Exploring the Possibilities of a New Relativistic Quantum Theory

[A. Neumaier]

1. Can I interpret \psi as a wave function? I.e., is the square of \psi the probability density?
2. What is the meaning of N? Is it the number of particles?
3. Why there are N time labels?
4. What is the t-dependence of \psi? Without such explicit t-dependence the Hamiltonian remains undefined.
5. How this form can be used to calculate the interacting time evolution of an initial 2-particle state?

1. psi is a state in the physical Hilbert space. No further interpretation is necessary. Asymptotic states get a interpretation as superpositions of bound state tensor products through Haag-Ruelle theory.
2. No. It is just a convenient label.
3. Because one needs to choose some N to write down a state. But as in Fock space, one can have superpositions of states with different values of N.
4. The t-dependence is defined as usual form the t-independent initial condition at t=0 by the Schroedinger equation. Defining the Hamiltonian itself needs no t.
5. This is your task to figure out, not mine. Nobody else needs it in any application of QED. But I'll guide you into a simplified exercise related to your question in the thread
https://www.physicsforums.com/showthread.php?p=3174961 . Working this out should give you enough intuition about the more complicated cases that you are interested in.

 

[meopemuk]

5. This is your task to figure out, not mine. Nobody else needs it in any application of QED.

In my opinion, studying the time evolution of initial 2-particle states is rather interesting and important. The most common thing calculated in QFT is the scattering amplitude in a 2-particle initial state. I hope you would agree that scattering is a dynamical time-dependent process. So, it would be very educational to follow the time evolution of the colliding system from distant past to distant future and see how scattering amplitudes appear from this time-dependent treatment.

Moreover the time resolved 2-particle scattering could be even investigated experimentally if sufficiently precise instruments are used.

Eugene.

 

[A. Neumaier]

In my opinion, studying the time evolution of initial 2-particle states is rather interesting and important. The most common thing calculated in QFT is the scattering amplitude in a 2-particle initial state. I hope you would agree that scattering is a dynamical time-dependent process. So, it would be very educational to follow the time evolution of the colliding system from distant past to distant future and see how scattering amplitudes appear from this time-dependent treatment.

One wouldn't see more than one sees it from the usual QM way of deriving scattering amplitudes. Some formulas that are discarded after one has the scattering angle.

I'll follow the educational route in the other thread. You think it is very instructive, so I'll enable you to do it. But I won't do it myself.

Moreover the time resolved 2-particle scattering could be even investigated experimentally if sufficiently precise instruments are used.

How would you do it? You claimed in the photon thread that one cannot say anything about what is not observed. But if you try to observe, you won't get the desired scattering. So the situation here is as in the case of the two slits...

 

[meopemuk]

One wouldn't see more than one sees it from the usual QM way of deriving scattering amplitudes. Some formulas that are discarded after one has the scattering angle.

Scattering amplitudes only connect states in the remote past and remote future. I am interested at intermediate time points.

How would you do it? You claimed in the photon thread that one cannot say anything about what is not observed. But if you try to observe, you won't get the desired scattering. So the situation here is as in the case of the two slits...

I don't quite understand your objection. Quantum mechanics tells us that any isolated physical system is described by a time-dependent wave function, which satisfies the Schroedinger equation. The wave function is generally a collection of complex numbers - amplitudes, whose squares are exactly the probabilities of certain measurements. It is tacitly assumed that these measurements can be performed and experimental probabilities will match the theoretical ones. So, there is nothing fancy in my suggestion to perform observations in a time-dependent state of colliding particles.

In reality, such observations are difficult to perform, because collisions occur in a very small region of space in a short time interval. This is why most people are completely satisfied with the S-matrix description. However "difficult" does not mean "impossible". A complete theory of physical event must be able to describe the unitary time evolution of interacting states.

Eugene.

 

[A. Neumaier]

The wave function is generally a collection of complex numbers - amplitudes, whose squares are exactly the probabilities of certain measurements. It is tacitly assumed that these measurements can be performed and experimental probabilities will match the theoretical ones.

It doesn't need to be assumed. The claim of the generally accepted minimal interpretation is only that _if_ you can perform a discrete and perfect projective measurement then you get results with probabilities conforming to Born's rule. (This assumption is practically realizable only in measurements of spin or polarization degrees of freedom, or of the ''presence'' of a particle at a detector.)

However "difficult" does not mean "impossible".

It is you who are claiming beyond the minimal consensus that it is not impossible. Thus you'd be able to tell us how it is possible.

