Baretti  LECTURES ON QUANTUM MECHANICS

by Reinaldo Baretti Machín   (reibaretti2004@yahoo.com)

Chapter 9 - The Electron Spin

From left to right Niels Bohr, Werner Heisenberg and Wolfgang Pauli.

Chapter 1 - Some experimental facts

Chapter 2 - The Old Quantum theory

Chapter 3 - The Postulates of Quantum Mechanics

Chapter 4- Potential Barriers

Chapter 5 - Potential Wells

Chapter 6 -  Periodic Potentials

Chapter 7 - Angular momentum

Chapter 8 - The Hydrogen Atom

Chapter 10 -The two electron atom

References:

1. Principles of quantum mechanics; nonrelativistic wave mechanics with illustrative applications.

2. Introduction to Quantum Mechanics with Applications to Chemistry by Linus Pauling and E. Bright Wilson Jr.

3. Quantum Mechanics: Non-Relativistic Theory, Volume 3, Third Edition (Quantum Mechanics) by L. D. Landau and L. M. Lifshitz

5.  Theoretical physics by A. S Kompaneets

6. Fundamentals of Modern Physics -  by Robert Martin Eisberg,  Chapter 8.

7.  http://en.wikipedia.org/wiki/Spin_(physics)

8.   http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment

9.  Hans C. Ohanian , What is spin?  , American Journal of Physics -- June 1986 -- Volume 54, Issue 6, pp. 500-505

# 10. http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/hydfin.html

Ample experimental evidence exists pointing  to an additional quantum number for the electron. In the hydrogen atom , for example, we have a wave function dependent on three quantum numbers  n , l , m. The emission lines are explained to a good approximation (perhaps three significant numbers in  λ ) by applying selection rules to the quantum numbers  n , l ,m .Highly refined spectroscopy  reveal that the lines have some kind of multiplicity and the rules for the changes ∆n , ∆l , ∆m  were not sufficient. A new  quantum number ,the spin, was postulated.

The Stern Gerlach experiment evidenced that the electron had a magnetic moment not related to the orbital motion.

Beams of neutral silver atoms with the outer electron in an s orbital were split in two by the inhomogeneous magnetic field along the Z axis.

There is ample evidence that a single electron in a magnetic field  B ,  will have just two eigenvalues of internal angular momentum (spin) .The passage of silver atoms splits in two when passing through a inhomogenous magnetic field . The significance of such observation in regards to the spin quantum number is discussed below.

It represents an extra degree of freedom. The consequences of this assignment  goes well beyond the electron . When particles like protons , neutrons and many others created in high energy physics are considered, a spin quantum number had to be assigned. It turned out that protons , neutrons and hundreds of other particles were not exactly elementary. Nevertheless they had a definite quantum number. More internal symmetries were recognized in the micro world carrying each new quantum numbers. Spin just had opened the gates for internal symmetries.

According to the modern viewpoint, the truly "elementary particles " are categorized in three classes. Quarks , leptons and force carriers. The electron has a special place being the lightest of the leptons when compared with the muon and tau leptons. Originally the neutrinos were viewed as massless.

Fig 1. Modern elementary particles.

A more detailed classification is

Quarks (spin =1/2)
Name   Symbol   Antiparticle   Charge
e
Mass (MeV/c2)
up u u +23 1.5–3.3
down d d 13 3.5–6.0
charm c c +23 1,160–1,340
strange s s 13 70–130
top t t +23 169,100–173,300
bottom b b 13 4,130–4,370

Leptons (spin=1/2)

Name   Symbol   Antiparticle   Charge
e
Mass (MeV/c2)
Electron e e+ −1 0.511
Electron neutrino νe νe 0 < 2.2 eV/c2
Muon μ μ+ −1 105.7
Muon neutrino νμ νμ 0 < 0.170
Tauon τ τ+ −1 1777
Tauon neutrino ντ ντ 0 < 15.5

Bosons ( spin= integer)

Name Symbol Antiparticle Charge (e) Spin Mass (GeV/c2) Interaction mediated Existence
Photon γ Self 0 1 0 Electromagnetism Confirmed
W boson W W+ −1 1 80.4 Weak interaction Confirmed
Z boson Z Self 0 1 91.2 Weak interaction Confirmed
Gluon g Self 0 1 0 Strong interaction Confirmed
Higgs boson H0 Self? 0 0 > 112 None Unconfirmed
Graviton G Self 0 2 0 Gravitation Unconfirmed

Notice that existence of at least two bosons is considered unconfirmed.

