1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum physics - time dependant changes

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data
    a particle of mass m is in the groundstate of a 1D box with width a. the box then expands rapidly and symmetrically to width 2a. i). What is the probability that the particle is in the new groundstate immediately after the expansion? and ii). how rapid must this expansion be for your calculation to be valid?

    2. Relevant equations
    timescale for change must be hbar/ |E_alpha -E_ beta|
    i think for probabilty you use that the statefunction evolves in time according to the Schrodinger equation and probabilty is the mod squared of the wave-function.

    3. The attempt at a solution
    • pre expansion i got the energy of the groundstate as h^2/8ma^2
    • and after expansion h^2/32ma^2
    • the wavefunction is SQRT(PI/2a) cos (PI.x/a) before and after the expansion.

    for part ii). i got (3/64PI) * (ma^2/h) for how rapid the expansion must be
    for part i). i am stuck! [We were given the hint to use the fourth postulate to compute the probability that given the state-function (above), a measurement of energy will give the ground state energy of the expanded box (which i worked out above). ]
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 2, 2010 #2


    User Avatar
    Homework Helper

    But not everywhere! The wavefunction only takes that form within a certain domain of x. What is that domain, before the expansion?

    For rapid expansion, you can assume that the wavefunction does not change during the expansion. But once the box expands, the eigenstates do change; even though the wavefunction was originally the ground state, it is no longer going to be the ground state. But it will have some component of the ground state in it, and you can figure out just how much by taking the inner product of the wavefunction with the new ground state. From that, you can find the probability.
  4. Jan 2, 2010 #3
    The wavefunction has that form between minus and plus a/2 i think.
    Is the new ground state wavefunction: sqrt(1/a) sin (PI. x/2a).
    which i then multiply by the 'original' wavefunction and integrate. Also what will then be the integration limits: minus and plus a/2?
    Then mod squared of the inner product would give the probability. Thanks
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook