Calculate the uncertainty in the particle’s momentum.

To calculate the uncertainty in the momentum,${\sigma }_p$one must first calculate the expectation value of the momentum and the expectation value of the square of the momentum. The momentum's expectation value is twave function's integral squared${\psi }^{*}\psi$multiplied by the momentum operator$\hat{p}$.

The wave function of the particle is$2\mathrm{\psi }$.

role="math" localid="1656090791837" $\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{2}\sqrt{{\mathbf{a}}^{\mathbf{3}}{\mathbf{xe}}^{\mathbf{-}\mathbf{ax}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{>}\mathbf{0}\\ \mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{<}\mathbf{0}\end{array}$

To find the expectation value of particle’s momentum.

The momentum's expectation value is twave function's integral squared${\psi }^{*}\psi$multiplied by the momentum operator$\hat{p}$.

$\begin{array}{rcl}\mathbf{⟨}\mathbf{p}\mathbf{⟩}& \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\mathbf{\left(}\hat{\mathbf{p}}\mathbf{\right)}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\mathbf{(}\mathbf{-}\mathbf{i}\mathbf{\hslash }\mathbf{)}\frac{\mathbf{d}}{\mathbf{dx}}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}\mathbf{i}\mathbf{\hslash }{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\left(e^{-\mathrm{ax}}-{\mathrm{axe}}^{-\mathrm{ax}}\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}\mathbf{i}\mathbf{\hslash }{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-2\mathrm{ax}}\right)\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{ax}\mathbf{)}\mathbf{dx}\end{array}$

Use integration by parts to evaluate the integral.

$\begin{array}{rcl}\mathbf{⟨}\mathbf{p}\mathbf{⟩}& \mathbf{=}& \mathbf{-}\mathbf{4}\mathbf{i}\mathbf{\hslash }{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left(e^{-2\mathrm{ax}}\right)\left(x-{\mathrm{ax}}^2\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}\mathbf{i}\mathbf{\hslash }{\mathbf{a}}^{\mathbf{3}}{\left[\left(e^{-2\mathrm{ax}}\right)\left(-\frac{x-{\mathrm{ax}}^2}{2a}-\frac{1-2\mathrm{ax}}{4a^2}+\frac{2a}{8a^3}\right)\right]}_{\mathbf{0}}^{\mathbf{\infty }}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}\mathbf{i}\mathbf{\hslash }{\mathbf{a}}^{\mathbf{3}}{\left[\left(e^{-2\mathrm{ax}}\right)\left(-\frac{4a^{2}x}{8a^3}+\frac{4a^3{x}^2}{8a^3}-\frac{2a}{8a^2}+\frac{4a^{3}x}{8a^3}+\frac{2a}{8a^3}\right)\right]}_{\mathbf{0}}^{\mathbf{\infty }}\\ & \mathbf{=}& \mathbf{0}\\ & & \end{array}$

Now calculate the expectation value of the square of the momentum${}^2>$

$\begin{array}{rcl}<p^2>& \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\left({\hat{p}}^2\right)\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\left(-{\hslash }^2\right)\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dx}}^{\mathbf{2}}}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-\mathrm{ax}}\right)\frac{\mathbf{d}}{\mathbf{dx}}\left(e^{-\mathrm{ax}}-{\mathrm{axe}}^{-\mathrm{ax}}\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left({\mathrm{xe}}^{-2\mathrm{ax}}\right)\left(a^{2}x-2a\right)\mathbf{dx}\end{array}$

Use integration by parts to evaluate the integral.

$\begin{array}{rcl}<p^2>& \mathbf{=}& \mathbf{-}\mathbf{4}{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\left(e^{-2\mathrm{ax}}\right)\left(a^2{x}^2-2\mathrm{ax}\right)\mathbf{dx}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{3}}{\left[e^{-2\mathrm{ax}}\left(-\frac{2a^{2}x-2a}{2a}-\frac{2a^2}{4a^2}\right)\right]}_{\mathbf{0}}^{\mathbf{\infty }}\\ & \mathbf{=}& \mathbf{-}\mathbf{4}{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{3}}\left(\frac{-1}{4a}\right)\\ & \mathbf{=}& {\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}\end{array}$

The uncertainty in the momentum ${\sigma }_p$is the square root of the difference between tile expectation value of the square of the momentum and the expectation value of the momentum squared, $⟨p{⟩}^2$.

$\begin{array}{rcl}{\mathbf{\sigma }}_{\mathbf{p}}& \mathbf{=}& \sqrt{{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{0}}^{\mathbf{2}}}\\ & \mathbf{=}& \mathbf{\hslash }\mathbf{a}\end{array}$

The uncertainty in the particle's momentum is ${\mathrm{\sigma }}_{\mathrm{p}}=\mathrm{ha}$.