It was logical to start conceiving the electron magnetic dipole with the aid of a classical picture. At the same time we can not expect a hundred percent correspondence with a classical construct since we alredy know that quantum mechanics is at variance with classical physics.

A classical picture of the magnetic dipole moment  μ  , consists in a charge q (coulombs) circulating in a circle of radius r(meters)  with a period T (seconds). The current  i=q/T , the area traced A= π r2 , and T=2πr/v ; thus

μ = i A  = (q/T) π r2    = {q v/(2πr)}π r2

= evr/2    ~ amp-m2 ~ joules/tesla ~ in SI units .                 (1)

Equation (1) can be expressed in terms of the  orbital angular momentum (see figure 1.)  The angular momentum L = rmv  . For a negative charge , the angular momenum and the magnetic moment point in opposite directions , so the vector expression is

μ =  - q L /(2m)                                                . (2)

In quantum mechanics L has dimensions of h' , so the basic scale of magnetic dipole moment for the electron is called the Bohr magneton, defined by

μB = q h'/(2m) = 9.27400915(23)×10−24  joules/tesla                  .  (3)

For the  electron , the vectors  L and μ have opposite directions since  q = -e .

If one wishes to postulate an elementary magnetic pole (qm ) , one  may find its magnitude. Consider the dimensional argument; let now L= length scale=h'/(mc)

μ ~  qm L = i Area  = (e/tau) L2   = (e mc2/h) L2  ,

q~  (e mc2/h) L = (e mc2/h) (h/mc) =  e c =(1.6E-19 coulombs)(3e8 m/s) = 4.8e-10 dirac

An argument suggestive of a spin  (h/2)   can be made by considering a relation between angular momentum L and some minimal electromagnetic radiation with energy U . When the the electromagnetic field is quantized , the energy for each frequency ν is that of a harmonic oscillator   U(n,ν) = ( n+1/2) ( h ν ) . If we assume that a single frequency is allowed and the electromagnetic field is in its ground state (n=0) , one has the minimal angular momentum dividing U by  ν ,  i.e. L = U(n=0 , ν ) / ν =  h/2  .

Current wisdom holds that  the magnetic moment  cannot be reasonably modeled by considering the electron as a spinning sphere or ascribing some structure to it. High energy scattering shows no "size" of the electron down to a resolution of about 10-3 fermis= 1.E-18 meters .

For some kind of intuitive picture read Hans C. Ohanian abstract ,

# What is spin?

### American Journal of Physics -- June 1986 -- Volume 54, Issue 6, pp. 500-505

According to the prevailing belief, the spin of the electron or of some other particle is a mysterious internal angular momentum for which no concrete physical picture is available, and for which there is no classical analog. However, on the basis of an old calculation by Belinfante [Physica 6, 887 (1939)], it can be shown that the spin may be regarded as an angular momentum generated by a circulating flow of energy in the wave field of the electron. Likewise, the magnetic moment may be regarded as generated by a circulating flow of charge in the wave field. This provides an intuitively appealing picture and establishes that neither the spin nor the magnetic moment are internal''—they are not associated with the internal structure of the electron, but rather with the structure of its wave field. Furthermore, a comparison between calculations of angular momentum in the Dirac and electromagnetic fields shows that the spin of the electron is entirely analogous to the angular momentum carried by a classical circularly polarized wave.

Example:   Picture an electron in a constant external magnetic field , say along the z axis, B =  Bz k.

