1
PHY4605
Problem Set 8
Solutions
1. Dipole transitions
V̂ = −qr · E
(1)
neglect spatial variation of E, neglect B (O(v/c) effects!). Then
V̂ = −qE0 ² · r cos ωt
(2)
Find angular distribution of electrons ejected from H-atom ground state
Identity: cos θ²r = cos θ0 cos θ + sin θ0 sin θ cos(φ0 − φ)
Want to calculate Pf (θ0 , φ0 )– probability for finding ejected electron at angles θ0 , φ0 with respect to ².
Initial and final states
|ψi i = |ψ100 i ∝ e−r/a0
|ψ 0 i = (told not to worry about normalization) exp ikF · r
(3)
(4)
So we’ll need matrix elements
·Z
¸
3 −ikF ·r
−r/a0
hf |V̂0 |ii ∝
d re
² · re
Z
Z
π
=
Z
0
∞
¡
¢
r2 dre−ikF r cos θ r cos θ²r e−r/a0
dφ
Z
π
Z
2π
dθ sin θ
0
0
2π
¢
dφ cos θ0 cos θ + sin θ0 sin θ cos(φ0 − φ)
0
0
µZ ∞
¶
Z 1
x=cos θ
= 2π cos θ0
drr3 e−r/a0
dxxe−ikF rx
=
¡
dθ sin θ
Z
+ sin θ
0
0
2π
−1
µZ
0
dφ cos(φ − φ)
0
∞
Z
3 −r/a0
0
dx
−1
∞
r3 dre−ikF r cos θ e−r/a0
0
p
sgnx 1 − x2 e−krx
|
{z
}
1
drr e
Z
zero since x-integrand antisymmetric
= 2πA cos θ0
(5)
where
Z
A=
∞
Z
drr3 e−r/a0
0
1
dxxe−ikF rx
(6)
−1
= mess ! (doable though)
(7)
The point is not to evaluate this but notice that hf |V̂0 |ii ∝ cos θ0 . So we have
0
0
Pf (θ , φ ; ω) ∝ q
2
cos2 θ0
E02
2
~
£
¤
sin2 (ω − ω0 )t
(ω − ω0 )2
(8)
~2 k2
where ~ω0 = 2mF − E0 where the first term is the kinetic energy of electron in final state and the last term is
the energy of electron in ground state(= −1 Ryd). Note in dipole approximation Pf is independent of φ0 it
only depends on angle between ² and kF !
2. Dipole selection rules
a) classify states by |n, l, m, πi consider transitions between states of different parity π induced by
V̂ = −E0 ² · d cos ωt
hπ 0 |V̂ |πi ∝ ² · hπ 0 |d|πi
(9)
2
Since
π̂d = −dπ̂ and π̂ 2 = 1,
d = −π̂dπ̂, so
hπ 0 |d|πi = hπ 0 | − π̂dπ̂|πi = −ππ 0 hπ 0 |d|πi ⇒
−ππ 0 = 1, π 0 = −π
since π, π 0 = ±1 only
(10)
(11)
(12)
(13)
(14)
Parity must change in electric dipole transition!
b) I’ll prove these for d = qr; if there is more than one charge, must sum over all charges:
i) [L̂z , dz ] = q[x∂y − y∂x , z] = 0.
ii) [L̂z , d± ] = −i~q[x∂y − y∂x , x ± iy] = ±~qx[∂y , y] + i~qy[∂x , x] = ±~q(x ± iy).
iii)
[L̂± , dz ] = q[L̂x ± iL̂y , z]
= −i~q[y∂z − z∂y ± iz∂x ∓ ix∂z , z]
£
¤
= −i~q y[∂z , z] ∓ ix[∂z , z]
= ∓~q(x ± iy) = ∓~d±
(15)
iv) [L̂± , d± ] = Cαβ α, β = +, −
[L̂α , dβ ] = [L̂x + iαL̂y , dx + iβdy ]
= −i~q[y∂z − z∂y + iα(z∂x − x∂z ), x + iβy]
= −i~q(−izβ + iαz) = ~qz(α − β) ⇒
(16)
[L+ , d− ] = 2~qz, [L− , d+ ] = −2~qz
(17)
and [Lα , dα ] = 0, α = ±
(18)
P
c) Want to show d+ |lmi is an admixture of various |l0 m + 1i, d+ |lmi = l0 al0 |l0 m + 1i. It is sufficient to
show d+ |lmi is in state of definite L̂z , eigenvalue ~(m + 1):
£
¤
£
√
L̂z d± |lmi = ± ~d± + d± Lz |lmi = ± ~ + m~]d± |lmi = ~(m ± 1)d± |lmi
(19)
P
Lz
because since |lmi are complete, any state
Pmay be written lm alm |lmi, and therefore any state with
quantum number m + 1 may be written l al |lm + 1i. Note you were not asked to prove which l0 are
contained in d+ |lmi yet!
¸
·
P (+)
P (−)
1
0 0
0 0
0 0 d+ +d−
0 0
d) hl m |dx |lmi = hl m | 2 |lmi = 2 hl m | l al |lm + 1i + hl m | l al |lm − 1i . This can be
nonzero only if m0 = m ± 1 .