A Forward-in-time analysis of delayed-choice entanglement swapping

[PeterDonis]

this should be distinguished from the standard Bell inequality violations due to entanglement."

This looks to me like a blatant misstatement by the authors of the paper, since the states they are analyzing are Bell states, i.e., entangled states.

 

[Morbert]

The projection postulate does not say that at all, and you have misapplied the idea of projection. What is says (as applies here): IF photons 2 & 3 are measured on the same basis as photons 1 & 4, THEN the results will be certainly |LL>₂₃. No one is disputing this. But that is a far cry from what happens - because those measurements were not performed. There statements by many authors in the literature (and actually 866,000 links with this exact quote, per Google) similar to the following famous quote from Peres (1978):

"Unperformed measurements have no results."

Sambuco is right. The projection postulate says Alice and Bob, if they observe RR, will project the tetraphoton system onto $\ket{R}_1\ket{L}_2\ket{L}_3\ket{R}_4$ whatever Victor does at a later time. This is because 1 & 2 are initially entangled, and 3 & 4 are initially entangled (really, photons 1 & 4 are destroyed by Alice's and Bob's measurement but we can put a pin in that for now). Alice and Bob can then determine possible outcomes for Victor if he decides to carry out a BSM by expanding 2 & 3 in a Bell basis. $$\ket{L}_2\ket{L}_3=\frac{1}{\sqrt{2}}(\ket{\phi^-}_{23} - i\ket{\psi^+}_{23})$$Hence, if Victor attempts a BSM and he is successful, he must obtain the result $\phi^-$. He cannot obtain $\phi^+$. This is consistent with the DCES experiment, where the $\phi^+$ set has anticorrelation records of 1 & 4 in the R/L basis.

 

[eloheim]

There is a very clear difference of opinion here that should be incredibly easy to solve:

We're doing a single run of Ma's experiment. Alice and Bob have measured and gotten |R> and |R> for their results. Victor has yet to take any action. If Victor does a BSM on photons 2 and 3, what result can he get?

DrChinese predicts he can get all four bell states.
Morbert predicts he can only get ∣Ψ+⟩ and ∣Φ−⟩

This has a definite answer does it not?

 

[Sambuco]

There is a very clear difference of opinion here that should be incredibly easy to solve:

We're doing a single run of Ma's experiment. Alice and Bob have measured and gotten |R> and |R> for their results. Victor has yet to take any action. If Victor does a BSM on photons 2 and 3, what result can he get?

DrChinese predicts he can get all four bell states.
Morbert predicts he can only get ∣Ψ+⟩ and ∣Φ−⟩

This has a definite answer does it not?

If Alice and Bob both measure R, the state after that is $\ket{\Psi(t_1)} = \ket{R}_1 \otimes \frac{1}{\sqrt{2}} (\ket{\phi^-}_{2,3} - i \ket{\psi^+}_{2,3}) \otimes \ket{R}_4$, so Victor only can measure $\ket{\phi^-}_{2,3}$ or $\ket{\psi^+}_{2,3}$ in those runs, as @Morbert explained. For more details, the derivation is in my post #146.

Lucas.

 

[Sambuco]

That's correct, but it misses a key point: it is possible to separate the 1&4 measurement outcomes into subsets according to the photon 2&3 measurement outcomes. It's just that those subsets do not show Bell correlations in the SSM case, but they do in the BSM case.

In other words, if you do what we normally do in any other area of science, and start with the experimental data, and apply a simple, well-defined procedure to that data in both cases, you get different results: BSM -> correlations; SSM -> no correlations. And in any other area of science, this kind of thing is taken to show that the choice BSM vs. SSM has some kind of real effect on photons 1 & 4. But somehow, when it's this particular QM experiment, people make strenuous efforts to avoid this obvious conclusion that in any other area of science would be commonplace.

Thanks @PeterDonis to mention this issue. I think that, in part, that is responsible for some common misunderstanding about DCES and the postselection vs. backward-in-time influence interpretations. Let me explain that with some detail. First, do the math and only after that, I will discuss how the results could be interpreted in different (complementary) ways.
For the analysis, I'll strictly follow what Ma's paper. I'll consider the sequence where at time $t_0$ the four photons were prepared, then at time $t_1$ Alice and Bob measure photons 1 and 4 in the $L/R$ basis, and, finally, at time $t_2$ Victor performs a BSM/SSM on photons 2&3, using the $H/V$ basis in his measurements.

Given the preparation procedure, the state of the system after preparation (time $t_0$), but before Alice and Bob measurements (time $t_1$) is:
$\ket{\Psi(t_0<t<t_1)} = \ket{\psi^-}_{12}\otimes \ket{\psi^-}_{34}$
Then, Alice and Bob perform measurements and obtain one out of the four possible combinations RR, RL, LR, LL. According to the projection postulate the possible states of the system after $t_1$ but before $t_2$ are:

$\ket{\Psi(t_1<t<t_2)}_A = \ket{R}_1 \otimes \ket{R}_4 \otimes \ket{L}_2 \otimes \ket{L}_3$
$\ket{\Psi(t_1<t<t_2)}_B = \ket{R}_1 \otimes \ket{L}_4 \otimes \ket{L}_2 \otimes \ket{R}_3$
$\ket{\Psi(t_1<t<t_2)}_C = \ket{L}_1 \otimes \ket{R}_4 \otimes \ket{R}_2 \otimes \ket{L}_3$
$\ket{\Psi(t_1<t<t_2)}_D = \ket{L}_1 \otimes \ket{L}_4 \otimes \ket{R}_2 \otimes \ket{R}_3$

Now, first consider the case where Victor perform a SSM. To facilitate the analysis, it is convenient to rewrite the previous states $\ket{\Psi(t_1<t<t_2)}_{A,B,C,D}$ with photons 2&3 in the $H/V$ basis. Since $\ket{R} = \frac{1}{\sqrt{2}} (\ket{H}+i\ket{V})$ and $\ket{L} = \frac{1}{\sqrt{2}} (\ket{H}-i\ket{V})$, it is straightforward that

$\ket{\Psi(t_1<t<t_2)}_A = \ket{R}_1 \otimes \ket{R}_4 \otimes \frac{1}{2} (\ket{H}_2-i\ket{V}_2) \otimes (\ket{H}_3-i\ket{V}_3)$
$\ket{\Psi(t_1<t<t_2)}_B = \ket{R}_1 \otimes \ket{L}_4 \otimes \frac{1}{2} (\ket{H}_2-i\ket{V}_2) \otimes (\ket{H}_3+i\ket{V}_3)$
$\ket{\Psi(t_1<t<t_2)}_C = \ket{L}_1 \otimes \ket{R}_4 \otimes \frac{1}{2} (\ket{H}_2+i\ket{V}_2) \otimes (\ket{H}_3-i\ket{V}_3)$
$\ket{\Psi(t_1<t<t_2)}_D = \ket{L}_1 \otimes \ket{L}_4 \otimes \frac{1}{2} (\ket{H}_2+i\ket{V}_2) \otimes (\ket{H}_3+i\ket{V}_3)$

Reordering the terms,

$\ket{\Psi(t_1<t<t_2)}_A = \ket{R}_1 \otimes \ket{R}_4 \otimes \frac{1}{2} (\ket{HH}_{2,3}-\ket{VV}_{2,3}-i\ket{HV}_{2,3}-i\ket{VH}_{2,3})$
$\ket{\Psi(t_1<t<t_2)}_B = \ket{R}_1 \otimes \ket{L}_4 \otimes \frac{1}{2} (\ket{HH}_{2,3}+\ket{VV}_{2,3}+i\ket{HV}_{2,3}-i\ket{VH}_{2,3})$
$\ket{\Psi(t_1<t<t_2)}_C = \ket{L}_1 \otimes \ket{R}_4 \otimes \frac{1}{2} (\ket{HH}_{2,3}+\ket{VV}_{2,3}-i\ket{HV}_{2,3}+i\ket{VH}_{2,3})$
$\ket{\Psi(t_1<t<t_2)}_D = \ket{L}_1 \otimes \ket{L}_4 \otimes \frac{1}{2} (\ket{HH}_{2,3}-\ket{VV}_{2,3}+i\ket{HV}_{2,3}+i\ket{VH}_{2,3})$

At time $t_2$, Victor perform a SSM on photons 2&3 and consider the runs where the outcomes are either HH or VV. According to the Born's rule applied to the previous states $\ket{\Psi(t_1<t<t_2)}_{A,B,C,D}$, the HH/VV outcomes occurs 50% of the runs in each one of the four states. This means that, when Victor communicate his results to Alice and Bob and they considered only the runs where Victor obtained HH or VV, this subset is formed by 50% of the runs where Alice and Bob measured RR, 50% of the runs where Alice and Bob measured RL, 50% of the runs where Alice and Bob measured LR, and 50% of the runs where Alice and Bob measured LL. From this result, we conclude that when Alice and Bob post-select those runs according to Victor outcomes HH/VV, photons 1 and 4 show no correlation.

Now, I'll switch to the case where Victor perform a BSM. In this case, I'll rewrite the states $\ket{\Psi(t_1<t<t_2)}_{A,B,C,D}$ using the $\psi^\pm/\phi^\pm$ basis for photons 2&3. Since $\ket{\psi^\pm}_{2,3} = \frac{1}{\sqrt{2}} (\ket{HV}_{2,3}\pm\ket{VH}_{2,3})$ and $\ket{\phi^\pm} = \frac{1}{\sqrt{2}} (\ket{HH}_{2,3}\pm\ket{VV}_{2,3})$, it is straightforward that

$\ket{\Psi(t_1<t<t_2)}_A = \ket{R}_1 \otimes \ket{R}_4 \otimes \frac{1}{2} (\ket{\phi^-}_{2,3}-i\ket{\psi^+}_{2,3})$
$\ket{\Psi(t_1<t<t_2)}_B = \ket{R}_1 \otimes \ket{L}_4 \otimes \frac{1}{2} (\ket{\phi^+}_{2,3}+i\ket{\psi^-}_{2,3})$
$\ket{\Psi(t_1<t<t_2)}_C = \ket{L}_1 \otimes \ket{R}_4 \otimes \frac{1}{2} (\ket{\phi^+}_{2,3}-i\ket{\psi^-}_{2,3})$
$\ket{\Psi(t_1<t<t_2)}_D = \ket{L}_1 \otimes \ket{L}_4 \otimes \frac{1}{2} (\ket{\phi^-}_{2,3}+i\ket{\psi^+}_{2,3})$

At time $t_2$, Victor perform a BSM on photons 2&3 and consider the runs where the outcomes are either HH or VV, and one photon is detected at $b^{''}$ and the other at $c^{''}$, because this requirement allows him to conclude that the state of 2&3 is $\ket{\phi^-}_{2,3}$. According to the Born's rule applied to the previous states $\ket{\Psi(t_1<t<t_2)}_{A,B,C,D}$, the $\ket{\phi^-}_{2,3}$ outcomes occurs 50% of the runs in the subset formed by $\ket{\Psi(t_1<t<t_2)}_A$ and 50% of the runs in the subset formed by $\ket{\Psi(t_1<t<t_2)}_D$. This means that, when Victor communicate his results to Alice and Bob and they post-select only the runs where Victor outcomes were consistent with $\ket{\phi^-}_{2,3}$, this subset shows perfect correlations between photons 1 and 4 in the $L/R$ basis.

