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

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.

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.