208
3 Quantum Mechanics – II
It is seen from the last column of the table that the degeneracy D is given by
the sum of natural numbers, that is, = n(n + 1)/2, if we replace n by N + 1,
D = (N + 1)(N + 2)/2.
3.61 As the time evolves, the eigen function would be
ψ(x, t) =
n=0.1 C n ψ n (x) exp(−i E n t/)
= C 0 ψ 0 (x) exp(−i E 0 t/) + C 1 ψ 1 (x) exp(−i E 1 t/)
The probability density
|ψ(x, t)|
2
= C
2
0 + C
2
1 + C 0 C 1 ψ 0 (x)ψ 1 (x)[exp(i(E 1 − E 0 )t/)
− exp(−i(E 1 − E 0 )t/)]
= C
2
0 + C
2
1 + 2C 0 C 1 ψ 0 (x)ψ 1 (x) cos ω t
where we have used the energy difference E 1 − E 0 = ω. Thus the probability
density varies with the angular frequency.
3.62 < E >=
n=1,2,3
|C n |
2 E n = C
2
0 E 0 + C
2
1 E 1 + C
2
2 E 2
=
1
2
·
ω
2
+
1
3
·
3ω
2
+
1
6
·
5ω
2
=
7ω
6
.
3.63 (a) ψ 0 (x) = A exp(−x
2
/2a
2 )
Differentiate twice and multiply by −
2
/2m
−
2
2m
d
2
ψ 0
dx 2 =
A
2
2ma 2
1 −
x
2
a 2
exp
−
x
2
2a 2
=
2
2ma 2
ψ 0 −
2 x
2
2ma 4
ψ 0
or −
2
2m
d
2
ψ 0
dx 2 +
2 x
2
2ma 4
ψ 0 =
2
2ma 2
ψ 0
Compare the equation with the Schrodinger equation
E =
2
2ma 2 =
ω
2
ω =
ma 2
(1)
or a =
mω
1/2
Same relation is obtained by setting
V =
2 x
2
2ma 4 =
mω
2 x
2
2
(b) ψ 1 = Bx exp
−
x
2
2a 2
Differentiate twice and multiply by −
2
2m
−
2
2m
d
2
ψ 1
dx 2 =
B
2 x
3 exp
x
2
a
2ma 4
+
3B
2 exp
−
x
2
2a 2
2ma 2