186 3 Quantum Mechanics – II
H=
(
−
^2
2 m)
∇^2 +a[
x^2 +y^2 +z^2 −5
6
x^2]
=
(
−
^2
2 m)
∇^2 +a[
x^2
6+y^2 +z^2]
(2)
The Schrodinger’s equation is
Hψ(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)=ψxψyψz (4)Hψ(x,y,z)=−^2
2 m(
∂^2
∂x^2+
∂^2
∂y^2+
∂^2
∂z^2)
ψxψyψz+a[
x^2
6+y^2 +z^2]
ψxψyψz=Eψxψyψz−^2
2 mψyψz∂^2 ψx
∂x^2−
^2
2 mψxψz∂^2 ψy
∂x^2−
^2
2 mψxψy∂^2 ψz
∂X^2+ax^2
6ψxψyψz+ay^2 ψxψyψz+az^2 ψxψyψz=Eψxψyψz
Dividing throughout byψxψyψz−^2
2 m1
ψx∂^2 ψx
∂x^2−
^2
2 m1
ψy∂^2 ψy
∂y^2−
^2
2 m1
ψz∂^2 ψz
∂z^2+
ax^2
6+ay^2 +az^2 =E(5)
−^2
2 m1
ψx∂^2 ψx
∂x^2+
ax^2
6=E 1
a
6=^1 / 2 k 1 (6)−
^2
2 m1
ψy∂^2 ψy
∂y^2+ay^2 =E 2 a=^1 / 2 k 2 (7)−
^2
2 m1
ψz∂^2 ψz
∂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 3ω 1 =√
k 1
m=
√
a
3 m;ω 2 =ω 3 =√
2 a
mE=E 1 +E 2 +E 3 =(
n 1 +1
2
)
ω 1 +(n 2 +n 3 +1)ω 2The lowest energy level corresponds ton 1 =n 2 =n 3 =0, withE=ω 1
2+ω 2 =(√
1
12
+
√
2
)
√(
a
m)
It is non-degenerate.
The next higher state is degenerate withn 1 = 1 ,n 2 = 0 ,n 3 =0;E=3
2
√
a
3 m+
√
2 a
m=
(√
3
4
+
√
2
)
√
a
m
This is also non-degenerate.