I Exploring Non-locality in Many-Worlds Interpretation (MWI)

[bhobba]

OK, is the wave function a function? If yes, what is the domain of that function? Without any reference to space, time, particles, fields, chairs.

Oh come on - you are an advanced enough mathematician to know about Rigged Hilbert Spaces and how that branch of math would answer such a question. As I said these things are not that easily pinned down. Even the fields of QFT are not as simple as usual fields like EM fields - they are quantum operators. The mind boggles about the 'reality' of those.

Thanks
Bill

 

[martinbn]

Yes, it is a function $\mathbb{R}^{3n+1} \rightarrow \mathbb{C}$, where $n\in \mathbb{N}$.

Now you are cheating. Why 3 and 1 in the 3n+1 and not say 7 and 2? And what are they anyway without saying anything about space and time?

 

[martinbn]

Oh come on - you are an advanced enough mathematician to know about Rigged Hilbert Spaces and how that branch of math would answer such a question. As I said these things are not that easily pinned down. Even the fields of QFT are not as simple as usual fields like EM fields - they are quantum operators. The mind boggles about the 'reality' of those.

Thanks
Bill

No, I am not asking about subtleties in the use of functional analysis in quantum mechanics. It's just that if space-time is to come after in some emergent way, then what is the wave function a function of? Or what Hilbert space (or rigged Hilbert space) it is an element of?

 

[Demystifier]

Now you are cheating. Why 3 and 1 in the 3n+1 and not say 7 and 2? And what are they anyway without saying anything about space and time?

Good question! Any physical theory has some free parameters that must be chosen such that the theory fits the observations. The numbers 3 and 1 fit the fact that we observe 3-dimensional space and 1-dimensional time.

 

[Demystifier]

No, I am not asking about subtleties in the use of functional analysis in quantum mechanics. It's just that if space-time is to come after in some emergent way, then what is the wave function a function of? Or what Hilbert space (or rigged Hilbert space) it is an element of?

Mathematical physicists tend to overlook the essence of physics and pay attention to mathematical subtleties. I like you for being the exact opposite of that, for being a physical mathematician.
(Something like Roger Penrose.)

 

[bhobba]

Now you are cheating. Why 3 and 1 in the 3n+1 and not say 7 and 2? And what are they anyway without saying anything about space and time?

Yes I was glib in my answer, and Demytifyer is right my math background sometimes smacks me in the face . I don't want to say I was a smart ass - but I was on rereading it.

Cosmology isn't really my thing but my reading of some of the latest stuff is it started from this thing called the false vacuum that has its own parameters - that's the parameters used in labeling however you want to expand the state. Just a thought. But I think its more QFT than QM which makes it harder to decipher. I guess right now its a bit of a mystery. Maybe someone more conversant in cosmology can chime in.

Thanks
Bill

 

[DrChinese]

Why on Earth do you say that? Please don't tell me Bell's Theorem!

You said they were "matched up" at spacetime separated spots. How?

 

[PeterDonis]

You said they were "matched up" at spacetime separated spots. How?

Based on my post #20, I think the answer is that there is no "matching" because there doesn't have to be.

It might also be worth pointing out that neither A nor B can know how their results match up until ordinary light-speed-or-slower information has traveled between them; nothing they can measure or observe before that will tell them what result was obtained at the other measurement, or even whether the other measurement was made or not. So although there is information "stored" in the wave function about which result was obtained, that information is not accessible until ordinary classical information arrives to allow it to be "decoded". Deutsch, in a paper on information flow in entangled quantum systems, calls this "locally inaccessible information":

https://arxiv.org/pdf/quant-ph/9906007.pdf

 

[DrChinese]

Based on my post #20, I think the answer is that there is no "matching" because there doesn't have to be.

It might also be worth pointing out that neither A nor B can know how their results match up until ordinary light-speed-or-slower information has traveled between them; nothing they can measure or observe before that will tell them what result was obtained at the other measurement, or even whether the other measurement was made or not. So although there is information "stored" in the wave function about which result was obtained, that information is not accessible until ordinary classical information arrives to allow it to be "decoded". Deutsch, in a paper on information flow in entangled quantum systems, calls this "locally inaccessible information":

https://arxiv.org/pdf/quant-ph/9906007.pdf

Thanks for these points.

I keep trying to figure out how MWI can be local only, and it still makes no sense to me. If Alice makes a measurement, she gets V in one world and H in another. In the new V world, every other spot knows it is to act as if Alice got V. So that makes Bob's later measurement match the appropriate statistics. In the new H world, every other spot knows it is to act as if Alice got H. So that also makes Bob's later measurement match the appropriate statistics.

