# Homework Assignment 5 Solutions

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```					15/03/2007 Particle Physics, Imperial College

Particle Physics Homework Assignment 5

Dr. Costas Foudas 16/3/07

Problem 1: As shown in class the Dirac matrices must satisfy the anti-commutator
relationships:
2
{  i ,  j } = 2  ij ,   {  i ,  } = 0 with     = 1

I. Show that the  i ,  are Hermitian, traceless matrices with eigenvalues ±1 and even
dimensionality.

II. Show that, as long as the mass term is not zero and the matrix  is needed, there is
no 2x2 set of matrices that satisfy all the above relationships. Hence, the Dirac

matrices must be of dimension 4 or higher. First show that the set of matrices 1 ;  
can be used to express any 2×2 matrix. That is the coefficients c 0 , c i always exist
such that any 2×2 matrix can be written as:

 
A B
C D
= c 0 1c i  i   (1)

Having shown this you can pick and intelligent choice for the  i in terms of the
Pauli matrices, for example  i = i which automatically obeys {  i ,  j } = 2 ij ,

and express  in terms of 1 ;   using (1) . Show then that there is no 2×2
 matrix that satisfies {  i ,  } = 0 .

Solution:

(a) The Dirac Hamiltonian is given by H = ⋅  m . Hence,  i ,  are required to
a p
be hermitian and as one can see in the standard representation the are indeed Hermitean.
2         2
(b)      i  =   ⇒ i  =                                          (1)

{  i ,  j } = 2  ij ⇒ 2  i 2 = 2 ii ⇒  i 2 = 1               (2)

where in (2) the double index does not imply summation in this case.

From (1) and (2) we have that:  = ±1

{  i ,  } = 0 ⇒  i     i = 0 ⇒  i = −  i                     (3)

Partcile Physics 2007, Lecturer Dr. C. Foudas      1
15/03/2007 Particle Physics, Imperial College

Hence using (3) and 2 = 1 we have that:

2
Tr  i  = Tr −  i  = −Tr   i  = −Tr  i  ⇒ Tr  i  = 0

(recall that tr  A⋅B = tr  B⋅A )

Similarly we have that:

{  i ,  } = 0 ⇒  = − i   i ⇒ Tr  = −Tr  i   i  ⇒

Tr  = −Tr  ⇒ Tr  =0

Since the Dirac matrices have eigenvalues +/- 1 and are traceless they must have even-
dimensionality. This is because one can diagonalise them and in this case the diagonal
elements are the eigenvalues. So the sum of the eigenvalues (trace) must be zero. If the
eigenvalues are only +/- 1 the only way this can happen is if there is an equal number of
+1 and -1 (even dimensionality)

(b) Any 2x2 matrix can be written as:

 
a b
c d
= c 0⋅1c i⋅ i

because for every set of a , b , c , d there is always a set of c i such that:

a = c 0 c 3 ; d = c 0−c 3 ; b=c 1−ic 2 ; c = c 1ic 2 ⇒

ad             a−d             bc            c−b
c0 =            ; c3 =          ; c1 =         ; c2 =          
2                 2                 2               2 i 

Now suppose that : 0 = c 0⋅1c i⋅ i and  i =  i if this set of gamma matrices is to
be a valid one it should satisfy the Dirac commutation relationships. Hence,

i    j                    j           i   j        j       ij      j
{ c 0⋅1c i⋅ ,  } = 0 = c 0 2  c i {  ,  } = 2 c 0  2 c i  = 2 c 0  2 c j

Partcile Physics 2007, Lecturer Dr. C. Foudas       2
15/03/2007 Particle Physics, Imperial College

The only way that this last expression can be zero is if both coefficients are always zero.
Hence, there is no non-trivial choice for 0 and a 2x2 representation for the Dirac
matrices is not possible. As we will see later, if the particle is massless then we don't need
the  matrix and then a 2x2 representation is possible. The solutions of such an equation
will be then two component spinors which in the past were the basis for the two-
component neutrino theory.

Problem 2:

I. Show that the Weyl matrices:


 =
     
− 0
0        =     
0 1
1 0

satisfy all the Dirac conditions of Problem 1. Hence, they form just another
representation of the Dirac matrices, the Weyl representation, which is different than
the standard Pauli-Dirac representation.

II. Show the the Dirac matrices in the Weyl representation are

 =

    0 
− 0


        0 =    
0 1
1 0

0 1 2 3
III. Show that in the Weyl representation 5 = i     =                   −1 0
0 1   
a p m]  = E  in the particle rest frame using the
IV. Solve the Dirac equation [⋅
Weyl representation.
1±5 
V. Compute the result of the chirality operators         when they are acting on the
2
Dirac solutions in the Weyl representation.

Solution:
(a)

{  i ,  j } = 2 ij ⇒
   − i 0
0 i       − j 0
0   j    

− j 0
0   j      − i 0
0 i         = 2 ij⋅1

Partcile Physics 2007, Lecturer Dr. C. Foudas     3
15/03/2007 Particle Physics, Imperial College

⇒
    i  j
0
0
i
 j      

 j  i
0
0
 j i      = 2 ij⋅1

   { i , j }
0         i
0

{  , j }
= 2 ij⋅1 ⇒

   { i , j }
0         i
0

{  , j }
= 2 ij⋅1 ⇒

   2 ij 0
0 2 ij        = 2 ij    
1 0
0 1
= 2 ij⋅1 ⇒

(b)       { i ,  } = 0 ⇒
     
− 0
0       
0 1
1 0

0 1
1 0
− 0

0       = 0

(c)       2 = 1 ⇒              
0 1
1 0
0 1
1 0
=
1 0
0 1

So the Weyl representation is just another representation which satisfies the same anti-
commutation relationships as the Pauli-Dirac (standard) representation.

