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.

 

[Entropia]

I can provide some insight into this topic. In relativistic quantum mechanics, the energy of a particle is described by the energy-momentum dispersion relation E(p) = \sqrt{p^2c^2 + m^2c^4}, where p is the momentum of the particle, m is its mass, and c is the speed of light.

In this particular scenario, we have a spherically symmetric potential V(r) that is equal to -V_0 for a certain distance range and 0 elsewhere. This potential creates a barrier for the particle, which means that the particle must have enough energy to overcome this barrier in order to escape.

For a non-relativistic particle, we would have bound states for -V_0 < E < 0, meaning that the particle has a negative energy but is still bound to the potential. However, in the case of a relativistic particle, the energy range for bound states is |E| < mc^2. This is because the energy-momentum dispersion relation tells us that the energy of the particle is not only dependent on its mass, but also on its momentum. As the momentum of the particle increases, so does its energy.

Since the particle is subject to a spherically symmetric potential, its momentum will be limited and therefore its energy will also be limited. This means that in order for the particle to have a negative energy and be bound to the potential, its energy must be less than its rest energy (mc^2). Hence, the energy range for bound states is |E| < mc^2.

In summary, the relativistic energy-momentum dispersion relation takes into account the momentum of the particle, resulting in a different range of energies for bound states compared to non-relativistic particles. This is a fundamental concept in relativistic quantum mechanics and is crucial for understanding the behavior of particles in potential wells. I hope this explanation helps clarify the concept for you.