Information transfer using entanglement?

  • #1
entropy1
Gold Member
889
53

Main Question or Discussion Point

I split this off a separate thread in response to a post of @Nugatory .
The matter was about the (im)possibility of transfering information using entanglement.
This is a basic thread, so I keep it simple: There are two particles/detectors, A and B. The particles are in the singlet state.
No. It does prove that measurements on an entangled system do not transmit information at any speed, whether faster than light or not.
My question:
Could we claim that, should we have chosen a different detector angle for A, measurement B would have had the same outcome?
If we can't claim that, can we even rule out causation, and in that context, information transfer?
Of course any change in A doesn't affect the randomness of the outcomes of B. So, if the random pattern in B is changed in a different random pattern by changing something in A, there is no information transfer. There is an effect however (in this supposition).

My train of thought went like this: suppose a single pair in singlet state does not facilitate information transfer. However, a change in angle of A could lead to a different result in B. Then would it be conceivable that an ensemble of pairs (in time of space) would lead to a whole pattern of deviations in B as an effect of changes in the angle of detector A? I'm thinking for instance that the pattern A shows correlates with the pattern B shows (which is manifest in singlet ensembles). There is, as it were, information about a relative relation.

It may be a long shot and I'm won't be too surprised if I overlook something. :biggrin:
 
Last edited:

Answers and Replies

  • #2
jfizzix
Science Advisor
Insights Author
Gold Member
753
346
Without receiving information from A, there is no way of telling from the ensemble of B, whether B is entangled with A or not. Information from both parties is required, which is why entanglement cannot be used to send FTL information.

Entanglement can still be used to send information, such as the case with quantum super-dense coding.
Using entanglement, one can encode twice as much information on a particle than is physically possible without entanglement.
This is because manipulating A alone can change the entangled state A and B share to be any possible Bell state.
Since there are more distinguishable Bell states of a pair of particles than there are eigenstates of a single particle, one can send more information using entanglement than one can without it.
 
  • #3
entropy1
Gold Member
889
53
Without receiving information from A, there is no way of telling from the ensemble of B, whether B is entangled with A or not. Information from both parties is required, which is why entanglement cannot be used to send FTL information.
I follow that. On one side (A or B) the measurements seem random. What I am suggesting is that the actual manifestation of such a random pattern may be dependent of the angle of the detector, for if that wouldn't be the case, correlations wouldn't be able to vary between A and B dependent of the angle. A and B don't know (can't verify) that they correlate, but they do! When A and B meet each other they could say: "Hey, we correlate!" as they were doing all along. This information is relative, like it would for instance be relative to find you both have a paper with letters that combined make a message (i.e. the correlation). The information in question is as it were distributed over A and B. What I mean is that changing one of the detector angles changes the correlation, and so something has to have changed in one or both random patterns.
 
Last edited:
  • #4
m k
Gold Member
11
7
How much we can delay light?

Let polarization be entangled(E) and be one part thru angled slit(S) and other go undisturbed(U).
If we increase time E-S can we "decrease" distance S-U?
 
  • #5
morrobay
Gold Member
711
113
The following is a case that outcomes at B are not affected as setting at A varies:
Trial 1.
A, α = 30o
B, β = 1700
In this trial θ = 1400 With the sin2 formula calculate P++ and P-- = .883
Trial 2.
A,α = 500
B,β = 1700
With data outcomes from trial 1 detector A (30o) and data outcomes at trial 2 detector B 1700 )
determine by inspection if P++ P-- also = .883
and therefore B outcomes at 1700 were not affected as A setting varied from 300 to 500

And furthermore if the above is valid and two more trials show that A outcomes were not affected as B
setting varied. Then this could be a case for E (a,b) = ∫ dλ C (a.a' b,b' λ)
 
Last edited:
  • #6
entropy1
Gold Member
889
53
With data outcomes from trial 1 detector A (30o) and data outcomes at trial 2 detector B 1700 )
determine by inspection if P++ P-- also = .883
and therefore B outcomes at 1700 were not affected as A setting varied from 300 to 500
I follow that, but I am not sure I agree. If you would substitute Trial 2 (α=50°) with Trial 1 (α=30°) then Trial 2 would get a different correlation. A specific correlation occurs between two quasi random binary patterns only at certain values at a certain alignment (and I have code to back this up). If you have a pair of (quasi random) strings of values and you replace one of the pair by a different string, you most probably get a non-correlating pair (ie random p=.5).
 
Last edited:
  • #7
morrobay
Gold Member
711
113
Ok, see post #21 by @Nugatory in my thread: Local hidden variable model that equals qm predictions. The second part of reply for my sidetrack question. Unless there was a misunderstanding @Nugatory agreed with these types of substitutions being valid.
 
