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Approaching the problem o 1D well that changes size

  1. Dec 16, 2014 #1
    1. The problem statement, all variables and given/known data

    You have a potential well, it's 1-dimensional and has a width of 0 to a. All of a sudden the wall of the well is pushed inward so that it's half as wide. Now the well is only extending from 0 to a/2.

    in the well is a particle (mass m) that is in the first excited state n=2.

    We want to know the following:

    - What are the allowed energies and eigenfunctions in our new (smaller) infinite well?
    - what is the probability of finding our particle in the ground state? The first excited state?

    2. Relevant equations
    We know the ground state of a particle (the wave function) is
    ## \psi_0 = \sqrt{ \frac{2}{L}} \sin \left( \frac{n \pi}{L} x \right) ## and that ##E_n = \frac{\pi^2 \hbar^2}{2mL^2}n^2##

    3. The attempt at a solution
    What I did here was look first at the equation for E. I can just plug in n=2 and a/2 = L and get

    [itex]E_2 = \frac{16 \pi^2 \hbar^2}{m}[/itex], which tells me the n=2 energy.

    So I want to know th probability of the ground state, though. For that I need the series expansion

    ## \sum c_n \psi_n ##

    where

    ## c_n = \sqrt{ \frac{2}{a}} \int_0^{\frac{a}{2}} \sin \left(\frac{n\pi x}{a} \right) \sqrt{ \frac{2}{L}} \sin \left(\frac{n\pi x}{L}\right) dx##

    and making sure that I plug in a/2 for L:
    ## c_n = \sqrt{ \frac{2}{a}} \int_0^{\frac{a}{2}} \sin \left(\frac{n\pi x}{a} \right) \sqrt{ \frac{4}{a}} \sin \left(\frac{2 n\pi x}{a}\right) dx##
    pull out the constants:
    ## c_n = \frac{\sqrt{8}}{a} \int_0^{\frac{a}{2}} \sin \left(\frac{n\pi x}{a} \right) \sin \left(\frac{2 n\pi x}{a}\right) dx##

    and we have here a pretty well-behaved function. A trig substitution / identity gives me:
    ## c_n = \frac{\sqrt{8}}{a} \int_0^{\frac{a}{2}} \frac{1}{2} \left[\cos \left(\frac{n\pi}{a}-\frac{2 n\pi}{a} x \right) -\cos \left(\frac{n\pi}{a}+\frac{2 n\pi}{a} x \right) \right] dx##
    ##= \frac{\sqrt{8}}{2a} \int_0^{\frac{a}{2}} \cos \left(\frac{ n\pi}{a} x \right) -\cos \left(\frac{3 n\pi}{a} x \right) dx##
    ##= \frac{\sqrt{8}}{2a} \left[ -\frac{a \sin \left(\frac{ n\pi}{a} x \right)}{n\pi} -\frac{ a \sin \left(\frac{3 n\pi}{a} x \right)}{3n\pi} \right]^{\frac{a}{2}}_0 ##
    ##= \frac{\sqrt{8}}{2a} \left[ -\frac{a \sin \left(\frac{ n\pi}{2} \right)}{n\pi} -\frac{ a \sin \left(\frac{3 n\pi}{2} \right)}{3n\pi}\right] = -\frac{\sqrt{8}}{2n\pi} \left[ \sin \left(\frac{ n\pi}{2} \right) +\frac{ \sin \left(\frac{3n\pi}{2} \right)}{3}\right] ##

    So I see that the sine terms are
    n=1 --> 2/3
    n=3 --> -2/3
    n=5 --> 2/3

    and so on. So I want to know the probability that the particle is in the first excited state, n=2. It turns out that it can't be there, because ##c_n## is zero wherever n is even. But at n=1 ##(c_1)^2 = \frac{8}{4\pi^2}\frac{4}{9} = \frac{32}{36\pi^2}##. That's my probability for that particular state. At ##(c_3)^2 = \frac{8}{36\pi^2}\frac{4}{9} = \frac{32}{324\pi^2}##, et cetera.

    Ayhow I am checking to see if I approached this thing right, and didn't make a dumb mistake.
     
  2. jcsd
  3. Dec 17, 2014 #2

    BvU

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    This tells you the wave function ##\psi##. There is a 2 in there, but no longer an n.
    To get the ##c_{n'}## you need to evaluate ##\int \psi^*_{n'}\; \psi##.

    I have a problem with this exercise, because pushing in the wall can't leave the wave function as it was: it has to remain normalized. And I think there's no telling what that does to the energy of the particle.

    (The reverse problem allows one to extend the wave function with 0 in the 'new' range, thus avoiding this renormalization issue)
     
  4. Dec 17, 2014 #3
    As I recall the problem was stated as an infinite potential is inserted at a/2, if that helps, and the idea was w could just ignore the part of the well on the right. But if I hear you right I should set up [itex]\int \psi^*_n' \psi [\itex] as ##\int \psi^*_n' \psi_2 ## ?
     
  5. Dec 17, 2014 #4
    As I recall the problem was stated as an infinite potential is inserted at a/2, if that helps, and the idea was w could just ignore the part of the well on the right. But if I hear you right I should set up [itex]\int \psi^*_n' \psi [\itex] as ##\int \psi^*_{n'} \psi_2 ## ? And then the issue is figuring out what n' is? (Or can I go with 1 since the well is half-size, and the wave is going to be cut in half -- foe purposes of this problem I don't think we had to renormalize).
     
  6. Dec 17, 2014 #5
    i just tried getting the dangd Latex to work and it's buggy today. but i think you ought to be able to see what i was asking.
     
  7. Dec 17, 2014 #6

    BvU

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    Yeah, you need curly brackets around {n'} :)

    Cutting the wave function in half makes the probablility the particle is somewhere equal to one half, which I find hard to swallow.
    Otherwise, yes, I agree: that's what I tried to bring across. n (for the full width well) is fixed to 2 by the exercise statement.
     
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