1

Real hydrogen atom Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton (Date: February 04, 2016)

Willis Eugene Lamb, Jr. (July 12, 1913 – May 15, 2008) was an American physicist who won the Nobel Prize in Physics in 1955 together with Polykarp Kusch "for his discoveries concerning the fine structure of the hydrogen spectrum". Lamb and Kusch were able to precisely determine certain electromagnetic properties of the electron (see Lamb shift). Lamb was a professor at the University of Arizona College of Optical Sciences.

http://en.wikipedia.org/wiki/Willis_Lamb

Fig. F

E The correThe enerwith the u

To the or

E

To the or

E

E

To the or

rom the noteE. Fermi, Not

ections to thrgy of the eluse of the pe

rder of 2 :

)0(

2

1mE en

rder of 4 :

)0()1( EE nn

)1( EE nn

rder of 5 :

e on quantumte on Quantu

he Bohr energectron in hyerturbation th

2

22

nce

,

2/1

1(

12

jn

1(12)0(

n

m mechanicsum Mechani

gy are calledydrogen can heory.

)4

3

n

)4

3

n

2

s (E. Fermi).ics (The Univ

d the fine strbe expresse

for 0l

for 0l

. p.117. versity of Ch

ructure. Theyed in terms o

hicago, 1961

y are all relaof the fine st

1).

ativistic in natructure con

ature. nstant,

3

the expression of the frequency for the Lamb shift (the non-relativistic case) (for n = 2), can be expressed as

)9.8

1ln(

12 22

35

c

cme

= 1038.27 MHz ≈ 1057 MHz (experimental value).

___________________________________________________________________________ 1. Bohr model

According to the Bohr model, the electron energy of the hydrogen atom is given by

2

22

20

2

2)0(

2

1

2 ncm

n

E

an

eE e

Bn

,

nlmEnlme

mnlmH n

e

)0(22

0 )ˆ2

ˆ(ˆ

r

p,

where

2222

42

2

42

0 2

1

222cm

c

ecmem

a

eE e

ee

B

E0 =13.60569253 eV

and the fine structure constant is

036.137

12

c

e

=7.2973525698 x 10-3

The Bohr radius is given by

2

2

ema

eB

= 0.5291772109217 Å

The velocity of the electron in the n-th state:

036.137

122 c

nn

c

n

c

c

e

n

evn

4

n: principal quantum number l: Azimuthal quantum number m: magnetic quantum number

((Note)) Fine structure constant (=c

e

2

)

In physics, the fine-structure constant, also known as Sommerfeld's constant, commonly denoted α (the Greek letter alpha), is a fundamental physical constant characterizing the strength of the electromagnetic interaction between elementary charged particles. It is related to the elementary charge (the electromagnetic coupling constant) e, which characterizes the strength of

the coupling of an elementary charged particle with the electromagnetic field, by the formula

(= )/(2 ce ). Being a dimensionless quantity, it has the same numerical value in all systems of

units. Arnold Sommerfeld introduced the fine-structure constant in 1916. https://en.wikipedia.org/wiki/Fine-structure_constant 2. Hydrogen fine structure

The Schrödinger solution gives a very good description of the hydrogen atom. The motion of electrons is treated as a non-relativistic particle. In reality it is not. There are corrections to these values of the energy. From the Dirac’s relativistic electron theory, the approximation of the Hamiltonian can be derived as follows.

EpEσp

p 22

2

2222

42

8)(

482

1

cm

e

cm

e

cme

mH

eeee

,

where e>0. (i) Third term: relativistic correction

...8

1

2

...]8

1

21[

1

23

42

2

44

4

22

22

2

22

2222242

cmm

cmcmcm

cm

cmcm

cmcmccm

ee

e

ee

e

e

e

eee

pp

pp

pp

(ii) The fourth term (Thomas correction): spin-orbit interaction

5

Thomas term = )(4 22 pEσ

cm

e

e

.

For a central potential

r

erVre

2

)()( , or )(1

)( rVe

r ,

where is the scalar potential, -e is the charge of electron (e>0 here).

reEdr

dV

erdr

dV

erV

e r

11)(

1 ,

LprpEdr

dV

erdr

dV

er

1)(

1 ,

where L is an orbital angular momentum. Then the Thomas term is rewritten as

LS

LS

LσpEσ

322

2

22

2222

1

2

1

2

1

)1

(4

)(4

rcm

e

dr

dV

rcm

dr

dV

ercm

e

cm

e

e

e

ee

(Spin-orbit interaction)

The spin angular momentum is defined by

σS2

,

which is an automatic consequence of the Dirac theory. (iii) The last term is called the Darwin term.

The Darwin term changes the effective potential at the nucleus. It can be interpreted as a smearing out of the electrostatic interaction between the electron and nucleus due to zitterbewegung, or rapid quantum oscillation, of the electron. Sir Charles Galton Darwin (18 December 1887 – 31 December 1962) was an English physicist, the grandson of Charles Darwin.

6

The Darwin term = E22

2

8 cm

e

e

For a hydrogen atom,

)(411 22 rE er

eVe

.

It gives rise to an energy shift (the diagonal matrix element)

rrrrrr dcm

ed

cm

enlm

e

nlm

e

2

22

222

22

22

)()(2

)()(48

which is non-vanishing only for the s state. ______________________________________________________________________________

3. Relativistic correction

According to special relativity, the kinetic energy of an electron of mass m and velocity v is:

23

42

82 cmmK

ee

pp

.

