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Dirac equation for the conjugated field

  1. Dec 12, 2009 #1
    This is probably a stupid question, but when I apply the Euler-Lagrange equation to the Lagrangian density of the Dirac field I get for the conjugate field

    [tex]\bar{\psi} (-i \partial_\mu \gamma^{\mu} -m) = 0[/tex] (derivative acts to the left).

    But when I take a hermitian conjugate of the Dirac equation for the field I get an extra [tex]\gamma^0[/tex]:

    [tex]0 = \left[ (i \partial_\mu \gamma^{\mu} -m)\psi \right]^\dagger = \psi^\dagger (-i \partial_\mu (\gamma^{\mu})^\dagger -m) = \psi^\dagger (-i \partial_\mu \gamma^0 \gamma^{\mu} \gamma^0 -m) = \psi^\dagger \gamma^0(-i \partial_\mu \gamma^{\mu} \gamma^0 -m) = \bar{\psi} (-i \partial_\mu \gamma^{\mu} \gamma^0 -m)[/tex].

    What am I missing?
     
  2. jcsd
  3. Dec 12, 2009 #2

    dextercioby

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    Taking the normal hermitean conjugate of the <original> Dirac eqn will not give you the <conjugated> equation. You need an extra [itex] \gamma_0 [/itex].

    Actually, you took out the [itex] \gamma_0 [/itex] from the paranthesis without it being there next to the <m>. That's wrong.
     
  4. Dec 13, 2009 #3
    Ah yes, of course. Then I can multiply with [tex]\gamma^0[/tex] from the right and get the same equation as from Euler-Lagrange. I'm such an idiot. Thanks a lot.
     
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