Is MWI Considered Local in Quantum Mechanics?

[PeterDonis]

your statement "all of the operations involved in these experiments commute" means nothing in QM

I agree that if you do the latter part, the operations no longer commute, since applying the collapse postulate is non-unitary

I realized that I should clarify this, because it's a very important point. My statement in response to yours (@DrChinese) quoted above is correct--but it also means that even doing "standard" QM too strictly, and applying the collapse postulate immediately after every measurement, regardless of time ordering, means you can't account for the correlations!

In order to get the right answer even with "standard QM", you need to, as you yourself have said before, take into account the entire context, including future measurements that will be made on at least some of the quantum systems involved in the current measurement. In this case, that means that you have to not apply the collapse postulate until you have applied the unitary operators corresponding to all of the measurements in the context--and then, once you have the final state after all of those operators are applied, that state will contain all of the possible final sets of results, each with its correct amplitude, and you can correctly use the Born rule to predict the probabilities for each of the sets of results, and the collapse postulate to reduce the state to the one that reflects the set of results that was actually observed.

In other words, when you insist on taking into account the entire future context in order to properly account for the results, you are doing the same thing that the MWI does! The only difference is that you are applying the collapse postulate at the end. Everything up to that point has to be entirely unitary, with no collapse, in order to properly take into account the entire context. And that is exactly what I did in post #79. If you look at the final state I come up with in post #79, and put back in the correct normalization factors, you will see that it reflects all of the possible sets of outcomes of the three measurements (BSM, Photon 1, Photon 4), each with its correct amplitude. And if you go back and apply the collapse postulate to one of the intermediate states, and then apply further measurement operators, you will see that you don't get the correct final state with the proper correlations in it (unless you are doing the easy case where the BSM happens first).

 

[DrChinese]

1) There is decoherence whenever a measurement is made. That includes the detection of whether or not a swap is made at the BSM, by the photon detections that happen at the detectors at the BSM. Whatever happens at those detections, once they have taken place, there is no longer any maximal entanglement between 1 & 2 (or between 3 & 4).

Either the maximal entanglement was swapped at the BSM (if a swap is detected), so it is now between 1 & 4, and 2 & 3, and the detection then ends the maximal entanglement between 2 & 3 by absorbing those photons at the detectors; or there was no swap and absorbing photons 2 & 3 at the detectors ends the maximal entanglement between 1 & 2 (and 3 & 4), and leaves no entanglement at all between any of the photons. (If one possibility in the "no swap" case involves no detection of photons 2 or 3, then decoherence occurs once those photons have interacted with the environment enough, which will inevitably happen--they can only stay coherent for a finite time.)

2) Decoherence of course does involve entanglement, but it is not maximal between any particular pair of degrees of freedom; it is spread out among a very large number of degrees of freedom in the environment that are not individually trackable, so the entanglement between any two particular degrees of freedom is very small.

1) I don't see how decoherence figures into this discussion at all. If there is decoherence upon measurement of Photon 1 (or at the BSM), which I have no particular objection to assuming for discussion purposes: then you are saying that the entangled state of Photon 2 is over. Were that true by your reasoning, then Photon 2 is now has a definite polarization and is not in a superposition of polarizations as it previously was.

Obviously such definite states (for 1 or for 2 if Photon 1 was measured first) can never later become entangled with Photon 4 under any circumstances. And if the BSM occurs first, we have decoherence of photons 2 & 3 (at essentially the same time). Since you have decoherence as a physical process involving entanglement dissipation into the environment, exactly how do you think 1 & 4 are to become correlated?2) No particular disagreement with this description either, again in which decoherence is a physical process. Under your line of thinking, order *should* matter. Of course we agree it doesn't. Further, we know that the total entanglement cannot exceed a certain threshold due to Monogamy considerations, right?And in reality: the swap actually occurs at the BSM at the precise moment (or small time window) in which Photons 2 & 3 become indistinguishable, right? After all, that's an irreversible operation. And yet... what really happens is that they become entangled. It doesn't matter where Photon 1 is, where Photon 4 is, or for that matter if either have been measured by a polarizer or if they do or don't exist at all. Decoherence matters not.

Of course, I insist that the point at which Photons 2 & 3 become indistinguishable is the swap, and it being irreversible means it is a physical process. When the BSM detectors click, we learn what kind of Bell state we got.

 

[DrChinese]

I realized that I should clarify this, because it's a very important point. My statement in response to yours (@DrChinese) quoted above is correct--but it also means that even doing "standard" QM too strictly, and applying the collapse postulate immediately after every measurement, regardless of time ordering, means you can't account for the correlations!

In order to get the right answer even with "standard QM", you need to, as you yourself have said before, take into account the entire context, including future measurements that will be made on at least some of the quantum systems involved in the current measurement. In this case, that means that you have to not apply the collapse postulate until you have applied the unitary operators corresponding to all of the measurements in the context--and then, once you have the final state after all of those operators are applied, that state will contain all of the possible final sets of results, each with its correct amplitude, and you can correctly use the Born rule to predict the probabilities for each of the sets of results, and the collapse postulate to reduce the state to the one that reflects the set of results that was actually observed.

In other words, when you insist on taking into account the entire future context in order to properly account for the results, you are doing the same thing that the MWI does! The only difference is that you are applying the collapse postulate at the end. Everything up to that point has to be entirely unitary, with no collapse, in order to properly take into account the entire context. And that is exactly what I did in post #79. If you look at the final state I come up with in post #79, and put back in the correct normalization factors, you will see that it reflects all of the possible sets of outcomes of the three measurements (BSM, Photon 1, Photon 4), each with its correct amplitude. And if you go back and apply the collapse postulate to one of the intermediate states, and then apply further measurement operators, you will see that you don't get the correct final state with the proper correlations in it (unless you are doing the easy case where the BSM happens first).

