4
Time-ind. Perturbation Theory II
We said we solved the Hydrogen atom exactly, but we lied. There are a
number of physical effects our solution of the Hamiltonian H = p2/2m −
e2/r left out. We already said motion of proton can be accounted for
by nearly trivial reduced mass correction–not important, since spectra of
original problem unchanged except for tiny overall energy scaling.
In fact there are a hierarchy of other effects we left out, each of which
can be treated in perturbation theory on top of the other. In the most
complete theory, almost all degeneracies are split. Classify formally in
powers of fine structure constant α ≡ e2/h̄c ' 1/137.
1. Bohr levels:
En = −R/n2
2. Fine structure:
2 2
− Rnα4
µ
n
j+1/2
−
O(α2mc2)
3
4
¶
O(α4mc2)
(relativistic effects–kinetic energy & spin-orbit coupling)
3. Hyperfine structure:
µ0 ge2
hSp
3πmp ma30
· Se i
O(m/mp)(α4mc2)
(spin-spin coupling of e− & p+ magnetic moments)
4. Lamb shift:
1 R
3π n2
µ
¶
2α 3
mc2
ln
n
(|Em −En |)avg
O(α5mc2)
(quantization of photon field–really larger than hyperfine)
4.1
Fine Structure—spin-orbit coupling
1st effect of interest is coupling of i) spin of electron to ii) its motion, 2
things which we certainly treated independently before. Full derivation of
proper V̂ is hairy, give only hand-waving derivation here, misses factor of
2.
Force on electron is −eE = −e2r/(4π²0r3). Recall that observer moving
in electric field E sees magnetic field
1
v
×E
c2
So electron with magnetic moment
e
~µ = −g S,
2m
should have energy
B=
v¿c
(1)
g'2
(2)
V̂ = −µ · B
e
e
v
= −g S ·
r
×
2m 4π²0r3
c2
(3)
(4)
so with L ' r × mv, get


ge2  1 
L·S
V̂ =
2m2c2 4π²0r3
(5)
Off by factor of 2 from full relativistic calculation (neglected Thomas precession!):
1
ge2
L·S
V̂ =
4m2c2 4π²0r3
= f (r)L · S
(6)
(7)
(8)
Now define total angular momentum
J=L+S
(9)
Jˆ2 = L̂2 + Ŝ 2 + 2L · S,
(10)
whose square is
so
V̂ =
f (r) ˆ2
(J − L̂2 − Ŝ 2)
2
(11)
Verify: Jˆ2, L̂2, Ŝ 2, and Jˆz now commute with H = H0 + V̂ . L̂z and Ŝz
no longer do. We choose therefore to label states as
2
Thus we aren’t quite following prescription we set out, by looking at expectation value of V̂ in original eigenstates of H0. Why? Because we can
find the new exact eigenstates with no work. So energy shift calculation
is exact—except for higher order relativistic effects. In this basis we have
(Jˆ2 − L̂2 − Ŝ 2)|n`sjmj i
3
= h̄2(j(j + 1) − `(` + 1) − )|n`sjmj i
4
(12)
so energy shift due to spin-orbit coupling is
δEso = hn`sjmj |V̂ |n`sjmj i
3
= h̄2(j(j + 1) − `(` + 1) − )hf (r)i
4
For hf (r)i we need
1
1
h 3i =
r
`(` + 1/2)(` + 1)n3a30
(` 6= 0)
(13)
(14)
(see Griffiths Probs. 6.34, 6.35). Spin-orbit energy shift is therefore


δEso
En2  n(j(j + 1) − `(` + 1) − 43 ) 


=
(15)
mc2
`(` + 12 )(` + 1)


En2
1
1




=
2n
−
(check for j = ` ± 1/2!) (16)
1
1
mc2
`+ 2 j+ 2
Note although (15) looks singular at ` → 0, proper relativistic treatment
3
by Dirac shows nonsingular result (16) is correct for all `–not obvious
though!
Relativistic kinetic energy corrections. Up to now we have neglected relativistic effects in kinetic energy term T̂ . Properly we should probably
write
r
p̂2
p̂4
2
2
2
2
4
T̂ = p̂ c + m c − mc '
−
+ ...
(17)
2m 8m3c2
Then we need to find the energy shift in |ψi to leading order in the per4
turbation V̂ = 8mp̂3c2 :
1
1
δErel = − 3 2 hψ|p̂4|ψi = − 3 2 hp̂2ψ|p̂2ψi
(18)
8m c
8m c
The rhs supposed to be evaluated using unperturbed wave fctns, so use
(p̂2/2m − e2/(4π²0r))|ψi = En|ψi to find
*
*
+
1 +
1
1
2
2
4
]
(19)
[E + 2Ene
+e
δErel = −
2mc2 n
4π²0r
(4π²0)2r2
The expectation values depend only on a0, n and ` as usual:
* +
*
1
1
1
1+
=
= 2 ;
r
n a0
r2
(` + 1/2)n3a20
So full relativistic kinetic energy shift is


En2  4n



δErel = −
−
3
2mc2 ` + 1/2
(20)
(21)
and full fine structure energy splitting is
δEf ine = δEso + δErel
(22)
yielding an energy spectrum for hydrogen up to fine structure corrections
of



α2  n
R


E = En + δEf ine = − 2 1 + 2 
− 3/4
(23)
n
n j + 1/2
4
Figure 1: Hydrogen fine structure—spin-orbit coupling
4.2
Hyperfine Interaction
Hyperfine splitting of atomic spectral lines arises from interaction of electron & nuclear dipole moments. Whereas ground state in hydrogen not
split by fine structure, it is split by hyperfine interaction, allowing for
transition between two levels, producing radiation of wavelength 21 cm
(∼ 10−7Ryd, allows tracking of hydrogen by astronomers, measurements
of proper motion of celestial objects (Doppler shifts of line), & construction of stable atomic clocks. Calculation is lengthy, but can be solved by
elementary methods to high accuracy (“real” quantum mechanics problem!).
Magnetic field due to dipole moment µ:


1  3r(~µ · r) ~µ 

B=
− 3 ,
4π²0
r5
r
(24)
whereas the interaction of a magnetic moment with external field is
U = −~µ · B
5
(25)
Figure 2: If µ
~ p k ẑ, r k x̂, B k −ẑ, then µe k −µp in lowest energy configuration. If µ
~ p k ẑ, r k ẑ, B k ẑ, then µe k µp
for lowest energy. Compare with (25).
so interaction is


1  3(~µe · r)(~µp · r) (~µe · ~µp) 
−

+
V̂ =
4π²0
r5
r3
(26)
where r = re − rp
Magnetic moment operators:
~µe = −ge
e
Se ;
2Me
~µp = gp
e
Sp
2Mp
(27)
with gyromagnetic ratios ge ' 2.00, gp ' 5.59.
Perturbation theory
Want to use V̂ in (25) as perturbation, write H = H0 + Hf ine + V̂ , look
for hyperfine energy shift δEhf in ground state |ψ0i.1 Can write ground
state as
|ψ0i = |ψ100(r)i|me mpi
(28)
−r/a0
q
where 1st factor is spatial part, old friend ψ100 = e
/ πa30, 2nd factor
represents spin state of electron and now proton, too (before we ignored
proton spin since it didn’t contribute to energy, now we must include it as
1 Note we will focus exclusively on the ground state of the H-atom, where the spin-orbit coupling has already been accounted for.
Although in general states of definite mp , me are not eigenstates because of the L · S coupling, for the S1/2 ground state, mj (e’value of
Jˆz /h̄) is the same as ms (e’value of Ŝz /h̄) always since ` = 0. We call ms of electron me here, and ms for proton mp .
6
dynamical variable). Note there are four degenerate states before taking
V̂ into account, since both me and mp can be ±1/2. A priori we expect
that since V̂ commutes with Ŝ 2 = (Se + Sp)2, eigenstates of new H with
hyperfine term should be eigenstates of Ŝ 2, i.e. either S = 1 (triplet) or
S = 0 (singlet).2 21cm spectral line we’re looking for will be difference
in
√
shifts of 3 degenerate triplet√states [| ↑↑i, | ↓↓i, | ↑↓i + | ↓↑i)/ 2], and
singlet state | ↑↓i − | ↓↑i)/ 2.
So we want to evaluate3
δEhf = hψ00 |V̂ |ψ00 i
Ã
=
3(Se ·r)(Sp ·r)
ge gp e2
hψ
(r)|hSM
|
100
16π²0 Me Mp
r5
=
−
(Se ·Sp )
r3
(29)
!
|SM i|ψ100(r)i
ge gp e2
αβ
16π²0 Me Mp I hSM |Ŝeα Ŝpβ |SM i
(30)
where the |SM i are e’states of total spin Ŝ 2 and total Ŝz , and


