T to T + 2∆T and so on; then the right-hand side of (13.3) requires information at t = 0, ∆t, 2∆t, and so on.] The interval 0 6 t 6 T comprises an infinite set of initial conditions, which are enough to specify A1,A2, ....

13.3 General solution of the initial value problem

[Ext]

Consider a driven, linear delay differential equation with driving term f(t) and constant coefficients of the form

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dx(t)

dt + αx(t) + βx(t−T) = f(t). (13.8)

Let the initial data be written as x(t) = g(t) on the interval 0 6 t 6 T. [Equation (13.8) is not a completely general, linear, first-order delay differential equation, but it gets close.] Can we write down a closed-form expression for the solution of (13.8), given the initial data?

One approach to solving (13.8) involves a useful mathematical tool called the Laplace transform. Given any function x(t) of time t, whether is it part of a delay differential equation or not, we can define its Laplace transform as

x̃(s) =

∫ ∞ 0

dte−stx(t). (13.9)

In this notation, x̃ is clearly a different function than x (sometimes the tilde is omitted for readability), and s is a complex variable in general. It is possible to invert x̃(s) to get back x(t) (see below), but we don’t worry about that yet. The integral in (13.9) is finite as long as Re(s) is large enough, e.g., Re(s) > 0 is x(t) ∝ t2, Re(s) > 1 if x(t) ∝ exp (t). [Note: things like x(t) ∝ exp (t2) are not allowed.]

Laplace transforms, like Fourier transforms, allow you to convert differential equations into algebraic equations. Let us multiply (13.8) by exp (−st) and integrate over t > T, the domain where we seek to solve for x(t). The first term integrates by parts:

∫ ∞ T

dte−st dx(t)

dt =

[ e−stx(t)

]t=∞ t=T

+ s

∫ ∞ T

dte−stx(t) (13.10)

= −x(T)e−sT + sx̃(s) −s ∫ T 0

dte−stg(t). (13.11)

The second and fourth terms are straightforward:

∫ ∞ T

dte−stαx(t) = αx̃(s) −α ∫ T 0

dte−stg(t), (13.12)

∫ ∞ T

dte−stf(t) = f̃(s) − ∫ T 0

dte−stf(t). (13.13)

The third term is simplified by changing the dummy variable of integration from t to t−T:

∫ ∞ T

dte−stβx(t−T) = βe−sT x̃(s). (13.14)

Upon combining (13.11)–(13.14), we arrive at

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x̃(s) = g(T)e−sT +

∫∞ T dte−stf(t) + (s + α)

∫ T 0 dte−stg(t)

s + α + βe−sT . (13.15)

Everything on the right-hand side of (13.15) can be calculated from the known functions f(t) and g(t). Hence, we can derive the unique solution for x(t) as long as we can invert x̃(s).

Luckily, for a wide range of conditions, the inverse of x̃(s) exists. It can be calculated from the following contour integral:

x(t) = 1

2πi lim R→∞

∫ γ+iR γ−iR

dsestx̃(s). (13.16)

In (13.16), γ is a real number chosen so that the contour lies in the region where x̃(s) is validly defined. [Recall that the integral in (13.9) is finite, as long as Re(s) is large enough.] The contour closes in a giant, counter clockwise arc with Re(s) < γ, along which we have |exp (st)x̃(s)|→ 0.

There are many ways to evaluate the contour integral in (13.16). One way is to find the poles of x̃(s), i.e., the values of s inside the contour, where x̃(s) “blows up.” For example, the complex function (1 + s2)−1 = (1 + is)−1(1 − is)−1 has simple poles at s = ±i, while the function (s + 3i)−2 has a double pole at s = −3i. One can show (we don’t prove it here) that (13.16) can be written as

x(t) = ∑ poles

Res [ estx̃(s)

] . (13.17)

The residue Res[h(s)] of any function h(s) at a pole s = s0 of order n (n = 1 for a simple pole) is

Res[h(s)] = 1

(n− 1)! lim s→s0

dn−1

dsn−1 [(s−s0)nh(s)] . (13.18)

Equation (13.15) features simple poles, wherever the denominator vanishes (the natural modes of the system). The numerator may also contribute simple poles, e.g., f(t) = 1 implies s−1 exp (−sT) for the first term, which features a simple pole at s = 0.

You will learn much more about Laplace transforms in your mathematical physics subjects in later years. When this happens, don’t forget that delay differential equations — and the cycles they produce — constitute a fascinating practical application!

