3.3 Solutions 233
First order correction is
ΔE=
∫
ψ 0 ∗H′ψdτ
∫
ψ 0 ∗ψdτ
∫
ψ 0 ∗H′ψdτ=K^2 W
∫ a 2
0
sin^2
(n 1 πx
a
)
dx
∫a/ 2
0
sin^2
(n 2 πy
a
)
dy
=K^2 W
[
x
2
−
(
a
4 n 1 π
)
sin
(
2 n 1 πx
a
)]a/ 2
0
[
y
2
−
(
a
4 n 2 π
)
sin
(
2 n 2 πy
a
)]a/ 2
0
=
K^2 Wa^2
16
∫a
0
ψ∗ 0 ψ 0 dτ=K^2
∫a
0
∫a
0
sin^2 n 1 πx
a
sin^2 n 2 πy
a
dxdy=
k^2 a^2
4
ThereforeΔE=
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.
Assume
ψs=
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)