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Sudden perturbation, particle in a box, ground state

  1. Dec 1, 2014 #1

    fluidistic

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    Gold Member

    Hi guys!
    Suppose there's a particle in a box, initially in its ground state. Suppose that one chooses a system of coordinates such that the potential V(x) is 0 from 0 to L.
    Suppose that one suddenly perturbate the system at a particular time so that V(x) becomes 0 from 0 to 2L.
    I've calculated the probability for the particle to be in the ground state of the twice-bigger box as ##\frac{2^{3/2}}{3\pi}##.

    Now suppose we solve a similar problem, where V(x) is initially worth 0 from -L/2 to L/2 and psi is in its ground state, and suddenly one expands the box so that V(x)=0 from -L to L.
    Now the probability for the particle to be in the ground state of the 2L sized box is... ##\frac{2^3}{3\pi}##. Which is different from the 1st case. (I may not recall the exact values but I remember they differed).

    However I'm having trouble understanding what's the difference physically in the 2 cases. To me it just looks like a 1 dimensional box of size L suddenly double its size, in both cases. So physically I don't understand why I get 2 different results as if there was a priviledged observer in the universe.
    Mathematically I do understand that the overlap of the 2 wavefunctions (before and after the perturbation) are different in the 2 cases. It's just that physically I'm not grasping what's going on.

    Could someone shed some light on this? Thanks.

    P.S.: ##\psi _0(x)=\sqrt {\frac{2}{L}} \sin (\frac{\pi x}{L} )## for the ground state of the L sized box according to an observer whose coordinate system's origin is at the left wall.
    If the origin of that coordinate system is set midway between the 2 walls, ##\psi _0(x)=\sqrt {\frac{2}{L}} \cos (\frac{\pi x}{L} )##.
    ##P=|\langle \psi _{\text {final}}| \psi _{\text {initial}} \rangle |^2##.
     
  2. jcsd
  3. Dec 1, 2014 #2

    Nugatory

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    They're different physical situations. You're right that in both cases we're dealing with a box of width 2L, but the initial conditions are different. In one case, the particle is localized in the left-hand side of the new larger box at the moment that the barriers go down; in the other case it is in localized in the center. The same Schrodinger's equation governs the evolution of the wave function either way, but (as with any differential equation) different initial conditions lead to different solutions.
     
  4. Dec 1, 2014 #3

    fluidistic

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    Ok thanks. I now see the difference physically.
    Extending the box from 1 wall is not the same than extending it from both walls (I find it easier to visualize when I consider a localized particle in the box, like say in classical mechanics).
     
  5. Dec 1, 2014 #4

    jtbell

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    Try expanding the second box from [-L/2, +L/2] to [-L/2, +3L/2]. This should give you the same result as with expanding the first box from [0, L] to [0, 2L].
     
    Last edited: Dec 2, 2014
  6. Dec 2, 2014 #5

    DrDu

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