Question

Show that the energy-time uncertainty principle reduces to the "your name" uncertainty principle (Problem 3.14), when the observable in question is x.

Short answer

The energy-time uncertainty principle reduces to${\sigma }_H^2{\sigma }_x^2=\frac{{\hslash }^2}{4m^2}⟨p{⟩}^2$

Step-by-step solution

The generalized uncertainty principle for two operators.

The generalized uncertainty principle for two operators is:

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

Show that the energy-time uncertainty principle reduces to the "your name" uncertainty principle.

From the energy-time uncertainty principle, an operator Q satisfies the equation:

for the position operator, $Q=x$, the time derivative of the position is zero, since $x$ and $t$ are independent variables,

$\frac{d}{dt}⟨x⟩=\frac{i}{\hslash }⟨[H,x]⟩$

Let $\hat{A}=\hat{H}$and $\hat{B}=\hat{x}$and use (2), to get:

$$\begin{gathered} {\sigma }_H^2{\sigma }_x^2⩾{\left(-\frac{\hslash }{2}\frac{d⟨x⟩}{dt}\right)}^2 \\ =\frac{{\hslash }^2}{4m^2}{\left(m\frac{d⟨x⟩}{dt}\right)}^2 \\ =\frac{{\hslash }^2}{4m^2}⟨p{⟩}^2 \\ {\sigma }_H^2{\sigma }_x^2=\frac{{\hslash }^2}{4m^2}⟨p{⟩}^2 \end{gathered}$$

Thus, the energy-time uncertainty principle reduces to${\sigma }_H^2{\sigma }_x^2=\frac{{\hslash }^2}{4m^2}⟨p{⟩}^2$.