Question

Prove the famous "(your name) uncertainty principle," relating the uncertainty in position$\mathbf{A}\mathbf{=}\mathbf{x}$ to the uncertainty in energy$\mathbf{B}\mathbf{=}{\mathbf{p}}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{v}$:

For stationary states this doesn't tell you much-why not?

Short answer

The uncertainty principle is

Step-by-step solution

Concept used

The generalized uncertainty principle for two observables A and B is given by:

Calculate the uncertainty principle

The generalized uncertainty principle for two observables A and B is given by:

The position-energy uncertainty relation is:

....(1)

So, we need to find the commutator$\left[\hat{\mathrm{x}},\hat{\mathrm{H}}\right]$as:

$$\begin{gathered} \left[\hat{\mathrm{X}},\hat{\mathrm{H}}\right]\mathrm{g}=-\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}\times \frac{{\partial }^2\mathrm{g}}{\partial {\mathrm{X}}^2}+\mathrm{xVg}+-\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}\frac{{\partial }^2}{\partial {\mathrm{x}}^2}\left(\mathrm{xg}\right)-\mathrm{x}\left(\mathrm{Vg}\right) \\ =\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}\left(-\mathrm{x}\frac{{\partial }^2\mathrm{g}}{\partial {\mathrm{X}}^2}+2\frac{\partial \mathrm{g}}{\partial \mathrm{x}}+\mathrm{x}\frac{{\partial }^2\mathrm{g}}{\partial {\mathrm{X}}^2}\right) \\ =\frac{{\sout{\mathrm{h}}}^2}{\mathrm{m}}\frac{\partial \mathrm{g}}{\partial \mathrm{x}} \\ =\frac{\mathrm{i}\sout{\mathrm{h}}}{\mathrm{m}}\left(\mathrm{pg}\right) \end{gathered}$$

Substitute in equation 1:

So, the uncertainty principle here becomes

For stationary states, this doesn't tell you much because the average position of the particle doesn't change,