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Wavefunction is cos(kx), probability of momentum?

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that ψ(x) = Acos(kx) is not an eigenfunction of the momentum operator. If you were to measure the momentum of a particle with this wavefunction, what possible values would you get and what would the probability be of obtaining these values?

    2. Relevant equations

    Momentum operator is;

    [itex] -i\frac {h}{2\pi} \frac {d}{dx} [/itex]

    3. The attempt at a solution


    It's obvious that it's not an eigenfunction of the operator, but how do I do the latter two questions? If it's not an eigenfunction, how could it be known? I could use the expectation value formula (<p> = ∫ψ*pψdx) but what would my integral values be? Could someone give me a hint?

    Someone told me elsewhere that I want the wave function expressed in the momentum representation which is obtained by taking the fourier transform. Whether that's right or not, this is just meant to be a series of revision questions from my last year quantum mechanics, and we never dealt with Fourier transforms.
     
  2. jcsd
  3. Sep 27, 2012 #2
    Remember that the Born Rule says that the measurement probability for a wavefunction to be in a given eigenstate is dependent on the coefficient of that eigenstate in the wavefunction. Since every wavefunction can be written as the weighted sum of one or more eigenstates, your first step should be to determine what the eigenstates of momentum actually are, and then try to express your wavefunction in terms of them.

    I'd highly suggest you invest some time in learning about Fourier transforms at some point--they're really common in quantum mechanics, so it'd be well worth the effort.
     
  4. Sep 27, 2012 #3
    Ah, now that you say it, it's come flooding back to me. I'll just post my thoughts in case I'm wrong.

    (NOTE: By h I mean h-bar, can't seem to put strikes through in itex)

    The operator

    [itex] -ih \frac {d}{dx} [/itex]

    has eigenfunctions of the form

    [itex] ψ = e^{\frac{ipx}{h}} [/itex]

    where p is the momentum and the eigenvalue of the operator.


    The wavefunction

    [itex] Bcos(kx) [/itex]

    Is a superposition of two wavefunctios;

    [itex] \frac {1}{2}(Be^{ikx} + Be^{-ikx}) [/itex]


    These seperate wavefunctions have definite momentums of -hk and +hk. Thus the possible values one could obtain for momentum from [itex] Bcos(kx) [/itex] is -hk and +hk, and the probability to obtain both of these is

    [itex] {\frac {B}{2}}^2 = \frac {B^2}{4} [/itex].

    Or, assuming the wavefunction is normalised, the probability to obtain either of these is just 1/2.
     
  5. Sep 27, 2012 #4
    Exactly!

    As you may have realized by now, the connection to Fourier transforms is that this process of splitting up the wavefunction into momentum eigenstates is precisely the act of taking the Fourier transform of the wavefunction. As you get further into QM, you'll be constantly going back and forth between position and momentum bases, so it's worth it to get familiar with the Fourier transform, so that when these sorts of problems arise, you don't even have to really think about it.

    (Oh, and you can make an h-bar in Tex by using the macro \hbar.)
     
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