Observation Independence - what and why?

1. Nov 1, 2005

Hurkyl

Staff Emeritus
Observation Independence -- what and why?

So, heuristically, observation independence is the condition that separated measurements are statistically independent.

I have lots of questions about this! (Actually, I have lots of questions about the theoretical foundation of using statistics at all in physics, but anyways...)

The first question is for a precise statement of this condition.

Presumably, you could say something like, for separated measurements A and B, we have P(A = a) = P(A = a | B = b), but that leads into my next question:

Classically, what is the theoretical justification for observation independence? Is there a proof that classical physics is observation independent?

The naive statement I gave above would seem to fail in certain simple cases. For example, if we packaged a red ball and a green ball, randomly mixed them up, and shipped them off to different cities, we would have that P(A = green) = 0.5, but P(A = green | B = red) = 1.

The usual answer is that the given scenario is not the complete description of the physical situation, in which the universe "knows" which ball went which way... but why would we ever be speaking about probabilities in the first place?

Deterministically, I suppose we have a trivial proof of observation independence. Does it make sense to ask about nondeterministic classical physics?

The impression I get is that, classically, we use statistics to describe our "lack of knowledge", but I cannot see any reason to think that lack of knowledge should obey the observation independence postulate.

Moving onto the topic I really wanted to discuss! (Though I would still like to have good answers to the previous questions)

When discussing hidden variable theories, the observation independence postulate is often invoked (indirectly, through the Bell-locality criterion). Isn't the point of a HVT that the probabilities describe lack of knowledge? If so, then why would we ever think that the observation independence postulate should be applicable?

Consider this thought experiment:

Suppose we have any system governed by some probability distribution.
We make a measurement of the system.
We write down two copies of the outcome of the measurement, and ship them off to different cities.
Now, consider the two observers in the different cities who open the envelopes and read the results...

If we apply the observation independence postulate, it is easy to see that the observations must have a definite outcome. (because they must be the same, and this is the only way they can be statistically independent)

So, it seems that observation independence forces the notion of a statistical collapse somewhere along the way, I suppose either when the original measurement occured, or when the duplication of information occurred.

My gut tells me that if we carry this thought experiment out to its logical conclusion, we must eventually conclude that all interactions are deterministic, and even if the original system somehow happened to be governed by an actual probability law, that it is quickly forced to choose one way or the other by the environment.

So, I shall be so bold as to claim that observation independence, as a fundamental postulate, is effectively equivalent to determinism!

Furthermore, it seems fruitless to attempt to develop any sort of statistical theory in which the observation independence postulate is satisfied!

(Ack, I didn't think I was headed in this direction! I wonder what others think of it)

2. Nov 1, 2005

-Job-

If i understand correctly you're saying that there's an initial observation which is performed by whomever will mail out the letters which collapses the first observable. Two letters are then sent out. Each of the two people then receiving the letter then make an observation, which will collapse into a definite state whatever was in the envelope. When the observable in one envelope collapses so does the other one, so even before it is open already has a preset value (determinism?).
Isn't it the case that your scenario is deterministic to begin with though? You are assuming nothing interacts with the envelopes such that it may alter the contents of the envelope (someone intercepts it and changes its contents). isn't that the same as saying that, given that anyone one of these given sequences of events which end up in the same outcome actually happens then that outcome will happen with 100% certainty.
The two observations may be independent without having the same value. Suppose that the envelope interacts with the environment changing its contents. Then when each person opens his/her respective envelope, the independence of observations is still retained, even though the observables are now different. In statistics two events are independeny if their intersection is empty set, the event that changes the contents of one of the envelopes is independent (under most scenarios) from the event where the other person opens the unchanged envelope.
If envelopes A and B are sent out and B is changed en route, then:
P(B = b) = P(B = b | A undergoes a change)
So ultimately:
P(B = b) = P(B = b | A not b) (non-deterministic)

Sorry if i misunderstood, this stuff is confusing.

