1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

QM L-box

  1. Aug 21, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle is in the ground state of a L-box. At t = 0 the wall at x = L is suddenly moved to 2L.

    (a) If an energy measurement is made after the wall is moved, what is the probability of measuring the energy to be that of the ground state of the new box? (given solution ≈ 0.36)

    (b) What is the probability of measuring the energy to be that of the first excited state of the new box? (given solution = 1/2)

    *L-box is the infinite square well with zero potential for 0≤x≤L.

    2. Relevant equations

    eigenstates ψ[itex]_{n}[/itex] = √(2/L) sin (n*pi*x/L)

    eigenvalues (energy) E[itex]_{n}[/itex] = ((n*pi*ℏ)^2)/(2mL^2)

    3. The attempt at a solution

    So initially the ground state (of L-box) ψ[itex]_{1}[/itex] = √(2/L) sin (pi*x/L) and E[itex]_{1}[/itex] = ((pi*ℏ)^2)/(2mL^2), so I think this energy is the maximum energy that the particle can have, before and after the change in size of the box.

    For the 2L-box its ground state ψ'[itex]_{1}[/itex] = √(1/L) sin (pi*x/2L) and E'[itex]_{1}[/itex] = ((pi*ℏ)^2)/(8mL^2); its 1st excited state ψ'[itex]_{2}[/itex] = √(1/L) sin (pi*x/L) and E'[itex]_{2}[/itex] = ((pi*ℏ)^2)/(2mL^2).

    So the particle can only be in either of these states (it doesn't have enough energy to move to other states). But since I wasn't given the wave function in this new box, how am I supposed to calculate the probablities?

    My guess was that there should be no preference on the 2 states in the new box, so the probabilities are both 1/2, but which are not exactly correct according to the solution
  2. jcsd
  3. Aug 21, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It's not the energy that remains constant; it's the wave function that's the same immediately before and after the box is lengthened.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted