# 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.