It was demonstrated that post-selection according to Victor results (strictly following the criteria in Ma's paper) leads to perfect correlations between photons 1 and 4 when Victor performs a BSM and no correlation between photons 1 and 4 when Victor performs a SSM. Since the previous analysis was always forward-in-time following the axioms of QM, no backward-in-time change of the state of the photons 1&4 was invoked. Please, note that in each case (BSM/SSM) the subsets were formed by different runs, so the selection criteria are not strictly the same.

Finally, I want to emphasize that QM predictions are the same irrespective of the order in which the Alice/Bob and the BSM/SSM measurements were performed. Therefore, the previous analysis does not prohibit that Victor (or anyone) decides to first projected out the state of the system giving priority to the BSM measurement he performed, in which case, he will conclude that for the runs where he obtained $\ket{\phi^-}_{2,3}$, the photons 1&4 will be in the entangled state $\ket{\phi^-}_{1,4}$, i.e. consistent with the perfect correlation between photons 1 and 4. Both approaches are complementary and, of course, both are right.

Lucas.

 

[PeterDonis]

in each case (BSM/SSM) the subsets were formed by different runs, so the selection criteria are not strictly the same.

In both cases you pick out subsets using exactly the same method. As I said before, that's how things are done in every other branch of science. Nobody says in any other branch of science, when comparing two sets of runs of an experiment, one with a certain manipulation by the experimenter and one without, that "the selection criteria are not strictly the same" because the two sets of runs are different, and that that somehow means you can't conclude, from the difference in results with the manipulation and without, that there is a real effect present.

 

[Morbert]

In both cases you pick out subsets using exactly the same method. As I said before, that's how things are done in every other branch of science.

If BSM/SSM subsets are picked using the exact same method, then that would be very strong evidence for nonlocal influence. The question is whether BSM/SSM subsets are picked using the exact same method. They are both picked using the exact same signature (HH or VV in separate spatial modes) but when the BiSA is configured to perform a BSM, the polarization of the incident 2 & 3 photons are modified before they are detected. Same polarization signature + modified polarization = different method.

[edit] - The "same signature + modification = different method" relation holds in classical contexts too.

Say you are going trick-or-treating down a street that has both mean and kind houses. The kind houses will give you a candy wrapped in red. The mean houses will give you a brussels sprout wrapped in blue.

When you get home, you want to discard the sprouts, so you use the wrapper as a signature condition: "If the wrapper is blue, discard". You have selected all the candies.

But alternatively, you could first modify the contents of your bag by exchanging wrappers so that sprouts are now wrapped in red and candies are now wrapped in blue. Now, the same signature condition "discard blue" will select all the sprouts.

Same signature + no modification = selection method that picks candies.
Same signature + modification = different selection method that picks sprouts.

 

[DrChinese]

There is a very clear difference of opinion here that should be incredibly easy to solve:

We're doing a single run of Ma's experiment. Alice and Bob have measured and gotten |R> and |R> for their results. Victor has yet to take any action. If Victor does a BSM on photons 2 and 3, what result can he get?

DrChinese predicts he can get all four bell states.
Morbert predicts he can only get ∣Ψ+⟩ and ∣Φ−⟩

This has a definite answer does it not?

Almost.

In the case you describe, Ma’s experiment, Morbert is correct. I’ve never said otherwise. The question is what happens when you arrange to start with |RR> for 1&4. This requires a simple modification to Ma’s setup, one they actually execute during setup but don’t report (my Ma-X). The result is a Product state for all 4 photons, just as Morbert claims. He says that leads to statistics matching a swap (because there is no actual swap). I say that’s impossible, and we already know that experimentally. I call this scenario Mjelva error #1.

The question then becomes: how did Mjelva get the right results using a wrong intermediate Product State? That’s what you are asking about, Mjelva error #2. And you are exactly correct that can be demonstrated experimentally as well, either way. I’ve searched for something that says this explicitly, without any luck. The problem is that no one is performing experiments with Product (separable) state pairs, everyone uses Entangled state pairs.

 

[DrChinese]

If BSM/SSM subsets are picked using the exact same method, then that would be very strong evidence for nonlocal influence. The question is whether BSM/SSM subsets are picked using the exact same method. They are both picked using the exact same signature (HH or VV in separate spatial modes) but when the BiSA is configured to perform a BSM, the polarization of the incident 2 & 3 photons are modified before they are detected. Same polarization signature + modified polarization = different …

As already pointed out: Ma’s plates do not alter the setup per your criticism. Fortunately, there’s another experiment that gets the same results using a technique that unambiguously does NOT modify any phase, polarization, or other potential observable. The only difference required to get an SSM is to make the 2&3 photons distinguishable. This is done by merely adding a short delay to the line of one of the photons.

This clearly explains why your criticism falls flat. But perhaps you reject all experimental work that disagrees with your viewpoint.

Megidish, figure 3c.

 

[Sambuco]

In both cases you pick out subsets using exactly the same method. As I said before, that's how things are done in every other branch of science. Nobody says in any other branch of science, when comparing two sets of runs of an experiment, one with a certain manipulation by the experimenter and one without, that "the selection criteria are not strictly the same" because the two sets of runs are different, and that that somehow means you can't conclude, from the difference in results with the manipulation and without, that there is a real effect present.

