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Momentum from position state vector

  1. Mar 26, 2008 #1
    Hi all, I'm a newbie to this forum and as a high school student, I only have a basic understanding of quantum mechanics, but here's something that I really want to know.

    My question is, if I know the state vector of a quantum particle in the position basis, how do I transform it to the momentum basis? From what I've read, it should involve the Fourier transform, and since I'm interested in (discrete, finite) state vectors rather than continuous wavefunctions, I think I would need the discrete Fourier transform matrix, which I'm familiar with. Also, if I know the state vector that specifies the probability that a particle will be in any given position out of a certain number of positions, can I determine the possible values of the momentum of that particle (that is, the spectrum of eigenvalues of the momentum operator)? On that note, how do I determine the momentum operator of a particle, as a matrix? (since observables of a quantum system correspond to operators, and their eigenvalues give the allowed measurable quantities)

    Here's an example of a question I'd like to be able to answer (very contrived):

    Say we have a particle, in two dimensions, confined to a two-dimensional box (or square). The square is divided into four equal quadrants. Each quadrant corresponds to a vector in the position basis. The probabilities that the particle will be found in quadrants 1, 2, 3, or 4 are, respectively, 1/3, 1/6, 1/4, and 1/4. Let's say that the state vector for the particle (which tells us about its possible positions) is

    [tex][\frac{1}{\sqrt{3}}e^{i\theta_1}, \frac{1}{\sqrt{6}}e^{i\theta_2}, \frac{1}{2}e^{i\theta_3}, \frac{1}{2}e^{i\theta_4}].[/tex]

    How do I determine the possible values for momenta that the particle can have, and more specifically, the matrix operator corresponding to momentum?


    Also, if you know of any books that treat this topic relatively simply, but clearly, I'd like to know of them. I have never taken a course in quantum mechanics.

    Thanks in advance.
     
  2. jcsd
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