Of course I agree that ordering does not matter, and collapse should not be considered (or applied) as occurring a piece at a time. You must look at the full nonlocal, nontemporal context as highlighted. As you say, you can't get the answer right otherwise.

And yet: MWI claims itself local AND deterministic. The past causes the future in MWI, and not the reverse. (We should agree that because we consciously experience life as proceeding from past to future, it would be difficult to assert things actually work the other way.)

If deterministic, then it is fair game to assert that the ordering of events either local or nonlocal will change outcomes. That doesn't happen in actual experiments.

 

[DrChinese]

1) The "random" effects come into play with collapse, and I am leaving out collapse because the MWI does not have it.

2) You need to read the rest of the posts I referred to. I address all of this in those posts. I said so in what you quoted; please take me at my word and don't assume that I must have failed when you haven't read what I wrote.

1) QM Collapse = MWI branching? What's the difference here?

2) You have this somewhat backwards. I *am* working through your series 79/81/82/83/84 in order, but it's a lot to cover. I'm a bit slow, and not nearly as fast as you are.

And I am assembling a new thread post discussing GHZ and MWI, but I won't be finishing that until I have worked with you through 79/81/82/83/84. You did a lot of work on these, and they deserve proper study.

PS And yes, I do sometimes answer posts in a LIFO manner rather than the more disciplined FIFO approach. At least I avoid NIFO (Next In First out) which would require a leap into the future.

 

[PeterDonis]

@DrChinese, I don't even want to respond to most of what is in your posts #107, #108, and #109, because until you have read through all of the previous posts of mine that I referenced, I think discussion is premature. I have no problem with it taking some time for you to work through those posts, they took me a fair bit of time to write and writing them is of course going to be easier for me than working through them will be for you. Take all the time you need, I'll still be here.

That said, there are a few things that it might help to clarify now:

If there is decoherence upon measurement of Photon 1 (or at the BSM), which I have no particular objection to assuming for discussion purposes: then you are saying that the entangled state of Photon 2 is over.

Not in the MWI, no. In the MWI, Photon 2 is still entangled after the Photon 1 measurement: it's just that the entanglement is no longer with Photon 1, but with all of the degrees of freedom that got involved in the Photon 1 measurement (and the degrees of freedom in the environment that that entanglement spreads to). What is true is that Photon 2 is no longer maximally entangled with any one of those individual degrees of freedom. But the overall entanglement of Photon 1 is still there. Note that none of the states I wrote down in my previous posts are product states of Photon 2 with something else; all of them are entangled.

In a collapse interpretation, yes, measuring one of a pair of entangled particles ends the entanglement, because the collapse forces the state to be a product state. But there is no collapse in the MWI, and the math in my posts reflects that.

Under your line of thinking, order *should* matter.

This is simply wrong, and is one of the things I really wish you would postpone discussion of until you have worked your way through all of my previous posts.

Further, we know that the total entanglement cannot exceed a certain threshold due to Monogamy considerations, right?

I already addressed this in post #103.

in reality: the swap actually occurs at the BSM at the precise moment (or small time window) in which Photons 2 & 3 become indistinguishable, right? After all, that's an irreversible operation. And yet... what really happens is that they become entangled.

If we want to break down the "swap/no swap decision" process, which I didn't do in my earlier posts, here is what I gather from your earlier posts and the papers you referenced. I am treating the idealized version where if a swap is possible at all, it always happens, i.e., the sole relevant variable is the experimenter's decision.

(1) The experimenter makes a decision that determines whether or not a swap occurs. We model this in the math as there being some amplitude $s$ for a swap to occur, and a corresponding amplitude $n$ for no swap to occur, such that $|s|^2 + |n|^2 = 1$. The operator that I called $U_{S/N}$ in my earlier posts can then be expressed as $s U_S + n I$, where $U_S$ is the unitary swap operator and $I$ is the identity.

(2a) If the experimenter decides that a swap will occur, photons 2 & 3 arrive at the BSM within a short enough time window to be indistinguishable, they go through the BSM, and one photon is detected in each output arm of the BSM. This provides the "event ready" indication that a swap has taken place. The state after the swap is given by the unitary operator $U_S$ applied to the state before the swap.

(2b) If the experimenter decides that a swap will not occur, photons 2 & 3 do not arrive at the BSM within a short enough time window to be indistinguishable, they go through the BSM, and a detection occurs either in just one output arm of the BSM, or no detection occurs at all in either output arm of the BSM. For our purposes we combine all of those possibilities into the "no swap" result. The state is unchanged in this case because the operator involved is just the identity.

In the short time between photons 2 & 3 going through the BSM and the detections (if any) in the output arms of the BSM, yes, photons 2 & 3 will be entangled if there is a swap. Once the detections take place, that entanglement spreads to all the degrees of freedom involved in the detections, and their environment. If there is no swap, the previous entanglements of photons 2 & 3 get transferred to either the detector degrees of freedom (if the photons are detected) or directly to the environment (if they aren't detected and just decohere naturally because of their finite coherence time).

Decoherence matters not.

It does if we want the results to be irreversible. Without decoherence, we could imagine, for example, recombining photons, as in a Mach-Zehnder interferometer, and undoing the swap operation.

If deterministic, then it is fair game to assert that the ordering of events either local or nonlocal will change outcomes.

In general it might, but in the case under discussion, it doesn't. You can't just assert that the ordering will change outcomes based on "determinism". You have to actually do the math and see. That's what I've done.

QM Collapse = MWI branching? What's the difference here?

MWI branching is unitary. Collapse is not.

