Bound states in relativistic quantum mechanics

[AxiomOfChoice]

Suppose a particle is subject to a spherically symmetric potential V(r) such that V(r) = -V_0, V_0 > 0, for 0\leq r \leq a and V(r) = 0 elsewhere. If we were considering a non-relativistic particle, we would have bound states for -V_0 < E < 0 (which I understand); however, since the particle is relativistic, apparently (or, at least, according to my professor), we have bound states for |E| < mc^2. 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 E(p) = \sqrt{p^2c^2 + m^2c^4}, but I can't wrap my head around it.Thanks.

 

[azntoon]

A:The reason for this is that the energy dispersion relation can be written as $$E^2 = p^2c^2 + m^2c^4,$$where $m$ is the particle's rest mass and $p$ is its momentum. This means that even if the momentum is zero, the energy is still equal to $mc^2$, which sets a lower bound on the energy for a relativistic particle. For a non-relativistic particle, the energy dispersion relation is just $E=\frac{p^2}{2m}$, so the particle can have arbitrarily small energy if the momentum is zero.