Let me explain why this is wrong even in a much simpler case. Let's consider a single EPR-like pair prepared in the state $\ket{\Psi}_{1,2} = \frac{1}{\sqrt{2}} (\ket{\uparrow_{1z}\uparrow_{2z}} + \ket{\downarrow_{1z}\downarrow_{2z}})$ at time $t_0$. Suppose particle 1 was delivered to Alice, while particle 2 was delivered to Bob. Now, let's considered the following two experiments:

1^st experiment: At time $t_1$, Alice measured particle 1 in the z-basis. At time $t_2$ Bob measure particle 2 in the z-basis. After the experiment, Alice and Bob reunite and they form a subset (post-selection) formed by particles 1 that correspond to the same runs where Bob measure spin-up ($\ket{\uparrow_{2z}}$). Obviously, all the particles in this subset while be spin-up, i.e. $\ket{\uparrow_{1z}}$.

2^nd experiment: At time $t_1$, Alice measured particle 1 in the z-basis. At time $t_2$ Bob measured particle 2 in the x-basis, and then at time $t_3$ he measured again particle 2 but in the z-basis. After the experiment, Alice and Bob reunite and they form a subset (post-selection) formed by particles 1 that correspond to the same runs where Bob measure spin-up ($\ket{\uparrow_{2z}}$). In this case, half of the particles in this subset will be spin-up, while the other half will be spin-down.

If you compare both experiments, what you call "the same criteria" was applied in both cases, and the results of the spins of the particles 1 in the subsets are different between the experiments. If we compare both experimental procedures, the only difference is that, in the 2^nd experiment, Bob made an additional first measurement on particle 2 in the x-basis. Does that mean that these measurements performed by Bob in the future-light cone of Alice's ones physically changed the spin of particles 1 in the subsets? Of course not! What it means is that the criteria is not the same, since the measurement that Bob performed on particles 2 in the x-basis detroyed the information about the spin value in z-axis.

Lucas.

 

[PeterDonis]

If BSM/SSM subsets are picked using the exact same method, then that would be very strong evidence for nonlocal influence. The question is whether BSM/SSM subsets are picked using the exact same method. They are both picked using the exact same signature (HH or VV in separate spatial modes) but when the BiSA is configured to perform a BSM, the polarization of the incident 2 & 3 photons are modified before they are detected.

As @DrChinese has already pointed out, this is an artifact of one particular experiment. It is not a general property of all entanglement swapping experiments. Hence, you can't rely on it as a refutation of my claim.

 

[PeterDonis]

Now, let's considered the following two experiments

Which are obviously different because Bob's measurements in the two cases are different. That's not what I'm talking about.

What I'm talking about, in the case of a single pair of qubits, would be described thus:

(1) The two qubits are entangled. Alice measures her qubit in the z-basis. Bob measures his in the z-basis. The measurements show perfect correlation (they either always agree or always disagree, depending on which entangled state the qubits were prepared in) when a large number of runs are done.

(2) The two qubits are not entangled. Alice measures her qubit in the z-basis. Bob measures his in the z-basis. The measurements show zero correlation (they are as likely to agree as to disagree) when a large number of runs are done.

This would be taken as evidence that the experimental intervention that entangles the qubits, which is done in (1) but not in (2), produces a real effect. The effect is nonlocal because the measurements can be spacelike separated and can show correlations that violate the Bell inequalities (as could be shown by choosing the Alice and Bob measurement bases in both cases to be at an appropriate angle to each other).

 

[DrChinese]

Given the preparation procedure, the state of the system after preparation (time $t_0$), but before Alice and Bob measurements (time $t_1$) is:
$\ket{\Psi(t_0<t<t_1)} = \ket{\psi^-}_{12}\otimes \ket{\psi^-}_{34}$
Then, Alice and Bob perform measurements and obtain one out of the four possible combinations RR, RL, LR, LL. According to the projection postulate the possible states of the system after $t_1$ but before $t_2$ are:

$\ket{\Psi(t_1<t<t_2)}_A = \ket{R}_1 \otimes \ket{R}_4 \otimes \ket{L}_2 \otimes \ket{L}_3$
$\ket{\Psi(t_1<t<t_2)}_B = \ket{R}_1 \otimes \ket{L}_4 \otimes \ket{L}_2 \otimes \ket{R}_3$
$\ket{\Psi(t_1<t<t_2)}_C = \ket{L}_1 \otimes \ket{R}_4 \otimes \ket{R}_2 \otimes \ket{L}_3$
$\ket{\Psi(t_1<t<t_2)}_D = \ket{L}_1 \otimes \ket{L}_4 \otimes \ket{R}_2 \otimes \ket{R}_3$

Lucas,

I've been unable to locate any reference supporting your above application of the Projection Postulate. This intermediate Product state is excluded theoretically ("Unperformed measurements have no results"). If I'm incorrect about this point, and you believe your projection is textbook QM, then there should be hundreds of examples for you to pull from sources you know I will gladly accept. On the other hand, here is what Vaidman (1998) says on this subject (counterfactual reasoning):

"I can not comprehend the meaning of the probability for the result M = mj given that the measurement M has not take place. As far as I can see Prob(M = mj|E[ψiI, ψkF]) has no physical meaning. ... In standard quantum theory unperformed experiments have no results, see Peres (1978)." And quoting further: "There is no reason to assign the same values to the following probabilities: the probability that an intervening measurement of M had the result mj given that such a measurement in fact took place, and the probability that intervening measurement would have had the result mj given that no such intervening measurement of M in fact took place."