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Chapter 14

Stochastic Processes [Ext]

In the previous chapters, we focus on systems that are deterministic. In this chapter, we take a brief first look at systems that are stochastic, i.e., systems whose state at time t + ∆t depends probabilistically on their state at time t. Quantum systems are inherently stochastic. Classical systems can exhibit stochastic behaviour in two situations:

• when they are subjected to fundamental, random fluctuations, e.g., thermal noise;

• when they involve extremely complicated microprocesses, which can be modelled deterministically in principle but are unpredictable in practice, e.g., die roll, births and deaths of living organisms.

14.1 Probability density function

Let x(t) be a random variable which is a function of time t. What does this mean in practice? It means that, if we run the system forward and stop it at any specific time t, x(t) has a definite value. But, if we “run the system again,” the value of x(t) is different in general — and unpredictably so.

To make this more precise, consider N identical copies of the system at t = 0. The copies form an ensemble. They are prepared under identical macroscopic conditions; only their microscopic states are different, which is where the randomness comes from. (Think, for example, of N sealed boxes of gas, all prepared at the same temperature and pressure, but with different molecular positions and velocities at the instant of preparation.) We now evolve the N systems identically up to time t. Let {x(1)(t), ...,x(N)(t)} be the values of x(t) in the N systems. We can plot a histogram of these values to see which ones are more and less likely than others. The histogram is called the probability density function (PDF) p(x,t) of the variable x.

If x(t) is a discrete variable, we have

p(x,t) = 1

N

N∑ i=1

δ [ x−x(i)(t)

] (14.1)

where δ(x) is the Dirac delta function with the properties δ(x) = 0 for x 6= 0, δ(x) = ∞ for x = 0, and

∫∞ −∞ dxδ(x) = 1.

If x(t) is a continuous variable, we say that p(x,t)dx equals the fraction of the N systems in the ensemble which end up with x 6 x(i)(t) 6 x + dx. Equivalently, p(x,t)dx is the probability of measuring x(t) in the range [x,x + dx] at time t. Equation (14.1) is a

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decent estimator of (or approximation to) p(x,t), even when x(t) is continuous, but it is not the only such estimator, nor is it the best one. Look up “kernel density estimators” in a statistics book if you would like to learn more about estimating PDFs in clever ways.

In the above discussion, the randomness “comes from” the N identically prepared systems in the ensemble. Of course, in most cases, x(t) also fluctuates randomly from one instant t to the next instant t + dt. These two sorts of randomness are related but different. (They are equivalent, when a system is ergodic.) In these lectures and in most practical applications, it is neater and safer to interpret random variables in terms of an ensemble, each member of which evolves in time, instead of a single system fluctuating in time.

14.2 Master equation

The mathematics of classical stochastic processes is too advanced to explain rigorously in this course. We content ourselves here with a simplified treatment which nonetheless works well in many practical situations.

Consider again the random variable x(t). Suppose that it defines the state of a system. Let T(x′ → x′′)dx′′ be the rate at which the system jumps from state x(t) = x′ at time t to state x(t+dt) = x′′ at time t+dt. In other words, T(x′ → x′′)dx′′dt equals the conditional probability that the system finds itself at x(t + dt) = x′′ if it starts at x(t) = x′. The rate T(x′ → x′′) may be calculated from first principles or it may be given; for now we don’t mind either way.

Now consider p(x′, t)dx′, the probability that the system is in the state x′ 6 x(t) 6 x′+dx′

at time t. How does it change, as time steps forward from t to t + dt?

• It increases, when the system jumps out of the state x′′ 6 x(t) 6 x′′ + dx′′ and into the state x′ 6 x(t+dt) 6 x′+dx′. The probability of this occurring equals p(x′′, t)dx′′

[i.e., the probability of x′′ 6 x(t) 6 x′′ + dx′′] multiplied by T(x′′ → x′)dx′dt [i.e., probability of the jump x′′ → x′]. There are many possible starting states x′′, so we need to integrate over all of them.

• It decreases, when the system jumps out of the state x′ 6 x(t) 6 x′ + dx′ into some other state x′′ 6 x(t+dt) 6 x′′ +dx′′. The probability of this occurring is p(x′, t)dx′

multiplied by T(x′ → x′′)dx′′dt, integrated over all possible end states x′′.

