Collapse of Wave: Effects on Energy & Position Measurements

[klw289]

How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?

 

[vanhees71]

There is no collapse of the state or wave function, although some flavors of the Copenhagen interpretation assume one, but it doesn't help to understand quantum theory nor has it any other merit. Thus the collapse assumption, which is not well defined anyway, should not be taken as a serious part of the physics of quantum theory. Thus your question does not make much sense. What happens to the measured system due to the interaction with the measurement apparatus depends on the choice of this measurement apparatus and thus cannot be answered in as general terms as you might expect.

 

[atyy]

Contrary to what vanhees71 says, the collapse is a central part of the orthodox interpretation of quantum mechanics. It is used in the Schroedinger picture to calculate joint probabilities.

Here is the rough statement: if a measurement of a particle's energy collapses its wave function into an energy eigenstate, then immediate repetition of the measurement will yield the same outcome. There are technical subtleties for continuous variables, but the rough statement is not misleading.

If you would like to see the collapse stated for continuous variables, see http://arxiv.org/abs/0706.3526 (Eq 3).

 

[vanhees71]

If you measure a photon's energy with a usual photodetector, it is absorbed. So there is not a single photon left with a state that's collapsed into an energy eigenstate (or a true state with a narrow energy spread). This is only one very simple example which shows that this collapse hypothesis is very misleading.

Further, it is inconsistent with quantum theory to assume that there is a classical dynamics on top of quantum dynamics, leading to the collapse. So far nobody could find a clear physical distinction when classical dynamics should be applicable and not quantum dynamics. Nowadays, one can demonstrate the superposition principle, particularly also entanglement, for macroscopic objects, and there is no natural scale (in whatever kind of system size) where you can make this "quantum-classical cut".

Last but most importantly an instantaneous nonlocal collapse of the state is incompatible with the very foundations of relativistic space time and its causality structure. If at all, you can take the collapse as a short-hand expression for "adapting ones knowledge because of new information on the system", i.e., in an epistemic instead of an ontological way of interpreting the state, but then you can as well simply adapt the minimal interpretation, taking Born's rule as another postulate rather than something derivable from the other postulates, particularly quantum dynamics. As nicely demonstrated by Weinberg in his "Lectures on Quantum Mechanics" such a derivation is anyway still not known.

 

[bhobba]

Vanhees is of course correct.

Collapse is not a part of QM - its only a part of some interpretations:
https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics

In some interpretations the wave-function is simply a state of knowledge. That it collapses is of zero concern.

Thanks
Bill

 

[atyy]

If you measure a photon's energy with a usual photodetector, it is absorbed. So there is not a single photon left with a state that's collapsed into an energy eigenstate (or a true state with a narrow energy spread). This is only one very simple example which shows that this collapse hypothesis is very misleading.

Further, it is inconsistent with quantum theory to assume that there is a classical dynamics on top of quantum dynamics, leading to the collapse. So far nobody could find a clear physical distinction when classical dynamics should be applicable and not quantum dynamics. Nowadays, one can demonstrate the superposition principle, particularly also entanglement, for macroscopic objects, and there is no natural scale (in whatever kind of system size) where you can make this "quantum-classical cut".

Last but most importantly an instantaneous nonlocal collapse of the state is incompatible with the very foundations of relativistic space time and its causality structure. If at all, you can take the collapse as a short-hand expression for "adapting ones knowledge because of new information on the system", i.e., in an epistemic instead of an ontological way of interpreting the state, but then you can as well simply adapt the minimal interpretation, taking Born's rule as another postulate rather than something derivable from the other postulates, particularly quantum dynamics. As nicely demonstrated by Weinberg in his "Lectures on Quantum Mechanics" such a derivation is anyway still not known.

This is wrong, because quantum mechanics is incompatible with the very foundations of relativistic spacetime and its causality structure. That is what Bell's theorem says.

 

[Nugatory]

Uh, guys... Anyone want to take a try at the original poster's question? Something about how two successive position or energy measurements behave?

 

[stevendaryl]

Vanhees is of course correct.

Collapse is not a part of QM - its only a part of some interpretations:
https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics

In some interpretations the wave-function is simply a state of knowledge. That it collapses is of zero concern.

Thanks
Bill

I don't see how the knowledge interpretation is tenable, in light of Bell's theorem, unless you allow nonlocal interactions.