The first term is the standard non-relativistic expression for kinetic energy. The second term is

the lowest-order relativistic correction to this energy. We apply the perturbation theory for this

system.

23

422

08

ˆ)

ˆ2

ˆ(ˆˆˆ

cm

e

mHHH

eeR

p

r

p

where

nlmEnlmH n)0(

0ˆ ,

and

7

r

pˆ2

ˆˆ22

0

e

mH

e

,

23

4

8

ˆˆcm

He

R

p

Noting that

)ˆ

ˆ(2ˆ2

02

rp

eHme

we calculate the matrix element

]11

2[4

)ˆ

ˆ)(ˆ

ˆ(4ˆ

24)0(22)0(2

2

0

2

024

avavnne

e

re

rEeEm

nlme

He

Hnlmmnlmnlm

rr

p

((Comment))

A. Das, Lectures on Quantum Mechanics 2nd edition (World Scientific, 2003)

As a result, even though the energy levels are degenerate, we can still apply non-degenerate

perturbation theory, since the potentially dangerous terms are zero because the numerator

vanishes.

Then we have the first order correction to the energy eigenvalue

]11

2[2

1

ˆ8

1ˆ

24)0(22)0(

2

4

23

avavnn

e

e

R

re

rEeE

cm

nlmnlmcm

nlmHnlm

p

Here we use

Bav anr 2

11

,

8

)2

1(

1123

2

lanrB

av

and

2

22)0(

2

1

ncmE en

,

where

2

2

ema

eB

(Bohr radius),

c

e

2

(fine structure constant).

Then we find that

)4

3

2

11

(1

)32/1

4(

2

)32/1

4(

8

)4

3

2/1

1(

2

]4

3

)2/1(

1[

2

1

ˆ

)0(2

2

2)0(

4

42

3

42

4342

)1(

nlnE

l

n

cm

E

l

n

n

cm

nln

cm

nlncm

nlmHnlmE

n

e

n

e

e

e

reln

where

4

4422)0(

4

1

ncmE en

.

2

22)0(

2

1

ncmE en

((Calculation of )1(nE )) Hydrogen

9

)1(

nE

1s (l = 0) -7.30456 cm-1

________________________________________

2s (l = 0) -1.18699 cm-1

2p (l = 1) -0.21305 cm-1

________________________________________

3s (l = 0) -0.378755 cm-1

3p (l = 1) -0.0901798 cm-1

3d (l = 2) -0.0324647

________________________________________

4s (l = 0) -0,165494 cm-1

4p (l = 1) -0.0437513 cm-1

4d (l = 2) -0.094027 cm-1

4f (l = 3) -0.00896766 cm-1

((Note)) Commutation relations

23

4

8

ˆˆcm

He

R

p

We note that

0]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆ[ 222 zyx LLL ppp

0]ˆ,ˆ[ 22 Lp

since the Hamiltonian of free particle e

free mH

2

ˆˆ2p

is invariant under the rotation of the system,

Then we have

0]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆ[ zfreeyfreexfree LHLHLH .

10

0]ˆ,ˆ[ 2 LfreeH

These relations lead to the commutation relations

0]ˆ,ˆ[ 24 Lp , 0]ˆ,ˆ[ 4 zLp

since

0ˆ]ˆ,ˆ[]ˆ,ˆ[ˆ]ˆ,ˆ[ 22222224 pLpLppLp

0ˆ]ˆ,ˆ[]ˆ,ˆ[ˆ]ˆ,ˆ[ 22224 ppppp zzz LLL

Thus we have

0]ˆ,ˆ[ 2 LRH , 0]ˆ,ˆ[ zR LH

Since

0]ˆ,ˆ[ 20 LH , 0]ˆ,ˆ[ 0 zLH

with

r

pˆ2

ˆˆ22

0

e

mH

e

We note that

0]ˆ,ˆ[ 2 LH , 0]ˆ,ˆ[ zLH

but

0]ˆ,ˆ[ HHR

((Perturbation theory-non-degenerate case))

11

is a simultaneous eigenket of 2ˆ ,ˆ LH , and zL ;

EH ˆ , )1(ˆ 22 llL , mLz ˆ

We apply the perturbation theory for the non-degenerate case;

...)...)((...))(ˆˆ( )1()0()1()0()1()0(0 nnnnnnR EEHH

with

nlmn )0(

0-th order

)0()0()0(0

ˆnnn EH

-th order

)0()1()1()0()1(0

)0( ˆˆnnnnnnR EEHH

Multiplying )0(n from the left-side of this equation

)1()1()0()0()1(

0)0()0()0( ˆˆ

nnnnnnnRn EEHH

or

)4

3

2

11

(1

1[

ˆ

ˆ

2)0(

)0()0()1(

nlnE

nlmHnlm

HE

n

R

nRnn

The energy depends on both n and l.

12

Fig. Effect of perturbation of RH . nlm is the simultaneous eigenket of { }ˆ ,ˆ ,ˆ 20 zLH L . We

note that 0]ˆ,ˆ[ 2 LH , 0]ˆ,ˆ[ zLH , where RHHH ˆˆˆ0 . Note that. 0]ˆ,ˆ[ 0 RHH . The

quantum numbers l and m remain unchanged when RH is added to 0H . RH is the

perturbation. nlm is the unperturbed eigenstate.

((Mathematica)) Commutation relation Proof of

0]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆ[ 444 zyx LLL ppp , 0]ˆ,ˆ[ 24 Lp , 0]ˆ,ˆ[ 04 Hp

n,l,m

H0,L2,Lz

H0 HR

HR,L2,Lz

13

Clear "Global`" ; ux 1, 0, 0 ; uy 0, 1, 0 ; uz 0, 0, 1 ;

r x, y, z ; p : Grad , x, y, z , "Cartesian" & Simplify;

px : ux.p &; py : uy.p &; pz : uz.p &;

L : Cross r, Grad , x, y, z , "Cartesian" & Simplify;

Lx : ux.L &; Ly : uy.L &; Lz : uz.L &;

Lsq : Lx Lx Ly Ly Lz Lz &;

Psq : px px py py pz pz &; Pf : Psq Psq &;

H1 :1

2 mPsq

q2

x2 y2 z2 1 2&;

Pf Lz x, y, z Lz Pf x, y, z Simplify

0

Pf Lx x, y, z Lx Pf x, y, z Simplify

0

Pf Ly x, y, z Ly Pf x, y, z Simplify

0

Pf Lsq x, y, z Lsq Pf x, y, z Simplify

0

Pf H1 x, y, z H1 Pf x, y, z Simplify

14

_____________________________________________________________________________ 4. Spin-orbit coupling (see the detail for the section of spin-orbit interaction)

We introduce a new Hamiltonian given by

soHHH ˆˆˆ0 ,

The total angular momentum J is the addition of the orbital angular momentum and the spin angular momentum,

SLJ ˆˆˆ , The spin-orbit interaction is defined by

)ˆˆˆ(2

1ˆˆˆ 222 SLJSL soH .

where

LLL ˆˆˆ i , SSS ˆˆˆ i

The unperturbed Hamiltonian ˆ H 0 commutes with all the components of L and S .

1

x2 y2 z2 5 2

4 q2 4 x2 y2 2 z2 0,0,2 x, y, z z x2 y2 z2 0,0,3 x, y, z

6 y z 0,1,1 x, y, z x2 y 0,1,2 x, y, z y3 0,1,2 x, y, z

y z2 0,1,2 x, y, z x2 0,2,0 x, y, z 2 y2 0,2,0 x, y, z

z2 0,2,0 x, y, z x2 z 0,2,1 x, y, z y2 z 0,2,1 x, y, z

z3 0,2,1 x, y, z x2 y 0,3,0 x, y, z y3 0,3,0 x, y, z

y z2 0,3,0 x, y, z 6 x z 1,0,1 x, y, z x3 1,0,2 x, y, z

x y2 1,0,2 x, y, z x z2 1,0,2 x, y, z 6 x y 1,1,0 x, y, z

x3 1,2,0 x, y, z x y2 1,2,0 x, y, z x z2 1,2,0 x, y, z

2 x2 2,0,0 x, y, z y2 2,0,0 x, y, z z2 2,0,0 x, y, z

x2 z 2,0,1 x, y, z y2 z 2,0,1 x, y, z z3 2,0,1 x, y, z

x2 y 2,1,0 x, y, z y3 2,1,0 x, y, z y z2 2,1,0 x, y, z

x3 3,0,0 x, y, z x y2 3,0,0 x, y, z x z2 3,0,0 x, y, z

15

0]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆ[ 02

02

0 zz LHLHH L ,

0]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆ[ 02

02

0 zz SHSHH S ,

and

0]ˆˆ,ˆ[]ˆ,ˆ[ 00 zzz SLHJH

We note that

0

)]ˆˆˆˆˆˆ(,ˆ[

]ˆˆ,ˆ[]ˆ,ˆ[

0

00

zzyyxx

so

SLSLSLH

HHH

SL

Then we have

0]ˆˆˆ,ˆ[ 2220 SLJH

or

0]ˆ,ˆ[ 20 JH

We also note that

0]ˆ,ˆˆ[2]ˆ,ˆˆ2ˆˆ[]ˆ,ˆ[ 222222 LSLLSLSLLJ

0]ˆ,ˆˆ[2]ˆ,ˆˆ2ˆˆ[]ˆ,ˆ[ 222222 SSLSSLSLSJ

0

ˆˆ2ˆˆ2ˆˆ2ˆˆ2

]ˆ,ˆ[ˆ2]ˆ,ˆ[ˆ2ˆ]ˆ,ˆ[2ˆ]ˆ,ˆ[2

0]ˆˆ,ˆˆ[2

]ˆˆ,ˆˆ2ˆˆ[]ˆ,ˆ[ 222

xyyxyxxy

zyyxzxyzyxxz

zz

zzz

SLiSLiSLiSLi

SSLSSLSLLSLL

SL

SLJ

SL

SLSLJ

16

Thus we conclude that is the simultaneous eigenket of the mutually commuting

observables{ ˆ H 0 , 2L , 2S , 2J , and ˆ J z }.

)0(0

ˆnEH

)1(ˆ 22 llL

)1(ˆ 22 ssS

)1(ˆ 22 jjJ

mJ z ˆ .

