Gamma Matrices - Identities - Trace Identities

Trace Identities

Num	Identity
0
1	trace of any product of an odd number of is zero
2	trace of times a product of an odd number of is still zero
3
4
5
6
7

Proving the above involves the use of three main properties of the Trace operator:

• tr(A + B) = tr(A) + tr(B)
• tr(rA) = r tr(A)
• tr(ABC) = tr(CAB) = tr(BCA)

From the definition of the gamma matrices,

We get

or equivalently,

where is a number, and is a matrix.

(inserting the identity and using tr(rA) = r tr(A))

(from anti-commutation relation, and given that we are free to select )

(using tr(ABC) = tr(BCA))

(removing the identity)

This implies

To show

First note that

We'll also use two facts about the fifth gamma matrix that says:

So lets use these two facts to prove this identity for the first non-trivial case: the trace of three gamma matrices. Step one is to put in one pair of 's in front of the three original 's, and step two is to swap the matrix back to the original position, after making use of the cyclicity of the trace.

This can only be fulfilled if

If an odd number of gamma matrices appear in a trace followed by, our goal is to move from the right side to the left. This will leave the trace invariant by the cyclic property. In order to do this move, we must anticommute it with all of the other gamma matrices. This means that we anticommute it an odd number of times and pick up a minus sign. A trace equal to the negative of itself must be zero.

To show

Begin with,

For the term on the right, we'll continue the pattern of swapping with its neighbor to the left,

Again, for the term on the right swap with its neighbor to the left,

Eq (3) is the term on the right of eq (2), and eq (2) is the term on the right of eq (1). We'll also use identity number 3 to simplify terms like so:

So finally Eq (1), when you plug all this information in gives

The terms inside the trace can be cycled, so

So really (4) is

or

To show

,

begin with

		(because )
		(anti-commute the with )
		(rotate terms within trace)
		(remove 's)

Add to both sides of the above to see

.

Now, this pattern can also be used to show

.

Simply add two factors of, with different from and . Anticommute three times instead of once, picking up three minus signs, and cycle using the cyclic property of the trace.

So,

.

For a proof of identity 6, the same trick still works unless is some permutation of (0123), so that all 4 gammas appear. The anticommutation rules imply that interchanging two of the indices changes the sign of the trace, so must be proportional to . The proportionality constant is, as can be checked by plugging in, writing out, and remembering that the trace of the identity is 4.

Denote the product of gamma matrices by Consider the Hermitian conjugate of :

		(since conjugating a gamma matrix with produces its Hermitian conjugate as described below)
		(all s except the first and the last drop out)

Conjugating with one more time to get rid of the two s that are there, we see that is the reverse of . Now,

		(since trace is invariant under similarity transformations)
		(since trace is invariant under transposition)
		(since the trace of a product of gamma matrices is real)