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Pseudoscalar current of Majorana fields

  1. Nov 22, 2015 #1
    Consider a Majorana spinor
    $$
    \Phi=\left(\begin{array}{c}\phi\\\phi^\dagger\end{array}\right)
    $$
    and an pseudoscalar current ##\bar\Phi\gamma^5\Phi##. This term is invariant under hermitian conjugation:
    $$
    \bar\Phi\gamma^5\Phi\to\bar\Phi\gamma^5\Phi
    $$
    but if I exploit the two component structure
    $$
    \bar\Phi\gamma^5\Phi=-\phi\phi+\phi^\dagger\phi^\dagger
    $$
    the invariance under hermitian conjugation seems lost
    $$
    -\phi\phi+\phi^\dagger\phi^\dagger\to\phi\phi-\phi^\dagger\phi^\dagger
    $$
    Where is the catch?
     
  2. jcsd
  3. Nov 22, 2015 #2

    Orodruin

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    You are forgetting that you are dealing with Grassmann numbers and therefore miss a minus sign (in addition to not separating ##\phi^\dagger## and ##\phi^c## ...).
     
  4. Nov 23, 2015 #3
    Writing down the spinor indices the product ##\phi\phi## becomes
    $$
    \phi\phi=\phi^\alpha\phi_\alpha=\phi^\alpha\epsilon_{\alpha\beta}\phi^\beta
    $$
    the component ##\phi^\alpha## are Grassmann numbers but the product ##\phi\phi## should commute, am I wrong?
    For the difference between ##\phi^\dagger## and ##\phi^c## I'm using the conventions of this review, so I have that
    $$
    \Psi=\left(\begin{array}{c}\chi_\alpha\\\eta^{\dot\alpha\dagger}\end{array}\right)\quad\quad\bar\Psi=\left(\eta^\alpha,\chi_{\dot\alpha}^{\dagger}\right)\quad\quad\Psi^c=\left(\begin{array}{c}\eta_\alpha\\\chi^{\dot\alpha\dagger}\end{array}\right)
    $$
     
    Last edited: Nov 23, 2015
  5. Nov 23, 2015 #4
    Ok, i found the (silly) error:
    $$
    \bar\Phi\gamma^5\Phi=\Phi^\dagger\gamma^0\gamma^5\Phi
    $$ so under hermitian conjugation this becomes
    $$
    \Phi^\dagger\gamma^5\gamma^0\Phi=-\Phi^\dagger\gamma^0\gamma^5\Phi=-\bar\Phi\gamma^5\Phi
    $$
    that imply
    $$
    \bar\Phi\gamma^5\Phi+h.c.=0
    $$
    the same result that we found exploiting the two component structure. Correct?
     
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