Solve for Eigenvalues in QFT using Matrix Algebra | Ryder's QFT p.44

[Jimmy Snyder]

Homework Statement

On page 44 of Ryder's QFT, near the bottom of the page, it says:

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,

Homework Equations

Equation (2.94) is
(\gamma^{\mu}p_{\mu} - m)\psi(p) = 0

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.

 

[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.

 

[Jimmy Snyder]

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.

 

[kakarukeys]

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

 

[George Jones]

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.

 

[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.