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Six gamma trace

  1. Apr 23, 2006 #1

    I think I need the following for a QFT problem.


    I know that

    and also, using




    on the first equation I inadvertently derived


    which while surprising doesn't bring me any closer to the resolution of my problem.

    Any suggestions would be great.

    Eoin Kerrane.
    Last edited: Apr 23, 2006
  2. jcsd
  3. Apr 23, 2006 #2


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    I don't have any of my QFT notebooks and textbooks with me right now :mad:

    I don't recall but what is the expression for the anticommutator of gamma 5 with a gamma matrix (it's been a long time since I have worked with gamma 5) ?
  4. Apr 24, 2006 #3


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    I had forgotten that gamma_5 anticommutes with all the gamma :surprised silly me...

    given this, Ihave this suggestion (maybe there is something wrong with this idea..maybe it gives a trivial result , I am not sure). You could do, say

    [tex]tr(\gamma^{5}\gamma^{\mu}\gamma^{\alpha}\gamma^{\beta}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma})= - tr(\gamma^{5}\gamma^{\alpha}\gamma^{\mu}\gamma^{\beta}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}) + 2 \eta^{\mu,\alpha} tr(\gamma^{5}\gamma^{\beta}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma})[/tex]

    the second case you know. For the first term, you keep pushing the [itex] \gamma^\mu [/itex] all the way through until you finally bring it back to its initial position, then you solve for your trace.

    However, I have the strange feeling that everything might cancel out and leave you with a trivial result of the form [itex]tr(\gamma^{5}\gamma^{\mu}\gamma^{\alpha}\gamma^{\beta}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}) = tr(\gamma^{5}\gamma^{\mu}\gamma^{\alpha}\gamma^{\beta}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma})[/itex]:frown:
  5. Apr 25, 2006 #4
    yes, your strange feeling is correct. after pushing the [tex]\gamma^{\mu}[/tex] all the way back to where it started you end up with a plus sign on the big trace meaning it cancels with the one on the far side leaving you with the funny identity

    [tex]\eta^{\mu\alpha}\epsilon^{\beta\nu\rho\sigma}-\eta^{\mu\beta}\epsilon^{\alpha\nu\rho\sigma}+\eta ^{\mu\nu}\epsilon^{\alpha\beta\rho\sigma}-\eta^{\mu\rho}\epsilon^{\alpha\beta\nu\sigma}+\eta ^{\mu\sigma}\epsilon^{\alpha\beta\nu\rho}=0[/tex]

    However, I think I have solved the problem. We have six indices for which there are four possibilities. Therefore, at the very least, two pairs of indices must be equal. These can be moved beside each other, multiplied and removed, leaving us with a trace over two [tex]\gamma[/tex]swhich is zero. Does this solution sound valid, or am I jumping to conclusions?

    Many thanks,
    Eoin Kerrane.
  6. Apr 26, 2006 #5

    On reflection it may not be zero. There are six indices with four possibilities for each. Therefore some must be equal. Remember that unequal indices anticommute. The possibilities are:

    1. 5 indices are equal. Four can be squared and removed, leaving


    2. 4 indices are equal, same as above.

    3. 3 indices are equal, two can be squared and removed, leaving a term like


    which is not zero as long as the three remaining indices are distinct.

    4. 2 indices are the same, in which case there must be two pairs of equal indices, they can all be squared and removed leavin [itex]tr(\gamma^{5}\gamma^{\mu}\gamma^{\nu})=0[/itex]

    Therefore the whole thing is proportional to some linear combination of the [itex]{6\choose 3}=20[/itex] terms like [itex]\eta^{\mu\nu}\epsilon^{\rho\sigma\alpha\beta}[/itex]. The next question is the relative sign of each term, and the overall proportionality constant.

    Last edited: Apr 26, 2006
  7. Apr 26, 2006 #6


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    By pushing it "all the way back", do you mean that you anticommute it back through the trace? Generally, in these problems, don't you commute the first gamma matrix once to the opposite end and then exploit the cyclic symmetry of the trace to move it back to the beginning (without change of sign)?
    Last edited: Apr 26, 2006
  8. Apr 28, 2006 #7
    Yes that is what I mean, commute the first gamma all the way to the end and then bring it back to the start using the cyclic property, then commute it through the gamma 5 to leave it back where it started. However, in this case, this process does not result in an overall factor of -1 on the big trace meaning it just cancels, leaving you with the aforementioned identity.

