# Trace of Pauli Matrices

1. Oct 12, 2009

### 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

2. Oct 13, 2009

### haushofer

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

$$Tr(\sigma^{\mu}) = 2\delta^{\mu}_{0}$$

?

So I would guess that your expression is

$$16\delta^{\alpha}_{0}\delta^{\beta}_{0}\delta^{\gamma}_{0}\delta^{\delta}_{0}$$

:)

3. Oct 13, 2009

### 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.

4. Oct 13, 2009

### 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.

5. Oct 13, 2009

### 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

6. Oct 13, 2009

### 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).

7. Oct 13, 2009

### 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:

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

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

8. Oct 13, 2009

### RedX

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.

9. Oct 13, 2009

### Parlyne

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 $\sigma_0$.