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Homework Help: Momentum from position state vector (quantum mechanics)

  1. Mar 28, 2008 #1
    I made a post in the quantum mechanics section, but it hasn't gotten any replies, so I'll try again here. This isn't strictly a homework/coursework problem, but something that I really want to know.

    1. The problem statement, all variables and given/known data
    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]\left[\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}\right].[/tex]

    How do I determine the possible values for momenta that the particle can have, and more specifically, the matrix operator corresponding to momentum for the particle? (And is this scenario even possible?)

    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. Relevant equations
    The time-independent Schrodinger equation, in matrix operator form?

    3. The attempt at a solution
    This is more of a conceptual issue and not a specific problem. (is my understanding completely wrong?)
  2. jcsd
  3. Mar 28, 2008 #2


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    I don't understand what you mean by "discrete and finite position state vectors". Since the vectors in the position basis are infinite, the wavefunction in the position basis is also infinite.

    If the wavefunction in the position basis is [tex]\psi(x,t)[/tex], the momentum space wavefunction is indeed the fourier transform.

    [tex] \phi(p,t) = \frac{1}{\sqrt{2 \pi \hbar}}\int_{-\infty}^{\infty} e^{-i p x/ \hbar} \psi(x,t) dx[/tex]
  4. Mar 28, 2008 #3
    So when I say "discrete and finite," I mean that I'm interested in the "state vector" ([tex]|\psi\rangle[/tex]) formulation of quantum mechanics rather than the "wavefunction" ([tex]\psi(x, t)[/tex]) formulation. A state vector is like a discrete version of a wavefunction that can be represented as a ray in finite-dimensional Hilbert space. A state vector must be such that the sum of the squares of the moduli (absolute values) of its complex entries is 1, which is analogous to the condition for wavefunctions that the integral from -infinity to infinity of [tex]|\psi|^2[/tex] must be 1 (as for any probability density function). I now know that to transform a state vector from the position to the momentum basis, I simply multiply it by the DFT (discrete Fourier transform) matrix of suitable dimension.

    What I'm mainly interested in, however, is how to find the momentum operator [tex]\hat{p}[/tex] for an arbitrary state vector, which is a matrix (and from this, I can find the kinetic energy operator as [tex]\hat{K} = \frac{\hat{p}^2}{2m}[/tex]). From what I've read, the momentum operator should be a diagonal matrix, and hence its eigenvalues should be on its diagonal. Then, its eigenvalues should give the allowed values for the momentum of the quantum system that is described by the state vector. How does one find the momentum operator?
    Last edited: Mar 28, 2008
  5. Mar 28, 2008 #4
    Note that state vectors can be vectors in infinite dimensional vector spaces. If your state is describe-able by a position, then there is some operator [tex]\hat{x}[/tex] on the space, with a continuum of observed values: [tex]\hat{x}\left|x\right>=x\left|x\right>[/tex], which is only possible if the vector space is infinite (uncountably so, in fact) dimensional. Thus to even talk about position or momentum basis, you posit a infinite dimensional Hilbert space.
  6. Mar 28, 2008 #5
    Hmmm... I think, though, that it is possible to have a finite state vector if there is some limit on the "precision" of the position of a particle described by that state vector. Take, for example, what I was saying before - we have a particle confined to a two-dimensional square, described by a state vector that specifies the probabilities it will be found in any of the four quadrants. We can't say anything more about the position of the particle other than that it is in one of the four quadrants. Hence, the state vector doesn't describe the probability that it will be found in any position in the square (in which case it would indeed be infinite), but only the probabilities that it will be found in certain portions of the square. Is it still possible to associate a concept of "momentum" with such a state vector, and to find an operator matrix for momentum in this case?

    Note: the main reason I'm asking this question regarding finite state vectors is that I'm studying some quantum computing right now, and quantum computers have a limit on their computational resources, and hence on their precision - they can't deal with infinite state vectors.
    Last edited: Mar 28, 2008
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