I Realizing a Quantum Mechanical Watch-Stop

[Someone_physics]

Question
---
I can show for a position eigenstate $| x \rangle$ if it evolves in time $U(\Delta t) | x\rangle$ (where $U$ is the unitary operator). 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 $\Delta x$) and knows in advance the rate of change of momentum $|\langle x | \dot p | x \rangle |$ then one can bound $\Delta t$

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

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

Is the proof correct? Is there any realistic experiment one can do to realize this quantum mechanical watch-stop?

Proof
---

Consider the following limit:

$$ \lim_{\delta x \to 0} \frac{| x + \delta x \rangle - | x \rangle}{\delta x} = \frac{i}{\hbar}\hat p |x \rangle$$

Similarly:

$$ \lim_{\delta t \to 0} \frac{U(\delta t)| x \rangle - | x \rangle}{\delta t} = \frac{-i}{\hbar}\hat H |x \rangle$$

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

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

Using Heisenberg's equation of motion:

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

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

$$ \frac{\langle x |U^\dagger (\Delta t)| x + \Delta x \rangle - \langle x + \Delta x|U (\Delta t)| x \rangle}{\Delta x \Delta t}
- \frac{\langle x |U^\dagger (\Delta t)| x \rangle - \langle x | U (\Delta t)| x \rangle}{\Delta x \Delta t} \approx \frac{i}{\hbar}\langle x | \dot p | x \rangle$$

Taking the modulus and using the triangle inequality:

$$ \Big | \frac{\langle x |U^\dagger (\Delta t)| x + \Delta x \rangle - \langle x + \Delta x|U (\Delta t)| x \rangle}{\Delta x \Delta t}
\Big | + \Big | \frac{\langle x |U^\dagger (\Delta t)| x \rangle - \langle x | U (\Delta t)| x \rangle}{\Delta x \Delta t} \Big | \geq \frac{i}{\hbar} | \langle x | \dot p | x \rangle |$$

Let us consider square of the first term:$$ T_1^2 = \Big | \frac{\langle x |U^\dagger (\Delta t)| x + \Delta x \rangle - \langle x + \Delta x| U (\Delta t)|x \rangle}{\Delta x^2 \Delta^2 t} \Big |^2 = \frac{2 | \langle x | U^\dagger (\Delta t) | x + \Delta x \rangle |^2 - \langle x |U^\dagger (\Delta t)| x + \Delta x \rangle^2 - \langle x + \Delta x| U (\Delta t) |x \rangle^2}{\Delta x^2 \Delta t^2} $$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 $z = a+ib$:

$$ \frac{4 | \langle x| x + \Delta x \rangle |^2 }{\Delta x^2 \Delta t^2} \geq T_1^2 $$Since all quantities are positive:

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

Similarly, we define:

$$ T_2 = \Big | \frac{\langle x | U^\dagger (\Delta t) | x \rangle - \langle x | U (\Delta t)|x \rangle }{\Delta x \Delta t} \Big | $$

Then,

$$ \frac{ 2 | \langle x | U^\dagger (\Delta t) | x \rangle| }{\Delta x \Delta t} \geq T_2$$

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

$$ \frac{2 | \langle x | U^\dagger (\Delta t) | x + \Delta x \rangle | }{\Delta x \Delta t} + \frac{ 2 | \langle x | U^\dagger (\Delta t)| x \rangle| }{\Delta x \Delta t} \geq \frac{1}{\hbar} |\langle x | \dot p | x \rangle | $$

Or in terms of $\Delta t$:

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

Similarly for momentum eigenkets:

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

 

[anuttarasammyak]

I appreciate your effort to pursuit complicated calculation. Saying that I would say a few challenges I observed.

$|x>$ and $|p>$ are basis for expansion and not physical states. I am afraid whether your calculation corresponds to physical phenomena. If it is OK then relativity tells us $|x>$ ;t=0 changes with time as expanding light sphere because indefinite p means the particle has almost speed c in any direction. Also it says many particle antiparticle pairs should appear. I am afraid such situations are not favorable to you.

 

[Someone_physics]

I appreciate your effort to pursuit complicated calculation. Saying that I would say a few challenges I observed.

