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Little issue in Relativistic Quantum Physics

  1. Sep 10, 2015 #1
    Hey!
    I wanted to prove that:

    $$ P_L \bar{ \psi } = \bar{ \psi } P_R $$

    And I want to know if I did it correctly.
    $$ --- $$
    Here is what I did:
    $$ P_L \bar{ \psi } = \frac{ ( 1 - \gamma_5 ) }{2} \psi^{ \dagger } \gamma_0, $$
    $$ = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 - \frac{ 1 }{2} \gamma_5 \psi^{ \dagger } \gamma_0 $$
    $$ = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 - \frac{ 1 }{2} \psi^{ \dagger } \gamma_5 \gamma_0 $$
    $$ = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 + \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 \gamma_5 $$
    $$ = \psi^{ \dagger } \gamma_0 \frac{ ( 1 + \gamma_5 ) }{2} $$
    $$ = \bar { \psi } P_R. $$
    $$ --- $$
    Where
    $$ \psi =
    \left(
    \begin{smallmatrix}
    \psi_1 \\
    \psi_2 \\
    \psi_3 \\
    \psi_4
    \end{smallmatrix}
    \right) $$

    And
    $$ P_L = \frac{ ( 1 - \gamma_5 ) }{2} \qquad P_R = \frac{ ( 1 + \gamma_5 ) }{2} $$

    And
    $$ \bar{ \psi } = \psi^{ \dagger } \gamma_0 $$

    And
    $$ \gamma_0 \gamma_5 + \gamma_5 \gamma_0 = 0 $$

    $$ \mbox{ Note: } \gamma_0 \mbox{ and } \gamma_5 \mbox{ are Dirac's matrices. }$$

    $$ --- $$

    Is there anything wrong with this?
     
  2. jcsd
  3. Sep 11, 2015 #2

    Avodyne

    User Avatar
    Science Advisor

    It is very wrong. ##\bar\psi## is a row vector; it must be to the left of all matrices. So ##P_L\bar\psi## is not a legal expression.
     
  4. Sep 11, 2015 #3

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Hint: Calculate
    $$\overline{P_L \psi}=(P_L \psi)^{\dagger} \gamma^0,$$
    using the usual "Diracology" of ##\gamma## matrices. I guess it's a homework problem and should be moved to the homework section. That's why I only give this hint and not a full solution!
     
  5. Sep 11, 2015 #4
    Thank you for the reply!
    It isn't homework, just need to calculate that because I want to understand the following:
    $$ J_L^{ \mu + } = \bar{ \psi_L } \gamma^{ \mu } \psi_L \equiv V - A $$
    Thanks for the hint, I'll try to solve it that way!
     
  6. Sep 11, 2015 #5
    Is this the correct way?

    $$ \overline{P_L \psi } = \left[ \frac{ ( 1 - \gamma_5 ) }{2} \psi \right]^{ \dagger } \gamma_0, $$

    We have

    $$ ( \gamma_5 )^{ \dagger } = \gamma_5 $$

    So:

    $$ = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 - \frac{ 1 }{2} \psi^{ \dagger } \gamma_5 \gamma_0 \\
    = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 + \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 \gamma_5 \\
    = \psi^{ \dagger } \gamma_0 \frac{ ( 1 + \gamma_5 ) }{2} \\
    = \bar { \psi } P_R. $$
     
  7. Sep 11, 2015 #6

    Avodyne

    User Avatar
    Science Advisor

    Yes, that's correct!
     
  8. Sep 11, 2015 #7
    Thank you! :)
     
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