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Charge Conjugate Dirac Field

  1. Jan 22, 2010 #1
    Hi,

    I'm trying to work my way through Halzen and Martin's section 5.4. I'd appreciate if someone could answer the following question:

    How does

    [tex]j^{\mu}_{C} = -e\psi^{T}(\gamma^{\mu})^{T}\overline{\psi}^{T}[/tex]

    become

    [tex]j^{\mu}_{C} = -(-)e\overline{\psi}\gamma^{\mu}\psi[/tex]

    ? Is there some identity I'm missing?

    Thanks in advance.

    -Vivek
     
  2. jcsd
  3. Jan 23, 2010 #2
  4. Jan 24, 2010 #3
    transpose the entire thing, then use the fact that the psi-bar contains a gamma_0 matrix
     
  5. Jan 24, 2010 #4
    Why? If I transpose the entire thing, I get the next line without a minus sign. But why do I transpose? Not sure I follow you..
     
  6. Jan 24, 2010 #5
    ok I can get this:



    [tex]
    -e\psi^T\gamma^0{\gamma^{\mu}}^*\psi^*
    [/tex]

    so performing a complex conjugation one gets

    [tex]
    {j_C^{\mu}}^* -e\overline{\psi}{\gamma^{\mu}}\psi
    [/tex]
     
    Last edited: Jan 24, 2010
  7. Jan 24, 2010 #6
    I don't think you understand my question here. The two expressions are equal. But are you asking me to transform one to the other by performing a transpose followed by a complex conjugation (in other words asking me to take the Hermitian adjoint)? That is, to prove A = B, I should take the Hermitian adjoint of A and find it to be equal to B. Is that what you're saying?

    (Do you intend to utilize the fact that the current density 4 vector is real? If so, we should merely be taking the complex conjugate.)
     
  8. Jan 24, 2010 #7
    can you just for completeness write down the four current and the C- transformation?

    there are a couple of conventions out there you know..
     
  9. Jan 24, 2010 #8
    ok if you go to peskin page 70, if you have it then you can work it you I think, with Halzens definitions I have no clue sorry
     
  10. Jan 25, 2010 #9
    the current is a spinor scalar, it has no spinor indecies, so do a transpose in spinor space and use that the \psi's anticommute.
     
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