I Clarification of the postulates of QM

[stevendaryl]

Well, then you can call putting a beam dump at all partial beams of a Stern-Gerlach apparatus except one with a definite spin component in direction of the B-field is a "collapse".

You're saying if I believe X, then I have to believe Y. Well, I believe X and I don't believe Y. So there.

The collapse hypothesis is simply the rule that if I have a system in state \psi and I perform a measurement M_1 and get result r_1, then afterwards, the appropriate state to use for subsequent measurements is \Pi_{M_1, r_1} \psi, the result of projecting \psi onto the subspace of those wave functions that are eigenstates of M_1 with eigenvalue r_1. Nothing is changed by calling the measurement a "preparation" instead of a "measurement". That seems like a completely ridiculous argument. You're using the collapse hypothesis at the same time you're denying it.

 

[stevendaryl]

Before A's measurement the system has been prepared in the given state. That's all. Since A knows this she updates her knowledge about B's measurement (no matter whether he measures before or after her). That happens locally at her place in her brain, but nothing happens instantaneously to B's particle. As I said countless times before, the 100% correlation between A's and B's outcomes of measurements is due to the state preparation not due to a mutual influence of A's and B's measurements. Great effort has been put into the "loop-hole free" setup of these measurements to demonstrate precisely this! This particular loophole is excluded by making sure that the measurement events (registrations of particles) at A's and B's place are truly space-like separated. Is there any reason for doubts that these experimental setups are somehow flawed, and the loop hole is still there? Do you think that there are still hidden correlations built up by faster-than-light influence of the apparati at the far distant places? Well, then you can never close that loophole, but I'd invoke Occam's razor here to say that it's the most simple explanation.

It seems to me that there is no substantial difference between your way of interpreting QM and the way that uses Von Neumann collapse. You just don't like to use that word.

 

[vanhees71]

You're saying if I believe X, then I have to believe Y. Well, I believe X and I don't believe Y. So there.

The collapse hypothesis is simply the rule that if I have a system in state \psi and I perform a measurement M_1 and get result r_1, then afterwards, the appropriate state to use for subsequent measurements is \Pi_{M_1, r_1} \psi, the result of projecting \psi onto the subspace of those wave functions that are eigenstates of M_1 with eigenvalue r_1. Nothing is changed by calling the measurement a "preparation" instead of a "measurement". That seems like a completely ridiculous argument. You're using the collapse hypothesis at the same time you're denying it.

Well, in the above example your rule is obviously not true, because either the particle runs further without much happening to it or it's absorbed by a wall. In the latter case it's far from the state predicted to be assiciated with it by your collapse hypothesis.

It's also clear that measuring something is not necessary also preparing the measured object. E.g., usually a photon gets absorbed by the detector, and it's not prepared for further experiments.

 

[stevendaryl]

It seems to me that there is no substantial difference between your way of interpreting QM and the way that uses Von Neumann collapse. You just don't like to use that word.

As I said, before Alice's measurement, Bob's particle is not in a state of having a definite value for spin in the z-direction. Afterwards, it is in the state of having a definite value for spin in the z-direction. How can you say that's not a change in the state of Bob's particle? The only way (it seems to me) is to deny that there is such a thing as "the state of Bob's particle".

 

[stevendaryl]

Well, in the above example your rule is obviously not true, because either the particle runs further without much happening to it or it's absorbed by a wall. In the latter case it's far from the state predicted to be assiciated with it by your collapse hypothesis.

Yes, measurement processes typically are destructive, so further measurements are impossible. But you can salvage that by considering composite systems:

|\Psi\rangle = \sum_{\alpha j} C_{\alpha j} |\phi_\alpha\rangle |\chi_\beta\rangle

A destructive measurement of a property of one subsystem would still allow followup measurements to be performed on the other subsystem.

 

[vanhees71]

For Bob his particle is always described by the state $\hat{\rho}_B=1/2 \mathbb{1}$. He just has a beam of unpolarized particles. This finding is unchanged no matter whether A made her measurement before B made his measurement or not.

