Simple Matrix algebra

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
[tex](\gamma^{\mu}p_{\mu} - m)\psi(p) = 0[/tex]

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

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



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

I got [tex]m^2 = E^2 - p^2[/tex].

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.

kakarukeys

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

George Jones

So, each of your distinct eigenvalues is repeated.

Jimmy Snyder

Thanks kakarukeys. I was making two errors. First of all, I had the wrong matrix for [tex]\gamma^0[/tex], and second of all, I was calculating the determinant incorrectly. With your help, I now get the following determinant:
[tex](E^2 - m^2 - p^2)^2[/tex] 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.