91影视

A particle of mass is represented by a wavefunction,

$\mathit{蠄}\left(x\right)\mathbf{=}\left\{\begin{array}{ll}2\sqrt{a^3}x{e}^{-ax}& x>0\\ 0& x<0\end{array}\right\}$

The expectation value of is defined as

The expression for momentum operator is

$\hat{p}=-i\mathrm{鈩�}\frac{鈭�}{鈭�x}$

The expectation value of momentum is

Solve further as:

The integral in the first term will be equal to one as $蠄\left(x\right)$is normalised. Solving integral in the second term separately as follows:

$\begin{array}{rcl}l& =& {鈭�}_0^{鈭�}\left(x{e}^{-2ax}\right)\\ & =& x\frac{e^{-2ax}}{-2a}-{鈭�}_{}^{}\left(\frac{dx}{dx}\right)\left({鈭�}_{}^{}{e}^{-2ax}\right)\\ & =& {\left.-\frac{x{e}^{-2ax}}{2a}-\frac{e^{-2ax}}{4a^2}\right|}_0^{鈭�}\\ & =& \frac{1}{4a^2}\end{array}$

Inserting this result in the above expression for momentum.

The expected value of momentum is $i\mathrm{鈩�}\left(\frac{1-a^2}{a}\right)$.