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Background

Let us assume I have a particle which is in the position basis. I was wondering what was the most probable trajectory taken by such a particle when constantly measured.

Let the particle be at position $x_0$ and evolve unitarily for time $\Delta t$ before being measured again and found to be at $x_1$. The probability of such an event is:

$$ P(x_0,x_1) =| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle |^2$$

After being measured at $x_1$ we again measure it at $x_2$ and again:

$$ P(x_1,x_2) =| \langle x_1| e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2$$

Hence, the probability of the trajectory $P$ is:

$$ P = P(x_0,x_1,\dots,x_n) = P(x_0,x_1) P(x_1,x_2) \dots P(x_{n-1},x_n) $$

Or:

$$ P(x_0,x_1,\dots,x_n) = | \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle |^2| \langle x_1| e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \dots | \langle x_{n-1}| e^{\frac{-i}{\hbar} H \delta t}|x_n \rangle |^2 $$

Question

How does one maximize the below?

$$ P(x_0,x_1,\dots,x_n) = | \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle |^2| \langle x_1| e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \dots | \langle x_{n-1}| e^{\frac{-i}{\hbar} H \delta t}|x_n \rangle |^2 $$

My Attempt

Now consider, $P' = \max P(x_0,x_1,\dots,x_n) \geq P(x'_0,x'_1,\dots,x'_n) $. If we perturb $P'$ by varying $x_1$ then:

$$P' - \delta P = | \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 + \delta x_1 \rangle |^2| \langle x_1 + \delta x_1| e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \dots | \langle x_{n-1}| e^{\frac{-i}{\hbar} H \delta t}|x_n \rangle |^2 $$

Substituting with $P'$:

$$- \delta P = \Big(| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 + \delta x_1 \rangle |^2| \langle x_1 + \delta x_1| e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 -| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle |^2| \langle x_1 | e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \Big) \dots | \langle x_{n-1}| e^{i H \delta t}|x_n \rangle |^2 $$

Let us focus on the factor: $\Big(| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 + \delta x_1 \rangle \langle x_1 + \delta x_1| e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 -| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle |^2| \langle x_1 | e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \Big)$

Using the translation operator $e^{-i\frac{p}{\hbar}.x}$ with the momentum operator $p$:

$$ \Big(| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}(1 - \frac{i}{\hbar} \frac{\delta x_1}{\hbar} .p)|x_1 \rangle \langle x_1 | (1+ i \frac{\delta x_1}{\hbar} .p)e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 -| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle |^2| \langle x_1 | e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \Big) $$

Ignoring $\delta x_1^2$ terms:

$$ \Big(|\langle x_0| ( e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle \langle x_1 | e^{\frac{-i}{\hbar} H \delta t} - i e^{\frac{-i}{\hbar} H \delta t} [\frac{\delta x_1}{\hbar} .p, |x_1 \rangle \langle x_1 | ] e^{\frac{-i}{\hbar} H \delta t})|x_2 \rangle |^2 -| \langle x_0| e^{\frac{-i}{\hbar} H \delta t}|x_1 \rangle |^2| \langle x_1 | e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \Big) $$

We define the projection operator:

$$ P_{x_1} = |x_1 \rangle \langle x_1 |$$

and,

$$\frac{\delta x_1}{\hbar} p_{\delta x_1} = \frac{\delta x_1}{\hbar} .p$$

where $p_{\delta x_1}$ is momentum operator along the translated direction of $\delta x_1$. Substituting the above:

$$ \Big(|\langle x_0| ( e^{\frac{-i}{\hbar} H \delta t} P_{x_1} e^{\frac{-i}{\hbar} H \delta t}| x_2 \rangle - i \langle x_0| e^{\frac{-i}{\hbar} H \delta t} [\frac{\delta x_1}{\hbar} p_{\delta x_1}, P_{x_1} ] e^{\frac{-i}{\hbar} H \delta t})|x_2 \rangle |^2 -| \langle x_0| e^{\frac{-i}{\hbar} H \delta t} P_{x_1} e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle |^2 \Big) $$

