Entanglement and teleportation

[vanesch]

I will now, based upon my first calculation, show how this is related to a Copenhagen view of things.

We had, at t1, the measurement of Bob ; this can partly be done again with the unitary evolution operator, but (if we apply the Heisenberg cut at the level of "Bob"):

This means that a time evolution operator U_b acts,
such that:

U_b |bob0> |thb+> -> |bob+> |sys0>
U_b |bob0> |thb-> -> |bob-> |sys0>

U_b acting only on H_bob x H_sys2.

Rewriting psi(t0):

|psi(t0)> = |alice0>|bob0>|cable0>(|z+>(-sin(thb) |thb+> + cos(thb) |thb->) -
|z->( cos(thb) |thb+> + sin(thb) |thb->) )/sqrt(2)

Applying U_b

|psi(t1)> = {- sin(thb)|alice0>|bob+>|cable0>|z+>|sys0>
+ cos(thb) |alice0>|bob->|cable0>|z+>|sys0>
- cos(thb) |alice0>|bob+>|cable0>|z->|sys0>
- sin(thb) |alice0>|bob->|cable0>|z->|sys0>}/sqrt(2)

we have to apply the projection postulate now.

With 50% chance, bob is bob+, and we have then (after renormalization):

|psi(t1+)> = - sin(thb)/sqrt(2) |alice0> |bob+> |cable0> |z+> |sys0>
- cos(thb)/sqrt(2) |alice0> |bob+> |cable0> |z-> |sys0>

(and with 50% chance, we had bob- and another state which I won't write out).

HERE YOU SEE THE NON-LOCALITY AT WORK.
Indeed, the decision to go to the bob+ state affected immediately the amount of |z+> and |z-> (the Alice particle) in the state !
This wasn't the case when we kept the entire state |psi(t1)>: you can verify that the total length of the vector containing |z+> was still 50% in that case.
So the mechanism of the projection introduces the non-locality, in that the length of tensor product components (hilbert spaces of remote components) has suddenly changed. The evolution with Alice will be similar, but the main EPR effect occurred right here, in the Copenhagen view.

Alice's measurement:
from t1 to t2, Alice measures system 1 along direction th_a, so we have
an evolution operator U_a which acts:

U_a |alice0> |tha+> -> |alice+>|sys0>
U_a |alice0> |tha-> -> |alice->|sys0>

U_a acts only on H_alice x H_sys1

Rewriting psi(t1+):

|psi(t1+)> = - sin(thb)|alice0>|bob+>|cable0>(cos(tha) |tha+> + sin(tha) |tha->)|sys0>
- cos(thb) |alice0>|bob+>|cable0>(-sin(tha) |tha+> + cos(tha) |tha->)|sys0>


and applying U_a:

|psi(t2)> = - sin(thb) cos(tha)|alice+>|bob+>|cable0> |sys0> |sys0>
- sin(thb) sin(tha)|alice->|bob+>|cable0> |sys0> |sys0>
+ cos(thb) sin(tha)|alice+>|bob+>|cable0> |sys0> |sys0>
- cos(thb) cos(tha)|alice->|bob+>|cable0> |sys0> |sys0>

or:

|psi(t2)> = {(-sin(thb) cos(tha) + cos(thb) sin(tha) ) |alice+>|bob+>
+(-sin(thb) sin(tha) - cos(thb) cos(tha) ) |alice->|bob+>}|cable0> |sys0>|sys0>

or:

|psi(t2)> = { sin(tha-thb) |alice+> |bob+>
-cos(tha-thb) |alice-> |bob+>} |cable0> |sys0>|sys0>

After this measurement, again we have to use the projection postulate: with sin^2(tha-thb) probability, alice will have measured a + state (we already know bob had a + state and this is taken into account: we have here a conditional probability for alice), and the state will be, after normalization:

|psi(t2+)> = |alice+>|bob+> |cable0> |sys0>|sys0>

The whole "mystery" resides then in 2 things:

1) what about this non-locality ? Clearly it is contained in the quantum formalism (a la Copenhagen) and clearly also it doesn't correspond to any specific dynamics. One cannot say that it is "due to a force yet to be discovered", because it *is* already present in the formalism, and it is NOT some dynamics of unknown sort.

2) How does it come that Bob changes the states in such a way at Alice's that a) this directly influences the probabilities of outcomes Alice will observe, but b) that the mixture of influences that Bob prepares for Alice (when repeating the experiment) is exactly such, that, when weighting with Bob's mixture of outcomes, Alice finds finally a 50/50 probability AS IF Bob didn't influence her stuff. That, to me, sounds like a serious conspiracy :-)

It is here that I see a certain superiority of the MWI view: we know why this has to remain 50/50 because after the unitary evolution at Bob, the length of the vectors at Alice weren't influenced.

cheers,
Patrick.

 

[Sherlock]

Well, because of some view that there should be an underlying unity to physics. You're not required to subscribe to that view, but I'd say that physics then looses a lot of interest - that's of course just my opinion.
The idea is that there ARE universal laws of nature. Maybe that's simply not true. Maybe nature follows totally different laws from case to case. But then physics reduces to a catalog of experiments, without any guidance. A bit like biology before the advent of its molecular understanding.
I think that the working hypothesis that there ARE universal laws has not yet been falsified. Within that frame, you'd think that ONE AND THE SAME theory must account for all experimental observations concerning optics. We have such a theory, and it is called QED. Of course we had older theories, like Maxwell's theory and even the corpuscular theory ; and QED shows us IN WHAT CIRCUMSTANCES these older theories are good approximations ; and in what circumstances we will get deviations from their predictions.
It just turns out that in EPR type experiments you are in fact NOT in a regime where you can use Maxwell's theory because it is exactly the same regime in which you have the anti-coincidence counts. In one case however, Maxwell gives you (I'd say, by accident) an answer which corresponds to the QED prediction, in the other case, it is completely off.

I don't think that adopting a more easily visualizable 'classical'
explanation when possible for some experiments destroys the idea
that there are universal organizing principles. I believe that
a coherent 'big picture' that is close to the 'deep' or 'true'
nature of the universe can eventually be developed. (I think that
it will be some sort of wave mechanics that will account
for both the orderly and the chaotic/turbulent aspects of
reality, and that it will provide a communicable 'picture'
in a way that current quantum theory doesn't.) But that
belief isn't why I study physics.

Yes, QED can account for the instrumentally produced
data. But that isn't a picture of the sub-microscopic,
sub-atomic reality. It's a picture of the experimental data.
There is no picture of what light actually is, just
sometimes paradoxical experimental results. Using the
same single-photon light source you can make light behave
as if it is composed of indivisible 'particles' or
divisible waves. (The same setups that produce
anti-coincidence counts can be modified to produce
interference effects.) This could be due to the
interference-producing setups analyzing indivisible units
in aggregate (via combined streams using interferometer
in the beamsplitter setups or long time exposure using
detection location data in the double-slit setups), or it
could be due to instrumental insensitivity to sub-threshold
(divisible) wave activity. The answer isn't clear yet, afaik.

