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given by

1 Z 1 e; k2 t+ikx dk :

I(x t) = 2 (15.9)

;1

This means that our problem is solved when the one-dimensional Fourier integral (15.9)

is solved. In order to solve this integral it is important to realize that the exponent in

the integral is a quadratic function of the integration variable k. If the integral would be

R1

of the form ;1 e; k2p the problem would not be di cult because it known that this

dk

integral has the value =R . The problem can be solved by rewriting the integral (15.9)

1

in the form of the integral ;1 e; k2 dk.

Problem h: Complete the square of the exponent in (15.9), i.e. show that

2 x2

t k 2ixt

k2 t + ikx = ; (15.10)

4t

; ; ;

and use this result to show that I(x t) can be written as:

1 e;x2 =4 t Z 1;ix=2 t e; k0 2 t dk0 :

I(x t) = 2 (15.11)

;1;ix=2 t

CR

Im k = 0

C

Im k = -x/2 Îºt

CC

Figure 15.1: The contours CR , CC and C in the complex k-plane.

CHAPTER 15. GREEN'S FUNCTIONS, EXAMPLES

194

;

With these steps we have achieved our goal of having an integrand of the form exp k2 ,

;

but have paid a price. In the integral (15.9) the integration was along the real axis CR ,

see gure (15.1). In the transformed integral the integration now takes place along the

integration path CC in the complex plane that lies below the real axis, see gure (15.1).

However, one can show that when the integration path CC is replaced by an integration

along the real axis the integral has the same value:

1 e;x2 =4 t Z 1 e; k2 t dk :

I(x t) = 2 (15.12)

;1

Problem i: When you have studied section (13.2) you have seen all the material to give

a proof that (15.12) is indeed identical to (15.11). Show that this is indeed the case

by using that the closed integral along the closed contour C in gure (15.1) vanishes.

Problem j: Carry out the integration in (15.12) and derive that

e;x2=4 t

I(x t) = (15.13)

4t

p

and show that this implies that the Green's function is given by

1 exp 2 =4 t :

G(r t) = (15.14)

(4 t)N=2

;r

Problem k: This result implies that the Green's function has in any dimension the form

of the Gaussian. Show that this Gaussian changes shape with time. Is the Gaussian

broadest at early times or at late times? What is the shape of the Green's function

in the limit t 0, i.e. at the time just after the heat forcing has been applied.

#

Problem l: Sketch the time-behavior of the Green's function for a xed distance r. Does

the Green's function decay more rapidly as a function of time in three dimensions

than in one dimension? Give a physical interpretation of this result.

It is a remarkable property of the derivation in this section that the Green's function

could be derived with a single derivation for every number of dimension. It should be

noted that this is not the generic case. In many problems, the behavior of the system

depends critically of the number of spatial dimensions. We will see in section 15.4 that

wave propagation in two dimensions is fundamentally di erent from wave propagation

in one or three dimensions. Another example is chaotic behavior of dynamical systems

where the occurrence of chaos is intricately linked to the number of dimensions, see the

discussion given by Tabor 59].

15.2 The Schrodinger equation with an impulsive source

In this section we will study the Green's function for the Schrodinger equation that was

introduced in section (6.4):

h @ (r t) = h2 2 (r t) + V (r) (r t) (6:13) again

i @t 2m

; ; r

15.2. THE SCHRODINGER EQUATION WITH AN IMPULSIVE SOURCE 195

Solving this equation for a general potential V (r) is a formidable problem, and solutions

are known for only very few examples such as the free particle, the harmonic oscillator and

the Coulomb potential. We will restrict ourselves to the simplest case of a free particle, this

is the case where the potential vanishes (V (r) = 0). The corresponding Green's function

satis es the following partial di erential equation:

h2

h @G(r t) 2 G(r

t) = (r) (t) : (15.15)

i @t 2m

; r

Before we compute the Green's function for this problem, let us pause to nd the meaning

of this Green's function. First, the Green's function is for r =0 and t6=0 a solution of

6

Schrodinger's equation. This means that 2 gives the probability density of a particle

jGj

(see also section 6.4). However, the right hand side of (15.15) contains a delta function

forcing at time t = 0 at location r = 0. This is a source term of G and hence this is a

source of probability for the presence of the particle. One can say that this source term

creates probability for having a particle in the origin at t = 0. Of course, this particle

will not necessarily remain in the origin, it will move according to the laws of quantum

mechanics. This motion is described by equation (15.15). This means that this equation

describes the time evolution of matter waves when matter is injected at t = 0 at location

r =0.

Problem a: The Green's function G(r t r0 t0 ) gives the wave-function at location r and

time t for a source of particles at location r0 at time t0 . Express the Green's function

G(r t r0 t0 ) in the solution G(r t) of (15.15), and show how you obtain this result. Is

this result also valid for the Green's function for the quantum-mechanical harmonic

oscillator (where the potential V (r) depends on position)?

In the previous section the Green's function gave the evolution of the temperature eld

due to a delta function injection of heat in the origin at time t = 0. Similarly, the Green's

function of this section describes the time-evolution of probability for a delta function

injection of matter waves in the origin at time t = 0. These two Green's functions are

not only conceptually very similar. The di erential equations (15.1) for the temperature

eld and (15.15) for the Schrodinger equation are rst order di erential equations in time

and second order di erential equations in the space coordinate that have a delta-function

excitation in the right hand side. In this section we will exploit this similarity and derive

the Green's function for the Schrodinger's equation from the Green's function for the heat

equation derived in the previous section rather than constructing the solution from rst

principles. This approach is admittedly not very rigorous, but it shows that analogies are

useful for making shortcuts.

The principle di erence between (15.1) and (15.15) is that the time-derivative for

Schrodinger's equation is multiplied with i = whereas the heat equation is purely

p

;1

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