S4+Hydrogen+Atom+2011

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SECTION 4: THE HYDROGEN ATOM

1. Fine Structure in the Hydrogen Atom 2

1.1 The fine structure 3 1.2 The hydrogen atom in the absence of fine structure 3 1.3 Relativistic mass correction 4

1.4 Spin-orbit interaction 5

1.5 The Darwin term 8

1.6 Total fine-structure correction 8 1.7 Application to the ground and first-excited states 9

2. Zeeman effect for Hydrogen 11

2.1 Weak-field case (Zeeman effect) 12 2.2 Strong-field case (Paschen-Back effect) 15

References

The original version of these notes was based on sections 7.4 and 7.5 of the textbook by Mandl, although it

has been considerably modified by revision, and much detail has added; see in particular sections 6.3 and 6.4

of the book by Griffiths. References are given as footnotes to sections that are closely based on other textbooks.

Original Version: July 1995; Last Major Revision: July 2003

Revision Date: 28 October 2009

Printed: 28 October 2009

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1. FINE STRUCTURE IN THE HYDROGEN ATOM

The simple formula for the energy levels of the hydrogen atom derived in earlier lectures ignores relativistic

effects, since it is based on the non-relativistic formulation of quantum mechanics.

Relativistic effects are clearly visible if the spectral lines that are emitted by hydrogen gas in a discharge

tube are investigated with instruments of sufficient sensitivity.

It is found that each of the spectral lines is actually a group of very closely spaced lines, called the fine

structure.

The splitting of spectral lines is caused by splitting of the energy levels of the hydrogen atom, as described

below.

Here we will use perturbation theory to estimate the effects on the energy levels of hydrogen of:

1) the relativistic mass-energy;

2) the inclusion of the intrinsic spin of the electron, which is also a relativistic effect.

To obtain the correct description of hydrogen or hydrogen-like atoms, rather than using the non-relativistic

Schrdinger equation, we should use the relativistic Dirac equation for an electron placed in the Coulomb

potential created by the nucleus; this problem can be solved exactly, but we do not follow this approach here.

The description based on perturbation theory, although only an approximation, is useful in that it

provides additional physical insight into the problem and its solution.

The approximation is also accurate, since the hydrogen atom is a very weakly relativistic system: the

kinetic energy of the electron is of the order of 10 eV, whereas its rest energy is 511 keV.

This approach also provides a guide to how we should include similar effects in other atomic systems for

which the exact approach based on the Dirac equation is not applicable.

The relativistic corrections that we consider are smaller than the energies predicted by the non-relativistic

theory by a factor of the order of 2 105 where the dimensionless fine structure constant is given by

2

0

1.

4 137

e

c

=

(1)

Note that there are two other smaller corrections to the spectrum of hydrogen, which we shall not consider in

this course:

The Lamb shift1: this is associated with the quantization of the Coulomb field and the self energy of the

electron, and is smaller than the relativistic effects discussed here by a further factor of the order of .

Hyperfine structure2: this is due to the magnetic interaction between the electron and proton dipole

moments. It is smaller than the fine structure by a factor of the order of me/mp, the ratio of the electron

mass to the proton mass, and smaller than the Lamb shift by approximately one order of magnitude.

1This is discussed briefly on pages 174-5 of the book by Mandl; an explanation of the effect requires the use of Quantum Electro-dynamics. 2This topic is the subject of section 6.5 of the book by Griffiths; the effect can be treated using techniques similar to those used here for the fine structure.

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1.1 The fine structure

To find the main corrections to the usual non-relativistic formula for energy levels, we look3 at the limiting

form of the Dirac description when the system is weakly relativistic. In this limit we get for the Hamiltonian

H an equation in which the potential energy V appears in the form of a power series expansion in the ratio v/c,

where v is the electrons speed and c is the speed of light. To first order in v/c:

2 0 Rel SO DH mc H V V V= + + + + + (2)

where

mc2 is the electron rest energy, which simply shifts the whole spectrum by a constant amount and can be

ignored.

H0 is the usual non-relativistic Hamiltonian

2

0 ( )2

C

pH V r

m= + (3)

where 2 0( ) /4CV r e r= is the static Coulomb potential due to the proton and p is the electrons

relativistic momentum.

VRel is the relativistic mass correction

4

Rel 3 28

pV

m c= . (4)

VSO is called the spin-orbit interaction

2 21 1

2C

SO

dVV l s

m c r dr=

(5)

written in term of the scalar product of the orbital angular momentum and the intrinsic spin of the

electron.

VD is the Darwin correction term

2

22 2 ( ).8

D CV V rm c

=

(6)

The last three terms, all of the same order of magnitude, collectively produce the fine structure splitting of the

energy levels of hydrogen; they can be treated as perturbations on the dominant contribution to the energy

due to H0, using first-order perturbation theory.

1.2 The hydrogen atom in the absence of fine structure

We first summarize the description of the hydrogen atom in the absence of the fine structure.

The energy of an unperturbed state of hydrogen depends on the quantum number n only, and is denoted

(0)nE :

( )

4 2(0)

2 22 20 00

1

2 42 4n

me eE

n an = =

(7)

where

3See page 1213 of Quantum Mechanics Vol. 2, by C Cohen-Tannoudji, B Diu and F Lalo, published by Wiley, 1977.

4

2

00 2

4a

me

=

(8)

is the Bohr radius.

The unperturbed states of the atom are eigenstates of the angular momentum operators 2 2, ,zl l s and sz so

that valid quantum numbers to label the states are n, l, ml, s and ms.

The atomic states are denoted l snlm sm

where ssm

represents the spin state of the electron (which is

either spin-up or spin-down ) and lnlm

is the spatial wave function, the product of the radial wave

function Rnl(r) and a spherical harmonic ( , )llmY .

The unperturbed energy levels of the atom have degeneracy 2n2.

We estimate separately the effect of each of the three fine-structure energy corrections using first-order

degenerate perturbation theory.

1.3 Relativistic mass correction

The relativistic expression for the total energy of a particle of rest mass m and relativistic momentum p is

2 2 42 2 2 2 2

2 2 3 21

2 8

p p pE c m c p mc mc

m c m m c= + = + + +

where we have used the binomial expansion to approximate the square root. Note that the momentum is

p mv= to first order. Hence the kinetic energy is

2 42

3 22 8

p pK E mc

m m c= = +

yielding the correction term VRel of eq.(4):

4

Rel 3 2 .8

pV

m c=

The unperturbed energy levels of the hydrogen atom are degenerate; however, the perturbation VRel is

automatically diagonal in the unperturbed basis l snlm sm

:

The basis states l snlm sm

are eigenfunctions of the operators 2, ,z zl l s (as well as H0).

These operators, taken together, have a unique set of eigenvalues {l, ml, ms} for each of the 2n2 degenerate

states of given energy.

The perturbation VRel commutes with each of the operators 2, ,z zl l s .

The set of operators 2, ,z zl l s therefore diagonalizes the perturbation.

The shift in energy of an unperturbed hydrogen state n produced by the interaction VRel is therefore, in

first-order perturbation theory,

(1)Rel Rel Rel( ) l s l sE nl nlm sm V nlm sm nl V nl= =

In the last line, we have used the fact that the interaction is independent of spin, and since the energy

correction is independent of ml we can use nl to represent the spatial wave function lnlm . It is easily shown4

that

4The proof of this result is the subject of a tutorial problem.

5

( )(1) (0) 2Rel Rel 12

1 1 3( ) all

4nE nl nl V nl E l

n l n

= = + (9)

where (0)nE is the unperturbed energy given by eq.(7) and is the fine-structure constant defined in eq.(1).

In contrast to the situation for the unperturbed states, where the energy of a state depends only on the

quantum number n, when this perturbation is included the energy depends on l as well as n; the relativistic

mass correction on its own would remove the accidental degeneracy of the Coulomb potential for hy-

drogen.

1.4 Spin-orbit interaction

This is essentially the interaction between the intrinsic spin of the electron and its orbital motion or

equivalently, between the intrinsic magnetic dipole moment of the electron and the magnetic field, internal to

the atom, created by the proton at the centre of the atom.

We can justify equation (5) as follows, using a semi-classical approach5.

In the protons reference frame the positively charged proton creates a static electric field E through

which the electron moves with velocity v. This field, which points radially outwards, can be expressed in

terms of the Coulomb potential energy VC(r):

1 ( )CdV r rEe dr r

=

(10)

where the electronic charge is e and /r r

is the radial unit vector.

In the frame of reference of the electron, the proton is moving relative to the electron with velocity v.

According to special relativity6, the electron therefore experiences a magnetic field intB due to the proton,

which is given to first order in v/c by

int 2 21 1

B v E p Ec mc

= =

(11)

where the momentum p mv= to first order. Note that intB

is a magnetic field internal to the atom.

Combining eqs.(10) and (11), we see that the magnetic field is given by

( )int 2 21 1 1 1C CdV dVB p r l

emc r dr emc r dr= =

(12)

where l r p=

is the orbital angular momentum of the electron.

The electron has an intrinsic magnetic dipole moment associated with its intrinsic spin; the corresponding

vector operator is given by Diracs relativistic wave equation:

es

m =

where s is the intrinsic spin of the electron. This dipole moment interacts with the magnetic field created

by the proton with interaction energy

int int.SOe

V B s Bm

= =

Substitution of the expression for the magnetic field eq.(12) into this yields

5Adapted from section 12.2 of Quantum Mechanics by R Liboff, 2nd edition, published by Addison-Wesley; see also p1215 of the textbook by Cohen-Tennoudji et al. For an alternative derivation see section 6.3.2 of the textbook by Griffiths. 6See page 532 of Introduction to Electrodynamics, 3rd edition, by D J Griffiths, published by Prentice Hall.

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( )2 21 1

.CSOdV

V l sm c r dr

=

However, the electrons frame of reference is not an inertial frame. It has been shown by Thomas (1926)

that changing to the protons frame (the laboratory frame) introduces an additional factor of 1/2 into the

result; this is called the Thomas precession effect7. We then obtain eq.(5):

2 21 1

2C

SO

dVV l s

m c r dr=

(13)

As already noted, the eigenstates l snlm sm

of H0 are degenerate, so we need to determine whether the per-

turbation VSO is diagonal in this basis.

The set of quantum numbers l, ml and ms taken together uniquely distinguishes the degenerate states, as

discussed earlier.

The corresponding angular momentum operators 2, ,zl l and sz obviously commute with the unperturbed

Hamiltonian, since these are the operators that define the angular momentum part of the basis.

Therefore, we must determine whether the angular momentum part of the interaction, ,l s

commutes

with the each of the operators 2, ,zl l and sz.

In order to do this, we must rewrite l s

in terms of other operators whose commutation relations are known.

We define the total angular momentum operator of the electron:

j l s= +

so that we have

( ) ( )2 2 22j l s l s l l s s= + + = + +

where we have used the fact that l and s

commute. Hence

( )2 2 21

2l s j l s =

(14)

The operator j2 contained in this expression does not commute with the operators lz and sz that define the

unperturbed basis states. It follows that l s

does not commute with lz and sz and therefore VSO does not.

Therefore the perturbation VSO is not diagonal in the basis l snlm sm .

It also follows that ml and ms are not good quantum numbers for eigenstates of the full Hamiltonian with

the spin-orbit interaction included; the Hamiltonian mixes states with different values of ml and ms.

However, the operators 2 2,j l and 2s contained in l s

all commute with 2j and jz (as well as with 2l and

2s ), so that a suitable set of commuting operators when VSO is included is 2 2 2, , ,zj j l s . The basis which

diagonalizes VSO is therefore ;nls jm , where, from the angular momentum addition theorem,

,

; ( , | )l s

l s l s

m m

nls jm C lm sm jm nlm sm= . (15)

In other words, the basis states we must use to compute the effect of the spin-orbit interaction are known linear

combinations of the original basis states l snlm sm

.

The term l s

in the Hamiltonian couples together the orbital and spin degrees of freedom of the electron;

we therefore require a basis that reflects the presence of this coupling.

7See the discussion in the second footnote to page 241 (1st edition) or page 272 (2nd edition) of the book by Griffiths.

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To first order of perturbation theory, the energy shift due to VSO is therefore, from eqs.(13) and (14),

( )

( )

(1)

2 2

2 2 22 2

( ) ; ;

1 1; ;

2

1 1; ; .

4

SO SO

C

C

E nlj nls jmV nls jm

dVnls jm l s nls jm

m c r dr

dVnls jm j l s nls jm

m c r dr

=

=

=

Now ;nls jm is an eigenstate of each of the operators 2 2,j l and s2 so that, with s = 1/2,

2

(1)2 2

3 1( ) ( 1) ( 1) ; ;

4 4C

SO

dVE nlj j j l l nls jm nls jm

m c r dr

= + +

(16)

We first consider the implications of the factor [j(j + 1) l(l + 1) 3/4] in this equation:

For l = 0 (for which j = 1/2), this factor is zero and there is no energy shift.

