Collapse of Wave: Effects on Energy & Position Measurements

[martinbn]

Before you usually have a thermal state of a beam of silver atoms coming from an oven with a little opening, which is well approximated by a Gaussian wave packet with a momentum spread given by the thermal width. After the Stern-Gerlach apparatus you have a superposition of two wave packets (for spin 1/2), where the spin component in direction of the magnetic field is entangled with position, i.e., the spin-up and spin-down are spatially separated. This can be treated (semi-)analytically, see e.g.,

G. Potel, F. Barranco, S. Cruz-Barrios, and J. Gómez-Camacho. Quantum mechanical description of Stern-Gerlach experiments. Phys. Rev. A, 71, 2005.
http://dx.doi.org/10.1103/PhysRevA.71.052106

Are you talking about the whole ensemble? I was asking about the state of an individual atom before and after it passes through the Stern-Gerlach.

 

[vanhees71]

The state of a single atom is given by its wave function (choosing the position representation of the state for convenience). Of course, the wave function has a probabilistic meaning only and thus, as with any probabilistic property, you can verify it only with an ensemble of stochastically independently prepared systems, i.e., an ensemble. I don't believe in any practical meaning of probabilities like qbists tend to suggest. To verify a probabilistic statement you have to use sufficiently large ensembles to test the hypothesis within a given confidence level.

 

[martinbn]

The state of a single atom is given by its wave function (choosing the position representation of the state for convenience). Of course, the wave function has a probabilistic meaning only and thus, as with any probabilistic property, you can verify it only with an ensemble of stochastically independently prepared systems, i.e., an ensemble. I don't believe in any practical meaning of probabilities like qbists tend to suggest. To verify a probabilistic statement you have to use sufficiently large ensembles to test the hypothesis within a given confidence level.

Ok but then, if we ignore everything else but spin, isn't the state of an atom before Stern-Gerlach given by a|spin up>+b|spin down> and after passing through, it is either |spin up> or |spin down>? So there is something collapsing here.

 

[atyy]

What do you mean by "Bayesian updating"? If A measures the polarization state of her single photon AND knowing that B measures a photon that was polarization entangled with hers (in the same polarization direction), she instantly knows what B must (have) measure(d) about his photon. Then she can update her knowledge about Bob's measurement outcome to certainty for the corresponding polarization. Whether this is Bayesian or not, I don't care, but it follows standard rules of probability theory, right?

It does not follow the standard rules of probability unless one postulates collapse. That is the important point about Bayesian updating - the standard rules of probability are not enough without collapse.

 

[atyy]

If the updating is the collapse, then I don't understand why it is still so fervently discussed, because then it's nothing what necessarily happens to the object under consideration, as in the here discussed example (at least if you accept standard QED as the valid description).

Again, I wish to stress that your interpretation is very non-minimal. How can you be sure that "nothing has happened to the object"?

You replied saying QED obeys signal causality. Sure, but as I have stressed repeatedly, signal causality being respected does not mean that "nothing has happened to the object". A simple way to see that you lack an argument for your assertion that the updating is purely informational with nothing happening to the object, is that if collapse is physical, then something has happened to the object and yet faster than light signalling is prevented.

The minimal interpretation is agnostic, not confidently assertive of things it cannot show, unlike your claim that "nothing has happened to the object". If you read Cohen-Tannoudji, Diu and Laloe's famous text, you will see that they are not so cavalier as you are at this point.

 

[stevendaryl]

According to QED, if the registration of A's and B's photons mark space-like separated events, then the one measurement cannot have affected the outcome of the other, and that's why A's update about B's photon doesn't do anything to B's photon.

I don't see that that's relevant. In an EPR-type experiment, QED (or QM--there's really nothing special brought in by QED that isn't already present in QM) allows us to compute probability amplitudes. But unless you're assuming Many-Worlds, amplitudes are not the end of the story. When someone makes a measurement, he doesn't get an amplitude, he gets an eigenvalue of whatever was being measured. The question of nonlocality in QM is precisely about the process by which one specific eigenvalue is selected from a set of possibilities. This process is NOT described by QED. So the fact that QED has no nonlocal interactions is irrelevant.

 

[stevendaryl]

• If it's [the claim that Bob has a 100% chance of measuring spin-down along axis $\vec{\alpha}$] 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.

It was true beforehand

Really? That doesn't sound like a "minimal interpretation". For context, let me repeat the background: You have an EPR experiment with anti-correlated twin pairs of spin-1/2 particles. Alice and Bob both choose to measure the spins of their particles along an axis $\vec{\alpha}$. Alice measures spin-up for her particle slightly before (in the lab frame) Bob measures his particle's spin. So Alice makes the claim "Bob has a 100% chance of measuring spin-down along axis $\vec{\alpha}$".

You're saying that Alice's claim was true BEFORE Alice made her measurement? Was it true before Alice even chose the axis for her Stern-Gerlach device?

 

[Mentz114]

Ok but then, if we ignore everything else but spin, isn't the state of an atom before Stern-Gerlach given by a|spin up>+b|spin down> and after passing through, it is either |spin up> or |spin down>? So there is something collapsing here.

Not collapsing - rotating. Your coefficients a and b are constrained so a²+b² = 1 but either can go to zero.

 

[stevendaryl]

Not collapsing - rotating. Your coefficients a and b are constrained so a²+b² = 1 but either can go to zero.

But in the EPR experiment with spin-1/2 particles, the statistics are:

• If Alice and Bob measure the spin along the same axis, they always get opposite answers.
• If they measure along different axes, then they get opposite answers with probability $cos^2(\theta/2)$ and same answers with probability $sin^2(\theta/2)$

The prediction doesn't involve time, at all. So if Alice measures her particle at one time, and Bob measures his at a different time, presumably any kind of rotational effect would be eliminated (unless the time difference they choose happened to be a multiple of the rotational frequency, which seems weird).

But even disregarding the timing question, I don't see how the rotating model could possibly work. Let's consider the case in which the rotation makes it so the spin at some moment is pointing in the y-direction. Alice at that moment measures the spin along an axis that makes a 90-degree angle, clockwise, from the y-axis. So she has a nonzero probability of measuring spin-up (you would think it would be $cos^2(90/2)=1/2$. Bob measures the spin along an axis that makes a 90-degree angle counter-clockwise from the y-axis. He would also have a chance of 1/2 of measuring spin-up. But it never happens that both Alice and Bob measure spin up, if their axes are 180 degrees apart. So how does a rotating model explain that?

 

[Mentz114]

But in the EPR experiment with spin-1/2 particles, the statistics are:

• If Alice and Bob measure the spin along the same axis, they always get opposite answers.
• If they measure along different axes, then they get opposite answers with probability $cos^2(\theta/2)$ and same answers with probability $sin^2(\theta/2)$

The prediction doesn't involve time, at all. So if Alice measures her particle at one time, and Bob measures his at a different time, presumably any kind of rotational effect would be eliminated (unless the time difference they choose happened to be a multiple of the rotational frequency, which seems weird).

But even disregarding the timing question, I don't see how the rotating model could possibly work. Let's consider the case in which the rotation makes it so the spin at some moment is pointing in the y-direction. Alice at that moment measures the spin along an axis that makes a 90-degree angle, clockwise, from the y-axis. So she has a nonzero probability of measuring spin-up (you would think it would be $cos^2(90/2)=1/2$. Bob measures the spin along an axis that makes a 90-degree angle counter-clockwise from the y-axis. He would also have a chance of 1/2 of measuring spin-up. But it never happens that both Alice and Bob measure spin up, if their axes are 180 degrees apart. So how does a rotating model explain that?

Did you mean to post this in the thread I started ?

I was not suggesting a rotating state here - just that the polarizer can rotate spin orientations.

 

[stevendaryl]

Did you mean to post this in the thread I started ?

I was not suggesting a rotating state here - just that the polarizer can rotate spin orientations.

I thought you were bringing up the same model again. Sorry. But in any case, you still have the second issue: Alice measures polarization along one axis, and suppose that that causes the polarization to rotate so that afterward, it is aligned with Alice's filter. Meanwhile, Bob measures the polarization of his photon along a different axis. Does his photon get rotated along with Alice's? If so, that's instantaneous action at a distance. If not, then how do you explain that Bob has zero chance of getting a photon with a polarization that is 90 degrees off from Alice's?

 

[Mentz114]

I thought you were bringing up the same model again. Sorry. But in any case, you still have the second issue: Alice measures polarization along one axis, and suppose that that causes the polarization to rotate so that afterward, it is aligned with Alice's filter. Meanwhile, Bob measures the polarization of his photon along a different axis. Does his photon get rotated along with Alice's? If so, that's instantaneous action at a distance. If not, then how do you explain that Bob has zero chance of getting a photon with a polarization that is 90 degrees off from Alice's?

I didn't mean anything as grand as that. Consider the state a|0> + b|1> which after being filtered is in state a'|0> + b'|1>. This means that we can find an angle a so that a'=cos(a) and b' = sin(b). High-school geometry.

My point being that 'collapse' is not an appropriate term for this process. OK ?

Be sure that I'm not pushing any 'explanation' for quantum correlations here or elsewhere.

 

[stevendaryl]

I didn't mean anything as grand as that. Consider the state a|0> + b|1> which after being filtered is in state a'|0> + b'|1>. This means that we can find an angle a so that a'=cos(a) and b' = sin(b). High-school geometry.

My point being that 'collapse' is not an appropriate term for this process. OK ?

Not by itself, but when you consider that afterward, distant measures on an entangled photon by will give results consistent with his photon being in the state $a'|0\rangle + b'|1\rangle$, the phrase "wave function collapse" seems more appropriate.

 

[Mentz114]

Not by itself, but when you consider that afterward, distant measures on an entangled photon by will give results consistent with his photon being in the state $a'|0\rangle + b'|1\rangle$, the phrase "wave function collapse" seems more appropriate.

OK. I don't believe any of that happens ( well, I am easily led). It seems more likely that quantum correlations become fixed during the preparation and that no non-locality is rquired.

 

[stevendaryl]

OK. I don't believe any of that happens ( well, I am easily led). It seems more likely that quantum correlations become fixed during the preparation and that no non-locality is rquired.

Well, it might seem more likely, but that's exactly what Bell's inequality (and the fact that actual experiments violate it) proves is not the case.

[edit] It depends on what you mean by "correlations". It certainly is fixed at the time of preparation that the two photons are correlated. But what isn't fixed at the time of preparation is the answers to questions along the lines of:

If Alice measures the polarization along axis $\vec{\alpha}$, will she get H or V?​

Bell proved that the predictions of QM for the EPR experiment are not consistent with the assumption that all such questions have predetermined answers. (That is, that the answers are determined at the time the two photons are created) But if they don't have predetermined answers, how can you guarantee that if Alice and Bob both set their filters at orientation $\vec{\alpha}$, they always get the same result?

 

[Mentz114]

Well, it might seem more likely, but that's exactly what Bell's inequality (and the fact that actual experiments violate it) proves is not the case.

[edit] It depends on what you mean by "correlations". It certainly is fixed at the time of preparation that the two photons are correlated. But what isn't fixed at the time of preparation is the answers to questions along the lines of:

If Alice measures the polarization along axis $\vec{\alpha}$, will she get H or V?​

Bell proved that the predictions of QM for the EPR experiment are not consistent with the assumption that all such questions have predetermined answers. (That is, that the answers are determined at the time the two photons are created) But if they don't have predetermined answers, how can you guarantee that if Alice and Bob both set their filters at orientation $\vec{\alpha}$, they always get the same result?

I don't believe absolutely all the inferences attributed to Bells theorem. We must differ for the time being. Maybe if I come across some 'cooler' equations ...

 

[stevendaryl]

I don't believe absolutely all the inferences attributed to Bells theorem. We must differ for the time being.

It's fine to agree to disagree, but I would like to know what, specifically, you're disagreeing with.

 

[Mentz114]

It's fine to agree to disagree, but I would like to know what, specifically, you're disagreeing with.

It's been pointed out already that we should use field theory to model the entangled scenarios. The kind of non-locality one encounters there would be acceptable because it does not involve communication of any kind. Unfortunately there seems to be a lack of consensus about locality in QFT, which let's me off the hook.

Bells theorem is a theorem, but physics needs symmetries. If Bells theorem could be related to conservation of angular momentum it could be stronger. The symmetry group of the polarizer projection operators is su(2) which helps.

 

[stevendaryl]

It's been pointed out already that we should use field theory to model the entangled scenarios.

I think that's a mistake. The main difference between QM and QFT is that the latter can deal with a variable number of particles, but that doesn't seem to be an important consideration in the EPR experiment. It's enough that you have (in the spin 1/2 case) a pair of particles that are in a spin-0 state.

