Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Non-negativity of the eigenvalues of the Dirac operator

  1. May 10, 2016 #1

    ShayanJ

    User Avatar
    Gold Member

    How can I prove that the eigenvalues of the operator ## i\gamma^\mu \partial_\mu ## are non-negative?
    I've tried using the ansatz ## \psi=u(p) e^{ip_\nu x^\nu} ## but it didn't help.
    I've also tried playing with the equation using the properties of gamma matrices but that doesn't seem to lead anywhere too.
    Thanks
     
  2. jcsd
  3. May 10, 2016 #2

    ShayanJ

    User Avatar
    Gold Member

    I think its because the Dirac equation is ## i\gamma^\mu\partial_\mu\psi=m \psi ## and so the eigenvalue of that operator is the mass of the Dirac field, so it should be non-negative.

    This was given to one of my friends to prove and she asked me for help, so I don't know what was the motivation.
     
  4. May 10, 2016 #3

    ChrisVer

    User Avatar
    Gold Member

    So OK:
    [itex] i \gamma^\mu \partial_\mu \psi = a \psi[/itex]
    [itex] [i \gamma^\mu \partial_\mu - a] \psi =0[/itex]
    Now the eigenvalues can be found by solving:
    [itex] det[i \gamma^\mu \partial_\mu - a \textbf{1}] =0[/itex]
    ?
    I have never tried that...
    But I think the expression is obtained much more easily by the ansatzes taken while deriving the Dirac equation.
     
  5. May 10, 2016 #4

    ShayanJ

    User Avatar
    Gold Member

    That's a differential operator so you can't just cancel the Dirac spinor.
     
  6. May 10, 2016 #5

    ChrisVer

    User Avatar
    Gold Member

    try just write it as the momentum...
     
  7. May 10, 2016 #6

    ShayanJ

    User Avatar
    Gold Member

    As I said in the OP, I tried the ansatz ## \psi=u(p) e^{i p_\nu x^\nu} ##, but it doesn't help because you'll get a quadratic equation in m which has both negative and positive solutions.
     
  8. May 10, 2016 #7

    ChrisVer

    User Avatar
    Gold Member

    how do you get a quadratic term by differentiating once the exponential?
     
  9. May 10, 2016 #8

    ChrisVer

    User Avatar
    Gold Member

    [itex] i \gamma^\mu \partial_\mu u(p) e^{i p x} = a u(p)e^{ipx}[/itex]
    [itex] - \gamma^\mu p_\mu u(p) e^{ipx} = a u(p) e^{ipx}[/itex]
    [itex] (a + \slash{p} ) u(p) e^{ipx} = 0[/itex]
    so just try the determinant of the matrix on the left?
     
  10. May 10, 2016 #9

    ShayanJ

    User Avatar
    Gold Member

    And that's exactly what gives you the quadratic equation! (Actually its fourth order, but you can solve it to get a quadratic equation.)
     
  11. May 10, 2016 #10

    ChrisVer

    User Avatar
    Gold Member

    In fact the more I think of it, the more I tend to say that by construction there is no reason to have positive EVs...
    It's again coming from the fact that negative masses wouldn't make much sense and would violate causality...
    And then, the rest comes from the way we derive the Dirac equation, by using the [itex]E^2= p^2+m^2[/itex] where [itex]m^2>0[/itex].
     
  12. May 10, 2016 #11

    Paul Colby

    User Avatar
    Gold Member

    I guess I would expect the operator, ##i\gamma^\mu\partial_\mu## to have a sign symmetric spectrum. Is there a reason to think otherwise? Take the special case ##v_ne^{-i\omega t}## where ##v_n## is an eigen vector of ##\gamma_t##. Both signs occur.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Non-negativity of the eigenvalues of the Dirac operator
  1. Waerden's Dirac (Replies: 6)

  2. Dirac Spinor (Replies: 1)

Loading...