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

Roger Griffith
Physics 137B
hw. # 9
Proffessor Clarke
Problem # 1
An infinite one-dimensional square well [V=0, for |x|≤
2 a , V=∞ elsewhere] contains a particle of mass
m in the ground state. At time t = 0 a pertubation H ′ = Aδ(x) is turned on: δ(x) is the Dirac delta function.
Calculate the initial rate of transition into the first (n=2) and (n=3) excited states using time dependent
pertubation theory.
[NOTE: Find coefficients a 2 (t) and a 3 (t) using the integral expressions we derived in class and appropriate
matrix elements. Look carefulle at the parity of the two matrix elements before evaluating them!
“Initial” means ωt ≪ 1, where ω is the relavent energy differece.
We are given
H ′ = Aδ(x)
ΔE = ω
and the wave functions for the one-dimensional infinite square well are given by
√
2
( nπx
)
√
2
( nπx
)
ψ odd =
a sin ψ even =
a
a cos a
The integral expression derived in class is given by
Z t
a k (t) = − i H k ′
j
eiω k jt ′ dt ′ ω k j = 1 [
E
0
0
k − En]
0
where k is the final state of the system and j is the initial state.
We need to find the matrix elements H
k ′
j
. (we also know that n = 1 is an even wave function and n = 2
is an odd wavefunction)
For the n : 1 → 2 we find
H ′
21 = 〈2|Aδ(x)|1〉 = 2A a
thus we know that
Z a/2
−a/2
cos
( πx
)
sin
a
( 2πx
a
)
δ(x)dx = 2A a
cos(0) · sin(0) = 0
For the n : 1 → 3 we find
H ′
31 = 〈3|Aδ(x)|1〉 = 2A a
we know what ω 31 is
a 2 (t) = − i
Z a/2
−a/2
cos
Z t
0
( πx
)
cos
a
H ′
21 eiω 21t ′ dt ′ = 0
( 3πx
a
)
δ(x)dx = 2A a cos(0) · cos(0) = 2A a
ω 31 = 1 [
E
0
3 − E1
0 ] 4π 2 =
ma 2
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

where
H ′
200,100 = 〈200| − er cos(θ)ε 0 e −t/τ |100〉 = −eε 0 e −t/τ 〈200|r cos(θ)|100〉
= − eε 0e −t/τ Z 2π Z π
Z ∞
(
√ dφ cos(θ)sin(θ)dθ r 3 1 − r )
e −3r/2a 0
dr
8πa
3
0
2a 0
0
0
0
but since we know that
causes
Z π
0
cos(θ)sin(θ)dθ = 0
H ′
200,100 = 0
and thus
a 200 (t) = − i Z t
H k ′
j
eiω k jt ′ dt ′ = 0
0
and if you square this number, you still get 0 thus the probability of this transition happening is nill.
(b) Show that after time t ≫ τ, the probability that the atom is in the 2p state Ψ nlm = Ψ 210 is
|a 210 (t)| 2 = 215
3 10 e 2 ε 2 0 a2 0
2 (ω 2 + 1/τ 2 )
we must first find the matrix element H
210,100 ′ which is given by
H ′
210,100 = 〈210| − er cos(θ)ε 0 e −t/τ |100〉 = −eε 0 e −t/τ 〈210|r cos(θ)|100〉
= − eε 0e −t/τ Z 2π
√ dφ
8πa
4
0
and so we find the integrals to be
Z 2π
0
Z π
dφ cos 2 (θ)sin(θ)dθ
0
0
Z π
0
cos 2 (θ)sin(θ)dθ
Z ∞
0
r 4 e −3r/2a 0
dr
letting u = cos(θ) du = −sin(θ)dθ
thus we find
Z 2π Z π
Z −1
dφ cos 2 (θ)sin(θ)dθ = 2π u 2 du = 4π
0 0
1 3
and for the other integral we find
putting this all together we get
Z ∞
0
(
r 4 e −3r/2a 2a0
0
dr = 4!
3
) 5
H 210,100 ′ = −eε 0e −t/τ ( )
4π
5
√
8πa
4
0
3 4! 2a0
= − 28 a 0
3 3 5√ 2 eε 0e −t/τ
3

now that we have this solution we can find the coefficient. Plugging this into the equation derived in
class we find
but since we know that
Z t
0
a 210 (t) = − i
e t′ (iω−1/τ) dt ′ =
and so we find the coefficient to be given as
and the probability is given by
= ieε 0
Z t
0
H ′
k j eiω k jt ′ dt ′
a
√ 0 2 8
2
1
iω − 1/τ
for t ≫ τ e t(iω−1/τ) → 0
a 210 (t) = − ieε 0
( )
a0 2
8
(
√
2 3 5
Z t
3 5 e t′ (iω−1/τ) dt ′
0
[
]
e t(iω−1/τ) − 1
)
1
iω − 1/τ
|a 210 (t)| 2 = a 210 (t)a 210 (t) ∗ = 215
3 10 e 2 ε 2 0 a2 0
2 (ω 2 + 1/τ 2 )
4