Expectation Value of Momentum in a Given State

A particle is in the state $\psi(x)=\left({1\over 2\pi\alpha}\right)^{1/4} e^{ik_0x} e^{-{x^2\over 4\alpha}}$. What is the expectation value of $p$?

We will use the momentum operator to get this result.

$$\begin{eqnarray*} \langle p\rangle_\psi &=&\langle\psi\vert p\vert\psi\rangle= \... ...2\alpha}}-{2x\over 4\alpha} e^{-{x^2\over 2\alpha}}\right)dx \\ \end{eqnarray*}$$

The second term gives zero because the integral is odd about $x=0$.

$$\begin{eqnarray*} \langle\psi\vert p\vert\psi\rangle &=& \left({1\over 2\pi\alph... ...pi\alpha}\right)^{1/2}{\hbar k_0}\sqrt{2\pi\alpha}=\hbar k_0 \\ \end{eqnarray*}$$

Excellent.