# What time does the particle reach the screen in this thought experiment?

Suppose a particle with a gaussian wavefunction moves to begin with towards a position detector screen. How do we obtain the 'Time of arrival' distribution, when time can't be an observable? Should the average time of arrival be inversely proportional to the mean of momentum distribution? Does having a wide initial location distribution also give us a wide time of arrival distribution?

Initially this wavefunction distribution will be Gaussian but as it moves towards the screen I imagine it will 'bunch up' at the screen thus causing the probability of arrival to tend rapidly towards one. This effect essentially represents classical sense of the velocity of the particle.

If we use the Copenhagen interpretation, we may integrate the wavefunction amplitude squared within a region and define this as the probability that eigenstates outside this region drop to zero. This allows us to model the screen as a measuring device with a discretized 'frame rate'. With a continuous frame rate the wavefunction will just be perpetually 0 at the screen which leads to the quantum zeno paradox. With a discrete frame rate, increasing the frame should either cause the wavefunction at the screen to remain smaller on average, thereby slowing down the arrival of the particle. However as probably_someone stated, it would also increase the number of 'trials' thereby actually speeding up the time of arrival, this is known as the quantum anti zeno effect. Logically speaking, the frame rate of the instrument is all we need in theory to mathematically predict the correct arrival time distribution.

However I am wondering whether this method produces an accurate distribution. If not, why not? How negligible is the effect of speeding up or slowing down detection rates? Why haven't we yet calculated the actual frame rate of a position detector yet? Why doesn't this method say anything to prevent superluminal travel?

• Note: "wavefunction collapse" is not necessarily an actual physical phenomenon. It's an interpretation of what happens during a measurement, one which conceals the actual fine-scale details of the interaction. There are ways to interpret quantum mechanics that don't have the wavefunction instantaneously "collapsing" in this way (for example, decoherence-based interpretations), and there are also experimentally-demonstrated "weak measurements" that don't cause "collapse" in the sense you mean here. Sep 4, 2020 at 17:33
• @BioPhysicist Yes, that much is true, and will be the substance of the answer I'm writing. Sep 4, 2020 at 17:45
• The answer by benrg is perfectly correct. If you want to avoid wasting the reputation you spent on the bounty, you should explain why you think it's wrong! Oct 17, 2020 at 19:36
• @lucky-guess But the detector is constantly making measurements. Every moment that it doesn't click causes a partial collapse of the wavefunction. Not only is this not in question, it is standard material without which many experimental results can't be explained. You can find it in textbooks about atomic and optical physics. Oct 17, 2020 at 22:59
• @knzhou "Every moment that it doesn't click causes a partial collapse of the wavefunction" - I know this to be not only unsupported by any experimental evidence, but also not true, as it leads to the quantum zeno effect and the prediction that the particle never arrives. This is in contradiction with many experiments, for example the double slit, in which particles do arrive, after some delay. Check out doi.org/10.1063/1.523304 which shows that under that hypothesis there is no arrival. Oct 19, 2020 at 23:15

I just finished a thesis on this subject and I'm happy to share. None of the linked papers are my own.

The time of arrival in quantum mechanics is actually a subject of ongoing research. It is certainly a question which begs for an answer, as experiments have been able to measure the distribution of arrival times for decades (see for example Fig. 3 of this 1997 paper by Kurtsiefer et. al). Note: If you do not have access to journals let me know and I will see if I can include the figure in this answer.

Part 1 of this answer describes why there is a problem with arrival time in quantum mechanics.

Part 2 outlines the modern situation in regards to this problem.

Part 3 gives, in my view, the best answers we currently have, which still need experimental verification.

1. New Ideas are Here Needed: The observable-operator formalism seems not to work for arrival times

Normally in QM you have operators $$A$$ corresponding to the variables used in classical mechanics. This lets you define a basis of eigenfunctions of that operator, which are found through the equation $$A|a\rangle = a |a\rangle$$. With such a basis in hand, the probability of finding the value $$a$$ in an experiment on a particle in state $$|\psi\rangle$$is $$|\langle a|\psi\rangle|^2$$.

