# Relativistic Energy Quantization for Particle in Box

1. Nov 7, 2006

### Mindscrape

How would you go about finding the energy for "the particle in a box" when the particle is relativistic? Since the energy is no longer p^2/2m, then the general quantization won't apply.

I know that the two principles that still apply even when a particle is relativistic are:
$$\lambda = \frac{h}{p}$$
and
$$E = h f = \frac {hc}{\lambda}$$
such that
$$E = c \sqrt{\hbar^2 k^2 +m_0 c^2}$$

From here, I am not really sure what to do with the wave-vector. I suppose that the wavefunction still has to satisfy the general solution that

$$\psi(x) = Asin(kx) + Bcos(kx)$$ for 0 < x < L

The upper bound will change from length contraction, but does that change anything about how the interior of the wave must vanish at x=0 and x=L? If not, then the wavefunction must still satify the equation that
$$Asin(kL) = 0$$
where the solution is that
$$kL = n \pi$$

or maybe...
$$Asin(\frac{kL}{\gamma}) = 0$$
in which
$$\frac{kL}{\gamma} = n \pi$$

Then depending on what value k is, I can substitute it into the relativistic energy equation, and get the equation. But which value is the right one for k?

Am I anywhere on the right track? I know that, ultimately, I need to regain that
$$E_n = \frac{\hbar^2 k^2}{2m} = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$$.

Last edited: Nov 7, 2006