Entanglement swapping and Bohmian mechanics

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PeterDonis said:
From what @Morbert posted in #40:

Simple algebra gives (leaving out the normalization factors):

ρ1234=[HH]14[VV]23+[VV]14[HH]23+[HV]14[VH]23+[VH]14[HV]23

If we know photons 2 and 3 are both vertically polarized, all but one of the above terms go away, and we are left with:

ρ1234=[HH]14[VV]23

which of course can be rewritten as

ρ1234=[HH]14[Φ+−Φ−]23
Note that in the math order of measured doesn't matter as long as you keep bookeeping. This also agrees with what @javisot pointed out in #54
 
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PeterDonis said:
I would choose ii). I would think that would be the natural choice for any advocate of Bohmian mechanics. :wink: The natural Bohmian account seems to me to be that the effective collapse due to the pre-measurement of two of the photons "steers" the other two, via the effective wave function/pilot wave, into the appropriate output channels to register a swap that is consistent with the pre-measured photon results. This is basically the account I gave in my earlier post (and have been elaborating in follow-up posts).
The main issue is - can an event in the future cause another event in the past? I say it can't. Let me explain.

The very notions of cause and effect assume an asymmetry; if A causes B, then B dos not cause A. That's an inherent part of meanings of the notions of cause and effect. Otherwise, we should not talk about cause and effect. For example, x=1 implies x+1=2, and also x+1=2 implies x=1, so we do not say that x=1 "causes" x+1=2.

But where does this asymmetry between cause and effect come from? The deterministic equations of motion (in both classical and quantum mechanics, and also in Bohmian mechanics) are essentially time-inversion invariant, hence the asymmetry cannot arise from the equations of motion. Instead, it is widely accepted that the asymmetry arises from the "thermodynamic" arrow of time, namely, from the fact that some kind of disorder (technically usually represented by some kind of entropy) grows in time, which, in turn, can be reduced to a past-hypothesis, according to which the Universe had a very low entropy at some "initial" time in the past. In this way the notions of cause and effect are emergent, making sense only at the macroscopic level, where we can talk about entropy. If we have two events, with one being cause of another the effect, then the cause is the one associated with a lower entropy, and the effect is the one associated with a larger entropy. In practice, this means that the cause is in the past and the effect is in the future.

Applying this general principle to the entanglement swapping experiment, it rules out the choice ii). Q.E.D.
 
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Demystifier said:
The main issue is - can an event in the future cause another event in the past? I say it can't. Let me explain.

The very notions of cause and effect assume an asymmetry; if A causes B, then B dos not cause A. That's an inherent part of meanings of the notions of cause and effect. Otherwise, we should not talk about cause and effect. For example, x=1 implies x+1=2, and also x+1=2 implies x=1, so we do not say that x=1 "causes" x+1=2.

But where does this asymmetry between cause and effect come from? The deterministic equations of motion (in both classical and quantum mechanics, and also in Bohmian mechanics) are essentially time-inversion invariant, hence the asymmetry cannot arise from the equations of motion. Instead, it is widely accepted that the asymmetry arises from the "thermodynamic" arrow of time, namely, from the fact that some kind of disorder (technically usually represented by some kind of entropy) grows in time, which, in turn, can be reduced to a past-hypothesis, according to which the Universe had a very low entropy at some "initial" time in the past. In this way the notions of cause and effect are emergent, making sense only at the macroscopic level, where we can talk about entropy. If we have two events, with one being cause of another the effect, then the cause is the one associated with a lower entropy, and the effect is the one associated with a larger entropy. In practice, this means that the cause is in the past and the effect is in the future.

Applying this general principle to the entanglement swapping experiment, it rules out the choice ii). Q.E.D.
In the mesoscopic and macroscopic regimes, there appears to be a general arrow of time, but I do not know whether such an arrow exists in the microscopic regime, nor whether its existence is necessary to explain the results.

As far as entanglement is concerned, the requirement is that the product of the states be non-factorizable, not that an arrow of time exists.
 
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As the conversation has continued I'll post here.
PeterDonis said:
Indeed, looking at this and the diagram of the experimental setup (Figure 2 in the paper) makes me even more confused, because the evolution starts before BS1, i.e., the evolution assumes that photons 2 and 3 are in one of the Bell states before BS1--but then what puts them into those states? The only things in the experiment before BS1 are the preparations of the initial entangled pairs (1 and 2, 3 and 4), the time delay, and a half wave and quarter wave plate, which as far as I can tell are the same for both photons, so they should not change the phase relationship between them. What in all this can put photons 2 and 3 into a Bell state?
In Ma's actual experiment, the BiSA apparatus is interpreted as performing a measurement at the incoming spatial modes. I.e. the detector clicks that occur near modes b'' and c'' are interpreted as a measurement at the prior modes b and c. This framing introduces a peculiar temporal character to the measurement, but it is reliable because even though the detectors are really measuring in a HV basis at b'' and c'', they establish a correlation between the Bell basis at b and c and the environment.

But we can formally get around this correct but peculiar convention. In post #14 I sketched an ideal nondestructive BSM yielding$$\rho_{1234} = \frac{1}{4}(\left[\Psi^+\right]_{14}\left[\Psi^+\right]_{23}+\left[\Psi^-\right]_{14}\left[\Psi^-\right]_{23}+\left[\Phi^+\right]_{14}\left[\Phi^+\right]_{23}+\left[\Phi^-\right]_{14}\left[\Phi^-\right]_{23})$$If instead we model Ma's BiSA-based measurement as a (still nondestructive) measurement of polarization and spatial modes of 2 and 3, the state post measurement would actually be$$\begin{aligned}
\rho_{\mathrm{post}}
={}&
\frac{1}{4}[VH]_{14}[HV]_{b''c''}
+
\frac{1}{4}[HV]_{14}[VH]_{b''c''}
\\
&+
\frac{1}{8}[\Phi^+]_{14}
\left(
[HV]_{b''b''}
+
[HV]_{c''c''}
\right)
\\
&+
\frac{1}{8}[\Phi^-]_{14}
\left(
[HH]_{b''c''}
+
[VV]_{b''c''}
\right)
\end{aligned}$$Here the "entanglement swap" (the preparation of 1&4 in a ##\Phi^\pm## Bell state) is induced by the corresponding H,V,b'',c'' measurement results. I kept the nondestructive simplification to keep the expression simple. Otherwise we would have to include detector degrees of freedom. Similarly, if Alice and Bob had already carried out their measurements in the HV basis, the final state is$$\begin{aligned}
\rho_{1234}
={}&
\frac{1}{4}[VH]_{14}[HV]_{b''c''}
+
\frac{1}{4}[HV]_{14}[VH]_{b''c''}
\\
&+
\frac{1}{16}
\left(
[HH]_{14}
+
[VV]_{14}
\right)
\Big(
[HV]_{b''b''}
+
[HV]_{c''c''}
+
[HH]_{b''c''}
+
[VV]_{b''c''}
\Big)
\end{aligned}$$An interesting key is this is the final state whether or not Alice and Bob carry out their measurements before or after Victor. Hence the statistics produced by the experiment are the same. Though to get the full statistics you would have to also model Alice and Bob's choice of basis and Victor's choice to actually perform the measurement with a quantum random number generator. This would be conceptually simple but arduous to actually write out.

Note that I'm trying to keep this as interpretation-free as possible and so these states can be interpreted as simply yielding measurement outcome statistics. This might itself still be an interpretation. It is possible to carry out an analysis including explicit collapses upon measurements but based on experience it quickly becomes messy.

[edit] - improved spatial mode notation
 
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PS to tie my last post to Bohmian mechanics you could presumably include detector and environmental degrees of freedom so the states becomes a pure states and hence pilot waves for the total system's trajectory.
 
javisot said:
In the mesoscopic and macroscopic regimes, there appears to be a general arrow of time, but I do not know whether such an arrow exists in the microscopic regime, nor whether its existence is necessary to explain the results.