For example, if I have two entangled photons in the singlet state, $H_1 V_2 - V_1 H_2$, and I measure Photon 1, collapse says the state becomes either $\bar{H}_1 V_2$ or $\bar{V}_1 H_2$; those are both non-unitary transformations from the original state. The MWI says the state becomes $\bar{H}_1 V_2 - \bar{V}_1 H_2$, which is a unitary transformation from the original state (the bars just mean spreading entanglement among the degrees of freedom in the Photon 1 detector and its environment). Those are different states. They are indistinguishable experimentally because the bars, indicating decoherence, mean that you can't interfere the terms any longer, so there is no way, for example, to build a Mach-Zehnder interferometer that undoes the Photon 1 measurement and allows us to distinguish the MWI state from either of the two collapse states.

 

[PeterDonis]

doing "standard" QM too strictly, and applying the collapse postulate immediately after every measurement, regardless of time ordering, means you can't account for the correlations!

Actually, after taking a further look at the math, I'm no longer sure even this is true, at least not for the cases we're discussing. But I'll postpone further posting along those lines for now since we have enough on the table at present.

 

[Morbert]

3. We hopefully agree that before the swap, we have $|\Psi^i\rangle_{12}\Psi^i\rangle_{34}$ and after we have $|\Psi^i\rangle_{14}\Psi^i\rangle_{23}$ except photons 2 & 3 no longer exist so it simplifies to $|\Psi^i\rangle_{14}$

Under Wallace's account of Everettian QM: Before the swap, we have $$|\Psi\rangle\langle\Psi| = |\Psi^1_{12},\Psi^1_{34},M^\mathrm{ready},\epsilon^0\rangle\langle \Psi^1_{12},\Psi^1_{34},M^\mathrm{ready},\epsilon^0|$$ after the swap (and now supposing the BSM is destructive but otherwise ideal), we have $$U|\Psi\rangle\langle\Psi|U^\dagger = \sum_{i,j}c_ic^*_j|\Psi^i_{14},M^i,\epsilon^i\rangle\langle \Psi^j_{14},M^j,\epsilon^j|$$

4. I am lost here. It seems we now have an extended spacetime region which is nonlocal in every respect - if you are saying the BSM is not spacelike separated from the rest of the system because it is now included in the overall system. I know your are channeling what Wallace would say, that's proper. But by my thinking it is circular to say there is no AAD as long as you consider a big enough subsystem. There is a person making a decision, and the entire extended system changes (or not) based on her decision.

So basically: if Wallace's definition were reasonable, we would need to include ALL potential sources in our equation. That essentially means a spacetime region that is not particularly limited in any way. Not the most useful of ways to avoid calling something "action at a distance".

The bit in bold needs to be made more precise. It is true that properties of the entire system can be affected by the BSM, but properties of subsystems, with spacelike separated instantiations, cannot. The BSM apparatus subsystem, for example, is spacelike separated from the [1,4]-photon subsystem. The state of the [1,4]-photon subsystem before BSM is $$\mathrm{Tr}_{2,3,M,\epsilon}|\Psi\rangle\langle\Psi| = \sum_i|c_i|^2|\Psi^i\rangle_{14}\langle\Psi^i|_{14}$$We can see that there is no entanglement. As you say, if the state was something like $|\Psi^i\rangle_{14}\langle\Psi^i|_{14}$, there would be entanglement. After the BSM, the state is $$\mathrm{Tr}_{2,3,M,\epsilon}U|\Psi\rangle\langle\Psi|U^\dagger = \sum_i|c_i|^2|\Psi^i\rangle_{14}\langle\Psi^i|_{14}$$We can see that the [1,4]-photon state has not changed. This isn't surprising since this subsystem is spacelike-separated from the BSM. This is what Wallace means by no AAD: The properties instantiated by the [1,4]-photon subsystem, represented by the above state, cannot be immediately affected by the BSM. We could introduce some nonrelativistic dynamics and break this constraint, but entanglement doesn't break this constraint.

You might not agree with this constraint being called "no AAD", but hopefully you can at least appreciate the distinction in the Everettian context. If AAD was permitted, then the properties instantiated by a subsystem could be affected by events outside the past light cone of the subsystem.

5. OK, if you want to say that separability/nonseparability defines a kind of definition of locality, that makes some sense. But for it to apply, you still have to say 1 & 4 started in a Product State and ended in an Entangled State.

The problem is an assertion about 1 & 4, unqualified, implies an assertion about properties of that subsystem alone. Before and after the BSM we have a mixed state, unaffected by the BSM. If we want to identify 1 & 4 entanglement. We have to e.g. consider a larger subsystem including the quasiclassical branches of the BSM, and identify the relative states: The state of 1 & 4 relative to each of the decoherent branches.

 

[vanhees71]

a. Global variables sound confusingly nonlocal to me. And the issue we are discussing is whether MWI is local, not the nonlocal elements of other interpretations. We already (mostly) agree that standard QM has nonlocal elements. The nonlocal extent of an entangled system is something even @vanhees71 agrees to.

I discuss the following, using the clear and solid language of HEP physics:

Definition: a relativistic quantum-field theory is called local, if the local observables obey the microcausality condition and the Hamilton density is a local observable-operator. This implies Einstein causality, i.e., that there's no causal connection between space-like separated events.

Non-relativistic physics has no notion of "locality" in this sense.

Now to the above statements.

I also don't know, what "global variables" should be too. I also have no clue, whether the many-worlds interpretation is local or not. For me MWI doesn't make sense at all.

Standard QM of course has "nonlocal elements", because Newtonian physics is nonlocal. There's no limiting speed in the Newtonian spacetime model, and instantaneous interactions are the standard way to describe interactions, and the assumption of the existence of instantaneous signal propagation is implicit in the assumption of an absolute time.

