# Most probable trajectory of a constantly measured particle?

## 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 $$x_0$$

• It might be worth looking at this paper Feb 21, 2020 at 16:45
• It's behind a pay wall ... Feb 24, 2020 at 4:07
• I think the answer would be dependent on how x_n are related to each other Feb 24, 2020 at 16:01
• Is the Hamiltonian known? In that case, what is it? Feb 26, 2020 at 8:13
• I was hoping for some kind of solution for an arbitrary Hamiltonian. But feel free to use the harmonic oscillator as an example? Feb 26, 2020 at 8:19

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.