 

[meopemuk]

It is you who are claiming beyond the minimal consensus that it is not impossible. Thus you'd be able to tell us how it is possible.

My understanding is that when in quantum mechanics we write the wave function $$\psi(x,t)$$ then we assume the following meaning: If we make a device that detects the presence of the particle in the volume V and turn on this device at time t, then the probability of this device actually clicking will be

$$\int_V |\psi(x,t)|^2$$

I don't know how exactly this device can be made or what are the problems with experimental errors associated with such measurements. This is the job of experimentalist to worry about such details.

Do you think that my understanding is incorrect?

Eugene.

 

[A. Neumaier]

My understanding is that when in quantum mechanics we write the wave function $$\psi(x,t)$$ then we assume the following meaning: If we make a device that detects the presence of the particle in the volume V and turn on this device at time t, then the probability of this device actually clicking will be

$$\int_V |\psi(x,t)|^2$$
Do you think that my understanding is incorrect?.

Yes; this is only the kindergarden version - it attaches to the abstract wave function a plausible (and never checked) interpretation, so that the young kids are not afraid and go on learning the weird stuff they are told. Those who need to work with it on a real life level must unlearn all the kindergarden tales...

Your formula cannot be correct since it takes time till the detector responds to the interaction, and a click needs time to be produced and measured. But your probability p_kindergarden is independent of the time. For a particle in a monochromatic beam, where psi(x,t) =e^i(itp_0-ix dot \p)psi_0, you get
p_kindergarden=V^3|psi_0|^2,
as long as V is so small that the plane wave approximation is valid.

To see how the real detection probability looks like, compare your formula with that given by Mandel & Wolf in (14.8-16). For the detection probability p_expert of a single monochromatic photon in a momentum eigenstate, take m=n=1, and remember that the waiting time interval T is supposed to be small, since O(T^2) effects are neglected. One gets the formula
p_expert=alpha c S T/L^3,
where alpha is the detector efficiency, c is the speed of light, S is the area of the detector, and L^3 is the volume V within which the photon energy is supposed to be uniformly distributed (because of the plane wave approximation).

Now this was derived for photon detection, but I doubt that it is very different for an electron.

 

[meopemuk]

Yes; this is only the kindergarden version - it attaches to the abstract wave function a plausible (and never checked) interpretation, so that the young kids are not afraid and go on learning the weird stuff they are told. Those who need to work with it on a real life level must unlearn all the kindergarden tales...

This just confirms my suspicion that you and I understand the basic quantum mechanics very very differently.

I may agree that measurements in real life cannot exactly reproduce the probability given in my kindergarten formula. But this is a purely technical problem. In principle (though, perhaps, not in practice), one can improve the experimental equipment so as to achieve the exact QM result. I have many issues with Mandel & Wolf interpretation of quantum measurements, and I prefer to remain in my happy kindergarten for now.

Eugene.

 

[A. Neumaier]

I may agree that measurements in real life cannot exactly reproduce the probability given in my kindergarten formula. But this is a purely technical problem. In principle (though, perhaps, not in practice), one can improve the experimental equipment so as to achieve the exact QM result.

The exact QM result is in Mandel & Wolf, not in the kindergarden world of introductory QM texts. In the latter, everything is heavily idealized, reducing the complexity of the real world to something that can be reproduced in an exam, even by the average student.

I have many issues with Mandel & Wolf interpretation of quantum measurements, and I prefer to remain in my happy kindergarten for now.

Well, Mandel and Wolf are world-famous experts on quantum optics, which is the basis for understanding quantum measurements and for testing the foundations of quantum mechanics. Learning from them is not a bad idea.

But it takes time to grow up.

I am teaching tough, for those who are prepared to grow.

Knowing that it is difficult, I offer help - but not compromises.

 

[meopemuk]

Well, Mandel and Wolf are world-famous experts on quantum optics, which is the basis for understanding quantum measurements and for testing the foundations of quantum mechanics. Learning from them is not a bad idea.

Well, Feynman is not a lesser expert, but his interpretation of the double-slit experiment is rather different from Mandel & Wolf's.

Eugene.

P.S. Feynman is actually the principal in my kindergarten.

 

[A. Neumaier]

Well, Feynman is not a lesser expert, but his interpretation of the double-slit experiment is rather different from Mandel & Wolf's.

But probably the last measurement he took was during his undergraduate studies.
And he probably never analyzed one on the quantum mechanical level in his whole career.