I just don't see how that is local, if Alice and Bob are far apart. They must be accessing the same information so that everything sorts into the proper world.

 

[PeterDonis]

I just don't see how that is local, if Alice and Bob are far apart.

In the description I wrote down in post #20, the non-locality is in the quantum state; each of the terms obviously entangles two subsystems that are spacelike separated.

However, the state I wrote down in #20 is in the Schrodinger picture. Deutsch's claim in the paper I linked to is basically that, if you use the Heisenberg picture instead, the non-locality disappears; all of the correlations go into the measurement operators that describe each qubit separately, and which get "carried along" by the qubits so that they can be viewed as local.

 

[DrChinese]

Deutsch's claim in the paper I linked to is basically that, if you use the Heisenberg picture instead, the non-locality disappears; all of the correlations go into the measurement operators that describe each qubit separately, and which get "carried along" by the qubits so that they can be viewed as local.

I saw that in the paper you cited, but it seems like some fancy footwork to call that "local". Seems global to me.

 

[Derek P]

Thanks for these points.

I keep trying to figure out how MWI can be local only, and it still makes no sense to me. If Alice makes a measurement, she gets V in one world and H in another. In the new V world, every other spot knows it is to act as if Alice got V. So that makes Bob's later measurement match the appropriate statistics. In the new H world, every other spot knows it is to act as if Alice got H. So that also makes Bob's later measurement match the appropriate statistics.

I just don't see how that is local, if Alice and Bob are far apart. They must be accessing the same information so that everything sorts into the proper world.

The correlations in each world are of the "pair of gloves" kind. There is no need to share information.

It's very simple. The original state of the entangled pair is
a|h>₁|v>₂ + b|v>₁|v>₂ + c|v>₁|v>₂ + d|h>₁|h>₂.
All four product terms are needed because Alice and Bob may use different orientations and there are four possible measurement results.

Consider just one of the product states, for example the first. Unitary evolution is deterministic. So by interacting with this state Alice will definitely detect H and Bob will definitely detect V. It's a predetermined matching pair of gloves! The result is an HV world. The other terms give the VH, VV and HH worlds.

 

[DrChinese]

The correlations in each world are of the "pair of gloves" kind. There is no need to share information.

It's very simple. The original state of the entangled pair is
a|h>₁|v>₂ + b|v>₁|v>₂ + c|v>₁|v>₂ + d|h>₁|h>₂.
All four product terms are needed because Alice and Bob may use different orientations and there are four possible measurement results.

Consider just one of the product states, for example the first. ...

There are no coefficients a/b/c/d that can reproduce the quantum expectation values because entangled pairs are not Product states. An entangled pair is hh + vv (or hv+vh). Now I realize that you are attempting to account for the basis being different for Alice and Bob, but their results are not independent of each other. When expressed in that manner, there are correctly still only 2 terms - and not 4 as you show.

It should be obvious that the "pair of gloves" analogy cannot work any better here than in other interpretations.

 

[PeterDonis]

There are no coefficients a/b/c/d that can reproduce the quantum expectation values because entangled pairs are not Product states.

I'm not sure what you're trying to say here. Entangled pairs are linear combinations of product states with appropriate coefficients--basically you choose two of a/b/c/d to be 1 (or you can throw in phase factors if needed) and the other two to be zero.

 

[PeterDonis]

The original state of the entangled pair is
a|h>₁|v>₂ + b|v>₁|v>₂ + c|v>₁|v>₂ + d|h>₁|h>₂.

I think you have a typo in here; shouldn't it be a|h>₁|v>₂ + b|v>₁|h>₂ + c|v>₁|v>₂ + d|h>₁|h>₂?

 

[Derek P]

There are no coefficients a/b/c/d that can reproduce the quantum expectation values

Sure there are. I can even tell you what they are within a phase factor
a = ½ cos(α-β)
b = ½ sin(α-β)
c = ½ cos(α-β)
d = ½ sin(α-β)

because entangled pairs are not Product states.

Umm, a|h>1|v>2 + b|v>1|v>2 + c|v>1|h>2 + d|h>1|h>2 isn't a product. It's the sum of products.

An entangled pair is hh + vv (or hv+vh).

Yes when measured in the same basis. And what @PeterDonis said. It's just that two coefficients are zero.

Now I realize that you are attempting to account for the basis being different for Alice and Bob, but their results are not independent of each other.

That's right. They are trigonometric functions of the two angles of measurement. The difference between them to be precise.

When expressed in that manner, there are correctly still only 2 terms - and not 4 as you show.

Only when α=β. To get from that case to the general case, one or other "h" must be resolved in the other observer's basis.

