# Expectation of Momentum Operator

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Science Advisor
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I need help getting started on this problem:

A free particle moving in one dimension is in the initial state $\Psi(x,0)$. Prove that $<\hat{p}>$ is constant in time by direct calculation (i.e. without recourse to the commutator theorem regarding constants of the motion).

Our professor strongly advised against doing it this way:

$$<\hat{p}> = \int{\Psi^* \hat{p} \Psi \, dx}$$

instead saying that we should try this:

$$<\hat{p}> = \int{p |\phi(k,t)|^2 \, dk}$$

Since we know that:

$$\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\phi(k)e^{ikx} \, dk}$$

and therefore

$$\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\Psi(x,0)e^{-ikx} \, dx}$$

Yay Fourier transforms :yuck:

My questions are:

1. Practically speaking, how does one go about expressing phi(k) using this method?

2. Even if I get a reasonable expression for phi(k), what about the time dependence? How do I then get phi(k,t)? I'm going to need that in order to find <p> and show that d<p>/dt = 0.

Last edited:

## Answers and Replies

dextercioby
Science Advisor
Homework Helper
Since the particle is free, then the whole time dependence of the state vector (it's actually a functional) can be factorized and, by virtue of being expressed in terms of a complex exponential, vanishes when you consider this baby

$$\langle \Psi (t) |\hat{p}| \Psi (t) \rangle$$

Daniel.