For an orbiting electron, the interaction potential energy will be given by the scalar product (writing m = me , for the electron mass to avoid confusion with the magnetic quantum number m)

U  =  - μ ∙B  = - {q /(2me)}  Lz Bz    .                                                                    (3)

Suppose an electron in the hydrogen atom has  a wave function  Ψ n , l ,m . The average of (3) is

Uave  = ∫ Ψn,l,m *  ( - {q Bz/(2me)}  Lz  ) Ψ n,l,mdτ

=   - {q Bz/(2me)} h' m      ,        m = l ,l-1,...0,-1,..-l             .       (4)

Spectral lines , of hydrogen , will show a multiplet structure  with corresponding energies energies

E (n,m) = Eground / n2  - {q Bz/(2me)} h' m      .                                 (5)

If the orbital quantum number l=0  , then m=0 and there would no splitting. As mentioned before, the Stern Gerlach experiment shows that the electron should have an internal spin and a corresponding magnetic dipole not related to an orbital motion or mechanical motion.

Example: Calculate the splitting given by (5) , if m=1 and n=2  and  Bz ~  1.0  tesla

c calculations 12 nov 2009
real mu0 ,me
pi=2.*asin(1.)
q=-1.609e-19
hbar=6.63E-34/(2.*pi)
me=9.11E-31
Eground=2.17e-18
Bz=1.
deltau=(q*Bz/(2.*me))*hbar
ratio=deltau/eground
print*,'Bz(T),deltau(eV),ratio=',Bz,deltau/1.602e-19,ratio
stop
end

RUN

Bz(T),deltau(eV),ratio= 1. -5.8167283E-005 -4.29419288E-006

We find that  {q Bz/(2me)} h' ~ 5.82E-005(eV) . Compared with the ground state energy ∆U/Eground ~ 4.29E-006 . If such a transition, is posible of about  6.0 E-5 eV , it would give rise to wavelength   λ ~ ch/∆U ~ 0.021 meters .

Going back to the Stern Gerlach experiment, there the magnetic force Fz  is measured and the gradient  ∂ Bz / ∂ z is known. They are given by

Fz = - (∂ Bz / ∂ z ) gs μB ms                                      (6)

The result shows two positions marked  by  the neutral silver after crossing the inhomogeneous B field.

This means that the only variable in (6),  ms , the projection of spin along the z axis has two possible values. If one imposes the condition that these two values are separated in absolute value by unity then    2 /ms / = 1    , and   /ms / = 1/2   .

If one writes instead of (6) ,   Fz = - (∂ Bz / ∂ z ) μB ms  , omitting the gs factor the observed splitting or force if you like ,would be twice as large as predicted. This meant that an extra factor , namely

gs = 2 , was needed to match theory and experiment.

Table of g (spin) factors.

Elementary Particle g-factor Uncertainty
Electron ge -2.002 319 304 3622 0.000 000 000 0015
Neutron gn -3.826 085 45 0.000 000 90
Proton gp 5.585 694 713 0.000 000 046
Muon gμ -2.002 331 8414 0.000 000 0012

Fig 2. The angular momentum undergoes a precession along Z axis. Its projection is Lz = mh.  The spin S , when not coupled to L, also undergoes  a precession  along the magnetic ( z-axis). Its component along Shas only two values , (+ /-) h'/2.

Moreover, the spin operator will operate on the spin variables , formed by a 2x1 column , not on space coordinates. The original recipe for writing operators where x operator → x  and px → - i h'∂ /∂ x   breaks down here.

Wolfgang Pauli introduced a method for the incorporation of  the electron spin into quantum mechanics ( see Ref. 7) . It does not explain the origin or predicts its magnitude. The internal angular momentum (spin) will not be represented by differential operators. Thus its eigenvalues , just two , do not emanate from the imposition of boundary conditions on a wave function. For another view point go to reference  9.

The original hypothesis  was to introduce a set of three (2x2) spin matrices that obey the same  conmutation rules as the angular momentum components  L x ,  Ly  and  Lz . These operators are given in chapter 3. They satisfy the conmutation relations  [Lx , Ly] = ih'Lz , [Ly , Lz] = ih'Lx , [Lz , Lx] = ih'Ly .

The spin operator will operate on the spin variables not on space coordinates. It will not be a differential operator, they will be matrices. Also in analogy with the angular momentum that satisfies,
L2  = h'2 l( l+1)    ,  Lz = m h'  ,  the spin S operator satisfies ,

S2 = h' s(s+1)    , Sz  =  (+/- ) ms h'   ,                     (7)

where s=1/2   and   ms =1/2 .

