Little issue in Relativistic Quantum Physics

[StephvsEinst]

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?

 

[Avodyne]

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.

 

[vanhees71]

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!

 

[StephvsEinst]

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!

 

[StephvsEinst]

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:

$$ = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 - \frac{ 1 }{2} \psi^{ \dagger } \gamma_5 \gamma_0 \\
= \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 + \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 \gamma_5 \\
= \psi^{ \dagger } \gamma_0 \frac{ ( 1 + \gamma_5 ) }{2} \\
= \bar { \psi } P_R. $$

 

[Avodyne]

Yes, that's correct!

 

[StephvsEinst]

Yes, that's correct!

Thank you! :)