II. Just as in the Dirac Pauli case one can define the Dirac gamma matrices as:

    
 =  =             
0 1
1 0

− 0
0        =
0 
− 0



0 =  =            
0 1
1 0


Since we have already proven that the  ,  satisfy the Dirac anti-commutation
relationships, it follows automatically as in the standard representation that the gamma
matrices of the Weyl representation also satisfy:

{  ,  } = 2 g  

Partcile Physics 2007, Lecturer Dr. C. Foudas         4
15/03/2007 Particle Physics, Imperial College

III.Again we have by definition that                  5 = i  0 1  2  3

                                                                  
1                     2                  3
0         I     0                       0                    0   
Hence,         5 = i                   1                        2                     3
⇒
I         0    −   0                   −   0                −   0

5 = i
   − 1 0
0   1       0   2
− 2 0                0   3
− 3 0              ⇒

5 = i
  0
− 1  2
− 1  2
0                    0   3
− 3 0               ⇒

                                            
3                      3
0     −i                   0   
5 = i          3                         3
⇒
−i     0                   −   0

                                  
3                    3
0            0                           −I          0
5 =                                            =
3 0            3
−   0                         0          I

IV. At the particle rest frame the momentum is zero so we have that:

[⋅
a p m] = E  ⇒ = E  ⇒                                      −E m
m −E              
1
2
= 0     (1)

By requiring that we have non-trivial (non-zero) solutions we get that the determinant
must be zero. This way there inverse matrix does not exist because if it did one could
always multiply with the inverse from the left side and prove that the solution is
identically zero.

This means that:                    E 2 = m2 ⇒ E = ±m

and again we have negative and positive solutions.

Positive Energy Solutions:

To obtain the positive energy spinors we substitute E = m in equation (1) :

Partcile Physics 2007, Lecturer Dr. C. Foudas                 5
15/03/2007 Particle Physics, Imperial College

   −m m
m −m    
1
2
= 0 = 0 ⇒             2 =1

 
1                 0
Hence:                    =



⇒  =
1           0
1
0
2
;  =
1
0
1

Negative Energy Solutions:

In a similar fashion the negative energy solutions can be obtained by substituting
E = −m in equation (1) :

   m m
m m    1
2
= 0 = 0 ⇒                    2=−11

and

 
1                   0
 =
 

−
⇒  =
1          0
−1
0
;  =
2        1
0
−1

V. The chirality operators are given by:

15 
2
=           0 0
0 1

1−5 
2
=           1 0
0 0

Partcile Physics 2007, Lecturer Dr. C. Foudas           6
15/03/2007 Particle Physics, Imperial College

Hence:

15  1,2
2
 =              
0 0
0 1


=
0


15  3,4
2
 =             
0 0
0 1

−
= −
0


1−5  1,2
2
 =             
1 0
0 0


=

0

1−5  3,4
2
 =             
1 0
0 0

−
=

0

Notice that the chirality operators in the Weyl representation do not mix upper and lower
components as they do in the Pauli-Dirac representation.

Problem 3: Use the Dirac Hamiltonian in the standard Pauli-Dirac representation,

H = ⋅
a p m

           
to compute [ H , L] and [ H ,  ] and show that they are not zero.
Use the results to show that:

  1 
[ H , L   ] = 0
2

where the components of the angular momentum operator is given by:

           
Li =  ijk x j p k

and the components of the spin operator are given by:


i =
 
i 0
0 i

Recall that the Pauli matrices satisfy  i  j =  ij i  ijk  k

Partcile Physics 2007, Lecturer Dr. C. Foudas     7
15/03/2007 Particle Physics, Imperial College

Solution:

In the Dirac Pauli representation we have that:

 p m =
H = ⋅
        p
m ⋅
 p
⋅ −m                   (1)

and

[ H , L j ]=[ i p i , L j ]= i [ p i , jlm x l p m ] =  i  jlm x l [ p i , p m ] i  jlm [ p i , x l ] p m ⇒

 p
[ H , L j ] =  i  jlm x l 0 i  jlm −i  il  p m = −i  jlm  l p m = −i   ×  j ⇒

 L]
[ H ,  = −i   × 
 p                                        (2)

(all the variables are QM operators and obey QM commutation relationships)

      a p m ,  ] =
[ H ,  ] = [⋅       
[        p
m ⋅
 p
⋅ −m
,

 0

0   ]              ⇒


[ H ,] =
    p
 p
m ⋅
⋅ −m          

 0

0 
−

 0

0 
m ⋅
 p
 p
⋅ −m             ⇒


[ H , ] =
       0
 p 
[ ⋅ ,  ]
 p 
[ ⋅ ,  ]
0                             (3)

However:               i  j =  ij i  ijk  k ⇒ [ i , j ] = 2 i  ijk  k                     (4)

Using (3) and (4) we get:

Partcile Physics 2007, Lecturer Dr. C. Foudas              8
15/03/2007 Particle Physics, Imperial College


[ H ,]j =
         0
[ i p i ,  j ]
[ i p i ,  j ]
0               = 2 i  ijk
   p
0
i k
pi  k
0      

[ H ,  ] j = 2 i  ijk
  0
i k
p
pi  k
0         = −2 i  jik p i
   0 k
k 0           = −2 i  jik p i  k ⇒

            p 
[ H ,  ] j = −2 i  ×                                        (5)

And from (3) and (5) we have that:

  1                    1
 p
[ H , L   ] = −i   ×  −2 i    = 0
p×
2                     2

Partcile Physics 2007, Lecturer Dr. C. Foudas               9

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