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!

TDSE, time evolution between states (Check my working please?)

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi guys, I've recently taken up quantum, so it's all very new to me, it would be greatly appreciated if someone could check my working!


    Let ψ1(x) and ψ2(x) be two orthonormal solutions of the TISE with corresponding
    energy eigenvalues E1 and E2. At time t = 0, the particle is prepared in the symmetric
    superposition state:

    and subsequently allowed to evolve in time. What is the average energy of the system
    as a function of time? What is the minimum time ¿ for which the system must evolve
    in order to return to its original state (up to an overall phase factor), when it starts in
    the state ψ(+) (x)

    Determine the probability to ¯nd the system in the antisymmetric superposition
    state

    as a function of time when it starts in the state ψ(+) (x)

    At time t1 the particle is found in the antisymmetric superposition state. What is
    the probability to ¯nd the particle in the symmetric superposition state at time t1 + T,
    where T is the time found above?


    21eol6h.png


    2. Relevant equations



    3. The attempt at a solution

    6j3tcg.png
    33xc0a9.png
    2qmnkg5.png
    2uqeni9.png

    Not sure if the last part is right, as it suggests that the probability of finding the particle in the left half of the box is independent of time! Then again, in an infinite square well the potential doesn't depend on time, so TDSE is reduced to TISE?
     
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I think you should express ##\Delta \omega## in terms of the energies.

    For the last part, why do you think those mixed terms are zero if you integrate them from 0 to L/2?
    The problem statement itself is bad here, I think. The answer depends on the phases used to define the first two states, and those phases are arbitrary.
     
  4. Jun 9, 2013 #3
    Because ψ1 and ψ2 are orthogonal states? And to get the answer in the previous parts they were zero...Is the reason why they are non-zero here because the limits of integration are not from -∞ to ∞?
     
  5. Jun 9, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right, orthogonality just means that the integral over the whole space is zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: TDSE, time evolution between states (Check my working please?)
  1. Please check my work (Replies: 0)

Loading...