Download Thirteenth quantum mechanics sheet

Universität Heidelberg
SS 2009
Thirteenth quantum mechanics sheet
To be handed in on 07.07.
All points in this sheet are awarded as bonus points.
Q 34: Hydrogen atom, fine structure
(16 points)
~ S
~ and J~ are angular momentum operators. The notationˆis omitted
Remark: In this question L,
for these operators.
We consider the Hamiltonian of the Hydrogen atom
~2
P̂
Ze2
Ĥ0 =
−
+ Ŵ ,
2m
R̂
Z=1
with the spin-orbit interaction (see the lectures)
Ŵ =
Ze2 1 ~ ~
S·L
2m2e c2 R̂3
The aim of this exercise is to treat Ŵ as a perturbation from Ĥ0 and from this calculate the
2
2
(0)
Z 2 e4 1
= − me2c α2 Zn2 (α = finestructure constant) at
corrections to the energy levels En = − me2h̄
2
n2
first order in perturbation theory. The energy En 0 depends only on the main quantum number
n and not on the angular momentum quantum numbers l and m. Therefore we must use the
~ 2 and L3 with |n, l, ml i.
degenerate perturbation theory. We describe the joint Eigenstates Ĥ0 , L
It follows:
Ĥ0 |n, l, ml i = En(0) |n, l, ml i
~ 2 |n, l, ml i = h̄2 l(l + 1)|n, l, ml i
L
L3 |n, l, ml i = h̄ml |n, l, ml i
ψnlml (~r) = h~r|n, l, ml i = Rnl (r)Ylml (Ω)
~ is given by S
~ = h̄ ~τ with the Pauli matrix τi . The state space of Ĥ = Ĥ0 + Ŵ
The spin operator S
2
~ 2 , L3 :
is spanned by the product of Eigenstates of Ŝ3 and the Eigenstates |n, l, ml i of Ĥ0 , L
|n, l, ml , ms i := |n, l, ml i|ms i
~ ·S
~ commutes with L
~ 2 and S
~ 2 , but not with L3 and S3 . (4 points)
a) Show that Ŵ ∼ L
This means that the product state |n, l, ml , ms i does not build an appropriate basis for the
perturbed theory, because of the energy level degeneracy of l, ml and ms . One must build for
a fixed l sub vector space, a new basis through linear combinations of |n, l, ml , ms i, in which
~ ·S
~ is diagonal (= angular momentum coupling). To do this one uses L
~ and S
~ to build a
Ŵ ∼ L
~
new operator J of the total angular momentum:
~ +S
~
J~ = L
b) Show that the components Ji satisfy the angular momentum commutation relation, i.e.
that
[Ji , Jj ] = ih̄
X
k
ǫijk Jk .
show furthermore that
h
i
h
i
h
i
i
h
i
h
i
~ 2 = J3 , J~2 = 0
~ 2 = J3 , S
J3 , L
h
~ 2 = J~2 , S
~2 = L
~ 2, S
~2 = 0
J~2 , L
(5 points)
~ 2 and S
~ 2 . We describe
From b) it follows that it is possible to form joint Eigenvectors of J~2 , J3 , L
these Eigenvectors |j, mj ; l, si with
J~2 |j, mj ; l, si = h̄2 j(j + 1)|j, mj ; l, si
J3 |j, mj ; l, si = h̄mj |j, mj ; l, si
~ 2 |j, mj ; l, si = h̄2 l(l + 1)|j, mj ; l, si
L
~ 2 |j, mj ; l, si = h̄2 s(s + 1)|j, mj ; l, si .
S
(1)
Here s = 1/2.
~ ·S
~ and hence Ŵ is diagonal
c) Show that the states |j, mj ; l, si are also Eigenstates of L
~ ·S
~ (in terms of j, l and s). (3
in this basis. Determine the corresponding Eigenvalues L
points)
d) Calculate now the energy corrections at first order in perturbation theory
En(1) = hn; j, mj ; l, s|Ŵ |n; j, mj ; l, si .
For this use without proof that
hn; j, mj ; l, s|
1
R̂3
|n; j, mj ; l, si =
Z3
,
a30 n3 l(l + 1)(l + 21 )
(0)
a0 =
h̄2
me e2
(1)
The degeneracy of the Energy Eigenvalues En are broken by the energy corrections En . How
large is the splitting ? Consider
(1)
En
(0)
En
. (4 points)