Transcript Introduction to Quantum Mechanics AEP3610 Professor Scott

Quantum theory of spin: algebraic approach
• in analogy with classical physics, we believe that a spinning mass
carries intrinsic angular momentum, along with the angular
momentum of its center-of-mass, which is L
• we call this electron spin angular momentum S
• Stern and Gerlach experimentally demonstrated, by putting a
beam of electrons in a non-uniform magnetic field, that electrons
seemed to carry two possible values of one spin component, so the
suggestion is that electrons have spin quantum number s = ½ and a
z-component of spin quantum number m = ± ½  ‘spin-up’ or
‘spin-down’
• we will work almost exclusively with operator algebra, and push
the analogy with what we have seen already to the hilt
• the explicit representation for spin will be 2 x 2 matrix algebra
• we have a new notation now for our humble electron’s spin
• | s m > is the ‘spin state’ for an electron with spin angular
momentum (√3ħ /2) and spin z-component ±ħ /2
What do our spin operators look like?
• start with commutation relations: assume verysame structure


by cyclicity Sˆx , Sˆ y  iSˆz
Sˆ , Sˆ  iSˆ Sˆ , Sˆ  iSˆ
y
z
x
z
x
y
• what does this tell us? Same operator behavior:
Sˆ 2 s m : s(s  1)  2 s m  34  2 s m
Sˆz s m : m s m   12  s m
• eigenstates (simultaneously measurable) of those two operators
• while we’re on the subject, we have raising and lowering operators
Sˆ : Sˆ x  iSˆ y defined in exact ly t eh same manner

Sˆ s m : s ( s  1)  m(m  1)  s m  1

 3  1  s m  1   1 1 for m   1
 4 4
2
2 2
 34  m(m  1)  s m  1   3 1
1
1
1


s
m

1



for
m


2
2
2
 4 4
• otherwise results are zero: can’t raise m = ½ or lower m =  ½
• that √ prefactor is exactly zero as a guarantee
• in fact, s may be an integer (like l was) or a ‘half-integer’ (as here)
What do our spin states look like, and are they eigenstates?
• what are our spin states? The general state is a spinor c
• we have an alternate, more explicit, column vector notation
a 
1 0
c :    ac   bc  : a    b 
b 
0 1
whichdefines thebasis spinors
• the basis spinors are eigenfunctions of S2 and Sz operators
• the eigenvalues are s(s+1) ħ2 and mħ, respectively
• in this notation, the operators are 2 x 2 matrices
• inner (scalar) products are row vectors times column vectors
• to get of S2 as a matrix, look at its effect on ‘up’ and ‘down’ spin

c d  1 3 2 1
3 2
3 2
2
ˆ
S c  : 4  c   
 4     c  4  and e  0 



 e f  0 
0 


c d   0  3 2  0 
3 2
3 2
2
ˆ
S c  : 4  c   



d

0
and
f







4
4

e f  1
1

0
3 2 1
2
s  4  
0 1 
What do more of our spin operators look like?
• to get Sz as a matrix, look at its effect on ‘up’ and ‘down’ spin
• we do it in class! Result:

c d  1 1 1
1
1
ˆ
S z c  : 2 c   
 2     c  2  and e  0




e
f
0
0
1 0 

 
 

1
 sz  2  

0

1
c
d
0
0

 
 


1
1 
Sˆ z c  :  12 c   




d

0
and
f



 
1
2
2 
e f  1
 

• now get the raising and lowering operators! Results:

 c d  0 
1
ˆ
S  c  : c   
     d   and f  0 



e f  1
0 


c
d
1
0

 
 

Sˆ c  : 0  



c

0
and
e

0
 0 
0 

e
f

 
 


 c d  0 
0 
ˆ
S  c  : 0  
     d  0 and f  0 



e f  1
0 


c d  1
0 

ˆ
S  c  : c   



c

0
and
e


 0 
1

e
f

 
 

0 1 
s    
0 0 
0 0 
s    
1 0
Now to get the last couple of them by working
backwards, and define the Pauli spin matrices
• to get Sx and Sy as a matrix recall that the raising and lowering
operators were expressed in terms of them
• we also define the 3 Pauli spin matrices {s}
• with the identity matrix I the four are said to span spin space,
since any complex 2 x 2 matrix can be expressed as a linear combo

Sˆ x :
1
2
Sˆ y :
1
2i

0 1  
S   S      : 2 s x
1 0
 i 
1
 0
ˆ
ˆ
S   S   S y : 2i S   S    2 
: 2 s y

i 0 
Sˆ  Sˆ  S x :