Fig. Perturbation due to the spin-orbit interaction. mjsln ,,,, is the simultaneous eigenkets

of ˆ H 0 , 2L , 2S , 2J , and ˆ J z without spin-orbit interaction. When the spin-orbit interaction

is switched on, SL ˆˆ mixes states of same mj, and different lm and sm . 0]ˆ,ˆ[ 2 JH .

0]ˆ,ˆ[ zJH . H is the total Hamiltonian.

n,l,s, j,m

H0,L2,S2,J2,Jz

S L

S L,J2,Jz

17

The eigenket can be described by

slmj ,;,

Note that the expression of the state can be formulated using the Clebsch-Gordan coefficient (which will be discussed later). The value of j is related to l and s (=1/2) as

2

1 lslj ,

2

1 lslj .

When the spin orbit interaction is the perturbation Hamiltonian, we can apply the degenerate theory for the perturbation theory.

slmjE

slmjsslljj

slmjslmjH

so

so

,;,

,;,)1()1()1([2

,;,)ˆˆˆ(2

,;,ˆ

)1(

2

222

SLJ

where

)]1()1()1([2

2)1( sslljjEso

with s = 1/2. Here we note that

32

42

3322

2

322

2

)1)(2/1(

1

2

1

)1)(2/1(

1

2

1

2

nlllcm

anlllcm

e

rcm

e

e

Be

ave

where

333 )1)(2/1(

11

Bav anlllr .

18

with

2

2

ema

eB

,

c

e

2

Then we have

)1)(2/1(2

]4

3)1()1([

2

1

)1)(2/1(

)]1()1()1([

4

1

)1()1()1([2

3

42

342

2)1(

lll

lljj

ncm

llln

sslljjcm

sslljjE

e

e

so

When 2

1 lj

)1)(2/1(2

)1)(2/1(22

1

2)0(

3

42)1(

llnl

lE

lll

l

ncmE

n

eso

When 2

1 lj

)1)(2/1(

1

2

)1)(2/1(2

)1(

2

1

2)0(

3

42)1(

llnl

lE

lll

l

ncmE

n

eso

where

2

22)0(

2

1

ncmE en

.

19

4. Darwin term

The additional perturbation of the Darwin term is given by

)]ˆ(,ˆ[,ˆ[8

1ˆ22 rpp V

cmH

e

D

The first-order energy shift due to the Darwin term is

Spin-orbit interaction

l = 0

l¥1

j = l+12

j = l-12

D2=d l

D1=d l+1

20

)()(8

)()]()[(8

1

)()](,[,)[(8

1

)()](,[,)[(8

1

,,'')]ˆ(,ˆ[,ˆ[,,'8

1

,,)]ˆ(,ˆ[,ˆ[,,8

1

,,ˆ,,

22

22

2

22*

22

*

22

*

22

22

22

)1(

rrr

rrrr

rpprr

rpprr

rrrpprrrr

rpp

Vdcm

Vdcm

rVdcm

rVdcm

mlnVmlnddcm

mlnVmlncm

mlnHmlnE

nlm

e

nlmnlm

e

nlmnlm

e

nlmnlm

e

e

e

DD

The potential energy is given by the Coulomb energy

r

eV

2

)( r .

Noting that

)(412 rr

,

we get

2

22

22

2

22

22

22

22

22)1(

)0(2

)()(2

)1

()(8

r

rrr

rr

nlm

e

nlm

e

nlm

e

D

cm

e

dcm

e

rd

cm

eE

We note that only l = 0 states are nonzero of the wave function at the origin. This implies that the

Darwin term contributes only for the s states.

21

43

2

3322

22

200

2

022

22

2

0022

22

)1(

2

4

8

),()0(2

)0(2

0,0,ˆ0,0,

n

cm

ancm

e

YrRcm

e

cm

e

nHnE

e

Be

n

e

n

e

DD

r

where

2

1),(0

0 Y ,

33

2

0

4)0(

Bn an

rR ,

and

2

2

ema

eB

.

c

e

2

((Mathematica)) Proof

The proof of the formula by using the Mathematica.

)()()()]}(,[,[)](,[,[)](,[,{[ 22 rrrrrr VVppVppVpp zzyyxx

where

xipx

, yi

py

, zi

pz

,

22

In the following mathematica, we use the unit of ħ = 1.

5. Summary The sum of the relativistic correction, the spin coupling for

0l is given by

)4

3

2/1

1(

1

]4

3

2/1

1[

2

1

])2/1)(1(2

4

3)1()1(

4

3

2/1

1[

2

1

)1)(2/1(

]4

3)1()1([

4

)4

3

2/1

1(

2

2)0(

3

42

3

42

23

4

23

4

)1()1()1(

njnE

njncm

lll

lljj

nlncm

lll

lljjcm

n

nlcm

n

EEE

n

e

e

e

e

soreln

where

2

1 lj

Clear"Global`";

SetCoordinatesCartesianx, y, z;

ux 1, 0, 0; uy 0, 1, 0; uz 0, 0, 1;

Px : ux.Grad & Simplify;

Py : uy.Grad & Simplify;

Pz : uz.Grad & Simplify;

f1 PxPxVx, y, z x, y, z PxVx, y, z Px x, y, z PyPyVx, y, z x, y, z PyVx, y, z Pyx, y, z PzPzVx, y, z x, y, z PzVx, y, z Pzx, y, z PxVx, y, z Pxx, y, z Vx, y, z PxPxx, y, z

PyVx, y, z Pyx, y, z Vx, y, z PyPyx, y, z PzVx, y, z Pzx, y, z Vx, y, z PzPzx, y, z

FullSimplify

x, y, z V0,0,2x, y, z V0,2,0x, y, z V2,0,0x, y, z

23

This result is intriguing in the sense that although both the spin-orbit interaction and the relativistic corrections individually remove the l-degeneracy in the hydrogen levels, when they are combined the total shift depends only on the quantum number j.