P.S. Feynman is actually the principal in my kindergarten.

Yes; he created the ''Feynman Lectures on Physics'' to teach kindergarden kids at university.

Even fairy tales told by the Brothers Grimm http://en.wikipedia.org/wiki/Brothers_Grimm don't change the fact that the stories are fairy tales.

 

[A. Neumaier]

For a particle in a monochromatic beam, where psi(x,t) =e^i(itp_0-ix dot \p)psi_0, you get
p_kindergarden=V^3|psi_0|^2,
as long as V is so small that the plane wave approximation is valid.

compare your formula with that given by Mandel & Wolf in (14.8-16). For the detection probability p_expert of a single monochromatic photon in a momentum eigenstate, take m=n=1, and remember that the waiting time interval T is supposed to be small, since O(T^2) effects are neglected. One gets the formula
p_expert=alpha c S T/L^3,
where alpha is the detector efficiency, c is the speed of light, S is the area of the detector, and L^3 is the volume V within which the photon energy is supposed to be uniformly distributed (because of the plane wave approximation).

Note that experiment agrees with p_expert, and contradicts p_kindergarden !

 

[meopemuk]

p_kindergarden=V^3|psi_0|^2,

p_expert=alpha c S T/L^3,

Note that experiment agrees with p_expert, and contradicts p_kindergarden !

The kindergarten formula guarantees that the probability is always between 0 and 1, which is kind of nice. Does the expert formula offer the same guarantee? I don't see it, even in the case when the detector efficiency alpha=1.


In general, I think we are separated by a big philosophical divide. I prefer to think that if there was a sparkle on a luminescent screen, this simply means that an electron (which is a tiny particle) hit exactly at this location.

Your (and M&W) interpretation is different. (Please correct me if I misrepresent your views. I surely misrepresent them, because I don't understand them) You represent the electron as a continuous extended field, which somehow excites atoms in the entire screen. This excitation conspires to produce a sparkle at a single location. The location of the sparkle is sort of unpredictable, because of the chaotic behavior of the atoms in the screen.

These are two completely different views on quantum mechanics and on the origin of quantum uncertainties. They are as different as the corpuscular and wave pictures of the world. I don't know much about real experiments, but I can believe that both these pictures can explain observations. I am working entirely in the corpuscular picture. If you know an experiment where this picture fails completely, this would be a big shock to me. I would like to see the exact reference.

Eugene.

Eugene.

 

[meopemuk]

Yes; he created the ''Feynman Lectures on Physics'' to teach kindergarden kids at university.

There are many other quantum mechanics textbooks teaching the same stuff: Landau & Livgarbagez, Ballentine, to name a few. Are you suggesting to through them away and read only Mandel & Wolf from now on?

Eugene.

 

[A. Neumaier]

There are many other quantum mechanics textbooks teaching the same stuff: Landau & Livgarbagez, Ballentine, to name a few.

It is stated in the beginning as an interpretation aid without proof, and never taken up again in the context of real measurements where the claim would have to be justified. It is very common to make this sort of idealized assumption to get started; but once the formalism is established, this assumption is never used again.

For example, Landau & Lifgarbagez begin in Section 2 of their Vol. 3 with such a statement, but immediately replace it in (2.1) and (3.10) by the more correct version about the interpretation of the expectation value <K> = Psi^* K Psi, where K is an arbitrary observable (linear integral operator) depending on the form and values of the measurement. From then on, only the latter interpretation is used; never the fictitious, idealized introductory remark.

And it cannot be different, since quantum mechanics is used in many situations where the state vectors used in the formalism have no interpretation as a function of position - the whole of quantum information theory and the whole of quantum optics belonging to this category.

 

[A. Neumaier]

The kindergarten formula guarantees that the probability is always between 0 and 1, which is kind of nice. Does the expert formula offer the same guarantee? I don't see it, even in the case when the detector efficiency alpha=1.

The expert formula is derived under the assumption that alpha T is small, since higher order terms are neglected in the derivation. Thus p<<1.

The kindergarden formula is underspecified since you are not saying which volume is the relevant volume. So it says nothing about the actual experimental situation.

In general, I think we are separated by a big philosophical divide. I prefer to think that if there was a sparkle on a luminescent screen, this simply means that an electron (which is a tiny particle) hit exactly at this location.

Yes, we are at opposite sides of the particle -wave spectrum of interpretations.

You represent the electron as a continuous extended field, which somehow excites atoms in the entire screen. This excitation conspires to produce a sparkle at a single location.

