Entanglement swapping and Bohmian mechanics

[PeterDonis]

Don't you see this has no useful meaning in the physics we are discussing?

No, we don't see that, because, as has already been pointed out multiple times, the fact that Ma's (1) is algebraically equivalent to Ma's (2), to us, means that physically they are the same state. So if Ma's (1) is the pre-swap state, so is Ma's (2).

You disagree; you think Ma's (2) is the post-swap state. That's a disagreement about physics, not about algebra. It's you not believing what I said in the previous paragraph. Which, to us, is like not believing that the different expressions in #238 are all describing the same thing.

Here's another analogy: suppose I have a vector on the Euclidean plane. I pick a basis $\vec{x}$, $\vec{y}$ such that the vector is $\vec{x}$. It's an arrow with a specific length (one) pointing in a specific direction (whichever direction the $x$ axis points).

Now I rewrite that vector in a new basis $\vec{u}$, $\vec{v}$, whose axes are rotated by 45 degrees, so that we have

$$
\vec{x} = \frac{1}{\sqrt{2}} \left( \vec{u} - \vec{v} \right)
$$

To me, and as far as I can tell to everyone in the thread except you, the expression on the RHS above is describing the same vector--the same arrow, with the same magnitude, pointing in the same direction--as the LHS. To you, it's apparently some meaningless math that has no relationship to anything. Or maybe to you it describes some other vector, which we obtained by doing something to the arrow described by $\vec{x}$. At this point it's hard for me to tell what your mental model of all this is.

That's the kind of disagreement we're having. All these states we've been writing down are vectors in a vector space. True, it's a vector space over the complex numbers instead of the reals, and it has eight dimensions (2 for each photon) instead of two, so it's not the same vector space as the 2-dimensional Euclidean plane. But it's still a vector space, and a general fact about all vector spaces is that you can rewrite a vector in a different basis--which is just algebra--without changing the vector. And when we use vector spaces in physics, the physical meaning of what I just said is that different rewritings of the same mathematical vector in different bases describe the same physical thing. In this case, the state of the 4-photon system pre-swap. If Ma's (1) describes that physical thing, and Ma's (2) is a rewriting of Ma's (1) in a different basis, i.e., just algebra, then Ma's (2) also describes that physical thing. It doesn't describe some other physical thing.

That's how I see things, and as far as I can tell, it's how everyone here except you sees things. One thing is certain: it's not just a mathematical disagreement. It's a physical disagrement, because we're using math here to do physics.

 

[PeterDonis]

I can multiply unconnected things together.

Nobody else is doing that. Nobody else is dragging in photons from all over the universe to clutter up the simple mathematical expressions in Ma's (1) and Ma's (2). All the rest of us are doing is algebraically refactoring Ma's (1) to obtain Ma's (2). I.e., the only photons the rest of us are describing are the four photons that are in the experiment. We're not the ones talking about other photons somewhere else. Only you are.

I agree that other photons elsewhere in the universe are unconnected (in the scenario we're discussing) with the 4 photons in the experiment. So I agree that we can ignore all those photons and not bother writing down states that include them. None of which changes the things I said in post #241 just now. That's what the disagreement is about. It's not about photons #17 or #33 or #googolplex or whatever. It's about Ma's (1) and Ma's (2) describing the same physical state of the 4 photons in the experiment--the pre-swap state.

 

[DrChinese]

What Ma actually says is (2) is arrived at by "rewriting Eq. (1) in the basis of Bell states of photons 2 and 3" (Ma's exact words). What you did is rewrite this to say (2) is "the newly swapped state" (your words, not Ma's).

And for the zillionth time: You keeping ignoring the crucial point, in the literal words of Ma:

State (1) -> Swap -> State (2)

Ma: "As schematically shown in Fig. 1, if [and this could be labeled iff] Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled and entanglement swapping is achieved."

DrChinese: Thereby creating "...the newly swapped state...".

What is in question here? If you don't like my words, just use Ma's! Fine with me, since they say the same thing.

