Download Isospin, Strangeness, and Quarks

The Dirac Equation: Introduction
The most natural candidate for a relativistic wave equation is the Klein-Gordon equation:
2
2

t 2
2
 2    m2 c 4 
or equivalently
H
2
 P 2    m2 
(1.1)
With the usual operator definitions of H  i  0 and Pk  i  k , this is equivalent to E 2  P 2  m 2 .
But just as this equation has the two roots E   P2  m2 , equation (1.1) yields two types of solution,
one for positive E and one for negative E(i.e.,   Ae ip x  Ae i ( Et  p x ) . The negative energy solutions
appear to be unphysical, but we cannot simply exclude them because we must require that the solutions
of (1.1) form a complete set.
The unwanted negative-energy solutions are produced because this is a second order equation
in  / t . To eliminate them, Dirac tried to find an equation first-order in  / t like the Schrodinger
equation, but first-order in space as well as in time so that space and time coordinates are on the same
footing. He tried
i
or equivalently


  k
   mc 2 
k
t
i x
H  =  i P i   m  
(1.2)
(1.3)
But if this is to be a relativistically invariant equation, then operating with H again on (1.2) must
reproduce (1.1)b. In order for this to happen, the cross-terms produced by squaring (1.3) must vanish,
so the  k and  must satisfy the following properties:
(b)
i2   2  1
 j   j i  0 (for i  j)
(c)
 i    i  0
(a)
(1.4)
Equation (1.4)b tells us that i and  cannot be ordinary numbers, they must be matrices in
order to anticommute. (Consequently the “1” in equation (1.4)a must be interpreted as the identity
matrix.) Since these matrices operate on  ,  must be a column matrix. This tells us that  has
multiple components which is encouraging since we know that the electron has two spin states and we
are used to representing these as column matrices.
From (1.3), the i and  must be Hermitian matrices, like H.
From (1.4)a they must have eigenvalues of 1 .
Equations (1.4)b and c tell us that these matrices must be traceless. The trace of a matrix is the
sum of its diagonal elements [ Tr(A)  Tr(Ai j )  Aii , with implied summation over i]. The trace of a
product of matrices is invariant under cyclic permutations of the matrices, e.g. Tr(ABC)=Tr(CAB).
(Proof: Tr ( ABC )  Tr ( Ai j B j k C k l )  Ai j B j k C k i  C k i Ai j B j k  Tr (CAB) ). So starting from (1.4)c, for
example, we can right-multiply both sides by  and take the trace to get (using   1 ):
1
Tr ( i )  Tr (  i  )  Tr (  i )  Tr ( i )
(1.5)
requiring Tr ( i )  0 . A similar operation, but right-multiplying by  i , shows Tr (  )  0 .
Since the trace of a hermitian matrix is just the sum of its eigenvalues, and since these
eigenvalues must be 1 for the  and  matrices, the dimension N of these matrices must be even.
The obvious choice is N=2, especially since we know the electron has two spin components. The Pauli
matrices are the clear choice for the  i ’s:
0 1
 0 i 
1 0 
, 2  
, 3  

1 0
i 0 
 0 1
1  
(1.6)
These satisfy the  -parts of equations (1.4), but the only linearly independent matrix satisfying (1.4)a
is the unit matrix, and this is not traceless. There is no set of 2  2 matrices that will work. The next
possibility is N=4, and here we find success. The canonical choice is
0
i  
 i
i 
1 0 
,   

0
 0 1
(1.7)
where each matrix element is a 2  2 matrix. You should check that these do in fact satisfy (1.4).
Current Conservation
If the Dirac Equation is to be the equation for an electron (or fermion, in general), then we must
be able to define a conserved current from it. To create this, we follow the same procedure that we
used to produce a conserved current from the Schrodinger Equation. First we take the hermition
conjugate of equation (1.3):
 †
 †
(1.8)
i
  k  k   †  mc 2
t
x
i
Note that since  is a 4-component column vector,  † is a 4-component row vector,
 †   1* ,  *2 ,  *3 ,  *4  . Now left-multiply (1.8) by  , right-multiply (1.2) by  , and subtract. This
gives



  † 
  †
i  †

      † k k  k  k  
t
t
i
x
x






 †     k   † k  

t
x
(1.9)
Identifying
  †
and
J k   † αk 
then equation (1.9) becomes the equation for current conservation:

 J  0
t
(1.10)
(1.11)
This is a very significant result, and must have been critical in indicating to Dirac that he was on the
right track.
2
Spin ½
If an operator is conserved, then it must commute with the Hamiltonian, since
d
i
   , H 
dt
But using the Dirac Equation in the form of (1.3), with L  R  P , we find
 L, H    R  P, H     k R P kˆ,  i Pi   kˆ  k  R , Pi  P  i

 
 

 kˆ  i   P   i kˆ P   i   P
 k
i

i k 
i
(1.12)
(1.13)
i
so orbital angular momentum is not conserved! But if we define spin angular momentum in the most
natural way, as
 0 
(1.14)
 

20  
then
, H   i   P
(1.15)
So in general, L and  are not separately conserved, but the total angular momentum J  L   is
conserved. This is exactly the behavior we expect for a spin-1/2 particle.
This must have been a second powerful indication to Dirac that he was on the right track.
Free Particle Solutions
If we apply the Dirac Equation to a particle at rest, p=0, then from equation (1.3)
I 0 
(1.16)
H  m  m

