Homework Helper

## Main Question or Discussion Point

Hello, i tried to evaluate this particular little guy:

$$\text{Tr} (\gamma ^0 p_\mu \gamma ^\mu \gamma ^0 q_\nu \gamma ^\nu )$$

using these identities:

$$\gamma^0 \gamma^0 = I$$

$$\text{Tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 4 (g^{\rho \sigma} g^{\mu \nu} - g^{\nu \sigma} g^{\mu \rho} + g^{\mu \sigma}g^{\nu \rho} )$$

$$\text{Tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}$$

$$g^{00} = 1, \quad g^{ii} = -1$$

using that second relation, I get:

$$p_\mu q_\nu \text{Tr} (\gamma ^0 \gamma ^\mu \gamma ^0 \gamma ^\nu ) = p_\mu q_\nu 4 (g^{0\mu} g^{0 \nu} - g^{0 0} g^{\mu \nu } + g^{\mu 0}g^{\nu 0} ) =$$

$$p_\mu q_\nu (8\delta ^{0\mu}\delta ^{0\nu} - 4g^{\mu \nu } ) = 4p^0q^0 + 4\vec{q}\cdot \vec{p}$$

Using the first and third, and the fact the traces are invariant under cyclic permutations of matrices.

$$p_\mu q_\nu\text{Tr} (\gamma^0 \gamma ^0 \gamma ^\mu \gamma ^\nu ) = p_\mu q_\nu 4g^{\mu \nu } = 4p^0q^0 - 4\vec{q}\cdot \vec{p}$$

What happened?

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Homework Helper

I found the solution to the paradox, I did not do a CYCLIC permutation ;-)

CarlB
Homework Helper

I think it is much easier to substitute $$p_\mu\gamma^\mu = p_0\gamma^0 + p_j\gamma^j$$. Then separate into two sums. You get two sums. One has $$(\gamma^0)^3$$ which you can reduce easily. The other has a product $$\gamma^0\gamma^j\gamma^0$$. To reduce this, note that $$\gamma^0$$ and $$\gamma^j$$ anticommute. Now do the same thing with the other gamma product.

I never could figure out why students are taught the hard way to do these problems. One should always rely on the facts of the Clifford algebra, that is, anticommutation, and squaring to +1 or -1 for the gammas.

Homework Helper

I don't know, we are taught to use anticommutation relations with the metric and using cyclic invariant of trace.

CarlB