Homework Help: Finding the momentum operator matrix of the harmonic oscillator

1. Nov 20, 2008

GuitarDean

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. Nov 20, 2008

tshafer

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
$$\hat{p}=i \sqrt{\frac{\hbar m\omega}{2}} \left(\hat{a}^\dagger - \hat{a} \right)$$

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

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

Tom

Last edited: Nov 20, 2008
3. Nov 20, 2008

Avodyne

Yep, that's how you do it.

4. Nov 20, 2008

naresh

That is correct. H and p don't commute since the creation and annihilation operators don't commute.