 0 I 
so the four eigenvectors are
1  0  0  0
       
 0,  1,  0,  0
(1.17)
 0  0  1  0
       
 0  0  0  1
with corresponding eigenvalues of m, m,  m,  m . So we still have negative-energy solutions! On the
other hand, we have the relativistic energy-momentum relation, spin ½, and a conserved current.
So let’s put aside the problem of negative energies, and investigate the behavior of the two
positive-energy solutions in the presence of an electromagnetic field.
Dirac Particle in an Electromagnetic Field
We can extend the free-particle Hamiltonian to include the case of a particle in a general
electromagnetic field using the minimal-coupling substitution p  p  eA / c   (and for this analysis,
we will explicitly include c as the speed of light). To simplify notation, we will write the 4-component
 
wavefunction in terms of two 2-component spinors,     . Then the Dirac equation becomes

 
 
  
2  
i
   c     e    mc   
t   



(1.18)

 
2  
 c     e    mc  
 

  
3
In the non-relativistic limit, the energy of the particle will be dominated by its rest mass, so we can
write
 
 
(1.19)
 (t )     exp  i  mc 2 /  t   


where  and  are relatively slowly-varying functions of time. The left-hand side of equation (1.18)
then becomes
  
  
2  
(1.20)
i
 i
   mc  
t   
t   


 i mc2 /
t
Moving the second term to the right side of equation (1.18), and dividing out e
gives

 
  
20
(1.21)
i
   c     e    2mc  
t   
 


This expression is really two coupled equations, one for the upper 2 components and one for the lower
two. The second of these can be approximated, in the case of interaction energies small compared to
mc 2 , by
c 


(1.22)
2mc 2
Note that  is small compared to  for this positive-energy solution, as we expect from the freeparticle solution in the particle rest frame (where   0 ). Substituting this back into the upper
equation gives

      
(1.23)
i

 e  
t 
2m

Now we can use the identities  i j  i ijk k and  i2 to show
 i ai   j a j   aibi
 ai b j  i ijk k 
a b

 i a  b

(1.24)
Applying this to (1.23) gives


2


  p   e / c  A
e

i


 B  e  
(1.25)
t 
2m
2mc


which is the same as the non-relativistic Pauli equation (or Schrodinger equation) for a spin-1/2
particle in an electromagnetic field.
This leads immediately to correct fine structure prediction for hydrogen, including the factor of
g  2 in the electrons magnetic moment. Simply expand the first term to first order in e , and use
A  12 B  r , giving
Interpretation of the Negative-Energy Solutions
(a) Dirac “Hole Theory”
(b) Feynman antiparticle interpretation
4
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The Gamma Matrices
In order to put E and p on the same footing, it usual to left-multiply multiply equation (1.3) by
 and to define the  -matrices by:
1
 0
0
i
0   
 ,   i   
 0 1
 i
i 

0
(1.26)
The gamma matrices have the following defining properties, easily derived from (1.26) and (1.4):
 ,          2 g
    ,    
    I,      I

0 †

 


k †
0
0 2

I
k
(1.27)
k 2
Note that the I in these equations is the identity matrix. Also note that equation (1.27)c is already
contained in (1.27)a; it is included only for emphasis.
In terms of the gamma matrices, the Dirac equation is now written as:
 
 
i   0   k k    m or equivalently   P  m    0
(1.28)
x 
 t
The summation   P , which is the contraction of a 4-vector with the gamma-matrices, occurs so often
that Feynman introduced a shorthand, called the “Feynman slash notation” that is universally used
now. We write P    P , so the Dirac equation (1.28) becomes
 P  m 
0
(1.29)
The Conserved Dirac Current
We have already seen that current conservation follows from the Dirac equation, and is critical
for the viability of the theory. The continuity equation can also be derived using the gamma-matrices,
of course, and this gives a very convenient and manifestly covariant way of writing the current. This
can be done as follows: first take the hermitian conjugate of (1.28), giving
i
 † 0†  † k †
  i k   m †
t
x
 i
 † 0  † k
  i k   m †
t
x
 0
(1.30)
where we have used equation (1.27) for the hermiticity of the gamma matrices. This no longer has the
form of equation (1.28) because of the relative minus sign between the first two terms. We can fix this
by right-multiplying by  0 and using the commutation relation (1.27) to obtain
5
 † 0 0  † k 0
i
  i
   m † 0
t
xk
 † 0 0  † 0 k
 i
  i
   m † 0
t
xk
 0
(1.31)
Now we define the adjoint wavefunction  by
   † 0
(1.32)
so the adjoint Dirac equation (1.31) becomes
i
 0  k
 
 i
  m   i
  m  0
t
xk
x
(1.33)
Right multiply (1.33) by  , left-multiply (1.28) by  , and subtract to get
 
 
i            0
x 
 x
i

     0
 
x
(1.34)
Consequently, the four quantities obtained from the matrix multiplication    must transform like a
4-vector. (Note that these 4 quantities are numbers, not matrices or column vectors; . Moreover, when
contracted with  / x  they give zero, so this four-vector is a conserved current. In order to give a
quantity that represents an electric current we multiply by e :
j     

 
j 0
x 
and this is the definition of a conserved current. This is an incredibly important step.
6
(1.35)