αβ
Z
3rα rβ δ 

− 3
(31)
I αβ ≡ d3r|ψ100(r)|2 
5
r
r
is a tensor we have to evaluate using the wave function ψ100. Tedious but
straightforward to show that I αβ = Iδαβ , with I = 8π|ψ100(0)|2/3 (see
below!). I diagonal =⇒ only spin matrix elements we need are
1
hSM |Se · Sp|SM i = hSM | (Ŝ 2 − Ŝe2 − Ŝp2)|SM i
2

2
h̄  1(1+1)− 12 (1+ 12 )− 21 (1+ 12 ) S = 1
=

0− 12 (1+ 21 )− 12 (1+ 12 )
S=0
2 


 h̄2 /4
S=1
= 
 −3h̄2 /4 S = 0
(32)
So energy shifts for singlet and triplet states are
that Ŝ 2 commutes with V̂ by first writing Ŝ 2 = (Se + Sp )2 = Ŝe2 + Ŝp2 +
2
since [Ŝe , Ŝe,α ] = 0, similarly for Ŝp2 . Now Se · Sp obviously commutes with second
[Se · Sp , (Se · r)(Sp · r)] = 0. To see this, use relation σi σj = δij + i²ijk σk .
3 Need to use states ψ 0 > in which V̂ is diagonal—these are precisely the |SM i!
0
2 Check
7
Se · Sp . First of all, Ŝe2 commutes with V̂
term of V̂ (see (25)), harder part is to see
δEhf


2 
 h̄2 /4
S=1
 8π|ψ100 (0)| 


2

 −3h̄ /4 S = 0
3
k


3 6
8M
e
e



6 
3h̄
g e g p e2
=
16π²0MeMp

 1
gegpe8Me2 
S=1
=
 −3 S = 0
24π²0Mph̄4 
(33)
So splitting between triplet and singlet (energy of photon emitted in triplet
decay) is
h̄ω = δEhf (S = 1) − δEhf (S = 0)
gegpe8Me2
=
6π²0Mph̄4
(34)
and frequency of transition is
gegpe8Me2
ν = ω/2π =
12π 2²0Mph̄5
= 1420405751.800 ± 0.028Hz
(35)
(36)
or λ = c/ν = 21.1 cm. Value given is current measurement!
Pf. of I αβ = 8π|ψ100(0)|2δαβ /3
Recall definition
Ã
Z
I
αβ
≡
3
d r|ψ100 (r)|
2
3rα rβ δαβ
− 3
r5
r
!
Quantity I αβ is singular integral and must be handled carefully. Note
8
(37)
1. Off-diagonal terms vanish because in ` = 0 state answer must be invariant under rotation of
coordinate system, so I αβ = Iδαβ .
2. Diagonal terms funky, since angular part of, e.g. I zz gives
Z
2
2
dΩ (3z − r ) = 2πr
2
Z 1
−1
d cos θ(3 cos2 θ − 1) = 0
(38)
but radial integral is proportional to
Z
dr
e−r/a0
=∞
r
(39)
So we have to find way to take limit 0 · ∞ sensibly. Note singularity in (38) comes entirely
from r = 0; if we were to exclude r = 0 from range of integration integral would vanish!. So
consider very small sphere around r = 0, of radius r0 so small we can approximate smoothly
varying fctn. ψ100 (r) by ψ100 (0). Consider diagonal elt. I zz (others must be same by rotation
symmetry), express in cylindrical coords:
I
zz
= |ψ100 (0)|
= |ψ100 (0)|
2
0
2
= |ψ100 (0)|2
= |ψ100 (0)|2
=
Z r0
Z r0
−r0
Z r0
Ã
1
3z 2
−
dr
r5
r3
√
Z
2
2
!
3
dz
r0 −z
0
"
2
dρ π
"
3z 2
1
−
(ρ2 + z 2 )5/2 (ρ2 + z 2 )3/2
2
−2z 2
+ 2
dz π
2
2
3/2
(ρ + z )
(ρ + z 2 )1/2
−r0
Z r0
−r0
Ã
dz 2π
z2
1
−
r0 r0
8π
|ψ100 (0)|2
3
!
#
#ρ2 =r2 −z 2
0
ρ2 =0
(40)
9