    I'm still working on the solution to this, as mentioned in post yesterday.

  9. Apr 27, 2010 #8
    apply the following identity
    [tex]\gamma\mu\gamma\nu\gamma\lambda = g\mu\nu\gamma\lambda+g\nu\lambda\gamma\mu-g\nu\lambda\gamma\mu-i\epsilon\sigma\mu\nu\lambda\gamma\sigma\gamma5[/tex]
  10. Apr 9, 2011 #9
    I know this is an old post, but I figured I'd reply since the question is still unanswered. I had to work this out recently for a project of my own, and I got,

    \text{Tr}[\gamma^5 \gamma^{\alpha} \gamma^{\beta} \gamma^{\gamma} \gamma^{\delta} \gamma^{\epsilon} \gamma^{\eta}] =
    4i \Big(
    g^{\alpha \beta} \epsilon^{\gamma \delta \epsilon \eta}
    - g^{\alpha \gamma} \epsilon^{\beta \delta \epsilon \eta}
    + g^{\alpha \delta} \epsilon^{\beta \gamma \epsilon \eta}
    - g^{\alpha \epsilon} \epsilon^{\beta \gamma \delta \eta}
    + g^{\alpha \eta} \epsilon^{\beta \gamma \delta \epsilon}
    + g^{\beta \gamma} \epsilon^{\alpha \delta \epsilon \eta}
    - g^{\beta \delta} \epsilon^{\alpha \gamma \epsilon \eta}
    + g^{\beta \epsilon} \epsilon^{\alpha \gamma \delta \eta}
    - g^{\beta \eta} \epsilon^{\alpha \gamma \delta \epsilon}
    + g^{\gamma \delta} \epsilon^{\alpha \beta \epsilon \eta}
    - g^{\gamma \epsilon} \epsilon^{\alpha \beta \delta \eta}
    + g^{\gamma \eta} \epsilon^{\alpha \beta \delta \epsilon}
    + g^{\delta \epsilon} \epsilon^{\alpha \beta \gamma \eta}
    - g^{\delta \eta} \epsilon^{\alpha \beta \gamma \epsilon}
    + g^{\epsilon \eta} \epsilon^{\alpha \beta \gamma \delta}

    Some notes about the notation:

    1. [tex]g[/tex] is the metric tensor diag[1, -1, -1, -1]. [tex]\epsilon[/tex] is the antisymmetric Levi-Civita tensor, defined so that [tex]\epsilon^{0123} = -1[/tex]. This seems backwards, but it means that [tex]\epsilon_{0123} = +1[/tex], which is useful.

    2. In the trace I used the first six letters of the Greek alphabet for the indices, except for the last letter I used [tex]\eta[/tex] instead of [tex]\zeta[/tex] because the latter is a pain to write.

    3. I apologize if it bothers you that [tex]\gamma[/tex] and [tex]\epsilon[/tex] are used for both tensors and indices.

    4. It should be easy to see the pattern on the right-hand side. Terms are written down in alphabetical order of the indices on [tex]g[/tex]. The remaining four indices on [tex]\epsilon[/tex] are written in alphabetical order. It would be possible to make all the terms positive by permuting indices, but in my opinion this way is easier to use when computing QFT traces.

    How to derive this: I used this definition of [tex]\gamma^5[/tex]:

    \gamma^5 = \frac{i}{4!} \epsilon_{abcd} \gamma^a \gamma^b \gamma^c \gamma^d

    This turns the trace into [tex]\epsilon[/tex] times the trace of ten gamma matrices. Without the Levi-Civita, this would give 945 terms. However, in this case only 15 unique terms survive. You quickly find that a lot of the terms are the same and a few terms vanish, so you only need to carry a few of the trace calculations to the very end. It's fairly mindless but you have to be organized or you'll get lost.

    Hope this is helpful to someone!
  11. Jun 1, 2011 #10
    This was very helpful, Thanks! :)
  12. Aug 3, 2011 #11
    just wanted to mention that all the properties of \gamma_5 traces
    can be easily seen, using the above mentioned prescription.

    and the identity obtained (without using the \gamma_5 prescription) among epsilon-tensors is known as 4-dim Schouten identity. It means that in 4-dim there can be only four independent
    vectors and therefore any fifth vector can always be expressed in those four vectors.

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