91Ó°ÊÓ

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$\mathrm{ψ}\left(x\right)$.

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

To find the uncertainty in the particle’s position.

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

$\mathbf{Δ³\ae }\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{\left(}\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{\right)}}^{\mathbf{2}}}$

The expectation value of the position.

$\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{∞}}{\mathbf{∫}}}{\mathbf{³\ae ψ}}^{\mathbf{2}}\mathbf{dx}$

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

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

Use the property ${∫}_0^{∞}{\mathrm{x}}^3{\mathrm{e}}^{-2\mathrm{ax}}\mathrm{dx}=\frac{\mathrm{n}!}{{\mathrm{a}}^{\mathrm{n}+1}}$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.

$\stackrel{\mathbf{¯}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{∞}}{\mathbf{x}}^{\mathbf{2}}{\mathbf{ψ}}^{\mathbf{2}}\mathbf{dx}$

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

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

Use the property ${∫}_0^{∞}{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{ax}}\mathrm{dx}=\frac{\mathrm{n}!}{{\mathrm{a}}^{\mathrm{n}+1}}$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.

$\mathit{Δ}\mathit{x}\mathbf{=}\sqrt{\stackrel{\mathbf{¯}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}{\stackrel{\mathbf{¯}}{\mathbf{x}}}^{\mathbf{2}}}$

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

$\begin{array}{rcl}\mathbf{Δ³\ae }& \mathbf{=}& \sqrt{\frac{\mathbf{3}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}{\left(\frac{3}{2a}\right)}^{\mathbf{2}}}\\ & \mathbf{=}& \frac{\sqrt{\mathbf{0}\mathbf{.}\mathbf{75}}}{\mathbf{a}}\\ & \mathbf{=}& \frac{\mathbf{0}\mathbf{.}\mathbf{866}}{\mathbf{a}}\end{array}$

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