# B Entanglement measurements timing

1. Feb 27, 2017

I have a trivial question: when measuring 2 entangled particles A and B, what should be the timing between the measurments to get correct results?

In other words after measuring A how long can we wait before measuring B before its state changes?

2. Feb 27, 2017

### entropy1

The experimental setup determines what is measured at which instant; it can be A before B or vice versa.

3. Feb 27, 2017

### Karolus

the B of particle status changes instantaneously to the change of the particle A. From a classical point of view this is incomprehensible, since this would involve action at a distance, or a message that is propagated with infinite speed. But MQ must abandon the classical view. In MQ, apply the principles of "non-locality", and above all, falls on the principle of "causality" according to which everything happens as an effect of a cause which precedes it.

4. Feb 27, 2017

### DrChinese

It doesn't make any observable different which is measured first, or generally the timing.

5. Feb 27, 2017

### Staff: Mentor

These experiments are aspecially interesting when the two measurements are very close together in time, so close that there is no way a signal moving at or below the speed of light and emitted when one measurement is made can get to the other station before it has made its measurement - what is properly called "space-like separated". In this case, relativity of simultaneity (if you're not familiar with that term, google for it and check out the many threads over in the relativity subforum) means that there is no "first" measurement, and there are observers who will correctly say that A was measured before B and others who will just as correctly say that B was measured before A.
However....
Pretty much forever, as long as we keep B isolated so that it doesn't interact with something else before we get around to measuring it.

6. Feb 27, 2017

### bahamagreen

Just wondering... Doesn't confirmation of both A and B states always occur "inside a light cone"?
If the status of B at A is not known until a measurement value of B is received by A via <c delivery,
how is "instantaneously" not counterfactual definite?

7. Feb 28, 2017

I think my question was misunderstood maybe the word "timing" was misleading, what I mean is time period between measurements, ex if I measure A, then should I measure B within 1 second to get an accurate result?

8. Feb 28, 2017

That is in theory, but in practice it is different story, do you know how they do it in practical experiments? do they usually measure B within a specific time period after measuring A?

9. Feb 28, 2017

### StevieTNZ

Makes no difference, although I am a bit perplexed by your usage of 'accurate result'. What do you mean by writing that?

10. Feb 28, 2017

I simply mean before the value at B get changed again, or will it stay in same state forever?

11. Feb 28, 2017

### entropy1

It is not required that A exists while measuring B. It is the measured value of A that correlates with (the measured value of) B. Besides that, if the measurements are spacelike-separated, it makes no sense to speak about one measurement "before"/"after" the other. Or maybe I still don't understand your question?

12. Feb 28, 2017

Lol no you still dont get my question (my bad actually I phrased it wrongly).

I measured A then I sent B to the moon and reflected it back then measured it.. will I still get the expected result?

In other words If I am making an experiment setup, how long can I wait between the 2 measurments?

13. Feb 28, 2017

### entropy1

Well, if A and B remain undisturbed, and the order doesn't care, then indefinitely!

Consider that you measure B that soon, that you measure it before A, then that would be soon enough. If in this case we observe that A is measured relatively longer after B, then this would pose no problem whatsoever in the opposite way.

So, as far as I know, it doesn't matter.

14. Feb 28, 2017

### Staff: Mentor

That depends on the nature of the interaction with the mirror that's doing the reflecting. But if you put a detector on the moon (or alpha centauri, or the Andromeda galaxy, or ....) and that detector is making the first interaction the entangled particle undergoes, then yes, you will find the correlations predicted by quantum mechanics.

15. Feb 28, 2017

### DrChinese

As I said before, generally the timing is not a factor. Experiments have been done with a 1 hour delay. That done using a special (obviously) holding device.

http://lanl.arxiv.org/abs/1006.4344
Entanglement is a striking feature of quantum mechanics and an essential ingredient in most applications in quantum information. Typically, coupling of a system to an environment inhibits entanglement, particularly in macroscopic systems. Here we report on an experiment, where dissipation continuously generates entanglement between two macroscopic objects. This is achieved by engineering the dissipation using laser- and magnetic fields, and leads to robust event-ready entanglement maintained for 0.04s at room temperature. Our system consists of two ensembles containing about 10^{12} atoms and separated by 0.5m coupled to the environment composed of the vacuum modes of the electromagnetic field. By combining the dissipative mechanism with a continuous measurement, steady state entanglement is continuously generated and observed for up to an hour.

16. Feb 28, 2017

### Karolus

I do not know if I understand the question. In any case, the measurement of A and B takes place simultaneously. In other words, reduced in an extremely simple terms there is a source that emits photons, and two polarizers / detectors, placed at a certain distance from the source, but aligned in opposite directions. Suppose that the states of polarization can be + or -. So each detector can measure "+" or "-". Suppose that the detector "A" measure a polarized photon "+". Then A is automatically sure, without checking, that B has measured "-". In other words, when A receives +, B receives -. Things are more complicated course of this scheme, because the correlation must in fact be verified on a statistical basis. In addition there is another complication due to the fact that the experiment, rather than directly measure a non-local correlation, refutes the Bell's Theorem, built as a description to "local variables." Let's say that if it were true Theorem would expect a certain statistical value of the correlations, we assume:> TOT. However, the experiments show that the correlation statistic is <= TOT. Then the theorem is refuted, then it is wrong the assumption that there are local variables, then there is a non-local correlation. really reduced to minimum terms ...

well accepted corrections or refinements of this extremely exemplifying exposure...

17. Feb 28, 2017

### DrChinese

Sidebar on terminology: Bell's Theorem usually leads to an inequality when one assumes Local Realism. The inequality is violated, which refutes Local Realism. Bell's Theorem is not refuted in that process.

18. Feb 28, 2017

### Karolus

Sidebar on terminology:what does that mean? (I was looking for a physicist named Sidebar ... )
Local Realism = local variables?
Can you explicate better the concept, what process?

19. Feb 28, 2017

### bahamagreen

"Sure, without checking" looks like a pretty good definition of counterfactual definite.
When the verification of the measurements' correlation is performed, the verifier and the measurement information about A and B all co-reside locally.

If entanglement may not be employed for faster than c communications, this would suggest that if distant B happened to become destroyed that A could not immediately know (A's measurement would not be "peculiar" in any way). If A proceeds with the experiment at his end and measures "+", then A is automatically sure, without checking, that B has measured "-", yet B was destroyed before its scheduled measuring time.
If one suggests that B was really "-" unmeasured before it was destroyed, or would have presented that measure if not destroyed, that looks counterfactual definite.
If one suggests that the destruction of B prior to its measurement breaks the entanglement, that cannot make any difference to A, in order to prevent FTL signaling, so A proceeds with his measurement and finds no evidence of broken entanglement, so assumes (is automatically sure) B's measure is "-"...?

20. Feb 28, 2017

### Staff: Mentor

A's measurement is not "peculiar" in any way no matter what happens with B; it is impossible for A to even know that he was a measuring particles from a bunch of spin-entangled pairs until after he compares his results with B's after the fact. (If there's not a bunch of pairs, just one, then there is no way for either A or B to ever know the pair was entangled).
And from the previous post...
That's not quite right. A is not automatically sure, without checking, that B has measured "-". A is sure that if B does decide to make an measurement on the same axis then B will measure "-" - but there's no reason to think that B will ever make that measurement or any other, and the only way to find out is to wait around for a while and see if we can get together with B and compare notes.

Last edited: Mar 1, 2017