Download Physics 137B: Solutions for Spontaneous & Stimulated Emission Rates, Dipole Transitions - and more Assignments Quantum Mechanics in PDF only on Docsity!
Physics 137B, Fall 2007, Moore
Problem Set 9 Solutions
Let a denote the hydrogen 1s state and let b denote the the 2p state with magnetic
quantum number m. In the notation |n`m〉, we have |a〉 = | 100 〉, and |b〉 = | 21 m〉.
The transition rate for spontaneous emission, calculated via the statistical argument
due to Einstein, is given by Eq. (11.76):
W
s ab =^
ω ba^3
3 πc^3 ℏ 0
|Dba| 2
. (1)
Here
ωba =
E 2 − E 1
E 1
3 μα 2 c 2
and
Dba = 〈b| − er|a〉 = −e 〈 21 m|r| 100 〉.
For a fixed value of the magnetic quantum number m, the dipole moment Dba can
be calculated most easily by expressing the vector r in terms of so-called spherical
tensor operators. To introduce this notion, first define three new basis vectors
e+1 = −
(ˆx − iyˆ) , e− 1 =
(ˆx + iyˆ) , e 0 = ˆz.
(The fact that we are using complex vectors to span real 3-dimensional space looks
fishy, but it turns out to be harmless. Note that these vectors are orthogonal in the
complex sense, e ∗ m ·^ em′^ =^ δm,m′^ .)^ Next we define new coordinates^ rm^ ≡^ e
∗ m ·^ r^ for m = 0, ±1. Explicitly
r+1 = −
(x + iy) , r− 1 =
(x − iy) , r 0 = z.
Using the orthogonality property mentioned above, we can write
|Dba| 2 = e 2
| 〈b|r+1|ψa〉 | 2
- | 〈ψb|r− 1 |a〉 | 2
- | 〈ψb|r 0 |ψa〉 | 2
= e 2
m′
| 〈 21 m|rm′ | 100 〉 | 2 .
The motivation for these definitions is that they allow us to express rm simply
in terms of the spherical harmonic functions that we all know and love. Looking at
Table 6.1, for instance, we see that with these definitions
rm =
4 π
3
rY 1 m(θ, φ) , m = 0, ± 1.
Now the hydrogen atom wavefunctions are ψnm(r, θ, φ) = RnY`m(θ, φ), so
〈 21 m|rm′^ | 100 〉 =
d 3 r ψ ∗ 21 m(r)
4 π
3
rY 1 m′^ (θ, φ)ψ 100 (r)
4 π
3
0
r 3 R 21 (r)R 10 (r) dr
Y 1 m(θ, φ) ∗ Y 1 m′^ (θ, φ)Y 00 (θ, φ) sin θ dθ dφ
δm,m′
0
r 3 R 21 (r)R 10 (r) dr ,
since Y 00 = √^1 4 π and the spherical harmonics have the orthogonality property
∫
Ym(θ, φ) ∗ Y′m′ (θ, φ) sin θ dθ dφ = δ,′ δm,m′.
Using the explicit forms of the radial wavefunctions, given by Eq. (7.140), we find
∫ (^) ∞
0
r 3 R 21 (r)R 10 (r) dr =
a 0 √ 6
0
x 4 e − 3 x/ 2 dx =
a 0 √ 6
Thus
〈 21 m|rm′ | 100 〉 =
a 0 δm,m′ ,
and
|Dba| 2 =
15
e 2 a 2 0
δm, 1 + δm,− 1 + δm, 0
Since only m = 0 contributes, we only want the component 0 to be nonzero. Thus
the incident radiation should be polarized along e 0 , i.e. ˆ = ˆz.
Calculating as in Problem 1, we find the rate of stimulated emission to be
W (^) ab =
πI(ωba)
ℏ^2 c 0
e 2 a 2 0
217 π^2
310
αa^20
ℏ
I(ωba)
= 4. 24 × 10
12 I(ωba) J s/m
To obtain a rate of W (^) ab = 1 per nanosecond = 10 9 s − 1 , we would need
I(ωba) = 2. 36 × 10 − 4 J/m
2 .
