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Gamma matrix identity (QFT)

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that [itex]\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a}[/itex] = [itex]2\left(\gamma^{e}\gamma^{b}\gamma^{c}\gamma^{d}+\gamma^{d}\gamma^{c} \gamma^{b}\gamma^{e}\right)[/itex]

    Each of the [itex]\gamma^{i}[/itex]s are as used in the Dirac equation.

    2. Relevant equations

    [itex]\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}\gamma_{a}[/itex] = [itex]-2\gamma^{d}\gamma^{c}\gamma^{b}[/itex]

    [itex]\gamma^{a}\gamma^{b} + \gamma^{b}\gamma^{a} = 2g^{ab}[/itex]

    3. The attempt at a solution

    [itex]\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a}[/itex] = [itex]2g^{ab}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a}[/itex] - [itex]\gamma^{b}\gamma^{a}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a}[/itex]

    = [itex]2\gamma^{c}\gamma^{d}\gamma^{e}\gamma^{b}[/itex] + [itex]2\gamma^{b}\gamma^{e}\gamma^{d}\gamma^{c}[/itex]

    Perhaps I mixed up something or there is a typo...
     
  2. jcsd
  3. Sep 7, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think what you wrote so far is correct. But, as you can see, it is not yet very close to what you want.

    Instead, start over and try commuting ##\gamma^{e}## with ##\gamma_{a}##. To do this, you will need to use the identity obtained by lowering ##a## in the identity [itex]\gamma^{a}\gamma^{b} + \gamma^{b}\gamma^{a} = 2g^{ab}[/itex]
     
  4. Sep 7, 2013 #3
    Thank you.
     
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