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.
Lectures on Quantum Mechanics
Introduction
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 9  The electron spin
Chapter 10 The two electron atom
FREE DOWNLOAD FORTRANFORCE 2.0.8
References:
1.
Principles of quantum mechanics;
nonrelativistic wave mechanics with illustrative applications.
by
W. V. Houston
2.
Introduction to Quantum Mechanics with
Applications to Chemistry by Linus Pauling and E. Bright Wilson Jr.
3.
Quantum Mechanics: NonRelativistic Theory,
Volume 3, Third Edition (Quantum Mechanics) by L. D. Landau and L. M.
Lifshitz
4.
Applied mathematics for engineers and physicists
, by
Louis Albert Pipes
5.
Theoretical physics
by A. S Kompaneets
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/c^{2}) 
up 
u 
u 
+^{2}⁄_{3} 
1.5–3.3 
down 
d 
d 
−^{1}⁄_{3} 
3.5–6.0 
charm 
c 
c 
+^{2}⁄_{3} 
1,160–1,340 
strange 
s 
s 
−^{1}⁄_{3} 
70–130 
top 
t 
t 
+^{2}⁄_{3} 
169,100–173,300 
bottom 
b 
b 
−^{1}⁄_{3} 
4,130–4,370 
Bosons ( spin= integer)
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 r^{2} , and T=2πr/v ; thus
μ = i A = (q/T) π r^{2} = {q v/(2πr)}π r^{2}
= evr/2 ~ ampm^{2} ~ 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
μ_{B} = 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 (q_{m} ) , one may find its
magnitude. Consider the dimensional argument; let now L= length scale=h'/(mc)
μ
~ q_{m }L = i Area = (e/tau) L^{2}
= (e mc^{2}/h) L^{2} ,
q_{m }~ (e mc^{2}/h) L = (e mc^{2}/h)
(h/mc) = e c =(1.6E19 coulombs)(3e8 m/s) = 4.8e10 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.E18 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. 500505
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 = B_{z}
k.
For an orbiting
electron, the interaction potential energy will be given by the scalar product
(writing m = m_{e} , for the electron mass to avoid confusion with the
magnetic quantum number m)
U =  μ ∙B =  {q /(2m_{e})} L_{z}
B_{z .
}(3)
Suppose an electron
in the hydrogen atom has a wave function Ψ _{n , l ,m} . The
average of (3) is
U_{ave} = ∫ Ψ_{n,l,m}^{ *} (  {q B_{z}/(2m_{e})}
L_{z} _{ }) Ψ^{ }_{n,l,m}dτ
=  {q B_{z}/(2m_{e})} h' m
, m = l ,l1,...0,1,..l
. (4)
Spectral lines , of
hydrogen , will show a multiplet structure with corresponding energies
energies
E (n,m) = E_{ground }/ n^{2}  {q B_{z}/(2m_{e})}
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 B_{z }~
1.0 tesla
c calculations 12 nov 2009
real mu0 ,me
pi=2.*asin(1.)
q=1.609e19
hbar=6.63E34/(2.*pi)
me=9.11E31
Eground=2.17e18
Bz=1.
deltau=(q*Bz/(2.*me))*hbar
ratio=deltau/eground
print*,'Bz(T),deltau(eV),ratio=',Bz,deltau/1.602e19,ratio
stop
end
RUN
Bz(T),deltau(eV),ratio= 1.
5.8167283E005 4.29419288E006
We find that
{q B_{z}/(2m_{e})} h' ~ 5.82E005(eV) .
Compared with the ground state energy ∆U/E_{ground}
~ 4.29E006 . If such a transition, is posible of about 6.0 E5 eV ,
it would give rise to wavelength λ ~ ch/∆U ~
0.021 meters .
Going back to the Stern Gerlach experiment,
there the magnetic force F_{z} is measured and the gradient
∂ B_{z} / ∂ z
is known. They are given by
F_{z} =  (∂ B_{z}
/ ∂ z ) g_{s }μ_{B}
m_{s}
(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),
m_{s} , 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 /m_{s} / = 1 ,
and /m_{s} / = 1/2 .
If one writes instead of (6) , F_{z} = 
(∂ B_{z}
/ ∂ z ) μ_{B}
m_{s} , omitting the g_{s} factor
the observed splitting or force if you
like ,would be twice as large as predicted. This meant that an extra factor ,
namely
g_{s }= 2 ,
was needed to match theory and experiment.
Table of g (spin) factors.