Nothing needs to conspire; each excitable atom acts independently of the other, according to the incident intensity.

The location of the sparkle is sort of unpredictable, because of the chaotic behavior of the atoms in the screen.

Yes. This is what the usual semiclassical analysis reveals. Under ordinary circumstances, quantum corrections are tiny and can be neglected.

These are two completely different views on quantum mechanics and on the origin of quantum uncertainties. They are as different as the corpuscular and wave pictures of the world.

Yes, and the modern view is that Nature is governed by quantum field theory, not by Schroedinger equations for many-particle systems.

If you know an experiment where this picture fails completely, this would be a big shock to me.

We are going in circles. I had mentioned spontaneously broken theories, etc.. But you discounted as pure speculation what others found worthy of a Nobel prize.

 

[meopemuk]

The expert formula is derived under the assumption that alpha T is small, since higher order terms are neglected in the derivation. Thus p<<1.

But when I shine one photon on the photographic plate I get the detection probability of almost 1. Is your theory applicable in this case?

The kindergarden formula is underspecified since you are not saying which volume is the relevant volume. So it says nothing about the actual experimental situation.

V^3 is the volume of the particle detector. This could be the volume of the photographic plate, for example.

We are going in circles. I had mentioned spontaneously broken theories, etc.. But you discounted as pure speculation what others found worthy of a Nobel prize.

Well, I've asked you about *experimental* refutation of the probability interpretation of wave functions. You are talking about *theories*, which are not confirmed yet, by the way.

Eugene.

 

[meopemuk]

For example, Landau & Lifgarbagez begin in Section 2 of their Vol. 3 with such a statement, but immediately replace it in (2.1) and (3.10) by the more correct version about the interpretation of the expectation value <K> = Psi^* K Psi, where K is an arbitrary observable (linear integral operator) depending on the form and values of the measurement. From then on, only the latter interpretation is used; never the fictitious, idealized introductory remark.

The probability

$$\int_{V} |\psi(x)|^2 dx$$


can be also regarded as expectation value. In this case the Hermitian operator of observable is the projection on the subset V of the position operator spectrum. This observable can be characterized as a "yes-no experiment" or a question "is the particle present in the volume V?"

Eugene.

 

[meopemuk]

It [the probabilistic interpretation of wave function] is stated in the beginning as an interpretation aid without proof, and never taken up again in the context of real measurements where the claim would have to be justified. It is very common to make this sort of idealized assumption to get started; but once the formalism is established, this assumption is never used again.

The probabilistic interpretation is the centerpiece of the quantum logic approach to quantum mechanics. See chapter 1 in my book. In this approach, quantum mechanics is nothing but a modified (rather generalized) version of the probability theory. The probabilistic interpretation is used everywhere where QM is applied. The most basic things calculated in QM are probabilities of measurements. How are you going to get them without wave function and its probabilistic interpretation? I am very surprised to learn that other opinions exist on this matter.

Eugene.

 

[meopemuk]

Nothing needs to conspire; each excitable atom acts independently of the other, according to the incident intensity.

No, there is a huge conspiracy there. In your theory, before approaching the screen the electron is represented by a continuous field. This means that the electron charge is spread over large area. Nevertheless, the "click" occurs only in one place. I hope you wouldn't deny the fact that after the "click" the full electron charge resides in the neighborhood of the "clicked" atom. So, somehow this charge density cloud has condensed at one point. Not at two points, not at three points - always at one point. What is the explanation of this mysterious behavior?

The idea of independent excitable atoms does not seem to be a good explanation. It seems that all atoms in the screen agree to choose one of them as the "condensation" point and to send their portion of the incident electron's charge density exactly to this point. I call it a conspiracy.

Eugene.

 

[A. Neumaier]

The idea of independent excitable atoms does not seem to be a good explanation. It seems that all atoms in the screen agree to choose one of them as the "condensation" point and to send their portion of the incident electron's charge density exactly to this point. I call it a conspiracy.

If that is a conspiracy, then all the experiments done with Alice and Bob point to the same sort of conspiracy between far away particles in your favored interpretation.
This only proves that this sort of conspiracy is an unavoidable feature of QM.

 

[meopemuk]

If that is a conspiracy, then all the experiments done with Alice and Bob point to the same sort of conspiracy between far away particles in your favored interpretation.
This only proves that this sort of conspiracy is an unavoidable feature of QM.

Are you talking about entanglement?

