# The Dirac Equation

1. Jan 3, 2008

### raintrek

[SOLVED] The Dirac Equation

I'm trying to understand the following property of the Dirac equation:

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

Acting twice with $$(i \gamma^{\mu}\partial_{\mu} - m)$$:

$$(i \gamma^{\mu}\partial_{\mu} - m)^{2} \Psi(x) = 0$$

$$= [ - \gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} - 2im\gamma^{\mu}\partial_{\mu} + m^{2}]\Psi = 0$$

But then somehow the book jumps to this step:

$$= [ 1/2 \left{\{\gamma^{\mu}, \gamma^{\nu}\right}\} \partial_{\mu}\partial_{\nu} + m^{2}]\Psi = 0$$

And I have no idea how it got there! I understand the { } denote an anticommutator, but I just can't see how the factor of 1/2 has appeared, where the minus has gone and where the middle term has gone. Help!!!

2. Jan 3, 2008

### Dick

$$\gamma^\mu \gamma^\nu \partial_\mu \partial _\nu=\gamma^\nu \gamma^\mu \partial_\mu \partial _\nu$$. Add those and divide by two, to get the anticommutator. Let the middle term act on $$\Psi$$ and use the dirac equation to see where it has gone. The minus will become an overall minus once you do all of this.

3. Jan 3, 2008

### raintrek

Ah, I understand now why the commutator arises, thanks Dick. However, I'm still confused about the middle term business. If I understand you right, you're saying that:

$$(-2im\gamma^{\mu}\partial_{\mu})\Psi$$

is analagous to:

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

and hence should disappear? I don't quite understand that part...

4. Jan 3, 2008

### Dick

It doesn't disappear. It's equal to -2m^2*psi. Do you see why?

5. Jan 3, 2008

### raintrek

^ Ah! Got it, $$i\gamma^{\mu}\partial_{\mu} = m$$ from the original eq. Yes, I see now! Thanks so much, Dick!