188
3 Quantum Mechanics – II
Going back to (3)
1
Y
∂
2 Y
∂ y 2 +
1
Z
∂
2 Z
∂z 2 = −α
2
1
Z
∂
2 Z
∂z 2 = −
1
Y
∂
2 Y
∂ y 2 − α
2
(6)
Each side must be equal to a constant, say −γ
2 for the same argument as
before.
−
1
Y
∂
2 Y
∂ y 2 − α
2
= −γ
2
Or
−
1
Y
∂
2 Y
∂ y 2 +
α
2
− γ
2
= 0
Or
∂
2 Y
∂ y 2 + μ
2 Y = 0
where μ
2
= α
2
− γ
2
(7)
Y = D sin μy
Y = 0 at y = b
This gives μ =
n y π
b
(8)
Going back to (6)
1
Z
d
2 Z
dz 2 = −γ
2
This gives Z = F sin γ z
where γ =
n z π
c
(9)
∴ ψ ∼ sin
n x π x
a
sin
n y π y
b
sin
n z π z
c
(b) Combining (4), (5), (7), (8) and (9)
μ
2
= α
2
− γ
2
= (2m E/
2 ) − β
2
− γ
2
Or
2m E
2 = μ
2
+ β
2
+ γ
2
=
n y π
b
2 +
n x π
a
2 +
n z π
c
2
Or
E =
h
2
8m
n
2
x
a 2 +
n
2
y
b 2 +
n
2
z
c 2
(10)
3.40 For a = b = c
E = (h/8ma
2 )
n
2
x + n
2
y + n
2
z
(Equation 10 of Prob 3.39)