Such matrices are called the Pauli matrices  ( http://en.wikipedia.org/wiki/Pauli_matrices  ) ,

$\sigma_1 = \sigma_x = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$
$\sigma_2 = \sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}$
$\sigma_3 = \sigma_z = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}.$                                                                            (8)

For all  sigmas  their square , σi 2 = I , where I is the (2x2) unit matrix.

They satisfy the commutation relations
[σa , σb ] =σa σb - σb σa = 2 i εabc σc , where  εabc  is the Levi-Civita symbol.

Their anti commutation relations give ,

{σa , σb }= σa σb + σb σa = 2 I δa,b  ;δa,b is the Kronecker delta.

Two important relations follow. First

when the vectors  a and b  commute with the Pauli matrices. Second
, for  a = a n .

The spin operators are ,

Sx = (h'/2) σx  ,   Sy = (h'/2) σy   ,     Sz = (h'/2) σz                .              (9)

Using the commutation rules for the σ  one gets ,

[Sx , Sy] = (h'/2)2  ( σx σy - σx σy ) = (h'2 /4) ( 2i ) σz = ih' (h'/2) σz

= ih' Sz .                                                                                (10)

One also can verify that [Sy , Sz] = ih' Sx  and  [Sz , Sx] = ih' Sy

The Si operate on 2x1 columns.

All S i operators have the same eigenvalues  equal to (+/-) h'/2. For example let

Sx u = λ u   where  u is the (2X1) column (c1,c2)transposed .  Then one has the system

( Sx -λ I ) u = 0. This leads to the  null determinant ,to be solved for the unknown λ eigenvalues .

det (  Sx -λ I  )         λ2 -(h'/2)2 =0 . Thus λ = h/2 , -h'/2.

The eigenvectors of Sx follow from the pair of equations that give rise to the null determinant. Actually only  one is needed ,

- λ c1 + (h'/2) c2 =0.   Letting  λ = h/2 and  c1≡ 1 makes c2=1.

If   λ = -h/2   and c1=1 ,then c2= -1.

Thus the eigenfunctions of Sx are   (1 ,1) transposed  and ( 1,-1)transposed . They are orthogonal.

In the case of Sy   the solution of  det ( Sy - λI )  leads to λ2 -(-i)2 (h'/2)2 =0 , which again produces the same eigenvalues h'/2 , -h'/2 .

The corresponding equations for c1 and c2 is  -λc1 -i(h'/2) c2=0  . Letting λ= h'/2  ,   and choosing c1=1 implies c2= i . If we take λ= - h'/2 then

c1=1 , c2= -i  .The two orthogonal eigenfunctions of Sy are   (1 ,i ) transposed  and ( 1,-i )transposed .

Finally the solution of  det ( Sz - λI )   gives  ( h'/2-λ) ( -h'/2 -λ) = 0 ,with solutions h'/2 , - h'/2 . The solutions to

( h'/2 - λ ) c1 + (-h'/2 - λ) c2 = 0  are chosen to be  c1=1 , c2=0 when λ=h'/2   and c1=0 , c2=1 when  λ= -h'/2 . The corresponding orthogonal eigenfunctions of Sz are

(1 ,0 ) transposed  and ( 0, 1 )transposed

Hamiltonian function for the spin in a magnetic field ( see chapter VIII ref. 1)

We are interested in that part of the hamiltonian that operates on the spin function ( the spinor). The characteristic energies and the functions are sought.

Let  B = Bx ex  +  By ey  + Bz ez  . The hamiltonian is

H = gs (e /2m) S ∙ B = (e'h'/(2m)) {σx Bx + σy By + σz Bz  }

= μBohrx Bx + σy By + σz Bz  }                         ,     (11)

where gs =2 as mentioned before.

One gets

μBohrx Bx + σy By + σz Bz  } ( c1,c2)transpose = E (c1,c2)transpose              .          (12)

The solution of the determinant equation is shown in the next steps ,

(13)

As expected the two possible values for the energies along the B are  , (+/-) μBohr B .

