Question

Prove the famous "(your name) uncertainty principle," relating the uncertainty in position $\mathbf{(}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathit{x}\mathbf{)}$to the uncertainty in energy$\left(B=p^2/2m+V\right)$: ${\mathit{\sigma }}_{\mathbf{x}}{\mathit{\sigma }}_{\mathbf{H}}\mathbf{⩾}\frac{\mathbf{\hslash }}{\mathbf{2}\mathbf{m}}\mathbf{|}\mathbf{⟨}\mathit{p}\mathbf{⟩}\mathbf{|}$

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

Short answer

The uncertainty principle is${\sigma }_x{\sigma }_H⩾\frac{\hslash }{2m}|⟨p⟩|$

Step-by-step solution

Concept used

The generalized uncertainty principle for two observables $\mathit{A}$and $\mathit{B}$is given by:

${\mathit{\sigma }}_{\mathbf{A}}^{\mathbf{2}}{\mathit{\sigma }}_{\mathbf{B}}^{\mathbf{2}}\mathbf{⩾}{\left(\frac{1}{2i}⟨[\hat{A},\hat{B}]⟩\right)}^{\mathbf{2}}$

Calculate the uncertainty principle

The generalized uncertainty principle for two observables $\mathrm{A}$and $B$ is given by:

${\sigma }_A^2{\sigma }_B^2⩾{\left(\frac{1}{2i}⟨[\hat{A},\hat{B}]⟩\right)}^2$

The position-energy uncertainty relation is:

${\sigma }_x^2{\sigma }_H^2⩾{\left(\frac{1}{2i}⟨[\hat{x},\hat{H}]⟩\right)}^2$ ...... (1)

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

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

Substitute in equation 1:

$$\begin{gathered} {\sigma }_x^2{\sigma }_H^2{\left.⩾(\frac{1}{2i}⟨\hat{x},\hat{H}]⟩\right)}^2 \\ =\frac{{\hslash }^2}{4m^2}⟨p{⟩}^2 \end{gathered}$$

So, the uncertainty principle here becomes

${\sigma }_x{\sigma }_H⩾\frac{\hslash }{2m}|⟨p⟩|$

For stationary states, this doesn't tell you much because the average position of the particle doesn't change,${\sigma }_H=0$and$⟨p⟩=0$