For non-zero l, the angular momentum addition theorem tells us that the allowed values of the quantum

number j are

, , 1/2 and 1/2j l s l s l l= + = +

where we have again used the fact that s = 1/2. Thus, for each l 0 there exists two values of the total

angular momentum, namely j = l 1/2.

Each energy level (nl) for l 0 is therefore split by the interaction VSO into two levels labelled (nlj). Since

the remaining expectation value in eq.(16) is positive, as indicated below, the level for which j = l + 1/2

is higher in energy than the level with j = l 1/2 .

The effect of the spin-orbit interaction on an energy level (nl) is shown in the figure below.

We now consider the remaining expectation value in eq.(16):

2

30

1 1; ; ; ;

4CdV enls jm nls jm nls jm nls jm

r dr r= (17)

where we have substituted

2 2

30 0

1 1

4 4CdV d e e

r dr r dr r r

= =

It can be shown that8

3 3 3 10 2

1 1

( 1)( )nl nl

r a n l l l=

+ +

8See the footnote on page 238 (1st edition) or the second footnote on page 269 (2nd edition) of the textbook by Griffiths for a ref-erence.

8

where the Bohr radius a0 is given by eq.(8). The operator 1/r3 has no angular or spin dependence, so that the

expectation value in eq.(17) is actually independent of the quantum numbers s, j and m. Combining these

equations we find

2

3 3 10 0 2

1 1 1; ;

4 ( 1)( )CdV enls jm nls jm

r dr a n l l l=

+ + (18)

Substituting this in eq.(16), we obtain the complete expression for the energy correction due to the spin-orbit

coupling term:

2 2 3(1) 4

2 3 3 10 0 2

( 1) ( 1)1( )

(2 ) 4 ( )( 1)SO

j j l leE nlj

mc a n l l l

+ + =

+ +

which can be simplified using eq.(7) for (0)nE and the definition eq.(1) for the fine-structure constant , to give

(1) (0) 2

1 12 2

1 1 1 1( ) , , 0

2

0, 0

SO nE nlj E j l ln l j

l

= = + +

= =

(19)

The energy correction does not depend on the magnetic quantum number m so that each level is still

(2j + 1) fold degenerate. This reflects the fact that both H0 and VSO and therefore the total Hamiltonian

H are rotationally invariant.

For the upper level (2j + 1) = 2l + 2 and for the lower level (2j + 1) = 2l, so that the total number of states

remains unchanged at 2(2l + 1).

1.5 The Darwin term9

The energy correction due to the Darwin term, eq.(6), is zero unless l = 0, in which case10 it is given by

(1) (0) 21

, 0 onlyD nE E ln

= = (20)

This is exactly equal to the expression for (1)SOE from eq.(19) with l = 0, j = 1/2. It follows that

(1) (1) (0) 21 12 2

1 1 1 1, ,

2SO D nE E E j l

n l j

+ = = + + (21)

valid for all orbital angular momenta.

1.6 Total fine-structure correction

The combined effect of all three contributions to the fine structure is

(1) (1) (1)Rel( ) ( ) ( ).FS SO DE nlj E nlj E E nl = + +

From eqs.(9) and (21) we get, after some manipulation,

(0) 212

1 3 1( ) , 1/2

4FS nE nlj E j l

n n j

= = + (22)

9 See pages 1216-17 of the book by Cohen-Tennoudji et al for a full discussion.

10This follows from the fact that ( )21

4 rr

=

leading to 2(1) (0)

DE ; the radial wave function is zero at the origin

unless l =0.

9

Note:

This equation is valid for all angular momenta; it is in fact independent of the orbital angular momentum

l but depends on j.

The exact energy of the hydrogen atom ground state, as predicted11 directly by the Dirac equation, is

( )

1/22

2

2 21 12 2

1 1 .E mcn j j

= + + +

It can be shown that eq.(22), supplemented by the electrons rest energy and the non-relativistic Bohr

energy, is actually equal to the result of Diracs theory to fourth order in .

1.7 Application to the ground and first-excited states

We present below calculations for the effect of the fine-structure interaction for the n = 1 and n = 2 states of

the hydrogen atom; the n = 3 level can be treated in a similar fashion. Figure 1, not drawn to scale, shows the

fine structure splitting of the lowest two excited levels of hydrogen.

Figure 1: Fine structure of the n = 2 and n = 3 levels of the hydrogen atom.

The n = 1 state

For this state only l = 0 is possible; therefore the state has total angular momentum j =1/2. Using the

conventional spectroscopic notation (nlj), in which the letters s, p, d, f, g, indicate orbital angular mo-

mentum l = 0, 1, 2, 3, 4, respectively, this state is designated 1s1/2. The first-order energy shift is, from

eq.(22),

(0) 2 41/2 1

1(1 ) 1.8 10 eV;

4FS nE s E

= = =

11For a reference see the footnote on page 244 (1st edition) or page 276 (2nd edition) of the textbook by Griffiths.

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i.e. the ground state is shifted downwards by 1.8 104 eV. It is still two-fold degenerate, corresponding to

m = 1/2.

The n = 2, l = 0 state

In the case of n = 2 both l = 0 and l = 1 are possible. The l = 0 state has total angular momentum j = 1/2

and is designated (nlj) = 2s1/2. The first-order energy shift is, from eq. (22),

(0) 2 51/2 2

5(2 ) 5.7 10 eV;

16FS nE s E

= = =

i.e. this state is shifted downwards by 5.7 105 eV. It is still two-fold degenerate, corresponding to

m = 1/2.

The n = 2, l =1 state

The n = 2, l = 1 state has total angular momentum j = l 1/2 = 1/2 or 3/2. For j = 1/2 the state is

designated 2p1/2 and the energy shift, to first order, is

(0) 2 51/2 2

5(2 ) 5.7 10 eV.

16FS nE p E

= = =

The level therefore remains doubly degenerate.

For j = 3/2 the state is designated 2p3/2 and the energy shift, to first order, is

(0) 2 53/2 2

1(2 ) 1.1 10 eV.

16FS nE p E

= = =

The level therefore remains four-fold degenerate.

In summary for the n = 2 states:

The 2s1/2 and 2p1/2 states are both shifted downwards by 5.7 105 eV and remain degenerate12.

The 2p3/2 state is shifted downwards by 1.1 105 eV.

The degeneracy of the n = 2 energy level is therefore only partially broken by the fine structure interaction;

the original eight degenerate states have been split into two groups of four states.

12This degeneracy is lifted by the Lamb shift, observed by Lamb and Retherford in 1947; the splitting is about 4 106 eV.

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2. ZEEMAN EFFECT FOR HYDROGEN

Consider a hydrogen atom13 placed in an external uniform magnetic field B. There is now a preferred di-

rection in space (that of the field) so that rotational invariance is destroyed; the system is invariant under

rotations about the direction of B only, so that the degeneracy with respect to m is broken.

The complete Hamiltonian for an atom in the magnetic field is written

0 FS magH H V V= + +

where the unperturbed Hamiltonian H0, given in eq.(3), includes the kinetic energy and Coulomb terms, and

VFS is the complete fine-structure interaction discussed above, and given by eqs.(4)(6). The additional term

Vmag is the interaction energy of the total magnetic dipole moment of the atom with the external magnetic

field B:

magV B=

The total atomic magnetic moment includes contributions from the orbital motion and intrinsic spin of the

electron14:

.L S = +

The orbital contribution is the same as that obtained classically for an orbiting charge; quantum mechanically

it is obtained by considering the Schrdinger equation for a charge e in a magnetic field:

2L

el

m =

where l is the orbital angular momentum of the electron. The intrinsic magnetic dipole moment of the

electron, as predicted by the Dirac equation, is given by

S

es

m =

where s is the intrinsic spin operator of the electron. Combining these equations, the interaction energy

becomes:

( )22

mag

eV l s B

m= +

We take the z direction to be that of the magnetic field, so that the final expression for the interaction energy

becomes

( )

( )

22

2

mag z z z

z z z

eV l s B

me

j s Bm

= +

= + (23)

where we have used the fact that jz = lz + sz.

13

Mandl considers the more general case of any atom placed in a magnetic field in section 7.5 of his book. The specific case of hydrogen is the topic of section 6.4 of the book by Griffiths, but the approach used in these notes is closer to that adopted by Mandl. 14

The nucleus of the atom, the proton, has an intrinsic magnetic dipole moment, but this is about three orders of magnitude smaller than the other two terms and can be neglected.

12

We can compare the order of magnitude of this interaction with the spin-orbit interaction, which is due to the

interaction of S with the internal magnetic field of the atom, which is given by eq.(12). It is found that

4

5 4

10 eV

10 10 eV

mag z

FS

V B

V

with the field strength Bz in tesla. Note that Bz = 1 T would be a reasonably strong field.

What is observed experimentally and how we describe the atom theoretically now depends critically on the

relative magnitude of these two interactions, which means on the strength of the external field in comparison

with the internal magnetic field that gives rise to the spin-orbit coupling.

If the magnetic field is sufficiently weak, i.e. Vmag VFS, the fine structure dominates and we would expect

the atomic energy levels to differ only slightly from those observed in the absence of the external field.

Then Vmag can be treated as a perturbation on the Hamiltonian H0 + VFS discussed in the last section. This

is the situation for fields of normal strength that was investigated experimentally by Zeeman; we consider

it first and in some detail.

As the strength of the external field is increased, the effect of the field will increase and eventually, when

Vmag VFS, overwhelm the fine structure. In 1912, before the existence of electron spin was known,

Paschen and Back suggested that if the field is made sufficiently strong the then unknown cause of

observed multiplet splitting would be over-powered, and all spectral patterns would revert to normal.

They subsequently found experimental evidence of this. Note that very strong fields are required for a

proper experimental study. The theoretical approach used for weak fields will no longer be valid in this

regime; the effect of the magnetic field in the absence of fine structure must first be calculated, as de-

scribed below.

We do not consider at all the more complicated situation where the two perturbations Vmag and VFS are

of comparable magnitude15.

2.1 Weak-field case (Zeeman effect)

We first derive an expression for the energy correction due to a weak magnetic field, and then apply the results

to the lowest states of the hydrogen atom16.

We treat Vmag as a perturbation on the remainder of the Hamiltonian, which we write as

0 0 FSH H V = +

As discussed in the previous section, this Hamiltonian has unperturbed energies

(0)( ) ( )n FSE nlsj E E nlsj= +

with ( )FSE nlsj given by eq.(22):

(0) 2

12

1 3 1( ) 1/2

4FS nE nlj E j l

n n j

= = +

15A discussion of the n = 2 states of the hydrogen atom for this intermediate situation is presented in section 6.4.3 of the book by Griffiths. 16P. Zeeman investigated the effect of a magnetic field on the spectrum of sodium atoms, beginning in 1896. It was partly to explain his results that the concept of intrinsic spin was introduced (Pauli 1924, Goudsmit and Uhlenbeck 1925).

13

and (0)nE given by eq.(7), or

(0)2

13.6 eV.nE

n=

The unperturbed state vectors are

; , , ,nls jm m j j= + (24)

The unperturbed energy levels of the Hamiltonian H0 are degenerate with respect to the quantum number m,

corresponding to the operator jz, with further degeneracy in l corresponding to the operator 2l . However, the

perturbative interaction Vmag is diagonal in the basis (24):

the eigenvalues of jz and 2l , taken together, uniquely distinguish the degenerate eigenstates (consider the

degenerate states in Fig.1 for example);

Vmag (lz + 2sz) commutes with the operators 2l and jz;

the set of operators that diagonalizes the perturbation is therefore 2, .zl j

Hence from eq.(23) the effect of the magnetic field is to produce a shifting and splitting of the energy levels

given to first order of perturbation theory by17

( )

2

mag mag

zz z

E lsjm lsjmV lsjm

eBlsjm j lsjm lsjm s lsjm

m

=

= +

The basis states ;nls jm are clearly eigenstates of jz so that

.zlsjm j lsjm m=

The states ;nls jm are not eigenstates of sz so that the calculation of the expectation value of this operator

is not so simple. However, it can be shown18 that

( 1) ( 1) ( 1)

2 ( 1)z

j j s s l llsjm s lsjm m

j j

+ + + +=

+

so that

( 1) ( 1) ( 1)( ) .

2 2 ( 1)z

mag

eB j j s s l lE lsjm m m

m j j

+ + + + = + +

Finally, introducing the Bohr magneton defined by

55.79 10 eV/T2

B

e

m = =

we obtain

( )mag z BE lsjm m g B = (25)

with the Land g factor defined as

17Do not be confused by two uses of the symbol m in this and subsequent equations, for both the mass of the electron and the pro-jection of jz. 18See page 183 of the book by Mandl. An alternative derivation is given in section 6.4.1 of the textbook by Griffiths.

14

( 1) ( 1) ( 1)1

2 ( 1)

11 for 1/2.

2 1

j j s s l lg

j j

j ll

+ + + += +

+

= = +

Eq.(25) indicates that a weak magnetic field splits the fine-structure multiplets into (2j + 1) equidistant

levels with spacing = g Bz B.

We apply eq.(25) to the ground and first-excited states of the hydrogen atom; the results are illustrated in Fig.

2 and Fig. 3, respectively, which also include the effect of the fine structure terms that were discussed earlier.

Figure 2: Effect of a weak magnetic field on the n = 1 level of hydrogen ( = BB).

We consider first the effect of a weak magnetic field on the n = 1 energy level of hydrogen. For this level

l = 0, j = 1/2, which yields g = 2. The total energy of the 1s1/2 state is then

( ) 2121

1 13.6 eV 1 .4

B zE s B = +

The last term results from the magnetic field, with the upper sign corresponding to m = +1/2 and the

lower sign to m = 1/2.

Thus the original doubly-degenerate level is split into two by the field, separated by energy 2Bz B, as

shown in Fig. 2.

Next we consider the l = 0 component of the n = 2 level. Again, l = 0, j = 1/2, yielding g = 2. The total

energy of the 2s1/2 level is therefore

( ) 21/25

2 3.40 eV 1 .16

B zE s B = +

As we found for the 1s1/2 level, the 2s1/2 level is split into two by the field, separated by energy 2Bz B, as

shown in Fig. 3.

Now we consider the l = 1 component of the n = 2 energy level. For this level there are two possible values

of j, namely j =1/2 with g = 2/3 and j =3/2 with g = 4/3. The total energy for the 2p1/2 state is

( ) 21/25 1

2 3.40 eV 1 .16 3

B zE p B = +

So the 2p1/2 level is also split into two by the field, separated by energy 23 B zB .

Note that the 2p1/2 level was originally degenerate with the 2s1/2 level, with a total degeneracy of four; the

effect of the magnetic field is to produce four non-degenerate levels.

For the 2p3/2 state

( ) 23/21 4

2 3.40 eV 116 3

B zE p B m = + +

with 1 32 2,m = .

15

Thus, the 2p3/2 level is split into four levels by the field, adjacent levels separated by energy 43 B zB , as seen

in the upper part of Fig. 3.

Note that the fine-structure interaction splits the n = 2 level of hydrogen into two sets of states, one with

j = 1/2 and the other with j = 3/2, each four-fold degenerate. This degeneracy is completely removed by the

weak magnetic field; the net result is two well-separated clusters each containing four non-degenerate states.

Figure 3: Effect of a weak magnetic field on the n = 2 level of hydrogen ( = BB).

Note also that the validity of perturbation theory requires that the energy shift caused by the magnetic field

be small compared with the fine-structure effect; this implies that Bz must be much less than about 2 T for

the n = 2 states considered here.

2.2 Strong-field case (Paschen-Back effect)

This applies only if the magnetic field is extremely strong. We now write the Hamiltonian as

0 mag FSH H V V= + +

with H0 given by eq.(3), VFS by eqs.(4)(6) and Vmag by eq.(23):

( ) ( )22 2

mag z z z z z z

e eV l s B j s B

m m= + = + (26)

We look first at the effect of the perturbation Vmag directly on H0, and then we consider VFS as an even

smaller perturbation on H0 = H0 + Vmag.

As discussed previously, the unperturbed Hamiltonian H0 has energy levels

(0)2

13.6 eVnE

n=

with degeneracy 2n2, and eigenstates

; l snls mm ,

which are eigenstates of the operators 2 2, , zl s l and sz.

16

The interaction Vmag due to the magnetic field is diagonal in this basis; this follows most directly from the fact

that the basis states are actually eigenstates of the perturbation Vmag, since they are eigenstates of the op-

erators lz and sz.

Thus the interaction Vmag causes a shifting and splitting of the unperturbed energy levels (0)nE given to first

order in perturbation theory by

( )

( )

; ;

; 2 ;2

2

mag l s mag l s

zl s z z l s

B z l s

E nlm sm V nlm sm

eBnlm sm l s nlm sm

m

B m m

=

= +

= +

where in the last line we have again used the fact that the unperturbed states ;l snlm sm are eigenstates of

lz and sz. So the eigenstates of the Hamiltonian H0 are the same as those of H0, namely ;l snlm sm , but with

eigenvalues

( )213.6 eV

( ; ) 2 .l s B z l sE n mm B m mn

= + +

The quantum numbers relevant to the calculation of this energy for the n = 1, l = 0 level of hydrogen are

shown in following table:

ml ms ml + 2ms

0 +1/2 +1

0 1/2 1

We see that the n = 1 level, which in the absence of the magnetic field and relativistic effects is two-fold

degenerate, is split into two non-degenerate states separated by an energy 2Bz B, corresponding to

ml + 2ms = 1 (or equivalently ms = 1/2). One state is shifted upwards and the other downwards by

the same amount.

The next table gives the quantum numbers of the n = 2 level of hydrogen. The resulting energy levels are

shown schematically in Fig. 4.

l ml ms ml + 2ms l ml ms ml + 2ms

0 0 +1/2 +1

0 0 1/2 1

1 +1 +1/2 +2 1 +1 1/2 0

1 0 +1/2 +1 1 0 1/2 1

1 1 +1/2 0 1 1 1/2 2

We see that:

The n = 2 level that was originally 8-fold degenerate is split into 5 levels, two of which are non-degenerate

and three of which are doubly degenerate.

All the degenerate states can be distinguished by specifying the two quantum numbers l and

m = ml + ms.

17

Figure 4: Effect of a strong magnetic field on the n = 2 level of the hydrogen atom ( = BB).

We now consider19 the effect of the additional perturbation VFS = VRel + VSO + VD. Because most of the

degeneracy of the original energy levels has been broken by Vmag the perturbation VSO is actually diagonal in

the basis ;l snlm sm .

The states ;l snlm sm are eigenfunctions of both 2l and jz..

As discussed for the n = 2 states above, the remaining degenerate states are uniquely identified by the

quantum numbers l and m (which are the quantum numbers associated with the operators 2l and jz).

The interaction VFS commutes with both 2l and jz.

Therefore the combination of operators that diagonalizes VFS is 2, zl j .

Thus the interaction VFS causes a further shifting of the energy levels given, in first order perturbation theory,

by

Rel; ;FS l s SO D l sE nlm sm V V V nlm sm= + + (27)

The calculation of the effect of the mass-energy correction is identical to the calculation in the absence of the

magnetic field (the expression for the interaction is identical and the basis states are the same in the two cases).

See the discussion leading up to the result, eq.(9):

(0) 2Rel 12

1 1 3

4nE E

n l n

= + (28)

The effect of the Darwin term is also identical to that in the absence of the field, yielding

(1) (0) 2

(1)

1for 0

0 for 0

D n

D

E E ln

E l

= =

=

(29)

We now calculate the correction due to the spin-orbit interaction, as given by eq.(5),

2 2

1 1; ; .

2C

SO l s l s

dVE nlm sm l s nlm sm

m c r dr=

19See page 247 (1st edition) or page 280 (2nd edition) of the textbook by Griffiths.

18

The calculation of the expectation value 1 CdV

r dr is identical to that carried out in the absence of the magnetic

field, yielding eq.(18):

2

3 30 0

1 1 1, 0

4 ( 1/2)( 1)CdV enl nl l

r dr a n l l l=

+ +

whereas the calculation of the expectation value of the operator l s

is modified since, as already noted, we

now require its value in the uncoupled basis. We note that

; ; ; ;

.

l s l s l s x x y y z z l s

x x y y z z

lm sm s l lm sm lm sm s l s l s l lm sm

s l s l s l

= + +

= + +

However, for a state that is an eigenfunction of lz and sz, we must have 0x x y ys l s l= = = = , so that

2; ; .l s l s z z s llm sm s l lm sm s l m m = =

Combining these results we obtain

(0) 21 for 0( 1/2)( 1)

0 for 0

s lSO n

SO

m mE E l

n l l l

E l

= + +

= =

(30)

where we have used eqs.(1) and (7). The final expression for the fine-structure correction is therefore, from

eqs.(28), (29) and (30):

(0) 2

(0) 2

1 3 ( 1)for 0

4 ( 1/2)( 1)

1 31 for 0

4

s lFS n

FS n

l l m mE E l

n n l l l

E E ln n

+ = + +

= =

It is easily seen that the correction term EFS removes the remaining degeneracies in the unperturbed

spectrum for the n = 2 states, except that the states with ml + 2ms = 0 remain degenerate.

The validity of perturbation theory for the strong field case requires that (0)0FS magE E E . For the lowest

levels of the atom this implies the field strength must be much greater than 1-2 T (the upper constraint does

not impose any real restriction).

S4+Hydrogen+Atom+2011

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