Entangled systems do not have a "non-local extent" but they describe correlations between the outcome of measurements on observables with indetermined properties, that are stronger than possible for a realistic local hidden-variable theory. It seems to me that the notion of "non-locality" in the "quantum-foundations community" usually has this meaning, but that is a "non-locality" which is NOT at odds with the above defined notion of locality in the HEP-community sense. That's why I'd prefer it very much, if one could find another term, and I consider Einstein's definition of "inseparability" as in his Dialectica article of 1948 as very clear.

b. I'm accepting this point as being a tenet of MWI. And this is actually its most appealing point, in my opinion. It would solve a lot of conceptual problems.

c. Whoa, that's completely impossible! Photons 1 and 4 aren't entangled yet! That doesn't occur until our distant observer chooses to perform the swap or not. And that can be done AFTER Photon 1 is already measured, and there has been a splitting into a V> branch and an H> branch. Photon 4 has no connection to Photon 1 whatsoever, any more than it has a connection with any other photon anywhere. There is nothing at this point that gives an indication that they will be entangled in the future.

I think, again the confusion is immediately gone when using the minimal statistical interpretation. In the entanglement swapping experiment you consider indeed the description of different ensembles (in the experiments realized as sufficiently large "statistical samples")

You start with two independently prepared entangled two-photon states, i.e., a state of the form
$$|\Psi \rangle=|\psi_{12} \rangle \otimes |\psi_{34} \rangle,$$
where for simplicity we may take the singlet states (it doesn't really matter, you can use either of the four Bell states you like, it's only important that you have maximally entangled two-photon states)
$$|\psi_{ab} \rangle=\frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{p}_a,\text{H}) \hat{a}^{\dagger}(\vec{p}_b,\text{V}) - \hat{a}^{\dagger}(\vec{p}_a,\text{V}) \hat{a}^{\dagger}(\vec{p}_b,\text{H})].$$
Here, of course the photon pairs 2 and 3 as well as 1 and 4 are unentangled, i.e., the reduced polarization state is
$$\hat{\rho}_{23} =\frac{1}{4} \hat{1}_2 \otimes \hat{1}_3,$$
i.e., you simply have two independent unpolarized photons.

Now you do a (local!) Bell measurement on photons 2 and 3, enabling you to select a subensemble of the above prepared photons in only considering the four photons if photons 2 and 3 have been found to be in the singlet state $|\psi_{23} \rangle$. This happens randomly for each single experiment with a probablity of 1/4. The so selected subensemble is then described by the state
$$|\Psi' \rangle=|\psi_{23} \rangle \otimes |\psi_{14} \rangle,$$
i.e., also the pair 14 is entangled. That's "entanglement swapping". It makes use of the fact that there's the strong correlation between photons 1 and 2 as well as the photons 3 and 4. Although each of the single photons is ideally unpolarized, there's the 100% correlation of the polarizations of photons 1 and 2 as well as of photons' 3 and 4, and this enables the selection of a subensemble in which photons 1 and 4 are entangled although these two photons never have been in "causal contact" with each other by selecting a subensemble due to measurement results on photons 2 and 3. The cause of the entanglement of the photons 1 and 4 due to selection based on measurements on photons 2 and 3 is that the four photons have been prepared in the above given specific state, and this "preparation procedure events" (i.e., the creation of each of the entangled pairs by parametric down conversion) are in the past light-cone of the measurement event on photons 2&3 finding them in the said entangled state, and this implies the entanglement also of photons 1&4 for this subensemble.

d. This is my point. It cannot be local because distant events have yet to occur that will change Photon 4's relationship with Photon 1 from Product State to Entangled State. That occurs in the future, and MWI is supposed to strictly reject anything which does not follow Einsteinian causality. (And please, don't ask me to define that as I think everyone understands that term the same way.)

Nothing is changed on photons 1&4, you only selected a subensemble due to an outcome of a local measurement on photons 2&3. The correlations between 1&2 as well as between 3&4, necessary for the entanglement of 1&4 in the subensemble, have been present already before the measurement on photons 2&3, i.e., due to the corresponding preparation of the four photons in the past light cone of this measurement.

There's nowhere any need for faster-than-light signal propgation or other "non-locality" violating Einstein causality as soon as you use the minimal statistical interpretation. I don't know, how MWI interprets the locality (in the HEP sense) though, i.e., what is considered observable in MWI. That's why I prefer the minimal interpretation, because it is an interpretation clarifying the meaning of the QT formalism in a way as applied by experimentalists when doing experiments in the real world. They never experience the splitting of the universe only because detector A makes a click and not detector B in their universe ;-)).

 

[Morbert]

@vanhees71 The relevant question is how this experiment is framed by MWI. E.g. "An experiment has a set of possible outcomes, one of which occurs" is replaced by "An experiment produces a set of decoherent branches".

So instead of the (ideal) BSM producing an outcome that we can use to select a subensemble. The BSM produces a branching event, and as observers enter the future light cone of this event, they self-locate themselves on one of the branches, and describe the world in terms of states relative to that branch.

 

[Morbert]

@DrChinese Maybe this approach will be helpful: Light cone diagrams used in Wallace's book, but applied to the entanglement swapping experiment. Shown below is a sketch of a spacetime diagram with two observers: Alice and Bob. Alice and Bob are both using frames of reference that agree on the simultaneity of events (though this is not needed)

At time t0, both alice and Bob say there is no entanglement in the distant [1,4]-photon subsystem (not shown).

At time t1, alice performs an ideal BSM, the light cone of which is shown as the black triangle. This event begins a branching process, and Alice will identify with one of the branches.

At time t2, Alice, identifying with one of the branches, will describe the [1,4]-photon subsystem with a state relative to her branch. She will say there is entanglement between photons 1 and 4. Bob, however, is outside the light cone of the BSM. He will continue to describe the [1,4]-photon subsystem the usual no-entanglement mixed state. He will also describe the Alice + BSM subsystem with a highly nonclassical macroscopic superposition: 4 branches with 4 Alices.

At time t3, Bob will have branched due to the BSM. Like Alice, he will now uniquely identify with one of the branches, and describe the [1,4]-system with a relative state. He will agree with the Alice on his branch that photons 1 and 4 are entangled.