 

[stevendaryl]

For Bob his particle is always described by the state $\hat{\rho}_B=1/2 \mathbb{1}$. He just has a beam of unpolarized particles. This finding is unchanged no matter whether A made her measurement before B made his measurement or not.

But Alice knows better. If she has measured her particle to have spin-up in the z-direction, then she knows for sure that Bob will measure spin-down in the z-direction. So Bob's description of his particle using $\hat{\rho}_B=1/2 \mathbb{1}$ is not the most accurate, for predicting Bob's future measurements.

 

[vanhees71]

It is. Why should it not be accurate? You can of course find the correlations described by the entangled state by comparing A's and B's measurement protocols with precise enough time stamps of their particle registrations (and provided the particles come well-separated enough to the detectors, so that you precisely know which two measurement events at A's and B's place belong to precisely the same entangled particle pair).

 

[stevendaryl]

It is. Why should it not be accurate?

Because, that density matrix is consistent with either result for Bob's measurement of spin along the z-axis, while Alice knows that he can only get spin-down along the z-axis.

 

[vanhees71]

But that doesn't make the assignment of the state $\hat{\rho}_B$ invalid. It still describes Bob's knowledge about the spin state of his particle correctly, no matter what Alice knows about what he will find. Of course, if A tells him what she has measured, B also knows what he will measure, but A can send her message only with at most the speed of light to B. Thus no instantaneous "collapse" happens from B's point of view but just information provided by A separately from the system measured, but I think we just turn in the usual circles here :-(.

 

[stevendaryl]

But that doesn't make the assignment of the state $\hat{\rho}_B$ invalid.

It's a perfectly valid description of Bob's knowledge about his particle, but it's not an accurate description of the particle itself. Saying it's spin-down in the z-direction is more accurate.

 

[ShayanJ]

But that doesn't make the assignment of the state $\hat{\rho}_B$ invalid. It still describes Bob's knowledge about the spin state of his particle correctly, no matter what Alice knows about what he will find. Of course, if A tells him what she has measured, B also knows what he will measure, but A can send her message only with at most the speed of light to B. Thus no instantaneous "collapse" happens from B's point of view but just information provided by A separately from the system measured, but I think we just turn in the usual circles here :-(.

Let me see if I get it right. You're saying that in B's view, the state of his particle is $\rho_B= \frac 1 2 \hat 1$ no matter before or after A's measurement. But in A's view, the state of B's particle is $\rho_B=\frac 1 2 \hat 1$ before A's measurement and either $|S_z;+ \rangle$ or $|S_z; - \rangle$(depending on A's result) after A's measurement. But this means there is a collapse, its just subjective!(Which also means you think the state vector only describes the observer's knowledge about the system, and not anything objective.)

 

[stevendaryl]

It's a perfectly valid description of Bob's knowledge about his particle, but it's not an accurate description of the particle itself. Saying it's spin-down in the z-direction is more accurate.

Let me give an analogy. If Bob is handed a deck of 52 cards, and he's asked to describe his probability of getting an ace of spades if he selects one at random, he might say: My chances are 1.9%

But if beforehand, Alice swapped the ace of spades for a Joker, then she knows that Bob has 0% chance of picking the ace of spades.

Bob's prediction was as accurate as he could possibly be, given his knowledge, but Alice's prediction is more accurate.

 

[vanhees71]

Sure, you can have such subjective ideas, but it has nothing to do with the physics of the two particles measured by A and B. I can as well say that for me the state of the German lottery drawing machine instantaneously collapses every Saturday as soon as the numbers are drawn, but that happens for me only if I take notice of the outcome of the experiment. It doesn't do anything on what's going on at the drawing of the actual numbers.

 

[ShayanJ]

Sure, you can have such subjective ideas, but it has nothing to do with the physics of the two particles measured by A and B. I can as well say that for me the state of the German lottery drawing machine instantaneously collapses every Saturday as soon as the numbers are drawn, but that happens for me only if I take notice of the outcome of the experiment. It doesn't do anything on what's going on at the drawing of the actual numbers.

