# Momentum of a state vector

1. Feb 3, 2012

### unchained1978

If we assume a system (pure for now) is in a state described by a single state vector, how can you determine the momentum? The momentum of a wavefunction is simply -i times the gradient, but that's for a continuous function. In the hilbert space representation of psi as a ket vector, what does the momentum operator look like?

2. Feb 3, 2012

### Ben Niehoff

In the position basis, it still looks like $-i \hbar \vec \nabla$. In the momentum basis, it's simply $p$. To obtain it in some other basis, you change basis in the standard way.

The best way to find the result of acting with $\hat p$ is to insert identity operators like so:

\begin{align} \hat{p} | \psi \rangle &= \int dp \; \hat{p} | p \rangle \langle p | \psi \rangle = \int dp \; p | p \rangle \langle p | \psi \rangle \\ &= \int dp \; p | p \rangle \int dx \; \langle p | x \rangle \langle x | \psi \rangle \\ &= \int dp \; | p \rangle \int dx \; p e^{-ipx} \psi(x) \\ &= \int dp \; | p \rangle \int dx \; i \frac{d}{dx} \big( e^{-ipx} \big) \psi(x) \\ &= \int dp \; | p \rangle \int dx \; e^{-ipx} \bigg( -i \frac{d}{dx} \psi(x) \bigg) \\ &= \int dx \; \int dp \; | p \rangle \langle p | x \rangle \bigg( -i \frac{d}{dx} \psi(x) \bigg) \\ &= \int dx \; | x \rangle \bigg( -i \frac{d}{dx} \psi(x) \bigg) \end{align}
Here I've done these steps:

1. Insert the resolution of the identity in the p basis.
2. Since we're in the p basis, the p operator can be replaced by its eigenvalue on each basis ket.
3. Insert the resolution of the identity in the x basis (since our goal is to get the p operator in the x basis).
4. Use the fact that $\langle p | x \rangle = e^{-ipx}$.
5. Do standard calculus tricks, integrate by parts.
6. Observe that we have an identity in the p basis we can remove.
7. Finally, we see the action of the p operator in the x basis.