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Trace of Pauli Matrices

  1. Oct 12, 2009 #1
    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:

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


    [tex] \sigma^{\alpha} = (\sigma^0, \sigma^1, \sigma^2, \sigma^3) [/tex]

    [tex] \sigma^0 = I_2 [/tex]

    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.

  2. jcsd
  3. Oct 13, 2009 #2


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    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


  4. Oct 13, 2009 #3
    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., [tex] \sigma_i \sigma_j=\delta_{ij}\sigma_0+i\epsilon_{ijk}\sigma_{k} [/tex], 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.
  5. Oct 13, 2009 #4


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    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.
  6. Oct 13, 2009 #5
    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
  7. Oct 13, 2009 #6


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    I believe your trace can be shown to be proportional to
    [tex]{\rm Tr}[\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\delta(1{-}\gamma_5)][/tex]
    where the [itex]\gamma[/itex]'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).
  8. Oct 13, 2009 #7
    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.
  9. Oct 13, 2009 #8
    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.
  10. Oct 13, 2009 #9
    This isn't quite the same. This gives [itex]2 {\rm Tr}[\sigma^\alpha\overline{\sigma}^\beta\sigma^\gamma\overline{\sigma}^\delta][/itex], where [itex]\sigma = (I_2,\vec{\sigma})[/itex] and [itex]\overline{\sigma} = (I_2,-\vec{\sigma})[/itex]. This has sign differences in terms involving [itex]\sigma_0[/itex].
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