3.3 Solutions
211
u
2
210 dτ =
1
8πa
3
0
∞
0
y
4 e
−y
dy
4a
2
0
a
5
0
cos
3
θ
3
1
−1
(2π )
=
1
24
× 4! = 1
Similarly u
2
21±1 dτ =
1
π(2a 0 ) 3
e
−2x
2
x
2 sin
2
θ e
±iϕ e
∓iϕ r
2 sin θ dθ dϕdr
=
1
8πa
3
0
∞
0
e
−
r
a 0
2
r
4 dr
4a
2
0
+1
−1
(1 − cos
2
θ ) d(cos θ )
2π
0
dϕ
=
a
5
0
64πa
5
0
∞
0
y
4 e
−y dy
4
3
(2 π)
=
1
192
(4!)(8) = 1
3.67
u 21±1 u 210 dr =
A
2
2
√
2
e
−x x cos θ e
−x x sin θ e
+ϕ r
2 sin θ dθ dϕdr
The integral
2π
0 e
±iϕ dϕ = 0
Therefore u 21±1 and u 210 are orthogonal.
Further the integral
u
∗
211 U 21−1 dτ involves the integral
2π
0
e
−iϕ e
−iϕ dϕ or
2π
0
e
−2iϕ dϕ = 0
So, the functions u 211 and u 21−1 are also orthogonal.
3.68 The degree of degenerating is given by 2n
2 . So for n = 1, degenerary is 2, for
n = 2 it is 8, for n = 3, it is 18 and for n = 4, it is 32.
3.69 Parity of the state is determined by the factor (−1)
l . For 1s, l = 0, parity= +1,
for 2 p, l = 1, parity = − 1 and for 3d, l = 2, parity = +1.
3.70 To show that the probability density of the 3d state is independent of the polar
angle θ. We form
u
∗ u = u
∗ (3, 2, 0)u(3, 2, 0) + 2 u
∗ (3, 2, 1)u(3, 2, 1) + 2 u
∗ (3, 2, 2)
u(3, 2, 2)
The factor 2 takes care of m values ±1 and ±2, as in Table 3.2. Inserting the
functions the azimuth part, e
iϕ or e
−iϕ drop off when we multiply with the
complex conjugate,
i.e.
e
iϕ
∗ e
iϕ
= 1 or (e
−iϕ )
∗ (e
−iϕ ) = 1
u
∗ u = A
2
3 e
−2x x
4
1
18
(3 cos
2
θ − 1)
2
+
2
3
sin
2
θ cos
2
θ +
1
6
sin
4
θ