P460 - real H atom1 The Real Hydrogen Atom Solve SE and in first order get (independent of L): can use perturbation theory to determine: magnetic effects

P460 - real H atom 1

The Real Hydrogen Atom• Solve SE and in first order get (independent of L):

• can use perturbation theory to determine:

magnetic effects (spin-orbit and hyperfine e-A)

relativistic corrections

• Also have Lamb shift due to electron “self-interaction”. 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....)

2

6.13

n

eVEn


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):

• 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

BSg

BF

Ldr

rdV

remcB

bs

)(112

EvB

rvAB

use

c

21

LSdr

dV

r

g

emcE bs

1

2

12


Spin-Orbit: Quantum Numbers • The spin-orbit coupling (L*S) causes ml and ms to

no longer be “good” quantum numbers

• spin-orbit interactions changes energy.

• 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

0,

,:

)(2

22

2

HLHLLH

LLionseigenfunctalso

EHandrVHif

zzz

z

m

commutedonotLLLas

LSLSLSLLSL

SLaHH

zyx

zzzyyxxz

,,0

,,


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)

• assume H wave function is ~eigenfunction of perturbed potential

)(

)(

2

0,

22221

22221

222

SLJSL

SLJSL

orSSLLJ

SLifSLJ

43

21

21

2

)1(,

:20

)1()1()1(2

sswithllj

valueslfor

sslljjSL


SL Expectation value • To determine the energy shift, also need the

expectation value of the radial terms using Laguerre polynomials

• put all the terms together to get spin-orbit energy shift. =0 if l=0

0)12)(1(

221*1

14

1

330

33

330

2

ldrr

needas

lllnarr

rre

drdV

r

)12)(1(

))1()1((3

432

0

llln

lljjEE

SL

n=2

L=0,1

J=3/2, L=1

J=1/2 L=0

J=1/2 L=1

with relativistic

j=3/2

j=1/2


Numerology • have

• but

• and so

300

2

22

2

42

1

a

e

cme

22

00

2

2

20

0

2

1

4

14

cmE

c

e

cmcma

e

ee

2

2

02

222

3

333

0

2

22300

2

22

22

2

4

12

42

1

Emc

cme

cma

e

cm ee


Spin Orbit energy shift • For 2P state. N=2, L=1, J= 3/2 or 1/2

• and so energy split between 2 levels is

L=1

J=3/2

j=1/2eVE

EEE

split

split

5105.4

1371

21

23

48

2

48

)2(

485218

)21(

)12)(1(

)1()1((

204

323

212

0

204

325

232

0

3432

0

21

23

EEE

EEE

llln

lljjEE

SL


Relativistic Effects

• Solved using non-relativistic S.E. can treat relativistic term (Krel) as a perturbation

• <V> can use virial theorem

23

4

23

42

24222

882 cm

pK

cm

p

m

p

mccmpcmET

rel

VEVEEEVEbut

VVEEK

VEK

ESEV

KEvEE

nn

mcrel

mcmp

mcrel

mp

relnn

,0,

)2(

)()(

.).(

22

22

21

2

212

221

2

2

2

2

2

2

nnaZe

rZe EV

o22

0

2

0

2

41

4


Relativistic+spin-orbit Effects• by integrating over the radial wave function

• combine spin-orbit and relativistic corrections

• energy levels depend on only n+j (!). Dirac equation gives directly (not as perturbation). For n=2 have:

)12(22

412

42

30

2

20

2

)()(

lna

Zer

Zeo

V

)( 83

1212

3

44

nlenZ

rel cmK

):()( 2

143

122

43

122

)12)(1()1()1(

3

20

43

3

20

ljuse

EE

njn

E

nlllllljj

n

E

relSL

8

5

8

3

12

2

8

1

8

3

12

2

21

23


Energy Levels in Hydrogen• Degeneracy = 2j+1

• spectroscopic notation: nLj with L=0 S=state, L=1 P-state, L=2 D-state

• also can note spin “doublet” is single electron with s=1/2

E

N=1

N=3

N=2

223,31,0

443,32,1

632

21

21

23

23

25

21

23

25

PSjl

DPjl

Djl

222,21,0

421

21

21

23

21

23

PSjl

Pjl

210212

1 Sjl

# states

21

2 S


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

• 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

......

)2(

2121 SSSandLLLwith

SLandSLJ b

)1(2

)1()1()1(1

jj

llssjjg

mBgBE jb

23

2P

23

21

21

23

j

j

j

j

m

m

m

mB=0 B>0


Zeeman Effect

L=+-1 m=0,+1,-1

strong field


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

212

21 sS

23

21

21

23

j

j

j

j

m

m

m

mB=0B>0

212 ,

23 sP

21

21

j

j

m

m

3

4

2

212

21

415

43

415

43

43

43

Ps gg

0n


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). G-factors 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)

• have added terms to energy. For S-states, L=0 and can ignore that term

Bp

epp

ppp m

mgI

g

58.5

eppnuc bLaE


Hyperfine Splitting• Electron spin couples to nuclear spin

• so energy difference between spins opposite and aligned. Gives 21 cm line for hydrogen (and is basis of NMR/MRI)

)2

3

2

122(

2,1

2

3

2

12

2,0

))1()1()1((2

2

2

2

ISfalignedspins

ISfoppositespins

iissffIS

SIFletISep

eVffE 6106)]0()1[(

Documents

P460 - real H atom1 The Real Hydrogen Atom Solve SE and in first order get (independent of L): can use perturbation theory to determine: magnetic effects