Consider a spin-1/2 EPR experiment in which Alice and Bob both choose to measure the spins of their respective particles along the same axis, \vec{\alpha}. Immediately after Alice measures spin-up (assuming some frame in which her measurement comes first), she knows with 100% certainty that Bob will measure spin-down (since the particles are perfectly anti-correlated). Bob's chances, as viewed by Alice, "collapse" from 50/50 chance of spin-up or spin-down to 0/100. If this change is simply a change in Alice's knowledge, then to me, that means that Bob's chances were 0/100 before Alice's measurement. That implies hidden variables determining the outcomes. And in light of Bell's theorem, any such hidden variables would have to involve nonlocal interactions.

It seems to me that Bell's theorem is all about whether the wave function can be considered just a state of knowledge, and while it doesn't definitively rule out such an interpretation, it places very tight constraints on it.

 

[atyy]

Uh, guys... Anyone want to take a try at the original poster's question? Something about how two successive position or energy measurements behave?

I did in post #3.

 

[stevendaryl]

How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?

The prediction of the usual QM formalism is that immediately after a measurement, the system is in an eigenstate of whatever operator was measured. So immediately after an energy measurement, the system will be in an eigenstate of energy, and so a second measurement, immediately afterward, should give the same energy. Similarly, measuring the position results in an eigenstate of position, so a second position measurement immediately afterward will give the same position.

If you alternate measurements of different, noncommuting operators, then you will get probabilistic results.

 

[stevendaryl]

Contrary to what vanhees71 says, the collapse is a central part of the orthodox interpretation of quantum mechanics. It is used in the Schroedinger picture to calculate joint probabilities.

Well, you could split it into two different assertions:

1. When you measure an observable, you get an eigenvalue, with probabilities given by the Born rule.
   
2. After a measurement, the wave function is in the eigenstate corresponding to the eigenvalue measured.

Only the second assertion is a "collapse". Of course, you might argue that you need collapse to give predictions for a sequence of measurements, but you don't, really; you can just view the sequence of measurements as a single, compound measurement.

 

[atyy]

Well, you could split it into two different assertions:

1. When you measure an observable, you get an eigenvalue, with probabilities given by the Born rule.
   
2. After a measurement, the wave function is in the eigenstate corresponding to the eigenvalue measured.

Only the second assertion is a "collapse". Of course, you might argue that you need collapse to give predictions for a sequence of measurements, but you don't, really; you can just view the sequence of measurements as a single, compound measurement.

But would you call the compound measurement a "sequence" of measurements? The Born rule applies to measurements made at one slice of simultaneity. In the compound measurement, one does not measure A followed by B, rather one measures AA and B simultaneously, and p(AA,B) has the same form as p(A,B) so that measuring AA is taken to be the "same" as measuring A. However, the difference is that AA and B are real in the compound measurement, whereas A then B is real in the sequence of measurements.

 

[martinbn]

Isn't a collapse needed for state preparatio?. If you need a system of half spin in the state |spin up>, how would one prepare it without collapse?

 

[bhobba]

Isn't a collapse needed for state preparatio?.

In modern times states are the equivalence class of preparation procedures. The concept of collapse need not even be introduced.

Thanks
Bill

 

[stevendaryl]

But would you call the compound measurement a "sequence" of measurements? The Born rule applies to measurements made at one slice of simultaneity. In the compound measurement, one does not measure A followed by B, rather one measures AA and B simultaneously, and p(AA,B) has the same form as p(A,B) so that measuring AA is taken to be the "same" as measuring A. However, the difference is that AA and B are real in the compound measurement, whereas A then B is real in the sequence of measurements.

In principle, you can get away with a single measurement (or a single application of the Born rule) in the following way:

You have some system, S_1, that you'd like to make repeated measurements on. So you create a second system, S_2, which consists of a device that makes repeated measurements on S_1 and records the results (say, on a DVD). Then the creation of the DVD does not, in principle, need to involve the Born rule at all. The various possible sequences of measurement results correspond to orthogonal states of the resulting DVD, and the Schrödinger equation for S_1 \otimes S_2 \otimes DVD would give you the probabilities for each sequence. So it's only necessary to invoke the Born rule once, and it's not necessary to consider what state the DVD is in after the measurement.

 

[stevendaryl]

In modern times states are the equivalence class of preparation procedures. The concept of collapse need not even be introduced.

But it seems to me that the idea that a certain "preparation procedure" results in a specific state (mixed or pure) of the system being prepared is equivalent to the collapse hypothesis. Or another alternative approach (which might be equivalent--I'm not sure) is just to use decoherence and pretend that an improper mixed state is a proper mixed state (which I think is sort of a collapse-like assumption, but the collapse takes place off-stage, which might make it more palatable).

 

[DrChinese]

How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?

Welcome to PhysicsForums, klw289!

Position (q) and momentum (p) are non-commuting observables, as you are probably aware (Heisenberg Uncertainty Principle).

a. A strong (tight) measurement of one makes the other indeterminate (regardless of any value it previously had). If you know a particle's momentum quite accurately, its position could be any of a range of places. It would not make sense to attempt to extrapolate position in this case (ie by using its previous position as a starting point).

b. On the other hand, repeated measurements of the same quantum observable generally yields the same value over and over again.To understand the rules on this, it is often convenient to consider particle spin; as with spin you don't get hung up in trying to construct physical models in your head. Such models have a tendency to throw you off.

 

[atyy]

In principle, you can get away with a single measurement (or a single application of the Born rule) in the following way:

You have some system, S_1, that you'd like to make repeated measurements on. So you create a second system, S_2, which consists of a device that makes repeated measurements on S_1 and records the results (say, on a DVD). Then the creation of the DVD does not, in principle, need to involve the Born rule at all. The various possible sequences of measurement results correspond to orthogonal states of the resulting DVD, and the Schrödinger equation for S_1 \otimes S_2 \otimes DVD would give you the probabilities for each sequence. So it's only necessary to invoke the Born rule once, and it's not necessary to consider what state the DVD is in after the measurement.

Yes, but there is a slight difference, at least within Copenhagen. If you don't invoke the Born rule when you make the DVD, then the record on the DVD is not necessarily real. So there is no real sequence of measurement outcomes, where reality is attributed only to measurement outcomes (ie. when one invokes the Born rule).

 

[bhobba]

But it seems to me that the idea that a certain "preparation procedure" results in a specific state (mixed or pure) of the system being prepared is equivalent to the collapse hypothesis.

It is - its just a change in how its viewed - but a very important one. Preparations do not happen instantaneously so you avoid this whole non-local instantaneous collapse with observation thing.

Thanks
Bill

 

[DuckAmuck]

How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?

Imagine you measure position, and collapse the position wave function. Now imagine you keep measuring it over and over every second. You'll just keep getting the same result, since the wave function will remain collapsed. But if you stop and give it some time, the collapsed wave function will evolve back into a non-collapsed wavefunction. So the next time you measure it after waiting, you will get a different result, in general.

 

[vanhees71]

This is wrong, because quantum mechanics is incompatible with the very foundations of relativistic spacetime and its causality structure. That is what Bell's theorem says.

Well, you might say that quantum field theory is mathematically not strictly well defined in the sense of an exactly solvable theory, but perturbative relativistic local QFT is a very successful description of nature. It describes all Bell experiments with very high accuracy. Since by construction this class of theories does not contain any nonlocal interactions, it is compatible with the relativistic space-time structure (by construction!). It's also not what Bell's theorem says, because Bell's theorem refers to local classically realistic theories, and quantum theory is not such a theory. That's the whole point of Bell's work: You have an objective way to decide between classical realistic local theories and quantum theory, with the result of $N$ experiments to 0, where $N$ is a very large number ;-)).

It is very important to distinguish between nonlocal interactions (which are by construction not present in standard relativistic QFT) and longranged correlations, which are a specific consequence of any quantum theory, including relativistic QFT. It is not that measurement of one photon's polarization in an entangled two-photon state that causes the correlation but the preparation of the entangled two-photon state in the very beginning of the experiment. Although the single-photon polarizations are maximally indetermined, the 100% correlation between them is there all the time due to the preparation procedure in an entangled state. There is no non-local interaction between the single photons and the polarizers and photo detectors at A's and B's far-distant locations whatsoever! Consequently, according to local relativistic QFT it cannot be nonlocal interactions caused by the local measurements at the local detectors. This shows that these experiments do not contradict standard QED and thus there's no need for nonlocal interactions!

 

[vanhees71]

Imagine you measure position, and collapse the position wave function. Now imagine you keep measuring it over and over every second. You'll just keep getting the same result, since the wave function will remain collapsed. But if you stop and give it some time, the collapsed wave function will evolve back into a non-collapsed wavefunction. So the next time you measure it after waiting, you will get a different result, in general.

