1. May 2, 2013

### PRB147

Dirac Equation as Example,
Dirac Equation: $$\left(i\gamma^\mu \partial_\mu -m \right)\psi(x)=0$$
Can I write it in the following way?
$$\left(i\gamma^0 \partial_0- i\gamma^j \partial_j -m \right)\psi^p(t,{\bf -x})=0$$

2. May 2, 2013

### andrien

under parity transformation,dirac spinor transforms as
ψ(x,t)-γ0ψ(-x,t),wich can be obtained from fact that under parity transformation left and right handed spinors exchange.So after knowing how chiral spinors transform we can go for dirac spinor.Now γ0ψ(-x,t) satisfies dirac eqn.you can then use some manipulation to send γ0 on the left to get the eqn for ψ(-x,t).

3. May 2, 2013

### PRB147

The second equation in my first thread is the direct consequence by taking parity
transformation over two sides of the first Dirac equation (of course, the right hand side of the Dirac equation is zero)?

4. May 3, 2013

### andrien

In writing the dirac eqn after parity transformation,you should change the sign of space variables which you have done.But so far you also have to know how dirac spinor will transform under parity.You can not simply put a minus sign with x to get parity transformed dirac spinor because
ψ(x,t)≠ψ(-x,t).the parity transformed dirac eqn reads
γ0(iγ00-iγii-m)ψ(-x,t)=0

5. May 3, 2013

### PRB147

please look at my equation carefully, it is $$\psi^p(t,-{\bf x})$$, not ψ(-x,t)

6. May 3, 2013

### andrien

then you should simply write ψp(t,x) which will be parity transformed spinor and then you should seek relation to original ψ.
edit-One more thing,Dirac eqn does not look like the original eqn with ψ(-x,t) but with γ0ψ(-x,t) it looks same.

Last edited: May 3, 2013
7. May 6, 2013

### PRB147

my derivation is as follows:
Dirac Equation as Example,
Dirac Equation: $$\left(i\gamma^\mu \partial_\mu -m \right)\psi(x)=0$$
Taking parity transform in both sides of the above Dirac Equation:
$$\left(i\gamma^0 \partial_0- i\gamma^j \partial_j -m \right)\psi^p(t,{\bf -x})=0$$
Then we multiply from left $$\gamma^0$$, then we get:
$$\left(i\gamma^0 \partial_0+ i\gamma^j \partial_j -m \right) \gamma^0\psi^p(t,{\bf -x})=0$$

that is to say, $$\gamma^0\psi^p(t,{\bf -x})$$ still obey the Dirac Equation

So

$$\gamma^0\psi^p(t,{\bf -x})=\eta_p\psi^p(t,{\bf x})$$

Then
$$\psi^p(t,{\bf -x})=\eta_p \gamma^0\psi^p(t,{\bf x})$$

and then:

$$\psi^p(t,{\bf x})=\eta_p \gamma^0\psi^p(t,{\bf -x})$$

This is my derivation, the result is the same as yours and that in textbook.

8. May 6, 2013

### andrien

your derivation is highly confusing.Please specify what is the meaning of$\psi^p(t,{\bf x})$ and $\psi^p(t,{\bf -x})$.when you write the parity transformed state $\psi^p(t,{\bf -x})$ then what is meaning of $\psi^p(t,{\bf x})$.

9. May 7, 2013

### PRB147

My previous derivation contains typo

the correct is as follows:

10. May 7, 2013

### PRB147

Thank you very much for your patience, I understand what you said.