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Someone_physics

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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

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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:

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