What is a "collapsed" or "non-collapsed" wave function? There's no such thing in standard quantum mechanics. The wave function describes a state in a certain basis, namely the (generalized) eigenbasis of the position operator (and perhaps other observables compatible with position like spin to have a complete basis). That's it. There's no way to say a wave-function is "collapsed" or "non-collapsed". It simply doesn't make any sense!

You can only say that the position of the particle is more or less determined, depending on the width of the probability distribution of position (which is given by the modulus of the wave function squared, due to Born's rule).

 

[atyy]

Well, you might say that quantum field theory is mathematically not strictly well defined in the sense of an exactly solvable theory, but perturbative relativistic local QFT is a very successful description of nature. It describes all Bell experiments with very high accuracy. Since by construction this class of theories does not contain any nonlocal interactions, it is compatible with the relativistic space-time structure (by construction!). It's also not what Bell's theorem says, because Bell's theorem refers to local classically realistic theories, and quantum theory is not such a theory. That's the whole point of Bell's work: You have an objective way to decide between classical realistic local theories and quantum theory, with the result of $N$ experiments to 0, where $N$ is a very large number ;-)).

It is very important to distinguish between nonlocal interactions (which are by construction not present in standard relativistic QFT) and longranged correlations, which are a specific consequence of any quantum theory, including relativistic QFT. It is not that measurement of one photon's polarization in an entangled two-photon state that causes the correlation but the preparation of the entangled two-photon state in the very beginning of the experiment. Although the single-photon polarizations are maximally indetermined, the 100% correlation between them is there all the time due to the preparation procedure in an entangled state. There is no non-local interaction between the single photons and the polarizers and photo detectors at A's and B's far-distant locations whatsoever! Consequently, according to local relativistic QFT it cannot be nonlocal interactions caused by the local measurements at the local detectors. This shows that these experiments do not contradict standard QED and thus there's no need for nonlocal interactions!

Let's take QED with just photons. It's a mathematically well-defined theory, and it has collapse.

It is wrong to say that collapse has to be rejected because it is not consistent with "relativistic causality". There are two possible meanings of "relativistic causality".

1. Signal causality - collapse does not violate signal causality, and it does not need to be rejected for being inconsistent with this type of relativistic causality, so this cannot be what you mean if your statement is to be correct.

2. Classical relativistic causality - collapse does violate classical relativistic causality, but if your statement is applied here, then your statement is misleading to wrong because there is no quantum mechanics that preserves classical relativistic causality.

 

[vanhees71]

Isn't a collapse needed for state preparatio?. If you need a system of half spin in the state |spin up>, how would one prepare it without collapse?

I'd use a magnetic field like in the Stern-Gerlach experiment. There's nothing collapsing there ;-).

 

[vanhees71]

Let's take QED with just photons. It's a mathematically well-defined theory, and it has collapse.

It is wrong to say that collapse has to be rejected because it is not consistent with "relativistic causality". There are two possible meanings of "relativistic causality".

1. Signal causality - collapse does not violate signal causality, and it does not need to be rejected for being inconsistent with this type of relativistic causality, so this cannot be what you mean if your statement is to be correct.

2. Classical relativistic causality - collapse does violate classical relativistic causality, but if your statement is applied here, then your statement is misleading to wrong because there is no quantum mechanics that preserves classical relativistic causality.

How can QED with only photons, which is a free-field theory have collapse? Since the photons don't interact with anything within this theory, how can their states propagate other than through the unitary time evolution according to the full Hamiltonian (take the Schrödinger picture here for ease of discussion)?

Concerning the causality debate. Ad 1) Collapse, taken as a physical process, violates "signal causality", because it claims the meausurement at A's photon instantaneously determines also the polarization of B's photon measured at a far distant place. Of course, that's a wrong interpretation, because the entanglement of A's and B's photon guarantees already the 100% correlation between their measurements of the single-photon polarizations without any necessity for this "action at a distance collapse".

I don't know, what (2) means. How do you define "classical relativistic causality"? In my opinion causality is the property of a physical theory or model describing dynamics, and so far all of physics (and all of natural sciences anyway) is based on the hypothesis that nature is describable by causal theories. Of course both classical and quantum theory are causal, i.e., if you know the state of the system at a time $t=t_0$ and you know the Hamiltonian of the system exactly, then you know its state at any time $t>t_0$ exactly (modulo the impossibility to integrate the quantum equations of motion, but that's a discussion about principles not practicalities).

