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Dirac spinors

  1. Oct 1, 2016 #1
    Hi,

    It's known that ## \bar{\psi}_L \psi_L = \bar{\psi} \psi_L## I tried to work this out but i do not reach that



    Here what I do : since ## \bar{\psi} = \psi^\dagger \gamma^0##, and ## \gamma_5 \gamma_0 = - \gamma_0 \gamma_5 ##

    then

    ## \bar{\psi}_L \psi_L = \frac{1}{4} (1-\gamma_5 ) \psi^\dagger \gamma^0 (1-\gamma_5 ) \psi = \frac{1}{4} \psi^\dagger \gamma^0 (1+\gamma_5 ) (1-\gamma_5 ) \to 0 ##

    I get this equals zero ! since (1+\gamma_5 ) (1-\gamma_5 ) = 0

    so what's wrong I made ?

    Best ..
     
  2. jcsd
  3. Oct 2, 2016 #2
    I knew the answer ## \bar{\psi}_L \gamma^\mu \psi_L ## which equals ## \bar{\psi} \gamma^\mu \psi_L ##
     
  4. Oct 7, 2016 #3
    Not sure I understand what you're trying to do. You want to prove this?
    $$\bar{\psi}_L\gamma^{\mu}\psi_L=\bar{\psi}_L\gamma^{\mu}\psi_L$$
     
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