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## Simple Matrix algebra

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:
 Quote by Ryder it is straightforward to show, by writing out all four components of (2.94), that the eigenvalues of E are: $$E = +(m^2 + p^2)^{1/2}$$ twice, $$E = -(m^2 + p^2)^{1/2}$$ twice,
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.

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

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 Quote by kakarukeys I got $$m^2 = E^2 - p^2$$.
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.

## Simple Matrix algebra

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

Mentor
 Quote by kakarukeys At one stage of my calculations, I had ()^2 = 0 implies () = 0 so a 4th order eq became quadratic eq
So, each of your distinct eigenvalues is repeated.

 Blog Entries: 1 Recognitions: Gold Member 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.