1
2

2

0
1 0  
3 2
3 2 1
2
and we also haveS z : 
: 2 s z and S  4  I : 4  


0

1
0
1





2
Effect of Sx and Sy on eigenvectors of Sz
• what is the effect of (say) Sx on an eigenspinor of Sz ?
1 1 0 
 0
S x c  : ?  2 
     2 c



1 0 0  2 


 
0
1
0





2
S x c  : ?  2 
    2 c



1 0 1 0
• so that’s interesting… it ‘flips’ the basis spinor
• what is the effect of (say) Sy on an eigenspinor of Sz ?
 i  1  0  i
 0
S y c  : ?  2 
  i   2 c 



 i 0  0  2 
 i


0

i
0





2
S y c  : ?  2 
 



 i 0  1  0 
i
2
c
• so that’s even more interesting… it ‘flips’ the basis spinor and it
does a strange sign/imaginary operation on it, depending
• note that we can normalize any spinor by taking its inner product
with its adjoint (in this representation, columnrow and c.c.)
• example of how this works:




0
0
2
2
i
i  

let c  A  i   c  A 0  2  1  c c  A 0  2  i   A
2
2

 2
2
 A  2
Eigenvectors for Sx
• what are the eigenspinors and eigenvalues of Sx ?
Sxc
( x)

: c
 
 det  
 2
( x)

1    0     
 0
2

 






1 0     0       2

 0
  

2

 2
2
   0
    
      0

2
0
0 1     
    choose :    
2 



1 0   
 
• but any multiple of an eigenvector is also an eigenvector of the same
eigenvalue, so let’s rescale by dividing by  (bad news if  = 0)
1 1   1    1 
 0
1 1
( x)
 2            1  c   2   when normalized



2 
1 0  
  1   
1

2

2

2
0 1 1 
1   
1 
( x)
  









1

c
 









1 0  
  1 
 

2
1
2
1
 1
 
Eigenvectors for Sy
Syc
( y)

: c
( y)

 i     0     
 0
2

  i






 i 0      0       2
   0
    
      0
 i
2
  2i 
 2


2
 det  i

0




0




choose

:




2
2
2


 2

 i   1    1    i   1 
1 1
 0
( y)
2
 2 
      i  c 




i 
i
0


i

2

 
  
  


0  i  1 
1   i    1 
( y)
  








i

c
 









 i 0   
   i    

2
1
2
1
 i 
 
• all operators have real eigenvalues and are observable
• the normalized eigenspinors for Sz are obvious building blocks for
an arbitrary spinor
• however, the normalized eigenspinors for Sx and for Sy are
equally good: we need a ‘Fourier’s trick’ in spin space!!
Decomposing an arbitrary spinor into a basis set
 
( z)
( z)
 an electronis in somespin statec 

c

c

1
 
 
• [notation modified for consistency with x and y]
• assume c is normalized:

1 
 
2
     
 



2
  is probability it's spin - up,
2
and  thatit's - down
2
• how to express c as a linear combination of (say) eigenspinors of Sx
 
 
( x)
( x)
c :      ac   bc1    
 
 
 1   
1  1

a  b    
2   
 1  1   
 
 
c :      cc ( y )  cc1( y )    
 
 
 1   
1  1


a

b


2   

 i    i    
1
2
1
2
a  b  a  12    

a  b  b  12    
• we can also express c as a linear combination of eigenspinors of Sy
1
2
i
2
a  b  a  12   i 

a  b  b  12   i 
Do these matrices obey the requisite commutation relations?
• for example, does Sx commute with Sz ?
0 1 1 0   2 1 0  0 1  2
S x , S z     
 4 
 4




1 0 0  1
0  1 1 0
  0  i  
 2
 2 0
  iS y
 4 
 i  2 
YES!



