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Bohr Quantization with linear potential

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Using Bohr's quantization rule find the energy levels for a particle in the potential: $$U(x) = \alpha\left|x\right|, \alpha > 0.$$

    2. Relevant equations
    ##\oint p\, dx = 2\pi\hbar (n + \frac{1}{2})##

    3. The attempt at a solution
    Okay so:
    \oint p\, dx &= \int \sqrt{2m(E-U(x))}\,dx\\
    &= \int\limits_{-\infty}^{+\infty} \sqrt{2m(E-\alpha\left|x\right|)}\,dx\\
    &= 2\pi\hbar (n+\frac{1}{2}
    So far, I believe this is correct, but the integral doesn't converge so either im missing something or I've done something wrong. I can't seem to see what it is. Any help is greatly appreciated.
  2. jcsd
  3. Jan 27, 2015 #2
    There's a turning point where the square-root becomes imaginary, that's your upper limit...
  4. Jan 27, 2015 #3
    Right I see, so I get the two turning points at $$\pm \sqrt{\frac{E^2}{\alpha^2}}$$ and now its just a matter of evalutating the integral.
  5. Jan 27, 2015 #4
    Yes, and the integral is easy to carry out.
    Also, note that it is symmetric about 0 so you can dispose of the absolute value and take twice the integral from 0 to the turning point.
  6. Jan 27, 2015 #5
    Thank you for your help. Got an answer of $$E_n = \alpha\left(\frac{3}{2}\frac{\pi\hbar}{\sqrt{2m}}(n+\frac{1}{2})\right)^{2/3}.$$ which seems okay.
  7. Jan 27, 2015 #6
    No problem... But i think you want to put this α inside the parenthesis ;)
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