)(2

1 2 r

L

tik

tik

rki ededekd

r

2)2()(

3

3

where I’ve let ħ,c=1.

†

Recalling that is a function of k, 22mkk

it should be primed when k is. Again, I’ve adopted the ħ,c = 1.

tik

tik

rki ededekd

ir

2)2()(

3

3†

i

)()(),( 3 rrirr You should be able to rewrite the left-hand side of in the form:

)(3

3

3

3

2)2(2)2(, rkrkie

kdkdirr

You will need to take the Fourier transform of both sides:

)(2)2(2)2(

3

3

3

3

3

rreekdkd

i rkirki

d3reiK·r eik·r d3reiK·r

I’ve already cancelled1/(2)3 from both sides

Integrating over d3r introduces a delta function that simplifies the integration over d3k

Next do yet ANOTHER Fourier transform:

)(2

333

rrerdekd

i riri

KK d3r'eiK'·r' eik·r d3r'eiK'·r'

Again already canceling1/(2)3 from both sides

Problem 5-3b Hint

You should be able to take it from here, completing the arguments to show

)(, 3 kk dd kk

†