Simplifying the above:

$$ \Big( - i \langle x_2| ( e^{\frac{i}{\hbar} H \delta t} P_{x_1} e^{\frac{i}{\hbar} H \delta t}| x_0 \rangle \langle x_0| e^{\frac{-i}{\hbar} H \delta t} [\frac{\delta x_1}{\hbar} p_{\delta x_1}, P_{x_1} ] e^{\frac{-i}{\hbar} H \delta t})|x_2 \rangle + i \langle x_2| e^{\frac{i}{\hbar} H \delta t} [\frac{\delta x_1}{\hbar} p_{\delta x_1}, P_{x_1} ] e^{\frac{i}{\hbar} H \delta t})|x_0 \rangle \langle x_0| ( e^{\frac{-i}{\hbar} H \delta t} P_{x_1} e^{\frac{-i}{\hbar} H \delta t}| x_2 \rangle \Big) $$

Substituting in $\delta P$ equation and using $\frac{\delta P}{\delta x_1} = 0$:

$$ \langle x_2| e^{\frac{i}{\hbar} H \delta t} [ p_{\delta x_1}, P_{x_1} ] e^{\frac{i}{\hbar} H \delta t}|x_0 \rangle \langle x_0| e^{\frac{-i}{\hbar} H \delta t} P_{x_1} e^{\frac{-i}{\hbar} H \delta t}| x_2 \rangle = \langle x_2| e^{\frac{i}{\hbar} H \delta t} P_{x_1} e^{\frac{i}{\hbar} H \delta t}| x_0 \rangle \langle x_0| e^{\frac{-i}{\hbar} H \delta t} [ p_{\delta x_1}, P_{x_1} ] e^{\frac{-i}{\hbar} H \delta t}|x_2 \rangle $$

Or another way of putting it is the imaginary part is $0$ and using $P_{x_0} = |x_0 \rangle \langle x_0 | $:

$$ \text{Im} \langle x_2| e^{\frac{i}{\hbar} H \delta t} [ p_{\delta x_1}, P_{x_1} ] e^{\frac{i}{\hbar} H \delta t} P_{x_0} e^{\frac{-i}{\hbar} H \delta t} P_{x_1} e^{\frac{-i}{\hbar} H \delta t}| x_2 \rangle = 0$$

But I still don't know the relation of $x_1$ and $x_0$

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  • $\begingroup$ It might be worth looking at this paper $\endgroup$ – march Feb 21 at 16:45
  • $\begingroup$ It's behind a pay wall ... $\endgroup$ – More Anonymous Feb 24 at 4:07
  • $\begingroup$ I think the answer would be dependent on how x_n are related to each other $\endgroup$ – Tejas Shetty Feb 24 at 16:01
  • $\begingroup$ Is the Hamiltonian known? In that case, what is it? $\endgroup$ – JezuzStardust Feb 26 at 8:13
  • $\begingroup$ I was hoping for some kind of solution for an arbitrary Hamiltonian. But feel free to use the harmonic oscillator as an example? $\endgroup$ – More Anonymous Feb 26 at 8:19
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At least for free particle and harmonic oscillator propagators have the form $$ \langle x_0|e^{-\frac{i}{\hbar}\hat{H}t}|x_1\rangle = A(t)\, e^{-\frac{i}{\hbar}S(x_0,x_1,t)}, $$ where $S(x_0,x_1,t)$ is a real-valued function and $A(t)$ doesn't depend on $x_0$, $x_1$. Hence, in this cases, the transition probabilty $$ P(x_0,x_1,t) = |\langle x_0|e^{-\frac{i}{\hbar}\hat{H}t}|x_1\rangle|^2 = |A(t)|^2 $$ doesn't depend on coordinates $x_0, x_1$. I am not sure if this means that quantum particle from fixed position $x_0$ with equal probability transits to any other position. I suppose such interpretation might be possible due to the Heisenberg uncertainty principle. Accurate knowledge of the coordinates of a quantum particle leads to infinite uncertainty in its momentum, which leads to the complete uncertainty of the coordinates in the future.

Free particle and harmonic oscillator propagators, for example, see at wiki.

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