Question

In an interesting version of the energy-time uncertainty principle31, $\mathbf{∆}\mathit{t}\mathbf{=}\mathit{\tau }\mathbf{/}\mathit{\pi }$where $\mathit{\tau }$is the time it takes$\mathit{\Psi }\left(x,t\right)$to evolve into a state orthogonal to$\mathit{\Psi }\left(x,0\right)$ . Test this out, using a wave function that is an equal admixture of two (orthonormal) stationary states of some (arbitrary) potential:$\mathit{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\left(1/\sqrt{2}\right)\left[{\Psi }_1\left(x\right)+{\Psi }_2\left(x\right)\right]$

Short answer

This is consistent with the original time-energy uncertainty principle:

$∆E.∆t=\frac{\sout{h}}{2}$.

Step-by-step solution

Wave function

In the energy-time uncertainty principle $\mathbf{∆}\mathit{t}\mathbf{=}\mathit{\tau }\mathbf{/}\mathit{\pi }$, where$\mathit{\tau }$is the time taken by $\mathit{\Psi }\left(x,t\right)$to grow into a state orthogonal to $\mathit{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}$, testing this is required. Start with a wave function that is a combination of two stationary states, that is:

$\mathit{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{(}{\mathbf{\psi }}_{\mathbf{1}}\left(x\right)\mathbf{+}{\mathbf{\psi }}_{\mathbf{2}}\left(x\right)\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}........\left(1\right)$

The wave function at the later time is, therefore:
$\Psi \left(x,\tau \right)=\frac{1}{\sqrt{2}}\left({\psi }_1\left(x\right)e^{-i{E}_1\tau /h}+{\psi }_2\left(x\right)e^{-i{E}_2\tau /h}\right)$ ........(2)

Multiply each term by the appropriate wiggle factor to get the position-space wave function at any future time t.

Substitute values from (1) & (2)

For the two functions in (1) and (2) to be orthogonal,

Substitute from (1) and (2) to get :

but:

The exponential function is never zero, thus:

$$\begin{gathered} \frac{1}{2}\left(e^{i{E}_{1}t/\sout{h}}+e^{i{E}_{1}t/\sout{h}}\right)=0.e^{i{E}_{2}t/\sout{h}} \\ =-e^{i{E}_{1}t/\sout{h}}{e}^i\left(E_2-E_1\right)t/\sout{h} \\ =-1 \end{gathered}$$

$$\begin{gathered} but{e}^{i\pi }+1=0,so: \\ {e}^{i\left(E_2-E_1\right)t/\sout{h}}=-1 \\ {e}^{i\left(E_2-E_1\right)t/\sout{h}}=e^{i\pi } \end{gathered}$$

Thus by the definition of $∆t$:

$\frac{\left(E_2-E_1\right)}{\sout{h}}=\pi$

Taking $\tau /\pi =∆t$and $E_2-E_1=∆E$giving the standard deviation in energy as:

$∆E∆t=\sout{h}>\frac{\sout{h}}{2}$

This confirms that the energy-time uncertainty principle is obeyed. This is consistent with the original time-energy uncertainty principle.