Calculate the uncertainty in the particle’s position.

Heisenberg's uncertainty principle states that it is impossible to measure or calculate exactly, both the position and the momentum of an object. This principle is based on the wave-particle duality of matter.

The wave function of the particle is$\psi \left(x\right)$.

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

To find the uncertainty in the particle’s position.

In order to calculate the uncertainty of the given particle's position, we first calculate$\overline{x}$and${\overline{x}}^2$.

$\mathbf{\Delta x}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}}^{\mathbf{2}}}$

The expectation value of the position.

$\overline{\mathbf{x}}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathbf{x\psi }}^{\mathbf{2}}\mathbf{dx}$

Substitute for$2\sqrt{{\mathrm{a}}^3}{\mathrm{xe}}^{-\mathrm{ax}}$for$\psi$

$\begin{array}{rcl}\mathbf{x}& \mathbf{=}& {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx}\\ & \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{x}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx}\end{array}$

Use the property ${\int }_0^{\infty }{\mathrm{x}}^3{\mathrm{e}}^{-2\mathrm{ax}}\mathrm{dx}=\frac{\mathrm{n}!}{{\mathrm{a}}^{\mathrm{n}+1}}$to solve above integral.

$\begin{array}{rcl}\mathbf{x}& \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}\frac{\mathbf{3}\mathbf{!}}{{\mathbf{\left(}\mathbf{2}\mathbf{a}\mathbf{\right)}}^{\mathbf{4}}}\\ & \mathbf{=}& \frac{\mathbf{3}}{\mathbf{2}\mathbf{a}}\end{array}$

The expectation value of the square of position.

$\overline{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{x}}^{\mathbf{2}}{\mathbf{\psi }}^{\mathbf{2}}\mathbf{dx}$

Substitute$2\sqrt{{\mathrm{a}}^3}{\mathrm{xe}}^{-\mathrm{ax}}$for$\psi$.

$$\begin{gathered} {\mathbf{x}}^{\mathbf{2}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{4}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx} \\ \mathbf{=}\mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{x}}^{\mathbf{4}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx} \end{gathered}$$

Use the property ${\int }_0^{\infty }{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{ax}}\mathrm{dx}=\frac{\mathrm{n}!}{{\mathrm{a}}^{\mathrm{n}+1}}$to solve above integral.

$\begin{array}{rcl}{\mathbf{x}}^{\mathbf{2}}& \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}\frac{\mathbf{4}\mathbf{!}}{{\mathbf{\left(}\mathbf{2}\mathbf{a}\mathbf{\right)}}^{\mathbf{5}}}\\ & \mathbf{=}& \frac{\mathbf{3}}{{\mathbf{a}}^{\mathbf{2}}}\end{array}$

The uncertainty in the position.

$\mathit{\Delta }\mathit{x}\mathbf{=}\sqrt{\overline{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}{\overline{\mathbf{x}}}^{\mathbf{2}}}$

Substitute $\frac{3}{a^2}$for $\overline{{\mathrm{x}}^2}$amd $\frac{3}{2a}$for$\overline{x}$

$\begin{array}{rcl}\mathbf{\Delta x}& \mathbf{=}& \sqrt{\frac{\mathbf{3}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}{\left(\frac{3}{2a}\right)}^{\mathbf{2}}}\\ & \mathbf{=}& \frac{\sqrt{\mathbf{0}\mathbf{.}\mathbf{75}}}{\mathbf{a}}\\ & \mathbf{=}& \frac{\mathbf{0}\mathbf{.}\mathbf{866}}{\mathbf{a}}\end{array}$

Hence, the uncertainty in the particle position is $\frac{0.866}{a}$.