Bells theorem is a theorem, but physics needs symmetries. If Bells theorem could be related to conservation of angular momentum it could be stronger. The symmetry group of the polarizer projection operators is su(2) which helps.

Well, in a certain sense, it's related to conservation of angular momentum, because the total state has angular momentum zero, so if you measure one particle to have spin $\vec{s}$, then you have to measure the other to have spin $-\vec{s}$.

But in another sense, entanglement doesn't have anything specifically to do with symmetries. If you start with any two-particle state, and allow the particles to interact, then they will become entangled.

 

[Mentz114]

I think that's a mistake. The main difference between QM and QFT is that the latter can deal with a variable number of particles, but that doesn't seem to be an important consideration in the EPR experiment. It's enough that you have (in the spin 1/2 case) a pair of particles that are in a spin-0 state.

The QFT model should/could predict the observed outcomes without leaving one looking for an explanation involving communication between space-like separated points.
If the Hamiltonian has terms like $a^{\dagger}b-ab^{\dagger}$ where a and b are spin-state destruction operators then a measurement by Alice affects the probabilities at Bobs side.

Was there any communication ? ( I really am asking - I don't have a view )

Well, in a certain sense, it's related to conservation of angular momentum, because the total state has angular momentum zero, so if you measure one particle to have spin $\vec{s}$, then you have to measure the other to have spin $-\vec{s}$.

Rotational symmetry.

But in another sense, entanglement doesn't have anything specifically to do with symmetries. If you start with any two-particle state, and allow the particles to interact, then they will become entangled.

Doing it with QFT makes symmetries important.

 

[stevendaryl]

The QFT model should/could predict the observed outcomes without leaving one looking for an explanation involving communication between space-like separated points.

The QFT explanation is no different than the QM explanation, which is either adequate or inadequate, depending on your tastes. There is no difference at all between the QFT case and the QM case. The point about entanglement is that you set things up so that distant subsystems are in a superposition of states of the form:

$C_1 |\psi_1\rangle |\phi_1 \rangle + C_2 |\psi_2\rangle |\phi_2 \rangle$

How you would get such states is no different in QFT than in QM. The nonlocality happens when you perform a measurement of the first subsystem to decide whether it is in state $|\psi_1\rangle$ or $|\psi_2\rangle$. Afterward, you know instantly that the other, distant subsystem is in the corresponding state:$|\phi_1\rangle$ or $|\phi_2\rangle$. Using QFT versus QM doesn't make any difference at all.

If the Hamiltonian has terms like $a^{\dagger}b-ab^{\dagger}$ where a and b are spin-state destruction operators then a measurement by Alice affects the probabilities at Bobs side.

Was there any communication ? ( I really am asking - I don't have a view )

Definitely there is no communication. The controversy is over whether there is a nonlocal interaction at all.

Doing it with QFT makes symmetries important.

I don't see how it is relevant to the issues raised by entanglement. As I said, the entanglement doesn't need to be connected with any obvious symmetry such as rotational or translation. Any interaction at all between systems will in general cause them to become entangled. Maybe every interaction has some associated symmetry, but I don't see how that is particularly relevant to entanglement.

 

[Mentz114]

The QFT explanation is no different than the QM explanation, which is either adequate or inadequate, depending on your tastes. There is no difference at all between the QFT case and the QM case.

I wouldn't expect a difference in the predictions. But in the QFT case there is more detail. To me that detail makes all the difference. Particularly the next point ...

Definitely there is no communication. The controversy is over whether there is a nonlocal interaction at all.

I position myself decisively on the fence !

I don't see how it is relevant to the issues raised by entanglement. As I said, the entanglement doesn't need to be connected with any obvious symmetry such as rotational or translation. Any interaction at all between systems will in general cause them to become entangled. Maybe every interaction has some associated symmetry, but I don't see how that is particularly relevant to entanglement.

probably redundant detail.

Thanks for the discussion which has helped me tidy up some ideas (believe it or not).

