Dear Sofia,

the continuity holds when the Schrödinger equation holds, i.e. when the time evolution is unitary. A measurement of an observable leads to a non-unitary change of the state vector of a system, but to the projection of the state onto some eigen-space of the observable under consideratuion. This means the continuity equation does not hold at the instance of measurement.

So the short answer to your question in the title: no, it isn't.

Best regards

Oliver

Dear Oliver,

Your answer is good - thanks for it.

Though, I want to challenge it with a question. As I see at your Info page that you are skilled in relativity, can you tell me in which frame of reference to write the Schrödinger equation?

You see, as I said in my question the two wave-packets ψL and ψR fly away from one another. Then, I can add a separate frame of reference to each one of them, FL and FR. If I write the Schrödinger equation according to FL, then, by the time of the measurement, tL,0, the center of the wave-packet ψR should have arrived at some position rL,0. Therefore at tL,0 and rL,0 the continuity equation of ψR should become invalid - more simply saying, ψR should disappear. However, if I write the Schrödinger equation according to FR, then, by the time of the measurement tR,0, the center of the wave-packet ψR should have arrived at some position rR,0. The 4-vector (tR,0, rR,0) is not the Lorentz transformation of the 4-vector (tL,0, rL,0).

Let me anticipate and say that some people believe that there exists a preferred frame. So, according to them, the Schrödinger equation which anyway is non-relativistic, should be written according to the preferred frame. But there is no physical phenomenon to confirm the existence of some preferred frame. Even the results of the measurements of quantum objects, are frame independent.

With best regards,

Sofia

Dear Sofia,

I think I'm not really sure what you're aiming at. Let me ask to clarify: are you describing a single-particle or a two-particle system? I was assuming your Gedankenexperiment describes a 1-particle system where the wave-function somehow gets split into two localized wave packets by means of a beam splitter or similar.

If so, then there is no continuity equation for either the one wave-packet or the other wave-packet, but only for the total wave function, consisting of 2 separately peaked partial wave packets, totalling up to a complete 1-particle wave package. So no matter if the particle is detected on the left or on the right side of the system, the whole wave-function collapses either way: ψ_L and ψ_R. In fact, it is misleading to introduce ψ_L and ψ_R in the first place, because in the end these are but parts of one and only one wave function, albeit in a possibly slightly idealized form.

And of course there is no preferred reference frame, neither in Galilei relativity (i.e. in this case), nor in relativity. But: transforming the state resp. the wavefunction from one reference frame to another needs to be done correctly, because one has to take into consideration a non-trivial, transformation-specific phase factor. This is due to the fact that there is no unitary representation of the Galilei group, only a projective one.

I am not sure if any of this answers your question, but then it's due to the fact I am still trying to comprehend where the catch is, my apologies in advance...!

Best regards

Oliver

Dear Oliver,

Please do not appologize. My questions are known as difficilt. They are part of my research, which goes very deeply into the issue what in fact the QM equations tell us.

My question refers to a wave-function describing single particles, see formula (1).

Now, is it clear to you why from the continuity equation (2) of the total wave-function, we get the formula (3) ? I will explain. Look please at (3), it does not contain the product ψL(r, t)ψR(r, t) (or derivatives of it) because if r sweeps the half-space on the left, the function ψR(r, t) and its derivatives vanish. Symmetrically, if r sweeps half-space on the right, ψL(r, t) and its derivatives vanish.

For the same reason we get a splitting of eq. (3) into two separate continuity equations, (4) and (5), because when the terms on the RHS of (3) are non-null, those on the LHS are null, and vice-versa. Please tell me, is my explanation clear?

Next: I indeed question the idea that if the detector CL clicks, the wave-packet ψR disappears. By the way, this idea is alien to the QM formalism, which follows the wave-function evolution as long as it is unitary. So, I try to attack the collapse idea with arguments of the quantum formalism. One argument is that it is not clear where in the space happens the disappearence of ψR. No matter which frame of coordinates we adopt, this disappearence should happen when CL clicks, i.e. at tL,0 according to the frame FL, or tR,0 according to the frame FR. But the place in space where ψR vanishes, i.e. rL,0 according to FL and rR,0 according to FR, is problematic, because the 4-vector (tR,0, rR,0) is not the Lorentz transformed of the 4-vector (tL,0, rL,0). In short, the event of disappearence of ψR is ambiguous.

