The Dirac equation and its conjugate

[Tomer]

This isn't really a homework problem, just a form of writing I don't quite understand.

The Dirac equation is: ("natural units")

(i\gamma^{\mu}\partial_{mu}-m)\Psi = 0

When I try to build the conjugated equation, where \bar{\Psi} := \Psi^{+}\gamma^{0}, I get:

i\partial_{\mu}\bar{\Psi}\gamma^{\mu}+m\bar{\Psi} = 0

Which I've then verified and it seems correct.

However, some sources show the conjugated equation in this form:

\bar{\Psi}(i\gamma^{\mu}\partial_{\mu}-m) = 0

Now, I know that the scalar product is an invariant, but what I don't understand, is how I can simply shove this \bar{\Psi} to the left side of the equation... how can the operator acting on it be situated *after* it and what does it mean?
And where does that "-m" come from? I get "+m" and so did other sources I saw...

I'm sorry if this question is dumb - this whole thing is rather new to me.

Thanks a lot!

Tomer.

 

[member 553083]

The Dirac equation is an example of a first order linear partial differential equation, which means that the solution to the equation can be found by rearranging the terms in the equation. The conjugated equation is just the equation rearranged so that the \bar{\Psi} is on the left side of the equation. To do this, you can multiply both sides of the original equation by the gamma matrix and its inverse, so that the terms with \Psi cancel out, leaving the rearranged equation with \bar{\Psi}. The minus sign in the equation comes from the fact that the gamma matrix and its inverse are both anti-symmetric.