But this inevitably means that you consider the state vector to be a subjective concept only describing the observer's knowledge about the system! This doesn't seem to be minimalistic in that one may ask "what's actually going on down there?", unless by minimalistic, you mean just not asking such questions.

Also, you're a proponent of the ensemble interpretation, so you shouldn't be able to talk about experiments not involving an ensemble. But now you're talking about a single pair of particles!

 

[vanhees71]

Let me give an analogy. If Bob is handed a deck of 52 cards, and he's asked to describe his probability of getting an ace of spades if he selects one at random, he might say: My chances are 1.9%

But if beforehand, Alice swapped the ace of spades for a Joker, then she knows that Bob has 0% chance of picking the ace of spades.

Bob's prediction was as accurate as he could possibly be, given his knowledge, but Alice's prediction is more accurate.

Well, sure. But what does this tell us about the QT example we are discussing? Of course, you'll put different probability distributions on the same situation if you have different knowledge about the system. That's very trivial, but in this case A and B have precisely the same knowledge, namely that each of them measure the spin of one of two spin-entangled particles. So the only thing both know is that each of them just has an unpolarized particle. Only bringing both measurements together in a way that you can compare what was measured for each of the particles prepared in the entangled two-particle state reveals the correlation implied by the entanglement. Of course, you can check it only with some significance at a large enough ensemble of so prepared two-particle states.

 

[stevendaryl]

Well, sure. But what does this tell us about the QT example we are discussing?

It seems perfectly analogous. Bob thinks he has a 50/50 chance of getting spin-up or spin-down, but Alice knows that he has 0% of getting spin-up.

 

[vanhees71]

But this inevitably means that you consider the state vector to be a subjective concept only describing the observer's knowledge about the system! This doesn't seem to be minimalistic in that one may ask "what's actually going on down there?", unless by minimalistic, you mean just not asking such questions.

Also, you're a proponent of the ensemble interpretation, so you shouldn't be able to talk about a experiments not involving an ensemble. But now you're talking about a single pair of particles!

No. The state is an objective description of the system, which is prepared in this state. Of course, any probabilistic statement can be checked only at an ensemble.

 

[vanhees71]

It seems perfectly analogous. Bob thinks he has a 50/50 chance of getting spin-up or spin-down, but Alice knows that he has 0% of getting spin-up.

So?

 

[ShayanJ]

No. The state is an objective description of the system, which is prepared in this state. Of course, any probabilistic statement can be checked only at an ensemble.

So before A's measurement, her spin is in the state $\frac 1 2 \hat 1$. Then she measures her spin and gets the result +, so she knows that her spin is pointing upward. But she's actually wrong and her measurement is meaningless because her spin is still in the state $\frac 1 2 \hat 1$ and not $|S_z;+\rangle$? This doesn't make sense!

 

[atyy]

So before A's measurement, her spin is in the state $\frac 1 2 \hat 1$. Then she measures her spin and gets the result +, so she knows that her spin is pointing upward. But she's actually wrong and her measurement is meaningless because her spin is still in the state $\frac 1 2 \hat 1$ and not $|S_z;+\rangle$? This doesn't make sense!

vanhees71 is a secret many-worlder, since that is the interpretation in which the state is objective, and neither hidden variables nor collapse :)

In fact, it is because he thinks the state is objective that he objects to collapse. If the state were subjective or just FAPP, there would be no problems with collapse.

Of course, in a sense, if MWI works it is certainly the minimal interpretation!

 

[martinbn]

To say that measurement "prepares" the state is the same thing as collapse.

Not the way I see it. Collapse asumes that the state decribes the system and that before and after the measurement you have the same system but in a different state. But if the state describes the set of equivalently prepatered systems (not an individual one), then after the measurement you have a diffrent system, not the same with collapsed state. You may say that this is just words, but that's what interpretations are.