Please tell me whether I explained my doubt clearly.

Best regards,

Sofia

Dear Sofia,

I think I am beginning to see what you are aiming at. Still I need some more thinking into it.

"My question refers to a wave-function describing single particles, see formula (1).

Now, is it clear to you why from the continuity equation (2) of the total wave-function, we get the formula (3) ? I will explain. Look please at (3), it does not contain the product ψL(r, t)ψR(r, t) (or derivatives of it) because if r sweeps the half-space on the left, the function ψR(r, t) and its derivatives vanish. Symmetrically, if r sweeps half-space on the right, ψL(r, t) and its derivatives vanish.

For the same reason we get a splitting of eq. (3) into two separate continuity equations, (4) and (5), because when the terms on the RHS of (3) are non-null, those on the LHS are null, and vice-versa. Please tell me, is my explanation clear?"

I think I understand your line of thought so far. But I am still having severe problems in understanding the point.

1.) What about dispersion (in the normal sense)?

What I do not see right now, however, is the conclusion that you get effectively 2 continuity equations: one for the left half-space, and one for the right half-space. Your argument "sweeping to the far left/right --> ergo LHS/RHS vanishes" only applies to one snapshot in time. What about dispersion of free particle wave packages due to unitary time evolution? Is the particle not free? But then how is the "free dispersion" eliminated? You know that only the harmonic oscillator potential allows for "dispersion-free" wave packages, so to speak.

I might not be able to precisely point to the flaw in the logic here, or even grasp what the actual statement is, but as I mentioned already anyway: the continuity equation is not compatible with the measurement process.

2.) Are we talking about Bohmian mechanics?

Are you somehow trying to follow the path of Bohm's interpretation of QM? I somehow have the feeling... but if so, then I must admit I may not be the ultimate expert on its formulation.

3.) Are we talking about quantum computing?

What I am somehow see though is you use operators like AL that model some circuitry element as in quantum computing, but apply it not to a state vector, but to some part of the wavefunction only. This seems highly questionable to me. Is this correct quantum-theoretical model building? In addition, I think you are somehow aiming at making a statement on quantum dynamics which quantum computing normally tries not to address wherever possible. Quantum logic formalism usually does not incorporate quantum dynamics.

4.) Why do you mention Lorentz transformations?

Also we are looking at non-relativistic QM, thus your mentioning of the Lorentz transformations and 4-vectors is also unclear to me.

5.) What is the actual statement?

So all in all I am still struggling to get the actual gist of the statement, and still I duly apologize. Maybe I could ask you this question quite directly: what is your actual statement?

Best regards

Oliver

Dear Oliver,

Thanks for your effort to understand my problem. I am trying then to answer your questions:

"1.) What about dispersion (in the normal sense)? . . . . . Your argument "sweeping to the far left/right --> ergo LHS/RHS vanishes" only applies to one snapshot in time. What about dispersion of free particle wave packages due to unitary time evolution?"

Please tell me, do you usually read experiments with down-conversion photons? The wave-packets of such photons have usually a very small dispersion. If you split the wave-packet of such a photon with a beam-splitter, and send one copy to the left and the other to the right, ψL and ψR, the widenning of the wave-packets due to dispersion is negligible in comparison to the distance the two wave-packets travel and separate from one another. Don't forget that these wave-packets travel with the velocity of light.

Therefore, the dispersion won't keep ψL and ψR overlapping. When r sweeps the half-space on the left, ψR and its derivatives are null, s.t. the RHS of eq. (3) vanishes. Thus, you get eq. (4). Symmetrically, when r sweeps the half-space on the right, ψL and its derivatives are null, s.t. the LHS of eq. (3) vanishes and you get eq. (5).

". . . . the continuity equation is not compatible with the measurement process."

I'd like first to be sure that you understand my mathematics - QM is rigorous science. As to your other questions, I don't deal here with Bohm's mechanics or quantum computing, I deal with the general quantum formalism. But, let's proceed step by step. So, please tell me whether so far I was clear.

