# Little issue in Relativistic Quantum Physics

1. Sep 10, 2015

### 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?

2. Sep 11, 2015

### 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.

3. Sep 11, 2015

### 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!

4. Sep 11, 2015

### StephvsEinst

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!

5. Sep 11, 2015

### 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:

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

6. Sep 11, 2015

### Avodyne

Yes, that's correct!

7. Sep 11, 2015

### StephvsEinst

Thank you! :)