 

[stevendaryl]

It is very important to distinguish between nonlocal interactions (which are by construction not present in standard relativistic QFT) and longranged correlations, which are a specific consequence of any quantum theory, including relativistic QFT. It is not that measurement of one photon's polarization in an entangled two-photon state that causes the correlation but the preparation of the entangled two-photon state in the very beginning of the experiment. Although the single-photon polarizations are maximally indetermined, the 100% correlation between them is there all the time due to the preparation procedure in an entangled state. There is no non-local interaction between the single photons and the polarizers and photo detectors at A's and B's far-distant locations whatsoever! Consequently, according to local relativistic QFT it cannot be nonlocal interactions caused by the local measurements at the local detectors. This shows that these experiments do not contradict standard QED and thus there's no need for nonlocal interactions!

Right. For the kind of nonlocality that people worry about from Bell's theorem, there is no big difference between QM and QFT. It's about the long-range correlations. Whether those correlations imply something nonlocal is what's in dispute.

In QM or QFT, Alice describes the universe (or the part of the universe under consideration) at a given moment (relativistically, I guess a "moment" means a spacelike hypersurface of spacetime) by means of a pure or mixed state \rho. This state describes both what's happening locally to Alice, but also what's happening far away, near Bob. When Alice makes an observation, or measurement, then she adjusts the state in light of what she's learned. So afterward, she uses a different state, \rho'. In switching from \rho to \rho', Alice not only updates the description of her own local neighborhood of the universe, but she also updates the description of Bob's neighborhood. The two descriptions are entangled, in general, which just means that you can't update one independently of updating the other.

The question really is: To what extent does Alice's \rho reflect something objective about Bob's neighborhood, and to what extent does it simply represent her knowledge about Bob's situation? In an EPR-type experiment, it might be that according to \rho, Bob has a nonzero chance of measuring a particle with spin-up along axis \vec{\alpha}, but according to \rho', it is certain that Bob will measure spin-down along that axis. That's definitely a change with Bob's situation. If that change is purely subjective, if it's purely about Alice's knowledge of Bob's future measurement results, then Alice could reason as follows:

• After updating the state from \rho to \rho', I know that Bob is certain to measure spin-down along axis \vec{\alpha}.
• The update is only a change in my knowledge; nothing really changed at Bob's location.
• Therefore, it must have been true before the update that Bob would measure spin-down along axis \vec{\alpha}.

So it seems to me that the assumption that Alice's measurement has no effect on Bob, only on her knowledge of Bob's future results, leads to the conclusion that Bob's future results are pre-determined. But that conclusion contradicts Bell's theorem (unless you invoke some weird loophole, like superdeterminism or retrocausality).

 

[vanhees71]

I think, a lot of confusion is because one says something like

In QM or QFT, Alice describes the universe (or the part of the universe under consideration) at a given moment (relativistically, I guess a "moment" means a spacelike hypersurface of spacetime) by means of a pure or mixed state \rho. This state describes both what's happening locally to Alice, but also what's happening far away, near Bob.

The state describes the preparation of the two photons and probabilities for measurements of observables concerning these photons, and nothing else. You can describe the probability that A measures the polarization (say H or V) of her photon that B measures the polarization (say also H or V) of his photon. Because of the preparation of the two photons in an polarization-entangled state, it is determined (!) that if A measures H then B must measure V and vice versa although the single-photon polarizations are described (according to the usual reduced statistical operators of subsystems) by the maximally indetermined state $\hat{\rho}_A = \hat{\rho}_B=\mathbb{1}/2$. There's also a certain probability of the registration of the photons at A's and B's place.

Nothing happens to B's photon due to A's measurement (which in fact could have been measured and thus absorbed even before A made her measurement, where I assume for simplicity that A and B are in a common inertial lab frame, i.e., at rest relative to each other, and we use the time of this common frame to make statements about temporal order of events). When A makes her measurement and if she knows that B could measure the second photon in the entangled two-photon state, of which she has measured one of the photons, she can adapt her description of the two-photon state or of the reduced state of B's photon, but that "collapse" is nothing what's happening to the photons.

