Dirac Lagr in Hermitian form

  1. May 11, 2014 #1

    ChrisVer

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    Is there any way to write the Dirac lagrangian to have symmetric derivatives (acting on both sides)? Of course someone can do that by trying to make the Lagrangian completely hermitian by adding the hermitian conjugate, and he'll get the same equations of motion (a 1/2 must exist in that case)...However same equations of motion, imply that there should be a connection between the two Lagrangians via a 4divergence, right? Unfortunately I cannot see what that could be.
    [itex]L_{D}= \bar{\psi} (i γ^{\mu}∂_{\mu}-m) \psi [/itex]
    In particular choosing
    [itex]L_{D}= \bar{\psi} (i γ^{\mu}∂_{\mu}-m) \psi + ∂_{\mu}K^{\mu}[/itex]
    seems to bring some result if I choose [itex]K^{\mu}=i\bar{\psi}\gamma^{\mu}\psi[/itex]
    but the initial lagrangian seems to get doubled...
     
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  3. May 11, 2014 #2

    ChrisVer

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    [itex]L_{D}=\frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi + \frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi -m \bar{\psi}\psi[/itex]

    [itex]L_{D}=\frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi + \frac{1}{2} i ∂_{\mu}(\bar{\psi}γ^{\mu}\psi) - \frac{1}{2}i (∂_{\mu}\bar{\psi})\gamma^{\mu}\psi -m \bar{\psi}\psi[/itex]

    [itex]L_{D}=\frac{1}{2} \bar{\psi} (i γ^{\mu}[∂_{\mu}^{→}-∂_{\mu}^{←}] -2m) \psi + \frac{1}{2} i ∂_{\mu}(\bar{\psi}γ^{\mu}\psi)[/itex]

    Should the - exist?
     
    Last edited: May 11, 2014
  4. May 11, 2014 #3
    Yes, your last expression is right.
     
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