Chapter I

Comments

Transcription

Chapter I
CH.III
Chapter VII
The Spin
Stern and Gerlach Experiment
Under the effect of a uniform magnetic field, the Hamiltonian of the central force problem is derived to be
H
 p  q c A2
2
 V r 
(1)
With the vector potential A is related to the magnetic filed B through the relation
  
B   A
(2)
If B is uniform then

 
A  12 B  r
(3)
To derive Eq.(3) we use the identity
  
     
A B  C  AC B  A B C

  
 
    
  A    12 B  r  12   r B    B r 

 

  
   
  12 3B  B B



Now Eq.(1) can be rewritten as
H
p2
q  
q2
 V r  
p A
A2
2
2
c
2c
Using Eq.(3) the above equation becomes
H



p2
q   
q2  
 V r  
p Br 
Br
2
2c
8c 2
2
(4)
If the magnetic field is weak enough, (which is the case in practice), to drop the quadratic term in B, Eq.(4) now
takes the form, after interchanging the dot and the cross
p2
q   
p2
q  




H
V r 
B r  p 
 V r  
LB
2
2c
2
2c
(5)
Defining the orbital magnetic dipole moment of the particle as

q 
ml  g l
L
2c
(6)
1
CH.III
With gl =1 is called the orbital gyromagnetic, or the g-factor. Eq.(5) can now be written as
H
 
p2
 V r   ml  B
2
(7)
The force acting on the magnetic dipole moment is



  
F  V   ml  B

(8)
In the famous experiment of Stern and Gerlach, a beam of silver atoms is allowed to pass through a nonuniform

magnetic field. If Bz is the predominant component of B and Bz is strongly depends on z so that

B
Bz  z zˆ 
z

B
eL Bz
F  m z z zˆ   z
zˆ
z
2 c z
(9)
Or
Fz  
Where  o 
B
em Bz
 m o z
2 c z
z
(10)
e
is the Bohr magneton.
2 c
From Eq.( 10) we would expect 2l+1 traces splitting in the Stern-Gerlach experiment. Yet only two traces are
detected and this leads to the conclusion that the dipole moment of the silver atoms couldn't have been
produced by orbital momentum. To interpret this observation it is hypothesized the existence of the electron
spin and intrinsic magnetic moment which assumed to have only two projections, i.e.,
ms  
e
2 c
(11)
2
CH.III
The Spin as a Dynamical Variable:
In addition to the 3-spatial variables we add now a fourth variable, the spin variable  with   1,  =1 for
spin up and  =-1 for spin down. Now the normalization condition is written as
2
2
2
 x, y, z,  dv   x, y, z,1 dv   x, y, z,1 dv  1
(12)
r,     r   
(13)
       if   1  1 &   1  0
(14)
       if   1  0 &   1  1
(15)
Letting
With
r,   r,1    r,1  

(16)
In matrix representation we have
1
 0
   ,
 0
   
1
(17)
And
   1
    1    1
  
   1
Or
c 
   1   c1  c2 
 c2 
(18)
With c1    1& c2    1 are complex numbers.
2
c1 is the probability of finding the particle with spin up.
2
c2 is the probability of finding the particle with spin down.
2
Now c1  c2
2

 c1

c 
c2  1   1
 c2 
Or
3
CH.III
 †  1
(19)
Note that  †   †   1 and  †   0   and  are two orthonormal vectors (spinors)  they form a basis
 any arbitrary spinors can be expanded in terms of  and .
It is postulated that every physical quantity cane be represented by a linear operator. Now let F be a linear
operator such that
F  F11  F21 and F  F12  F22 
(20)
  F  c1F  c2 F  c1F11  c2 F12   c1F21  c2 F22   d1  d 2 
(21)
And let
With d1  c1F11  c2 F12
d 2  c1F21  c2 F22
and
Or in matrix form we write
 d1   F11
   
 d 2   F21
F12  c1 
 
F22  c2 
(22)
The adjoint of the operator F is
 F
F †   11
F
 21
F12 
 
F22

And the expectation value of F is

F   F   c1

F
c2  11
 F21
F12  c1 
 
F22  c2 
Since F is observable  its expectation value must be real 
F  F
F  F†


 c1

F
c 2  11
 F21
F12  c1 
   c1
F22  c 2 
 F is Hermitian.
4
 F
c 2  11
F
 21
F12  c1 
   
F22
 c 2 

CH.III
The spin Operator:
Since any physical quantity must be represented by a a Hermitian operator, we denote the spin angular
momentum by the operator S. In analogy to the orbital angular momentum L, the components of S is expected
to satisfy the commutational relations
Si , S j   i S k
(23)
The eigenket of S is  sms s, ms such that
S 2 s, ms   2 ss  1 s, ms
(24)
S z s, ms  ms s, ms
(25)
And
Introducing the raising and lowering operators
S   S x  iS y
(26)
S   S x  iS y
(27)
S  s, ms   ss  1  ms ms  1 s, ms  1
(28)
S  s, ms   ss  1  ms ms  1 s, ms  1
(29)
And
with
and
Note that s 