 

[PeterDonis]

Alice will identify with one of the branches.

More precisely, there will be a branch of Alice corresponding to each of the branches that result from the BSM. Similarly, when Bob branches, there will be a branch of Bob corresponding to each of the branches that result from whatever event caused Bob to branch.

At time t3, Bob will have branched due to the BSM.

We need to be very careful here: as you state this, it can't be correct.

There are two possibilities: either Bob is entangled with any of the degrees of freedom involved, or he isn't.

If Bob is entangled with any of the degrees of freedom involved, then he branches instantaneously when the BSM is done, because the wave function is nonlocal and all entangled subsystems branch when any branching event affects any one of them. Bob entering the future light cone of the BSM (assuming that information about the BSM result is signaled to him at the speed of light) is when the Bob in each branch knows the branch he is in (because he now knows the BSM result), and updates his model accordingly. But Bob updating his model is not the same as Bob branching.

If Bob is not entangled with any of the degrees of freedom involved, then he does branch when the information about the BSM result reaches him: but then what causes his branching is not the BSM itself, but his observation of the light signals carrying the information about the BSM result, which entangles him with the source of the light. And in this case, Bob is not analogous to, for example, the other photons in the experiment, because those photons are entangled with the degrees of freedom involved in the BSM.

 

[DrChinese]

1) @DrChinese, I don't even want to respond to most of what is in your posts #107, #108, and #109, because until you have read through all of the previous posts of mine that I referenced, I think discussion is premature. I have no problem with it taking some time for you to work through those posts, they took me a fair bit of time to write and writing them is of course going to be easier for me than working through them will be for you. Take all the time you need, I'll still be here.

2) Not in the MWI, no. In the MWI, Photon 2 is still entangled after the Photon 1 measurement: it's just that the entanglement is no longer with Photon 1, but with all of the degrees of freedom that got involved in the Photon 1 measurement (and the degrees of freedom in the environment that that entanglement spreads to). What is true is that Photon 2 is no longer maximally entangled with any one of those individual degrees of freedom. But the overall entanglement of Photon 1 is still there. Note that none of the states I wrote down in my previous posts are product states of Photon 2 with something else; all of them are entangled.

In a collapse interpretation, yes, measuring one of a pair of entangled particles ends the entanglement, because the collapse forces the state to be a product state. But there is no collapse in the MWI, and the math in my posts reflects that.3) If we want to break down the "swap/no swap decision" process, which I didn't do in my earlier posts, here is what I gather from your earlier posts and the papers you referenced. I am treating the idealized version where if a swap is possible at all, it always happens, i.e., the sole relevant variable is the experimenter's decision.

(1) The experimenter makes a decision that determines whether or not a swap occurs. We model this in the math as there being some amplitude $s$ for a swap to occur, and a corresponding amplitude $n$ for no swap to occur, such that $|s|^2 + |n|^2 = 1$. The operator that I called $U_{S/N}$ in my earlier posts can then be expressed as $s U_S + n I$, where $U_S$ is the unitary swap operator and $I$ is the identity.

(2a) If the experimenter decides that a swap will occur, photons 2 & 3 arrive at the BSM within a short enough time window to be indistinguishable, they go through the BSM, and one photon is detected in each output arm of the BSM. This provides the "event ready" indication that a swap has taken place. The state after the swap is given by the unitary operator $U_S$ applied to the state before the swap.

(2b) If the experimenter decides that a swap will not occur, photons 2 & 3 do not arrive at the BSM within a short enough time window to be indistinguishable, they go through the BSM, and a detection occurs either in just one output arm of the BSM, or no detection occurs at all in either output arm of the BSM. For our purposes we combine all of those possibilities into the "no swap" result. The state is unchanged in this case because the operator involved is just the identity.

In the short time between photons 2 & 3 going through the BSM and the detections (if any) in the output arms of the BSM, yes, photons 2 & 3 will be entangled if there is a swap. Once the detections take place, that entanglement spreads to all the degrees of freedom involved in the detections, and their environment. If there is no swap, the previous entanglements of photons 2 & 3 get transferred to either the detector degrees of freedom (if the photons are detected) or directly to the environment (if they aren't detected and just decohere naturally because of their finite coherence time).4) It does if we want the results to be irreversible. Without decoherence, we could imagine, for example, recombining photons, as in a Mach-Zehnder interferometer, and undoing the swap operation.5) In general it might, but in the case under discussion, it doesn't. You can't just assert that the ordering will change outcomes based on "determinism". You have to actually do the math and see. That's what I've done.6) They are indistinguishable experimentally because the bars, indicating decoherence, mean that you can't interfere the terms any longer, so there is no way, for example, to build a Mach-Zehnder interferometer that undoes the Photon 1 measurement and allows us to distinguish the MWI state from either of the two collapse states.

1) All good, thanks.

2) I can accept all of this, no objections as presented.

3) Thanks for clarifying that we are using the ideal model where the experimenter is making the swap/no swap decision. Note that the clicks on the BSM detector occur in exactly the same manner whether there is a swap or not (we are still looking for 2 clicks). The only difference is that since they are distinguishable when there is no swap, one click occurs a bit outside the time window. That of course identifies the path in which a delay was added. For our discussion, we can combine all variations of "no swap" (either through experimenter choice or less than ideal conditions) as you say.

4) I agree that there is the ability to reverse the measurements of Photons 1 & 4 (by re-combining the outputs of a PBS). So I can accept your perspective about decoherence on this side from the QM perspective. Not sure how that fits in to the MWI perspective, but I am guessing I will understand that better shortly.

Not sure it matters for our discussion, but I am unaware of any manner in which 2 indistinguishable particles (the ones in the BSM) can later have distinguishability restored later. Seems like a contradiction in terms. Of course, after the detectors click, no reversibility is possible anyway.

5) Good.

6) Good with this view.

