Energies of a particle in a box vs. free particle?

[fhqwgads2005]

I think I'm trying to reconcile quantum mechanics and special relativity . . . or whatever I'm doing I'm pretty confused.

Ok, so the allowed energy states for a particle in a box are

$$E_n = \frac{\hbar^2 \pi^2}{2 m L^2} n^2$$.

This seems to mean, as you increase the length L, the particle's energy will tend towards zero. When L becomes very large, the particle will be essentially free, and according to the above equation will have an energy of E~0.

But the minimum energy of a free particle should be its rest-mass energy, E = mc^2, not zero. Also, the ground state (n = 0) energy of a particle in a box is inversely proportional to mass, while the ground state energy of a free particle is directly proportional to its mass.

How do you reconcile these ideas from quantum mechanics and relativity?

 

[Gordianus]

The allowed energy levels you have written are derived from the non-relativistic quantum mechanics formalism. Therefore, you won't find relativistic terms.

 

[fhqwgads2005]

Hmm, so is there a relatively simple way to think about the particle in a box including relativity?

 

[Gordianus]

Try relativistic quantum mechanics; it's the way to go.

 

[tom.stoer]

The problem can already be understood classically; w/o taking into account a potential V(x) the relativistic energy is

$$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\simeq mc^2 + \frac{1}{2}v^2$$

in non-rel. mechanics you forget about the mc² term; in non-rel. QM you quantize only the v² term

 

[fhqwgads2005]

The problem can already be understood classically; w/o taking into account a potential V(x) the relativistic energy is

$$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\simeq mc^2 + \frac{1}{2}v^2$$

in non-rel. mechanics you forget about the mc² term; in non-rel. QM you quantize only the v² term

Interesting... maybe one day I'll learn relativistic QM...