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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


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    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


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    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.
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