Or from Wechsler and Motzkin (2021), "The Quantum Mechanics Needs the Principle of Wave-Function Collapse—But This Principle Shouldn’t Be Misunderstood": "This interpretation [consistent histories] is bound to disagree with the QM in additional issues. It assumes that a quantum system has at each time during its evolution, a certain definite property. In relation with such an assumption let’s mention the famous dictum of A. Peres “Unperformed experiments have no results”.

-DrC

 

[DrChinese]

If you compare both experiments, what you call "the same criteria" was applied in both cases...

Your idea of the "same criteria" is radically different than mine. Completely different tests are performed between your 1 and 2 experiments, resulting in different samples.

In the Megidish experiment, there is no difference in the tests performed. And the only difference in setup between BSM and SSM is overlap in a beam splitter. This is basic scientific method, and it is difficult to imagine what there is to debate.

 

[Sambuco]

Which are obviously different because Bob's measurements in the two cases are different.

Yes, that's my point. The measurement procedure Victor applies to photons 2&3 in each case (BSM/SSM) are different, i.e. the Hamiltonian depends on Victor decision. In one case (SSM), the interferometer acts as a 0/100 beam splitter (just as a mirror), and photons 2&3 remain in a separable state after the beam splitter. In the other case, the interferometer acts as a 50/50 beam splitter allowing the interaction between 2&3 photons that leads to their state to changed from a separable to an entangled one.

What I'm talking about, in the case of a single pair of qubits, would be described thus:

(1) The two qubits are entangled. Alice measures her qubit in the z-basis. Bob measures his in the z-basis. The measurements show perfect correlation (they either always agree or always disagree, depending on which entangled state the qubits were prepared in) when a large number of runs are done.

(2) The two qubits are not entangled. Alice measures her qubit in the z-basis. Bob measures his in the z-basis. The measurements show zero correlation (they are as likely to agree as to disagree) when a large number of runs are done.

What you mention has nothing in common with what I'm saying about Ma's experiment. That's not a good analogy, because (i) it does not involve post-selection, and (ii) the systems are prepared in different initial states in cases (1) and (2).

Anyway, we disagree about the analogy I proposed in post #160, but don't worry about that. Please, get back to the central point of our discussion about Ma's experiment. I think that both @DrChinese and you claim that the change of the state of photons 1&4 when Victor performed a BSM can ONLY be interpreted as a (backward-in-time) "real physical effect" on photons 1&4, and no other interpretation is possible. This claim is not supported by QM theory, as I demonstrated in post #155, where I strictly follow textbook QM postulates and the detailed calculations predict the measurement outcomes in Ma's paper. If you have any objection against my treatment of the problem, please let me know.

As an aside, my treatment of the problem is in agreement with what Cohen said in the first paper where this kind of experiments were discussed (https://arxiv.org/abs/quant-ph/9907109). Note that the title of the paper is "Conterfactual entanglement and nonlocal correlations in separable states". As I mentioned in other post, this work was cited in Ma's paper, together with Peres's one as the references for DCES.

Lucas.

 

[Sambuco]

In the Megidish experiment, there is no difference in the tests performed. And the only difference in setup between BSM and SSM is overlap in a beam splitter. This is basic scientific method, and it is difficult to imagine what there is to debate.

I agree with you that this is the only difference between BSM and SSM in the Megidish's experiment. In fact, the delay is what makes possible the interaction between photons 2&3. When there is delay and the photons can be distinguished, there is no interaction, while when the photons cannot be distinguished, an interaction occurs leading to entanglement. This interaction/no interaction depends on the delay through the Hamiltonian, because to make an interaction possible we need that both photons coincide in a narrow window of space and time.

Lucas.

 

[Sambuco]

You won't find any reference anywhere that supports the above application of the Projection Postulate. This intermediate Product state is excluded theoretically ("Unperformed measurements have no results"). If I'm incorrect about this point, and you believe your projection is textbook QM, then there should be hundreds of examples for you to pull from sources you know I will gladly accept.

A nice treatment about the application of the projection postulate to entangled states is in Zweibach's book "Mastering quantum mechanics". In chapter 18 (Multiparticle states and tensor products), section 18.6 (Bell basis states), subsection "Partial measurement", he explains that. I copy here the relevant part:

"Suppose we have a general entangled state $\ket\Psi \in V \otimes W$ of two particles. Alice has access to the first particle and decides to measure along the basis $\ket{e_1},...,\ket{e_n}$ of $V$. This is analyzed with measurement axiom A3, using a complete set of mutually orthogonal projectors $M_i$:

$M_i = \ket{e_i}\bra{e_i} \otimes 1$

To simplify the writing of probabilities, consider the measurement of a general state $\ket\Psi$ written with the help of the basis vectors $\ket{e_i}$ as

$\ket\Psi = \sum_i \ket{e_i} \otimes \ket{w_i}$ (18.6.11)

If Alice finds her particle in $\ket{e_i}$ the state of the system after measurement is $M_i \ket\Psi$, suitably normalized:

$M_i \ket\Psi = (\ket{e_i}\bra{e_i} \otimes 1) \sum_i \ket{e_i} \otimes \ket{w_i} = \ket{e_i}\ket{w_i}$ (18.6.13)"

As shown, my application of the projection postulate in post #155 is correct.

Lucas.