I think that the conspiracy you are suggesting is much more troubling than entanglement. Your approach needs a real flow of charge density to one point from around the entire area of the detector. This is despite the fact that significant electrostatic repulsion needs to be overcome in order to concentrate the charge in one place. It seems that this flow of charge density can happen equally effectively independent on whether the detector is a conductor or an insulator.

On the other hand, nothing of that sort is needed in the particle interpretation. The "click" happens simply because a point-like electron hits the atom. End of story.

Eugene.

 

[strangerep]

[...]
So, somehow this charge density cloud has condensed at one point.
Not at two points, not at three points - always at one point.

It's misleading to say "always at one point". One must take account of the
time interval in the formula. There's a nonzero probability for more than
one click in a given time interval (assuming the incident beam remains "on").
And if one makes the time interval very small there's a very good
chance that nothing at all happens at any given point on the detector.

Such information involving time intervals is missing if one relies
only on the over-simplified picture that $$|\psi(x)|^2$$ is the
probability for finding the particle at x.

 

[Duderonimous]

This is an argument you don't get to witness first hand too often. Take notes kids.

 

[meopemuk]

It's not "always at one point". One must take account of the time interval
in the formula. There's a nonzero probability for more than one click in
a given time interval (assuming the incident beam remains "on").
And if one makes the time interval very small there's a very good
chance that nothing at all happens at any given point on the detector.

Such information involving time intervals is missing if one relies
only on the over-simplified picture that $$|\psi(x)|^2$$ is the
probability for finding the particle at x.

But I have arranged my experiment so that only one electron was emitted. So, if the detector's efficiency is 1, I must get one and only one click, provided that I've waited long enough to allow the electron to reach the detector.

Eugene.

 

[strangerep]

But I have arranged my experiment so that
only one electron was emitted.

Where did you "arrange" that? I scanned back through the earlier posts,
but didn't see the experimental arrangement described.

So, if the detector's efficiency is 1, I must get one and only one click,
provided that I've waited long enough to allow the electron to reach the detector.

Then you don't have an ensemble, so an interpretation involving probability
becomes problematic. If you do set up many repetitions of the experiment,
it's indistinguishable from a very low intensity beam striking a target.

 

[meopemuk]

Where did you "arrange" that? I scanned back through the earlier posts,
but didn't see the experimental arrangement described.

Sorry if it wasn't clear. In our discussions with Arnold we went back and forth between different threads, so this piece of info could be lost. In my posts I've always assumed emission of particles one-by-one. Here I've intentionally switched from photons (whose one-by-one emission can be problematic) to electrons, which can be easily emitted one-by-one, carry a unit charge and cannot be divided into smaller pieces.

Then you don't have an ensemble, so an interpretation involving probability
becomes problematic. If you do set up many repetitions of the experiment,
it's indistinguishable from a very low intensity beam striking a target.

I shoot one electron and get one dot on the luminescent screen or whatever detector was chosen. This is one member of the ensemble. Then I shoot the second electron and get the second dot in a different place. This is the second member of the ensemble. Then I repeat this procedure as many times as needed in order to form a representative ensemble of measurements. I am sure that each electron has produced a single dot for me. I also see that the distribution of dots on the screen forms a characteristic interference pattern. (I am doing the double-slit experiment here.)

From these observations I make a few conclusions:

1. Electrons are point-like particles, and luminescent sparkles are created by direct hits of incident electrons.

2. For each area of the detector I can measure the probability of it being hit by electrons. This is the number of electrons that have landed in this area divided by the total number of electron emitted.

3. For this double-slit setup I can calculate 1-electron quantum-mechanical wavefunction in the vicinity of the screen. Taking square of this wavefunction and integrating over area I get exactly the same probability as the one measured in 2.

So, standard 1-electron quantum mechanics gives a perfect description of the double-slit experiment in the corpuscular picture.

Eugene.

 

[A. Neumaier]

I think that the conspiracy you are suggesting is much more troubling than entanglement. Your approach needs a real flow of charge density to one point from around the entire area of the detector.

Unobservable fictions of the imagination need not be conserved.

The observable charge flow is the expectation of the current operator defined by the electron field. It satisfies the continuity equation rigorously and hence fully accounts for charge conservation.

 

[meopemuk]

The observable charge flow is the expectation of the current operator defined by the electron field. It satisfies the continuity equation rigorously and hence fully accounts for charge conservation.

I don't dispute the fact that the total charge is conserved in your model. But I don't understand what is the mechanism that forces this charge density to condense to one point against the force of Coulomb repulsion.

