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 $x_0$