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Simple Matrix algebra

  1. Aug 4, 2007 #1
    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.
    Last edited: Aug 4, 2007
  2. jcsd
  3. Aug 6, 2007 #2
    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].
  4. Aug 6, 2007 #3
    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.
  5. Aug 6, 2007 #4
    At one stage of my calculations, I had ()^2 = 0 implies () = 0
    so a 4th order eq became quadratic eq
  6. Aug 7, 2007 #5

    George Jones

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    So, each of your distinct eigenvalues is repeated.
  7. Aug 7, 2007 #6
    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.
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