# Question regarding Dirac matrices

1. Feb 10, 2005

### peperone

Hey there.

In an exercise I was trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation.

Now I have two very simple questions:

Is $$\gamma_\nu \gamma^\mu = \gamma^\nu \gamma_\mu$$ ?

And is $$\partial_\nu \partial^\mu = \partial^\nu \partial_\mu$$ ?

Thanks for the help.

2. Feb 10, 2005

### dextercioby

Not really.U should pay attention with this "Lorentz covariance" and always try to see these expressions as tensor products of 4-vectors.

Daniel.

3. Feb 10, 2005

### peperone

I kinda felt it couldn't be that simple. :p

Thanks anyway.

4. Feb 10, 2005

### peperone

Can anybody verify this?

$$\gamma^\nu\partial_\nu\gamma_\mu\partial^\mu = \partial_\mu\partial^\mu$$

Just as clarification: I am trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation. So I started with the Dirac equation:

$$(i\gamma_\mu\partial^\mu - m)\Psi(x) = 0$$

Then, I multiplied it with the operator $$-(i\gamma^\nu\partial_\nu + m)$$ (from left) which gave me:

$$(\gamma^\nu\partial_\nu\gamma_\mu\partial^\mu + m^2)\Psi(x) = 0$$

This surely looks somewhat like the Klein-Gordan equation if the above assumption is correct. I just fail to see why it is true.

Any hints appreciated!

Last edited: Feb 10, 2005
5. Feb 10, 2005

### dextercioby

Yes,it can be shown easily once you manipulate the Lorentz indices and the anticommutation relation between the gamma's.
$$\gamma_{\nu}\partial^{\nu}\gamma_{\mu}\partial^{\mu}=\gamma_{\nu}\gamma_{\mu}\partial^{\nu}\partial^{\mu}=\frac{1}{2}\gamma_{(\nu}\gamma_{\mu)}\partial^{\nu}\partial^{\mu}=g_{\mu\nu}\partial^{\nu}\partial^{\mu}=d'alembertian$$

Daniel.

6. Feb 10, 2005

### dextercioby

Why's the KG equation:

$$(\partial_{\mu}\partial^{\mu}-m^{2})\Psi (\vec{r},t) =0$$

$$-(\partial_{\mu}\partial^{\mu}+m^{2})\Psi (\vec{r},t)=0$$

What metric are u using...?

Daniel.

7. Feb 10, 2005

### peperone

Sorry, it was just a typo. I edited my post. Thanks for all the help. I think I got it now...

$$(i\gamma_\mu\partial^\mu - m)\Psi(x) = 0$$

Multiply the operator $$-(i\gamma_\mu\partial^\mu + m)$$ (from left):

$$(\gamma_\mu\partial^\mu\gamma_\mu\partial^\mu + m^2)\Psi(x) = (\gamma_\mu\gamma_\mu\partial^\mu\partial^\mu + m^2)\Psi(x) = 0$$

Because of the anticommutation relation $$\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu\nu}$$ it is possible to simplify $$\gamma_\mu\gamma_\mu$$ to $$g_{\mu\mu}$$ which proves $$\gamma_\mu\gamma_\mu\partial^\mu\partial^\mu = \partial^\mu\partial_\mu$$.

Thanks again.

Last edited: Feb 10, 2005
8. Feb 10, 2005

### dextercioby

Note that u have 4 identical (Lorentz) indices in the same tensor equation and that's not good,not good at all.
Pay attention with these indices,as u might confuse them.

Daniel.

9. Feb 10, 2005

### peperone

Hm, that's true. Do you know a better way? I'm highly interested in alternative solutions.

10. Feb 10, 2005

### peperone

To be honest, I don't understand what you're doing here (step 2 -> step 3)

11. Feb 10, 2005

### dextercioby

You multiply with the same operator,but with index "nu" (instead of "mu") and then use the trick i've used in post #5.It should come out nicely.

Daniel.

12. Feb 10, 2005

### dextercioby

$$\gamma_{\nu}\gamma_{\mu}T^{\nu\mu}=\frac{1}{2}(\gamma_{\nu}\gamma_{\mu}+\gamma_{\mu}\gamma_{\nu})T^{\mu\nu}$$

where T is a symmetric second rank tensor and to get the second term from the bracket i turned "mu"--->"nu" and "nu"---->"mu" and to factor the same tensor i used the fact that under this renotation,the tensor T is symmetric.

Daniel.

13. Feb 10, 2005

### peperone

This surely looks slick! :) But why is $$\partial^\nu\partial^\mu$$ a symmetric second rank tensor?

14. Feb 10, 2005

### dextercioby

Because wrt bosonic coordinates,the partial derivatives commute.

Daniel.