Question

In Exercise 45, the case is made that the position uncertainty for a typical macroscopic object is generally so much smaller than its actual physical dimensions that applying the uncertainty principle would be absurd. Here we gain same idea of how small an object would have♦ to be before quantum mechanics might rear its head. The density of aluminum is $\mathbf{2}\mathbf{.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{\text{kg}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{3}}$, is typical of solids and liquids around us. Suppose we could narrow down the velocity of an aluminum sphere to within an uncertainty of$\mathbf{1}\mathbf{\mu }\mathbf{\text{m}}$per decade. How small would it have to be for its position uncertainty to be at least as large as$\frac{\mathbf{1}}{\mathbf{10}}\mathit{\%}$of its radius?

Short answer

The radius of the microscopic object to obtain the required uncertainty in position is $r\approx {10}^{-5}\mathbf{m}$.

Step-by-step solution

Concept of Heisenberg's Uncertainty Principle. 

The expression for density is given by,

$\begin{array}{l}\mathbf{D}\mathbf{=}\frac{\mathbf{m}}{\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi R}}^{\mathbf{3}}}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi R}}^{\mathbf{3}}\mathbf{D}\end{array}$

Here, mis mass, Dis the density and Ris the radius

The expression for Heisenberg’s uncertainty principle in terms of mass and velocity is given by,

$\begin{array}{c}{\mathbf{m\Delta v}}_{\mathbf{\text{x}}}\mathbf{\Delta x}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}}\\ \left(\frac{4}{3}{\mathrm{\pi R}}^{3}D\right){\mathbf{\Delta v}}_{\mathbf{\text{x}}}\mathbf{\Delta x}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}}\\ \mathbf{8}{\mathbf{\pi DR}}^{\mathbf{3}}{\mathbf{\Delta v}}_{\mathbf{\text{x}}}\mathbf{\Delta x}\mathbf{\ge }\mathbf{3}\mathbf{h}\\ \mathbf{\Delta x}\mathbf{\ge }\frac{\mathbf{3}\mathbf{h}}{\mathbf{8}{\mathbf{\pi DR}}^{\mathbf{3}}{\mathbf{\Delta v}}_{\mathbf{\text{x}}}}\end{array}$

The expression for uncertainty in position as a fraction of radius is given by,

$\begin{array}{c}\frac{\mathbf{\Delta x}}{\mathbf{R}}\mathbf{\ge }\mathbf{k}\\ \frac{\left(\frac{3\delta }{8{\mathrm{eDR}}^3{\mathrm{\Delta v}}_{\text{x}}}\right)}{\mathbf{R}}\mathbf{\ge }\mathbf{k}\\ \frac{\mathbf{3}\mathbf{h}}{\mathbf{8}{\mathbf{\pi DR}}^{\mathbf{4}}{\mathbf{\Delta v}}_{\mathbf{\text{x}}}}\mathbf{\ge }\mathbf{k}\\ \mathbf{R}\mathbf{\le }\sqrt{\frac{\mathbf{3}\mathbf{h}}{\mathbf{8}{\mathbf{\pi kD\Delta v}}_{\mathbf{\text{x}}}}}\end{array}$

Here, ${\mathbf{\Delta v}}_{\mathbf{\text{x}}}$the change in velocity, Dis the density, h is Planck’s constant, $\mathbf{\Delta x}$

is the uncertainty in position and Ris the radius.

Use Heisenberg’s Uncertainty Principle for calculation. 

Consider volume for the object is$\left(V=\frac{4}{3}\pi r^5\right)$and the uncertainty in velocity, consequently the momentum, is known.

Substitute given values we can get an estimate of the value of$r$(note$\Delta x=0.10\%$of $r$).

$\begin{array}{l}\mathrm{\Delta x\Delta p}\ge \frac{\mathrm{\hslash }}{2}\\ \mathrm{\Delta xm\Delta v}\ge \frac{\mathrm{\hslash }}{2}\end{array}$

Here,$\Delta x$is the uncertainty in position, $\mathrm{\Delta p}$is uncertainty in momentum, m is the mass and $\Delta v$is velocity change and $\hslash$is Planck’s constant.

$\begin{array}{c}\Delta x=0.001r,\\ m=\rho V\\ m=\rho \times \frac{4}{3}\pi r^3\end{array}$

Now substitute the value in formula

$\begin{array}{c}\mathrm{\Delta xm\Delta v}\ge \frac{\mathrm{\hslash }}{2}\\ 0.001\mathrm{r}\times \mathrm{\rho }\times \frac{4}{3}{\mathrm{\pi r}}^3\mathrm{\Delta v}\ge \frac{\mathrm{\hslash }}{2}\end{array}$

Solve further as:

$\begin{array}{l}\mathrm{r}\approx \sqrt[4]{\frac{\mathrm{\hslash }}{2\times 0.001\times \mathrm{\rho }\times \frac{4}{3}\mathrm{\pi }\times \mathrm{\Delta v}}}\\ \mathrm{r}=\sqrt[4]{\frac{1.054\times {10}^{-34}\frac{\text{kg}\cdot {\text{m}}^2}{\text{s}}}{2\times 0.001\times 2.7\times {10}^3\text{kg}/{\text{m}}^3\times \frac{4}{3}\mathrm{\pi }\times \frac{{10}^{-6}\text{m}}{10\times 365\times 24\times 60\times 60\text{s}}}}\\ \mathrm{r}=\sqrt[4]{1.469\times {10}^{-20}}\text{m}\\ \mathrm{r}\approx {10}^{-5}\mathbf{m}\end{array}$

Thus, the radius of the microscopic object to obtain the required uncertainty in position is$\mathrm{r}\approx {10}^{-5}\mathbf{m}$ .