Question

Calculate the uncertainty in the particle’s position.

Short answer

The uncertainty in the particle position is $\frac{0.866}{\mathrm{a}}$.

Step-by-step solution

Step 1:Understandingthe concept of uncertainty.

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.

Given information.

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.

Use formula to calculate the expectation values of x¯and x2¯.

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.

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

Use formula to calculate the uncertainty of the position of the particle.

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 $\frac{0.866}{a}$.