In any case, I don't think the fact that the cos^2 theta
formula works in the standard two-detector optical EPR/Bell
setup, and the fact that it's a 200 year old optics formula is
just a coincidence. (Remember all that stuff about
an "underlying unity to physics" above? :) )

There do seem to be organizing principles that are peculiar
to certain scales and contexts. The phenomenology
of, say, human social interactions is certainly different
than the phenomenology of quantum interactions.

It seems unlikely to me that there will ever be anything
like a quantum gravity. Gravitational behavior (in accordance
with the equivalence principle by the way) can be thought of
as emerging via complex wave interactions many orders of
magnitude greater in complexity than the simpler interactions
that are characterized as quantum. This isn't to say that
there aren't quantum interactions happening in and between
gravitating bodies -- they just aren't important in that
context, they don't *determine* gravitational behavior.

String theory, on the other hand, by positing the existence
of an underlying universal particulate medium, seems very well
motivated, though obviously a contrivance. I think it's
sort of the wrong approach, and even if they get it to
work mathematically for everything that won't necessarily
mean that it's a 'true' description of reality.

... for me the essence of physics is the identification of an objective world with the Platonic world (the mathematical objects), in such a way that the subjectively observed world corresponds to what you can deduce from those mathematical objects. MWI, CI and Bohmian mechanics are different mappings between an objective world and the Platonic world ; only they lead to finally the same subjectively observed phenomena. Now if physics would be "finished" then it is a matter of taste which one you pick out. But somehow you have to choose I think.
However, physics is not finished yet. So this choice of mapping can be more or less inspiring for new ideas.

For me, the essence of physics is the recognition of associations
or connections wrt natural and experimentally observed phenomena
and the ability to quantify those (intuitive?) associations.
(For example, I'll bet you've wondered why there is any motion
at all. Most people just take it as a given. There's motion,
now proceed to Newton's Laws and so on. But, there are
observations that indicate that the universe is expanding
omnidirectionally. Could these observations be the basis
for a new fundamental, universal law?)

I agree that physics is not only not finished, it's pretty
much just getting started. I also think that MWI, CI and Bohmian
mechanics *are* a matter of taste, and not very inspiring. :)

I think that the perfect understanding is a fully coherent mapping between a postulated objective world and the platonic world of mathematical objects, in such a way that all of our subjective observations are in agreement with that mapping. There may be more than one way of doing this. I am still of the opinion that there exists at least one way.
Apart from basing the meaning of "explanation" on intuition (and we should know by now that that is not a reliable thing to do), I don't know what else can it mean, to "explain" something.

If there's more than one way of doing it (and using the
method that you advocate almost assures that there will
always be more than one way) then why would you consider
any one of those ways to be the 'perfect' understanding?

One's 'intuition' changes as one learns and observes.

My intuition tells me that, for example, MWI, CI and
Bohmian mechanics are *not* providing us with a true
picture of the real world -- regardless of how
'coherently' they 'map'. I think that most scientists'
intuitions would tell them this, and I think that
scientists intuitive judgements about things should
be taken seriously.

If you have a theory which makes unambiguous, correct predictions of experiments, then in what way is there still something not "understood" ? I can understand the opposite argument: discrepancies between a theory's prediction and an experimental result can point to a more complex underlying "reality". But if the theory makes the right predictions ? I would then be inclined to think that the theory already possesses ALL the ingredients describing the phenomenon under study, no?

Well, yes and no. :) For example, quantum theory makes
correct predictions. But, the *phenomena* under study are
experimental results, not an 'underlying reality' that
the results are, as presumed by some, about. So, you
sometimes get incomprehensible results. From this, the
CI view is that the 'quantum world' is simply
incomprehensible, and that analogies from the world of
our sensory experience are simply inapplicable. And, I
consider that to be a very wrongheaded view.

As for my statement regarding GR as simplistic:
if gravitational behavior is complex wave
interactions, then GR is an oversimplification.
Lots of people think that GR, and even the
Standard Model, won't be up to the task of
handling recent astronomical observations.

And, regarding MWI, I don't consider it to be a
physical theory -- even though it might be
a very clean mapping. :)

 

[Sherlock]

The whole "mystery" resides then in 2 things:

1) what about this non-locality ? Clearly it is contained in the quantum formalism (a la Copenhagen) and clearly also it doesn't correspond to any specific dynamics. One cannot say that it is "due to a force yet to be discovered", because it *is* already present in the formalism, and it is NOT some dynamics of unknown sort.

2) How does it come that Bob changes the states in such a way at Alice's that a) this directly influences the probabilities of outcomes Alice will observe, but b) that the mixture of influences that Bob prepares for Alice (when repeating the experiment) is exactly such, that, when weighting with Bob's mixture of outcomes, Alice finds finally a 50/50 probability AS IF Bob didn't influence her stuff. That, to me, sounds like a serious conspiracy :-)

It is here that I see a certain superiority of the MWI view: we know why this has to remain 50/50 because after the unitary evolution at Bob, the length of the vectors at Alice weren't influenced.

I don't think this clearly states the essence of the real physical
mystery, which I view as concerning whether all of the light
incident on a polarizer during a certain coincidence interval
associated with a photon detection is being transmitted by the
polarizer or not (there are similar considerations for the two-slit
and beamsplitter setups -- is the emitted light associated with a
photon detection going through both slits when they are both open,
and is the emitted light associated with a photon detection being
both reflected and transmitted after interacting with a beamsplitter?).
That is, it's known what photons *are* theoretically and to a certain
extent instrumentally, but the actual physical nature of photons
isn't known. Hence, there are some interpretational problems.

As for the projection, it's based on the idea that Alice and Bob
are analyzing in the joint context the same value of some physical
property during a certain interval associated with the production of
that value. The projected axis is taken as the axis of
maximum probability of detection because it produced a
detection. This in itself doesn't imply a nonlocal
physical connection between Alice and Bob. The nonlocal
stuff comes from people thinking that Bell proved
that the light incident on the polarizers couldn't
have a common motional property.

But, this is the essence of what Schroedinger called entanglement --
that two objects which have interacted, or have been produced
by the same process (like being emitted via one and the same
atomic transition), carry with them in their subsequent motion
information of the motion imparted via the interaction or
the process that created them. This shared property of
motion will stay with the objects no matter how far apart
they travel, as long as no external torques are introduced
which might modify the value of the shared property.

