Roger Griffith Physics 137B hw. # 9 Proffessor Clarke Problem # 1 ...

we will need to use the following identity for the next part
we find the coefficient to be
2isin(ω 31 t/2) = e iω 31t/2 − e −iω 31t/2
so we get
a 3 (t) = − i
Z t
H ′
0
= − 2A
a eiω 31t/2
The rate of transition is given by
31 eiω 31t ′ dt ′ = − 2Ai Z t
e iω 31t ′ dt ′ = − 2A [ e
iω 31 t ]
− 1
t
a 0
a ω 31 t
[
]
e iω31t/2 − e −iω 31t/2
t = − 2A [ isin(ω31 t/2)
ω 31 t
a eiωt/2 ω 31 t/2
a 3 (t) = − 2A
a eiω 31t/2
[ isin(ω31 t/2)
ω 31 t/2
]
t
]
t
|a 3 (t)| 2
t
=
( ) 2A 2 ( )
a t sin(ω31 t/2) 2
since ω 31 t ≪ 1
ω 31 t/2
sin(ω 31 t/2)
ω 31 t/2
⇒ 1
we get
|a 3 (t)| 2
t
= 4A2
2 a 2 t
Problem # 2
A hydrogen atom in its ground state is placed between the parallel plates of a capacitor. The z-axis
of the atom in perpendicular to the capacitor plates. At time t = 0, a uniform electric field⃗ε =⃗ε 0 e −t/τ is
applied to the atom. The perturbing Hamiltonian is thus
H ′ = −e⃗r ·⃗ε 0 e −t/τ = −ezε 0 e −t/τ
and since z is given by z = r cos(θ)
we find the Hamiltonian to be
H ′ = −er cos(θ)ε 0 e −t/τ
and the three hydrogen wave functions needed for this problem are given as
ψ 100 = √
1 e −r/a 0
ψ 200 = 1 (
1
√ 1 − r )
e −r/2a 0
ψ 210 = 1
πa 3 2πa0 2a 0 2a 0
0
√
2π
1
4a 5/2
0
cos(θ)e −r/2a 0
(a) Show that the electron has zero probability of being excited into the 2s state Ψ nlm = Ψ 200
what we are looking for is |a 200 (t)| 2 so we must use the formula
a 200 (t) = − i
Z t
0
H ′
k j eiω k jt ′ dt ′
2