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Finding the momentum operator matrix of the harmonic oscillator

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Given a particle is confined in a one dimensional harmonic oscillator potential, find the matrix representation of the momentum operator in the basis of the eigenvectors of the Hamiltonian.

    2. Relevant equations

    Potential: V(x) = 0.5 m w^2 x^2 where m is the mass of the particle and w is the angular frequency

    Schrodinger equation: H |psi> = E |psi> where H is the Hamiltonian operator, E are its eigenvalues (or the energies of the system), and |psi> are its eigenvectors (or the stationary states of the system).

    E = h w (n + 0.5) where h is the reduced Planck constant, w is the same as in V(x), and n = 1, 2, 3, 4, ...

    |psi> = http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Hamiltonian_and_energy_eigenstates (look under topic 1.1: Hamiltonian and energy eigenstates)

    3. The attempt at a solution

    This was a question on my midterm exam, and I had absolutely no clue how to tackle it. I still don't. It won't be necessary to do the algebra, but can someone just sketch out the method of obtaining the solution for me?
     
  2. jcsd
  3. Nov 20, 2008 #2
    Somebody feel free to tell me I'm crazy...

    We know the momentum operator can be represented in terms of the creation/annihilation operators as
    [tex]\hat{p}=i \sqrt{\frac{\hbar m\omega}{2}} \left(\hat{a}^\dagger - \hat{a} \right)[/tex]

    so we can find the matrix representing [tex]\hat{p}[/tex] in the usual way of inner products. To wit:
    [tex]A_{m,n} = \left<m\right|i \sqrt{\frac{\hbar m\omega}{2}} \left(\hat{a}^\dagger - \hat{a} \right)\left|n\right> = i \sqrt{\frac{\hbar m\omega}{2}}\left[ \left<m\right|\hat{a}^\dagger\left|n\right> - \left<m\right|\hat{a}\left|n\right> \right][/tex].

    You can see from here that the diagonal elements are zero, since [tex]\hat{a}\left|n\right> \propto \left|n-1\right>[/tex] and [tex]\hat{a}^\dagger\left|n\right> \propto \left|n+1\right>[/tex]. In fact, you would expect to have two off-diagonal elements per stationary state [tex]\left|j\right>[/tex], one corresponding to [tex]\left|j-1\right>[/tex] and one corresponding to [tex]\left|j+1\right>[/tex]. I suppose this implies a measurement of [tex]\hat{p}[/tex] yields a linear combination of stationary states (but don't quote me on that one), which would make some sense if [tex]\left[\hat{H},\hat{p}\right] \neq 0[/tex] (don't remember if they do...).

    Tom
     
    Last edited: Nov 20, 2008
  4. Nov 20, 2008 #3

    Avodyne

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    Science Advisor

    Yep, that's how you do it.
     
  5. Nov 20, 2008 #4
    That is correct. H and p don't commute since the creation and annihilation operators don't commute.
     
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