Probabilities are not explanations. They're descriptions of
behavior at the level of instrumental detection, which to
a certain extent can't be controlled.

 

[vanesch]

In any case, I don't think the fact that the cos^2 theta
formula works in the standard two-detector optical EPR/Bell
setup, and the fact that it's a 200 year old optics formula is
just a coincidence. (Remember all that stuff about
an "underlying unity to physics" above? :) )

Well, it doesn't really work. It works for ONE specific correlation, under the assumption (which is semi-classically correct) that you have a probability of clicking proportional to incident intensity, namely A+ and B-. Now if you use ABSORBING polarizers, that's all you get, so there it is ok. But if you use *polarizing beam splitters*, it DOESN'T work for some of the other correlations, as I tried to point out in post number 42 in this thread.

Now, if, within the same experiment, a certain way of reasoning explains SOME results, and is in contradiction with OTHERS, then that way of reasoning IS WRONG.

Like my old physics teacher used to say: we know that many solids have a dilatation as a function of temperature. Now, in summer, days are longer, and they are hotter too... (but it doesn't work for the summer nights...)

cheers,
Patrick.

 

[vanesch]

That is, it's known what photons *are* theoretically and to a certain
extent instrumentally, but the actual physical nature of photons
isn't known. Hence, there are some interpretational problems.

Que veut le peuple ?

If you know what they are "theoretically" and you know what they mean instrumentally, what else is there to know ?
A "mechanical" picture (like the discussions people had in the 19th century about *in what matter* the E and B fields had to propagate) ?

cheers,
Patrick.

 

[Sherlock]

Regarding cos^2 theta correlation curve in EPR/Bell experiments
you wrote:

Well, it doesn't really work. It works for ONE specific correlation ...

It describes the data curves for a class of setups. Which have
some things in common with the setup from which it was originally
gotten.

You disappoint me if you don't see at least the possibility
of some connection between the two.

 

[Sherlock]

Regarding photons, you wrote:

Que veut le peuple ?
If you know what they are "theoretically" and you know what they mean instrumentally, what else is there to know ?
A "mechanical" picture (like the discussions people had in the 19th century about *in what matter* the E and B fields had to propagate) ?

I know what 'gods' are 'theoretically'. And I know how people
react to the word. But I have no idea what gods *are*.
That is, I have no way of knowing how (in what form) or if
they exist outside those contexts.

It's sort of the same with photons, except that photons
are a much more interesting subject -- especially entangled
ones.

So, yes, I'd say that there's a lot more to be known
about photons, about light, than is currently known.
Some sort of mechanical picture of the deep reality
would be nice. Do you think that's impossible?

I think that not being curious in this way would
make physics a lot less interesting.

 

[vanesch]

Regarding cos^2 theta correlation curve in EPR/Bell experiments

It describes the data curves for a class of setups. Which have
some things in common with the setup from which it was originally
gotten.

You disappoint me if you don't see at least the possibility
of some connection between the two.

Sorry to disappoint you :-)

The link is however, rather clear. In the QED picture, the AVERAGE photon count rate is of course equal to the classical intensity, and we know that the classical intensities are related with a cos^2 theta curve.
So if you consider that the light beams are made up of classical *pulses* with random orientation, and you look at the intensities per pulse that get through the polarizers, then you get the cos^2 theta relationship. On average, then, the photon counting rates must also be related by a cos^2 theta relationship.
So *A* way to respect this constraint is just to have a correlation PER EVENT which is given by cos^2 theta. But that doesn't NEED to be so. For A+ and B-, it is so, agreed. But for A+ and A-, they have, in the same classical picture, intensities which vary from 50-50 to 0-100 (namely 50-50 when the incoming classical pulse is under 45 degrees with the polarizing BS orientation, and 0-100 when the classical pulse is parallel (or perpendicular) to the BS orientation). So you would expect a certain correlation rate (about 50%: you have EQUAL intensities in the 50-50 -> full correlation and you have anti-correlation in the 0-100 case).
Well, this IS NOT THE CASE. You find perfect anticorrelation. So this illustrates that the picture of a classical pulse with a random polarization, and a probability of triggering PER CLASSICAL PULSE of the photodetector, proportional to the classical intensity of the individual pulse, DOES NOT WORK IN THIS SETUP. If it doesn't work for certain aspects of the set-up, it doesn't work AT ALL.
The proportionality of detections and classical intensitis only works ON AVERAGE, not nessesarily PULSE PER PULSE.

The ONLY picture which gives you a consistent view on all the data is the photon picture, with a SINGLE DETECTABLE ENTITY PER "PULSE" in each arm. And if you accept THAT, you appreciate the EPR "riddle", and you do not explain it with the old cos^2 theta law, because that SAME cos^2 theta law would also give us SIMULTANEOUS HITS in A+ and A-, which we don't have. The EPR problem is only valid in the case where you do not have simultaneous
YES/NO answers, of course, otherwise you have, apart from a +z and a -z answer, also a (+z AND -z) answer, which changes Bell's ansatz.

But I repeat my question: people do experiments with light because of 2 reasons: it is feasable, and they *assume* already that we accept the photon picture. If you do not do so, then doing the EPR experiment with light is probably not very illuminating (-:.
However, (at least on paper), you can do the same thing WITH ELECTRONS. Now, I take it that you accept that a single electron going onto two detectors will only be detected ONCE, right ? Well, according to quantum theory, you get exactly the same situation (the cos^2 theta correlation) there. So how is this now explained "classically" ?

(ok, the angle is now defined differently because of the difference between spin-1 and spin-1/2 particles).

Do you:
a) think that QM just makes a wrong prediction there ?
b) do not accept that a single electron can only be detected in 1 detector ?
c) other ?

cheers,
Patrick.

 

[vanesch]

So, yes, I'd say that there's a lot more to be known
about photons, about light, than is currently known.
Some sort of mechanical picture of the deep reality
would be nice. Do you think that's impossible?

No, it is not impossible, Bohm's theory does exactly that.
The main objection I have against the view that we need a mechanical picture as an explanation, is: what MORE does a mechanical picture explain ? Isn't it simply because we grew up with Newton's mechanics, and the associated mathematics (calculus) and we develloped more "gut feeling" for it ? What is so special about some mechanical view of things ? I have nothing *against* a mechanical view, but I don't think a mechanical view is worth sacrifying OTHER ideas. And that's what, for instance, Bohm's theory does: it sacrifices locality (and so does the projection postulate).

