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Computing Expectation values as functions of time.

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data
    6) A particle in the infinite square well has the initial wave function
    Ψ(x,0)= Ax when 0<=x<=a/2
    Ψ(x,0)= A(a-x) when a/2<=x<=a

    a) Sketch Ψ(x,0), and determine the constant A.
    b) Find Ψ(x,t)
    c) Compute <x> and <p> as functions of time. Do they oscillate? With what frequency?


    2. Relevant equations

    Schrodinger Equation, Equations for expectation values

    3. The attempt at a solution

    Part (a) wasn't a problem. I solved for the normalization constant A by setting the integral of the norm of the wave function equal to zero.
    A = 2 * sqrt(3/a^3)

    Part (b) also wasn't a problem as Ψ(x,t)=Ψ(x,0)exp(-iEt/h)

    Part (c) confuses me a little. I know how to calculate <x> and <p> using the equations <x> = ∫ψ*(x)xψ(x)dx and <p> = ∫ψ*(x)(h/i)d/dx(ψ(x))dx. But the as functions of time confuses me. Won't the exp(-iEt/h) cancel out when multiplied by its conjugate. If this is true, then how can <x> and <p> oscillate? How could they oscillate with any frequency? Any help would be appreciated.
     
  2. jcsd
  3. Feb 19, 2007 #2

    Dick

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    psi(x,0) is not an energy eigenfunction. So there is no single value of E to use for it. You'll have to represent it as a sum of energy eigenfunctions (a fourier series) to determine its time evolution.
     
  4. Feb 19, 2007 #3
    Ok. I wasn't expecting that. I am not sure what you mean. Do you mean a series that looks exp(-iE1t/h) + exp(-iE2t/h) + exp (-iE3t/h) +...
     
  5. Feb 20, 2007 #4
    What are the eigenfunctions of the infinite square well, and how would you represent your initial function Ψ(x,0) in terms of them?
     
  6. Feb 20, 2007 #5
    I am not following at all. I don't really know what you are getting at. The only energy eigenfunction I have seen for the infinite square well is E=n^2(pi^2)(h^2)/(2ma^2).
     
  7. Feb 20, 2007 #6

    nrqed

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    Those are the energies which are eigenvalues (of the Hamiltonian). Corresponding to each of those eigenvalue is an eigenfunction. What are they?
     
  8. Feb 20, 2007 #7
    exp(-iEnt/h)
     
  9. Feb 20, 2007 #8

    Dick

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    That's only the time dependence. How about the space dependence?
     
  10. Feb 20, 2007 #9

    nrqed

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    That is just the time dependence. What is the x dependence?
     
  11. Feb 20, 2007 #10
    I think it would be Ψ(x)=Asin(n*pi*x/a)
     
  12. Feb 20, 2007 #11

    nrqed

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    Right. *These* are the energy eigenfunctions, so only these wavefunctions vary in time with an exponential [itex] e^{-iE_n t/ \hbar}[/itex]. You can't stick a time exponential to any function of x!!

    So first you must expand your wavefunction over those eigenfunctions using orthonormality. If you haven't seen an example of that in class, it will be long to explain and I have to go teach right now. Maybe I can help more later.
     
  13. Feb 20, 2007 #12

    nrqed

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    :biggrin: We wrote almost exactly the same sentence! :smile:
     
  14. Feb 20, 2007 #13
    Ok, so what happens to my initial wave function, Ψ(x,0)?
     
  15. Feb 20, 2007 #14

    Dick

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    For that reason, I feel confident you can continue here. I've gotta get to work.
     
  16. Feb 20, 2007 #15

    nrqed

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    You have to write
    [tex] \Psi(x,0) = \sum_n c_n {\sqrt{\frac{2}{a}} sin(\frac{n \pi x}{a}) [/tex]
    and find the the value of the c_n using orthonormlaity of the sin wavefunctions. Once you have the c_n , plug them back into
    [tex] \Psi(x,t) = \sum_n c_n {\sqrt{\frac{2}{a}} sin(\frac{n \pi x}{a}) e^{-iE_n t/\hbar} [/tex]
    Your answer will have to stay this way..as an infinite sum.

    Patrick
     
  17. Feb 20, 2007 #16
    Thanks my main man.
     
  18. Feb 20, 2007 #17

    nrqed

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    You're welcome. The only difficulty is in finding th ec_n which will require doing an integral of the sin wavefunction times your original wavefunction (leaving "n" an arbitrary parameter).

    I haven't addressed your question c) though but I have to go.

    Best luck

    Patrick
     
  19. Feb 20, 2007 #18
    Ok, I think I'm cool with (b). Help with part (c) anyone?
     
  20. Feb 21, 2007 #19

    dextercioby

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    I don't see the problem, now that you have the [itex] \Psi (x,t) [/itex].
     
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