√

1 ’ ’»|x’y|

√e ’√ e . (5.128)

2 ’» »

Notice that this decays away as |x ’ y| ’ ∞. The square root retains a

positive real part when » is shifted to » ’ iµ, and so the decay is still present:

i +i√»|x’y|’µ|x’y|/2√»

√

1 ’ ’»|x’y|

√e ’ ’√ e . (5.129)

2 ’» »

In each case, with » either immediately above or immediately below the

cut, the small imaginary part tempers the oscillatory behaviour of the Green

function so that χ(x) = G(x, y) is square integrable and remains an element

of L2 [R].

We now take the trace of R by setting x = y and integrating:

L

Tr R»+iµ = iπ . (5.130)

2π |»|

Thus,

L

ρ(») = θ(») , (5.131)

2π |»|

which coincides with our direct calculation.

Example: Let

L = ’i‚x , D(L) = {y, Ly ∈ L2 [R]}. (5.132)

This has eigenfunctions eikx with eigenvalues k. The spectrum is therefore

the entire real line. The local eigenvalue density of states is 1/2π. The

resolvent is therefore

1 1

∞

(’i‚x ’ »)’1 = eik(x’x ) dk. (5.133)

x,x

k’»

2π ’∞

To evaluate this, ¬rst consider the Fourier transforms of

θ(x)e’κx ,

F1 (x) =

F2 (x) = ’θ(’x)eκx , (5.134)

where κ is a positive real number.

5.5. ANALYTIC PROPERTIES OF GREEN FUNCTIONS 141

1

x x

’1

The functions F1 (x) = θ(x)e’κx and F2 (x) = ’θ(’x)eκx .

We have

11

∞

θ(x)e’κx e’ikx dx = , (5.135)

i k ’ iκ

’∞

11

∞

’θ(’x)eκx e’ikx dx = . (5.136)

i k + iκ

’∞

Inverting the transforms gives

1 1

∞

’κx

eikx dk,

θ(x)e =

k ’ iκ

2πi ’∞

1 1

∞

’θ(’x)eκx eikx dk.

= (5.137)

2πi k + iκ

’∞

These are important formul¦ in their own right, and you should take care

to understand them. Now we apply them to evaluating the integral de¬ning

R» .

If we write » = µ + iν, we ¬nd

1 1

∞

iθ(x ’ x )eiµ(x’x ) e’ν(x’x ) , ν > 0,

ik(x’x )

e dk = (5.138)

’iθ(x ’ x)eiµ(x’x ) e’ν(x’x ) , ν < 0,

k’»

2π ’∞

In each case, the resolvent is ∝ ei»x away from x , and has jump of +i at

x = x so as produce the delta function. It decays either to the right or to

the left, depending on the sign of ν. The Heaviside factor ensures that it is

multiplied by zero on the exponentially growing side of e’νx , so as to satisfy

the requirement of square integrability.

Taking the trace of this resolvent is a little problematic. We are to set x =

x and integrate ” but what value do we associate with θ(0)? Remembering

that Fourier transforms always give to the mean of the two values at a jump

1

discontinuity, it seems reasonable to set θ(0) = 2 . With this de¬nition, we

142 CHAPTER 5. GREEN FUNCTIONS

have ±

i

L, Im » > 0,

2

Tr R» = (5.139)

’ i L, Im » < 0.

2

Our choice is therefore compatible with Tr R»+iµ = πρ = L/2π. We have

been lucky. The ambiguous expression θ(0) is not always safely evaluated as

1/2.

5.6 Locality and the Gelfand-Dikii equation

The answers to many quantum physics problems can be expressed either as

sums over wavefunctions or as expressions involving Green functions. One

of the advantages writing the answer in terms of Green functions is that

these typically depend only on the local properties of the di¬erential oper-

ator whose inverse they are. This locality is in contrast to the individual

wavefunctions and their eigenvalues, both of which are sensitive to the dis-

tant boundaries. Since physics is usually local, it follows that the Green

function provides a more e¬cient route to the answer.

By the Green function being local we mean that its value for x, y near

some point can be computed in terms of the coe¬cients in the equations

evaluated near this point. To illustrate this claim, consider the Green func-

2

tion G(x, y) for the di¬erential operator ’‚x + q(x) + » on the entire real

line. We will show that there is a, not exactly obvious but easy to obtain

once you know the trick, local gradient expansion for the diagonal elements

G(x, x). We begin by recalling that we can write

G(x, y) ∝ u(x)v(y)

2

where u(x), v(x) are solutions of (’‚x + q(x) + »)y = 0 satisfying suitable

boundary conditions to the right and left. Suppose we set R(x) = G(x, x)

and di¬erentiate three times with respect to x. We ¬nd