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Relativistic Energy Quantization for Particle in Box

  1. Nov 7, 2006 #1
    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:
    [tex] \lambda = \frac{h}{p}[/tex]
    [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].
    Last edited: Nov 7, 2006
  2. jcsd
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