For l = 0, the spin-orbit contribution is zero and instead of it, Darwin term is added to

)1(nE ,

)4

31(

1

)4

31(

2

1

2

1)

4

32(

2

1

2)0(

3

42

3

42

3

42

)1()1()1(

nnE

nncm

ncm

nncm

EEE

n

e

ee

Dreln

6. Numerical calculation

(a) n = 2

l = 1, ml= 1, 0, -1

s= 1/2 ms = 1/2, -1/2

We note that

2/12/32/11 DDDD

Then we have

j = 3/2 (m = 3/2, 1/2, -1/2, -3/2), 22P3/2

j = 1/2 (m = 1/2, -1/2), 22P1/2

(b) n = 2

l = 0, ml= 0

s= 1/2 ms = 1/2, -1/2

24

We note that

2/12/10 DDD

Then we have

j = 1/2 (m = 1/2, -1/2), 22S1/2

We apply the above formula to the n = 2 levels of hydrogen.

)0(2

2)0(2

22/3

2

16

1)

2

1

2

1

4

1

4

3()2( EEPE

)0(2

2)0(2

22/1

2

16

5)1

2

1

4

1

4

3()2( EEPE

)0(2

2)0(2

22/1

2

16

13)1

4

1

4

3()2( EESE

When the Darwin term for l = 0 is taken into account,

)0(2

2)0(2

22/1

2

16

5)1

2

1

4

1

4

3()2( EEPE

The energy levels for the n = 2 (one electron) are as follows.

25

Fig. Hydrogen fine structure in the n = 2 level.

____________________________________________________________________________ 7. Exact solution from Dirac relativistic theory

The exact fine-structure formula for hydrogen (obtained from the Dirac equation without recourse to the perturbation theory) is

}1])

)2

1()2(

(1{[),( 2/12

22

2

jjn

cmjnE e .

Note that the energy depends only on n and j.

H so+H relH0 HDarwin

n=2

-11.320meV 22 P32

-56.603meV 22 P12

-147.17 meV

22S12

D1=45.283 meV

D2=90.565 meV

-3.4014 eV0 meV

Fig. E

coqu

Energy levels

omponent truantum num

s of the hydr

ransition of Hmbers n and j

rogen atom a

H line. Notej.

26

according to

e that each e

the Dirac th

energy level

heory showin

depends onl

ng the

ly on the

27

Fig. Fine structure of n = 2 levels of hydrogen atom according to the Dirac theory. The dotted line indicates the position of the n = 2 level according to the Bohr theory.

((Series expansion))

)4

3

2

11

(

.......)4

3

2/1

1(

2

),(),(

2)0(

3

42

)0(

njnE

njn

cm

EjnEjnE

n

e

n

((Numerical values))

n j E(n,j) [eV] 1 1/2 -181.135 ________________________________________________________ 2 1/2 -56.6048 2 3/2 -11.3207 ________________________________________________________ 3 1/2 -20.1261 3 3/2 -6.70853 3 5/2 -2.23616 ________________________________________________________ 4 1/2 -9.19826 4 3/2 -3.53779 4 5/2 -1.65099 4 7/2 -0.707533 ________________________________________________________ 5 1/2 -4.92689 5 3/2 -2.02867 5 5/2 -1.06263 5 7/2 -0.579613 5 9/2 -0.289824 __________________________________________________________

((Mathematica))

28

8. The Lamb Shift

The Lamb shift, named after Willis Lamb (1913–2008), is a small difference in energy

between two energy levels 2/12S and 2/1

2P (in term symbol notation) of the hydrogen atom in

quantum electrodynamics (QED). According to Dirac, the 2/12S and 2/1

2P orbitals should have

the same energies. However, the interaction between the electron and the vacuum causes a tiny

energy shift on 2/12S . Lamb and Robert Retherford measured this shift in 1947, and this

measurement provided the stimulus for renormalization theory to handle the divergences. It was the harbinger of modern quantum electrodynamics developed by Julian Schwinger, Richard Feynman, and Shin-ichiro Tomonaga. Lamb won the Nobel Prize in Physics in 1955 for his discoveries related to the Lamb shift. Theoretical and experimental values of the Lamb shift in MHz.