The pre-swap state |Ψ-〉12⨂|Ψ-〉34 (1) obviously features no entanglement whatsoever between photons 2 and 3, which could belong to any entangled systems anywhere in the entire universe at any time in history. The post-swap state |Ψ〉14⨂|Ψ〉23 * features entanglement between 2 and 3 subsequent to their interaction at a common BS. There is literally nothing to question here as to Ma's meaning.

And according to you (I guess): The choice of measurement to make at the BS - entangled or separable - doesn't even matter. Since the formula you keep presenting is the Entangled state outcome, and not the Product state outcome. In the Separable/Product state, there is never any correlation when the 1/4 basis is different than the 2/3 basis - that's clearly shown in Megidish figure 3.

* |Ψ〉14⨂|Ψ〉23 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)

 

[PeterDonis]

|Ψ〉14⨂|Ψ〉23 * features entanglement between 2 and 3 subsequent to their interaction at a common BS.

No, it doesn't. We have already explained to you why, multiple times. And I showed you in #231 the computation that demonstrates zero correlation between photons 2 and 3 in that state.

according to you (I guess): The choice of measurement to make at the BS - entangled or separable - doesn't even matter. Since the formula you keep presenting is the Entangled state outcome, and not the Product state outcome.

Nobody has made any such claim. We are not the ones claiming that Ma's (2) is the post-swap state. Only you are. We are saying that Ma's 2 is the pre-swap state written in a different basis. Of course the pre-swap state will be the same regardless of whether Victor chooses to do a BSM or an SSM. So the "problem" you see here is not a problem for us at all, because we are not incorrectly interpreting Ma's (2) the way you are.

As has already been pointed out multiple times, @Morbert in #40 wrote down what the post-swap state looks like for the case where Victor selects to do a BSM. I don't know if anyone has posted in this thread what the state would be post-SSM if Victor selects that instead, but if not, it would be straightforward to do if it would help.

 

[Morbert]

And for the zillionth time: You keeping ignoring the crucial point, in the literal words of Ma:

I'm not moving off this point until you acknowledge it. Ma explicitly says (2) is a rewriting of (1) in a Bell basis. You keep writing paragraphs to argue that what Ma actually meant to say was something different. I.e. You are arguing that Ma misspoke, and that (2) is not in fact a rewriting of (1).

 

[PeterDonis]

If you don't like my words, just use Ma's! Fine with me, since they say the same thing.

No, they don't. Ma never says his (2) is the post-swap state. Only you are saying that. You are interpreting other words of Ma's to mean that, but we do not agree with your interpretation.

 

[DrChinese]

Nobody else is doing that. Nobody else is dragging in photons from all over the universe to clutter up the simple mathematical expressions in Ma's (1) and Ma's (2). All the rest of us are doing is algebraically refactoring Ma's (1) to obtain Ma's (2). I.e., the only photons the rest of us are describing are the four photons that are in the experiment. We're not the ones talking about other photons somewhere else. Only you are.

I agree that other photons elsewhere in the universe are unconnected (in the scenario we're discussing) with the 4 photons in the experiment. So I agree that we can ignore all those photons and not bother writing down states that include them. None of which changes the things I said in post #241 just now. That's what the disagreement is about. It's not about photons #17 or #33 or #googolplex or whatever. It's about Ma's (1) and Ma's (2) describing the same physical state of the 4 photons in the experiment--the pre-swap state.

What is the initial connection between 2 and 3? None, as I believe you have previously agreed. The reason I am discussing it is precisely because there is no initial connection.

Why write a formula for these that doesn't include other potential future partners? You realize, I hope, that literally 3 photons could be presented at the BS (one each from 3 different initial entangled pairs). And in fact there are many possible permutations in that case. But only 2 of the 3 pairs would be maximally entangled after a swap. You cannot pick random pairs of entangled photons (1) and present them as a jointly entangled state (2) unless you first execute a swap.

 

[DrChinese]

No, it doesn't. We have already explained to you why, multiple times. And I showed you in #231 the computation that demonstrates zero correlation between photons 2 and 3 in that state.

