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

Charge conjugation matrix and Dirac equation's solutions

  1. Sep 20, 2015 #1
    I saw this somewhere but I think it is wrong...

    I already read Griffiths' "Introduction to Particle Physics" (the 1st edition) from the page 216 to the page 222 (chapter of Quantum Electrodynamics - section "Solution to the Dirac Equation") and I didn't understood why was there the imaginary number in the equation:

    $$ ( i \gamma \cdot \mathbf{p} + m ) \psi ( \mathbf{p} ) = 0. $$

    And what is the C matrix?

    Can anyone help?
     
  2. jcsd
  3. Sep 21, 2015 #2
    EDIT: In the quote I meant:

     
  4. Sep 21, 2015 #3
    He basically just fourier transformed the dirac equation, try it yourself and see what you get.
     
  5. Sep 21, 2015 #4

    phyzguy

    User Avatar
    Science Advisor

    In the Dirac equation, psi is not a four-vector. It is a spinor, and they are very different physical objects.
     
  6. Sep 23, 2015 #5
    I was thinking that psi could be described by two different relativistic fields, with each one having a (1 0) or (0 1) (those two states must be read in a COLUMN vector). Wouldn't this give us the four states: 2 different states of spin (up and down) of the particle and 2 different states of spin of the ANTIparticle?
     
  7. Sep 23, 2015 #6
    How the hell did I not think for once of doing the fourier transform...
    Thank you, I will work on that!
     
  8. Sep 23, 2015 #7

    phyzguy

    User Avatar
    Science Advisor

    The Dirac spin ψ does in fact have four components, but that does not make it a four-vector. A four-vector and a Dirac spinor transform very differently under coordinate transformations.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Charge conjugation matrix and Dirac equation's solutions
Loading...