Download The Scattering Green`s Function: Getting the Signs Straight

The Scattering Green’s Function:
Getting the Signs Straight
Jim Napolitano
April 2, 2013
Our starting point is (6.2.8) in Modern Quantum Mechanics, 2nd Ed, on page 392. The
problems begin in (6.2.9), so let’s take this over slowly. Just work with the “outgoing”
Green’s function. The first step, converting the summation to an integral, is fine. That is
Z
0
0
ik0 ·(x−x0 )
1
1 X eik ·(x−x )
3 0 e
0
dk 2
−→
G+ (x, x ) = 3
L k0 k 2 − k 0 2 + iε
(2π)3
k − k 0 2 + iε
Work the integral in spherical coordinates, measuring angles with respect to the k0 direction.
Z 2π
Z ∞
Z π
0
0
eik |x−x | cos θk0
1
0
02
0
0
0
0
dφ
G+ (x, x ) =
k
dk
sin
θ
dθ
k
k
k
(2π)3 0
k 2 − k 0 2 + iε
0
0
The integral over φk0 contributes a factor of 2π and nothing else. For the θk0 integral, use a
change of variables. Put µ ≡ cos θk0 . Then dµ = − sin θk0 , and θk0 = {0, π} corresponds to
µ = {1, −1}. So, reverse the order of integration on µ to get
Z ∞
Z 1
0
0
1
eik |x−x |µ
0
02
0
k dk
dµ 2
G+ (x, x ) =
(2π)2 0
k − k 0 2 + iε
−1
Here is a mistake in the textbook. The limits of integration on the µ integral are reversed
in the second line of (6.2.9). Now do the (easy) integral over µ to get
ik0 |x−x0 |
Z ∞
0
0 1
1
− e−ik |x−x |
0
0 e
0
G+ (x, x ) =
k dk
(2π)2 i|x − x0 | 0
k 2 − k 0 2 + iε
Now notice that the k 0 integrand is even. This allows to integrate over the whole real axis
and divide by two:
ik0 |x−x0 |
Z ∞
0
0 1
1
− e−ik |x−x |
0
0
0 e
G+ (x, x ) = 2
k dk
(1)
8π i|x − x0 | −∞
k 2 − k 0 2 + iε
1
This should be the last line of (6.2.9), but the sign error in the limits of integration for µ
carries down to here, reversing the order of the two terms in the integrand.
As described in the text, this last integral can be done with contour integration by making
k 0 a complex variable. Important: The Residue Theorem follows from the Cauchy Integral
Formula, which states that
I
1
f (z)
f (z0 ) =
2πi C z − z0
where the curve C contains the pole at z0 and is followed counter clockwise. Note the order
of the quantities z and z0 in the denominator! If this order is reversed, as you tend to do
when working from starting with (6.2.8) where k 0 → z and k → z0 , then a minus sign is
y 19, 2010 10:41
o05-main
Sheet number 12 Page number 393
AW Physics Macros
introduced.
rd REVISED PAGE PROOFS
Now return to Equation (1) above. We have poles when k 2 − k 0 2 + iε = 0, that is
k = +k + iε and k 0 = −k − iε, once again, redefining ε but keeping it small and positive.
These two poles
represented
by the black dots in Figure 6.1 from the text,
6.2are
The
Scattering Amplitude
393reprinted here:
0
Im(k′)
Re(k′)
two terms
(6.2.9)
contours.
The dotswe want to use
Call the upperFIGURE
contour6.1CUIntegrating
and thethelower
CL . inFor
theusing
firstcomplex
time in
(1) above,
0|
(crosses)
mark the positions
of the two poles for the + (−) form of G ± (x, x" ). We replace
ik0 |x−x
0
CU , since e the integral
→ 0 over
as ka real-valued
→ +i∞.k " The
contour is followed counter-clockwise,
so there is
in (6.2.9) with one of the two contours
in the figure,
0
"
"
no sign flip in choosing
the residue
poletoiszero
at along
k =thek semicircle
+ iε. The
the one theorem.
on which theThe
factorenclosed
e±ik |x−x | tends
at denominator
0
large
Im(k0 "−
). Thus,
to theto
contour
integral
is along
the real axis.
in (1) factors as
−(k
k)(kthe
+only
k),contribution
taking care
order
k 0 and
k appropriately
with respect
to the Residue Theorem. In this case, the first factor contributes the pole. Therefore, the
denominator contributes −2k to the residue, and we have
The integrand in (6.2.9) contains two terms, each with poles in the complex k "
Z ∞
I
0 |x−x
0|
plane. ik
That
is,0 |the denominator
of the terms
in0 |brackets becomes zero
when
k "2 =
ik0 |x−x
ik|x−x
e
e
"
"
0 k02 ± e
0
0
again,
we redefine
its ik|x−x0 |
k dk 2 i ε, or0 2k = k ±=i ε and kk =
dk−k 2∓ i ε. (Once
= +2πi
k " ε, keeping
= −πie
2
0
) axis and then
sign
k intact.)
− k Imagine
+ iε an integration
kcontour
− k running
+ iε along the Re(k−2k
−∞
CU
closed, with a semi-circle in either the upper or the lower plane. See Figure 6.1.
For the first term, close the contour in
2 the lower plane. In this case, the contribution to the integrand along the semicircle goes to zero exponentially with
"
"
e−ik |x−x | as Im(k " ) → −∞. Closing in the lower plane encloses the pole at
"
k = −k − i ε (k " = k − i ε) when the sign in front of ε is positive (negative). The
integral in (6.2.9) is just 2πi times the residue of the pole, with an overall minus
sign because the contour is traced clockwise. That is, the integral of the first term
in brackets becomes
Next, consider the second term in Equation (1) above. This time use the lower contour,
CL , followed clockwise, so an overall − sign. The relevant pole is at k 0 = −k − iε, and the
denominator contributes −(−k − k) = 2k. We have
0
∞
0
e−ik |x−x |
k dk 2
=−
k − k 0 2 + iε
−∞
Z
0
0
0
I
k 0 dk 0
CL
0
0
e−ik |x−x |
ei(−k)|x−x |
ik|x−x0 |
=
−2πi
(−k)k
=
+πie
2
2k
k 2 − k 0 + iε
Finally, put the two terms together in Equation (1), noting that the second is subtracted
from the first. That is
0
G+ (x, x0 ) =
h
i
1 eik|x−x |
1
1
ik|x−x0 |
ik|x−x0 |
−πie
−
πie
=
−
8π 2 i|x − x0 |
4π |x − x0 |
(2)
It is worth taking a little time to discuss the statement that G+ (x, x0 ) is the Helmholtz
Equation Green’s Function, namely (6.2.12), and use this to verify the overall minus sign.
Take k = 0, in which case G+ (x, x0 ) is just −1/4π times the electrostatic potential at x for
a unit charge located at x0 . Poisson’s Equation for the electrostatic potential (in Gaussian
units) is ∇2 Φ(x) = −4πρ(x) = −4πqδ(x − x0 ) for a point charge q located at x0 . It is well
known that in this case, Φ(x) = q/|x − x0 |, that is
∇2
1
= −4πδ(x − x0 )
0
|x − x |
which is completely consistent, including the sign, with Equation (2) above for k = 0.
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