Whoa there Nelly!

I said "|Ψ〉14⨂|Ψ〉23 * features entanglement between 2 and 3 subsequent to their interaction at a common BS." You say there is "zero correlation". Two completely different points. The post-swap state contains terms for 4 possible Bell states, all 4 of which are entangled. Yes, it's also correct that the post-swap state will not show any correlation if those states are not distinguished.

But of course, none of these experiments would have any meaning if they weren't, n'est-ce pas?

 

[DrChinese]

I'm not moving off this point until you acknowledge it. Ma explicitly says (2) is a rewriting of (1) in a Bell basis.

No, actually, Ma explicitly says this - and not just your cherry-picked sentence:

... (1)

"...As schematically shown in Fig. 1, if [and this could be labeled iff] Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled and entanglement swapping is achieved. This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3:"

...(2)

You (or any reader) can continue to look at the semantic content of a Nobel level paper. Maybe you would realize that they could have used the word "writing" instead of "rewriting" and it would make the same sense. Or read the dozens of other papers saying the same thing - also in words you will likely pick apart. I've already quoted Megidish ("When the two photons of time τ (photons 2 and 3) are projected onto any Bell state, the first and last photons (1and 4) collapse also into the same state and entanglement is swapped."). But you don't seem to care about them, or to understand the relevant meaning of the underlying physics. Which is:

State (1 - no 2/3 entanglement) -> Swap -> State (2 - 2/3 Bell state entanglement)

---

I am moving on to other relevant questions for this thread. One of which, as I have said, concerns whether you think the photons going into the BS are in the singlet state or are in a pure state (Bohmian or orthodox views).

 

[DrChinese]

I've found an intuitive pictorial way to answer one of the @DrChinese 's central questions: How to describe the timeline of the full entanglement swapping experiment, step by step, in BM? Here is how.

Initially at time $t_1$ the wave function is the product $|\psi_{AB}\rangle|\psi_{CD}\rangle$. In BM I represent it with
$$ t_1: \; A \mathrel{-} B \;\;\;\; C \mathrel{-} D$$
The link "$\mathrel{-}$" between two particles is not a "minus" sign. Instead, this link denotes the Bohmian instantaneous mutual influence between the particles. $A$ is linked with $B$, and $C$ is linked with $D$, but there is no link between $B$ and $C$. Note that the link does not have an arrow, because at the fundamental microscopic level there is no time arrow, i.e., there is no "cause" and "effect".

Next, at time $t_2$, we independently measure particles $A$ and $D$, so they interact with the macroscopic measuring apparatuses $M_A$ and $M_D$, respectively. In BM I represent this with
$$ t_2: \; M_A \mathrel{-} A \mathrel{-} B \;\;\;\; C \mathrel{-} D \mathrel{-} M_D$$
or equivalently
$$ t_2:\; (M_A,A)_2 \mathrel{-} B \;\;\;\; C \mathrel{-} (M_D,D)_2$$
The notation $(M_A,A)_2$ denotes that the system $M_A+A$ is viewed as one system at time $t_2$.

Finally, at time $t_3$, we measure the system $B+C$, so
$$ t_3:\; (M_A,A)_3 \mathrel{-} B \mathrel{-} M_{BC} \mathrel{-} C \mathrel{-} (M_D,D)_3$$
or equivalently
$$ t_3:\; (M_A,A)_3 \mathrel{-} (B,M_{BC},C)_3 \mathrel{-} (M_D,D)_3$$
This finishes the description at the fundamental microscopic level.

