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Harmonic oscillator in matrix form

  1. Jan 28, 2010 #1
    I know that the HO hamiltonian in matrix form using the known eigenvalues is
    <i|H|j> = E^j * delta_ij = (j+1/2)hbar*omega*delta_ij, a diagonalized matrix.

    How do I set up the non-diagonalized matrix from the potential V=1/2kx^2?
  2. jcsd
  3. Jan 28, 2010 #2
    \langle x'|H|x\rangle = \delta(x-x')\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}kx^2\right]

    Not very interesting, it's what you might think of as the Hamiltonian anyway.
  4. Jan 28, 2010 #3
    Nice, peter@cam,

    Can you cite a reference for further study?
  5. Jan 28, 2010 #4
    Well I can't think of a book reference, although I'm sure some people on this board can.

    The hilbert space for a 1-d quantum particle is infinite dimensional. In the energy basis, it is a discretely infinite thing, whereas in the position basis it is continuously infinite.

    You just need to get into a habit of thinking about functions as infinite dimensional vectors, and operators as matrices.
    It might help you visualise it if you do a discretised case writing out differentiation as a finite difference matrix (eg using Matlab). The hamiltonian above is non-diagonal because differential operators aren't diagonal in the position basis. (even though paradoxically they come attached with a delta function).

    Its important to remember that *all* operators should come with two indices. Just as a matrix has two subscripts, an operator on a function can always be thought of as a matrix multiplication by representing it with a kernel:
    (\mathsf A \mathbf{f})(x) = \int dx' A(x,x')f(x')
    (\mathsf A v)_i = \sum_j A_{ij}v_j

    Once you realise that the differential operator is really
    and that the position operator is really
    then the analogy with finite-d linear algebra becomes very clear, and constructing matrix elements doesn't feel as unnatural.
  6. Jan 28, 2010 #5
    Thanks for your valuable help. It's clear now.
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