186
3 Quantum Mechanics – II
H =
−
2
2m
∇
2
+ a
x
2
+ y
2
+ z
2
−
5
6
x
2
=
−
2
2m
∇
2
+ a
x
2
6
+ y
2
+ z
2
(2)
The Schrodinger’s equation is
H ψ(x, y, z) = Eψ(x, y, z)
( 3 )
This equation can be solved by the method of separation of variables.
Let
ψ(x, y, z) = ψ x ψ y ψ z
(4)
H ψ(x, y, z) = −
2
2m
∂
2
∂ x 2 +
∂
2
∂ y 2 +
∂
2
∂z 2
ψ x ψ y ψ z + a
x
2
6
+ y
2
+ z
2
ψ x ψ y ψ z = Eψ x ψ y ψ z
−
2
2m
ψ y ψ z
∂
2
ψ x
∂ x 2 −
2
2m
ψ x ψ z
∂
2
ψ y
∂ x 2 −
2
2m
ψ x ψ y
∂
2
ψ z
∂ X 2
+
ax
2
6
ψ x ψ y ψ z + ay
2
ψ x ψ y ψ z + az
2
ψ x ψ y ψ z = Eψ x ψ y ψ z
Dividing throughout by ψ x ψ y ψ z
−
2
2m
1
ψ x
∂
2
ψ x
∂ x 2 −
2
2m
1
ψ y
∂
2
ψ y
∂ y 2 −
2
2m
1
ψ z
∂
2
ψ z
∂z 2 +
ax
2
6
+ ay
2
+ az
2
= E
(5)
−
2
2m
1
ψ x
∂
2
ψ x
∂ x 2 +
ax
2
6
= E 1
a
6
=
1
/ 2 k 1
(6)
−
2
2m
1
ψ y
∂
2
ψ y
∂ y 2 + ay
2
= E 2
a =
1
/ 2 k 2
(7)
−
2
2m
1
ψ z
∂
2
ψ z
∂z 2 + az
2
= E 3
a =
1
/ 2 k 2
(8)
E 1 = (n 1 +
1
/ 2 )ω 1 ; E 2 = (n 1 +
1
/ 2 )ω 2 ; E 3 = (n 3 +
1
/ 2 )ω 3
ω 1 =
k 1
m
=
a
3m
; ω 2 = ω 3 =
2a
m
E = E 1 + E 2 + E 3 =
n 1 +
1
2
ω 1 + (n 2 + n 3 + 1)ω 2
The lowest energy level corresponds to n 1 = n 2 = n 3 = 0, with
E =
ω 1
2
+ ω 2 =
1
12
+
√
2
a
m
It is non-degenerate.
The next higher state is degenerate with n 1 = 1, n 2 = 0, n 3 = 0;
E =
3
2
a
3m
+
2a
m
=
3
4
+
√
2
a
m
This is also non-degenerate.