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

Neglected solutions to the (free) Dirac equation?

  1. May 12, 2012 #1
    So it is said that a basis for the plane wave solutions to the Dirac equation are of the form (p denotes the four-momentum vector) [itex]e^{-i p \cdot x} u^{(s)}[/itex] (for particles) and [itex]e^{i p \cdot x} v^{(s)}[/itex] (for antiparticles), with s = 1 or 2 (and u and v having predetermined structure).

    I'm reading in Griffiths' Introduction to Elementary Particles and there he derives the above, and in doing so he says (p233, on top) that e.g. the solution [itex]e^{-i p \cdot x} v^{(s)}[/itex] isn't allowed since it blows up as the three-space momentum [itex]\mathbf p \to 0[/itex]. However, why is this a sufficient reason to simply neglect the solution? It seems like a valid solution as long as the (three-space) momentum is not zero... And why isn't this state observed?
     
  2. jcsd
  3. May 12, 2012 #2

    strangerep

    User Avatar
    Science Advisor

    I don't have Griffiths, but it seems rather unphysical for a rest frame for a massive particle not to exist. It kinda conflicts with SR, etc...
     
  4. May 13, 2012 #3
    True, but then that means the solutions of the Dirac equation aren't consistent with relativity?
     
  5. May 13, 2012 #4

    Bill_K

    User Avatar
    Science Advisor

    I don't have Griffiths either, but the page you mention is available on Google books. Positive frequency solutions correspond to particles and negative frequency solutions to antiparticles. All he's saying is that you must choose the sign in Eq. 7.41 so that in both cases the energy comes out positive. E > 0 insures (among other things!) that the denominators E + mc2 in Eq. 7.42 will never be zero, even when the particle is at rest.
     
    Last edited: May 13, 2012
  6. May 13, 2012 #5
    I still don't understand on what ground the solution that blows up for [itex]\mathbf p \to 0[/itex] is excluded; maybe it's in your post Bill, in which case I'm not seeing it.
     
  7. May 13, 2012 #6

    Bill_K

    User Avatar
    Science Advisor

    He hasn't excluded anything, in Eq 7.42 he lists four independent solutions. It's just that to write them in their final form he had to replace k0 by ±p0, and in each case there was a sign choice to make. He used + for the particle states and - for the antiparticle states so p0 = E always comes out positive.

    The only reason he mentions what happens when p → 0 is to make this choice of sign seem plausible.
     
  8. May 13, 2012 #7
    I don't follow exactly what you're saying. It seems like you're implying that the sign business is just a convention? But surely a convention wouldn't have an effect on what would happen for [itex]\mathbf p \to 0[/itex]?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Neglected solutions to the (free) Dirac equation?
Loading...