Elementary Particle 
gfactor 
Uncertainty 
Electron
g_{e} 
2.002 319 304 3622 
0.000 000 000 0015 
Neutron
g_{n} 
3.826 085 45 
0.000 000 90 
Proton
g_{p} 
5.585 694 713 
0.000 000 046 
Muon g_{μ} 
2.002 331 8414 
0.000 000 0012 
Fig 2. The angular momentum undergoes a
precession along Z axis. Its projection is L_{z} = mh. The spin S
, when not coupled to L, also undergoes a precession along the
magnetic ( zaxis). Its component along S_{z }has 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 p_{x} →  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 },
L_{y} and L_{z} . These operators are given in
chapter 3. They satisfy the conmutation relations [L_{x} , L_{y}]
= ih'L_{z} , [L_{y} , L_{z}] = ih'L_{x} , [L_{z}
, L_{x}] = ih'L_{y} .
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,
 L^{2 } = h'^{2}
l( l+1) , L_{z} = m h'
, the spin S operator satisfies ,
S^{2} = h' s(s+1) , S_{z}
= (+/ ) m_{s} h' ,
(7)
where s=1/2 and
m_{s} =1/2 .
Such matrices are
called the Pauli matrices (
http://en.wikipedia.org/wiki/Pauli_matrices ) ,



(8)
For all sigmas their square ,
σ_{i }^{2} = I
, where I is the (2x2) unit matrix.


They satisfy the commutation relations
 [σ_{a}
, σ_{b} ] =σ_{a} σ_{b}  σ_{b}
σ_{a} = 2 i ε_{abc} σ_{c} , where
ε_{abc } is the LeviCivita symbol.
Their anti commutation relations give ,
{σ_{a}
, σ_{b} }=
σ_{a} σ_{b}
+ σ_{b} σ_{a} = 2 I δ_{a,b}
;δ_{a,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 ,
S_{x} = (h'/2)
σ_{x} , S_{y}
= (h'/2) σ_{y} ,
S_{z} = (h'/2) σ_{z}
.
(9)
Using the
commutation rules for the σ one gets ,
[S_{x} , S_{y}]
= (h'/2)^{2} ( σ_{x}
σ_{y}  σ_{x}
σ_{y} ) = (h'^{2} /4) ( 2i ) σ_{z}
= ih' (h'/2) σ_{z}
= ih' S_{z} .
(10)
One also can verify
that [S_{y} ,
S_{z}] = ih' S_{x} and [S_{z} ,
S_{x}] = ih' S_{y} .
The S_{i} operate on 2x1 columns.
All S_{ i }operators have the same
eigenvalues _{ }equal to (+/) h'/2. For example let
S_{x }u =
λ u where u is the (2X1)
column (c1,c2)^{transposed} . Then one has the system
( S_{x }λ
I ) u = 0. This leads to the null determinant ,to be solved
for the unknown λ eigenvalues .
det ( S_{x }λ
I )
= λ^{2}
(h'/2)^{2} =0 . Thus λ = h/2 , h'/2.
The eigenvectors of S_{x}
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 S_{x} are (1 ,1) ^{transposed}
and ( 1,1)^{transposed} . They are orthogonal.
In the case of S_{y}
the solution of det ( S_{y}  λI ) leads to λ^{2}
(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 S_{y} are (1 ,i ) ^{transposed}
and ( 1,i )^{transposed} .
Finally the solution
of det ( S_{z}  λI ) gives ( h'/2λ) (
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 S_{z} 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 =
B_{x} e_{x} + B_{y }e_{y}
+ B_{z} e_{z . }The hamiltonian is
H = g_{s} (e /2m) S ∙ B = (e'h'/(2m)) {σ_{x} B_{x}
+ σ_{y} B_{y} + σ_{z} B_{z} }
= μ_{Bohr} {σ_{x} B_{x} + σ_{y} B_{y}
+ σ_{z} B_{z} }
, (11)
where g_{s}
=2 as mentioned before.
One gets
μ_{Bohr} {σ_{x} B_{x} + σ_{y} B_{y}
+ σ_{z} B_{z} } ( 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 , (+/) μ_{Bohr}
B .
For the first
eigenfunction let E = μ_{Bohr}
B in the above matrix.
(B_{z}  B)
c1 + (B_{x}  i B_{y} ) c2 = 0 . Choose c_{1}=1 , then c_{2}
= (BB_{z} )/(B_{x}  i B_{y}) .