 

[PeterDonis]

I agree that there is the ability to reverse the measurements of Photons 1 & 4 (by re-combining the outputs of a PBS).

I don't think we can reverse the measurements, because "measurement" says that decoherence has occurred, so now it's not enough just to recombine PBS outputs, you would have to somehow find and reverse all of the entanglements that have spread through the environment, and that's not possible in practice (although since decoherence is unitary it is possible in principle, at least according to current theory).

Also, what you would need to recombine to undo the swap operation is Photons 2 & 3; in other words, instead of putting detectors in the two output arms of the BSM beam splitter, you would need to make that the first beam splitter in a Mach-Zehnder interferometer, where the second beam splitter in the interferometer recombines the beams. That (I think--I haven't done the math) would restore Photons 2 & 3 to the states they had before the first beam splitter, and therefore would undo the swap. But all that requires that there is no decoherence anywhere between the beam splitters, which means you can't detect Photons 2 & 3 there (you could detect them after the second beam splitter, and you would then expect to see the appropriate correlations with Photons 1 & 4 that are predicted by the originally prepared state, where 1 & 2 and 3 & 4 are the entangled pairs).

 

[DrChinese]

@DrChinese Maybe this approach will be helpful: Light cone diagrams used in Wallace's book, but applied to the entanglement swapping experiment. Shown below is a sketch of a spacetime diagram with two observers: Alice and Bob. Alice and Bob are both using frames of reference that agree on the simultaneity of events (though this is not needed)

View attachment 336903

At time t0, both alice and Bob say there is no entanglement in the distant [1,4]-photon subsystem (not shown).

At time t1, alice performs an ideal BSM, the light cone of which is shown as the black triangle. This event begins a branching process, and Alice will identify with one of the branches.

At time t2, Alice, identifying with one of the branches, will describe the [1,4]-photon subsystem with a state relative to her branch. She will say there is entanglement between photons 1 and 4. Bob, however, is outside the light cone of the BSM. He will continue to describe the [1,4]-photon subsystem the usual no-entanglement mixed state. He will also describe the Alice + BSM subsystem with a highly nonclassical macroscopic superposition: 4 branches with 4 Alices.

At time t3, Bob will have branched due to the BSM. Like Alice, he will now uniquely identify with one of the branches, and describe the [1,4]-system with a relative state. He will agree with the Alice on his branch that photons 1 and 4 are entangled.

I agree with everything you say above, but it leaves out the part where Photons 1 & 4 are affected by Alice's actions. That occurs completely outside Alice's light cone. On the other hand, Bob's agreeing with Alice at t3 is simply a result of getting a message from Alice telling him what she observed.

So once you add in Chris and Dale, who at t3 remain outside Alice's light cone and are observing photons 1 & 4 (they can be at the same spot or at different locations), it seems everything falls apart there.

But your diagram is how I understand MWI to work.

 

[PeterDonis]

your diagram is how I understand MWI to work.

It's not entirely correct as a description of how MWI works, at least not as @Morbert is interpreting it. See my post #116.

 

[DrChinese]

1) I don't think we can reverse the measurements, because "measurement" says that decoherence has occurred...

2) Also, what you would need to recombine to undo the swap operation is Photons 2 & 3; in other words, instead of putting detectors in the two output arms of the BSM beam splitter, you would need to make that the first beam splitter in a Mach-Zehnder interferometer, where the second beam splitter in the interferometer recombines the beams. ...

1) Fine.

2) Recombining them would not (I believe) produce distinguishable outputs as to the source. But I don't think this matters, we'll use your definition as you have it. If it is measured, there is decoherence.

 

[DrChinese]

It's not entirely correct as a description of how MWI works, at least not as @Morbert is interpreting it. See my post #116.

Your #116 is fine too. Of course we are still missing 1 & 4, and that is the $64,000 question.

 

[Morbert]

If Bob is entangled with any of the degrees of freedom involved, then he branches instantaneously when the BSM is done, because the wave function is nonlocal and all entangled subsystems branch when any branching event affects any one of them. Bob entering the future light cone of the BSM (assuming that information about the BSM result is signaled to him at the speed of light) is when the Bob in each branch knows the branch he is in (because he now knows the BSM result), and updates his model accordingly. But Bob updating his model is not the same as Bob branching.

As per usual, assuming Wallace's account of MWI: If Bob is entangled with the relevant degrees of freedom, Alice will use the appropriate relative state to describe Bob, but this doesn't mean a branching event is initiated at Bob's site when Alice performs her BSM. The nonlocal property of entanglement between microscopic degrees of freedom at Alice and Bob can only have determinable effects at the intersection of Alice and Bob's light cones.

Instead, a local branching event is initiated by Bob when he entangles himself with the relevant degrees of freedom, and a separate local branching event is initiated when Alice does the same, and a third event occurs when these two events intersect. It is this third event where the initial entanglement can have an effect. [edit] -removed the "branching" qualifier because it's not the same event as the events initiated by Alice and Bob. It merely notes the point of intersection.

Wallace provides such an account of two spacelike separated measurements on an entangled particle pair in chapter 8.7 of his book, and the picture below is from that account. I have added the names Alice and Bob.

If Bob is not entangled with any of the degrees of freedom involved, then he does branch when the information about the BSM result reaches him: but then what causes his branching is not the BSM itself, but his observation of the light signals carrying the information about the BSM result, which entangles him with the source of the light.

I don't see this as meaningfully different from what I said before so I will drop this "branch" of the conversation for now unless a meaningful difference is made clear.

 

[DrChinese]

1) I discuss the following, using the clear and solid language of HEP physics:

Definition: a relativistic quantum-field theory is called local, if the local observables obey the microcausality condition and the Hamilton density is a local observable-operator. This implies Einstein causality, i.e., that there's no causal connection between space-like separated events.

