Neglected solutions to the (free) Dirac equation?

[nonequilibrium]

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) $e^{-i p \cdot x} u^{(s)}$ (for particles) and $e^{i p \cdot x} v^{(s)}$ (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 $e^{-i p \cdot x} v^{(s)}$ isn't allowed since it blows up as the three-space momentum $\mathbf p \to 0$. 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?

 

[strangerep]

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...

 

[nonequilibrium]

True, but then that means the solutions of the Dirac equation aren't consistent with relativity?

 

[Bill_K]

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.

 

[nonequilibrium]

I still don't understand on what ground the solution that blows up for $\mathbf p \to 0$ is excluded; maybe it's in your post Bill, in which case I'm not seeing it.

 

[Bill_K]

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.

 

[nonequilibrium]

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 $\mathbf p \to 0$?