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).(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Neglected solutions to the (free) Dirac equation?

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