2 0 
 i 0 
2
4
0  0 0  1   2
1  0 0  0  4


0  0 1  0 
 0  1 0  0


• for example, does S2 commute with Sz ?
S , S  
1 0 1 0  33 1 0  1 0
z
0 1 0  1  8 0  1 0 1 






0  0 0 
3  3 0
 8 
  0 0 YES!
0
0

 

2
3 3
8
3 3
8
1  0 0  0 33 1  0 0  0
0  0 0  1   8  0  0 0  1 




How does one combine angular momentum?
• this might be a situation where two electrons need to have the
total spin added
• or it might be a situation where you have one electron and you
want to add its spin to its orbital angular momentum
• or you might want to combine an electron’s angular momentum
with the angular momentum of the nucleus…
• classically, it is merely the vector sum of each part – changes with
time, if torques act, etc.
• quantally, it is a more probabilistic cloud of somewhat quantized
possibilities
• the classical picture of adding two vectors is helpful, but one must
realize that in the end the sum is quantized both in length and in zcomponent, and that there will usually be more than one way to
achieve a given length/z-component combo
• we have two angular momenta: S(1) and S(2) represented by
operators of the usual form (operators, matrices..)
• to start, assume two s = ½ ‘spins’
What will our notation be? What can we cook up?
• a naïve attempt to create a couple of angular momenta together
 : both spins ' lined up upwards' ; m  12  12  1; s  maxedout  1
 : spins ' oppositelyaligned'; m   12  12  0; s  cancelledout  0
 : spins ' oppositelyaligned'; m   12  12  0; s  cancelledout  0
 : both spins ' lined up downwards' ; m   12  12   1; s  maxedout  1
in s m language we havethestates 1 1 ; 0 0 ; 0 0 ; 1 1
• if this is really a s = 1 situation, then where is s =1, m = 0 ?
• problem lies in the two middle states, which appear twice
• NOT eigenstates of the operator S2 = (S(1) + S (2) )·(S(1) + S(2))
• realize that the ‘1’ operators only operate on the first argument of
the ket, and the ‘2’ operators only on the second argument
Check out whether one of those middle states works




S 2 : S (1)  S ( 2 )  S (1)  S ( 2 )  S 2 (1)  S 2 ( 2 )  2 S (x1)S (x2 )  S (y1)S (y2 )  S (z1)S (z2 )

2

S  
 
   2 4    i 4i      ( 4  )   


3 2
3 2
2 
2
2 


  4  4  2      2  2      2     2    s ( s  1)  




2
3 2
4
3 2
4
• obviously, the other ‘naïve’ state will fail too
• Are either of them eigenstates of Sz= Sz (1) + Sz(2) ?
• NO
• are the first and last states eigenstates of either operator?




S 2 : S (1)  S ( 2)  S (1)  S ( 2)  S 2 (1)  S 2 ( 2)  2 S (x1)S (x2)  S (y1)S (y2 )  S (z1)S (z2 )
2
i i 
2

S  
 
   2 4    4    4  

3 2
3 2
2 
2
2 


  4  4  2      2  2     2 2    s ( s  1)  




2
3 2
4
3 2
4







with s  1
S z : S(z1)  S(z2)  S z    2             m   with m  1
What to do with the middle states, and recover the
missing combination to boot
• what can we cook up? Linear combinations of those two states!!
• strategy: try lowering that ‘top’ state


S : S (1)  S ( 2)  S    S (1)  S ( 2)  






  s (1) s (1)  1  m(1) (m(1)  1)     s ( 2) s ( 2)  1  m( 2) (m(1)  1)  

3

4
 12 ( 12 )    
   
3
4
 12 ( 12 )  
 lowered but not normalized(needs1/
2 factor) s  1
                      


 2
               
     


        
      
 






            2       
 s ( s  1)        with s  1 !!
S
3 2
4 2
1
2
2
2
4 2
i  i
4 2
3 2
3 2
2


4 2
4 2
2 2
2
2
3

4
3 2
4 2
3
4
1
2
2
1
2
1 

2
()
4 2
2
2 2
2
2
2 2
Another, singlet state: m = s = 0
• the new state had a + sign in it.. Try a ..
                      


 2
                
     


       
      
 





        (0)         0

 
s ( s  1)              with s  0
 S2
3 2
4 2
1
2
2
4 2
3 2
4 2
i  i
4 2
()
4 2
3 2
3 2
2


4 2
4 2
2 2
2
2
3

4
3
4
1 1 1 
  
2 2 2
2
2 2
2
2
2
2 2
1
2
1
2
• the triplet states have s = 1, m =  1, 0, 1
s m  11    ; s m  1 0 
1
2
• the singlet state has s = 0, m = 0
 

   ; s m  1 1   
sm  00 
1
2
     
• all four are orthornormal to one another
• all are eigenstates of Sz= Sz (1) + Sz(2) as well as of S2
• the Clebsch-Gordan coefficients tell how to assemble more
general linear combinations of two angular momenta