Gamma matrix trace Paradox?

  • #1
malawi_glenn
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Main Question or Discussion Point

gamma matrix trace Paradox??

Hello, i tried to evaluate this particular little guy:

[tex]\text{Tr} (\gamma ^0 p_\mu \gamma ^\mu \gamma ^0 q_\nu \gamma ^\nu )[/tex]

using these identities:

[tex]\gamma^0 \gamma^0 = I[/tex]

[tex]\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} ) [/tex]

[tex] \text{Tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu} [/tex]

[tex] g^{00} = 1, \quad g^{ii} = -1 [/tex]


using that second relation, I get:

[tex] 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} ) = [/tex]

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

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

[tex]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}[/tex]

What happened?
 

Answers and Replies

  • #2
malawi_glenn
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I found the solution to the paradox, I did not do a CYCLIC permutation ;-)
 
  • #3
CarlB
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I think it is much easier to substitute [tex]p_\mu\gamma^\mu = p_0\gamma^0 + p_j\gamma^j[/tex]. Then separate into two sums. You get two sums. One has [tex](\gamma^0)^3[/tex] which you can reduce easily. The other has a product [tex]\gamma^0\gamma^j\gamma^0[/tex]. To reduce this, note that [tex]\gamma^0[/tex] and [tex]\gamma^j[/tex] 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.
 
  • #4
malawi_glenn
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I don't know, we are taught to use anticommutation relations with the metric and using cyclic invariant of trace.

I see the strength in your advice although. Thank you
 
  • #5
CarlB
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I think the reason they teach it to you that way is because it is more general. That is, they are teaching you things according to the principles of symmetry and this will work for other spin cases than spin 1/2. The trick I showed you only works for spin 1/2.

By the way, there is a whole nother method that uses density matrix principles instead of spinors and I like those methods too.
 

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