 

[atyy]

Not the way I see it. Collapse asumes that the state decribes the system and that before and after the measurement you have the same system but in a different state. But if the state describes the set of equivalently prepatered systems (not an individual one), then after the measurement you have a diffrent system, not the same with collapsed state. You may say that this is just words, but that's what interpretations are.

The point is that collapse allows you to calculate the conditional probability of the second measurement outcome conditioned on a sub-ensemble of outcomes from the the first measurement.

In calculating the conditional probability, you need the state of a sub-ensemble.

The collapse rule assigns the state of a sub-ensemble. So collapse is still needed within the ensemble interpretation.

 

[martinbn]

The point is that collapse allows you to calculate the conditional probability of the second measurement outcome conditioned on a sub-ensemble of outcomes from the the first measurement.

In calculating the conditional probability, you need the state of a sub-ensemble.

The collapse rule assigns the state of a sub-ensemble. So collapse is still needed within the ensemble interpretation.

I know all this, but my point is that what you call the second measurment of the sub-ensemble is in fact the first measurment of a new ensemble. Before the measurment you didn't have that ensemble i.e. you cannot talk about a sub-ensemble. So what allows me to calculate conditional probability is the preparation of a new ensemble.

 

[atyy]

I know all this, but my point is that what you call the second measurment of the sub-ensemble is in fact the first measurment of a new ensemble. Before the measurment you didn't have that ensemble i.e. you cannot talk about a sub-ensemble. So what allows me to calculate conditional probability is the preparation of a new ensemble.

Sure, collapse is a form of state preparation. The point of collapse is that it links measurement and state preparation. Without the collapse, you do not have that link.

 

[vanhees71]

So before A's measurement, her spin is in the state $\frac 1 2 \hat 1$. Then she measures her spin and gets the result +, so she knows that her spin is pointing upward. But she's actually wrong and her measurement is meaningless because her spin is still in the state $\frac 1 2 \hat 1$ and not $|S_z;+\rangle$? This doesn't make sense!

If you have a usual polarization filter and the detection of the particle doesn't destroy the corresponding spin state then of course A will associate the state represented by $|\sigma_z=1/2 \rangle$ with it. What else? I only say that this is an association of the state that Alice does to her particle and I don't think that anything instantaneous happens to Bob's particle, because this contradicts the dynamics of relativistic quantum fields, which by definition is microcausal, i.e., local field operators that represent observables commute at space-like distances and thus the local interaction of the particle with A's measurement apparatus doesn't change immediately B's particle in any way. That's math and not subject to any interpretational issue!

 

[ShayanJ]

If you have a usual polarization filter and the detection of the particle doesn't destroy the corresponding spin state then of course A will associate the state represented by $|\sigma_z=1/2 \rangle$ with it. What else? I only say that this is an association of the state that Alice does to her particle and I don't think that anything instantaneous happens to Bob's particle, because this contradicts the dynamics of relativistic quantum fields, which by definition is microcausal, i.e., local field operators that represent observables commute at space-like distances and thus the local interaction of the particle with A's measurement apparatus doesn't change immediately B's particle in any way. That's math and not subject to any interpretational issue!

But saying "this is an association of the state that Alice does to her particle", means you think the wave-function is subjective and only describes the observer's knowledge about the system!

 

[vanhees71]

Sure, what else should it describe?

 

[ShayanJ]

Sure, what else should it describe?

Its just that in post #68, you insisted that "the state is an objective description of the system".

 

[stevendaryl]

Sure, what else should it describe?

The wave function of an electron might describe facts about an electron, maybe?

 

[vanhees71]

Sure, it describes objective facts about the electron; objective facts the physicist knows due to the (equivalence class of) preparation procedures defining the state. I think, all this is a pretty empty debate about semantics, as usual in such discussions about "interpretation".

I think it's pretty clear that A in our example knows something different than B, because she has done a "preparation procedure" with her particle by determining its spin component. Due to the $\sigma_z$-entangled state the two-particle system was prepared before, she knows also B's $\sigma_z$, but B doesn't know it, before he has measured it and just finds with 50% probability the one or the other outcome. So everything is consistent without any necessity to envoke "spooky action at a distance", which is indeed not implemented in the dynamics of the theory by construction, since we use a relativistic, local, microcausal QFT.

