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

  • #1
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

[tex]E_n = \frac{\hbar^2 \pi^2}{2 m L^2} n^2 [/tex].

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?
 

Answers and Replies

  • #2
271
23
The allowed energy levels you have written are derived from the non-relativistic quantum mechanics formalism. Therefore, you won't find relativistic terms.
 
  • #3
Hmm, so is there a relatively simple way to think about the particle in a box including relativity?
 
  • #4
271
23
Try relativistic quantum mechanics; it's the way to go.
 
  • #5
tom.stoer
Science Advisor
5,766
161
The problem can already be understood classically; w/o taking into account a potential V(x) the relativistic energy is

[tex]E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\simeq mc^2 + \frac{1}{2}v^2[/tex]

in non-rel. mechanics you forget about the mc² term; in non-rel. QM you quantize only the v² term
 
  • #6
The problem can already be understood classically; w/o taking into account a potential V(x) the relativistic energy is

[tex]E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\simeq mc^2 + \frac{1}{2}v^2[/tex]

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...
 

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