I think that's more or less the same as you stated as "A updates only her knowledge about B's local situation due to her measurement (and the knowledge about the two-photon state to begin with!)", and this is the minimal interpretation. Nowhere do you need a collapse as a physical process to describe what A and B measure, finding the 100% correlation of completely indetermined single-photon polarizations due to the entanglement of the prepared 2-photon state!

 

[atyy]

IThe state describes the preparation of the two photons and probabilities for measurements of observables concerning these photons, and nothing else. You can describe the probability that A measures the polarization (say H or V) of her photon that B measures the polarization (say also H or V) of his photon. Because of the preparation of the two photons in an polarization-entangled state, it is determined (!) that if A measures H then B must measure V and vice versa although the single-photon polarizations are described (according to the usual reduced statistical operators of subsystems) by the maximally indetermined state $\hat{\rho}_A = \hat{\rho}_B=\mathbb{1}/2$. There's also a certain probability of the registration of the photons at A's and B's place.

Nothing happens to B's photon due to A's measurement (which in fact could have been measured and thus absorbed even before A made her measurement, where I assume for simplicity that A and B are in a common inertial lab frame, i.e., at rest relative to each other, and we use the time of this common frame to make statements about temporal order of events). When A makes her measurement and if she knows that B could measure the second photon in the entangled two-photon state, of which she has measured one of the photons, she can adapt her description of the two-photon state or of the reduced state of B's photon, but that "collapse" is nothing what's happening to the photons.

I think that's more or less the same as you stated as "A updates only her knowledge about B's local situation due to her measurement (and the knowledge about the two-photon state to begin with!)", and this is the minimal interpretation. Nowhere do you need a collapse as a physical process to describe what A and B measure, finding the 100% correlation of completely indetermined single-photon polarizations due to the entanglement of the prepared 2-photon state!

That is not the minimal interpretation. The minimal interpretation is agnostic about whether something does or does not happen. How can you be sure that nothing happens faster than light? Also, the updating of the knowledge is alone without anything happening is problematic, because the updating does not follow Bayesian updating.

 

[martinbn]

I'd use a magnetic field like in the Stern-Gerlach experiment. There's nothing collapsing there ;-).

What are the states before and after passing through the Stern-Gerlach aparatus?

Classical relativistic causality.

What is classical relativistic causality?

 

[stevendaryl]

The state describes the preparation of the two photons and probabilities for measurements of observables concerning these photons, and nothing else.

I didn't say it described anything else. But Alice, when she makes her measurement, updates her state from \rho (in which Bob has a 50/50 chance of measuring spin-up or spin-down along axis \vec{\alpha}) to \rho' (in which Bob is certain to measure spin-down along axis \vec{\alpha}). The question is: Is this change in probabilities a physical change of Bob's particle, or is it merely a change in Alice's knowledge? Either way runs into problems. If it's merely a change in Alice's knowledge, then Bob's particle isn't changed by Alice's measurement. So if that particle is definitely spin-down along \vec{\alpha} after Alice's measurement, then it must have been spin-down beforehand.

Nothing happens to B's photon due to A's measurement

That leads to the conclusion that B's particle (I was thinking spin-1/2 particles, but it doesn't matter) was a definite state of spin before Alice's measurement.

I think that's more or less the same as you stated as "A updates only her knowledge about B's local situation due to her measurement (and the knowledge about the two-photon state to begin with!)", and this is the minimal interpretation. Nowhere do you need a collapse as a physical process to describe what A and B measure, finding the 100% correlation of completely indetermined single-photon polarizations due to the entanglement of the prepared 2-photon state!

The updating IS the collapse! The only issue is: what is the nature of that collapse? Is it purely a change in Alice's knowledge, or is it a change in Bob's particle's state? The first possibility contradicts Bell's theorem, and the second contradicts locality.

Let's forget about particles, and just speak in terms of measurements and probabilities of future measurement results. That's the "minimal interpretation", I think.

• Before Alice performs her measurement, she would say that Bob has a 50/50 chance of measuring spin-up or spin-down along axis \vec{\alpha} (or H and V, if you're using photons).
• After Alice's measurement, she knows with 100% certainty what Bob will measure. She can now say: "Bob will definitely measure X" (whatever the prediction is).
• Is her statement a physical claim about the state of Bob and his local environment? It sure seems to me that it is.
• If it's a statement about Bob and surroundings, then you have the two possibilities: either it became true when Alice did her measurement, or it was true beforehand. The first violates locality, and the second violates Bell's theorem.