2) Entangled systems do not have a "non-local extent" but they describe correlations between the outcome of measurements on observables with indetermined properties, that are stronger than possible for a realistic local hidden-variable theory. It seems to me that the notion of "non-locality" in the "quantum-foundations community" usually has this meaning, ...

3) Now you do a (local!) Bell measurement on photons 2 and 3, enabling you to select a subensemble of the above prepared photons in only considering the four photons if photons 2 and 3 have been found to be in the singlet state $|\psi_{23} \rangle$. This happens randomly for each single experiment with a probablity of 1/4. The so selected subensemble is then described by the state
$$|\Psi' \rangle=|\psi_{23} \rangle \otimes |\psi_{14} \rangle,$$
i.e., also the pair 14 is entangled. That's "entanglement swapping". It makes use of the fact that there's the strong correlation between photons 1 and 2 as well as the photons 3 and 4. Although each of the single photons is ideally unpolarized, there's the 100% correlation of the polarizations of photons 1 and 2 as well as of photons' 3 and 4, and this enables the selection of a subensemble in which photons 1 and 4 are entangled although these two photons never have been in "causal contact" ...

1) The issue we have with this centers around the usage of the word "causal". As you use the word, there is no causal action at a distance (AAD) possible - and I agree with that usage of the word "causal". This definition is perfectly fine, and I agree that AAD that involves doing something at one spot cannot deterministically ("causally") affect something at a far away location (i.e. outside the relevant light cone).

But no one in the general physics community is saying otherwise! What is being asserted is that there is a kind of AAD - called "quantum nonlocality" or just "nonlocality" in which indeterministic (random) effects propagate superluminally. I won't quote experiments, but simply quote the 2022 Nobel committee: "...[Zeilinger's] research group has demonstrated a phenomenon called quantum teleportation, which makes it possible to move a quantum state from one particle to one at a distance [i.e. outside a light cone]."

So if you define such indeterministic AAD to violate "Einsteinian causality" (as I do), fine. If you choose to say it violates something else (so you can maintain "Einsteinian causality"), that's fine - use whatever term you like for what is being violated by experimentally demonstrated indeterministic AAD. 2) If you choose to say that entangled particles exhibit correlations stronger than local realism allows, I agree with that. If you choose to say that such correlations can occur without long distance entanglement, I would disagree strongly.

The most common viewpoint for entanglement of 2 photons is that they DO have spatiotemporal extent. And in fact such a system is defined as one biphoton. (Which violates conventional locality by definition.)3) We have already well settled the fact that there are no subensembles of (1&2)x(3&4) in which 1&4 share any entanglement (or correlations) whatsoever. From our swapping example: "We confirm successful entanglement swapping by testing the entanglement of the previously uncorrelated photons 1 and 4." But sure, go ahead and ignore the results of Nobel winning experiments.

 

[PeterDonis]

Of course we are still missing 1 & 4

Once you've worked through the rest of my earlier posts that I referenced, my take on that should be clear.

 

[PeterDonis]

Wallace provides such an account of two spacelike separated measurements on an entangled particle pair in chapter 8.7 of his book

His book is pop science as far as I know, which means it is not a valid reference here. If his book contains references to peer-reviewed papers of his that expound the same viewpoint, then references to those papers would be very helpful.

a branching event is initiated by Bob when he entangles himself with the relevant degrees of freedom

And if this is what causes Bob to branch, then, as I said before, he wasn't entangled with any relevant degrees of freedom before that event, which means Bob is irrelevant to the questions we are actually discussing, questions like: if the swap/no swap choice is made by the experimenter at the BSM that operates on Photons 2 & 3, when do Photons 1 & 4 branch into "swap" and "no swap" branches? Photons 1 & 4 are entangled with 2 & 3 prior to the swap/no swap decision, so Bob is not analogous to them and nothing Wallace or anyone else says about Bob is relevant to them. Or, if Photon 1 is measured before the swap/no swap decision is made, when does Photon 2 branch? Photon 2 is entangled with Photon 1, so anything said about Bob is irrelevant. And so on.

I'm not aware of any peer-reviewed paper, by Wallace or anyone else, that addresses the case where Bob is entangled with the degrees of freedom that Alice operates on. But if you have a reference to one, by all means post it.

 

[gentzen]

His book is pop science as far as I know

I highly doubt that his book can be classified as "pop science". From a review:

While Albert [1988] was afraid to scare the reader with the concept of a spin and complicated his book by simulating it with colors, Wallace uses positive operator valued measures (POVMs), C-star algebras, Borel measure, decoherence functional, decision-theoretic representation theorems and many other concepts that most physics graduates never encountered. The reader is assumed to have a significant philosophical background too, so the book is fully accessible only to those few who can, like Wallace, have Ph.D. both in physics and philosophy.

Of course, it is not a textbook, and not peer reviewed either. So I am not disputing that it is not a valid reference here. But if you really want to claim that it is pop science, I would like to see a reference for your definition of pop science.

 

[PeterDonis]

if you really want to claim that it is pop science, I would like to see a reference for your definition of pop science.

Pop science is anything that's not a textbook or peer reviewed paper. Yes, of course the quality of such works varies widely. But the point is that whatever the author claims in the book is not constrained by having to pass peer review by other experts. That's what makes it pop science: the author can get away with claims that they would not be able to get away with in a peer-reviewed work, because there are no other experts in the loop to challenge them.

 

[gentzen]

Pop science is anything that's not a textbook or peer reviewed paper.