I will agree with you that quantum theory, or general relativity, or whatever, doesn't give us a "final view" on how nature "really" works ; for the moment however, it is the best we have. 300 years from now, I'm pretty sure that our paradigms will have changed completely, and people will look back on our discussions with a smile in the same way we could look back on people develloping a "world view" based upon a Newtonian picture. And they are being naive, because 600 years from now, their descendants will again have changed their views :-)

So for short I think it is a meaningless exercise to try to say what nature "really" looks like. But what you can try to do is to build a mental picture that gives you the clearest possible view on how nature is seen using things that we KNOW right now. It is in that context that I see MWI. I do not know/think/hope that the MWI view is the "real" view on the world (which, I outlined, I don't think we'll ever have). I think that MWI is about the purest mental picture of quantum theory, because *it respects most of all its basic postulates*. That's all. If you do formally ugly things, such as the projection postulate, to get "closer to your gutfeeling about nature" I think you miss the essential content of quantum theory, and as such I think you're in a bad shape to see where it could be extended, because you already mutilated it !

cheers,
Patrick.

 

[DrChinese]

That's the point. There are no hidden variables, and everything is local. So what gives, in Bell ? What gives is that, from Alice's point of view, Bob simply didn't have a definite result, and so you cannot talk about a joint probability, until SHE "decided" which branch to take. But when she did, information was present from both sides, so the Bell factorisation hypothesis is not justified anymore.

...

As I said, it is much less spectacular this way, because you only have Alice having a "superposition" of states of her photon. It's more spectacular to have her have a superposition of states of Bob.

cheers,
Patrick.

Thanks, that helps me to understand this perspective better!

 

[DrChinese]

In any case, I don't think the fact that the cos^2 theta
formula works in the standard two-detector optical EPR/Bell
setup, and the fact that it's a 200 year old optics formula is
just a coincidence. (Remember all that stuff about
an "underlying unity to physics" above? :) )

There are definitely TWO ways to look at that statement. Some of the vocal local realists argue that the cos^2 law isn't correct! They do that so the Bell Inequality can be respected; and then explain that experimental loopholes account for the difference between observation and their theory.

Clearly, classical results sometimes match QM and sometimes don't; and when they don't, you really must side with the predictions of QM. Even Einstein saw that this was a steamroller he had to ride, and the best he could muster was that QM was incomplete.

 

[vanesch]

I said the following:

And if you accept THAT, you appreciate the EPR "riddle", and you do not explain it with the old cos^2 theta law, because that SAME cos^2 theta law would also give us SIMULTANEOUS HITS in A+ and A-, which we don't have. The EPR problem is only valid in the case where you do not have simultaneous
YES/NO answers, of course, otherwise you have, apart from a +z and a -z answer, also a (+z AND -z) answer, which changes Bell's ansatz.

and I would like to illustrate WHERE it changes Bell's ansatz.

Consider again 3 directions, a, b and c, for Alice and Bob.

Alice has an A+ and an A- detector, and Bob has a B+ and a B- detector.
Usually people talk only about the A+ hit or the "no-A+ hit" (where it is understood that the no-A+ hit is an A- hit).

We then take as hidden variable a bit for each a, b and c:

If we have a+ this means that Alice will have A+ and bob will have no B+ in the a direction, if we have a b+ that means that Alice will have an A+ and bob will have no B+ in the b direction, and ...

So we can have: a(+/-) b(+/-) c(+/-) as hidden state. But that description already includes the anti-correlation: if A+ triggers, then A- does NOT trigger, and if A- triggers, then A+ does not trigger. When A+ and A- do not trigger, that is then assumed to be due to the finite quantum efficiencies of the detector, which lead to the "fair sampling hypothesis".

But if we accept the possibility that A+ AND A- trigger together, then each direction has, besides the + and - possibility, a THIRD possibility namely X: double trigger. So from here on, we have 27 different possible states. This changes completely the "probability bookkeeping" and Bell's inequalities are bound to change. The local realist cloud even introduces a fourth possibility: A+ and A- do not trigger, and this is not due to some inefficiency, with symbol 0.

So we have a(+/-/X/0), b(+/-/X/0), c(+/-/X/0) which gives us 64 possibilities.
You can then easily show that Bell's inequalities are different and that experiments don't violate them.

The blow to this view is that whenever you make up a detector law as a function of intensity which allows you to consider the 0 case, you also have to consider the X case. The X case is never observed, so there are reasons to think that the 0 case doesn't exist either, especially because QED tells us so, and that you do get out the right results (including the observed number of 0 cases) when applying the quantum efficiency under the fair sampling hypothesis.

cheers,
Patrick.

 

[Sherlock]

That's sort of funny, you know. Application of classical optics' formula cos^2\theta is incompatible with hidden variables but consistent with experiment.

The cos^2 theta formula isn't incompatible with hidden
variables.

For the context of individual results you can write,

P = cos^2 |a - lambda|,

where P is the probability of detection, a is the
polarizer setting and lambda is the variable
angle of emission polarization.

This doesn't conflict with qm. If you knew
the value of lambda, or had any info about
how it was varying (other than just that
it's varying randomly), then you could more
accurately predict individual results (by
individual results I mean the data streams
at one end or the other).

How do we know that there *is* a hidden
variable operating in the individual measurement
context? Because, if you keep the polarizer
setting constant the data stream varies
randomly.

Now, this hidden variable doesn't just
stop existing because we decide to
combine the individual data streams wrt
joint polarizer settings.

However, the *variability* of lambda
isn't a factor wrt determining coincidental
detection.

a b and c are the hypothetical settings you could have IF local hidden variables existed. This is what Bell's Theorem is all about. The difference between any two is a theta. If there WERE a hidden variable function independent of the observations (called lambda collectively), then the third (unobserved) setting existed independently BY DEFINITION and has a non-negative probability.

Bell has nothing to do with explaining coincidences, timing intervals, etc. This is always a red herring with Bell. ALL theories predict coincidences, and most "contender" theories yield predictions quite close to Malus' Law anyway. The fact that there is perfect correlation at a particular theta is NOT evidence of non-local effects and never was. The fact that detections are triggered a certain way is likewise meaningless. It is the idea that Malus' Law leads to negative probabilities for certain cases is what Bell is about and that is where his selection of those cases and his inequality comes in.

Suppose we set polarizers at a=0 and b=67.5 degrees. For the a+b+ and a-b- cases, we call that correlation. The question is, was there a determinate value IF we could have measured at c=45 degrees? Because IF there was such a determinate value, THEN a+b+c- and a-b-c+ cases should have a non-negative likelihood (>=0). Instead, Malus' Law yields a prediction of about -10%. Therefore our assumption of the hypothetical c is wrong if Malus' Law (cos^2) is right.

Bell demonstrated that using the variability of lambda
to augment the qm formulation for coincidental
detection gives a result that is incompatible
with qm predictions for all values of theta
except 0, 45 and 90 degrees.