There'd be only a problem with this "interpretation" (which is just what the formalism, particularly Born's rule tells us, nothing else, and in this sense it's "minimal") if it would make a difference for B whether or note A measures her particle's $\sigma_z$ first or not, but it doesn't. So it's all consistent.

 

[stevendaryl]

Sure, it describes objective facts about the electron; objective facts the physicist knows due to the (equivalence class of) preparation procedures defining the state. I think, all this is a pretty empty debate about semantics, as usual in such discussions about "interpretation".

It's an empty debate because there is no real distinction between your position and those who believe that observation collapses the wave function.

What is a "preparation procedure"? Can you define it without invoking either the macroscopic/microscopic distinction, or the observer/observed distinction?

 

[stevendaryl]

So everything is consistent without any necessity to envoke "spooky action at a distance", which is indeed not implemented in the dynamics of the theory by construction, since we use a relativistic, local, microcausal QFT.

I don't see how the dynamics of QFT is relevant. There are two aspects to quantum theory: (1) Smooth evolution of the wave function (or smooth evolution of the field operators, in QFT), and (2) the Born interpretation of quantum amplitudes.

The issue is whether the combination is local, not whether the smooth evolution part is local.

 

[atyy]

Sure, it describes objective facts about the electron; objective facts the physicist knows due to the (equivalence class of) preparation procedures defining the state. I think, all this is a pretty empty debate about semantics, as usual in such discussions about "interpretation".

I think it's pretty clear that A in our example knows something different than B, because she has done a "preparation procedure" with her particle by determining its spin component. Due to the $\sigma_z$-entangled state the two-particle system was prepared before, she knows also B's $\sigma_z$, but B doesn't know it, before he has measured it and just finds with 50% probability the one or the other outcome. So everything is consistent without any necessity to envoke "spooky action at a distance", which is indeed not implemented in the dynamics of the theory by construction, since we use a relativistic, local, microcausal QFT.

There'd be only a problem with this "interpretation" (which is just what the formalism, particularly Born's rule tells us, nothing else, and in this sense it's "minimal") if it would make a difference for B whether or note A measures her particle's $\sigma_z$ first or not, but it doesn't. So it's all consistent.

But if collapse is a preparation, and the preparation prepares an objective state, then collapse would seem to be "objective" in your terminology.

 

[stevendaryl]

But if collapse is a preparation, and the preparation prepares an objective state, then collapse would seem to be "objective" in your terminology.

It occurs to me that there is (perhaps) a subtle distinction between measurement and preparation. For example, when you send a stream of electrons through a Stern-Gerlach device, those with spin-up are deflected one way and those with spin-down are deflected the other way. You haven't measured the spin of any electron, though.

Then you can send those electrons deflected in one direction to go on to a second Stern-Gerlach device, with a different orientation. Those electrons split into two groups, as well.

Eventually, you will (presumably) do a real measurement, by checking for the presence/absence of an electron. But all the selection prior to this doesn't involve measurement, and presumably doesn't collapse the wave function.

So I think that vanhees71 might be saying that it is possible to do experiments so that there is only one real measurement/observation, at the very end. So you don't need collapse (because you don't do any further experiments on the electron after the actual observation).

 

[atyy]

It occurs to me that there is (perhaps) a subtle distinction between measurement and preparation. For example, when you send a stream of electrons through a Stern-Gerlach device, those with spin-up are deflected one way and those with spin-down are deflected the other way. You haven't measured the spin of any electron, though.

Then you can send those electrons deflected in one direction to go on to a second Stern-Gerlach device, with a different orientation. Those electrons split into two groups, as well.

Eventually, you will (presumably) do a real measurement, by checking for the presence/absence of an electron. But all the selection prior to this doesn't involve measurement, and presumably doesn't collapse the wave function.

