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Simple chirality question

  1. Jul 30, 2015 #1
    Hi!
    Can anyone explain to me the math behind this simple step:
    $$ P_L \psi = \psi P_R $$ where $$ P_L = \frac{1}{2} ( 1 + \gamma_5 ) $$ and $$P_R = \frac{1}{2} ( 1 - \gamma_5 )$$.

    And why is $$ \bar{\psi }P_R \gamma^{\mu } \psi = \bar{\psi } \gamma^{\mu } P_L \psi ,$$
    where $$ \gamma_5$$ and $$\gamma_\mu $$ are Dirac matrices.


    Can anyone help??


    Edit: The psi's $$ \psi $$ represent Dirac spinors.
     
    Last edited: Jul 30, 2015
  2. jcsd
  3. Jul 30, 2015 #2

    strangerep

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    Where did that come from? It doesn't look quite right.

    Hint: what is the anticommutation relation between ##\gamma_5## and ##\gamma_\mu## ?
     
  4. Jul 31, 2015 #3
    It's true that ## \bar{\psi }_L \gamma^{ \mu } \psi_L = \bar{\psi }P_R \gamma^{\mu } P_L \psi ## but I can't prove mathematically this step.

    I see now that [ ## \gamma_{\mu }, \gamma_5 ] = 0 ## so it's true that ## \gamma_5 \gamma_{\mu } = \gamma_{\mu } \gamma_5 ## .

    I am still not understing the first step, though.
     
  5. Jul 31, 2015 #4

    ChrisVer

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    what does the bar notation stand for? If you apply its meaning on the expression below you will have it.
    [itex] \bar{\psi_L} = \bar{(P_L \psi)} = ...[/itex]
     
  6. Jul 31, 2015 #5

    ChrisVer

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    That is wrong. The anticommutation is zero... {A,B}= AB+BA
     
  7. Jul 31, 2015 #6

    Hepth

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    Gold Member

    First one should have a bar over the second Psi. You get it by taking the complex conjugate of the left hand side, then inserting 1=GAMMA0.GAMMA0, where needed, then commuting them through the P.
     
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