Now, there's at least two ways to interpret Bell's
analysis. Either (1) lambda suddenly stops existing when we
decide to combine individual results, or (2) the variability
of lambda isn't relevant wrt joint detection.

I think the latter makes more sense, and in fact
it's part of the basis for the qm account which
assumes that photons emitted by the same atom
are entangled in polarization via the emission
process. This is why you have an entangled
quantum state prior to detection. So, all you
need to know to accurately predict the
*coincidental* detection curve is the angular
difference between the polarizer settings. And,
as in all such situations where you're analyzing,
in effect, the same light with crossed linear
polarizers the cos^2 theta formula holds.

 

[vanesch]

The cos^2 theta formula isn't incompatible with hidden
variables.

For the context of individual results you can write,

P = cos^2 |a - lambda|,

where P is the probability of detection, a is the
polarizer setting and lambda is the variable
angle of emission polarization.

Ok, that's the probability for the A+ detector to trigger. And what is the probability for the A- detector to trigger, then ? P = sin^2 |a - lambda| I'd say...

cheers,
Patrick.

EDIT:

I played around a bit with this, and in fact, it is not so easy to arrive at a CORRELATION function which is cos^2(a-b). Indeed, let's take your probability which is p(a+) = cos^2(lambda-a).
Assuming independent probabilities, we have then that the correlation, which is given by p(a+) p(b+) = cos^2(lambda-a) sin^2(lambda-b) for an individual event. (the b+ on the other side is the b- on "this" side)

Now, by the rotation symmetry of the problem, lambda has to be uniformly distributed between 0 and 2 Pi, so we have to weight this p(a+) p(b+) with this uniform distribution in lambda:

P(a+)P(b-) = 1/ (2 Pi) Integral (lambda=0 -> 2 Pi) cos^2(lambda-a) sin^2(lambda-b) d lambda.

If you do that, you find:

1/8 (2 - Cos(2 (a-b)) ) = 1/8 (3-2 Cos^2[a-b])

And NOT 1/2 sin^2(a-b) !

I checked this with a small Monte Carlo simulation in Mathematica and this comes out the same. Ok, in the MC I compared a+ with b+ (not with b-), and then the result is 1/8 (2+cos(2(a-b)))

So this specific model doesn't give us the correct, measured correlations...

cheers,
Patrick.

I attach the small Mathematica notebook with calculation...

 

[DrChinese]

The cos^2 theta formula isn't incompatible with hidden
variables.

For the context of individual results you can write,

P = cos^2 |a - lambda|,

where P is the probability of detection, a is the
polarizer setting and lambda is the variable
angle of emission polarization.

This doesn't conflict with qm. If you knew
the value of lambda, or had any info about
how it was varying (other than just that
it's varying randomly), then you could more
accurately predict individual results (by
individual results I mean the data streams
at one end or the other).

How do we know that there *is* a hidden
variable operating in the individual measurement
context? Because, if you keep the polarizer
setting constant the data stream varies
randomly.

Now, this hidden variable doesn't just
stop existing because we decide to
combine the individual data streams wrt
joint polarizer settings.

Now, there's at least two ways to interpret Bell's
analysis. Either (1) lambda suddenly stops existing when we
decide to combine individual results, or (2) the variability
of lambda isn't relevant wrt joint detection.

I think the latter makes more sense, and in fact
it's part of the basis for the qm account which
assumes that photons emitted by the same atom
are entangled in polarization via the emission
process. This is why you have an entangled
quantum state prior to detection. So, all you
need to know to accurately predict the
*coincidental* detection curve is the angular
difference between the polarizer settings. And,
as in all such situations where you're analyzing,
in effect, the same light with crossed linear
polarizers the cos^2 theta formula holds.

Or Lambda=LHV does not exist, a possibility you consistently pass over. It is a simple matter to show that with a table of 8 permutations on A/B/C, there are no values that can be inserted that add to 100% without having negative values at certain angle settings.

A=___ (try 0 degrees)
B=___ (try 67.5 degrees)
C=___ (try 45 degrees)

Hypothetical hidden variable function: __________ (should be cos^2 or at least close)

1. A+ B+ C+: ___ %
2. A+ B+ C-: ___ %
3. A+ B- C+: ___ %
4. A+ B- C-: ___ %
5. A- B+ C+: ___ %
6. A- B+ C-: ___ %
7. A- B- C+: ___ %
8. A- B- C-: ___ %

It is the existence of C that relates to the hidden variable function. What you describe is just fine as long as we are talking about A and B only. (Well, there are still some problems but there is wiggle room for those determined to keep the hidden variables.) But with C added, everything falls apart as you can see.

You can talk all day long about joint probabilities and lambda, but that continues to ignore the fact that you cannot make the above table work out. If you are testing something else, you are ignoring Bell. After you account for the above table, then your explanation might make sense. Meanwhile, the Copenhagen Interpretation (and MWI) accounts for the facts that LHV cannot.

 

[kleinwolf]

I would like to point out, in a previous round against Vanesh about EPR and many worlds, the following point (1) :

Usual "orthodox Copenhagen QM" contains

1) a local hidden variable that corresponds to the specification of the PRECISE endstate when the latter is degenerate. The "standard" Copenhagen QM is a special configuration of the endstate that corresponds to it's maximum.

However, there is more :

2) a NON-LOCAL hidden variable.

Let see the latter : a non-local measurement is obtained by the operator : \sigma_z\otimes\(\sigma_z\cdot\vec{n}_b)...hence Both side are measured, and there is no 1 operator on the other (non disturbing operator).

Let consider \theta_b=0

Hence : both directions of measurement are the same. The clearly the only 2 possible endstates are :

|+-> or |-+>, with p(+-)=|<+-|\Psi>|^2=\frac{1}{2}=p(-+)

This sounds very like more than intuitive and easy to understand.

However, one can see the things in an other way, by looking that :

M=\sigma_z\otimes\sigma_z=\left(\begin{array}{cccc} 1 &&&\\&-1&&\\&&1&\\&&&-1\end{array}\right)

Hence, then eigenvalues of M are 1,-1 and are both degenerate. 1 corresponds to |A=B> and -1 to |A<>B> (same or different results in A and B).

Here again, the eigenSPACE can be parametrized :

|same&gt;=\left(\begin{array}{c}\cos(\chi)\\0\\\sin(\chi)\\0\end{array}\right)
|different&gt;=\left(\begin{array}{c}0\\cos(\delta)\\0\\\sin(\delta)\end{array}\right)
|\Psi&gt;=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\1\\-1\\0\end{array}\right)

So that : p(different)=|&lt;different|\Psi&gt;|^2=\frac{1}{2}\cos(\delta)^2

p(same)=|&lt;same|\Psi&gt;|^2=\frac{1}{2}\sin(\chi)^2

Where \chi,\delta are GLOBAL HIDDEN VARIABLES...

So that in fact 2p(same)=1 at MAX...what is the interpretation of this, if there is no mistake of course...??

 

[vanesch]

So that in fact 2p(same)=1 at MAX...what is the interpretation of this, if there is no mistake of course...??

To me the interpretation is that your chi and delta are just variables that parametrize the eigenspaces of the operator sigma_z x sigma_z.

However, I don't understand your calculation. When you write out sigma-z x sigma-z, I presume in the basis (++, -+,+-,--), then I'd arrive at a diagonal matrix which is (1,-1,-1,1)... You seem to have taken the DIRECT SUM, no ?


cheers,
Patrick.

 

[kleinwolf]

Yes, you're entirely right...my mistake is unforgivable, since this will change all the afterwards calculation and interpretation of \delta.

Then the result is p(same)=0\quad p(diff)=\frac{1}{2}(1-\sin(2\chi))

However, you admit there are 2 visions of computing the probabilities with your correct M :

locally : p(+-)=p(-+)=1/2

globally, the endstate |->_g=(0,cos(a),sin(a),0), gives the prob :

p(+-)=cos(a)^2, p(-+)=sin(a)^2...hence on average or special values of a, the same as locally...but a infinite of possibilities more are allowed.

Can this be measured on the statistical results in an experiement, and how to find how to change the value of a experimentally ??

 

[vanesch]

Yes, you're entirely right...my mistake is unforgivable, since this will change all the afterwards calculation and interpretation of \delta.

Then the result is p(same)=0\quad p(diff)=\frac{1}{2}(1-\sin(2\chi))

Well, don't you find this funny that the sum of the probabilities for the two possible outcomes don't add up to 1 ? You could think that for each event, you have two possible results: they are the same, or they are different. And if you add up their probabilities, you don't find 1.
It's like: throw up a coin: 25% chance you have head, 30% you have tail :-)

cheers,
Patrick.

 

[kleinwolf]

It's just because we don't understand QM. But QM is omnipotent for everyone, just put : \chi=-\frac{\pi}{4}\Rightarrow p(diff)=1

In the other calculation, the sum add up to 1 in every case...

So what does it mean that the prob of the possible outcomes don't add up to 1 in everycase for the other calculation ?

Just because the correlation, even if measured along the same directions, of the singlet state, is not always perfect, remind : there is a non-local part and a local one...here it's just the non-local one.

Best regards.

 

[vanesch]

It's just because we don't understand QM.

I'd rather say: because the way you want do modify QM doesn't work :-)

But QM is omnipotent for everyone, just put : \chi=-\frac{\pi}{4}\Rightarrow p(diff)=1

Yeah, that's the projection as is proposed in standard QM :-) So then it works...

But you claim that one should have a kind of "equal distribution" or so of outcomes (which clearly is NOT standard QM). And then you get silly results such as that the sum of the probabilities of all possibilities is not equal to 1.

In the other calculation, the sum add up to 1 in every case...

So what does it mean that the prob of the possible outcomes don't add up to 1 in everycase for the other calculation ?

It means that you have been cheating :-) You have in fact used normal quantum mechanics, except for the fact that you have been rotating the |-+> and the |+-> vectors in the "different" eigenspace. When you then calculate the total length (squared) of the original vector, projected on each of those and add it together, you obtain of course the correct QM prediction. Indeed, total length is invariant under a rotation of the basis (in the "different" eigenspace). But that's not what you were proposing in the first place. What you proposed was that the probability of having the "different" result should be the projection on ONE SINGLE arbitrary direction in the "different" eigenspace, not the sum of all the possibilities (which corresponds to finding the total length of the projection, as prescribed by standard QM). And then you're back to your first formula, where the sum of probabilities of all the possible outcomes is not equal to 1. THAT was the technique you used for the EPR stuff. You didn't sum over the different projections (because then you'd have found the same predictions as standard QM: you'd just have been rotating the basis vectors in the eigenspace to calculate the total projection length, something you are of course allowed to do).

cheers,
Patrick.

 

[kleinwolf]

Yes, basically I wrote you, it's just completely normal Copenhagen QM, there is nothing new in what I said...just trying to be more precise.

Anyway, for myself already gave the answer...but this, like always, is only my opinion...you have yours of course, but why say yours is the right one ??

Let's take the definition of the correlation : the following calculation is really old and well-known...but this maybe explains a bit more...I learn like you.

C(A,B)=&lt;AB&gt;-&lt;A&gt;&lt;B&gt;

then we have in fact a correlation operator given by the superposition of non-local and local opertators :

M_{non-local}=\sigma_z\otimes\sigma_z
M_{local A}=\sigma_z\otimes\mathbb{I}
M_{local B}=\mathbb{I}\otimes\sigma_z

So that the correlation operator is :

C=M_{non-local}-M_{local_A}|\Psi\rangle\langle\Psi|M_{local_B}

So that the correlation operator depends on the state we measure, hence this operator is non-linear.

We have also the correspondance : M_{non-local}=M_{local_A}M_{local_B}

Now the fact is that the eigenstate of \mathhbb{I} are degenerate. So if we look at the spectral decomposition of the identity operator, then we are lead to a more general equivalence that can be solved by doing some operations on the parametrization of the eigenstates describing the eigenspace, in other words : we should not only work with orthogonal bases. If we look nearer, then :

Let 2 eigenstates of 1 be :
|\phi\rangle=\left(\begin{array}{c}\cos(\phi)\\\sin(\phi)\end{array}\right)
|\chi\rangle=\left(\begin{array}{c}\cos(\chi)\\\sin(\chi)\end{array}\right)

Hence, this allows for non-orthogonal bases of R^2, the generalized spectral decomposition is :

\mathbb{I}_{decomp}=|\phi\rangle\langle\phi|+|\chi\rangle\langle\chi|=\left(\begin{array}{cc}\cos(\phi)^2+\cos(\chi)^2&amp;\cos(\phi)\sin(\phi)+\cos(\chi)\sin(\chi)\\\cos(\phi)\sin(\phi)+\cos(\chi)\sin(\chi)&amp;\sin(\phi)^2+\sin(\chi)^2\end{array}\right)

Hence we have the relationships :

\mathbb{I}=\langle\mathbb{I}_{decomp}\rangle_{\phi,\chi}

and the other precise decomposition that specifies the parameters :

\exists\phi_0,\chi_0|\mathbb{I}_{decomp}(\phi=\phi_0,\chi=\chi_0)=\mathbb{I}

Now of course we can compute the complete correlation operator, that will give you expressions up to the 4th power in the cos and sin of the parameters...