When introducing the
angles θ and φ , the components of B are B_{z} =
B cos(θ) , B_{x} = B sin( θ ) cos( φ ) , B_{y}
= B sin( θ ) sin ( φ ).
Thus c_{1} =1
, c_{2} = { 1cos( θ ) }exp(i φ) / sin ( θ ) or c_{1}= 1 ,
c_{2} = tan(θ/2) exp(i φ) .
So u_{1}
= ( 1 , tan(θ/2) exp(i φ) ) ^{transposed} is the
eigenvector corresponding to E = μ_{Bohr}
B .
A vector orthogonal
to ( 1 , tan(θ/2) exp(i φ) ) ^{transposed} is , u_{2}
= ( 1, cot(θ /2) exp(i φ) )^{transposed} and corresponds to the
energy
E =  μ_{Bohr}
B . The orthogonality relation is readily verified , (1 , tan(θ/2) exp(i
φ) )^{*} ( 1, cot(θ /2) exp(i φ) )^{transposed} = 11=0.
Motice that u_{1}
and u_{2} have not been normalized.
Example :
Find the expectation value <1/S_{x}/ 1>
/<1/1> = u_{1}^{t*} S_{x}
u_{1} / (1+tan^{2} (θ/2))
= (1, tan (θ/2) exp(iφ) ) (h'/2) σ_{x}
(1, tan (θ/2) exp(iφ) ) /sec^{2} ( θ/2)
= (h'/2){tan( θ/2) (2 cos(φ) ) } / sec^{2}_{ }(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 , B_{internal}
,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 =
μ_{0} I /(2 r) , (μ_{0} =
4π x10^{7} , Tmeter/amp, the permeabilty of vacuum) , I = Z ev/(2πr)
and therefore
B = μ_{0} Ze v/(4 π r^{2}) .
Letting
v= L/(m_{e} r) we get
B = μ_{0} Z e L /(4 π m_{e} r^{3})
(14)
Writing the
expression for the interaction hamiltonian as a dot product one gets
H_{s}_{pinorbit}
= g_{s} (e /2m) S ∙ B =g_{s} (e /2m_{e}) S
∙ { μ_{0} Z e L /(4 π m_{e} r^{3})
}
= g_{s} [1/ (8π m_{e}^{2} )] μ_{0}
(Z e^{2} /r^{3} ) ( S∙ L)
. (15)
But the electrostatic potential
energy is Φ(r) =
(1/(4πε_{0} ))( Ze^{2} /r) and dΦ/dr = (1/(4πε_{0}
))(Ze^{2} /r^{2} ) , thus
Z e^{2} /r^{3}
= ( 4πε_{0} /r) dΦ/dr . Equation (15) becomes
H_{s}_{pinorbit} = g_{s }(1/(2m_{e}^{2}
))(μ_{0}ε_{0} ) (1/r) (dΦ/dr) ( S∙ L)
=(1/(m_{e} 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 spinorbit operator is
H_{s}_{pinorbit} = (1/2) (1/(m_{e} 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} Js ,
r ~ 4 a_{Bohr} ~ 2x10^{10} meters_{ }, m_{e}
~ 10^{30 } kg , c=3 x10^{8} m/s,
(1/r)(dΦ/dr) ~ (1/4πε_{0}
) e^{2} /r^{3 }~ (10^{10} ) (10^{38} )/(8 x10^{30}
) =.125x10^{2} . Inserting all the factors in (16) gives
E_{s}_{pinorbit} ~ (+/) 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.phyastr.gsu.edu/Hbase/quantum/hydfin.html .
The SchrodingerPauli 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 v^{2 } + q v ∙ A  q
φ .
( 17 )
The canonical momentum is
p = ∂
L / ∂
v = m v + q A = p_{0} + q
A , (with p_{0} =mv)
( 18 )
and
therefore
v = (1/m) ( p  q A ) = (1/m)(p + eA) ,
where q= e= 1.602E19 coulombs.
Checking the dimensions, one has that A
~ teslameter ~ newton s/coulombs . Hence qA ~ newtonsecond ~ kg (m/s) ~
linear momentum.
Substitution of v
in the hamiltonian ,
H = (1/2) m v^{2} + 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 }10^{6 } ~ 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) { p^{2} Ψ +e p ∙ ( AΨ )
+ e
A∙ p 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 ≈ ( p^{2} /(2m) ) Ψ + (e/m) A∙ (p
Ψ ) . (23)
The second term in (19) is examined
further, to get
A∙
p =(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 = p^{2} /(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 = p^{2}
/(2m) +
(e/2m) (L + 2S) ∙ B  e φ .
(26a)
Tha factor of 2 multiplying S comes from
the g_{s} factor which is approximately equal to 2. Introducing the
total angular momentum operator J = L+S ,
(22a) becomes ,
H = p^{2}
/(2m) +
(e/2m) (J + S) ∙ B  e φ .
(26b)
The magnetic term H_{m} = +
(e/2m) (L + 2S) ∙ B may also be written as
H_{ m} =  μ_{L} ∙ B  μ_{S}
∙ B .
(27)
Again , remember that here e=1.602E19 C
and not 1.602E19coul. These contributions are usually treated as a
perturbation. One has solved
The problem
E_{n,l,m} = < Ψ_{n,l ,m} / p^{2}
/(2m)  e φ / Ψ_{n,l
,m} > has been solved .
To apply the operator S which
depends on the matrices the spatial wave function Ψ_{n,l ,m} (r,θ,φ)
is multiplied by the spinor function u ,formed by a (2x1) column.
Hence Ψ_{total} (r,θ,φ,s) = Ψ_{n,l
,m} (r,θ,φ) u(m_{s}) . For example one writes m_{s}
=1/2 and u(1/2) = (1,0)^{transposed } to characterize an
electron spin oriented along the +Z axis or m_{s} = 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 H_{m} simplifies to (see
Atoms and Molecules
by
Mitchel Weissbluth , chapter 17
)
H_{m} =
(e/2m) (L_{z} + 2S_{z}) B
. (28)
The expectation value of ( 28) would be
(eh'/(2m_{electron}) ( m_{orbital} +2(+/)(1/2) ) ,
or μ_{Bohr} ( m_{orbital} (+/) 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 B_{internal
}is proportional and has the direction J. _{ }It will
have a corresponding associated ω_{i} =  e B_{internal}/(2m)
. The ω_{i} is the precession frequency of the triangle L
and S around J.
For an external field there is
ω_{e} =  e B_{external }/(2m) , the frequency of precession of
J about the external B_{external} .
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
<H_{m}> = μ_{Bohr} B J_{z} ( 1 +(S∙J)/J^{2}
) ≡μ_{Bohr} B J_{z} g_{ Lande} ,
(29)
where the famous Lande factor enters in
quantum mechanics ,
g_{ Lande} = 1 +(S∙J)/J^{2} = 1+ {J(J+1)
+S(S+1) L(L+1) }/[2J(J+1)] . (30)
The number of components in (29) is
governed by J_{z} which runs through the values, J_{z}
, J_{z} 1 , ... , J_{z} +1,  J_{z}
.
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} S_{1/2} , b)^{
2} P_{1/2}^{ }^{ }, c)
^{2} P_{3/2}
a) ^{2} S_{1/2}
≡^{ 2s+1}L_{ j} ._{
}Here l = 0 , s=1/2 , j = l+1/2 = 1/2 , j_{z}
= (+/) 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 .
E_{magnetic} = μ_{Bohr} B J_{z} g_{ Lande
= }μ_{Bohr} B (+/ 1/2) 2 = (+/) μ_{Bohr} B_{
. }
b) ^{2} P_{1/2} =
^{2s+1}L_{ j } . Here l = 1 , s=1/2 ,
j = l1/2 = 1/2 , j_{z} = (+/) 1/2 . The Lande
factor is
g _{Lande} = 1 +
{(1/2)(3/2) + (1/2)(3/2) (1)(2) }/ [ 2(1/2)(3/2)] =
11/3 = 2/3 .
E_{magnetic} = μ_{Bohr} B J_{z} g_{ Lande}
= μ_{Bohr} B (+/ 1/2) ( 2/3) = (+/) μ_{Bohr} B
/3
c)^{2} P_{3/2}
= ^{2s+1}L_{ j }Now
l = 1 , s=1/2 , j = l + 1/2 = 3/2 , j_{z}
= (+/) 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 , μ_{Bohr} B J_{z} g_{ Lande}
= μ_{Bohr} B (+/ 1/2) (4/3) = (+/) (2/3) μ_{Bohr} B
.
The other splitting is μ_{Bohr} B J_{z} g_{ Lande}
= μ_{Bohr} B (+/ 3/2) (4/3 ) = (+/) 2 μ_{Bohr} B
.
END OF CHAPTER