Eugene.

 

[A. Neumaier]

I don't dispute the fact that the total charge is conserved in your model. But I don't understand what is the mechanism that forces this charge density to condense to one point against the force of Coulomb repulsion.

The continuity equation not only tells that the total charge is conserved but also that at _every_ point in space and _every_ moment in time the inflow and outflow of charge balance exactly in the ensemble mean.

Nowhere in quantum mechanics one can have better conservation laws for energy and momentum - not even in the quantum mechanics of two nonrelativistic particles.

Thus requiring it of charge is unreasonable (and in fact untestable).

 

[meopemuk]

The continuity equation not only tells that the total charge is conserved but also that at _every_ point in space and _every_ moment in time the inflow and outflow of charge balance exactly in the ensemble mean.

Nowhere in quantum mechanics one can have better conservation laws for energy and momentum - not even in the quantum mechanics of two nonrelativistic particles.

Thus requiring it of charge is unreasonable (and in fact untestable).

Sorry, I don't understand your point here. I was not questioning the validity of continuity equation in your model. I can even close my eyes on non-conservation of energy and momentum. But what is really strange is the ability of the charge density to shrink to a point in some instances. This is reminescent of the QM wave function collapse, but much more troubling, because in your case the collapsing thing is not the imaginary probability density amplitude (as in QM), but real physical charge density.

Eugene.

 

[A. Neumaier]

Sorry, I don't understand your point here. I was not questioning the validity of continuity equation in your model.

You raised the objection that charge must flow to the firing point, and I replied that the continuity equation proves that nothing can be wrong with the flow of charge. (Tiny flows of charge happen all the time in a macroscopic body.)

I can even close my eyes on non-conservation of energy and momentum.

You have to close your eye on non-conservation of momentum already if you consider a particle picture of a double slit experiment!

But what is really strange is the ability of the charge density to shrink to a point in some instances. This is reminiscent of the QM wave function collapse, but much more troubling, because in your case the collapsing thing is not the imaginary probability density amplitude (as in QM), but real physical charge density.

The charge flows to the point where it is needed. It doesn't need to contract miraculously and instantly from everywhere. The fired electron removes charge locally, and this local deficiency is corrected for by inflow of a little bit of charge density from the neighborhood. (It is not really different from the flow of water in a bucket that has a little hole at the bottom from which drops leak out at random times, while the compensating inflow happens steadily at some other place.) The covariance and locality of quantum field theory ensures that nothing happens instantly.

 

[meopemuk]

The charge flows to the point where it is needed. It doesn't need to contract miraculously and instantly from everywhere. The fired electron removes charge locally, and this local deficiency is corrected for by inflow of a little bit of charge density from the neighborhood. (It is not really different from the flow of water in a bucket that has a little hole at the bottom from which drops leak out at random times, while the compensating inflow happens steadily at some other place.) The covariance and locality of quantum field theory ensures that nothing happens instantly.

I don't find your explanation convincing. I don't know of any physical force that would suck the whole widely distributed charge density to a single point. Charge densities normally have the tendency of self-repulsion. Here we have an example of a completely opposite effect. What kind of Maxwell equation can be written to demonstrate such an abnormal behavior?

Eugene.

 

[A. Neumaier]

I don't find your explanation convincing. I don't know of any physical force that would suck the whole widely distributed charge density to a single point. Charge densities normally have the tendency of self-repulsion. Here we have an example of a completely opposite effect. What kind of Maxwell equation can be written to demonstrate such an abnormal behavior?

There is a nontrivial and very rugged charge distribution already in the detector before anything arrives at it. Even in a single molecule like water you have a nontrivial charge distribution - one of the reasons water behaves as it does. This charge distribution changes and adapts continuously while the arriving wave reaches the body, and it keeps doing so during the time it takes to emit the electron.

Thus everything happens continuously, in conformance with the continuity equation. Nothing needs to be sucked into a point - the emitted electron is itself a moving charge distribution, radially expanding in its rest frame.

 

[meopemuk]

Nothing needs to be sucked into a point - the emitted electron is itself a moving charge distribution, radially expanding in its rest frame.

According to you, before the measurement the electron had the form of a charge density cloud extending over the range of several centimeters. After the measurement we see that one CCD pixel has its charge changed by -e, while all other pixels stay with the same (=0) charge. This looks like "sucked into a point" to me. How this behavior can be achieved by a "moving charge distribution, radially expanding in its rest frame"?

Eugene.