What I basically want to know is if you consider this exchange about science as a game and you want to win...I feel a kind of something unhealthy in the air...because I don't really see what the game or the competition is...and you ?

Best regards.

 

[Sherlock]

Ok, that's the probability for the A+ detector to trigger. And what is the probability for the A- detector to trigger, then ? P = sin^2 |a - lambda| I'd say...

cheers,
Patrick.

EDIT:

I played around a bit with this, and in fact, it is not so easy to arrive at a CORRELATION function which is cos^2(a-b). Indeed, let's take your probability which is p(a+) = cos^2(lambda-a).
Assuming independent probabilities, we have then that the correlation, which is given by p(a+) p(b+) = cos^2(lambda-a) sin^2(lambda-b) for an individual event. (the b+ on the other side is the b- on "this" side)

Now, by the rotation symmetry of the problem, lambda has to be uniformly distributed between 0 and 2 Pi, so we have to weight this p(a+) p(b+) with this uniform distribution in lambda:

P(a+)P(b-) = 1/ (2 Pi) Integral (lambda=0 -> 2 Pi) cos^2(lambda-a) sin^2(lambda-b) d lambda.

If you do that, you find:

1/8 (2 - Cos(2 (a-b)) ) = 1/8 (3-2 Cos^2[a-b])

And NOT 1/2 sin^2(a-b) !

I checked this with a small Monte Carlo simulation in Mathematica and this comes out the same. Ok, in the MC I compared a+ with b+ (not with b-), and then the result is 1/8 (2+cos(2(a-b)))

So this specific model doesn't give us the correct, measured correlations...

cheers,
Patrick.

I attach the small Mathematica notebook with calculation...

I'm not sure what you're saying above.

The statement in question was that the cos^2 theta formula
is incompatible with hidden variables. It isn't. If you consider just
the individual results, then you can write the probability of
detection as P = cos^2 |a - lambda|, where a is the setting
of an individual polarizer and lambda is the emission polarization.
But of course you don't know the value of lambda ... ever. So,
the actual probability of individual detection is simply .5.
The reason it's .5 is because, presumably, lambda is varying
randomly -- and the data streams indicate that for any set
of n emissions you'll get, in the ideal, .5n detections.
Of course the actual number is modified enormously due
to efficiency considerations.

It can be shown, empirically, that there is a local hidden variable
determining individual detections.

However, while this local hidden variable exists, it does
not determine joint detection.

The combined context, when you're considering detection
at both ends during a given interval, is different in that, while
the local hidden variable is still determining individual results,
its *variability* isn't a factor in determining correlations.
The only thing about the emitted light that matters wrt
joint detection is that during any given detection interval
the light incident on the polarizers is the same at both
ends -- that any two opposite moving photons emitted
by the same atom are polarized identically. This is what
the *entanglement* is based on. If it wasn't assumed
that the two polarizers are analyzing light with the same
physical properties, then what would be the basis for the
projection along the detection axis?

So, during any given coincidence interval, the separated
polarizers are, in effect, analyzing the *same* light.
Hence, the applicability of the cos^2 theta formula in
the joint context involving crossed linear polarizers.

 

[Sherlock]

Or Lambda=LHV does not exist, a possibility you consistently pass over. It is a simple matter to show that with a table of 8 permutations on A/B/C, there are no values that can be inserted that add to 100% without having negative values at certain angle settings.

A=___ (try 0 degrees)
B=___ (try 67.5 degrees)
C=___ (try 45 degrees)

Hypothetical hidden variable function: __________ (should be cos^2 or at least close)

1. A+ B+ C+: ___ %
2. A+ B+ C-: ___ %
3. A+ B- C+: ___ %
4. A+ B- C-: ___ %
5. A- B+ C+: ___ %
6. A- B+ C-: ___ %
7. A- B- C+: ___ %
8. A- B- C-: ___ %

It is the existence of C that relates to the hidden variable function. What you describe is just fine as long as we are talking about A and B only. (Well, there are still some problems but there is wiggle room for those determined to keep the hidden variables.) But with C added, everything falls apart as you can see.

You can talk all day long about joint probabilities and lambda, but that continues to ignore the fact that you cannot make the above table work out. If you are testing something else, you are ignoring Bell. After you account for the above table, then your explanation might make sense. Meanwhile, the Copenhagen Interpretation (and MWI) accounts for the facts that LHV cannot.

I've considered the idea that the lhv
doesn't exist and rejected it.

There's a difference between the lhv not existing
and the lhv being irrelevant in a certain context.
I agree with you that the lhv is not determining
joint detection -- but, that doesn't mean that
it doesn't *exist*.

The above table is irrelevant to the
argument of whether or not the lhv *exists*.

The individual data streams are *direct*
evidence of the existence of the lhv.

 

[Sherlock]

There are definitely TWO ways to look at that statement. Some of the vocal local realists argue that the cos^2 law isn't correct!

Then I think they're wrong about that.

They do that so the Bell Inequality can be respected; and then explain that experimental loopholes account for the difference between observation and their theory.

From what I know of the experiments, they're ok. However,
I think that Bell's analysis and the physical meaning of experimental
violations of the inequality are being misinterpreted.

Clearly, classical results sometimes match QM and sometimes don't; and when they don't, you really must side with the predictions of QM. Even Einstein saw that this was a steamroller he had to ride, and the best he could muster was that QM was incomplete.

I *am* siding with the predictions of qm. Where have I
said otherwise? But it's certainly not a complete description
of the physical reality. It's not designed to be.

 

[vanesch]

I'm not sure what you're saying above.

The statement in question was that the cos^2 theta formula
is incompatible with hidden variables. It isn't. If you consider just
the individual results, then you can write the probability of
detection as P = cos^2 |a - lambda|, where a is the setting
of an individual polarizer and lambda is the emission polarization.
But of course you don't know the value of lambda ... ever.

Well, yes, that's exactly what I did. But apparently now you assume EQUAL polarizations (lambda) at both sides, and not OPPOSITE polarizations. So be it.

For a GIVEN lambda (unknown, I agree), you say that, if we put up a polarizer at Alice in direction a, it has a probability equal to cos^2(a-lambda) of clicking (assuming it "100% efficient" ; we'll come to that later). This means then also, I would think, that we have a probability cos^2(a-lambda) of clicking at Bob's place if he also puts his polarizer in direction a, right ?
And if Bob puts his polarizer in direction b, I assume that his probability of clicking for the same lambda is cos^2(b-lambda), right ?
Of course, specifying individual probabilities doesn't give us the joint distribution, except if you say that these probabilities are independent. But normally, what happens at Bob is independent of what happens at Alice, once lambda is given. So the JOINT PROBABILITY that the detector (in direction a) at Alice clicks, and that within the same time interval, the detector (in direction b at Bob) clicks is then given by

P(a,b,lambda) = cos^2(a-lambda) cos^2(b-lambda)

If that is not the case, then give me your joint probability for a given lambda.