Theoretical 1057.70 ± 0.15 MHz Experimental 1057.77 ± 0.10 MHz

E1n_, j_ : me c2 1

n j 12 j 1

22

2

2 12

1 ;

SeriesE1n, j, , 0, 6 Simplify, j 0 &

c2 me 2

2 n2

c2 me 3 6 j 8 n 4

8 1 2 j n4

c2 me 5 1 2 j2 3 6 j 8 n 32 1 2 j2 n 24 n2 3 6 j 2 n 6

48 1 2 j3 n6 O7

29

4.372 eV 1057.14 MHz 0.0352626 cm-1

((Experiment)) Zeeman effect Landé g-factor:

)1(2

)1()1(

2

3

jj

llssg

For 2 2P3/2

BgmEmE B11)0(

111 )(

where

)0(1E = -11.320 eV

3

41 g ,

2

3,

2

1,

2

1,

2

31 m

For 2 2S1/2

BgmEmE B22)0(

222 )(

22S12

22 P12

4.372meV

30

where

)0(2E = -56.603 eV

22 g , 2

1,

2

12 m

For 2 2P1/2

BgmEmE B33)0(

333 )(

where

)0(3E = -56.603 - 4.372 eV

3

23 g ,

2

1,

2

13 m

Fig. The Zeeman effect. The energy levels of 2/32P (blue), 2/1

2S (red) and 2/12P (green). The

Lamb shift is taken into account. ((Experimental results)) W. Lamb, Novel lecture (1955).

500 1000 1500 2000 2500 3000BGauss

-80

-60

-40

-20

EmeV

Fig.

31

32

9. Stark effect in Lamb shift Problem 5-13 (Sakurai and Napolitano)

33

Sakurai 5-13

Compute the Stark effect for the 2 S1/2 and 2 P1/2 levels of hydrogen for an electric field sufficiently weak that 0ae is small compared to the fine structure, but take the Lamb shift ( (=

1,057 MHz) into account (that is, ignore 2 P3/2 in this calculation). Show that 0ae , the

energy shifts are quadratic in , whereas for 0ae , they are linear in . (The radial integral

you need is 03322 aprs ) Briefly discuss the consequences (if any) of time reversal for this

problem. This problem is from Gottfried 1966, Problem 7-3.

Fig. Energy level of real hydrogen atom. There is a small energy difference between between

2 2S1/2 and 2 2P1/2 (Lamb shift). In Dirac relativistic electron theory of hydrogen, the energy level depends only on the value of j.

H so+H relH0 HDarwin

n=2

-11.320meV 22 P32

-56.603meV 22 P12

-147.17 meV

22S12

D1=45.283 meV

D2=90.565 meV

-3.4014 eV0 meV

34

Fig. Lamb shift. There is an energy difference between 2 2S1/2 and 2 2P1/2. Lamb shift: the energy level of 2S1/2 (doubly degenerate states) is higher than that of 2P1/2

(doubly degenerate) by (= 4.372 x 10-6 eV = 1057 MHz). The perturbation (Stark effect)

zeH ˆˆ1

where e>0. This Hamiltonian is invariant under the time reversal. We now consider the state with n = 2.

n = 2 state (4 states degenerate) l = 1 (m = ±1, 0): p-state l = 0 (m = 0): s-state.

),()(cos),()(sinˆ'' ''

''2 m

lnlm

lnl YrRerYrRdddrrnlmzemnl

Addition of spin and orbital angular momentum (i)

l = 1, s = 1/2

22S12

22 P12

4.372meV

35

2/12/32/11 DDDD

2P3/2

j = 3/2 m = 3/2, 1/2, -1/2, and -3/2

2P1/2

j = 1/2 m = 1/2, -1/2

We use the Clebsch-Gordan co-efficient;

Fig. Clebsch-Gordan co-efficient. 2

1j . 1l .

2

1s .

zmlzmlmj ll 1,13

20,1

3

1

2

1,

2

123

and

ml

ms

101

1 2

1 2

2345

2 3

1 3

1 3

2 3

36

zmlzmlmj ll 0,13

11,1

3

2

2

1,

2

145

or

zmlnzmlnmP 1,1,23

20,1,2

3

12/1;2 2/1

zmlnzmlnmP 0,1,23

11,1,2

3

22/1,2 2/1

(ii)

l = 0, s = 1/2

2/12/10 DDD

2S1/2

j = 1/2 m = 1/2, -1/2

Then we have

zmlnmS 0,0,22/1,2 2/1

zmlnmS 0,0,22/1,2 2/1

We calculate the matrix elements of the perturbation.

02/1,2ˆ2/1,2 2/12/1 mSzmS

02/1,2ˆ2/1,2 2/12/1 mSzmS

37

0,0,2ˆ0,1,23

1

2/1,2ˆ2/1,2 2/12/1

mlnzmln

mSzmP

00,0,2ˆ1,1,23

2

2/1,2ˆ2/1,2 2/12/1

mlnzmln

mSzmP

_______________________________________________________________________

02/1,2ˆ2/1,2 2/12/1 mSzmS

02/1,2ˆ2/1,2 2/12/1 mSzmS

0

0,0,2ˆ1,1,23

2

2/1,2ˆ2/1,2 2/12/1

mlnzmln

mSzmP

0,0,2ˆ0,1,23

1

2/1,2ˆ2/1,2 2/12/1

mlnzmln

mSzmP

________________________________________________________________________

Note that the operator z is Hermitian. Then we have the matrix of )ˆ(ˆ1 zeH and Lamb shift

( H ).