For the  first eigenfunction let E = μBohr B in the above matrix.

(Bz - B) c1 + (Bx - i By ) c2 = 0 . Choose c1=1 , then c2 = (B-Bz )/(Bx - i By) .

When introducing the angles  θ and  φ , the components of B are   Bz = B cos(θ)  , Bx = B sin( θ ) cos( φ  )  , By = B sin( θ ) sin ( φ  ).

Thus c1 =1 , c2 = { 1-cos( θ ) }exp(i φ) / sin ( θ )  or c1= 1 , c2 = tan(θ/2) exp(i φ) .

So   u1 = ( 1 , tan(θ/2) exp(i φ) ) transposed    is  the eigenvector corresponding to E = μBohr B .

A vector orthogonal to ( 1 , tan(θ/2) exp(i φ) ) transposed  is ,  u2 = ( 1, -cot(θ /2) exp(i φ) )transposed  and corresponds to the energy

E = -  μBohr B . The orthogonality relation is readily verified ,  (1 ,  tan(θ/2) exp(i φ) )* ( 1, -cot(θ /2) exp(i φ) )transposed = 1-1=0.

Motice that u1 and u2 have not been normalized.

Example :  Find the expectation value     <1/Sx/ 1> /<1/1> =     u1t* Sx  u1 / (1+tan2 (θ/2))

=    (1, tan (θ/2) exp(-iφ) )  (h'/2) σx  (1, tan (θ/2) exp(iφ) ) /sec2 ( θ/2)

=  (h'/2){tan( θ/2) (2 cos(φ) ) } / sec2 (θ/2 ) = (h'/2) sin(θ) cos(φ)

Electron with Spin in a central field

The electron may in general interact with two magnetic fields. One is external customarily labeled B but in the the rest frame of the electron there is a magnetic field , Binternal ,due to the revolving positive nuclear charge.

Consider a hydrogenic atom of charge Z. The internal B is proportional to the angular momentum L ( see ref.  10) . For at the center of a curent loop

the field  B = μ0 I /(2 r) , (μ0 = 4π x10-7 , T-meter/amp, the permeabilty of vacuum)  , I = Z ev/(2πr)  and therefore

B = μ0 Ze v/(4 π r2)  .

Letting   v= L/(me r)   we get

B =  μ0 Z e L /(4 π me r3)                               (14)

Writing  the expression for the interaction hamiltonian as a dot product one gets

Hspin-orbit = gs (e /2m) S ∙ B =gs (e /2me) S ∙   { μ0 Z e L /(4 π me r3)  }

=  gs [1/ (8π me2 )] μ0  (Z e2 /r3 ) (  S∙ L)                               .  (15)

But  the electrostatic potential   energy  is  Φ(r)  = (1/(4πε0 ))(- Ze2 /r) and dΦ/dr = (1/(4πε0 ))(Ze2 /r2 )  , thus

Z e2 /r3 = ( 4πε0 /r) dΦ/dr . Equation (15) becomes

Hspin-orbit = gs (1/(2me2 ))(μ0ε0 ) (1/r)  (dΦ/dr) (  S∙ L)

=(1/(me c)2 ) (1/r) (dΦ/dr) (  S∙ L) .

L. H. Thomas showed in 1926 based on special relativity transformations that the precession frequency had to be corrected. This introduces a factor of (1/2) in the above formula , so that the correct spin-orbit operator is

Hspin-orbit = (1/2) (1/(me c)2 ) (1/r) (dΦ/dr) (  S∙ L) .        (16)

Example : Give an order of magnitude calculation of (16).  Let Z=1 and suppose the electron is in the 2p state.

The factors involved are , /S∙ L/ ~  h'2 ~ 10-68 J-s  ,   r ~ 4 aBohr ~ 2x10-10 meters  , me ~ 10-30  kg  , c=3 x108 m/s,

(1/r)(dΦ/dr) ~ (1/4πε0 ) e2 /r3 ~ (1010 ) (10-38 )/(8 x10-30 ) =.125x102 .   Inserting all the factors in (16) gives

Espin-orbit ~ (+/-) 2 x10-24    joules   =   (+/-) 1 x10 -5    eV    . The splitting is twice the given value , hence

∆ E ~ 2 x10-5  eV    .