Though the probability distribution of arrival times can be measured in experiment, predicting it in theory is less straightforward. There are two theorems I am aware of which indicate that the textbook observable formalism above will not work for arrival times:

1. Pauli's Theorem: In 1933, Wolfgang Pauli published a book on Quantum Mechanics called The General Principles of Wave Mechanics. In a footnote of this book, Pauli notes that if you have the commutation relation $$[T,H]=i\hbar$$ for some supposed self-adjoint time operator $$T$$, then $$H$$ would have to have all eigenvalues $$[-\infty, \infty]$$, which is not possible because systems could not have a ground state. His is an early variant of the theorem which has since been made more precise (modern proofs can be found in section 2 of this 1981 paper).
2. Allcock's Theorem: In 1969, Allcock gave another proof that the usual formalism won't work with time. He shows that it is impossible to have a complete set of orthonormal arrival time eigenstates which transform properly under change of coordinates $$(t,\vec{r}) \to (t+\Delta t,\vec{r})$$ - and thus that there cannot be an adequate self-adjoint time operator, since this would result in such eigenstates. The proof begins just before Equation 2.18 with "The reader...".

A number of authors have tried to define a time operator anyway, yet none of the variants I have seen were able to subvert both of the above theorems, rendering them unphysical.

2. Arrival time approaches outside of the textbook formalism

Because of the issues in Part 1 of this answer, many authors have tried to come up with ways to derive a distribution for the arrival time of a particle outside of the usual formalism. The distribution we seek is usually notated $$\Pi(t)$$ and should of course have the property that

$$\int_a ^b \Pi(t) \text{dt} = \text{Probability that the particle arrives at time } t \in [a,b]$$

There is no lack of proposals for this, actually the problem is that there are very many proposals which do not agree with one another. You can see a non-exhaustive summary of some of those proposals in this review paper by Muga (2000). It contains about half of the proposals I am aware of today.

Having gone through many of the existing proposals in detail, I will give my opinion: they are, for the most part, grotesquely unscientific. Problems with some these proposals (in peer-reviewed papers!) include:

• Not normalizable even for reasonable $$\psi$$ like gaussian wave packets
• Predicts negative probabilities
• Only works in 1 dimension
• Only works when $$V(x)=0$$

3. The best answers we have today

In recent months, an effort has accumulated to actually do experiments to rule out many of these proposals. An experiment is planned for the near future. Until the results come out, any conclusions on which proposal is best are subject to being proven wrong. That being said, some proposals are clearly very ad-hoc and inspire little confidence, while I cannot find objective flaws in others. According to my own, always-possibly-flawed understanding after working in this field, the best proposals we have today are

3.1 Bohmian Mechanics / The Quantum Flux

Bohmian Mechanics is a quantum theory in which particles follow definite trajectories (see the double slit trajectories for example). The predictions of Bohmian Mechanics agree with standard QM for position measurements. For each individual trajectory the arrival time is the moment when it first hits the detector. Since the initial position is unknown, many different trajectories are possible, and this defines a distribution of different possible arrival times.

It has been proven that typically, the arrival time distribution in Bohmian Mechanics is exactly equal to the (integrated) flux of probability across the detector $$D$$:

$$\Pi_{BM}(t) = \int_{\partial D} \vec{J}(\vec{r},t)\cdot \hat{n} \text{ dA}$$

where $$\vec{J}$$ is the flux as described in any QM textbook, and $$\hat{n}$$ is a unit vector pointing into the detector surface. This is the rate at which probability enters the detector, and so it very nicely correlates the arrival time statistics with position statistics.

However, the quantity $$\vec{J}\cdot \hat{n}$$, and therefore the entire integral, may be negative. In this case that the flux clearly does not work as a probability density, and it has been shown that it is exactly in this case (negativity for some point on the detector) that the Bohmian Mechanics prediction differs from the flux. The prediction made by Bohmian Mechanics, obtained by averaging over many trajectories, is always nonnegative. Negative flux corresponds to Bohmian Trajectories which loop around and leave the detector region.

