3.3 Solutions 233
First order correction isΔE=∫
ψ 0 ∗H′ψdτ
∫
ψ 0 ∗ψdτ
∫
ψ 0 ∗H′ψdτ=K^2 W∫ a
2
0sin^2(n 1 πx
a)
dx∫a/ 20sin^2(n 2 πy
a)
dy=K^2 W
[
x
2−
(
a
4 n 1 π)
sin(
2 n 1 πx
a)]a/ 20[
y
2−
(
a
4 n 2 π)
sin(
2 n 2 πy
a)]a/ 20=K^2 Wa^2
16
∫a0ψ 0 ∗ψ 0 dτ=K^2∫a0∫a0sin^2 n 1 πx
asin^2 n 2 πy
adxdy=k^2 a^2
4ThereforeΔE=K^2 W 0 a^2
16/
K^2 a^2
4=
W 0
4
3.3.8 Scattering (Phase Shift Analysis) .....................
3.104 Let the total wave function be
ψ=ψi+ψs (1)
whereψirepresents the incident wave andψsthe scattered wave.
In the absence of potential, the incident plane wave
ψi=Aeikz=eikz (2)
where we have dropped off A to choose unit amplitude.
Assumeψs=f(θ)eikr
r(3)
which ensures inverse square r dependence of the scattered wave from the
scattering centre.
σ(θ)=|f(θ)|^2 (4)
f(θ) being the scattering amplitude.
We can write (1)ψ=eikrcosθ+f(θ)eikr
r(5)
or
fθ=re−ikr(ψ−eikrcosθ)(6)
Ltr→∞
The azimuth angleφhas been omitted inf(θ) as the scattering is assumed
to have azimuthal symmetry. In the absence of potentialψiis the most
general solution of the wave equation.
∇^2 ψi+k^2 ψi=0(7)