Collapse of Wave: Effects on Energy & Position Measurements

[atyy]

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)?

The photons collapse when they interact with a classical measuring apparatus.

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".

Collapse does not violate signal causality, because it does not allow classical information to be transmitted faster than light. Just as Bell's theorem uses classical relativistic causality to generate inequalities, signal causality also generates inequalities. This is illustrated in Fig 2 of http://arxiv.org/abs/1303.2849, where they use the term "no signaling" for signal causality.

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).

In classical relativitistic causality, the real objects are the invariant events, and the causal structure defined by the light cones, eg. http://visualrelativity.com/LIGHTCONE/lightcone.html.

Bell's theorem says that relativistic quantum field theory violates classical relativistic causality.

 

[atyy]

What is classical relativistic causality?

I am referring to what is often called "local causality". The concept is diagrammed in Fig 1 of http://www.scholarpedia.org/article/Bell's_theorem.

 

[vanhees71]

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.

You can never be sure of anything. What I meant is that all Bell experiments so far can be well understood by standard QED, and as a local microcausal QFT it is a model where by construction two space-like separated events cannot be causally connected. So by construction there's no superluminal signal propagation within QED. 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?

 

[vanhees71]

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



What is classical relativistic causality?

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

 

[vanhees71]

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.

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. Since there's no known flaw of QED, I think it's a good hypothesis to think that it is the correct description. Of course, it's not a model-independent proof. One day one may find a non-local classically realistic theory which is as good as QED, and then you may interpret the experiment differently. The question, however, is if you can somehow create such a model. I'm pretty sceptical.

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.

Why? According to the preparation in a polarization-entangled pair to the contrary the polarization of the single photons is maximally indetermined, i.e., they are perfectly unpolarized photons.

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.

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).

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.

• 
  Indeed it is, but it's not due to her measurement that Bob's photon's/particle's polarization is now known (by Alice not necessarily Bob!) but it's due to her measurement and the preparation of the photon/particle pair in a polarization-entangled state in the very beginning of the experiment.
  

  [*]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.

It was true beforehand, and it doesn't violate Bell's theorem, because QED is not a locally realistic theory, which is the assumption going into the derivation of Bell's inequality and that's the very merit of Bell to have found this tricky way to test local realism against the prediction of entanglement within QT. Entanglement is precisely the most quantum feature at all, and it cannot be explained within a locally realistic theory!

 

[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.