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Infinite Square Well

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    consider a particle of mass m in the ground state of an infinite square well potential width L/2. What is the probability of detecting the particle at x=L/4 in a range of [tex]\Delta[/tex]x=0.01L (d not integrate)? Assume that the particle is in the normalized state [tex]\Psi[/tex](x,0)=c1[tex]\Psi[/tex]1+c2[tex]\Psi[/tex]2, what is the time dependent?

    2. Relevant equations

    3. The attempt at a solution
    Help, i dont know from where to start even...
  2. jcsd
  3. Nov 24, 2008 #2
    If you WERE allowed to integrate to get the probability, would you know what to do?
  4. Nov 24, 2008 #3
    i would integrate the wavefunction (multiplied by its complex part) to find <x> and <x^2>, and then use the uncertainty. isnt <x> always zero though?
  5. Nov 24, 2008 #4
    Close, but not quite. If you wanted to know the probability of finding the particle between [tex]x[/tex] and [tex]\Delta x[/tex] you would get it from
    [tex]P = \int_x^{x+\Delta x} \Psi^* \Psi dx[/tex].
    Now think about what a definite integral "means" and find an approximation to the integral if [tex]\Delta x[/tex] is really small. Maybe drawing a picture would help. You do not have to integrate.
  6. Nov 24, 2008 #5
    So I need from 0 to L/4 in increments of 0.01L, if delta x is really small isn't just going to be the wavefunction multiplied by itself with x=L/4. I mean i know that a definite integral is the area under the curve, maybe just the function where x=L/4 multiplied by 0.01L?
  7. Nov 24, 2008 #6
    You're very close. But your not interested in the particle between 0 and L/4. You are interested in a small interval AROUND L/4. Catch my drift?
  8. Nov 24, 2008 #7
    delta and expsilon? ===>limit?
  9. Nov 24, 2008 #8
    Wait!. I should've read your reply to the end. Your last statement
    "maybe just the function where x=L/4 multiplied by 0.01L?"
    is 99 percent right. Just use the probability AMPLITUDE instead of the wavefunction itself.
  10. Nov 24, 2008 #9
    ohhh...thanks. I've already started taking the limit....lol. I'll do that instead
  11. Nov 24, 2008 #10
    so for the first part i got 0.04 (4%), and the second part is just an infinite series.
    now, I'm asked to find the expectation values of energy <E>. Is that just--
  12. Nov 25, 2008 #11
  13. Nov 25, 2008 #12
    actually I got 8% on the first part. for the second part i got <E>=E.
    Another question- "now that the wave function returned to the ground state. at t=0 the well suddenly changes to an infinite square well of width L without affecting the wave function. find the probability that a measurement of energy right after the expansion will yield E=(h bar*Pi)^2/(2mL^2)?

    Isnt it a stationary state===> thus the probability never changes and stays 1?
  14. Nov 25, 2008 #13
    It is in a stationary state for the OLD well. The NEW well has different stationary states. You have to write your state as a linear combo of the NEW stationary states.
  15. Nov 25, 2008 #14
    but it states in the question that the width changes without affecting the wave function
  16. Nov 25, 2008 #15
    Right. But when the width changes, you have NEW eigenfunctions (i.e. stationary states). So your old state is no longer one of the stationary states for the new problem.
  17. Nov 25, 2008 #16
    where [tex]\Sigma[/tex]C^2n=1
    P1=C^21=1, thus, the probability of E didnt change b/c the expectation value of E didnt change.
  18. Nov 25, 2008 #17
    You can't tell much about the probability from the expectation value. They aren't really related.
  19. Nov 25, 2008 #18
    Ok. You are partly right. It seems to me that the state IS still a stationary state, but now it is NOT the ground state anymore. The state corresponds to an eigenvalue that is not the one for n=1.
  20. Nov 25, 2008 #19
    I don't know why the heck my last reply didn't show up.
    It turns out that your state is actually a stationary state (even after the expansion), but the catch is that it is not the "ground" state for the new well. You diggin' what I'm buryin'?
  21. Nov 25, 2008 #20
    lol.....got it. Thanks. Why your earlier replies didnt show up? weird
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