Expectation Value of Momentum in a Given State

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

We will use the momentum operator to get this result.

\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 \\

The second term gives zero because the integral is odd about \bgroup\color{black}$x=0$\egroup.

\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 \\


Jim Branson 2013-04-22