The Real Hydrogen Atom
• Solve SE and in first order get (independent of L):
− 13.6eV
En =
n2
• can use perturbation theory to determine:
magnetic effects (spin-orbit and hyperfine e-A)
relativistic corrections
• Also have Lamb shift due to electron “selfinteraction”. Need QED (Dirac eq.) and depends on
H wavefunction at r=0 (source of electric field).
Very small and skip in this course (first calculated
by Bethe using perturbation theory on train from
Long Island to Ithaca. Bethe also in film Fat Man
and Little Boy....)
P460 - real H atom
1
Spin-Orbit Interactions
• A non-zero orbital angular momentum L produces
a magnetic field
• electron sees it. Its magnetic moment interacts
giving energy shift
• in rest frame of electron, B field is (see book/ER):
r
B=
1 1 dV (r ) r
L
2
emc r dr
r r − g s µb r r
∆F = − µ • B =
S•B
h
use
r
r r
 B = Av × r 
r
r
r
 B = −21 v × E 
c


• convert back to lab frame (Thomas precession due
to non-inertial frame gives a factor of 2 – Dirac eq
gives directly). Energy depends on spin-orbit
coupling
1 g s µ b 1 dV r r
∆E =
S•L
2
2emc h r dr
P460 - real H atom
2
Spin-Orbit: Quantum Numbers
• The spin-orbit coupling (L*S) causes ml and ms to
no longer be “good” quantum numbers
if
H=
−h2
2m
∇ 2 + V (r ) and Hψ = Eψ
ψ also eigenfunctions : Lz , L2
[H , Lz ] = HLz − Lz H = 0
• spin-orbit interactions changes energy.
r r
H ′ = H + aL • S
r r
L • S , L z = L x S x + L y S y + Lz S z , L z
[
] [
≠ 0 as
L x , L y , Lz
]
donot commute
• In atomic physics, small perturbation, and can still
use H spatial and spin wave function as very good
starting point. Large effects in nuclear physics (and
will see energy ordering very different due to
couplings). So in atomic just need expectation
value of additional interaction
P460 - real H atom
3
SL Expectation value
• Determine expectation value of the spin-orbit
interaction using perturbation theory. Assume J,L,S
are all “good” quantum numbers (which isn’t true)
r r r
J = L + S if [L, S ] = 0 ⇒
r r
2
2
J = L + 2 L • S + S 2 or
r r
L • S = 12 ( J 2 − L2 − S 2 )
r r
< L • S >= 12 (< J 2 > − < L2 > − < S 2 >)
• assume H wave function is ~eigenfunction of
perturbed potential
r r h2
L • S ≅ ( j ( j + 1) − l (l + 1) − s ( s + 1) )
2
for l ≠ 0 2 values :
j = l + 12 , l − 12
with s ( s + 1) =
P460 - real H atom
3
4
4
SL Expectation value
• To determine the energy shift, also need the
expectation value of the radial terms using
Laguerre polynomials
1 dV
r dr
as
1
r3
=
e2
need
4 πε 0 r 3
= ∫ψ ψ
*
r dr =
1 2
r3
1
r3
2
a 03n 3l ( l +1)( 2 l +1)
l≠0
• put all the terms together to get spin-orbit energy
shift. =0 if l=0
∆E
SL
E0α 2 ( j ( j + 1) − l (l + 1) − 34 )
=
n 3l (l + 1)(2l + 1)
J=3/2, L=1
n=2
J=1/2 L=0
L=0,1
J=1/2 L=1
with relativistic
j=3/2
j=1/2
P460 - real H atom
5
Numerology
• have
• but
1
e2 2
2 me2 c 2 4πε 0 a03
4πε 0h 2 1 h
a0 =
=
2
me c
α me c
e2
1
α=
4πε 0 hc
me c 2 2
E0 = −
α
2
• and so
1
1
e2 2
e 2 α 3m 3c 3
= 2 2
2 2
3
2 me c 4πε 0 a0 me c 4πε 0 h 3
2 mc 2α 2 α 2
α2
=
= 2 E0 2
2
2
h
h
P460 - real H atom
6
Spin Orbit energy shift
• For 2P state. N=2, L=1, J= 3/2 or 1/2
∆E
∆E
∆E
SL
3
1
2
2
E0α 2 ( j ( j + 1) − l (l + 1) − 34
=
n 3l (l + 1)(2l + 1)
E0α 2 ( 32 52 − 1 • 2 − 34 ) E0α 2
=
=
8 •1 • 2 • 5
48
E0α 2 ( 12 32 − 2 − 34 ) − 2 E0α 2
=
=
48
48
• and so energy split between 2 levels is
α = 1137
J=3/2
L=1
j=1/2
Esplit = ∆E
3
− ∆E
2
1
Esplit = 4.5 ×10 −5 eV
P460 - real H atom
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2
Relativistic Effects
T = E − m = c 2 p 2 + m 2 c 4 − mc 2
p2
p4
p4
≈
− 3 2 ⇒ K rel = − 3 2
2 m 8m c
8m c
• Solved using non-relativistic S.E. can treat
relativistic term (Krel) as a perturbation
E ′ = E + vnn = E + K rel
(
p2
2m
)
+ V ψ = Eψ
K rel =
−1
2 mc 2
( S .E.)