In general, dipole transitions are only allowed if Dba = 〈b|ˆ · (−er)|a〉 is nonzero. Con-
sider the 3-dimensional harmonic oscillator with |a〉 = |nxnynz 〉 and |b〉 = |n′ xn′ yn′ z 〉,
and let ˆ = ˆx. Then
Dba = 〈n ′ xn
′ yn
′ z | −^ ex|nxnynz^ 〉^ =^ −e^ 〈n
′ x|x|nx〉 〈n
′ y|ny〉 〈n
′ z |nz^ 〉^.
The 1-dimensional harmonic oscillator states |n〉 satisfy the orthogality relation 〈n ′ |n〉 =
δnn′^ , and x can only raise or lower a state by one, since in terms of ladder operators
x =
ℏ 2 mω (a^ +^ a
†). Thus only transitions with
∆nx = ± 1 , ∆ny = 0 , ∆nz = 0
are allowed.
The oscillator strength is defined as
fka =
2 μωka
3 ℏe^2
|Dka| 2 =
2 μ(Ek − Ea)
3 ℏ^2
〈a|r|k〉 · 〈k|r|a〉.
Now we make the following observation:
(Ek − Ea) 〈a|r|k〉 · 〈k|r|a〉 = (Ek − Ea) 〈a|ri|k〉 〈k|ri|a〉
〈a|(Ek − Ea)ri|k〉 〈k|ri|a〉 + 〈a|ri|k〉 〈k|(Ek − Ea)ri|a〉
〈a|riH − Hri|k〉 〈k|ri|a〉 + 〈a|ri|k〉 〈k|Hri − riH|a〉
〈a|[ri, H]|k〉 〈k|ri|a〉 − 〈a|ri|k〉 〈k|[ri, H]|a〉
where summation over the repeated index i is implied. (This is Einstein’s summation
convention.) In the third line we used the fact that H is Hermitian, so 〈a| H = 〈a| Ea,
etc. We now sum over all states k, (including the continuum states!), and use the
completion relation
k |k〉 〈k|^ =^1. This gives ∑
k
fka =
2 μ
3 ℏ^2
〈a|[ri, H]ri|a〉 − 〈a|ri[ri, H]|a〉
μ
3 ℏ^2
〈a|
[
[ri, H], ri
]
|a〉.
For the commutator [ri, H], note that the only term in H that doesn’t commute
with ri is the momentum term p^2 / 2 μ. So
[ri, H] =
[
ri,
pj pj
2 μ
]
2 μ
pj [ri, pj ] + [ri, pj ]pj
2 μ
pj (iℏδij ) + (iℏδij )pj
iℏ
μ
pi ,
(again with summation over j implied). Finally, we have
[
[ri, H], ri
]
[
iℏ
μ
pi, ri
]
iℏ
μ
(− 3 iℏ) =
3 ℏ^2
μ
so ∑
k
fka =
μ
3 ℏ^2
〈a|
3 ℏ^2
μ
|a〉 = 1.
(a) (For part (a) of this problem we work in the lab frame. See Figure 1.)
conservation, after scattering the particles will maintain the same speed, but will be
deflected in a different direction.
To obtain the corresponding picture in the lab frame, we must add the constant
vB zˆ to all velocities. (This choice guarantees that the target particle will initially be
at rest.) The velocity of the target particle after scattering is pictured in Figure 2b.
Simple geometry shows that the scattering angles are related by θ = 2θL.
Suppose the scattering is isotropic in the CM frame, so the differential cross section
is given by dσ
dΩ
σtot
4 π
Recalling that the infinitesimal solid angle is dΩ = sin θ dθ dφ = d(cos θ) dφ, we can
use the relation θ = 2θL to change to lab frame variables:
( dσ
dΩ
L
d(cos θ)dφ
d(cos θL)dφL
dσ
dΩ
d(cos 2θL)
d(cos θL)
σtot
4 π
Now cos 2θL = 2 cos 2 θL − 1, so ( dσ
dΩ
L
= 4 cos θL
σtot
4 π
cos θL
π
σtot.
Note that θL is restricted to the range 0 ≤ θL ≤ π/2, since 0 ≤ θ ≤ π. Therefore
the total cross section, calculated in the lab frame, is
∫ (^2) π
0
dφL
0
d(cos θL)
dσ
dΩ
L
= 2π
0
d(cos θL)
cos θL
π
σtot = 2σtot
0
x dx = σtot ,
as it must be.
