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Problem getting a product of traces out of a single trace in a chiral perturbation theory computation

+ 4 like - 0 dislike
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I am stuck at a computation and I would appreciate any help. $U$ is the pion matrix in chiral perturbation theory

$$U=e^{i\sigma_a\phi_a/f}$$

where $\sigma_a$ are Pauli matrices, $\phi_a$ are three real scalar fields and $f$ is just a constant with mass dimensions. It is well known that this matrix is unitary, that is

$$U^{\dagger}U=I$$

Now comes the question. I want to compute this

$$Tr(\partial_{\mu}U^{\dagger}\partial_{\nu}U\partial^{\mu}U^{\dagger}\partial^{\nu}U-\partial_{\mu}U^{\dagger}\partial_{\nu}U\partial^{\nu}U^{\dagger}\partial^{\mu}U)$$

where $Tr$ denotes trace. I know the that answer is proportional to

$$Tr(\partial_{\mu}U^{\dagger}\partial_{\nu}U)Tr(\partial^{\mu}U^{\dagger}\partial^{\nu}U)$$

(if you want to know how I know this, it is claimed in page 9 of this http://arxiv.org/pdf/hep-ph/9403202v2.pdf paper). In any case, I am very puzzled here because a trace of something gives a product of traces. So, how does this follow?


  EDIT::

I still haven't solved it but I think that equation (D.7) of this review http://arxiv.org/pdf/hep-ph/0210398v1.pdf might might help to solve it

asked Jul 21, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]
edited Jul 21, 2015 by Dmitry hand me the Kalashnikov

I would have to say that answer is a typo. If you add to the term you have above the Tr(D^2) you get the "answer" you put, but without that inner trace. I believe it should just be \(Tr(\partial_{\mu}U^{\dagger}\partial_{\nu}U\partial^{\mu}U^{\dagger}\partial^{\nu}U)\)

I don't think it is a typo, and I think that equation (D.7) of this review http://arxiv.org/pdf/hep-ph/0210398v1.pdf might be the clue even though I still havent solved it. See for example how he gets two traces out of one just below D.7

I think you're right, it seems there are some neat trace properties that occur from the fact that these are 2x2 matrices.

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