It should be obvious that the "pair of gloves" analogy cannot work any better here than in other interpretations.

Well no. The pair of gloves analogy works just fine in any interpretation which manages to coax a mixed state out of the superposition, whether it's a proper one as in Copenhagen or an improper one as in Many Worlds..

 

[DrChinese]

I'm not sure what you're trying to say here. Entangled pairs are linear combinations of product states with appropriate coefficients--basically you choose two of a/b/c/d to be 1 (or you can throw in phase factors if needed) and the other two to be zero.

Yes, 2 are zero. And with phase factors, you could also expand those to get 4 terms. Since the phase term is dependent on the specific difference between the angle settings (basis), how does the world know to split in a fashion that reproduces the statistics?

In other words: the coefficients can reproduce the statistics (if you vary them suitably) but you must know both Alice and Bob's settings first. Then the formula is no longer local.

 

[DrChinese]

Sure there are. I can even tell you what they are within a phase factor
a = ½ cos(α-β)
b = ½ sin(α-β)
c = ½ cos(α-β)
d = ½ sin(α-β)

Now the terms are dependent on something which is non-local, which is what we sought to avoid. Sorry I don't think I was clear on this point.

Looking at it another way: Bob's setting might not have even been determined at the time, so no way the values could be determined.

 

[PeterDonis]

the coefficients can reproduce the statistics (if you vary them suitably) but you must know both Alice and Bob's settings first. Then the formula is no longer local.

Ah, got it.

 

[Derek P]

I think you have a typo in here; shouldn't it be a|h>₁|v>₂ + b|v>₁|h>₂ + c|v>₁|v>₂ + d|h>₁|h>₂?

Yes, thanks. Though to be consistent with my later post, it's the third state that should be changed.

 

[DrChinese]

Ah, got it.

My fault, reading my post it was not clear. I want a/b/c/d to be a constant that can be determined at the time there is a split.

 

[Derek P]

Now the terms are dependent on something which is non-local, which is what we sought to avoid. Sorry I don't think I was clear on this point.
Looking at it another way: Bob's setting might not have even been determined at the time, so no way the values could be determined.

On that basis you could say the distance between Alice and Bob is non-local. There's nothing wrong with having non-local functions, the only problem with non-locality is when cause and effect are non-local.

 

[PeterDonis]

On that basis you could say the distance between Alice and Bob is non-local.

Yes, it is.

 

[Derek P]

Ah, got it.

My fault, reading my post it was not clear. I want a/b/c/d to be a constant that can be determined at the time there is a split.

As I recall I explained that worlds only split in the future decoherence cone of a measurement. And since then I've said that deciding exactly when the split happens is a bit arbitrary because it's just a picture of the end result after decoherence. I'll make your point even stronger for you. a/b/c/d can be known precisely as soon as you have the angles. Knowing the coefficients makes no difference whatsoever to the state, it just tells you how to calculate the outcomes.
edited

 

[Derek P]

Yes, it is.

Thanks, I might never have been sure :)

 

[Derek P]

Ah, got it.

Well come on, share it with the rest of us, then.

 

[PeterDonis]

Well come on, share it with the rest of us.

I just meant I understood what @DrChinese was saying, and I don't have anything useful to add to it.

 

[Derek P]

But according to MWI, there is no 3-dimensional world. According to MWI, the only object that exists is the wave function, which does not live in a 3-dimenional world. ... According to MWI, the 3-dimensional world is an illusion

On thinking about it a bit more, I don't think that's entirely true. Certainly it is often said that in MWI, the only reality is the wavefunction. Which of course can be expressed in multidimensional phase space. But it can also be expressed in three dimensions as a sum of products. So you can interpret it as a multidimensional object with a curious penchant for three dimensional observables or a three dimensional object which can exist in superposition. I don't think MWI is dogmatic about which. In fact there is no reason why the wavefunction itself need be ontic at all, it could be an abstraction that maps onto something else via an unknown transformation. MWI will work just as well with any of them.

 

[akvadrako]

One thing to add to this discussion, which think it quite important, is that I think a recent paper from Sean Carroll does a better job than Deutsch at giving an easy to understand and less rigorous explanation of why WMI can be considered local:

https://arxiv.org/pdf/1801.08132.pdf

Vaidman also has some to-the-point insights on the topic.

 

[akvadrako]

In fact there is no reason why the wavefunction itself need be ontic at all, it could be an abstraction that maps onto something else via an unknown transformation.

The reason the wavefunction must be ontic (which that I mean maps 1-to-1 onto reality) is given by the PBR and Colbeck and Renner theorems.