  • #8
entropy1
Gold Member
889
53
In other words is it valid for this particular model to combine an outcome at setting at A , 800 from one pair. And then from another pair an outcome at B , at 3330 . To say this way: in this model could the results from two different pairs, A at 800 and B at 3330 be equal to outcomes for one pair measured at A, 800 and B 3330 ? If this is invalid then all 3602 measurements could be made.
I am not sure what you are saying here...

If you are suggesting taking measurements at one side of an entangled pair (say, A), and later on combine that with an independent measurement on the other side (B), I think you get two uncorrelated measurements.
 
Last edited:
  • #9
morrobay
Gold Member
711
113
I am not sure what you are saying here...
Just the same scenario as in post 5 above : trials 1 , 2
 
  • #10
entropy1
Gold Member
889
53
Since B1 and B2 have a different dataset, replacing A2 with A1 in set 2 will affect the correlation. The set 1 correlation is characteristic for A1 together with B1.

That's how I see it. :smile:
 
  • #11
DrChinese
Science Advisor
Gold Member
7,220
1,030
The following is a case that outcomes at B are not affected as setting at A varies:
Trial 1.
A, α = 30o
B, β = 1700
In this trial θ = 1400 With the sin2 formula calculate P++ and P-- = .883
Trial 2.
A,α = 500
B,β = 1700
With data outcomes from trial 1 detector A (30o) and data outcomes at trial 2 detector B 1700 )
determine by inspection if P++ P-- also = .883
and therefore B outcomes at 1700 were not affected as A setting varied from 300 to 500

And furthermore if the above is valid and two more trials show that A outcomes were not affected as B
setting varied. Then this could be a case for E (a,b) = ∫ dλ C (a.a' b,b' λ)
I am essentially in agreement with entropy1's comments on the above. Clearly the observed statistics change (a suitable set of trials) as theta changes for entangled pairs. This is not in question, is it? And if it's not, I don't follow the point either.

As to whether B outcomes are affected by varying the angle A is measured at: this is interpretation dependent. No one understands the precise mechanics.
 
  • #12
entropy1
Gold Member
889
53
As to whether B outcomes are affected by varying the angle A is measured at: this is interpretation dependent. No one understands the precise mechanics.
Ok. My question primarily targeted the possibility that for a single pair, if we we would have chosen a different measurement angle for A, the outcome could have been different for B. I think this is not exactly CFD because we do measure B. There is however no verifying both of A's angles in one trail. But, as you point out, the statistics (of the combination of A and B) are dependent of the theta angle, and I observe that the theta angle can only change if one or both angles of A and B change, so when we change angle alpha there should be a change in either measurement ensemble A and/or measurement ensemble B. You could argue that if you change angle alpha, that only measurements of A change. However, that change should be in accordance with the correlation, and thus with theta, and thus A's measurements should depend on B's measurements, so then we have non-locality. Since the situation is symmetric, you could also associate the change in theta to B's measurements. So indeed we can't pinpoint the underlying mechanics. But having said that, in my view it is possible that angle alpha has 'an' effect on B's measurement outcomes.
 
  • #13
DrChinese
Science Advisor
Gold Member
7,220
1,030
... Since the situation is symmetric, you could also associate the change in theta to B's measurements. So indeed we can't pinpoint the underlying mechanics. But having said that, in my view it is possible that angle alpha has 'an' effect on B's measurement outcomes.
Sure, probably so, in fact how could it not if the stats change? But again, there are interpretations in which there is no physical collapse. And in those, essentially, there may be no mutual (symmetric) collapse occurring.
 
  • #14
631
132
Ok. My question primarily targeted the possibility that for a single pair, if we we would have chosen a different measurement angle for A, the outcome could have been different for B. I think this is not exactly CFD because we do measure B. There is however no verifying both of A's angles in one trail. But, as you point out, the statistics (of the combination of A and B) are dependent of the theta angle, and I observe that the theta angle can only change if one or both angles of A and B change, so when we change angle alpha there should be a change in either measurement ensemble A and/or measurement ensemble B. You could argue that if you change angle alpha, that only measurements of A change. However, that change should be in accordance with the correlation, and thus with theta, and thus A's measurements should depend on B's measurements, so then we have non-locality. Since the situation is symmetric, you could also associate the change in theta to B's measurements. So indeed we can't pinpoint the underlying mechanics. But having said that, in my view it is possible that angle alpha has 'an' effect on B's measurement outcomes.
Hopefully making things more definite will add clarity.

1) From B's perspective he will see ±1 with probability ½ for each no matter what A is doing. Same for A. We know if they both measure at 0º they get the same value.