0302/1,2

0032/1,2

30002/1,2

03002/1,2

2/1,22/1,22/1,22/1,2

02/1

02/1

02/1

02/1

2/12/12/12/1

aeemS

aeemS

aeemP

aeemP

mSmSmPmP

38

where is the Lamb shift energy

030,0,2ˆ0,1,2 amlnzmln

030,0,2ˆ0,1,23

1amlnzmln

030,0,2ˆ0,1,23

1amlnzmln

The eigenvalue problem We use the Mathematica

Energy eigenvalue and eigenket with 03 aep

2

4

2

22

1

pE

]2/1,22/1,22

4[

2

4(2

12/12/1

22

2

221

mSmPp

p

p

p

2

4

2

22

2

pE

]2/1,22/1,22

4[

2

4(2

12/12/1

22

2

222

mSmPp

p

p

p

2

4

2

22

3

pE

]2/1,22/1,22

4[

2

4(2

12/12/1

22

2

223

mSmPp

p

p

p

39

2

4

2

22

4

pE

]2/1,22/1,22

4[

2

4(2

12/12/1

22

2

223

mSmPp

p

p

p

The energy levels (E1=E2, E3=E4) are degenerate. The perturbation (Stark effect) zeH ˆˆ1 is

invariant under the time reversal. Since j = 1/2, each level are still doubly degenerate (Kramers doublet). In the limit of p

3

42

21 pp

EE

3

42

43 pp

EE

The change of energy is quadratic in . In the limit of p

3

42

21 12882 pppEE

3

42

43 12882 pppEE

The change of energy is linear in . ((Mathematica))

40

Sakurai 5 13

p 3 e a0

M 0, 0, p, 0 , 0, 0, 0, p , p, 0, , 0 , 0, p, 0, ;

M MatrixForm

0 0 p 00 0 0 pp 0 00 p 0

eq1 Eigensystem M Simplify

12

4 p2 2 ,12

4 p2 2 ,

12

4 p2 2 ,12

4 p2 2 ,

0,4 p2 2

2 p, 0, 1 ,

4 p2 2

2 p, 0, 1, 0 ,

0,4 p2 2

2 p, 0, 1 ,

4 p2 2

2 p, 0, 1, 0

41

Normalization

E1 eq1 1, 1 ;

E2 eq1 1, 3

12

4 p2 2

Simplify Series E1, p, 0, 5 , 0

p2 p4

3 O p 6

Simplify Series E2, p, 0, 5 , 0

p2 p4

3 O p 6

Simplify Series E1, , 0, 5 , p 0

p2

2

8 p

4

128 p3 O 6

42

REFERENCES S.S. Schweber, QED and the men who made it (Princeton Univ. Press, 1994). E. Fermi, Note on Quantum Mechanics (The University of Chicago, 1961). A. Das and A.C. Melissinos, Quantum Mechanics A Modern Introduction (Gordon and Breach

Science, 1986). J. Pade, Quantum Mechanics for Pedestrians 2: Applications and Extensions (Springer, 2014). C.E. Burkhardt and J.J. Leventhal, Foundations of Quantum Physics (Springer, 2008). ______________________________________________________________________________

Simplify Series E2, , 0, 5 , p 0

p2

2

8 p

4

128 p3 O 6

E11 E1 . 1

12

1 1 4 p2

E21 E2 . 1

12

1 1 4 p2

Plot Evaluate E11, E21 , p, 0, 4 ,

PlotRange 0, 4 , 5, 5 , PlotStyle Hue 0 , Hue 0.7 ,

Prolog AbsoluteThickness 1.5 , AxesLabel "p ", "E "

1 2 3 4p

4

2

0

2

4

E

43

APPENDIX-A

Relativistic correction for Na (Z = 11)

According to special relativity, the kinetic energy of an electron of mass m and velocity v is:

23

42

82 cmmK

ee

pp

.

The first term is the standard non-relativistic expression for kinetic energy. The second term is

the lowest-order relativistic correction to this energy. We apply the perturbation theory for this

system.

23

422

08

ˆ)

ˆ2

ˆ(ˆˆˆ

cm

Ze

mHHH

eerel

p

r

p

where

nlmEnlmH n)0(

0ˆ

and

r

pˆ2

ˆˆ22

0

Ze

mH

e

,

23

4

8

ˆˆcm

He

rel

p

Noting that

)ˆ

ˆ(2ˆ2

02

rp

ZeHme

we calculate the matrix element

]11

2[4

)ˆ

ˆ)(ˆ

ˆ(4ˆ

242)0(22)0(2

2

0

2

024

avavnne

e

reZ

rEZeEm

nlmZe

HZe

Hnlmmnlmnlm

rr

p

44

Then we have the first order correction to the energy eigenvalue

]11

2[2

1

ˆ8

1ˆ

242)0(22)0(

2

4

23

avavnn

e

e

rel

reZ

rEZeE

cm

nlmnlmcm

nlmHnlm

p

Here we use

Bav an

Z

r 2

1

,

)2

1(

123

2

2

lan

Z

rB

av

and

2

222)0(

2

1

n

ZcmE en

where

2

2

ema

eB

(Bohr radius),

c

e

2

(fine structure constant)

Then we find that

)3

2/1

4(

2),(

2

2)0()1(

l

n

cm

ElnEE

e

nrelrel

We calculate the wavelength of emitted light between the states of n = 3, l = 1 and n = 3, l = 0 for Na (Z = 11);

45

)0,3()1,3( relrel EE

hc 23668.5 Å = 4 x 5890 Å.