Best regards,

Sofia

I agree (roughly) with Oliver. The measurement process colapse is more of a mental transformation you do with the wave function,

not a physical process as per the continuity equation. So intricate experimental thought experiments will not really help with this.

regards, JW

Dear Sofia,

I am afraid I am not getting clearer on this. Here's why:

"Please tell me, do you usually read experiments with down-conversion photons? The wave-packets of such photons have usually a very small dispersion. If you split the wave-packet of such a photon with a beam-splitter, and send one copy to the left and the other to the right, ψL and ψR, the widenning of the wave-packets due to dispersion is negligible in comparison to the distance the two wave-packets travel and separate from one another. Don't forget that these wave-packets travel with the velocity of light."

I am afraid I am not an experimentalist, and surely not versed in the experiments you mention. What strikes me however that you speak of the wave-function of a photon. I must admit I do not know whether there is any justification in introducing a photon wave function, and how this should exactly been done, but even if there is, this is far from being "standard QM".

In any case, even if there is a way to introduce a photon wave function, it does not strike me as an ideal case for asking fundamental questions about QM in its "rigorous" sense as you are saying. The correct formulation of any relativistic quantum theory is quantum field theory, mainly due to the fact that the conservation of particle number is no more, as is the case with photons. So, if you want to remain rigorous, stretching the non-relativistic Schrödinger equation and its implications to photons of all entities, is by no means compliant with that will. Tell me: are you familiar with the limits of relativistic QM, formulated in a "Schrödinger" form?

So I am still struggling with the ultimate purpose of this exercise, to be honest. Your mathematics above is not overly complicated, so I have no trouble following it. What I do struggle with is the whole formalism you apply, and especially the justification of it.

Also, I am still not understanding what you are eventually aiming at. Why don't you just try to make a case without photons, using "standard" 1-particle wave-packets in non-relativistic QM, and actually try to make a concrete point?

Best regards

Oliver

Dear Oliver,

First of all, I noticed that you asked why I speak of Lorentz transformation while I speak of QM. Yes, you are right, the expression of the current of probability in (3), (4) and (5) is not correct for relativistic QM and for photons. So, I changed it - please see the modified form of the equations (3), (4) and (5). I hope that the change relaxes your objection.

Now, physicists currently operate with the wave-packet of a photon in QM. The photon is a wave-packet of a certain length equal with the coherence length. Why do you have a problem with this? If you prefer the formalism of the quantum optics, then, eq. (1) becomes

(1) |ψ(r, t)> = 2-½[ |1>ψL(r, t) + |1>ψR(r, t)].

where |1> is the standard notation of a Fock state, while ψL(r, t) and ψR(r, t) describe the space-time form of the wave-packet. You don't need QFT when you work with a fix number of particles, and in fact, the proper formalism for photons is quantum optics (QO). But, again, I work with a fixed number of photons, one photon, s.t. I don't need QFT or QO.

If you prefer a mass-possessing particle instead of a photon, a heavy particle, e.g. an atom, its wave-function would also behave as in the equation (1), and also would have a small dispersion.

The ultimate purpose is of my question is to show that claiming the disappearence of ψR as a consequence of the detection of ψL suffers from ambiguity. This disappearence has to take place when the detector CL clicks. But at which point in the 4D space-time happens the disappearence, is unlear, because, I repeat, the 4-vector (tR,0, rR,0) is not the Lorentz transformed of the 4-vector (tL,0, rL,0). You know what is a Lorentz transformation from one frame to another.

Moreover, we have problems with the continuity equation of ψR . If this wave-packet vanishes all of the sudden, an extraordinary big flux has to burst. There is no experimental report of such a phenomenon.

Juan, my friend,

You do too much high phylosophy for my mind. I don't understand it. "The measurement process colapse is more of a mental transformation". What you mean? We try to understand what happens with ψR when the counter CL clicks. I mentioned in my question two problems about the claim that ψR disappears: the relativistic incompatibility, and the continuity equation. The latter does not allow ψR to disappear suddenly, unless a huge flux from ψR outwards, appears. No experiment reported such a flux.