As far as entanglement is concerned, the requirement is that the product of the states be non-factorizable, not that an arrow of time exists.
There is no arrow of time in the microscopic regime. But the entanglement swapping involves measurements, and measurement is always a macroscopic event. The arrow of time is relevant because some people here claim (and I refute) that a measurement in the future causes something in the past. Causation from the future to the past violates the arrow of time.
 
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Demystifier said:
Since @Sambuco talks about electron-positron annihilation, I would like to draw his attention to the Hardy setup
https://en.wikipedia.org/wiki/Hardy's_paradox#Setup_description_and_the_results
where an entangled electron-positron pair is created by postselection in which the annihilated pair is removed from the ensemble. Without postselection there is no entanglement, but after the postselection, that is, after the decision to ignore the cases when the pair annihilates and turns into photons, the resulting state ends up in a very interesting entangled state, called Hardy state, that Hardy used to show quantum nonlocality without using any inequalities.
Thanks @Demystifier! :smile: Indeed, I had never stopped to think about how important the post-selection role is in preparing Hardy state.

Lucas.
 
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Sambuco said:
Thanks @Demystifier! :smile: Indeed, I had never stopped to think about how important the post-selection role is in preparing Hardy state.

Lucas.
This, in essence, is very simple. Consider a product state
$$|\psi\rangle = |A\rangle |B\rangle$$
Any state can be written as a superposition in some basis. For instance, we can always write
$$|A\rangle = |A_1\rangle + |A_2\rangle$$
$$|A\rangle = |B_1\rangle + |B_2\rangle$$
(for simplicity I suppress the overall normalization of the states, because it's not important for the argument) so
$$|\psi\rangle = |A_1\rangle|B_1\rangle + |A_2\rangle|B_2\rangle + |A_1\rangle|B_2\rangle + |A_2\rangle|B_1\rangle$$
This superposition is still the product state, even if it does not look so at first sight. But if we do a postselection, we effectively remove some terms from the superposition. For example, if we do a postselection that removes the last term ##|A_2\rangle|B_1\rangle##, we end up in the post-selected state
$$|\psi'\rangle = |A_1\rangle|B_1\rangle + |A_2\rangle|B_2\rangle + |A_1\rangle|B_2\rangle$$
This post-selected state is not a product. It is an entangled state, known as Hardy state. This is arguably the simplest way to see how entanglement can result from post-selection.

Of course, a post-selection on an ensemble requires a measurement. In the Hardy setup the measurement is a detection of the pair of photons created by electron-positron annihilation. That corresponds to the case in which the removed term ##|A_2\rangle|B_1\rangle## describes electron and positron brought to the same position, so that they can interact. This means that from the ensemble we remove electrons and positrons that interacted, thus leaving only those electrons and positrons that never interacted. In this way we get entangled particles that never interacted (because the interacting ones are removed from the ensemble), which is conceptually very similar to entanglement swapping. And yet it is much simpler, because it involves only 2 particles.
 
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Demystifier said:
can an event in the future cause another event in the past? I say it can't.
I agree, and I certainly agree that that's the position Bohmian mechanics takes, which is why I chose ii). :wink:
 
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Demystifier said:
rules out the choice ii).
Are you sure you don't have the choices mixed up? Here they are again:

DrChinese said:
i) Whether the future swap - which the universe knows is going to occur? - causes the perfect A/D correlation, thereby creating the illusion of action to the past; or
ii) The outcome of the perfect A/D correlation causes the swap to succeed.
i) seems like the one that is problematic if you don't want backwards in time causation. ii) seems like the one that is straightforward forward in time causation: the A/D measurements happen first, before the swap, so that's the direction whatever causation there is would have to go.

Note: The wording of ii) assumes an experimental setup where all possible swaps can be distinguished, so we will be able to confirm experimentally that the swap succeeds. In some setups (like the Ma et al one), that's not the case, and ii) would have to include the possibility that the perfect A/D correlation leads to a swap that can't be distinguished, so the results as reported would not show that a swap occurred for that case. But the causation would still be forward in time, the A/D measurement happens first and has a causal effect on what happens at the swap.
 