However, at the emergent macroscopic level things look slightly different. The measurement outcomes encoded in $M_A$ and $M_D$ are stored on a computer, so they don't change much during the time. Also, after the measurement, the macroscopic system $M_A+A$ is practically indistinguishable from $M_A$ alone, i.e., the state of the particle $A$ after the measurement is practically irrelevant. The particle $A$ can even be destroyed, it doesn't matter as long as we keep the measurement outcome encoded in $M_A$. Thus, for practical purposes, we can use the approximation $(M_A,A)_2 \simeq (M_A,A)_3$. Likewise, we have $(M_D,D)_2 \simeq (M_D,D)_3$. Hence, effectively, the microscopic state at $t_3$ implies an effective macroscopic description
$$ (M_A,A)_2 \mathrel{-} (B,M_{BC},C)_3 \mathrel{-} (M_D,D)_2$$
At the statistical level, this explains why the results of measurements in the past at time $t_2$ are mutually correlated when combined with measurement results in the future at time $t_3$: $(M_A,A)_2$ is correlated with $(B,M_{BC},C)_3$, and $(B,M_{BC},C)_3$ is correlated with $(M_D,D)_2$, so it is not surprising that $(M_A,A)_2$ can be correlated with $(M_D,D)_2$. Furthermore, at the macroscopic level we also have a time arrow, due to which the past causes the future, so the correlation can also be interpreted causally as
$$ (M_A,A)_2 \rightarrow (B,M_{BC},C)_3 \leftarrow (M_D,D)_2$$
Thus we see that, at the macroscopic level, we can say that the results of measurements in the past cause the result of measurement in the future.

Finally, in the instrumental version of BM (see the paper in my signature and references therein), the Bohmian trajectories of the measured particles can be completely eliminated from the description, leaving only particle trajectories constituting the measuring apparatuses. In this description the last causal diagram reduces to
$$ (M_A)_2 \rightarrow (M_{BC})_3 \leftarrow (M_D)_2$$

This finishes my description of the entanglement swapping experiment from the Bohmian point of view, as an intriguing interplay between the fundamental microscopic and emergent macroscopic descriptions.

Great stuff, Demystifier. I will have a series of questions as I digest this.

A question off the top of my head. :

How important is it (in Bohmian terms) to include the measurement apparatus as a factor? I get that it might be technically a factor, but it seems like the measurement systems (detectors or whatever) always cancel out their effect to a net observable zero on the result. After all, you wouldn't have perfect correlations on typical entangled pairs if the measurement apparatus was itself a significant factor. (Obviously, the settings themselves matter. I am referring to when their settings are the same.)

 

[Morbert]

No, actually, Ma explicitly says this - and not just your cherry-picked sentence:

... (1)

"...As schematically shown in Fig. 1, if [and this could be labeled iff] Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled and entanglement swapping is achieved. This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3:"

Ma says "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3.". I am not picking apart anything. I, unlike you, am in full agreement with Ma's choice of words.

I am moving on to other relevant questions for this thread.

All of your misunderstandings of Ma's experiment stem from your misreading and misunderstanding of (2). Until you understand how you are misreading Ma's (2), you'll be stuck spinning your wheels.

 

[DrChinese]

From @Demystifier:

Furthermore, at the macroscopic level we also have a time arrow, due to which the past causes the future, so the correlation can also be interpreted causally as
$$ (M_A,A)_2 \rightarrow (B,M_{BC},C)_3 \leftarrow (M_D,D)_2$$
Thus we see that, at the macroscopic level, we can say that the results of measurements in the past cause the result of measurement in the future.

---

OK, I follow this. And presumably if the experimental ordering were reversed, we'd have:

$(M_A,A)_2 \leftarrow (B,M_{BC},C)_3 \rightarrow (M_D,D)_2$

And the results must be the same, correct? (I know this is a simple question.)

 

[DrChinese]

1. Ma says "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3.". I am not picking apart anything.

2. All of your misunderstandings of Ma's experiment stem from your misreading and misunderstanding of (2). Until you understand how you are misreading Ma's (2), you'll be stuck spinning your wheels.

1. It's literally the definition of cherry picking. Read and quote Ma's entire paragraph. Or Megidish.

2. I am quite comfortable with my understanding of Ma, Megisdish, Hensen, Weihs, etc. I am not stuck anywhere. It is slow going getting answers to some of my questions. Here's one for you:

Photon 1 has been observed to be in the state |H>. This implies that an H/V measurement on photon 2 would certainly yield a |V> result, given the biphoton's initial |Ψ−〉12 state. That photon is instead later fed into the Beam Splitter (BS) for a swap. When photon 2 enters the BS, is it still in a singlet state? Or is it in a pure |V> polarization state?

