Finding the Trace of a Product of 4 SL(2,C) Matrices | Helpful Guide

[div curl F= 0]

Dear All

I'd be very grateful if someone could help me out with finding the trace of a product of 4 SL(2,C) matrices, namely:

\mathrm{Tr} \left[ \sigma^{\alpha} \sigma^{\beta} \sigma^{\gamma} \sigma^{\delta} \right]

where:

\sigma^{\alpha} = (\sigma^0, \sigma^1, \sigma^2, \sigma^3)

\sigma^0 = I_2

I'm hoping this is a bunch of kronecker delta's but I can't seem to derive the correct expression needed for my work.

Regards

 

[haushofer]

Can't you just use the properties of a trace and

<br /> Tr(\sigma^{\mu}) = 2\delta^{\mu}_{0}<br />

?

So I would guess that your expression is

<br /> 16\delta^{\alpha}_{0}\delta^{\beta}_{0}\delta^{\gamma}_{0}\delta^{\delta}_{0}<br />

:)

 

[RedX]

I recall needing to calculate the trace of 4 Pauli matrices for calculating the Lagrangian of the pions. However, that was awhile ago, and I can't find the scrap paper where I wrote my result.

The way you do it is to note that the product of two Pauli matrices can be written as a single Pauli matrix, i.e., \sigma_i \sigma_j=\delta_{ij}\sigma_0+i\epsilon_{ijk}\sigma_{k}, which works for i,j not equal to 0: I separated cases for when one or both of the Pauli matrices is the zeroth one.

So basically you can reduce the 4 Pauli matrices to 1 Pauli matrix, and the trace of one Pauli matrix is either 2 or 0.

 

[haushofer]

Ah, I made a stupid mistake; I used that the trace of a product is the product of traces, but this is obviously not true; this goes only for the determinant.

 

[div curl F= 0]

Dear RedX and haushofer,

I have infact done the calculation in this way; separating out the zero cases and i,j not equal to zero cases but this has a big knock on effect on the next part of the calculation, making a very large equation out of a very small number of terms. I just thought there may be a general result for this product of 4 pauli matrices which included all the combinations in one rather than splitting it up into 4 separate cases for each pair (0-0, 0-i, i-0, i-j).

Cheers for the replies

 

[Avodyne]

I believe your trace can be shown to be proportional to
{\rm Tr}[\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\delta(1{-}\gamma_5)]
where the \gamma's are the Dirac gamma matrices. This can then be calculated using gamma-matrix trace rules. See Srednicki's QFT book for details (draft copy free online, google to find it).

 

[RedX]

I don't remember it being too difficult splitting it up. I can't think of an expression that encompasses everything. I mean something like this:

<br /> \sigma_\mu \sigma_\nu=\delta_{\mu \nu}\sigma_0+i\epsilon_{0 \mu \nu \rho}\sigma _{\rho}+ (\sigma_\mu \delta_{0 \nu}+ \sigma_\nu \delta_{0 \mu})(1-\delta_{\mu \nu})<br />

is certainly true, but I don't know if this is artificial: the equation still exhibits a split.

 

[RedX]

I believe your trace can be shown to be proportional to
{\rm Tr}[\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\delta(1{-}\gamma_5)]
where the \gamma's are the Dirac gamma matrices. This can then be calculated using gamma-matrix trace rules. See Srednicki's QFT book for details (draft copy free online, google to find it).

That's pretty clever. The product of 4 gamma matrices (in the Weyl basis) is block diagonal, and the left projection operator isolates the upper left block.

 

[Parlyne]

I believe your trace can be shown to be proportional to
{\rm Tr}[\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\delta(1{-}\gamma_5)]
where the \gamma's are the Dirac gamma matrices. This can then be calculated using gamma-matrix trace rules. See Srednicki's QFT book for details (draft copy free online, google to find it).

This isn't quite the same. This gives 2 {\rm Tr}[\sigma^\alpha\overline{\sigma}^\beta\sigma^\gamma\overline{\sigma}^\delta], where \sigma = (I_2,\vec{\sigma}) and \overline{\sigma} = (I_2,-\vec{\sigma}). This has sign differences in terms involving \sigma_0.