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Demystifier said:
But the entanglement swapping involves measurements, and measurement is always a macroscopic event.

It is an interpretation. "Never" and "always" are very big concepts.
 
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PeterDonis said:
I agree, and I certainly agree that that's the position Bohmian mechanics takes, which is why I chose ii). :wink:
You are right, in the meantime I forgot what i) and ii) have actually been saying. :D

The reason I ruled out ii) is because there was no correlation in the past before the post-selection, so that correlation could not have caused anything. Sure, something in the past have caused the future, and that something is related to the correlation observed later, but that something is not the correlation itself.
 
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Demystifier said:
there was no correlation in the past before the post-selection
"Correlation" might not be the best word to convey what has a causal effect. The point is that the A and D measurement results happen first, and those have a causal impact on what happens at the swap. In Bohmian mechanics, this is straightforward: those measurements cause an effective collapse of the wave functions of B and C, which steers them through the BSM in such a way as to produce a swap that is consistent with the A and D measurement results. That is what produces the final data with subensembles that show the correlations.

That's basically the story that ii) is pointing to in Bohmian terms.
 
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Demystifier said:
something in the past have caused the future
That's why I ruled out i); because it talks about the future swap causing the A/D correlation. As i) is phrased it says this creates the illusion of causation into the past, but the A/D measurement results are already known before the swap, and the swap can't change them. It can only adapt the B/C measurement results to them. I don't think "causing the A/D correlation" is the best way to describe that.

The other thing that caused me to rule out i) is the part about the universe knowing that the future swap is going to occur. The Bohmian account does not contain anything like that. There is no causal effect between the A/D measurements and the configuration of the swap machinery, nor is it necessary to have any kind of predetermined correlation between them; the universe does not have to "know" when A/D are measured that a swap is going to occur in the future. All that's necessary is for the pilot wave that "steers" B/C through the swap machinery to "know", via the effective collapse when A/D are measured, what A/D measurement results the swap has to be consistent with.
 
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PeterDonis said:
The other thing that caused me to rule out i) is the part about the universe knowing that the future swap is going to occur. The Bohmian account does not contain anything like that. There is no causal effect between the A/D measurements and the configuration of the swap machinery, nor is it necessary to have any kind of predetermined correlation between them; the universe does not have to "know" when A/D are measured that a swap is going to occur in the future. All that's necessary is for the pilot wave that "steers" B/C through the swap machinery to "know", via the effective collapse when A/D are measured, what A/D measurement results the swap has to be consistent with.
Yes, I basically agree. As a further support for your position, I would like to add something about determinism. Even though BM is deterministic, that's not at all the essential property of BM. (Instead, the essential property is realism, i.e., the existence of variables having values even when they are not measured.) There is an interpretation of QM with particle trajectories very similar to the Bohmian one, called Nelson interpretation, in which the particle trajectories are not deterministic but stochastic, and this interpretation resolves most conceptual questions of QM in essentially the same way as the Bohmian interpretation, thus demonstrating that determinism is not important.
 
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Morbert said:
we can formally get around this correct but peculiar convention.
Yes, I see what you're doing, and it makes it clearer what is actually being correlated in the Ma et al setup. I would say, based on your analysis, that the Ma et al setup makes everything behave as if photons 2 and 3 were in the appropriate Bell state as they come into the beam splitter marked BS1 in that setup. That part, at least if "behave" just refers to the observed correlations between measurement results, seems to me to be unproblematic, it's just a statement of the prediction the math makes and which is experimentally confirmed.

The part that I would say is interpretation dependent is the claim that, because of the above, photons 2 and 3 are actually in the appropriate Bell state as they come into BS1. If the BSM is supposed to be a projector, on this view it would have to project backwards in time--the results measured in output channels b'' and c'' would have to make photons 2 and 3 be in the appropriate Bell state at the input channels b and c. The interpretation the paper is using might be comfortable with that claim (since the paper has no problem with "quantum steering into the past"), but I don't think oither interpretations would.

