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Wavefunction of infinite square well potential between -L<x<L

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve for the wavefunctions and energy levels of an infinite square well potential extending between -L<x<L.

    Hint: It may be worth noting that for a potential symmetric in x, then the observed probability density must also be symmetric in x, i. |ψ(x)|2 = |ψ(-x)|2.

    2. Relevant equations


    Time independant Schroedinger equation:

    [itex][-\frac{\hbar^2}{2m}\frac{∂^2}{∂x^2} + U(x)]ψ(x) = Eψ(x)[/itex]

    3. The attempt at a solution

    For areas outside of the square with infinite potential, the potential is zero, so


    I take it that I only need to find the wavefunction for L<x<∞, as the symmetry means that the wavefunction for -∞<x<-L is going to be the same.

    The solutions to the above differential equation are

    [itex]ψ(x) = Asin(kx)+ Bcos(kx)[/itex]

    But from here I am stuck. I have the conditions that

    [itex]ψ(L)=0[/itex] and [itex]lim_{x → ∞} ψ(x) = 0[/itex], otherwise there is an infinite area under the curve for the probability amplitude.

    But my solution for ψ(x) can't satisfy both of these without both A and B being zero.
  2. jcsd
  3. Apr 25, 2013 #2


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    Certain potentials will not sufficiently constrain a particle to have either a discrete energy spectrum or a normalizable wavefunction. If the potential vanishes everywhere, so we have a free-particle, the energy spectrum is continuous and the wavefunctions are nonnormalizable plane waves.

    In this case, you refer to the potential as an infinite square well, but you seem to describe it as an infinite barrier. The well is usually ##V=0## on some interval, ##V=\infty## everywhere else. The barrier is ##V=\infty## on an interval, ##V=0## everywhere else. In either case we can discuss solutions, but they are different problems.

    For the barrier, we have a free particle that must satisfy [itex]ψ(L)=0[/itex], but like the free-particle, we cannot impose [itex]lim_{x → ∞} ψ(x) = 0[/itex].
  4. Apr 25, 2013 #3
    I think I might just have gotten the wrong impression. I initially thought that it was just a typical particle-in-the-box question where the well is from -L<x<L, but then the location of the well doesn't really matter, so I can just use the same results from the case where the well is from 0<x<L and substitute in 2L in place of L. But that seems way too easy, especially since the preceding question on the sheet was essentially a one-line calculation.

    But I did a google search on the infinite barrier you were talking about, and I've found something that might help me. I'll give it another crack and see what I can get. Thanks :)
  5. Apr 25, 2013 #4


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    You could do something like that. Replacing L by 2L will account for the difference in the width of the well, but you also need to take into account that the center of the well is at a different location, i.e. at x=0 instead of x=L/2.

    So is the potential indeed an infinite barrier? It's kind of a boring problem if so, and the hint suggests you should have a well.
  6. Apr 27, 2013 #5
    Oh, right, yes :P I just have to replace x with (x - L).

    Actually, I think it's the infinite well. But I'd like to know about the barrier as well. I don't really know how to approach the problem mathematically, but I remembered that if the length of the box is much larger than the wavelength of the particle, then the system behaves classically. So in the case of the barrier, that's basically two boxes of infinite length separated by a barrier of infinite potential. So in each 'box', the probability density is uniform. Am I on the right track here?

    And thanks for the reply :)
  7. Apr 28, 2013 #6


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    With barrier problems like this, you end up with plane wave solutions on either side of the barrier, so the probabilities are uniform. The interesting question is the probability that a particle can tunnel through a barrier. With an infinite barrier (of non-zero width), the particle can't get through, so it's not a very interesting situation.
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