 

[DrChinese]

1. "Suppose we have a general entangled state $\ket\Psi \in V \otimes W$ of two particles. Alice has access to the first particle and decides to measure along the basis $\ket{e_1},...,\ket{e_n}$ of $V$. This is analyzed with measurement axiom A3, using a complete set of mutually orthogonal projectors $M_i$:

$M_i = \ket{e_i}\bra{e_i} \otimes 1$

To simplify the writing of probabilities, consider the measurement of a general state $\ket\Psi$ written with the help of the basis vectors $\ket{e_i}$ as

$\ket\Psi = \sum_i \ket{e_i} \otimes \ket{w_i}$ (18.6.11)

2. If Alice finds her particle in $\ket{e_i}$ the state of the system after measurement is $M_i \ket\Psi$, suitably normalized:

$M_i \ket\Psi = (\ket{e_i}\bra{e_i} \otimes 1) \sum_i \ket{e_i} \otimes \ket{w_i} = \ket{e_i}\ket{w_i}$ (18.6.13)"

Thanks for this, but I don't read it the same as you do.

1. I'm looking for a starting entangled state; this one looks like a product state, but perhaps the full context clarifies how it is entangled. Note the use of "⊗" vs. "+" as might be expected. But I will assume his Ψ is entangled for the sake of discussion.

2. It is not clear whether further evolution is occurring, without performing a measurement. That is the important question. There is no question that if Alice measures and then Bob measures, his outcome is certain if they are measured on the same basis. This is not the question at hand though, is it? The question is about that "unperformed measurements" thing, intermediate states.

In our example: We initially have 2 entangled systems (T=0) and we measure one particle of each system (T=1). The unmeasured partner particles are allowed to interact (T=2), and we are discussing the predicted state of those partner particles at T=3. Got something like that?

 

[eloheim]

In the case you describe, Ma’s experiment, Morbert is correct. I’ve never said otherwise. The question is what happens when you arrange to start with |RR> for 1&4. This requires a simple modification to Ma’s setup, one they actually execute during setup but don’t report (my Ma-X). The result is a Product state for all 4 photons, just as Morbert claims. He says that leads to statistics matching a swap (because there is no actual swap). I say that’s impossible, and we already know that experimentally. I call this scenario Mjelva error #1.

Okay this really has to be at the root of this all:

We do an Ma-X experiment run. It's the same as the real Ma setup but each initial pair of photons (1+2 and 3+4) are anticorrelated in accordance with |Ψ-> but NOT entangled. At the current time Alice and Bob have measured and both got |R>. When Victor puts photons 2 and 3 through the BSM apparatus, what bell states can he get?

@DrChinese predicts he can get any of the 4 bell states
Does anyone predict he can only get |Ψ-> and Φ+? (@Sambuco, @Morbert) (because it looks to me like Mjelva/Sambuco/Morbert's math would suggest this)

 

[PeterDonis]

The measurement procedure Victor applies to photons 2&3 in each case (BSM/SSM) are different, i.e. the Hamiltonian depends on Victor decision

You're confusing the measurement procedure with the experimental manipulation. The BSM/SSM choice is the experimental manipulation; of course that's different in the two cases, and of course the Hamiltonian changes, since that's the whole point of the experimental manipulation.

But the measurement procedure is not the same as the experimental manipulation. The measurement procedure is "record in which output channels photons appear and what their polarizations are". That's the same regardless of what is done at the BSM/SSM.

 

[DrChinese]

Okay this really has to be at the root of this all:

We do an Ma-X experiment run. It's the same as the real Ma setup but each initial pair of photons (1+2 and 3+4) are anticorrelated in accordance with |Ψ-> but NOT entangled. At the current time Alice and Bob have measured and both got |R>. When Victor puts photons 2 and 3 through the BSM apparatus, what bell states can he get?

@DrChinese predicts he can get any of the 4 bell states
Does anyone predict he can only get |Ψ-> and Φ+? (@Sambuco, @Morbert) (because it looks to me like Mjelva/Sambuco/Morbert's math would suggest this)

Yes, I agree 100%.

Now add one thing, the bomb: IF you start with the easily created correlated/anti-correlated but Product State pairs - THEN you wont get ANY 4-fold entanglement after a BSM. That is the entire premise of the Mjelva paper that is being contested. There are 4 possible combinations, and none produce the swapping statistics in different bases (i.e. 1 & 4 measured H/V, 2 & 3 measured +/-, or whatever).

All of those Product State pairs can be created at will by a number of simple methods. Both Type I PDC (correlated) and Type II PDC (anti-correlated) crystals generate such pairs natively (usually denoted on the H/V basis). Type I creates product state HH from a V input. Type II generates product state HV. But... both Type I and Type II PDC outputs require significant extra treatment and collection to generate entangled pairs for actual swapping experiments. Obviously, that treatment would not be necessary if the original Product State pairs would suffice.

Type I PDC (Product State with 1 crystal, Entangled State with 2 crystals)

Because there is no correlation between 1 & 4 photon (Product State source) outcomes on one basis; and between 2 & 3 outcomes on any unbiased basis: there is therefore no particular restriction on which of the 4 Bell states are produced by a BSM*. Ergo, no swap correlations without real entanglement. Swap correlations require a physical swap**; swaps are real, they are not an artifact of mathematical coincidence. Hopefully I have explained this satisfactorily.


*Note that a BSM on 2 & 3 can produce a Bell state when the 4-fold source is initially a Product state. But that's true only for photons 2 & 3, because there is no swap related to photons 1 & 4.

**This is something of a corollary to what @PeterDonis is telling us. In a scientific experiment when a physical variable is switched from some A to B, then any changed outcomes are a result of that physical change. It must be physical, there's only one variable.

 

[Sambuco]

1. I'm looking for a starting entangled state; this one looks like a product state, but perhaps the full context clarifies how it is entangled. Note the use of "⊗" vs. "+" as might be expected. But I will assume his Ψ is entangled for the sake of discussion.