______________________________________________________________________________ APPENDIX-B B-1 Lamb’s report

The Lamb Shift and the Magnetic Moment of the Electron (S.S. Schweber, QED and the men who made it: Dyson, Feynman , Schwinger, and Tomonaga, Chapter 5, p.206).

Molecular hydrogen is thermally dissociation in a tungsten oven, and a jet of atoms emerges

from a slit to be cross-bombarded by an electron stream. About one part in a hundred million of

the atoms is thereby excited to the metastable 2/122 S state. The metastable atoms (with small

recoil deflection) move on out of the bombardment region and are detected by the process of electron ejection from a metal target. The electron current is measured with an FP-54 electrometer tube and a sensitive galvanometer. If the beam of metastable atoms is subjected to

any perturbing fields which may cause a transition to any of the P22 states, the atom will decay while moving through a very small distance. As a result, the beam current will decrease, since the detector does not respond to atoms in the ground state. Transitions may be induced by radiofrequency radiation for which h corresponds to the energy difference between one of the

Zeeman components of 2/122 S and any component of either 2/1

22 P or 2/322 P . Such

measurements provide a precise method for the location of 2/122 S state relative to the P states, as

well as the distance between the later states. We have observed an electrometer current of the order of 10-14 A which must be ascribed to

metastable hydrogen atoms. We have also observed the decrease in the beam of metastable atoms caused by microwaves in the wave length range 2.4 to 18.5 cm in various magnetic fields. The results indicate clearly that, contrary to theory but in essential agreement with Pasternack’s

hypothesis, the 2/122 S state is higher than 2/1

22 P by about 1000 MHz (0.033 cm-1).

[October 1946 to March 1948]. _____________________________________________________________________________ B-2 Beth’s calculation

(S.S. Schweber, QED and the men who made it: Dyson, Feynman , Schwinger, and Tomonaga, Chapter 5, p.231).

The actual calculation of the nonrelativistic Lamb shift was made on a train ride from New

York to Schenectady. Bethe had stayed in New York after the Shelter Island Conference to visit his mother, and had gone on to Schenectady to consult for General Electric. The calculation is

46

straightforward (Beth 1947). The self-energy of an electron in a quantum state m, due to its interaction with the radiation field, is

avemn EE

K

n

ZRy

c

eW

ln

3

83

432

where Ry is the Rydberg energy 22

2

1mc ,

c

e

2

. This is the expression that Bethe had

obtained on his arrival at Schenectady. He was not quite confident of its accuracy, because he

was not quite sure of the correctness of a factor of 2 in his expansion of the radiation operators in terms of creation and annihilation operators. This he checked on Monday morning in Heitler’s

book. He also got Miss Steward and Dr. Stehn from GE to evaluate numerically avemn EE

for the 2s state. It was found to be 17.8 Ry, “an amazingly high value. Inserting this into the

above equation Bethe found sW2 = 1040 MHz, “in excellent agreement with the observed value

of 1000 MHz” (Bethe, 1947). ((Note)) The expression of the frequency for the Lamb shift (the non-relativistic case) (for n = 2), can be expressed as

)9.8

1ln(

12 22

35

c

cme

= 1038.27 MHz.

http://quantummechanics.ucsd.edu/ph130a/130_notes/node476.html ((Mathematica))

Clear "Global` " ;

rule1 c 2.99792 1010,

1.054571628 10 27,

me 9.10938215 10 28,

7.2973525376 10 3, MHz 106 ;

5 me c3

12 2 cLog

1

8.9 2 MHz . rule1

N

1038.27

47

______________________________________________________________________________ B-3 Feynman’s Nobel Lecture:

R.P. Feynman, Nobel Lecture, December 11, 1965 The development of the space-time view of quantum electrodynamics.

Then Lamb did his experiment, measuring the separation of the 2/122 S and 2/1

22 P levels of

hydrogen, finding it to be about 1000 megacycles of frequency difference. Professor Bethe, with whom I was then associated at Cornell, is a man who has this characteristic: If there’s a good experimental number you’ve got to figure it out from theory. So, he forced the quantum electrodynamics of the day to give him an answer to the separation of these two levels. He pointed out that the self-energy of an electron itself is infinite, so that the calculated energy of a bound electron should also come out infinite. But, when you calculated the separation of the two energy levels in terms of the corrected mass instead of the old mass, it would turn out, he thought, that the theory would give convergent finite answers. He made an estimate of the splitting that way and found out that it was still divergent, but he guessed that was probably due to the fact that he used an un-relativistic theory of the matter. Assuming it would be convergent if relativistically treated, he estimated he would get about a thousand megacycles for the Lamb-shift, and thus, made the most important discovery in the history of the theory of quantum electrodynamics. He worked this out on the train from Ithaca, New York to Schenectady and telephoned me excitedly from Schenectady. APPENDIX-C

48

Fig. Fine and hyperfine structure of the hydrogen atom. Abbreviations: g = mc2α4 =1.45 x

10−3 eV, A = 1,420 MHz. The finite size of the nucleus is not taken into account. Scales are not preserved. [J. Pade, Quantum Mechanics for Pedestrians 2: Applications and Extensions (Springer, 2014)]. The lamb shift is 1057 MHz.