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A The commutations relations for left/right handed fermions

  1. Apr 15, 2016 #1
    I have a problem where I have to know the commutation relations for left handed fermions. I know
    ##\psi_L=\frac{1}{2}(1-\gamma^5)\psi##

    ##\psi^\dagger_L=\psi^\dagger_L\frac{1}{2}(1-\gamma^5)##

    and

    ## \left\{ \psi(x) , \psi^\dagger(y)\right\} = \delta(x-y)##

    So writing

    ## \left\{P_L\psi(x) , \psi^\dagger(y)P_L\right\} ##

    ##=P_L\psi(x) \psi^\dagger(y)P_L + \psi^\dagger(y)P_LP_L\psi(x)##

    And I don't see how I can go any further.
     
  2. jcsd
  3. Apr 15, 2016 #2

    ChrisVer

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    Gold Member

    it looks fine up to here...
     
  4. Apr 16, 2016 #3
    I need to write the left handed commutator in terms of the unprojected commutator I don't see a nice way to do this? How can I commute the P_L operators past the psi's?
     
  5. Apr 16, 2016 #4

    Hepth

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    Gold Member

    I want to say its the same, and will just be ##P_L \delta(x-y)##, but I'm not sure.

    Maybe what you can do is write out the spinor indices explicitly. Normally you have :
    $$
    \left\{\psi^a , \psi^{\dagger b} \right\} = \psi^a \psi^{\dagger b}+\psi^{\dagger b} \psi^a = \delta(x-y) \delta^{ab}
    $$
    now we want to understand
    $$
    \left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ai} \psi^i \psi^{\dagger j}P_L^{jb} + \psi^{\dagger j} P_L^{jb}P_L^{ai} \psi^i
    $$

    we can combine and use the full commutator:

    $$
    \left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ai}P_L^{jb} (\psi^i \psi^{\dagger j} + \psi^{\dagger j} \psi^i)\\
    \left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ai}P_L^{jb} \delta(x-y) \delta^{ij}
    $$
    you can change the indices with the delta function and use ##P_L^{ai}P_L^{ib} = P_L^{ab}## to combine the projections and get
    $$
    \left\{\psi_L^a , \psi_L^{\dagger b} \right\} = P_L^{ab} \delta(x-y)
    $$
    If ##a=b## then you get(in D then 4 dimensions)
    $$P_L^{aa} = \frac{D}{2} = 2$$
    This is because by definition the trace of gamma_5 is zero, and the trace of the identity matrix is D(4).

    I think this is right....
     
  6. Apr 16, 2016 #5
    Thanks, that makes sense, the whole dropping spinor indexes for simplicity makes things less clear sometimes.
     
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