$|x>$ and $|p>$ are basis for expansion and not physical states. I am afraid whether your calculation corresponds to physical phenomena.

I don't see why $|x>$ or $|p>$ can't be expressed as a superposition of energy-eigenstates and be plugged into the Schrodinger equation. Honestly, I was thinking of doing this in quantum mechanics.

The main motivation of this calculation is, It is known (see https://quantumcomputing.stackexcha...d-to-time-energy-uncertainty-true-or-testable):

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.

 

[anuttarasammyak]

I don't see why |x> or |p> can't be expressed as a superposition of energy-eigenstates and be plugged into the Schrodinger equation. Honestly, I was thinking of doing this in quantum mechanics.

We can exchange vector basis among |x>, |p> and |E>. As an example for free particle H=p^2/2m momentum eigenstates are also energy eigenstates. Energy eigenstate of harmonic oscillator is expanded as wave function <E|x> as Hermite polynomials.

 

[vanhees71]

The socalled "energy-time uncertainty relation" has a completely different meaning than the uncertainty relations between observables. The latter is a relation between the standard deviations of two observables for the quantum system prepared in a specific state $\hat{\rho}$, i.e., for all states
$$\Delta A \Delta B \geq |\langle \mathrm{i} [\hat{A},\hat{B}] \rangle|,$$
where
$$\Delta A=\sqrt{\langle \hat{A}^2 \rangle-\langle A \rangle^2}$$
and
$$\langle \hat{A} \rangle=\mathrm{Tr} (\hat{\rho} \hat{A}).$$
This means if you prepare the system in the state $\hat{\rho}$ and measure on a first sufficiently large ensemble the observable $A$ accurately to statistically determine the standard deviation $\Delta A$ and then measure on a second sufficiently large ensemble the observable $B$ accurately to statistically determine the standard deviation $\Delta B$, then the so determined standard deviations fulfill the given uncertainty relation.

A special case are position and momentum, because there the right-hand side of the unceratinty relation doesn't depend on the state, i.e., you get
$$\Delta x \Delta p_x \geq \hbar/2,$$
independent of the state. The operational meaning of this uncertainty relation is of course as discussed above. It has nothing to do with the ability to measure the one or the other observable accurately nor with the disturbance of the system by the measurement (as Heisenberg thought in his very first paper on the uncertainty relation and as was immediately corrected by Bohr). The question about how accurately can measure an observable is answered by the practical limitations of your technology to construct an accurate measurement device (basically a question of the available money ;-)). The question about how much the system is disturbed by a measurement must be analyzed in detail for each measurement setup, and there's a lot of fundamental research on this in recent years, also in the context of extended measurement descriptions for "weak measurements" and the socalled POVM formalism (POVM=positive operator valued measures), which extends the standard von Neumann filter measurement description to more realistic descriptions of reeal-world measurements with real-world apparati.

The energy-time uncertainty relation has a different meaning, which is because time is not an observable in quantum mechanics but a parameter. The energy-time uncertainty relation is also completely differently derived: You consider two systems, which are only weakly coupled, i.e., the interaction Hamiltonian of the two systems can be treated by perturbation theory. Now you ask how much does energy $E_1$ of one of the subsystem's change (of course $E_1+E_2=\text{const}$, i.e., you have $\Delta E_1 = E_1'-E_1=-\Delta E_2$. Time-dependent perturbation theory for a time-independent small interaction leads to the transition probability that after time $\Delta t$ the (accurately measured!) value of the energy of the subsystems
$$P(\Delta E_1)=P(\Delta E_2)=\sin^2(\Delta E_1 \Delta t/\hbar)/\Delta E_1^2.$$
The most probable change is of the order $\Delta E_1 \simeq \hbar/\Delta t$.

The consequence is that you can verify the energy conservation law, i.e., $\Delta E=0$ by measuring $E_1'$ and $E_2'$ only with an accuracy $\Delta E \simeq \hbar/\Delta t$, i.e., you need to wait long between the energy measurements to verify the energy conservation law accurately. For further discussion see Landau and Lifshitz vol. 3.