3. Nov 2, 2005

Hurkyl

Staff Emeritus
No -- I did assumed the contents of the letters didn't change in transit, but that's about it. The particle could've been in a nondeterministic state, the measurement could have been probabilistic, and even resulting in the measuring device being in a nondeterminstic state!

I think it's a fair assumption that the contents didn't change, based on experimental evidence. But, we can consider what if the contents could change.

Given that the measurement had a binary outcome, the observers are symmetric, the probability of an envelope's contents changing to the opposite value, and observer independence, we have:

P(A = B) = P(A = 0 and B = 0) + P(A = 1 and B = 1) = P(A = 0)² + (1 - P(A=0))²
P(A = B) = P(neither changed) + P(both changed) = (1-p)² + p²

If we suppose that A = 1 is the more likely of the two outcomes, then we have the following:

The odds that observer A got a "0" in the mail is exactly equal to the odds that the contents of an envelope can spontaneously change value.

The odds that the particle was actually in state zero can be no greater than twice p.

However, we can make p as small as we wish through error correction. (E.G. we could send three envelopes, and do a majority vote! The odds of more than one envelope spontaneously changing contents is much smaller) So, by the principle of exhaustion, we again have that the original outcome is deterministic.

Even without exhaustion, though, we have empirical evidence that it is extraordinarily rare for an envelope to spontaneously change contents, whereas seems to be a great deal of nondeterminism at the subatomic level.

4. Nov 3, 2005

Sherlock

Observations at A and B are independent if the sampling of one doesn't affect the sampling of the other. In Bell tests, the observations at A and B are not independent, because the results are matched up according to certain criteria. The criteria for pairing are that members of a pair must refer to physical disturbances that had a common source, or had interacted with each other, or had their motions altered by some other common influence. The pairings are done via coincidence circuitry that's gated when a detection occurs at A or B.

The statistics accumulated at A and B are independent if the observations that produced the statistics are independent. In Bell tests, the statistics accumulated at A and B are not independent.

I don't understand exactly what your Independence notation means. (I'm just learning this stuff so excuse my ignorance and any stupid statements I might make or questions I might ask.)

The justification would be that the sampling of A doesn't affect the sampling of B and vice versa, wouldn't it?
There's a definition in probability theory of observation independence which is something like P(AB) = P(A)P(B).
But it involves defining a probability space in terms of several components, and I haven't memorized all that stuff yet.
Because, at the outset, we don't know which ball went which way? In your example, the observations aren't independent.
I don't know what you mean by "proof of observation independence". The labeling of observations or data sets as independent or not just depends on whether they meet certain criteria, doesn't it?
I don't know, but I'll be interested to see what you and others come up with.
I would say that we use probabilistic formulations to provide an abstract accounting of what is known about something. The statistical results then confirm or not the probabilistic formulation.

The assumption is that if Nature is local, then the observations and statistics at A and B should be independent of each other. But this assumption is incorrect, because A and B are dependent on each other by other criteria which conform to the principle of locality. These criteria include the sampling techniques, which are based on the assumption of a common cause.
The results of deterministic interactions might not be precisely determinable experimentally. It's wrt experimental determinations, not the quantum-level interactions themselves, that probability formulations apply.
You lost me here. :-)
I'm not sure what you mean here either. Physics is interested in how things are related to each other, isn't it? In the Bell tests, A and B are related to each other. This relationship can be generally understood, if not precisely described, in terms of local physical interactions.

I probably don't understand the point of your post. But it looks interesting. Maybe if you refine your argument, perhaps numbering the statements, I could understand it.

Last edited: Nov 3, 2005
5. Nov 3, 2005

DrChinese

Hurkyl, I don't know if this helps or not (especially since this was early in your post and I am trying to absorb it all...), but hopefully it won't hurt either...