So I think that vanhees71 might be saying that it is possible to do experiments so that there is only one real measurement/observation, at the very end. So you don't need collapse (because you don't do any further experiments on the electron after the actual observation).

Not all preparations involve measurements, but some measurements are preparations.

 

[vanhees71]

I don't see how the dynamics of QFT is relevant. There are two aspects to quantum theory: (1) Smooth evolution of the wave function (or smooth evolution of the field operators, in QFT), and (2) the Born interpretation of quantum amplitudes.

The issue is whether the combination is local, not whether the smooth evolution part is local.

Well, if you only consider non-relativistic QT, there's no problem with a collapse concerning causality. You an assume an instantaneous action at a distance without any problem, and then there's no debate to begin with, i.e., you can use collapse arguments without contradictions.

The final sentence doesn't make sense to me. What do you mean by "local" here. Of course, in relativistic QFT by construction all interactions are local in space and time (you write down a Lagrangian with field operators multiplied at the same space-time point only). What's in some sense "non-local" in QT is related to our debate and entanglement, but that I'd not call "non-local" but long-range correlations between parts of a quantum system. One must not misunderstand long-range correlations with non-local interactions at a distance!

 

[stevendaryl]

Not all preparations involve measurements, but some measurements are preparations.

Yes. The tricky thing about applying the collapse hypothesis is that it's rare that you can do nothing more than measure an observable. To measure the position of an electron, you might have the electron collide with a photographic plate and see where the dot is. But that's a destructive measurement. The electron is gone afterward (absorbed by the photographic material).

That's what's special about entangled systems: You can perform a destructive measurement on one subsystem and that counts as a non-destructive measurement of the other subsystem.

 

[vanhees71]

Yes, and this is the truly interesting feature. One thing you can do nowadays, which sounds trivial first, is that you can prepare heralded single photons, i.e., a true single-photon Fock state by creating an entangled photon pair by parametric downconversion and measure one of the photons (the "trigger photon"), and then you now with certainty that you also have another photon (the "idler photon"), even with a specific polarization when the trigger photon's polarization state was determined by the measurement (putting a usual polarizer or other "optical elements" like quarter-wave plates before the detector). As usual, here you have preparation by filtering, i.e., you consider only a subensemble of many prepared photons selecting the wanted ones by a measurement. The point here is that you sort out the photons you want by this preparation procedure, but the measurement of the idler is not the cause of the idlers state, but the cause is that in the very beginning the two photons were prepared in the entangled two-photon state via parametric downconversion, and then you just sort out what's unwanted!

 

[atyy]

Well, if you only consider non-relativistic QT, there's no problem with a collapse concerning causality. You an assume an instantaneous action at a distance without any problem, and then there's no debate to begin with, i.e., you can use collapse arguments without contradictions.

The final sentence doesn't make sense to me. What do you mean by "local" here. Of course, in relativistic QFT by construction all interactions are local in space and time (you write down a Lagrangian with field operators multiplied at the same space-time point only). What's in some sense "non-local" in QT is related to our debate and entanglement, but that I'd not call "non-local" but long-range correlations between parts of a quantum system. One must not misunderstand long-range correlations with non-local interactions at a distance!

But the interesting point is that there is no problem with collapse in relativistic QFT.

If you think there is a problem, then you are using the wrong definition of causality.

 

[stevendaryl]

The final sentence doesn't make sense to me. What do you mean by "local" here. Of course, in relativistic QFT by construction all interactions are local in space and time (you write down a Lagrangian with field operators multiplied at the same space-time point only).

And that has nothing to do with the reason that people suspect that QM is nonlocal. So it's a distraction to bring it up.

What's in some sense "non-local" in QT is related to our debate and entanglement, but that I'd not call "non-local" but long-range correlations between parts of a quantum system. One must not misunderstand long-range correlations with non-local interactions at a distance!

I would say that because QT has nonlocal correlations that do not reduce to local interactions on local variables, QT is inherently a nonlocal theory.

