# Homework Help: Simple Matrix algebra

1. Aug 4, 2007

### Jimmy Snyder

1. The problem statement, all variables and given/known data
On page 44 of Ryder's QFT, near the bottom of the page, it says:
2. Relevant equations
Equation (2.94) is
$$(\gamma^{\mu}p_{\mu} - m)\psi(p) = 0$$

3. The attempt at a solution
Writing out all four components, and then taking the determinant and setting to zero, I get:
$$m^4 - (E^2 - p^2)^2 = 0$$ or $$m^4 = (E^2 - p^2)^2$$
Taking the square root once:
$$\pm m^2 = E^2 - p^2$$ or $$E^2 = p^2 \pm m^2$$.
And taking the square root again:
$$E = \pm(p^2 \pm m^2)^{1/2}$$
and I end up with different eigenvalues than I am supposed to.

Last edited: Aug 4, 2007
2. Aug 6, 2007

### kakarukeys

How did you calculate the determinant? I calculated it with some formulae, and I got only two eigenvalues. Maybe you should try using a software like mathematica to calculate it by brute force.

Let just hope that your Dirac matrices are same as mine. I used

http://www.stochasticsoccer.com/Clipboard01.jpg

then I used
$$(\sigma_\mu p^\mu)(\sigma'_\mu p^\mu) = p_\mu p^\mu$$
where $$\sigma^\mu = (1, \sigma^i)$$
where $$\sigma'^\mu = (1, -\sigma^i)$$
$$\sigma^i$$ are Pauli matrices.

I got $$m^2 = E^2 - p^2$$.

3. Aug 6, 2007

### Jimmy Snyder

Thanks for taking a look at this kakarukeys. I don't think your equation could be the determinant since there are supposed to be 4 eigenvalues, and your equation is only quadratic in E.

4. Aug 6, 2007

### kakarukeys

At one stage of my calculations, I had ()^2 = 0 implies () = 0
so a 4th order eq became quadratic eq

5. Aug 7, 2007

### George Jones

Staff Emeritus
So, each of your distinct eigenvalues is repeated.

6. Aug 7, 2007

### Jimmy Snyder

Thanks kakarukeys. I was making two errors. First of all, I had the wrong matrix for $$\gamma^0$$, and second of all, I was calculating the determinant incorrectly. With your help, I now get the following determinant:
$$(E^2 - m^2 - p^2)^2$$ and setting this to zero gives the correct eigenvalues. Thanks George to you as well. Actually, I gathered the same meaning from message #4 as you did, but it's good to know that you have my back.