For more details see Hydrogen Fine Structure in,  http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/hydfin.html .

The Schrodinger-Pauli equation

The non relativistic lagrangian of a particle of mass m , charge q, and  velocity v, moving in a region with potentials A and  φ is

L =  (1/2) m v2  + q v ∙ A   - q φ    .                ( 17 )

The canonical momentum is

p = L / v   =  m v + q A  = p0 + q A  , (with p0 =mv)                          ( 18  )

and   therefore                                     v = (1/m) ( p - q A ) = (1/m)(p + eA) ,   where q= -e= -1.602E-19 coulombs.

Checking the dimensions, one has that  A ~ tesla-meter ~ newton s/coulombs  . Hence qA ~ newton-second ~ kg (m/s) ~ linear momentum.

Substitution of  v in the  hamiltonian ,  H = (1/2) m v2  + q φ , gives the operator in quantum mechanics ,

H = (1/2m) ( p + e A )2 + q φ      .  (19)

When the square in (19) is taken, it is assumed that the B field makes   qA << p average . We can verify that for example in hydrogen for the first "orbit"  ,     p ~ m v ~ m (c/100)  ~ 10-30 106  ~ 10-24  kg m/s .On the other hand

qA ~ q B Length ~10-19 (1)(10-10 ) ~  10-29  kg m/s  .In this case , qA is five orders of magnitude smaller than p (average), and the ratio (q A/p)2 ~ 10-10  . What appears in the hamiltonian is  the square of these terms.

A heuristic way to derive the Pauli equation , is to consider the kinetic operator acting on a function provisionally called Ψ.

The nature of Ψ will be specified later.

(1/2m) ( p +e A )2  Ψ ≈  ( 1/2m) { p2 Ψ  +e p ∙ ( AΨ ) + e Ap Ψ  } ,            (20 )

where we have neglected the last term  of order  O ( (q A)2 Ψ  . The field B is assumed to be homogenous and the electromagnetic potential is the cross product

A = (1/2) ( B x r )                                         .     ( 21)

It follows that  the divergence of A is zero ;  div (A) = del ∙ A = 0 . The second term is,

e p ∙ ( AΨ ) =  e (p ∙ A )Ψ  + e A ∙ ( pΨ )  = 0  + e A ∙ ( pΨ )            .    (22)

Substitution of (18) in (16) gives

(1/2m) ( p + e A )2  Ψ ≈ ( p2 /(2m) ) Ψ   + (e/m) A(p Ψ )     .   (23)

The second term in (19) is examined further,  to  get

Ap =(1/2) (B x r)∙ p =  (1/2)( r x p) ∙ B = (1/2) L ∙ B           ,     (24)

where L is the angular momentum of the particle.

The second term in (23) is equal to   (e /(2m) ) L ∙ B Ψ  , and from eq (2) it also equals  - μ ∙ B  . In consequence  the hamiltonian is

H = p2 /(2m)   - μ ∙ B   - e φ                                                .      (25)

In a hydrogenic atom φ(r) = Ze/r (in C.G.S. units).

After all that has been explained about the electron spin , a  generalization  of (21) is given by

H = p2 /(2m)   + (e/2m) (L + 2S)  ∙ B   - e φ .                           (26-a)

Tha factor of 2 multiplying S comes from the gs factor which is approximately equal to 2. Introducing the total angular momentum  operator   J = L+S  , (22-a) becomes ,

H = p2 /(2m)   + (e/2m) (J + S)  ∙ B   - e φ .                        (26-b)

The magnetic term  Hm = + (e/2m) (L + 2S)  ∙ B   may also be written as

H m =  - μL ∙ B  - μS ∙ B           .                                                     (27)

Again , remember that here  e=1.602E-19 C and not -1.602E-19coul. These contributions are usually treated as a perturbation. One has solved

The problem                                 En,l,m  = < Ψn,l ,m  /   p2 /(2m)   - e φ         / Ψn,l ,m   >   has been solved .