I was more looking for a published reference, like this definition from wikipedia

Popular science (also called pop-science or popsci) is an interpretation of science intended for a general audience. While science journalism focuses on recent scientific developments, popular science is more broad ranging. It may be written by professional science journalists or by scientists themselves. It is presented in many forms, including books, film and television documentaries, magazine articles, and web pages.

or a listing of its common characteristics, like (from the same article)

Some usual features of popular science productions include:

• Entertainment value or personal relevance to the audience
• Emphasis on uniqueness and radicalness
• Exploring ideas overlooked by specialists or falling outside of established disciplines
• Generalized, simplified science concepts
• Presented for an audience with little or no science background, hence explaining general concepts more thoroughly
• Synthesis of new ideas that cross multiple fields and offer new applications in other academic specialties
• Use of metaphors and analogies to explain difficult or abstract scientific concepts

That definition distinguishes between "science journalism" and "pop science", even so both fall under your definition of being neither a textbook, nor peer reviewed. Also note that a monograph is not necessarily a textbook (even so some people may disagree). In fact, Wallace's book can be seen as a pretty normal monograph in philosophy.

 

[PeterDonis]

a monograph is not necessarily a textbook (even so some people may disagree)

A monograph is kind of a gray area; they aren't typically (as far as I know) peer-reviewed, but they are constrained more by existing research.

Wallace's book can be seen as a pretty normal monograph in philosophy.

Possibly, but that doesn't mean it would also be considered a pretty normal monograph in physics.

 

[Morbert]

And if this is what causes Bob to branch, then, as I said before, he wasn't entangled with any relevant degrees of freedom before that event

You did not say this before. You supposed that he was entangled with the relevant degrees of freedom before Alice's BSM, not his own interactions that induced his entanglement.

which means Bob is irrelevant to the questions we are actually discussing, questions like: if the swap/no swap choice is made by the experimenter at the BSM that operates on Photons 2 & 3, when do Photons 1 & 4 branch into "swap" and "no swap" branches? Photons 1 & 4 are entangled with 2 & 3 prior to the swap/no swap decision, so Bob is not analogous to them and nothing Wallace or anyone else says about Bob is relevant to them. Or, if Photon 1 is measured before the swap/no swap decision is made, when does Photon 2 branch? Photon 2 is entangled with Photon 1, so anything said about Bob is irrelevant. And so on.

Bob is relevant to the questions we are discussing because observer-dependence is an interesting angle to explore.

 

[PeterDonis]

You did not say this before.

Yes, I did:

If Bob is not entangled with any of the degrees of freedom involved, then he does branch when the information about the BSM result reaches him: but then what causes his branching is not the BSM itself, but his observation of the light signals carrying the information about the BSM result, which entangles him with the source of the light. And in this case, Bob is not analogous to, for example, the other photons in the experiment, because those photons are entangled with the degrees of freedom involved in the BSM.

 

[PeterDonis]

You supposed that he was entangled with the relevant degrees of freedom before Alice's BSM

That was one of two possibilities that I described. But I also described the other.

 

[Morbert]

Yes, I did:

That was one of two possibilities that I described. But I also described the other.

You are not following your own train of thought. You confused the former scenario for the latter.

 

[PeterDonis]

Bob is relevant to the questions we are discussing because observer-dependence is an interesting angle to explore.

Then you need to formulate a scenario that includes all of the measurements that @DrChinese specified, not just the BSM, and includes whatever kind of observer dependence you want to explore. What you have given so far does not do that.

 

[PeterDonis]

You are not following your own train of thought. You confused the former scenario for the latter.

Where did I do that? I don't see any such confusion in any of my posts, or in anything you quoted from my posts.

 

[Morbert]

Where did I do that? I don't see any such confusion in any of my posts, or in anything you quoted from my posts.

You said
"And if this is what causes Bob to branch, then, as I said before, he wasn't entangled with any relevant degrees of freedom before that event" which is not true, as he would have been entangled with any relevant degrees of freedom before the event, namely the BSM by Alice.

 

[Morbert]

Then you need to formulate a scenario that includes all of the measurements that @DrChinese specified, not just the BSM, and includes whatever kind of observer dependence you want to explore. What you have given so far does not do that.

My post was fine.

 

[PeterDonis]

You said
"And if this is what causes Bob to branch, then, as I said before, he wasn't entangled with any relevant degrees of freedom before that event" which is not true, as he would have been entangled with any relevant degrees of freedom before the event, namely the BSM by Alice.

You're the one who is missing the train of thought. Let's go back and start again.

There are two logical possibilities: either Bob is entangled with any relevant degrees of freedom before the event, or he isn't.

If he is entangled, then what causes him to branch is Alice making the BSM. Bob's branching happens instantaneously, because branching happens to the wave function, not in spacetime, and Bob, being entangled, is part of the wave function that all branches instantaneously when the BSM is done.

Therefore, if what causes Bob to branch is the light signals from the BSM reaching him, then, by simple contraposition, Bob wasn't entangled with any relevant degrees of freedom before the light signals reached him--because if he had been, he would have branched instantaneously when the BSM happened, not when the light signals from it reached him.

If the specification of the scenario that you intended is that Bob was entangled with relevant degrees of freedom before the event, that just means that Bob receiving the light signals cannot be what causes Bob to branch, by the reasoning I have just given. So if you are claiming both that Bob was entangled and that what causes him to branch is the light signals reaching him, then you are simply being inconsistent. You have to pick one or the other. You can't have both.

My post was fine.

Not for what I asked for. See above.

 

[Morbert]

If he is entangled, then what causes him to branch is Alice making the BSM.

The above statement is what is being challenged by Wallace's account of branching. I.e. Whether or not it is possible that Bob is entangled with the relevant degrees of freedom and also doesn't undergo instantaneous branching induced by Alice making the BSM. Your contraposition only follows if the above statement survives the challenge.

Bob's branching happens instantaneously, because branching happens to the wave function, not in spacetime.

Wallace discusses spacetime branching, and references authors like Nuel Belnap and Guido Bacciagaluppi, and I will need some time to see which references are relevant. (Though a preliminary scan suggests Bacciagaluppi is most relevant, and I will share when I can confirm this). Ultimately The relation between branching and the wave function is more subtle than the above quote.