Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shifting of the wall problem

  1. Aug 11, 2014 #1
    One of the problems in QM i frequently encounter in all textbooks is the shifting of the wall problem which goes like this.
    Assume a particle is in the ground state (or any stationary state) of an infinite potential well between 0<x<a. If the wall at a is suddenly shifted to 2a, then what is the probability of finding the particle in the ground state (or any other stationary state) of the new well.
    The way I understood it, the solution involves the assumption that since the wall is shifted suddenly, the wave function does not change. However, since the system itself has changed, the new system has different stationary states. The original wave f is a linear combo of these eigenstates and the probability of finding it in one of these states is the corresponding coefficient mod-squared.

    My question is since the wave function is unchanged, it looks like
    ψ= sin(∏x/a), 0<x<a
    ψ= 0 elsewhere
    Then at x=a, dψ/dx is discontinuous, even though V≠∞.
    How is such a wave function allowed?
    Or is my understanding of the solution to the problem wrong?
  2. jcsd
  3. Aug 11, 2014 #2


    User Avatar

    Staff: Mentor

    An "infinite potential well" is not posslble, strictly speaking; nor is the discontinuous dψ/dx at its boundary. It is an idealization similar to frictionless surfaces or massless strings in classical mechanics. Nevertheless, it is useful to consider such idealized situations because they are easier to work with mathematically, than non-infinite wells with continuous dψ/dx. They give results that are "close enough" to some physically realizable situations, and convey the essential physical principles (e.g. linear combinations of eigenstates).
  4. Aug 11, 2014 #3
    I understand that. I am not talking about the discontinuous dψ/dx at the boundary. In this case, the discontinuity appears in the middle of the new potential well, where V(x)= 0. Even theoretically, discontinuous 1st derivatives are allowed only at points where V(x)= ∞, according to Introduction to QM by Griffiths.
  5. Aug 11, 2014 #4


    User Avatar
    Science Advisor
    Gold Member

    The discontinuity in the derivative should very quickly be smoothed out by the Schroedinger evolution of the wave function. Consider that it is just there for "an instant", and that has to do with the wall being "suddenly" expanded, which is, again, another approximation! (Though a useful one, otherwise one would have to work with the full machinery of the time dependent perturbation theory.)
  6. Aug 11, 2014 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    That's true for solutions of the time-independent Schrodinger's equation. For time-dependent solutions, you can find a solution with a discontinuous first-derivative with respect to x. For example, for a free particle (the potential is zero everywhere), let [itex]F(x)[/itex] be the function satisfying:

    [itex]F(x) = 1 [/itex] if [itex]|x| < L[/itex]
    [itex]F(x) = 0 [/itex] if [itex]|x| > L[/itex]

    Let [itex]\tilde{\psi}(k)[/itex] be the Fourier transform of [itex]F(x)[/itex]. Then a solution to the time-dependent Schrodinger equation would be [itex]\psi(x,t)[/itex] defined by:

    [itex]\psi(x,t) = \int dk\ \tilde{\psi}(k) e^{i(kx - \omega t)}[/itex]

    where [itex]\omega = \dfrac{\hbar k^2}{2 m}[/itex]

    This solution has both [itex]\dfrac{\partial \psi}{\partial x}[/itex] and [itex]\dfrac{\partial \psi}{\partial t}[/itex] undefined at [itex]t=0, x= \pm L[/itex].
  7. Aug 11, 2014 #6


    User Avatar
    Science Advisor

    The infinite square well is an artificial problem; the sudden shift is even more artificial; so the discontinuity is the artificial artifact of an artificial problem; why worry?
  8. Aug 11, 2014 #7
    I think an interesting problem would be if it is expanded to something different to an integer of a . What would happen then?
    I dont think this an artificial problem as it can be achieved quite readily in waveguides.
  9. Aug 11, 2014 #8


    User Avatar
    Science Advisor

    I guess you would study a fast but not instantaneous moving wall, i.e [0,L(t)].

    Then I would propose to expand a wave function ψ in terms of the "eigenfunctions"

    [tex](H_{L(t)} - E_{n,L(t)})u_{n,L(t)}(x) = 0[/tex]

    [tex]u_{n,L(t)}(x) = \sin k_n(t)\,x;\;k_n(t) = n\pi/L(t)[/tex]

    [tex]\psi(x,t) = \sum_n \psi_n(t)\,u_{n,L(t)}(x)[/tex]
  10. Aug 11, 2014 #9
    Suppose though the walls were fixed and had a step in one of them?
  11. Aug 11, 2014 #10


    User Avatar
    Science Advisor

    What has this to do with the original question
  12. Aug 11, 2014 #11


    User Avatar
    Science Advisor

    To make the problem well defined, I would map the original problem to one where the length is fixed by introducing a new coordinate y=2x, and then renaming y back to x. Hence, the wavefunction get's scaled and the mass of the particle changes but everything is defined on the same hilbert space.
    Take also in mind that a discontinuous first derivative of the wavefunction means still a well defined wavefunction in hilbert space. However, this function is not in the domain of the Hamiltonian. For the discussion of time dependence this is not important as U=exp iHt is unitary and thus defined on the whole Hilbert space.
  13. Aug 11, 2014 #12
    I get it. I was afraid my understanding of the problem was wrong. Thanks.
  14. Aug 11, 2014 #13
    I see. thanks.
  15. Aug 12, 2014 #14


    User Avatar
    Science Advisor

    This works for my proposal
    as well. You need to introduce a rescaled coordinate

    [tex]\xi(t) = \frac{x}{L(t)};\;\xi \in [0,1][/tex]

    which creates a time-dependent mass

    [tex]m \to m\,L^2(t)[/tex]
  16. Aug 12, 2014 #15


    User Avatar
    Science Advisor

    no, unfortunately it does not work b/c the i∂t acts on the new coordinate and creates a term which violates the boundary condition; but I don't want to hijack this thread, so please continue
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook