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.(adsbygoogle = window.adsbygoogle || []).push({});

I know that the two principles that still apply even when a particle is relativistic are:

[tex] \lambda = \frac{h}{p}[/tex]

and

[tex] E = h f = \frac {hc}{\lambda} [/tex]

such that

[tex] E = c \sqrt{\hbar^2 k^2 +m_0 c^2} [/tex]

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

[tex]\psi(x) = Asin(kx) + Bcos(kx)[/tex] 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

[tex]Asin(kL) = 0 [/tex]

where the solution is that

[tex]kL = n \pi [/tex]

or maybe...

[tex]Asin(\frac{kL}{\gamma}) = 0 [/tex]

in which

[tex]\frac{kL}{\gamma} = n \pi[/tex]

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

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

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# Homework Help: Relativistic Energy Quantization for Particle in Box

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