As you deal with the phylosophy of science, you can understand my way of thinking: the collapse postulate is not part of the quantum formalism, it was added by von Neumann in absence of an appropriate theory of measurement of a quantum object with a classical apparatus. This postulate says that at the interaction with the classical detector, part of the wave-function disappears. The quantum formalism does not admit such a non-unitary behavior. So, what I thought was that since this postulate is at odds with the formalism, it is likely to come to a contradiction with the formalism. This is the background idea of my question.

You see, despite the fact that the quantum formalism does not answer satisfactorily some questions, it seems to refuse to be supplemented with additional assumptions (meant to solve those questions). For instance, both Bohm's mechanics and the GRW interpretation, add new assumptions. Both these interpretations come to contradictions with the formalism. I am writing an article on this issue, and this is why I open discussions.

So, the formalism, although leaving some questions not answered, seems to be a closed formalism, in the sense that it does not permit supplements. Interesting situation, isn't it?

With kind regards

My dear Sofia,

You said:

" the collapse postulate is not part of the quantum formalism, it was added by von Neumann in absence of an appropriate theory of measurement of a quantum object with a classical apparatus. This postulate says that at the interaction with the classical detector, part of the wave-function disappears. "

What is the meaning of probability to find the particle in location Pl if you, in fact, found it in location Pr?

In other words, what is the meaning of the wave-function after you really detect the particle?

ِI see that you deal with the wave-function as something real and independent from our knowledge of particle position.

I think after we detect the particle the all old wave-function will become meaningless.

With best regards.

It means that since you either do not have a detailed theory of measurement, or have a rather complicate one,

you just assume that when the measurement is made you no longer have a wave function. That is your mental transformation.

Regards, JW

Juan,

You have to read Feynman and Hibbs "The Quantum Mechanics and Path Integral" - the classical limit. The fact that one does not have anymore a wavefunction after the measurement, is a property of the macroscopic objects.

You measure the quantum system with a macroscopic object, and your quantum system becomes unseparable and even undistinguishable part of that macroscopic object. The result of the measurement is indicated by the final macroscopic state of the macroscopic object. Since the macroscopic object cannot be in a quantum superposition of states, it settles itself, quasi-randomly, on some macroscopic state. This state corresponds to one of the eigenvalues of the operator that we measure on the quantum system. I say quasi-randomly because the macroscopic object does not pick the macroscopic state arbitrarily. The probabilities provided by the wave-function for the eigenvalues have a hand in the choice.

This is the qualitative description of the measurement. For processing this description into a mathematical one, somebody younger than I is needed - or somebody who has more time than I.

However, my problem in this thread is what happens with ψR when CL clicks.

My Dear Sofia,

You said:

" However, my problem in this thread is what happens with ψR when CL clicks. "

From a mathematical point of view ψR and ψL is a one wave function, not two or not three or ... (only these functions give us the values of one function in different locations that we separate the space based on left and right ...)

for example : y = x^3 + x^2 + x + 5

we can said : y= (x^3) + ( x^2 + x + 5) = ( x^3 + x^2) +( x + 5 ) = ... etc..

this separation is from us, the function sees itself as one function that has one physical mean, the separation is in our mind only.

With best regards.

Dear Mazen,

ψR is on the right and CL is on the left. For instance ψR may travel in America and CL may be in Europe. ψR is a quantum object while CL a classical object, constructed for detecting.

As an analogy you have eyes, on the head, and legs on the lower part of the body. Both belong, of course, to the same body, however, with the eyes you see and with the legs you walk. Each organ has a different function and position. You can't re-arrange the functions of these two organs, e.g. see with an eye and a leg, and walk with the other eye and the other leg.

My Dear Sofia,

Sorry I mean ψL not CL ,

so I want to say that ψL and ψR belong to a one function.

then what happens for ψR is exactly what happens for ψL .

With best regards.

1，Insert a basic physical problem

***

The system is isothermal and exchanges heat with a large heat source to keep the temperature constant. Only volume changes are discussed.

The problem here is that the actual system is balanced and the second law of thermodynamics judges it to be unbalanced.