I am interested in an answer in terms of your own preferred interpretation. You don't need to tell me what interpretation it is (or you can - your call). I just want to know your opinion.

And ditto for anyone who would care to answer the question in bold. I recognize there is not an exact right/wrong answer (at least I don't think so), but any thoughts are appreciated, as well as any rationale you'd care to share.

 

[PeterDonis]

What is the initial connection between 2 and 3? None, as I believe you have previously agreed.

Of course. Nobody has disputed that.

You cannot pick random pairs of entangled photons (1) and present them as a jointly entangled state (2)

And we are not doing any such thing. You are not arguing with us at all at this point. You are arguing with some straw man you have in your head. What you are saying here simply has nothing to do with the arguments any of us have been making.

 

[PeterDonis]

I said "|Ψ〉14⨂|Ψ〉23 * features entanglement between 2 and 3 subsequent to their interaction at a common BS."

I know you did. That claim of yours is wrong.

You say there is "zero correlation".

I didn't just say it. I computed it. I showed you the computation.

Two completely different points.

If two particles are entangled, the correlation between them will not be zero. Hence, since the correlation between photons 2 and 3 in the state Ma's (2) described is zero, they cannot be entangled in that state.

The post-swap state contains terms for 4 possible Bell states, all 4 of which are entangled.

Sure. So does the pre-swap state, if you write it in the Bell state basis, as Ma's (2) does.

The difference is that the pre-swap state, Ma's (2), is a superposition of 4 such terms, so they can interfere with each other. And that interference cancels out the entanglement between photons 2 and 3, so in the overall state, they are not entangled. They're separable. That has been stated more than once already. I even showed you the math that demonstrates it.

The post-swap state, the one @Morbert wrote down in #40, is a mixture of 4 such terms. It describes a state where the swap has taken place but we don't yet know which outcome we got. The fact that it is a mixture, not a superposition, means it does not describe a single state that is a superposition of terms involving the 4 Bell states; there is no interference. It describes a state which could be any one of the four Bell states, with equal probability, but we don't know which one.

 

[PeterDonis]

Ma explicitly says this

No, he doesn't. Again, this is the meaning you are putting on other words Ma says. None of Ma's actual words that you quoted, or anywhere else in the paper, say "(2) is the post-swap state". That's an interpretation you're putting on other words in the paper.

Ma does, however, say, his actual words: "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3: <equation (2)>". We are simply looking at those explicit words and assuming that he meant what he said. You are not.

Now let's go back to what Ma says just before what I just quoted:

"if Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled. Consequently photons 1 (Alice) and 4 (Bob) also become entangled and entanglement swapping is achieved."

This is what he says "can be seen" by rewriting (1) as (2). How can it be seen?

You gave the answer to that--almost:

<Ma's (2)> contains terms for 4 possible Bell states, all 4 of which are entangled.

The only mistake you made, as I said in #255 just now, is that you called Ma's (2) the post-swap state. It's not. Nor does it need to be for us to use it to see what Ma says "can be seen" from it. All we need for that, even in the pre-swap state, is what you said in the quote above, but which I'll augment with a crucial clarification: Ma's (2), because it is rewritten in the Bell state basis, allows us to see what possible results would happen if we took that pre-swap state and did a BSM on it. Just look at which Bell state is in each term, square the amplitude of that term, and you have the probability to get that Bell state if you do a BSM. That's why Ma takes (1) and rewrites it as (2).

 

[PeterDonis]

Photon 1 has been observed to be in the state |H>. This implies that an H/V measurement on photon 2 would certainly yield a |V> result, given the biphoton's initial |Ψ−〉12 state. That photon is instead later fed into the Beam Splitter (BS) for a swap. When photon 2 enters the BS, is it still in a singlet state? Or is it in a pure |V> polarization state?

I am interested in an answer in terms of your own preferred interpretation.

I don't have a single preferred interpretation, but I can give what I think are the answers a couple of interpretations would give:

(1) I've given what I think is the Bohmian answer earlier in this thread, but I'll give it again: since photon 1 has been measured $H$, the effective wave function for photon 2 is $\ket{V}$. That is what will "steer" photon 2 when it goes through the BSM. What it will get "steered" into, if everything else is as we've been assuming (i.e., photons 3 and 4 were prepared in the singlet state and nothing else has been done), I posted earlier.

(2) The statistical/ensemble interpretation a la Ballentine would also assign $\ket{V}$ to photon 2 for the purpose of computing probabilities, because that's the rule for computing the probabilities of future measurement results on photon 2, conditioned on the fact that photon 1 was measured $H$. But that intepretation would not say that that "is" the state of photon 2, because in that interpretation the state describes ensembles, not individual systems. Ballentine would say that we are assigning $\ket{V}$ to photon 2 in the subensemble where photon 1's measurement result is known to be $H$.

Off the top of my head I can't think of an interpretation which would say that photon 2 is in the singlet state after photon 1 has been measured $H$.

 

[DrChinese]

That claim of yours is wrong.

The 2 and 3 photons emerge from the Bell-state measurement entangled in one of 4 Bell states. Oh, am I making unsubstantiated claims again? Twisting the words of authors?

Ma et al: "If Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled."

Megidish et al: "When the two photons of time τ (photons 2 and 3) are projected onto any Bell state, the first and last photons (1 and 4) collapse also into the same state and entanglement is swapped."

Riedmatten et al: "When photons B [2] and C [3] are measured in the Bell basis (Eq. 3), i.e. projected onto one of the four Bell states via a so-called Bell state measurement, photons A [1] and D [4] are projected onto the corresponding entangled state."

Pre-swap (2 & 3 not entangled) -> Bell state measurement -> Post-swap (2 & 3 entangled)

I'm always interested to find out what specific semantical approach you will take to convince yourself you are right and I am wrong. Please just step back and look at what you are saying. If you ignore the specific Bell state signatures, of course there are net zero correlations. There would be no experiments to write about if there were zero correlations to measure. So I guess they look at those signatures, yes?

PS You might be interested to know that Riedmatten uses the same phrase you and several others are using: "rewritten".

 

[PeterDonis]

The 2 and 3 photons emerge from the Bell-state measurement entangled in one of 4 Bell states.

Nobody is disputing that. We all agree that this is true. So piling quote on quote that says this contributes nothing to the discussion; you are trying to bash in a door that is already open.

What we don't agree with you about is your claim that Ma's (2) represents the post-swap state. You seem to think that, because it is written in the Bell state basis, it must. That is simply faulty logic.

If you ignore the specific Bell state signatures, of course there are net zero correlations.

It's not a matter of "signatures". "Signatures" are things we look for in measurement results. But if Ma's (2) describes the pre-swap state, it's not describing measurement results.

You have posed a number of questions for us. Here is one for you:

Ma's (2) has four terms in it, one for each of the Bell states. But two of those terms have minus signs. You'll note that in the state @Morbert wrote in post #40, the state that we are saying is the post-swap state, there are no minus signs. @Morbert pointed this out quite a few posts ago, not in response to you, but it's a significant clue to the meaning of Ma's (2).

Our explanation of why the minus signs are there in Ma's (2) is simple: because Ma's (2) is the pre-swap state, in which the different possibilities for Bell states can interfere with each other, and minus signs mean destructive interference, so that, even though the state is written in a basis of entangled Bell states, photons 2 and 3 are actually not entangled in that state. Not just "net zero correlations". Not entangled. Period.

In @Morbert's state in post #40, all the terms have plus signs. That's because, unlike Ma's (2), it is a mixture of the four possible Bell states, not a superposition. It does not describe one state, pre-swap, that could go one of four ways post-swap depending on the results at the BSM. It describes the BSM having already taken place--one of the four possible Bell states has already been swapped into--but we don't yet know which one it was. There is no interference between the terms; that doesn't even make sense for a mixture.

But you claim that Ma's (2) describes the post-swap state, where we've done the BSM, we've projected photons 2 and 3 into one of the four entangled Bell states, but we don't yet know which one. So given that claim of yours, what is your explanation of the minus signs in Ma's (2)?

 

[DrChinese]

No, he doesn't.

Why won't you give the full quote? Now you are doing the cherry-picking thing too.

Ma: "...As schematically shown in Fig. 1, if [and this could be labeled iff] Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled and entanglement swapping is achieved. This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3:"

I deduce that if Victor does NOT subject 2 & 3 to a Bell-state measurement, they won't be entangled. Is that a bridge too far? I would therefore deduce the Bell-state measurement causes entanglement. Is that a bridge too far? Really, what does this have to do with the physics?

Initial preparation (2 & 3 have no relationship)
-> Victor's Bell state measurement on 2 & 3
-> Post-swap (2 & 3 entangled in one of 4 Bell states)

 

[PeterDonis]

Why won't you give the full quote?

Because it doesn't say what you claim.

I deduce

Exactly. Your claim that Ma's (2) is the post-swap state is based on your deductions, not Ma's words. And your deductions are based on false premises, so they are leading you astray.

I deduce that if Victor does NOT subject 2 & 3 to a Bell-state measurement, they won't be entangled.

Of course. Nobody is disputing that.

I would therefore deduce the Bell-state measurement causes entanglement.

Of photons 2 and 3, certainly. They're the photons being operated on. I don't think any interpretation would disagree (though the ensemble interpretation would give the usual caution about the state not describing individual systems but ensembles, so the word "entanglement" has to be viewed with that in mind).

Whether the BSM causes entanglement of photons 1 and 4 is not quite so straightforward, some interpretations would not make that claim, at least not in the delayed choice case (because such a claim would imply backwards in time causation), but some (at least the one used in the Ma and Megidish papers) would, yes.

None of this has anything to do with the issue we are actually disagreeing about.

what does this have to do with the physics?

Your claim that Ma's (2) describes the post-swap state is a (wrong) claim about the physics.

If you discard that claim and just describe things in words, like these..

Initial preparation (2 & 3 have no relationship)
-> Victor's Bell state measurement on 2 & 3
-> Post-swap (2 & 3 entangled in one of 4 Bell states)

...then you would find us agreeing with these words as a description of the physics if Victor chooses to do a BSM.

 

[DrChinese]

... even though the state is written in a basis of entangled Bell states, photons 2 and 3 are actually not entangled in that state. Not just "net zero correlations". Not entangled. Period.

If you want to say that (2) is pre-swap algebra, and "photons 2 and 3 are actually not entangled in that state": that's what I have been saying all along. There is no physical significance to (2) unless and until and BSM is performed. The BSM causes entanglement of 2 & 3, and of 1 & 4.

 

[PeterDonis]

If you want to say that (2) is pre-swap algebra, and "photons 2 and 3 are actually not entangled in that state": that's what I have been saying all along.

Hm. So you actually have been agreeing with us all this time? Then these...

State (1) -> Swap -> State (2)

I said "|Ψ〉14⨂|Ψ〉23 * features entanglement between 2 and 3 subsequent to their interaction at a common BS."

State (1 - no 2/3 entanglement) -> Swap -> State (2 - 2/3 Bell state entanglement)

...were saying what, exactly?

If we were all misinterpreting what you've been saying, and if you agree that Ma's (2) describes the pre-swap state, that's great, we've made progress.

 

[DrChinese]

1. Because it [the full quote from a Nobel level paper] doesn't say what you claim.

2. If you discard that claim and just describe things in words, like these..

Initial preparation (2 & 3 have no relationship)
-> Victor's Bell state measurement on 2 & 3
-> Post-swap (2 & 3 entangled in one of 4 Bell states)

...then you would find us agreeing with these words as a description of the physics if Victor chooses to do a BSM.

