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Background

It is known:

In all physical systems in which energy is bounded below, there is no self-adjoint observable that tracks the time parameter t.

However I don't think this forbids any inequality on how much time has passed can be inferred between the time evolution of an eigenstate (position or momentum) $| x \rangle $ and $U(\Delta t)| x \rangle$ where $| x \rangle $ is the position eigenket and $U$ is a unitary operator for a generic Hamiltonian. In light of that I constructed the following inequality.

Question

I can show for a position eigenstate $| x \rangle $ if it evolves in time $U(\Delta t) | x\rangle$ (where $U$ is a unitary operator for a generic Hamiltonian). Then one can bound the time elapsed by finding the probability amplitude $| \langle x | U^\dagger(\Delta t)| x + \Delta x \rangle |^2 $ and $| \langle x | U^\dagger(\Delta t)| x \rangle |^2 $ (where $\Delta x$ is the translation in space by an amount $\Delta x > 0$) and knows in advance the rate of change of momentum $|\langle x | \dot p | x \rangle |$ then one can bound $\Delta t$

\begin{align} 2 \hbar \frac{ | \langle x | U^\dagger(\Delta t)| x + \Delta x \rangle | + | \langle x | U^\dagger(\Delta t) | x \rangle| }{\Delta x|\langle x | \dot p | x \rangle | } \geq \Delta t \tag{1} \end{align} Similarly for a momentum eigenstate $|p \rangle $ and the rate of change of position $|\langle p | \dot x | p \rangle |$:

\begin{align} 2 \hbar \frac{ | \langle p | U^\dagger(\Delta t) |p + \Delta p \rangle | + | \langle p | U^\dagger(\Delta t) | p \rangle| }{\Delta p|\langle p | \dot x | p \rangle | } \geq \Delta t \tag{2} \end{align}

Is there any realistic experiment one can do to realise this quantum mechanical watch-stop?

Proof

Consider the following limit:

\begin{align} \lim_{\delta x \to 0} \frac{| x + \delta x \rangle - | x \rangle}{\delta x} = \frac{i}{\hbar}\hat p |x \rangle \tag{3} \end{align}

Similarly:

\begin{align} \lim_{\delta t \to 0} \frac{U(\delta t)| x \rangle - | x \rangle}{\delta t} = \frac{-i}{\hbar}\hat H |x \rangle \tag{4} \end{align}

Hence, we multiply the first two equations (adjoint of one times the other) and substract:

\begin{align} & \lim_{\delta x,\delta t \to 0 }\frac{\langle x |U^\dagger (\delta t)| x + \delta x \rangle - \langle x + \delta x| U(\delta t)| x \rangle-\langle x |U^\dagger (\delta t)| x \rangle + \langle x | U (\delta t) |x \rangle}{\delta x \delta t} \\ &= - \frac{1}{\hbar^2} \langle x | [ \hat H,\hat p] | x \rangle \tag{5} \end{align}

Using Heisenberg's equation of motion:

\begin{align} \lim_{\delta x,\delta t \to 0 }\frac{\langle x | U^\dagger (\delta t)| x + \delta x \rangle - \langle x + \delta x| U (\delta t)|x \rangle}{\delta x \delta t} - \frac{\langle x | U^\dagger (\delta t) | x \rangle - \langle x |U (\delta t)|x \rangle}{\delta x \delta t} = \frac{i}{\hbar}\langle x | \dot p | x \rangle\tag{6}\end{align}

Replacing $\delta x$ and $\delta t$ with finite but small values:

\begin{align} \frac{\langle x |U^\dagger (\Delta t)| x + \Delta x \rangle - \langle x + \Delta x|U (\Delta t)| x \rangle}{\Delta x \Delta t} - \frac{\langle x |U^\dagger (\Delta t)| x \rangle - \langle x | U (\Delta t)| x \rangle}{\Delta x \Delta t} \approx \frac{i}{\hbar}\langle x | \dot p | x \rangle\tag{7}\end{align}