 

[vanhees71]

But the interesting point is that there is no problem with collapse in relativistic QFT.

If you think there is a problem, then you are using the wrong definition of causality.

No, there's no problem in relativistic QFT, as I stress all the time. It's only a problem if you assume a litteral collapse, where instaneously Bob's particle is affected by Alice's measurement. In the "minimal interpretation" there's no such assumption and thus no such problem.

 

[stevendaryl]

No, there's no problem in relativistic QFT, as I stress all the time. It's only a problem if you assume a litteral collapse, where instaneously Bob's particle is affected by Alice's measurement. In the "minimal interpretation" there's no such assumption and thus no such problem.

Well, I don't see any difference between a literal and nonliteral collapse unless you have a clear idea of the separation between what is real and what is subjective. To say that there are no nonlocal influences is to say that no change of a physical quantity here can affect a physical quantity at a spacelike separation. But what properties are physical, in quantum theory? Copenhagen says that only macroscopic properties are real, or only observed properties are real. But that requires a distinction between macroscopic/microscopic or between observer/observed which isn't made clear in the theory.

 

[atyy]

No, there's no problem in relativistic QFT, as I stress all the time. It's only a problem if you assume a litteral collapse, where instaneously Bob's particle is affected by Alice's measurement. In the "minimal interpretation" there's no such assumption and thus no such problem.

1) But if there is no difference in predictions between a literal and a non-literal collapse, why would one who holds to a minimal interpretation object to a literal collapse?

2) Why do you object to the collapse in Cohen-Tannoudji, Diu and Laloe's book? As far as I can tell, they are agnostic as to whether collapse is literal. (In fact, I have never heard of a collapse as literal, except from people who object to it.)

 

[vanhees71]

1) Well, if you take the state as a physical entity and claim that when A measures $\sigma_z$ instantaneously the state collapses, this is a real instaneous effect in the entire universe. This contradicts the very construction of local microcausal QFT. I don't see, why I should buy a self-contradictory postulate, which in fact I never need to describe observations using Q(F)T.

2) I'm a bit surprised, how inaccurate these authors (Nobel laureat included) state the fundamental postulates. I guess, they are pretty uninterested in "interpretation" and rather present the applications of the theory to observable phenomena, and this they do very well. So I don't say that it's a bad book, but, e.g., the formulation that if the system is prepared in state $|\psi \rangle$ the probability to find the system in state $|\phi \rangle$ is $|\langle \phi|\psi \rangle|^2$ is misleading. It made me crazy when I learned QT from another book (I don't remember which one it was), because I couldn't get how in this formulation anything can be independent of the picture of time evolution chosen, and that should be true, because how you choose the picture is quite arbitrary. The resolution is, of course, easy if you put it in the right way: If the system is prepared in the state $|\psi \rangle$ (more precisely the state is reprsented by $|\psi \rangle \langle \psi|$ or equivalently by the corresponding ray, which is another glitch in the chapter on the postulates), then the probability to measure the value $a$ of an observable $A$ is given by $P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2$, where $|a,\beta \rangle$ is the orthonormal basis of the subspace $\mathrm{Eig}(\hat{A},a)$ (modulo the possibility of continuous spectral values, where you have an integral instead of the sum).

 

[atyy]

1) Well, if you take the state as a physical entity and claim that when A measures $\sigma_z$ instantaneously the state collapses, this is a real instaneous effect in the entire universe. This contradicts the very construction of local microcausal QFT. I don't see, why I should buy a self-contradictory postulate, which in fact I never need to describe observations using Q(F)T.

The local microcausal construction is used for the Hamiltonian, which determines the unitary evolution between measurements. The collapse occurs at a measurement. There is no contradiction.