 

[jimgraber]

The QFT explanation is no different than the QM explanation, which is either adequate or inadequate, depending on your tastes. There is no difference at all between the QFT case and the QM case. The point about entanglement is that you set things up so that distant subsystems are in a superposition of states of the form:

$C_1 |\psi_1\rangle |\phi_1 \rangle + C_2 |\psi_2\rangle |\phi_2 \rangle$

How you would get such states is no different in QFT than in QM. The nonlocality happens when you perform a measurement of the first subsystem to decide whether it is in state $|\psi_1\rangle$ or $|\psi_2\rangle$. Afterward, you know instantly that the other, distant subsystem is in the corresponding state:$|\phi_1\rangle$ or $|\phi_2\rangle$. Using QFT versus QM doesn't make any difference at all.
Definitely there is no communication. The controversy is over whether there is a nonlocal interaction at all.
I don't see how it is relevant to the issues raised by entanglement. As I said, the entanglement doesn't need to be connected with any obvious symmetry such as rotational or translation. Any interaction at all between systems will in general cause them to become entangled. Maybe every interaction has some associated symmetry, but I don't see how that is particularly relevant to entanglement.

@stevendaryl " the nonlocality happens when" http://arxiv.org/abs/1512.01443

The QFT explanation is no different than the QM explanation, which is either adequate or inadequate, depending on your tastes. There is no difference at all between the QFT case and the QM case. The point about entanglement is that you set things up so that distant subsystems are in a superposition of states of the form:

$C_1 |\psi_1\rangle |\phi_1 \rangle + C_2 |\psi_2\rangle |\phi_2 \rangle$

How you would get such states is no different in QFT than in QM. The nonlocality happens when you perform a measurement of the first subsystem to decide whether it is in state $|\psi_1\rangle$ or $|\psi_2\rangle$. Afterward, you know instantly that the other, distant subsystem is in the corresponding state:$|\phi_1\rangle$ or $|\phi_2\rangle$. Using QFT versus QM doesn't make any difference at all.
Definitely there is no communication. The controversy is over whether there is a nonlocal interaction at all.
I don't see how it is relevant to the issues raised by entanglement. As I said, the entanglement doesn't need to be connected with any obvious symmetry such as rotational or translation. Any interaction at all between systems will in general cause them to become entangled. Maybe every interaction has some associated symmetry, but I don't see how that is particularly relevant to entanglement.

@stevendaryl, you say "the nonlocality happens when..."
http://arxiv.org/abs/1512.01443 Here is a reference to a short paper by Griffiths which was just posted on the arXiv tonight,
which states that nonlocality never happens in QM.
I would be interested in your reaction to it.
TIA,
Jim Graber

 

[stevendaryl]

@stevendaryl " the nonlocality happens when" http://arxiv.org/abs/1512.01443@stevendaryl, you say "the nonlocality happens when..."
http://arxiv.org/abs/1512.01443 Here is a reference to a short paper by Griffiths which was just posted on the arXiv tonight,
which states that nonlocality never happens in QM.
I would be interested in your reaction to it.
TIA,
Jim Graber

Thank you, I'll read it.

I think Griffiths is one of the advocates of the "consistent histories" interpretation of quantum mechanics, which is like Many-Worlds in that it considers the world that we experience to be just one of many, equally real, alternatives. The argument for nonlocality from QM is assuming that every measurement produces a definite result. In Many-Worlds, and I think consistent histories, you don't assume that measurements produce definite results.

 

[jimgraber]

Thank you, I'll read it.

I think Griffiths is one of the advocates of the "consistent histories" interpretation of quantum mechanics, which is like Many-Worlds in that it considers the world that we experience to be just one of many, equally real, alternatives. The argument for nonlocality from QM is assuming that every measurement produces a definite result. In Many-Worlds, and I think consistent histories, you don't assume that measurements produce definite results.

I just referred to Wikipedia and the Stanford Encyclopedia of Philosophy article on consistent histories (which is by Griffith himself).
It appears that there are multiple histories on the microscale, but only a single history on the macro scale, due to decoherence, if I understand it correctly.
For what it's worth, I am not a fan of many worlds, and I have struggled with understanding consistent histories.

 

[jimgraber]

I think you are right that this is not new, but a restatement of an existing established position.

 

[Mentz114]

It's fine to agree to disagree, but I would like to know what, specifically, you're disagreeing with.

Thanks to @jimgraber for the reference to the Griffith paper (still wet from the printer !).

Steven, my objections to Bell are stated there

However the derivation of the CHSH version of a Bell inequality, which is Eq. (1) in [1], has as one of its assumptions that S_z and S_y, and other components of spin can be replaced by classical, which is to say commuting, quantities, ...

