1
PHY4605
Problem Set 7
Solutions
1. H-atom in t-dependent electric field.
Ψn`m = Rn` Y`m . From Tables 4.3 and 4.7:
ψ100 = p
1
πa30
1
ψ210 = p
e
−r/a0
32πa30
;
ψ200 = p
1
8πa30
r −r/2a0
e
cos θ;
a0
µ
r
1−
2a0
¶
e−r/2a
1
ψ21±1 = ∓ p
64πa30
r r/2a0
e
sin θe±iφ
a0
But r cos θ = z and r sin θe±iφ = r sin
φ ± i sin φ) = r sin θ cos φ ± ir sin θ sin φ = x ± iy. So |ψ|2 is an even
R θ(cos
2
function of z in all cases! Therefore z|ψ| dx dy dz = 0, so Hii0 = 0. Also ψ100 , ψ200 , ψ21±1 are even in z, so the
matrix elements between them also vanish due to the oddness of the integrand. The only nonzero element is
Z
1
1
1
0
H100,210
= eE p 3 p
d3 re−r/a0 e−r/2a0 r2 cos2 θ
πa0 32πa30 a0
µ
µ 8 ¶
¶5
Z ∞
Z π
Z 2π
eE
eE
2a0
2
2
4 −3r/2a0
2
√
4!
2π = −
eEa0 .
= √
r e
dr
cos θ sin θdθ
dφ = √
4
4
5
3
3
4 2πa0 0
4 2πa0
3 2
0
0
2. Exponential Pulse. Since atom spherically symmetric, we can take perturbation, without loss of generality,
to be
(
0
t<0
V̂ = −eEz (t)z = −ez
(1)
−γt
E0z e
t>0
Z ∞
eE0z
c1s→2p =
hψ210 |z|ψ100 i
dteiω0 t e−γt
(2)
i~
0
where ~ω0 = E2 − E1 = 1 Ryd ( −1
4 − (−1)) = 3/4 Ryd. Need
Z ∞
−1
dte(iω0 −γ)t =
iω0 − γ
0
2
e2 E0z
⇒ P1s→2p = 2 2
|hψ210 |z|ψ100 i|2
~ (ω0 + γ 2 )
(3)
(4)
Now
|hψ210 |z|ψ100 i|2 =
a20 216
215
= 10 a20 = 0.555a20
10
2 3
3
P1s−2p '
2
.555a20 e2 E0z
2
2
2
~ (ω0 + γ )
Check the dimensions to make sure you believe this is a probability!
(5)
(6)