2) I'm a bit surprised, how inaccurate these authors (Nobel laureat included) state the fundamental postulates. I guess, they are pretty uninterested in "interpretation" and rather present the applications of the theory to observable phenomena, and this they do very well. So I don't say that it's a bad book, but, e.g., the formulation that if the system is prepared in state $|\psi \rangle$ the probability to find the system in state $|\phi \rangle$ is $|\langle \phi|\psi \rangle|^2$ is misleading. It made me crazy when I learned QT from another book (I don't remember which one it was), because I couldn't get how in this formulation anything can be independent of the picture of time evolution chosen, and that should be true, because how you choose the picture is quite arbitrary. The resolution is, of course, easy if you put it in the right way: If the system is prepared in the state $|\psi \rangle$ (more precisely the state is reprsented by $|\psi \rangle \langle \psi|$ or equivalently by the corresponding ray, which is another glitch in the chapter on the postulates), then the probability to measure the value $a$ of an observable $A$ is given by $P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2$, where $|a,\beta \rangle$ is the orthonormal basis of the subspace $\mathrm{Eig}(\hat{A},a)$ (modulo the possibility of continuous spectral values, where you have an integral instead of the sum).

But the Cohen-Tannoudji formulation on their p220 of volume 1 looks exactly the same as what you wrote (ie. one has to specify the measurement observable), except that they add that after the measurement, the state of the system is different from before the measurement.

And yes, if they only care about applications of the theory to observable phenomena, then they are agnostic as to whether the wave function and collapse are real or not. Isn't that the minimal interpretation?

 

[vanhees71]

The measurement is due to interaction of the measured system with a measurement apparatus. I also never bought the argument that this is something different than the interactions described by quantum theory. This doesn't make sense either! On the one hand we have to use quantum theory, driven by observations that tell us that the classical theory is only an approximation. So to claim a measurement doesn't follow the laws of QT is not very satisfactory, and I don't see, why one should use this assumption nowadays, where we have understood much better the emergence of classical behavior of macroscopic systems from quantum theory than the "founding fathers" of QT could have known in the beginning. The interaction of a particle with a detector follows the rules of quantum theory and thus is described as a local interaction between the measured system.

I don't want to bash the textbook by Cohen-Tanoudji et al, but I think you should get the postulates as precise as possible, because this helps tremendously to study the subject.

 

[ShayanJ]

The measurement is due to interaction of the measured system with a measurement apparatus. I also never bought the argument that this is something different than the interactions described by quantum theory. This doesn't make sense either! On the one hand we have to use quantum theory, driven by observations that tell us that the classical theory is only an approximation. So to claim a measurement doesn't follow the laws of QT is not very satisfactory, and I don't see, why one should use this assumption nowadays, where we have understood much better the emergence of classical behavior of macroscopic systems from quantum theory than the "founding fathers" of QT could have known in the beginning. The interaction of a particle with a detector follows the rules of quantum theory and thus is described as a local interaction between the measured system.

If you're talking about quantum decoherence, then what you've described in this thread till now, is in contradiction with it.
You say that a measurement doesn't change the wave-function at all, i.e. there is no collapse.
But when collapse is assumed, its assumed as a blackbox. No one says it has to come from somewhere else than the Schrodinger equation, its just assumed and the possibility of explaining it is left open. And decoherence has been able to explain it partially. So decoherence has been able to explain something(partially), that you've always denied. How can you consider it as a support for your arguments? Because it actually rules out what you suggest!

 

[vanhees71]

No, I don't say that measurements change the wave function (or better the state, because we discuss relativistic QT here, and there is no consistent descriptions of it by wave functions a la Schrödinger). I only say that the change of the state is due to quantum dynamics and not an instantaneous action at a distance leading to some collapse thing that is somehow outside of the general dynamical laws of QT. The emergence of classical behavior of macroscopic systems, among them measurement apparati, is quite well understood nowadys within quantum many-body theory, but this is just relativistic local microcausal QFT. So there are only local interactions, no actions at a distance by construction. So this cannot rule out what I suggest!

 

[AlexCaledin]

In the relativistic QFT, instead of "instantaneous action at a distance",

"it is impossible to keep a particle from traveling faster than light"

https://en.wikipedia.org/wiki/Path_integral_formulation#Quantum_field_theory