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Rectangular Potential well if hbar -> 0

  1. Feb 16, 2010 #1
    Rectangular Potential well if hbar --> 0


    I have a very simple question: My professor said that if [tex]\hbar \rightarrow 0[/tex] in the transmission coefficient does not yield 0 (case E < V0) as classical results would expect.

    I think the correct transmission coefficient is solved in Wikipedia: http://en.wikipedia.org/wiki/Rectangular_potential_barrier#E_.3C_V0

    But if I calculate [tex]\lim_{\hbar \rightarrow 0} T[/tex] I get 0.

    Can anybody tell me what is wrong about it?

    Thank you,
    Last edited: Feb 16, 2010
  2. jcsd
  3. Feb 16, 2010 #2


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    Re: Rectangular Potential well if hbar --> 0

    Your transmission coefficient is more correct. Wikipedia has assumed that k1 and k2 on the left and right side of the barrier is the same. Your transmission coefficient would be necessary if the barrier did not fall all the way back to the original position (or fell more).

    I think hbar going to 0 should yield classical results...I don't know why it wouldn't yield classical results as your professor suggests.
  4. Feb 16, 2010 #3
    Re: Rectangular Potential well if hbar --> 0

    Hi Matterwave,

    Thank you. you replied while I was editing my post. Sorry. Yes, I saw that and therefore re-edit my post.

    The reason is that the coefficient from my professor is not for a rectangular potential but for abrupt change. In case of the rectangular potential I have in the first and in the third region k1 and in the middle k2 in the lecture notes, that means that the result is the same in this case.

    Concerning the main problem: This is the statement in the lecture notes:

    But indeed, also Wikipedia says this but no further comments on it. The article describing the semiclassical approach:


  5. Feb 16, 2010 #4


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    Re: Rectangular Potential well if hbar --> 0

    Hmm, interesting. I have not heard about this before. I don't know why this would be the case, other than the fact that square wells are unphysical since they are not differentiable everywhere (classically, that would lead to a discontinuous, infinite, force at the turning points).

    Perhaps a more knowledgeable member can answer your question. :)
  6. Feb 16, 2010 #5
    Re: Rectangular Potential well if hbar --> 0

    Maybe I just calculate the limit the wrong way?

    In limes I can disregard any constants, can't I?


    [tex]T= \frac{1}{1+\frac{V_0^2\sinh^2(k_1 a)}{4E(V_0-E)}}[/tex]


    [tex]T= \frac{1}{1+\sinh^2(k_1 a)}[/tex]

    and as [tex]k_1=\sqrt{2m (V_0-E)/\hbar^{2}}[/tex]

    [tex]T= \frac{1}{1+\sinh^2(\sqrt{2m (V_0-E)/\hbar^{2}} a)}[/tex]

    and in the limit for [tex]\hbar[/tex]:

    [tex]T= \frac{1}{1+\sinh^2(\frac{1}{\hbar})}[/tex]

    Is this true for here?

    But it is obvious that this must be zero. Or am I completely wrong??

  7. Feb 17, 2010 #6
    Re: Rectangular Potential well if hbar --> 0

    Sorry, again me.

    Maybe I should not take the simple limit of T itself?

    On the one hand, this text suggests that I need to take the limit of T directly:
    http://img690.imageshack.us/img690/7789/rectpot.png [Broken]

    On the other hand I found this source: http://physics.usask.ca/~dick/tunnel.pdf, page 4 takes first the limit E --> U0.
    But I do not understand the result nor why he does this :-(

    But if [mm]\hbar[/mm] goes to zero the result is zero anyway ...

    Last edited by a moderator: May 4, 2017
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