3.2. The Kijowski Distribution

The second-most reasonable candidate I have seen is the Kijowski distribution. In this 1974 paper, Kijowski postulated it for the free particle by declaring a series of axioms. These axioms yield nicely a unique distribution, but as Kijowski notes,

Our construction is set up for free particles in both the non-relativistic and relativistic case and cannot be generalized for the non-free wave equation

Nonetheless the approach is well-liked as it yields a priori reasonable results and has a tendency to resemble the quantum flux. For this reason, Muga began calling it & its generalizations the "standard distribution".

By abandoning the axiomatic approach, a variant inspired by Kijowski's distribution has been created which works for other potentials, see paper here (2000). However there is a spacial nonlocality to this distribution, i.e. the position statistics don't correspond to the arrival time statistics. Basically it predicts that a particle can be found after a finite time at a location where, according to standard quantum mechanics, there is a 0% chance of finding it - this seems unphysical. A critique is given by Leavens in this paper (2002).

Final Remarks

Arrival time proposals are a dime a dozen at the moment, and even having done research in this field it is infeasible to rigorously go through every approach anyone has used in the literature. In addition, an experiment has not yet been done, so in some sense, science does not have an answer for you yet. To remedy this I have given what I can, namely my own understanding of the state of things after having spent a fair amount of time on the subject. If things go as I hope they do, there will be a scientific answer to this question in the coming years. In addition to the aforementioned experiment, there is for example an experimental proposal, possible to implement with modern-day technology, which could test arrival times in the most "juicy" regime: where the flux is negative. To be clear about potential biases, I know the authors of this paper. My thesis was not on the Bohmian approach.

• Comments are not for extended discussion; this conversation has been moved to chat. Oct 23, 2020 at 14:16
• although 1) is true, I think Holevo proposed in his book 'Probabilistic and Statistical Aspects of Quantum Theory' to use a POM for time, in the same sense as when having a reflecting boundary p is not self-adjoint anymore. I didnt fully understand the argument though. Mar 23, 2021 at 10:51

EDIT: After some discussion, the OP made it clear that they were actually asking about a more fundamental issue: given a time-dependent probability density $$p(x,t)$$, and given that we are observing a fixed spatial interval, when do we expect to first observe the event?

(Only the first observation is important, because the detection of a particle is an interaction that changes its wavefunction, and so we stop wondering when we'll detect the particle once we actually do detect the particle).

Let's ask a simpler question first, that might guide our intuition. Let's roll a die. The outcomes are 1 to 6, all equally probable, and each roll of the die is a discrete time interval (let's say we roll once per second). Let's ask the question: how long will it take, on average, for us to roll a 4?

The probability of rolling a 4 on the first roll is $$1/6$$. The probability of rolling your first 4 on the second roll and not on the first roll is $$1/6\times(1-1/6)$$. Likewise, the probability of rolling a 4 on the third roll but not on the first or second is $$1/6\times(1-1/6)^2$$. And the probability of rolling a 4 on the $$n$$th roll but not on any previous roll is $$1/6\times (1-1/6)^{n-1}$$. So, from our original probability distribution of outcomes per time interval, we can assemble a probability distribution of the amount of time it will take us to see a 4:

$$P(t_n)=1/6\times(1-1/6)^{n-1}$$

where $$t_n$$ is the $$n$$th time interval. The mean value of $$t_n$$, the expected time interval in which we'll see our first 4, is:

$$\bar{t}=\sum_{n=1}^\infty nP(t_n)=\sum_{n=1}^\infty n\times 1/6\times (1-1/6)^{n-1}=6$$

So we should expect it to take roughly 6 seconds to see our first 4.