Yes, it is an entangled state. If you look at eq. (18.6.11), you will see that the state is non-separable.

2. It is not clear whether further evolution is occurring, without performing a measurement. That is the important question. There is no question that if Alice measures and then Bob measures, his outcome is certain if they are measured on the same basis. This is not the question at hand though, is it? The question is about that "unperformed measurements" thing, intermediate states.

Thanks for your answer! Now, I see why you deny the "intermediate states", and why you thought my analysis of the Ma's experiment had to be wrong.
Let me explain how the projection postulate applied to entangled systems is not problematic in any way. When a partial measurement was performed on an entangled system formed by two particles, the state of the system after that is only one of the terms in the superposition, so the entire state collapses, not just the part of the state corresponding to the measured particle. This does not mean that the other particle will be measured, so there is no counterfactual reasoning (like the "elements of reality" in the EPR paper). The change in the state only means that the unmeasured particle is in an eigenstate corresponding to a certain observable. If it helps to understand, you could think of that process as a kind of (non-local) preparation procedure. When you prepare a quantum system in a given energy eigenstate this does not mean that an energy measurement neccesarily will be performed, it only says that, if you will measure the energy of the system, you have 100% certainty about the measurement outcome you will obtain.

In our example: We initially have 2 entangled systems (T=0) and we measure one particle of each system (T=1). The unmeasured partner particles are allowed to interact (T=2), and we are discussing the predicted state of those partner particles at T=3. Got something like that?

Yes, this is exactly how I analyzed the evolution of the system in my post #155.

Lucas.

 

[Sambuco]

But the measurement procedure is not the same as the experimental manipulation. The measurement procedure is "record in which output channels photons appear and what their polarizations are". That's the same regardless of what is done at the BSM/SSM.

When you say that the selection criteria depends on "record in which output channels photons appear and what their polarizations are" you are disregarding the quantum state of photons 2&3 before they were measured by Victor. In a (forward-in-time) theoretical analysis of the measurement according to textbook QM, these states are not the same if you only look at the runs that were post-selected, as I demonstrated in my post #155 for the runs where Alice and Bob measure photons 1&4 in the $L/R$ basis. Please, if you have any doubts about that, check it.
This point is the key because the state of the 2&3 photons before Victor's measurement depends on the measurement outcomes obtained by Alice and Bob, so if you don't consider that, it is natural that you conclude that a physical effect on photons 1&4 is mandatory to explain the measurement outcomes.
The consideration of the output channels and the polarization excluding the state before Victor measurement is what makes me say that the "selection criteria" are not the same.

Despite any minor disagreement, the important thing is that (I repeat myself, sorry) the state of photons 1&4 does not change upon Victor measurement in any run of the experiment if you follow a forward-in-time analysis of the Ma's experiment.

Lucas.

 

[PeterDonis]

When you say that the selection criteria depends on "record in which output channels photons appear and what their polarizations are"

That's not what I said. I said that what you quoted is the measurement procedure.

The selection criteria are "post-select the subsets of the runs that correspond to whatever combinations of output channels/polarizations for photons 2&3 signal a distinguishable Bell state if a swap is done".

The above are the same whether a swap is done or not. What changes if a swap is done is that the post-selected subsets of runs show Bell state correlations between photons 1 & 4, whereas they don't if no swap is done.

We've already been over this multiple times.

you are disregarding the quantum state of photons 2&3 before they were measured by Victor.

We're just going around in circles here. You're not responding to what I'm actually saying. You're just restating the same things I've already responded to.

 

[Sambuco]

Okay this really has to be at the root of this all:

We do an Ma-X experiment run. It's the same as the real Ma setup but each initial pair of photons (1+2 and 3+4) are anticorrelated in accordance with |Ψ-> but NOT entangled. At the current time Alice and Bob have measured and both got |R>. When Victor puts photons 2 and 3 through the BSM apparatus, what bell states can he get?

@DrChinese predicts he can get any of the 4 bell states
Does anyone predict he can only get |Ψ-> and Φ+? (@Sambuco, @Morbert) (because it looks to me like Mjelva/Sambuco/Morbert's math would suggest this)

It depends on the initial state. If the only information you have is that the system is in a separable (non-entangled) four-photon state, where 1&2 are anticorrelated and the same for 3&4, then you cannot know what are the Bell states Victor could measure, because you don't know what is the state of photons 2&3 before entering the BSM. You have to provide more information about the initial state to make a specific calculation.

 

[Sambuco]

We're just going around in circles here. You're not responding to what I'm actually saying. You're just restating the same things I've already responded to.

Ok, don't worry. Maybe I misinterpreted something you said.

[edit] - I read it again and you're right. Thanks for the clarification

 

[Sambuco]

This is something of a corollary to what @PeterDonis is telling us. In a scientific experiment when a physical variable is switched from some A to B, then any changed outcomes are a result of that physical change. It must be physical, there's only one variable.

That's exactly why what you called "intermediate states" are important. Let's see why there's no single variable that changes in the subsets formed according to the outcomes obtained by Victor on photons 2&3.

First, I'd like to recap the things we all agree on. Since I like how @PeterDonis summarized the experiment and its results in post #174, I'll use that:

The selection criteria are "post-select the subsets of the runs that correspond to whatever combinations of output channels/polarizations for photons 2&3 signal a distinguishable Bell state if a swap is done".
The above are the same whether a swap is done or not. What changes if a swap is done is that the post-selected subsets of runs show Bell state correlations between photons 1 & 4, whereas they don't if no swap is done.