2) If A measures at 45º and gets -1, while B measures at 0º and gets 1 what would they get if A measured at 0º instead? Would they both get 1? -1? If you insist that A must get 1 you are assuming she already had a value at 0º - CFD. However that experiment was not made, so we don't know.

How do you suggest in this context information might be exchanged?
 
  • #15
morrobay
Gold Member
711
113
You could argue that if you change angle alpha, that only measurements of A change. However, that change should be in accordance with the correlation, and thus with theta,
Following is data set/model suggesting that for two spin 1/2 particles in superposition with spin state l Ψ > = 1/√2 [ l +,-> -l -, + > ]
Setting/outcome changes at A do not have effect on B outcomes.* While outcomes at A are independent of setting at B , outcomes at A are not independent of outcomes at B ( the correlations ) Measurement order can be A , B or B, A
A 00 ..................... B 1200_______________ * (sin θ/2)2 P--, P++ = .75
- ................................. -
+..................................-
+..................................+
-...................................-
-...................................+
+..................................+
+..................................+
+..................................+


A 600..................B 1200_______________* (sinθ/2)2 = P--. P++ =.25 * (1/2 omitted for simplicity)
+................................-
+................................-
-................................+
-................................-
-................................+
-................................+
+...............................+
-................................+

Voila !

*
the only way information could be transferred is if outcome at A is dependent on setting at B and visa versa
 
Last edited:
  • #16
entropy1
Gold Member
889
53
@morrobay Your example is a perfect illustration of your quote of my text. So indeed that is a possibility. However, as I pointed out, A's outcomes must conform to theta and to B's outcomes, and hence can't be local. And my point was that since they can't be local, changes in A might just as well bring about changes in B, that being non-local too.
 
  • #17
entropy1
Gold Member
889
53
How do you suggest in this context information might be exchanged?
To clear this up: The title of my topic may be misleading - I am not advocating information transfer via entanglement, to the contrary: I am only suggesting the information in play, being the correlation between A's and B's outcomes, is dispersed (distributed), and as such dependent of A's and B's outcomes. If theta has a value of θ, the A's an B's outcomes have to 'conspire' to produce the corresponding correlation for θ. So this 'conspiration' I see as a form of 'information exchange', or perhaps better: 'information self-consistency'.
 
  • #18
morrobay
Gold Member
711
113
That outcomes at A and B are dependent or conform only relates to the distant correlations. This is not the same as one electron being affected by the measurement of the other
 
  • #19
entropy1
Gold Member
889
53
That outcomes at A and B are dependent or conform only relates to the distant correlations. This is not the same as one electron being affected by the measurement of the other
What if I propose that in your table, given θ (p=.75), on one line we flip one sign on whichever side. To keep the correlation intact, we have to flip the sign on the other side too, right?
 
  • #20
morrobay
Gold Member
711
113
The tables are just abreviations//averages for long stream ensembles . Can QM even address single interactions in these non classical correlations? You have to ask some of the talent here at A level going any farther with this.
 
  • #21
entropy1
Gold Member
889
53
The tables are just abreviations//averages for long stream ensembles .
Right! But I can see that as a greater-ensemble variation of the same thing (what I have called: 'conspiracy'): the conspiracy being dispersing information about theta between A and B.
 
  • #22
entropy1
Gold Member
889
53
2) If A measures at 45º and gets -1, while B measures at 0º and gets 1 what would they get if A measured at 0º instead? Would they both get 1? -1? If you insist that A must get 1 you are assuming she already had a value at 0º - CFD. However that experiment was not made, so we don't know.
I see your point: in your case, if B would be -1, A cannot other than be -1 too, so the outcome of A would have been unchanged. And how can something that did not change have an effect, right? But I'm not speaking of outcomes of A, but of settings of A. Given a different setting at A, would that allow for a different outcome of B?
 
Last edited:
  • #23
1,116
72
I would suggest mathematical outcomes are reliant on symmetries. However, measurement breaks these symmetries.
 
  • #24
m k
Gold Member
11
7
If partial signal is measured can any of the rest remain undisturbed?

Is there a difference if part through a slit is later measured or not?
 
  • #25
entropy1
Gold Member
889
53
But again, there are interpretations in which there is no physical collapse. And in those, essentially, there may be no mutual (symmetric) collapse occurring.
I am not necessarily speaking of collapse - I am speaking of "measurements". So the issue may 'come down' to "the measurement problem", which essentially is the same thing in a different guise I think...
 

Related Threads for: Information transfer using entanglement?

Replies
10
Views
360
Replies
12
Views
8K
Replies
1
Views
2K
Replies
2
Views
4K
  • Last Post
2
Replies
43
Views
3K
  • Last Post
2
Replies
29
Views
4K
Replies
6
Views
2K
Replies
4
Views
1K
Top