1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted