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!

Expectatoon value particle in superposition of momentum states

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Demonstrate the relation between the expectation value and the measurement outcomes of an observable of a particle by conisdering as an observable the kinetic energy operator
    E=p^/2m when the particle is in a superposition of 2 momentum eigenstates
    2. Relevant equations

    <O> = Int (from -inf -> inf) [(Psi*)O(Psi)] dx


    3. The attempt at a solution

    I am taking the superposition of 2 momentum eigenstates as

    Psi= square root (1/L) [ A*exp(ikx)exp(-iEt/Hbar) +B*exp(ikx)exp(-iEt/Hbar) ]

    And then putting this into the integral

    <O> = Int (from 0->L) [(Psi*)(-hbar/2m*d2/dx2(Psi)] dx

    However I end up with a very long equation for the expectation value whereas I thought the expectation value would be something along the lines of
    A2hbar2k12/2m + B2hbar2k22/2m as this looks like an eigenvalue
     
  2. jcsd
  3. Jan 14, 2012 #2
    I'd take out the [itex]\sqrt{1/L}[/itex] since there's no reason to consider any sort of box here (and besides you can absorb it into the [itex]A[/itex] and [itex]B[/itex] terms). Get rid of the time-dependent bit (since you're going to ignore it anyway - you're integrating over [itex]x[/itex]) and make sure you label your two [itex]k[/itex]s differently, like you have in your final suggestion: [itex]k_1[/itex] and [itex]k_2[/itex]. And then the approach you're using should work! You seem to have some idea what you expect to find, which is good - if you can't get there post where you get up to.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook