Question

To determine the classical expectation value of the position of a particle in a box is $\frac{\mathbf{L}}{\mathbf{2}}$ , the expectation value of the square of the position of a particle in a box isrole="math" localid="1658324625272" $\frac{{\mathbf{L}}^{\mathbf{2}}}{\mathbf{3}}$ , and the uncertainty in the position of a particle in a box is$\frac{\mathbf{L}}{\sqrt{\mathbf{12}}}$ .

Short answer

The uncertainty in the position of a particle in a box of length$L$is$\Delta x=\sqrt{\frac{-}{x^2-{\overline{x}}^2}}=\sqrt{\frac{L^2}{3}-\frac{L^2}{4}}=\frac{L}{\sqrt{12}}$.

Step-by-step solution

Step 1:

Given information: The classical probability per unit length of finding a particle in a box of length$L$ is$\frac{1}{L}$ along the entire length of the box i.e.$\frac{\mathrm{dP}}{\mathrm{dx}}=\frac{1}{\mathrm{L}}$ .

Step 2:

The classical expectation value of the position of a particle in a box of length $L$is

$\overline{\mathrm{x}}={\int }_0^{\mathrm{L}}\mathrm{x}\mathrm{dP}={\int }_0^{\mathrm{L}}\mathrm{x}\frac{\mathrm{dP}}{\mathrm{dx}}\mathrm{dx}=\frac{1}{\mathrm{L}}{\int }_0^{\mathrm{L}}\mathrm{x}\mathrm{dx}=\frac{1}{\mathrm{L}}\times \frac{{\mathrm{L}}^2}{2}=\frac{\mathrm{L}}{2}.$

The expectation value of the square of the position of a particle in a box of length role="math" localid="1658325211080" $L$is

$\overline{{\mathrm{x}}^2}={\int }_0^{\mathrm{L}}{\mathrm{x}}^2\mathrm{dP}={\int }_0^{\mathrm{L}}{\mathrm{x}}^2\frac{\mathrm{dP}}{\mathrm{dx}}\mathrm{dx}=\frac{1}{\mathrm{L}}{\int }_0^{\mathrm{L}}{\mathrm{x}}^2\mathrm{dx}=\frac{1}{\mathrm{L}}\times \frac{{\mathrm{L}}^3}{3}=\frac{{\mathrm{L}}^2}{3}$

The uncertainty in the position of a particle in a box of length $L$is

$\Delta x=\sqrt{\frac{-}{x^2-{\overline{x}}^2}}=\sqrt{\frac{L^2}{3}-\frac{L^2}{4}}=\frac{L}{\sqrt{12}}$