Putting together the two contributions above, the first positive and the second negative, we obtain

p(x′, t + dt)dx′ −p(x′, t)dx′ = ∑

states x′′

p(x′′, t)dx′′T(x′′ → x′)dx′dt

− ∑

states x′′

p(x′, t)dx′T(x′ → x′′)dx′′dt, (14.2)

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or, equivalently, in the limit dt → 0,

∂p(x′, t)

∂t =

∫ dx′′p(x′′, t)T(x′′ → x′) −

∫ dx′′p(x′, t)T(x′ → x′′). (14.3)

If x(t) is a discrete rather than a continuous variable, such that p(x′, t) equals the proba- bility that one has x(t) = x′ (without any need to multiply by dx′), then (14.3) becomes

∂p(x′, t)

∂t = ∑ x′′

p(x′′, t)T(x′′ → x′) − ∑ x′′

p(x′, t)T(x′ → x′′). (14.4)

Equation (14.3) [or (14.4)] is called a master equation.

14.3 Worked example: Poisson process

Consider some repeating event which occurs randomly at a mean rate λ(t). The rate may be constant or it may vary with time, but it is exogenous, i.e., it does not depend on the state of the system. For example, think of radioactive decays, tram or email arrivals, earthquakes, etc. [In many of these real systems, λ(t) does depend on the state of the system, but we often ignore the state dependence in a simplified model.]

Let N(t) be the number of events that occur from time t′ = 0 until t′ = t. Clearly N(t) is a random variable; it has a different value for every realisation in the ensemble. Common sense tells us that we should measure N(t) ≈ λt, if t is large and λ(t) = λ is constant, but even then one expects fluctuations around the mean number of events λt. How are these fluctuations distributed?

To answer this question, we write down a master equation. Let p(N,t) be the probability that N events occur up to time t. N is discrete (an integer), so there is no “dN” to worry about. When time advances from t to t + dt:

• p(N,t) increases, if N − 1 events occurred up to time t [probability = p(N − 1, t)] and another one occurs during the interval [t, t + dt] [probability = λ(t)dt];

• p(N,t) decreases, if N events occurred after time t [probability = p(N,t)] and an- other one occurs during the internal [t, t + dt] to give a total of N + 1 events at t + dt.

Master equation:

∂p(N,t)

∂t = λ(t)p(N − 1, t) −λ(t)p(N,t). (14.5)

Equation (14.5) is not as easy to solve as it looks, because it is nonlocal: ∂p/∂t at N depends on p at N −1 as well as N. In fact, it should remind you of the delay differential

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equations in the previous chapter, except that the delayed variable is N instead of t, and there are no N derivatives.

To solve equations like (14.5), mathematicians use various tricks. Usually (not always), we try to transform the equation in some way. One common approach applied to master equations is to define a generating function, e.g.,

G(s,t) = ∞∑ N=0

sNp(N,t). (14.6)

Upon multiplying both sides of (14.5) by sN and summing over N = 0, 1, 2, ..., we obtain

∂G(s,t)

∂t = −λ(t)G(s,t) + λ(t)

∞∑ N=1

sNp(N − 1, t) (14.7)

= −λ(t)G(s,t) + λ(t) ∞∑

N′=0

ssN ′ p(N ′, t) (14.8)

= λ(t)(s− 1)G(s,t), (14.9)

where we use p(−1, t) = 0 in line (14.7) and replace the dummy variable in the sum by N ′ = N − 1 to go from lines (14.7) to (14.8). Equation (14.9) is easily solved to give

G(s,t) = exp

[ (s− 1)

∫ t 0

dt′λ(t′)

] . (14.10)

The constant of integration is determined by the normalisation condition

G(1, t) = ∞∑ N=0

p(N,t) = 1, (14.11)

i.e., the sum of the probabilities for all the possible values of N equals one.

In order to convert (14.10) back to p(N,t), we observe from (14.6) that p(N,t) is just the coefficient of the N-th term in the Taylor expansion of G(s,t) about s = 0, viz.

p(N,t) = 1

N!

∂NG

∂sN

∣∣∣∣ s=0

(14.12)

= 1

N!

[∫ t 0

dt′λ(t′)

]N exp

[ − ∫ t 0

dt′λ(t′)

] . (14.13)

Equation (14.13) reduces to the familiar result p(N,t) = (N!) −1

(λt)N exp (−λt) for λ(t) = λ = constant.

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