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Bound states in relativistic quantum mechanics

  1. Feb 28, 2009 #1
    Suppose a particle is subject to a spherically symmetric potential [itex]V(r)[/itex] such that [itex]V(r) = -V_0[/itex], [itex]V_0 > 0[/itex], for [itex]0\leq r \leq a[/itex] and [itex]V(r) = 0[/itex] elsewhere. If we were considering a non-relativistic particle, we would have bound states for [itex]-V_0 < E < 0[/itex] (which I understand); however, since the particle is relativistic, apparently (or, at least, according to my professor), we have bound states for [itex]|E| < mc^2[/itex]. Could someone please explain why this is? I'm pretty sure it has something to do with the fact that the relativistic energy-momentum dispersion relation is [itex]E(p) = \sqrt{p^2c^2 + m^2c^4}[/itex], but I can't wrap my head around it.Thanks.
  2. jcsd
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