With a few tweaks, we can apply that logic to our current situation. Suppose we're observing over the spatial interval $$a. First, we need to calculate the probability of observing our outcome as a function of time:

$$P(t)=\int_{a}^b p(x,t) dx$$

Now, we discretize our continuous time parameter. Our detector interacts with the environment, but those interactions are not instantaneous: every interaction that would allow a detection has some associated timescale $$\Delta t$$ (for example, detectors based on ionization would have a timescale associated with the amount of time an incoming particle takes to ionize an atom). So we can model our detector as a device that periodically "checks" to see whether it interacted with a particle. So now we have a set of discrete time intervals, $$t=0, \Delta t,2\Delta t,...$$ during which the metaphorical dice are rolled.

But this time, each time these metaphorical dice are rolled, the probability is different. And it's clear that we can't actually use the probability at a particular instant, either, because that would imply that we know what the "phase" of the detector's interactions are, which we don't. So instead, we average the probability over one interaction timescale. Let $$P_n$$ be the probability that a detector detects a particle in the interaction timescale interval $$(n\Delta t, (n+1)\Delta t)$$:

$$P_n=\frac{1}{\Delta t}\int_{n\Delta t}^{(n+1)\Delta t} P(t)dt$$

So we can now play the same game as before: the probability that we detect a particle on the very first interaction timescale is $$P_0$$. The probability that we detect a particle on the second interaction timescale but not the first one is $$P_1(1-P_0)$$. The probability that we detect a particle on the third interaction timescale but not the second or first is $$P_2(1-P_1)(1-P_0)$$. And so on, generating our formula for the probability of seeing our particle on the $$n$$th interaction timescale:

$$P(\text{detection after }n\text{ interaction timescales})=P_n(1-P_{n-1})(1-P_{n-2})...(1-P_1)(1-P_0)$$

Now that we have our distribution for arbitrary $$n$$, this means that the expected number of interaction timescales that we'll have to wait to detect the particle is:

$$\bar{n}=\sum_{n=0}^\infty nP_n(1-P_{n-1})(1-P_{n-2})...(1-P_0)$$

Once we have numerically calculated $$\bar{n}$$, then we can easily get the expected wait time before detecting a particle:

$$\bar{t}=\bar{n}\Delta t$$

With that out of the way, let's calculate the actual probability density function.

Let's suppose that you prepare your Gaussian wavepacket in a minimum-uncertainty configuration. What I mean by that is described below.

The Heisenberg uncertainty principle states:

$$\sigma_x\sigma_p\geq\frac{\hbar}{2}$$

It turns out that the situation where the product $$\sigma_x\sigma_p$$ is minimized is actually a Gaussian wavefunction (proofs of this can be found elsewhere on the internet), so for that particular Gaussian wavefunction, we have:

$$\sigma_x\sigma_p=\frac{\hbar}{2}$$

The momentum probability distribution is also Gaussian, with some mean $$\bar{p}$$ and a standard deviation $$\sigma_p=\frac{\hbar}{2\sigma_x}$$.

So if we start with our Gaussian momentum wavefunction $$\psi(k)=e^{-\alpha(k-k_0)^2}$$, where $$\alpha=\frac{\hbar^2}{2\sigma_p^2}=\sigma_x^2$$, we can follow this procedure to find the position wavefunction as a function of time (and then normalize said wavefunction, because the authors of that source apparently didn't bother to do so):

$$\psi(x,t)=\left(\frac{\alpha}{2\pi}\right)^{1/4}\frac{1}{\sqrt{\alpha+i\beta t}}e^{i(k_0x-\omega_0 t)}e^{\frac{-(x-v_g t)^2}{4(\alpha+i\beta t)}}$$

where $$v_g=\frac{d\omega}{dk}$$ evaluated at $$k_0=\frac{\bar{p}}{\hbar}$$, and $$\beta=\frac{1}{2}\frac{d^2\omega}{dk^2}$$, also evaluated at $$k_0$$.

As you can see, in order to proceed, we need a relation between $$\omega$$ and $$k$$. This is called the dispersion relation, and for a relativistic electron, the dispersion relation is:

$$\omega=c\sqrt{k^2+(m_ec/\hbar)^2}$$

This means that:

$$\omega_0=c\sqrt{k^2+(m_ec/\hbar)^2}$$

$$v_g=\frac{ck_0}{\sqrt{k_0^2+(m_ec/\hbar)^2}}$$

$$\beta=\frac{c}{2\sqrt{k_0^2+(m_ec/\hbar)^2}}-\frac{ck_0^2}{2(k_0^2+(m_ec/\hbar)^2)^{3/2}}$$

Then, figuring out the probability that the electron will be at the screen position $$x_s$$ as a function of time is as simple as evaluating $$|\psi(x_s,t)|^2$$:

$$|\psi(x_s,t)|^2=\sqrt{\frac{\alpha}{2\pi(\alpha^2+\beta^2t^2)}}\exp\left(\frac{-\alpha(x_s-v_gt)^2}{2(\alpha^2+\beta^2t^2)}\right)$$

Obviously, this general solution doesn't tell us mere mortals very much in terms of intuition, so there are two special cases that are helpful to develop some understanding of the situation:

The ultra-relativistic limit

In the case where $$k\gg m_ec/\hbar$$, the dispersion relation reduces to:

$$\omega=ck$$

which means:

$$\omega_0=ck_0$$

$$v_g=c$$

$$\beta=0$$

Plugging these into the general solution, we find that:

$$|\psi(x_s,t)|^2=\frac{1}{\sqrt{2\pi}\sigma_x}\exp\left(-\frac{(x_s-ct)^2}{2\sigma_x^2}\right)$$

As you can see, the wavefunction simply travels to the right at velocity $$c$$ over time, with a constant width $$\sigma_x$$ as a function of time. So the uncertainty in detection time depends only on the uncertainty in initial position of the electron.

The non-relativistic limit

In the limit where $$k\ll m_ec/\hbar$$, the dispersion relation reduces to:

$$\omega\approx \frac{m_ec^2}{\hbar}+\frac{\hbar k^2}{2m_e}$$

which means that:

$$\hbar\omega_0=m_ec^2+\frac{p^2}{2m_e}$$

$$v_g=\frac{\hbar k_0}{m}=\frac{\bar{p}}{m}$$

$$\beta=\frac{\hbar}{2m}$$

Plugging these into the original formula, we find that the center of the wavepacket travels with a velocity $$v_g$$, as you would expect, and that the wavepacket also spreads out quite a bit over time: the width of the wavepacket is $$\sqrt{\alpha^2+\left(\frac{\hbar t}{2m}\right)^2}$$. So the uncertainty in the detection time depends both on the initial uncertainty in position and on the distance from the mean initial position to the screen. Generally, the further away the screen is, the more uncertain the detection time will be.

With these two extremes, we can now interpolate between them to say something about what happens to a relativistic (but not ultra-relativistic) electron: increasing the distance to the screen still increases the uncertainty in detection time, but not by as much as in the non-relativistic case (which makes sense - at relativistic speeds, changing your momentum doesn't actually change your velocity very much).

Incidentally, this is why time-of-flight detectors in particle physics experiments only work well at lower energies: determining momentum by measuring velocity gets more and more difficult as energy increases.

• Comments are not for extended discussion; this conversation has been moved to chat.
– Chris
Sep 5, 2020 at 6:01
• The issue with the FTs came up in another post I was answering, so I thought I would go ahead and address it. It is a little hand-wavy, but I think it gets the job done. Sep 17, 2020 at 15:39
• Really nice answer. Oct 26, 2020 at 21:02

This is similar to your last question, and I think it's answered by the answer I wrote to that one, but I'll try to explain it in a slightly different way.

The short version is that whenever a detector is switched on and actively waiting to detect something, a measurement (perhaps interaction-free) and an associated collapse takes place at every time.