From EPR and Bell Locality (2005)(a fave of ttn), equation (18):

(1)

$$P(A,B | a, b, \lambda) = P(A|B, a, b, \lambda) \times P(B | a, b, \lambda)$$

$$= P(A|B, a, \lambda) \times P(B | b, \lambda)$$

$$= P(A| a, \lambda) \times P(B | b, \lambda)$$

Where a and b are the "parameter" settings, and A and B are the "outcomes". So going from the 1st to 2nd line is parameter independence (PI) and going from the 2nd to 3rd line is outcome independence (OI). Both become your observer independence or "Bell Locality".

This definition is not exactly universal either. For example, Vaidman ([URL [Broken] on the Theme of the Greenberger-Horne-Zeilinger Proof (1998)
[/url]) mentions both of these:

(2)
"Action at a space-like separated region does not change the
outcome of a local measurement."

and

(3)
"...The ultimate goal of the project, as I understand it, is to show that quantum theory invariably leads to the failure of locality principle L. Two relevant results are well known.

( i) According to quantum theory, action at a space-like separated
region does not change the probability of an outcome of a local
measurement.

( ii) If the outcomes of quantum measurements are governed by hidden
variables, then there is an action at a space-like separated region
which does change the outcome of a local measurement.

I argue that there is no meaning for L beyond ( i) and ( ii)."

Last edited by a moderator: May 2, 2017
6. Nov 3, 2005

Hurkyl

Staff Emeritus
I'm still trying to absorb my post. I haven't articulated my questions before, and the conclusion that OI is equivalent to determinism really surprised me!

I was trying to remember what that formula was, thinks for bringing it back up.

Rephrasing my thought experiment... I'm moving away from the idea of a bell-type test, and considering only a single particle (or whatever).

But now, we have a device that makes one measurement that returns a binary result (0 or 1), and transmits the result to two spatially separated researchers.

Let A (B) denote the events where researcher A (B) received a '0' result.

My thought experiment has no parameters, so OI reduces to:

$$P(A, B | \lambda) = P(A | \lambda) P(B | \lambda)$$

(For Sherlock, P(X | Y) means the probability of X given Y, and is algebraically equal to P(X and Y) / P(Y))

If we assume that the machine flawlessly transmits the results to each of the researchers, we have that $P(A | B, \lambda) = 1$, which gives us $P(A, B | \lambda) = P(B | \lambda)$, and thus either $P(B | \lambda) = 0$ or $P(A | \lambda) = 1$.

In either case, it follows that the result of the experiment is deterministic.

In the case that the transmission of results may be faulty, the above equalities become approximate, but (in principle), we can apply error correction to make the error arbitrarily close to zero, and again determinism follows. Even if we deny that we can arbitrary error correction in principle, it still follows that the measurement must be incredibly close to deterministic.

7. Nov 3, 2005

-Job-

But arbitrarily close to 0 isn't 0, wouldn't you require 0 to have actual determinism? Also the error corrections have some requirements of their own to ensure that they are carried out properly, so the probability that all of the error corrections are successful is not 100%, so there's residual probability hanging around

8. Nov 3, 2005

Hurkyl

Staff Emeritus
Theorem: Suppose for all e > 0, |x - L| < e. Then x = L.

If the actual probability (call it p) was close to zero, but nonzero, then with a sufficiently good error-correction scheme, we can show that OI would require the probability to be less than p, and thus OI would be false.

9. Nov 4, 2005

Sherlock

Determinism means that any and all events are caused by an unbroken chain of prior events. The assumption that Nature is deterministic isn't experimentally falsifiable, is it? There are reasons to believe it, but it's a metaphysical preference, isn't it? Nevertheless, it underlies all of science, doesn't it?

Anyway, you're saying that if two sets of statistics, A and B, are independent of each other, then ... what?

In Bell tests, A and B aren't independent of each other. But they could be made independent by simply removing the coincidence circuitry and randomizing the pairings. So what would that tell you?