Get the current and the EQs of Motion of the Dirac-Lagrangian density

[Gaussian97]

Is $\gamma^5\bar{\psi}_L^T=+\bar{\psi}_L^T$ meant to be equivalent to $\gamma^5\psi_R=+\psi_R$? If yes, we also proved it at #62

Well, you have to prove this and you are done

 

[JD_PM]

Well, you have to prove this and you are done

We have

$$\gamma^5\bar{\psi}_L^T=+\bar{\psi}_L^T$$

And we want to end up with

$$\gamma^5\psi_R=+\psi_R $$

We recall that

$$\psi_{L,R} := \frac{1 \mp \gamma^5}{2} \psi$$

$$\bar{\psi} := \psi^{\dagger}\gamma^0$$

$$\psi^{\dagger} := (\psi^*)^T$$

The LHS can be written as follows

$$\gamma^5\bar{\psi}_L^T=\gamma^5(\psi^{\dagger}_L\gamma^0)^T=\gamma^5 \Big( (\psi_L^*)^T\gamma^0\Big)^T= \gamma^5 \psi_L^* \gamma^0= \frac 1 2 \gamma^5 (1+\gamma^5)\psi \gamma^0=\gamma^5 \psi_R \gamma^0$$

Mmm but

$$\gamma^5 \psi_R \gamma^0 \neq \bar{\psi}_L^T$$

So one of the following assumptions I made has to be wrong:

1) $(\gamma^0)^T= \gamma^0$

2) $(\gamma^5)^* = -\gamma^5$

3) $(\psi)^*=\psi$

 

[Gaussian97]

The same problem as always, in the third equality, $\psi_L \gamma^0$ makes no sense.
Also assumptions 2) and 3) are not true.

 

[JD_PM]

Alright so the issue is that this equality doesn't hold

$$\gamma^5 \Big( (\psi_L^*)^T\gamma^0\Big)^T= \gamma^5 \psi_L^* \gamma^0$$

This is because the $4 \times 1$ matrix $\psi$ (contained in $\psi_L^*$) cannot be multiplied by $\gamma^0$ (a $4 \times 4$ matrix).

So my mistake had to be made before reaching that equality. Do you agree? I still do not see it though...

 

[Gaussian97]

No, the LHS is correct, the problem is precisely this equality, I could tell you why is wrong, but I think would be more instructive that you try to show all the hidden steps and then write explicitly what properties are you using in each step.

 

[JD_PM]

I think would be more instructive that you try to show all the hidden steps and then write explicitly what properties are you using in each step.

I agree. You are doing right in not telling me what the solution is straight away.

Let's start over. This time I am going to use the fact that $\bar{\psi}=\bar{\psi_L}+\bar{\psi_R}$

$\gamma^5\bar{\psi}_L^T=\gamma^5\Big(\bar{\psi}-\bar{\psi}_R \Big)^T=\gamma^5\Big(\psi^{\dagger} \gamma^0-\psi^{\dagger}_R \gamma^0 \Big)^T=\gamma^5\Big(\psi^{\dagger}\gamma^0-\frac 1 2 \psi^{\dagger} \gamma^0-\frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0\Big)^T=\gamma^5\Big(\frac 1 2\psi^{\dagger}\gamma^0-\frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0\Big)^T$

Do you agree? At point I think everything makes sense: before taking the transpose, inside the parenthesis we have $1\times 4$ matrices; once we take the transpose we get a $4 \times 1$ matrix. Thus the product $\gamma^5(...)$ is OK.

Mmm now the issue is to show that

$$\Big(\frac 1 2\psi^{\dagger}\gamma^0-\frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0\Big)^T=\psi_R$$

Could you please give me a hint on how to do so?