In introductory quantum mechanics courses, measurements are generally treated as "complete": you measure position, for example, and the wave function is a delta function (or at least a narrow Gaussian) in position space after the collapse. No real detector works like that – it would have to fill all of space.

The simplest realistic example of a measurement device is a position detector that measures the value of the is-the-particle-here operator, which has two eigenvalues, 0 and 1, whose associated eigenstates are wave functions that are zero inside the detector and wave functions that are zero outside the detector. At each moment, if the detector detects the particle, the particle's wave function afterwards is zero outside, and at each moment, if it doesn't, the particle's wave function afterwards is zero inside. Both of these "disappearances" of part of the wave function are collapses associated with measurement. In the latter case, it's an interaction-free measurement. You will end up randomly (with probabilities dictated by the Born rule) in one of the futures where the detector measured 1 at a particular time, or in the future where it measured 0 at all times, and in each case the wave function will have "updated" to be consistent with what you know about when it was and wasn't detected.

Instead of thinking about this in the collapse picture, you can think about it in the many-worlds picture. At any given time, you can write the wave function as a weighted sum of a part where the electron is in the detector and a part where it's outside. By linearity it will be the same weighted sum of time-evolved versions of those states at any later time. The inside state evolves into a state where the environment differs from the outside state's environment in a complicated way, perhaps involving an audible click or an electrical impulse. They're different enough that there is no chance of future wavelike interference between them, so they can be treated as separate classical worlds.

Although measurements happen all the time, they don't happen continuously. There is a quantization of measurement times, associated with quantum interference in the early stages of detection, so the number of outcomes/worlds is finite. (Don't ask me for more details because I don't know them – but I'm pretty sure this is true.)

You can think of your screen as being made from a bunch of position detectors glued together, and the analysis is the same.

If the particle has zero chance of being at a detector at a given time, then no measurement or collapse happens, but it isn't necessary to treat this as a separate case – it's equivalent to the general case with the probability of one outcome being 0.

how is that every moment it doesn't click affects the wavefunction? the wavefunction evolves as per schrodingers equation which has nothing to do with the apparatus

The detector's failure to click tells you that the particle isn't in the detector, which is information about its location and so necessarily causes a collapse. This is called interaction-free measurement.

Possibly you're thinking that this can't be true because if the particle was being measured all the time then its behavior would become classical. The reason that doesn't happen is that failure to click usually doesn't tell you much about the particle's location, therefore the collapse doesn't change the wave function very much. If before the negative measurement the particle was spread out over a large spatial area (which includes the small detector), then after the negative measurement there's a small "hole" at the detector where the wave function is zero, while the rest of the wave function is completely unaffected (except that it's slightly rescaled to renormalize it). The small hole doesn't cause a large change in the particle's behavior.

• Comments are not for extended discussion; this conversation has been moved to chat. Oct 20, 2020 at 15:48

The following is a failed attempt (at best with an extra assumption it can only work with cases where momentum is conserved) and too long for comment. Hopefully it illustrates the difficulty of the problem.

Let us solve in a one dimensional universe (but can be further generalised) and let the last possible time the electron can hit the detector be $$T$$ and the earliest possible time be $$t_0$$. The probability of the event at time $$t_0$$ the electron will be measured at $$x$$ is given by $$p(t_0) \delta t$$ and that at time $$t+\delta t$$ being $$p(t_0 +\delta t) \delta t$$ and so on. Let $$U$$ be the unitary operator.

Now let us make use of the density matrix formalism to specify the density matrix after it is measured at time $$T$$.

$$\rho = p(T ) |x \rangle \langle x| + p(T - \delta t) U(\delta t) |x \rangle \langle x| U^\dagger(\delta t) + \dots$$

In the limit $$\delta t \to 0$$

$$\rho = \int_{t_0}^{T} p(t )U(T-t) |x \rangle \langle x| U^\dagger(T-t) dt$$

Let the distance between the electron gun and the screen be $$a$$. Now, lets slightly shift the screen away by a displacement along the x-axis by $$\delta a$$. Then the new density matrix will be:

$$\rho + \delta \rho = \int_{t_0 + \delta t_0 }^{T + \delta T} (p(t ) + \delta p(t) )U(T-t) |x + \delta a \rangle \langle x + \delta a | U^\dagger(T-t) dt$$