1
1
2
and
1
2, 2
ms   12 for electrons. (our case) 
and  
1
1
2, 2
(30)
Now since  †   †   1 and  †   0 
5
CH.III
s, ms s, m' s   mm's
(31)
Now using Eq.(24) we have
S 2  S 2
 34  2
1,1
2 2
&
S 2  S 2
1 , 1
2
2
&
Sz  Sz
1 , 1
2
2
 34  2 
and from Eq.( 25) we have
S z  S z
1,1
2 2
 12 
  12 
Similarly from Eqs.(27 & 28) we obtain
  ss  1  ms ms  1 12 , 32  0
S   S 
1,1
2 2
S   S 
1 , 1
2
2
S   S 
1,1
2 2
S   S 
1 , 1
2
2
  ss  1  ms ms  1 12 , 12  
And
  ss  1  ms ms  1 12 , 12  
  ss  1  ms ms  1 12 , 32  0
Now let us find the matrix representations of S.
Using Eq.(25) we have
s, ms S z s, ms  ms s, m' s s, ms  ms mm's   12  mm's
Sz 

 1 0 


2  0  1
(32)
Now adding Eq. (26 & 27) we have
Sx 
1
2
S

 S


s , m' s S x s , m s 
1
2
s , m' s S x s , m s 

2
s , m' s S   S  s , m s
3
4
Now using Eqs.(21 & 22) 
 ms ms  1 s, m' s s, ms  1 
6

2
3
4
 ms ms  1 s, m' s s, ms  1

CH.III
s , m' s S x s , m s 
Sx 

2
3
4
 ms ms  1 m's ,ms 1 

2
3
4
 ms ms  1 m's ,ms 1

 0 1


2  1 0 
(33)
Similarly subtracting Eq. (26 & 27) we have
Sy 
i
2
S

 S


s , m' s S   S  s , m s
Now using Eqs.(21 & 22) 
s , m' s S y s , m s 
i
2
s , m' s S y s , m s 
i
2
3
4
 ms ms  1 s, m' s s, ms  1 
s , m' s S y s , m s 
i
2
3
4
 ms ms  1 m's ,ms 1 
Sy 
i
2
3
4
i
2
3
4
 ms ms  1 s, m' s s, ms  1
 ms ms  1 m's ,ms 1
i  0  1


2  1 0 

(34)
Now S 2  S x2  S y2  S z2 . Using Eqs.( 32, 33, & 34) we get
S2 
3 2
4
1 0


0 1
(35)
It is left as an exercise to show that
0 1

S   
 0 0
(36)
 0 0

S   
1 0
(37)
And
It is, sometimes convenient to express the spin operator as
S
1
2

(38)
7

CH.III
With, from eqs.( 32, 33 & 34),
0 1
 ,
 x  
1 0
0  i

0 
 y  
i
1
0

 z  
 0  1
(39)
Which are known as the Pauli spin matrices. They obey the properties:
 i , j   2i k
 i , j   2 ij
3
  i2  3I
i 1
Tr  i  0
det  i  1
In the presence of the spin-orbit interaction, the angular momentum L is no longer constant of motion and this
assures that L is not the total angular momentum. The total angular momentum is now denoted by J and defined
as
With
  
J  LS
(40)
J i , J j   iJ k
(41)
Denoting the eigenkets of J2 and Jz by jm 
With
J 2 jm  j  j  1 2 jm
(42)
J z jm  m jm
(43)
j  0, 12 ,1, 32 ,, and  j  m  j
Spin_Orbit Interaction
Any particle with a spin posses a magnetic dipole moment. In a nalogy to Eq.(6), this dipole moment is given
by
8
CH.III

q 
ms  g s
S
2c
(44)
With g s  2.0023 is the spin g-factor. The energy acquired by this dipole due to the magnetic field produced
by the orbiting electron is
 
Vso  ms  Borbit
(45)
But according to an observer fixed with the electron, the induced m.field is given by


v 
Borbit    E
c
(46)


d
E  r    r rˆ
dr
(47)
With
Now Eq(45) becomes


v 
Vso   m s  12   E 
c

(48)
The factor 1/2 is added due to relativistic justification (Thomas precession).
Now from Eq.(480 and Eq.(47) we get

 v
 dr  
d

1
v  rˆ 
Vso  2 m s     r rˆ    12 ms
dr
dr
c

(49)
But
V  q
Vso  
&

r
rˆ 
r
 Eq.(49) now becomes
 
1 1 dV r  
1 1 dV r   
ms  v  r  
ms  L
2qc r dr
2qc r dr
(50)
Using Eq.(44) we now write the spin-orbit interaction energy as
Vso  
g s 1 dV r   
 
1 1 dV r  
m s  v  r  
SL
2qc r dr
4c 2 r dr
Now if an external magnetic field is applied we expect the Hamiltonian to be given as
9
(51)
CH.III
H  H cf  Vso  VlB  VsB
H  H cf 
gs
2c 2
1 dV r    qg l   qg s  
SL
LB
S B
2c 
2c 
r dr
(52)
For the H-atom we have g1  1, g s  2.0023 , V   e 2 r , so Eq.952) becomes
H  H cf 
e2
2c 2
1  
e  
e  
S

L

L

B

SB
2c 
c 
r3
(53)
The above equation can't be solved exactly. However, there are two cases of practical importance. If the
external magnetic field B is weak compared with the induced m.field we can neglect the last two terms of
Eq.(53) (anomalous Zeeman effect). In the other hand, if the external magnetic field B is strong compared with
the induced m.field we can neglect the second terms of Eq.(53) (Paschen-Back effect).
10

Similar documents