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Hermitian conjugate of Dirac field bilinear

  1. Apr 29, 2014 #1
    In the standard QFT textbook, the Hermitian conjugate of a Dirac field bilinear
    [itex]\bar\psi_1\gamma^\mu \psi_2[/itex] is [itex]\bar\psi_2\gamma^\mu \psi_1[/itex].

    Here is the question, why there is not an extra minus sign coming from the anti-symmetry of fermion fields?
     
  2. jcsd
  3. Apr 29, 2014 #2
    You don't need to anti-commute anything. The transposition automatically change the position of the fields:
    $$
    (\bar \psi_1\gamma_\mu\psi_2)^\dagger=\psi_2^\dagger\gamma_\mu^\dagger \gamma_0\psi_1=\psi_2^\dagger\gamma_0\gamma_\mu\psi_1=\bar \psi_2\gamma_\mu\psi_1
    $$
     
  4. Apr 29, 2014 #3
    This is exactly what i don't understand, so in the transposition there is a change of the postion of the fermions fields, but according to the anti-commutation rule of them, shouldn't there be a minus sign? I know there is something wrong in my understanding, but just cannot figure it out.

     
  5. Apr 29, 2014 #4

    Bill_K

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    Science Advisor

    There IS an extra minus sign. [itex]\gamma^\mu[/itex] is Hermitian, but [itex]\bar\psi\gamma^\mu \psi[/itex] is anti-Hermitian. The Hermitian quantity is [itex]i \bar\psi\gamma^\mu \psi[/itex].
     
  6. Apr 29, 2014 #5
    I'm not anti-commuting the fields. It's just the definition of transpose. Anti-commuting means, for example, to take [itex]\bar \psi_1\gamma_\mu\psi_2[/itex] and move [itex]\psi_2[/itex] on the other side, i.e. you are writing the same operator in a different way. When you take the transpose you are not rearranging the same operator, it's a new one (the transpose) and it is defined with the inverse order of operators.
     
  7. Apr 29, 2014 #6

    ChrisVer

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    Gold Member

    Is [itex]\gamma^{\mu}[/itex] hermitian?
    I am not so sure...
    [itex]\gamma^{0}[/itex] is
    but [itex]\gamma^{i}[/itex] is antihermitian.
    The conjugate of the gamma matrices, defined by Clifford Algebra [itex]\left\{ \gamma^{\mu},\gamma^{\nu}\right\} = 2 n^{\mu \nu} I_{4}[/itex] is given by:
    [itex] (\gamma^{\mu})^{\dagger} = \gamma^{0}\gamma^{\mu}\gamma^{0} [/itex]

    In fact you have:
    [itex] (\bar{\psi_{1}} \gamma^{\mu} \psi_{2} )^{\dagger}=\psi_{2}^{\dagger} (\gamma^{\mu})^{\dagger} (\gamma^{0})^{\dagger} \psi_{1} = \psi_{2}^{\dagger} \gamma^{0}\gamma^{\mu}\gamma^{0}\gamma^{0} \psi_{1} = \bar{\psi_{2}} \gamma^{\mu} \psi_{1} [/itex]
    The first is by definition of any "matrix", whether commuting or not, [itex] (AB)^{\dagger}= B^{\dagger}A^{\dagger}[/itex].
    If you now want to anticommute the AB: [itex] (AB)^{\dagger}=-(BA)^{\dagger}= -A^{\dagger}B^{\dagger}=B^{\dagger}A^{\dagger}[/itex].
     
    Last edited: Apr 29, 2014
  8. Apr 29, 2014 #7
    [itex] \gamma_\mu[/itex] is not Hermitian, but the additional $\gamma_0$ that you get from its conjugate goes together to [itex]\psi^\dagger[/itex] to form [itex]\bar \psi[/itex]. I honestly don't think there should be any extra minus
     
  9. Apr 29, 2014 #8
    One can chose all gamma matrices to be hermitian, it is merely a convention which one adopts and can be found in Sakurai, Mandl and Shaw. On the other hand one can chose only γ0 to be hermitian which is most common used convention.
     
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