Let's try to find what was the "cause" behind the difference in the statistical data for the subsets of photons 1&4 in the two experiments (BSM/SSM). As the subsets were formed according to the mearused outcomes obtained by Victor on 2&3, we have to find what causes in each case the 2&3 measurement outcomes. The obvious difference between the experiments is the BSM or SSM. But the measurement outcomes on 2&3 do not depend exclusively on whether the swap was done or not (BSM or SSM), but also on the state of photons 2&3 when entering the interferometer. If these states are the same in both subsets, irrespective of the Victor's decision to perform or not a swap, then, the only independent variable is, as @DrChinese said, Victor's decision, so that we are forced to accept that this is the cause of the different statistical data observed in those subsets of photons 1&4. But, are we sure that the state of photons 2&3 before the interferometer is the same in both subsets? What does this state depends on? Well, as (i) photon 2 was entangled with 1 and photon 3 was entangled to 4, and (ii) Alice and Bob performed measurements on photons 1&4 in the past light cone of Victor measurements, the state of 2&3 before the BSM/SSM depends on the measurement outcomes obtained by Alice and Bob on photons 1&4. Since these measurement outcomes are different in the post-selected subsets of each experiment (BSM/SSM), we can conclude that the difference in the state of photons 2&3 before BSM/SSM was "caused" by that. Do you see the circular reasoning? As we cannot avoid that, the claim that a "physical change" of outcomes on photons 1&4 was caused by Victor decision to perform or not a BSM is, at least, problematic.

In my view (what follows is interpretation-dependent), the problem arises when we tried to understand measurement outcomes in terms of cause-and-effect. If we assume that there are only correlations, the interpretation problems related to causation suddenly dissapear.

Lucas.

 

[eloheim]

It depends on the initial state. If the only information you have is that the system is in a separable (non-entangled) four-photon state, where 1&2 are anticorrelated and the same for 3&4, then you cannot know what are the Bell states Victor could measure, because you don't know what is the state of photons 2&3 before entering the BSM. You have to provide more information about the initial state to make a specific calculation.

Sorry I may have underspecified for sake of brevity. Can we just say we create unentangled photon pairs of one |L> and one |R>, so everybody knows the prepared basis, and if Alice or Bob measure |R> then we know photons 2+3 both started in |L>. Also the bell measurement of 2+3 is done in the H/V basis.

Will Victor get all 4 bell states or only 2?

 

[Morbert]

In the case you describe, Ma’s experiment, Morbert is correct. I’ve never said otherwise. The question is what happens when you arrange to start with |RR> for 1&4. This requires a simple modification to Ma’s setup, one they actually execute during setup but don’t report (my Ma-X). The result is a Product state for all 4 photons, just as Morbert claims. He says that leads to statistics matching a swap (because there is no actual swap). I say that’s impossible, and we already know that experimentally. I call this scenario Mjelva error #1.

Replacing Ma's initial state with the Ma-X initial state leads to statistics matching a swap only in the R/L basis. Total statistics of course disagree, and you lose Bell/perfect correlations/anticorrelations you see across all three bases in the Ma experiment. I described and illustrated a comparison between Ma and Ma-X in #90. Here I will repost the relevant image.

You can see that the initial state of the Ma-X experiment will enforce correlation in the R/L basis, but not in the other two mutually unbiased bases.

In both Ma and Ma-X experiments, or any similar experiment, following Mjelva's prescription, or equivalently, physicsforum's prescription will get you the right predictions.

As already pointed out: Ma’s plates do not alter the setup per your criticism. Fortunately, there’s another experiment that gets the same results using a technique that unambiguously does NOT modify any phase, polarization, or other potential observable. The only difference required to get an SSM is to make the 2&3 photons distinguishable. This is done by merely adding a short delay to the line of one of the photons.

This clearly explains why your criticism falls flat. But perhaps you reject all experimental work that disagrees with your viewpoint.

Megidish, figure 3c.

After passing through the PBS, the photons are rotated by HWPs to the $|p/m\rangle = (\ket{h}+\ket{v})/\sqrt{2}$ polarization basis.

If there was no such manipulations, then polarization detection could not be used for complementary measurements.

If we are to get into the weeds of Megidish's apparatus, a separate thread might be in order. What is relevant for Mjelva's paper is that, like before, the simple prescription of "evolve the initial state to the first measurement, then project onto a reduced state, then evolve to the next measurement etc" will get us all the right predictions so long as all possible outcomes are accounted for.

---

Ultimately, whether the experiment is Ma's or Megidish's, tetraphotons prepared in certain states and measured at different times will reproduce the correlations seen in measurements on entangled biphotons. Mjelva offers an interpretation of this fact based on evolution and projection onto states. Ma offers* an interpretation of this fact based on temporally nonlocal entanglement. Both procedures get you the right statistics, always.

* I say Ma offers even though I do not know if this is the focus of the paper, as opposed to the experimental feat itself. I have unfortunately not found any literature where Ma comments on alternative prescriptions.

 

[Sambuco]

Sorry I may have underspecified for sake of brevity. Can we just say we create unentangled photon pairs of one |L> and one |R>, so everybody knows the prepared basis, and if Alice or Bob measure |R> then we know photons 2+3 both started in |L>. Also the bell measurement of 2+3 is done in the H/V basis.

Will Victor get all 4 bell states or only 2?

If photons 2&3 were prepared in the state $\ket{\Psi}_{2,3} = \ket{L}_2 \otimes \ket{L}_3$, Victor will get only two possible states, $\ket{\phi^-}_{2,3}$ or $\ket{\psi^+}_{2,3}$.