1. You have no idea how funny that sounds - can't use full quote. I'll have to try using partial quotes in the future.

2. That's good enough for me!

Initial preparation (2 & 3 have no relationship)
-> Victor's Bell state measurement on 2 & 3
-> Post-swap (2 & 3 entangled in one of 4 Bell states)

 

[PeterDonis]

There is no physical significance to (2) unless and until and BSM is performed.

And of course when I said in #263 that we've made progress if you agree that Ma's (2) describes the pre-swap state, I was probably being way too optimistic.

The physical significance of Ma's 2 (2) is that it describes the pre-swap state. It does not describe the post-swap state. When the BSM is performed, it changes the state.

I'll stop here to see whether you agree or disagree.

 

[DrChinese]

if you agree that Ma's (2) describes the pre-swap state, that's great, we've made progress.

I agree that as a mathematical exercise, you can go from (1) to (2). There is no physical meaning to it until the swap is executed. Once the swap executes, I can perform an experiment on the remaining photons and report on it.

By way of analogy: If I have a single |V> 401 nm photon headed into a suitable PDC crystal apparatus, I would not say it is in the state |VV>+|HH> (i.e. 2 x 802 nm entangled photons) - even if I could prove it algebraically. And yet it could emerge in exactly that state.

I have said all along: (2) is the proper representation post-swap. In your view (I guess), it is also the representation pre-swap. I don't think the point you are making is really significant, but apparently it is more significant than I realize.

 

[DrChinese]

The physical significance of Ma's 2 (2) is that it describes the pre-swap state. It does not describe the post-swap state. When the BSM is performed, it changes the state.

Yes, the swap changes the state.

By way of ongoing analogy: If I have a single |V> 401 nm photon headed into a suitable PDC crystal apparatus, I would not say it is in the state |VV>+|HH> (i.e. 2 x 802 nm entangled photons) - even if I could prove it algebraically. And yet it could emerge in exactly that state.

And... if it so emerges, the 2 photons are in the entangled state |VV>+|HH>. That is the physical meaning of the down conversion. What's odd here?

 

[PeterDonis]

By way of analogy: If I have a single |V> 401 nm photon headed into a suitable PDC crystal apparatus, I would not say it is in the state |VV>+|HH> (i.e. 2 x 802 nm entangled photons) - even if I could prove it algebraically. And yet it could emerge in exactly that state.

In other words, the PDC changes the state. Yes.

Similarly, the BSM changes the state of photons 2 and 3. Yes?

 

[PeterDonis]

If I have a single |V> 401 nm photon headed into a suitable PDC crystal apparatus, I would not say it is in the state |VV>+|HH> (i.e. 2 x 802 nm entangled photons) - even if I could prove it algebraically.

Of course you could not prove it algebraically, since it's obvious just by inspection (I always hated it when my math teachers would throw that phrase at me in school, but it suits here) that there's no way to algebraically rewrite $\ket{V}$ as $\ket{VV} + \ket{HH}$. The second doesn't even have the same number of photons as the first. So of course those two expressions can't describe the same physical state.

But in the case of Ma's (1) and (2), they are both expressions with the same number of photons (four), and we've already proven that you can rewrite (1) as (2) algebraically. Which, according to how we understand how QM uses vector space math to represent physical states, means that Ma's (1) and (2) describe the same physical state. So since Ma's (1) describes the pre-swap state, Ma's (2) must also describe the pre-swap state. It can't be some kind of mathematical jiggery-pokery. It must describe the actual physical pre-swap state, just as Ma's (1) does. Which means that, since the BSM changes the state of photons 2 and 3, Ma's (2) can't also describe the post-swap state.

 

[PeterDonis]

(2) is the proper representation post-swap.

We disagree.

In your view (I guess), it is also the representation pre-swap.

Not "also", see above. But yes, it is the representation pre-swap.

I don't think the point you are making is really significant, but apparently it is more significant than I realize.

The point is that we disagree that Ma's (2) describes the post-swap state.

Whether that is "really significant" might depend on what other points you want to discuss. We agree on the measurement results, and that's normally what we would want to compute from the post-swap state.