Taking the modulus and using the triangle inequality:

\begin{align} \left | \frac{\langle x |U^\dagger (\Delta t)| x + \Delta x \rangle - \langle x + \Delta x|U (\Delta t)| x \rangle}{\Delta x \Delta t} \right | + \left | \frac{\langle x |U^\dagger (\Delta t)| x \rangle - \langle x | U (\Delta t)| x \rangle}{\Delta x \Delta t} \right | \geq \frac{i}{\hbar} | \langle x | \dot p | x \rangle | \tag{8} \end{align}

Let us consider square of the first term:

\begin{align} T_1^2 = \left | \frac{\langle x |U^\dagger (\Delta t)| x + \Delta x \rangle - \langle x + \Delta x| U (\Delta t)|x \rangle}{\Delta x^2 \Delta^2 t} \right |^2 = \frac{2 | \langle x | U^\dagger (\Delta t) | x + \Delta x \rangle |^2 - \langle x |U^\dagger (\Delta t)| x + \Delta x \rangle^2 - \langle x + \Delta x| U (\Delta t) |x \rangle^2}{\Delta x^2 \Delta t^2} \tag{9} \end{align}

We use the following inequality which uses $| \langle x |U^\dagger (\Delta t)| x + \Delta x \rangle| = |z|$, $ \langle x |U^\dagger (\Delta t)| x + \Delta x \rangle = z$ and $ \langle x + \Delta x |U(\Delta t)| x \rangle = z^*$ with $z = a+ib$:

\begin{align} \frac{4 | \langle x| x + \Delta x \rangle |^2 }{\Delta x^2 \Delta t^2} \geq T_1^2 \tag{10} \end{align}

Since all quantities are positive:

$$ \frac{2 | \langle x| U^\dagger (\Delta t) | x + \Delta x \rangle | }{\Delta x \Delta t} \geq T_1 \tag{11} $$

Similarly, we define:

\begin{align} T_2 = \left | \frac{\langle x | U^\dagger (\Delta t) | x \rangle - \langle x | U (\Delta t)|x \rangle }{\Delta x \Delta t} \right | \tag{12} \end{align}

Then,

\begin{align} \frac{ 2 | \langle x | U^\dagger (\Delta t) | x \rangle| }{\Delta x \Delta t} \geq T_2 \tag{13}\end{align}

Hence, substituting the $T_1$ and $T_2$ inequality:

\begin{align} \frac{2 | \langle x | U^\dagger (\Delta t) | x + \Delta x \rangle | }{\Delta x \Delta t} + \frac{ 2 | \langle x | U^\dagger (\Delta t)| x \rangle| }{\Delta x \Delta t} \geq \frac{1}{\hbar} |\langle x | \dot p | x \rangle | \tag{14} \end{align}

Or in terms of $\Delta t$:

\begin{align} 2 \hbar \frac{ | \langle x | U^\dagger (\Delta t)| x + \Delta x \rangle | + | \langle x | U^\dagger (\Delta t) | x \rangle| }{\Delta x|\langle x | \dot p | x \rangle | } \geq \Delta t \tag{15} \end{align}

Similarly for momentum eigenkets:

\begin{align} 2 \hbar \frac{ | \langle p | U^\dagger(\Delta t) |p + \Delta p \rangle | + | \langle p | U^\dagger(\Delta t) | p \rangle| }{\Delta p|\langle p | \dot x | p \rangle | } \geq \Delta t \tag{16} \end{align}

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  • $\begingroup$ I have begun making edits to improve the post, and I suggest you continue finishing that work for the rest of the question. $\endgroup$ – user1271772 Dec 28 '20 at 1:44
  • $\begingroup$ @user1271772 I've completed labelling the equations :) $\endgroup$ – More Anonymous Dec 28 '20 at 4:42
  • $\begingroup$ Nicely done, the other edit I made was changed \Big to \left and \right. See the equation between 8 and 9, versus Eq. 11 for example. Also it's good to label every line, so that I don't have to say "the equation between 8 and 9", this also goes for the equation between 10 and 11, and the second line of Eq. 7. Finally, another edit I had made was that I changed a $$ with \begin{align} which allowed med to line-up the equals signs. This would be useful for Eq. 9 and Eq. 9,5, for example. $\endgroup$ – user1271772 Dec 28 '20 at 4:45

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