And note that in the other papers @DrChinese referenced, nothing like this is necessary or claimed. The Bell state projection is done by a polarizing beam splitter, and the photons are only claimed to be in the appropriate Bell state coming out of the PBS, not going in.
 
PeterDonis said:
Do you think it's the same kind of physical change as if we measured the z-spin of particle 2?
Again, we're referring to Norsen's view on Bohmian mechanics:

Yes, of course Norsen means that the nonlocal change is physical. He's a Bohmian, and BM is all about nonlocal and instantaneous influences via a physical pilot wave* ("the pilot-wave theory adds something to the state descriptions of ordinary quantum mechanics"). That's the entire point of BM. Yes, one needs to use their powers of induction to put the Norsen quotes together to get the word "physical" out of it. If there's no physical influence, what's the point of the interpretation in the first place?

But a simple read - even a relatively cursory read - of the paper makes Norsen's position clear. Norsen specifically criticizes your earlier comment/caveat: "So the fact that we had an effective [?] collapse in particle 2's wave function does not mean its spin changed, or anything else about it. Indeed, we can't even define its spin, according to BM, until it's measured...

Well, Norsen rips that viewpoint explicitly! Norsen on Norsen: "...the physical state of (in particular) the waveguiding the particle is different before, and after, the intermediate SGx device. But in the pilot-wave theory, this difference – the physical influence of the measurement on the properties of the system in question – is a natural and straightforward consequence of the usual – the universal, the exceptionless – dynamical laws". Conclusion: It's physical; and I am fairly representing Norsen. And you are flat out wrong in your analysis of Norsen, not me.

You don't need to agree with what Norsen's saying, nor does anyone. But you don't need to keep parsing words relentlessly. But there is no point in keeping this going in circles. Just drop it and let's continue on with actual physics, there's plenty more juicy stuff in here that deserves our attention.


*Or whatever you choose to call it.
 
Demystifier said:
No.
Just to be clear, and mine is a serious question precisely because I thought there was a correspondence between the Hilbert representation and the system's "state" representation: I said "There's still 2 Bell states before the swap, and there's 2 completely different Bell states after. Presumably the Hilbert space changed too." But you say there's no change.

Norsen says: "...the physical state of (in particular) the waveguiding the particle is different before, and after, the intermediate SGx device. But in the pilot-wave theory, this difference – the physical influence of the measurement on the properties of the system in question – is a natural and straightforward consequence of the usual – the universal, the exceptionless – dynamical laws".

So is Norsen agreeing with you? Because: I would think the the Hilbert representation is meaningless if a swap occurs, and nothing changed. I an definitely missing the significance of bringing the Hilbert representation into the discussion in the first place, when state representations seem to suffice - including for most of the referenced authors.
 
DrChinese said:
we're referring to Norsen's view on Bohmian mechanics:
With regard to the question I asked that you quoted, which is not "is the change a physical change", but "is it the same kind of physical change as if we measured the z-spin of particle 2".

Which the quotes you gave from Norsen do not address at all.

I asked because I think the question is relevant to the physics we want to discuss. But if you just refuse to see that, and refuse to try to engage with my question, fine, I'll drop it for now. And when we get to a point where it matters, I'll bring the point up again, and we can argue about it then.
 
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DrChinese said:
you say there's no change.
There's no change in the Hilbert Space. There is a change in the wave function, or perhaps a better term if we're talking about the Hilbert Space formuiation would be "state vector" (what kets represent). The Hilbert Space is the vector space of all possible state vectors; that doesn't change. The state vector is the particular vector in that vector space that is relevant at a particular point in the experiment. That does change as the experiment proceeds and things happen.
 
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DrChinese said:
I thought there was a correspondence between the Hilbert representation and the system's "state" representation
There is: The Hilbert Space representation treats states as vectors in a vector space. The wave function representation treats states as functions on configuration space. It turns out that these are mathematically equivalent because the space of all possible wave functions turns out to be a vector space (with the individual vectors being functions) with the right properties to be isomorphic to the Hilbert Space in that representation.

