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Probability problem help

  1. Jun 6, 2006 #1
    Please help me with this problem:

    "A particle of mass m moves in one dimension in the infinite square well. Suppose that at time t = 0 its wave function is

    PSI(x,t=0) = A(a^2 - x^2)

    where A is a normalisation constant.

    Find the probability P_n of obtaining the value E_n of the particle energy, where E_n is one of the energy eigenvalues."


    I know how to find A and I know how to find the time-dependent wave function, but what then?
     
  2. jcsd
  3. Jun 6, 2006 #2

    nrqed

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    I notice that you are using a well going from x=-a to x=+a, right?

    You have to expand your wavefunction over the eigenstates of the Hamiltonian (which are simply the sine and cos functions with their normalization factors).
    [tex] \Psi(x) = \Sum c_n \psi_n(x) [/tex]
    where the [itex] \psi_n[/itex] are the normalized sin/cos functions.
    To find the c_n, just use the orthonormality of the psi_n,

    [tex] c_n = \int_{-a}^a dx \, \psi_n^*(x) \Psi(x) [/tex]

    and, finally, the probability of the energy being measured to be a specific value E_n is given by |c_n|^2.
     
  4. Jun 7, 2006 #3
    Yeah that does it, thank you. =)
     
  5. Jun 7, 2006 #4
    I use the sin function for even n and the cos function for odd n, right? But then c_n for even n gets zero..? Is that correct?

    For odd n I get P(E_n) = 15/((n^4)(pi^4))
     
  6. Jun 7, 2006 #5

    nrqed

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    I did not check your result for odd n but it makes sense that for even n you would get zero. Your wavefunction is symmetric function (it is even in the sense of Psi(-x) = Psi(x)) so that it does not "contain" any of the sin functions which are odd themselves.

    Your answer seems also to make sense since, as expected, the probabibility goes down with increasing n. Also, your function ressembles the most the ground state so one expects a very quick decrease with n.
    Lastly, one double check would be to verify that the sum to infinity over odd n of 1/n^4 gives pi^4/15.

    Good job!

    Pat
     
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