To apply the operator S  which depends on the matrices the spatial wave function  Ψn,l ,m (r,θ,φ) is multiplied by the spinor function  u ,formed by a (2x1) column.

Hence Ψtotal (r,θ,φ,s) = Ψn,l ,m (r,θ,φ)  u(ms)   . For example one writes ms =1/2  and u(1/2) = (1,0)transposed  to characterize an electron spin oriented along the +Z axis or ms = -1/2 , u(-1/2) = (0,1)transposed , for the electron oriented along -Z axis.

For some computations , one assumes that the direction of strong external field B is that of the  Z axis.  The magnetic interaction Hm simplifies to (see Atoms and Molecules by Mitchel Weissbluth , chapter 17    )

Hm =  (e/2m) (Lz + 2Sz)   B                           .           (28)

The expectation value of ( 28) would be   (eh'/(2melectron) ( morbital +2(+/-)(1/2) )   , or   μBohr ( morbital  (+/-) 1) .  In a strong field , the vectors L and S precess  about B more rapidly than they precess about the total angular momentum  J = L + S.

Larmor frequency    ω = - e B/(2m) .....  significance the precession is dominated by the strength of B.

An internal field Binternal  is proportional and has the direction J.  It will have a corresponding  associated ωi = - e Binternal/(2m)  . The ωi is the precession frequency  of the triangle L and S around J.

For an external field there is   ωe = - e Bexternal /(2m) , the frequency of precession of J about the external  Bexternal .

But in a weak field B formula (23) is not correct. Now L and S precess  about J more rapidly than about the external field B.  More detailed considerations ( see ref. 5 , Part III ) where the precession of the vector J around the external B is taken into account lead to the modified  formula

<Hm> =  μBohr B Jz ( 1 +(S∙J)/J2 ) ≡μBohr B Jz  g Lande  ,              (29)

where the famous Lande factor enters in quantum mechanics ,

g Lande =  1 +(S∙J)/J2  =  1+ {J(J+1) +S(S+1)- L(L+1) }/[2J(J+1)]   .  (30)

The number of components in (29) is governed by Jz which runs through  the values, Jz  ,   Jz -1 , ...  , -Jz +1,  - Jz .

Figure 3  . Shows  the relative precession of  L, S and J .

Fig 4. Projection of S upon J

Example: Employing (29) and (30)  ,calculate the splitting of the states    a) 2 S1/2  , b)   2 P1/2   , c) 2 P3/2

a)  2 S1/2 2s+1L j  .    Here l = 0 , s=1/2  , j = l+1/2 = 1/2   , jz = (+/-) 1/2 . Inserting these quantum numbers in (29)-(30)  gives ,

g Lande =  1  +   {(1/2)(3/2) + (1/2)(3/2) }/ [ 2(1/2)(3/2)] = 2 .

Emagnetic  =   μBohr B Jz  g Lande = μBohr B (+/- 1/2)  2 =   (+/-) μBohr B .

b) 2 P1/2 = 2s+1L j   . Here l = 1 , s=1/2  , j = l-1/2 = 1/2   , jz = (+/-) 1/2 . The Lande factor is

g Lande =  1  +   {(1/2)(3/2) + (1/2)(3/2)- (1)(2) }/ [ 2(1/2)(3/2)]  = 1-1/3 = 2/3  .

Emagnetic = μBohr B Jz  g Lande = μBohr B (+/- 1/2) ( 2/3) = (+/-) μBohr B /3

c)2 P3/2 = 2s+1LNow l = 1 , s=1/2  , j = l + 1/2 = 3/2   , jz = (+/-) 1/2 , ( +/-) 3/2  . The Lande factor is

g Lande =  1  +   {(3/2)(5/2) + (1/2)(3/2)- (1)(2) }/ [ 2(3/2)(5/2)]  =  4/3  .

One pair of energies is ,   μBohr B Jz  g Lande = μBohr B (+/- 1/2) (4/3) = (+/-) (2/3) μBohr B .

The other splitting is  μBohr B Jz  g Lande = μBohr B (+/- 3/2) (4/3 ) = (+/-) 2 μBohr B  .

END OF CHAPTER