Most QM textbooks discuss this at least to some extent. Ballentine does so in Chapter 1.
 
Working with an interpretation that includes an arrow of time is more intuitive for us, given that we are accustomed to the concept of an arrow of time.

So, what is the downside of BM? Non-locality.
 
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PeterDonis said:
So at this point I think the paper must be leaving out some crucial elements either of the experiment or of the logic they are using.

Are you saying this as a moderator, or a participant? We're talking about PeterDonis criticizing the work of a Nobel level team. I would expect an independent moderator might want to come in and parse your entire comment to death. Instead of course, here you are moderating your own statements.

PeterDonis said:
$$
\rho_{1234} = \left[HH\right]_{14}\left[VV\right]_{23}
$$

which of course can be rewritten as

$$
\rho_{1234} = \left[HH\right]_{14}\left[\Phi^+ - \Phi^-\right]_{23}
$$

And please, if you can ask for an authoritative reference, I guess I should be able to as well - and as moderator you should support me. Without saying "everyone knows this to be true except you", and otherwise quoting yourself, please show by explicit reference with quote that the following must be true:

$$
\left[VV\right]_{23} = \left[\Phi^+ - \Phi^-\right]_{23}
$$

Note: It would be clearer if you would state which portions are pre-swap and which are post-swap. I think that is relevant here to what you are trying to say.
 
DrChinese said:
Are you saying this as a moderator, or a participant?
I've already said I'm taking no moderation actions in this thread. I was speaking as a participant. And the paper in question was the Ma paper, which you already agreed was difficult to follow. Also, there have been a number of posts about that paper since the one of mine that you quoted, which have helped me to understand (at least for some value of "understand" :wink:) the approach they are taking. In particular see @Morbert #124 and my #139.

DrChinese said:
please show by explicit reference with quote that the following must be true
The post you quoted that from was based on math that @Morbert posted. It looked like basic QM math to me, applying the rule that when you know the results of a measurement, you update the effective wave function you're using to reflect that result. I gave a reference to Ballentine supporting that approach, since that was the textbook I had handiest. @Morbert used notation that's a little different from what I've seen in textbooks, but he explained it clearly in post #40.

As for the particular mathematical expression you asked about, it's basic algebra (leaving out constant normalization factors, which I said I was doing).
 
DrChinese said:
It would be clearer if you would state which portions are pre-swap and which are post-swap.
Everything I wrote down in that post (and in the post by @Morbert I was quoting from) is after the A/D measurement and before B/C enter the swap apparatus.
 
javisot said:
So, what is the downside of BM? Non-locality.
Not everyone would see this as a downside. 😉
 
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PeterDonis said:
1. I would say, based on your analysis, that the Ma et al setup makes everything behave as if photons 2 and 3 were in the appropriate Bell state as they come into the beam splitter marked BS1 in that setup. That part ... seems to me to be unproblematic...

2. And note that in the other papers @DrChinese referenced, nothing like this is necessary or claimed. The Bell state projection is done by a polarizing beam splitter, and the photons are only claimed to be in the appropriate Bell state coming out of the PBS, not going in.
1. I agree with this.

2. The various authors would say 2 and 3 enter the BS in the state you call "singlet", even though a polarization measurement could be made which would yield a certain outcome. Where do they say that, you ask? Nowhere. It's implied by their representation of the pre-swap state, combined with the general QM concept that one should not make assertions about particle behavior between measurements/interactions. They assume the reader will fill that in on their own, or perhaps view it in light of a particular interpretation.

Now, precisely because they are silent on this point: I am trying to nail this point down one way or the other. Seems like you are too! Nailing it down is clearly difficult, but I am trying my best to find a way to confirm. The approach I am making is to have the input 2 and 3 photons in a known pure state such as |VV> rather than an assumption of them being singlet at that point.

If we could rule out them being in state |VV> going in, then we might see that as a proof they are in fact in the singlet state. That would be useful, and perhaps might relate somehow to the Bohmian perspective we've discussed here (a la Norsen at least). Again, I don't know.
 
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