Z
ψ
Ã
x, t
ð Þψ x, t
ð Þdτ ¼
Z
ψ x, t
ð Þ
j
j
2 dτ
¼
1
2
Z
ϕ 1s
ð Þ
½
2 þ ϕ 2p z
À Á
Â
à 2
n
o
dτ þ cos ωt
Z
ϕ 1s
ð Þϕ 2p z
À Á
dτ ¼
1
2
þ
1
2
¼ 1,
where we used normalized functional forms of ϕ(1s) and ϕ(2p z ) together with
orthogonality of them. Note that both of the functions are real.
Next, we calculate the matrix element. For simplicity, we denote the matrix
element simply as P
ε e
ð Þ only by designating the unit polarization vector ε e . Then,
we have
P
ε e
ð Þ
¼ ϕ 1s
ð Þjε e Á Pjϕ 2p z
À Á
,
ð4:37Þ
where
P = ex = e e 1 e 2 e 3
ð
Þ
x
y
z
0
B
@
1
C
A:
ð4:38Þ
We have three possibilities of choosing ε e out of e 1 , e 2 , and e 3 . Choosing e 3 , we have
P
e 3
ð Þ
z,jp z i
¼ e ϕ 1s
ð Þjzjϕ 2p z
À Á
¼
e
4
ffiffi ffi
2
p πa 4
Z 1
0
r
4 e
À3r=2a dr
Z π
0
cos
2
θ sin θdθ
Z 2π
0
dϕ ¼
2
7
ffiffi ffi
2
p
3
5
ea % 0:745ea: ð4:39Þ
In (4.39), we express the matrix element as P
e 3
ð Þ
z,jp z i
to indicate the z-component of
position vector and to explicitly show that ϕ(2p z ) state is responsible for the
transition. In (4.39), we used z ¼ r cos θ. We also used a radial part integration
such that
Z 1
0
r
4 e
À3r=2a dr ¼ 24
2a
3
5
:
Also, we changed a variable cos θ ⟶ t to perform the integration with respect to θ.
We see that a “leverage” length of the transition moment is comparable to Bohr
radius a.
With the notation P
e 3
ð Þ
z,jp z i
we need some explanation for consistency with the latter
description. Equation (4.39) represents the transition from jϕ(2p z )i to jϕ(1s)i that is
accompanied by the photon emission. Thus, j p z i in the notation means that jϕ(2p z )i
134
4 Optical Transition and Selection Rules
ψ
Ã
x, t
ð Þψ x, t
ð Þdτ ¼
Z
ψ x, t
ð Þ
j
j
2 dτ
¼
1
2
Z
ϕ 1s
ð Þ
½
2 þ ϕ 2p z
À Á
Â
à 2
n
o
dτ þ cos ωt
Z
ϕ 1s
ð Þϕ 2p z
À Á
dτ ¼
1
2
þ
1
2
¼ 1,
where we used normalized functional forms of ϕ(1s) and ϕ(2p z ) together with
orthogonality of them. Note that both of the functions are real.
Next, we calculate the matrix element. For simplicity, we denote the matrix
element simply as P
ε e
ð Þ only by designating the unit polarization vector ε e . Then,
we have
P
ε e
ð Þ
¼ ϕ 1s
ð Þjε e Á Pjϕ 2p z
À Á
,
ð4:37Þ
where
P = ex = e e 1 e 2 e 3
ð
Þ
x
y
z
0
B
@
1
C
A:
ð4:38Þ
We have three possibilities of choosing ε e out of e 1 , e 2 , and e 3 . Choosing e 3 , we have
P
e 3
ð Þ
z,jp z i
¼ e ϕ 1s
ð Þjzjϕ 2p z
À Á
¼
e
4
ffiffi ffi
2
p πa 4
Z 1
0
r
4 e
À3r=2a dr
Z π
0
cos
2
θ sin θdθ
Z 2π
0
dϕ ¼
2
7
ffiffi ffi
2
p
3
5
ea % 0:745ea: ð4:39Þ
In (4.39), we express the matrix element as P
e 3
ð Þ
z,jp z i
to indicate the z-component of
position vector and to explicitly show that ϕ(2p z ) state is responsible for the
transition. In (4.39), we used z ¼ r cos θ. We also used a radial part integration
such that
Z 1
0
r
4 e
À3r=2a dr ¼ 24
2a
3
5
:
Also, we changed a variable cos θ ⟶ t to perform the integration with respect to θ.
We see that a “leverage” length of the transition moment is comparable to Bohr
radius a.
With the notation P
e 3
ð Þ
z,jp z i
we need some explanation for consistency with the latter
description. Equation (4.39) represents the transition from jϕ(2p z )i to jϕ(1s)i that is
accompanied by the photon emission. Thus, j p z i in the notation means that jϕ(2p z )i
134
4 Optical Transition and Selection Rules