I don't think I ever understood the maths Bell wrote, but the distinction between quantum and classical reckoning is brought into focus by his multicolored socks.

 

[zonde]

I don't think I ever understood the maths Bell wrote, but the distinction between quantum and classical reckoning is brought into focus by his multicolored socks.

Yes, quantum particle states are described using Hilbert space, but arguments based on this is just attacks on strawmen as detection events are classical facts. There is no need to make assumptions about microscopic reality when we derive Bell type inequality just from measurement results.

 

[DuckAmuck]

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).

In the case of position (or any observable), you may start out with a state distributed in a sinusoidal fashion over x. If you measure the position, the wavefunction "collapses" to a delta function centered on the position x₀ you measured. Over time it will evolve back to a sinusoidal state.

 

[Mentz114]

Yes, quantum particle states are described using Hilbert space, but arguments based on this is just attacks on strawmen as detection events are classical facts. There is no need to make assumptions about microscopic reality when we derive Bell type inequality just from measurement results.

I have to say that I cannot discern what you mean. Mea culpa.

 

[zonde]

I have to say that I cannot discern what you mean.

Say you don't know anything about QM. You just have measurement results at two distant locations (produced by black-box). You can still derive Bell type inequalities for these measurements assuming they are produced by process that respects relativistic locality.

 

[Mentz114]

Say you don't know anything about QM. You just have measurement results at two distant locations (produced by black-box). You can still derive Bell type inequalities for these measurements assuming they are produced by process that respects relativistic locality.

The inequalities serve to define statistics (functions of the data) which are sensitive to the things we are looking for. But any number of different ones can be used if they are suitable for the experimental setup.

 

[vanhees71]

Ok but then, if we ignore everything else but spin, isn't the state of an atom before Stern-Gerlach given by a|spin up>+b|spin down> and after passing through, it is either |spin up> or |spin down>? So there is something collapsing here.

In the Stern-Gerlach experiment you cannot ignore position. It's the crucial point of this experiment that after running through the inhomogeneous magnetic field, you have a spin-position entangled state, namely
$$\psi^j(\vec{x})=\psi_{\text{left}}(\vec{x}) \chi_{\text{up}}^j + \psi_{\text{right}}(\vec{x}) \chi_{\text{down}}^j.$$
If you tailor your magnetic field right, the $\psi_{\text{left}}$ and $\psi_{\text{right}}$ functions peak practically in separated regions of space, and you can filter out one part by just blocking the corresponding partial beam of particles. Then you have prepared particles with a determined spin component in direction of the magnetic field, i.e., up or down.

The blocking is also due to the interactions of the particles with the medium used to block them. There's no other dynamics than quantum dynamics and no collapse hypothesis is necessary.

 

[vanhees71]

It does not follow the standard rules of probability unless one postulates collapse. That is the important point about Bayesian updating - the standard rules of probability are not enough without collapse.

I still don't understand this terminology. You have a well defined initial state, describing the statistical properties of the system, including the correlation between spins, and then it is clear to A what B must measure after A has measured her photon. This is not due to some mystical collapse but just due to the usual rules of quantum kinematics and dynamics (without any necessity for an additional collapse hypothesis).

 

[vanhees71]

Again, I wish to stress that your interpretation is very non-minimal. How can you be sure that "nothing has happened to the object"?

You replied saying QED obeys signal causality. Sure, but as I have stressed repeatedly, signal causality being respected does not mean that "nothing has happened to the object". A simple way to see that you lack an argument for your assertion that the updating is purely informational with nothing happening to the object, is that if collapse is physical, then something has happened to the object and yet faster than light signalling is prevented.

The minimal interpretation is agnostic, not confidently assertive of things it cannot show, unlike your claim that "nothing has happened to the object". If you read Cohen-Tannoudji, Diu and Laloe's famous text, you will see that they are not so cavalier as you are at this point.

Within QED if A's and B's registration events are spacelike separated, the registration event at A cannot have affected B's photon before he registered it and vice versa. Since QED describes all these experiments with very high significance right, I tend to say that's the correct description. If one day one finds a reproducible violation of any prediction within QED, one has to think about a new theory, assuming non-local actions at a distance and maybe even the very foundation of the space-time model, but before this is not the case, I don't see any necessity to give up the very successful standard paradigm of local microcausal relativistic QFT.