p2 2
2m
( ) =
−1
2 mc 2
(E − V )2 ⇒
K rel = − 2 mc1 2 ( E 2 − 2 VE + V 2 )
but [E , V ] = 0 ⇒ E 2 = En2 , VE = En V
• <V> can use virial theorem
V =−
Ze2 1
4πε0 r
=−
Ze2
4πε0aon2
P460 - real H atom
= 2En
8
Relativistic+spin-orbit Effects
• by integrating over the radial wave function
V =(
2
=(
Ze2 2 1
4πε0
r2
)
⇒ K rel = −
Z 4α 4
n3
Ze2 2
2
4πε0ao
n3 (2l+1)
)
me c ( 2 l1+1 − 83n )
2
• combine spin-orbit and relativistic corrections
∆E
SL
=
+ ∆E
E0α 2
n3
=−
[
j ( j +1) −l ( l +1) − 3 4
l ( l +1)( 2 l +1)
E0α 2
n
rel
3
− 2 l2+1 + 43n
]
( 2 j2+1 − 43n ) (use : j = l ± 12 )
• energy levels depend on only n+j (!). Dirac
equation gives directly (not as perturbation). For
n=2 have:
2
3 1
2
3 5
−
2 +1 8
3
2
=
8
−
2 +1 8
P460 - real H atom
1
2
=
8
9
Energy Levels in Hydrogen
• Degeneracy = 2j+1
• spectroscopic notation: nLj with L=0 S=state,
L=1 P-state, L=2 D-state
# states
N=3
l=2 j=
5
2
l = 1, 2 j =
l = 0 ,1 j =
E
N=2
l =1 j =
⇒ 3D 5
3
2
⇒ 3 P3 ,3 D 3
1
2
⇒ 3 S 1 ,3 P1
3
2
⇒ 2 P3
l = 0 ,1 j =
N=1
6
2
2
4+4
2
2
2+2
2
4
2
1
2
⇒ 2 S 1 , 2 P1
2
l = 0 j = 12 ⇒1S1
2+2
2
2
2
• also can note spin “doublet” is single electron with
s=1/2
2
S1
2
P460 - real H atom
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Zeeman Effect:External B Field
• Energy shift depends on mj and removes any
remaining degeneracy. Now two fields (internal
and external) and details of splitting depends on
relative strengths
r
r r r
r
µb r
J ′ = L′ + S ′ and µ = − ( L′ + 2 S ′)
h
r r r
r r r
with L′ = L1 + L2 ... and S ′ = S1 + S 2 ...
• Unless S=0, the magnetic moment and the total
angular momentum are not in the same direction
(and aren’t in B direction). For weak external field,
manipulating the dot products gives
r
∆E = − µ • B = − µ b Bgm′j
r
j ′( j ′ + 1) + s′( s′ + 1) − l ′(l ′ + 1)
g = 1+
2 j ′( j ′ + 1)
B=0
2
B>0
m j = 32
mj =
P3
1
2
m j = − 12
2
P460 - real H atom
m j = − 32
11
Zeeman Effect
strong field
∆L=+-1
∆m=0,+1,-1
P460 - real H atom
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Zeeman Effect:External B Field
• Assume that weak B field (if strong then L and S
won’t couple)
• B field off 1 photon energy
B field on 6 photon energies (with their energy
depending on the g factor and on the B field
• One of the first indicators that the electron had
intrinsic angular momentum s=1/2
∆n = 0
+ 43
gs = 1 + 3 = 2
24
3
4
B=0
2
P3 , s =
2
2
S1 s =
2
gP = 1 +
B>0
15
4
+ 43 − 2 4
=
15
24
3
m j = 32
mj =
1
2
m j = − 12
1
2
m j = − 32
mj =
1
2
P460 - real H atom
1
2
m j = − 12
13
Hyperfine Splitting
• Many nuclei also have spin
• p,n have S=1/2. Made from 3 S=1/2 quarks (plus
additional quarks and antiquarks and gluons). Gfactors are 5.58 and -3.8 from this (-2 for electron).
• Nuclear g-factors/magnetic moments complicated.
Usually just use experimental number
• for Hydrogen. Let I be the nuclear spin (1/2)
g pµ p r
I
µp =
h
r
me
g p = 5.58 µ p =
µB
mp
• have added terms to energy. For S-states, L=0 and
can ignore that term
∆Enuc
r
r r
= aµ p • L + bµ p • µ e
r
P460 - real H atom
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Hyperfine Splitting
• Electron spin couples to nuclear spin
r r
r r r
r
r
µ p • µ e ∝ S • I let F = I + S
r r
h2
S•I =
( f ( f + 1) − s ( s + 1) − i ( i + 1))
2
r r
h2 1 3
spins opposite ⇒ f = 0 , S • I = −
2
2 2 2
r r
h2
1 3
spins aligned ⇒ f = 1, S • I =
(2 − 2
)
2
2 2
• so energy difference between spins opposite and
aligned. Gives 21 cm line for hydrogen (and is
basis of NMR/MRI)
∆E[( f = 1) − ( f = 0)] ≈ 6 ×10 −6 eV
P460 - real H atom
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