Using the translation operator and keeping the lower order terms:

$$\rho + \delta \rho = \int_{t_0 + \delta t_0 }^{T + \delta T} (p(t ) + \delta p(t) )U(T-t) (1 -\frac{\delta a \cdot \hat p}{\hbar})|x \rangle \langle x |(1 +\frac{\delta a \cdot \hat p}{\hbar}) U^\dagger(T-t) dt$$

The above R.H.S's expansion can be expressed as the sum of the below terms:

$$\tilde \rho = \int_{t_0 + \delta t_0 }^{T + \delta T} p(t )U(T-t) |x \rangle \langle x| U^\dagger(T-t) dt$$

$$\delta \tilde A = \int_{t_0 + \delta t_0 }^{T + \delta T} \delta p(t )U(T-t) |x \rangle \langle x| U^\dagger(T-t) dt$$

$$\delta \tilde B = \int_{t_0 + \delta t_0 }^{T + \delta T} p(t )U(T-t) (\frac{\delta a \cdot \hat p}{\hbar} |x \rangle \langle x| - |x \rangle \langle x| \frac{\delta a \cdot \hat p}{\hbar} )U^\dagger(T-t) dt$$

Hence,

$$\rho + \delta \rho = \tilde \rho + \delta \tilde A + \delta \tilde B$$

Focusing on $$\tilde \rho - \rho$$

$$\delta \tilde \rho = \tilde \rho- \rho = \int_{t_0 + \delta t_0 }^{t} p(t )U(T-t) |x \rangle \langle x| U^\dagger(T-t) dt + \int_{T }^{T + \delta T} p(t )U(T-t) |x \rangle \langle x| U^\dagger(T-t) dt$$

Hence,

$$\delta \rho = \delta \tilde \rho + \delta \tilde A + \delta \tilde B$$

Taking the trace:

$$\text{Tr } \delta \rho = \text{Tr } \delta \tilde \rho + \delta \tilde A + \delta \tilde B = 0$$

Additional I'd be willing to bet in scenarios where momentum is conserved $$[H, \hat p ] =0$$ then $$\delta T$$ and $$\delta t_0$$ increase linearly with $$\delta a$$

• Hi More Anonymous, you asked for a comment. The time-of-arrival problem takes a given initial state $\psi_0(x)$ and a location of the detector (for example a large wall at $\vec{r}=(L,y,z)$), and gives a probability distribution of the arrival times $\Pi(t)$ at which the particle arrives at the detector. The distribution necessarily depends on both. It is not clear to me where or how you define this distribution. If you gave an example calculation for something simple like a gaussian initial wave packet with momentum it would be clearer. Oct 23, 2020 at 14:52
• @doublefelix thanks for replying.Your comment made me realise somethings. My idea focuses more on how the distribution changes as I change $\vec r \to \vec r + \vec \delta r$. The initial distribution $\vec r = \vec a$ will have to be given you by experiment but then if the latest it can possibly arrive is earliest at $t=t_0$ and latest at $t=T$. I guess with conserved momentum shifting the screen will affect $t+ \lambda_1 \delta r$ and $T + \lambda_2 \delta r$ (Again $\lambda$'s would be determined by experiment). But we can determine the new probability distribution using my equation. Oct 23, 2020 at 15:15

If we know the wave function then we also know the time of arrival in a statistical sense. Consider a laser pulse. Suppose that the electric field is a wave package travelling at speed v, say a 3D gaussian. Let's assume the spread to be constant for simplicity. The probability of a transition in the sensor is proportional to E$$^2$$ by Fermi's golden rule. E is known at every position at every point in time and so is probability of detecting a photon. The arrival time will be a gaussian distribution centered at d/v.