Now, you say that we don't know lambda (which is the random polarization direction of the light sent out to both detectors in any event).
But we know that the distribution, whatever it is, must be rotation-invariant if we consider many events. Indeed, this is the only way to have, on one side, a probability equal to 1/2 averaged over the entire sample for ALL values of a. This means that the DISTRIBUTION of the different lambda values must be uniform over the 0 - 2Pi interval. Otherwise, we'd have on average MORE clicks in one direction than in another (on one single side).

Now, if we know that lambda, over different trials, is distributed uniformly, then we can calculate, over this population, what will be the average correlation P(a,b):
It is simply given by:

P(a,b) = 1/2Pi integral(lambda=0 -> 2 Pi) of P(a,b,lambda) d lambda

And if you do that, well, then you find:

P(a,b) = 1/8 (2 + cos(2 (a-b)) )

What changes now when the detectors are "inefficient" ? Well, this changes normally only the probability of clicking: instead of your "cos^2(a - lambda), we have a scale factor: epsilon cos^2(a - lambda). BTW, that is exactly what you get out of the semi-classical approach for the photo-electric effect.

The only thing it changes for P(a,b) is that you multiply with epsilon^2 (assuming detectors of identical quality on both sides).

The combined context, when you're considering detection
at both ends during a given interval, is different in that, while
the local hidden variable is still determining individual results,
its *variability* isn't a factor in determining correlations.
The only thing about the emitted light that matters wrt
joint detection is that during any given detection interval
the light incident on the polarizers is the same at both
ends -- that any two opposite moving photons emitted
by the same atom are polarized identically. This is what
the *entanglement* is based on. If it wasn't assumed
that the two polarizers are analyzing light with the same
physical properties, then what would be the basis for the
projection along the detection axis?

Well, that's exactly what I do. You give me the individual probabilities for a given value of the hidden variable, and from that, and a symmetry argument, I calculate the joint probability over the entire population of the hidden variable.

And comes out... something that is different than what QM predicts ! Ok, the "right" form is there, namely cos^2, but the "modulation depth" is much lower: you do not reach high enough correlations, but more importantly, you do not reach LOW ENOUGH correlations for certain angles either.
You can intuitively see that too.

If the polarizers at Bob and Alice are perpendicular, then according to the photon picture, you will have perfect ANTICORRELATION. Namely whenever the photon gets through Bob's, it is blocked for sure at Alice's and vice versa. Now, with classical light that cannot happen, because certain light pulses will get in under 45 degrees. That means that there is a reduced, but finite probability at Alice's of clicking, and also at Bob's, so the correlation will not be 0. Nevertheless, experimentally, it is 0, and according to the photon picture, it should also be 0.

cheers,
Patrick.

 

[kleinwolf]

Gosh...I should never have started with non-linear operators...anybody knows about the eigenvalues, or something like that...

For I get that the eigenvalues of the non-linear operator of the local part of the correlation of a bipartite system given by : M_{local}=(\sigma_z\otimes\mathbb{I})|\Psi&gt;&lt;\Psi|(\mathbb{I}\otimes\sigma_z)

are negative definite, continuous apparently, and need to solve a 4th order equation system of the style :

b^4+(2c-1)b^3-2c^2b^2-2c^2b+c^4=0
a^2=c^2-b^2
d^2=-w
a^2+b^2+c^2+d^2=1

where the eigenstate is given by : |\Psi&gt;=(a,b,c,d)

In fact the notation M_{local} is abusive, since the wavefunction is included, so that's why the average correlation is still non-local (I privately exchanged messages with RandallB before about why I came up with a non-perfect correlation for the singlet-state, giving CHSH=2.47 at maximum (hence non-local w.resp. to Bell's Ansatz, but still neare to experimental results 2.25)...

It's clear that if the total correlation operator is taken (with the local and the non-local parts), then, the eigenvalues should be in [-1;1]...but maybe the spectrum is a special one, with an infinite distribution, I cannot make any bet about that...What is your opinion ?

 

[DrChinese]

I've considered the idea that the lhv
doesn't exist and rejected it.

There's a difference between the lhv not existing
and the lhv being irrelevant in a certain context.
I agree with you that the lhv is not determining
joint detection -- but, that doesn't mean that
it doesn't *exist*.

The above table is irrelevant to the
argument of whether or not the lhv *exists*.

The individual data streams are *direct*
evidence of the existence of the lhv.

That isn't so... it is just a question of seeing what you want to see.

Suppose the random value is inserted when the observation is made - supplied by some randomizer which we cannot access and never can. That is a reasonable explanation and completely consistent with the facts. You see the randomness as evidence that there is more to know. Well, maybe that is so. But... that random value DID NOT exist prior to the observation, as Bell clearly tells us.

Now suppose the "random" value is not random at all - it is completely determined by some complex stochastic process having to do with the state of the entire universe. Because it is non-local, the same information is available to the entangled photons. That could be a reasonable explanation and consistent with the facts. Bell would still apply! And thus we know that the observations were still fundamental to the process, and there is no locally hidden variable that explains the observed results.

The observer settings are fundamental to the results; and there are no local hidden variables that completely determine the outcome independent of the observer settings. As Bell states, the purpose of LHV theories is to restore locality and causality to the description, and this cannot be done.

 

[vanesch]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true. But as many people stated that, I believed it for a while myself. However, as I tried to show, if you assume the following:

1) Intensity of (classical) radiation when incident radiation has polarization direction lambda and the polarizer has direction a is given by the original intensity multiplied by cos^2(lambda-a).

2) Identical incident radiation (during one "pulse") at Alice's and Bob's, with identical (or opposite, pick your choice) polarization directions

3) photodetection probability (clicking probability) proportional to intensity, and, for a given intensity, statistically independent of any other photodetection somewhere else.

If we assume that the source sends out pulse trains, each with an (unknown) value of polarization, and equal intensity, then:

The probability at Alice (polarizer at angle a) of clicking is:
P(a,lambda) = eps cos^2(a - lambda)

The probability at Bob (polarizer at angle b) of clicking is:
P(b,lambda) = eps cos^2(b - lambda)

and the probability of clicking together, these probabilities being considered independent, is given by:

P(a,b,lambda) = eps^2 cos^2(a-lambda) cos^2(b-lambda)

From a symmetry argument, one can deduce that lambda must be drawn from a uniform distribution between 0 and 2 Pi, and so, the observed overall probabilities of clicking on N trials are:

P(a) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon \cos^2(a - \lambda) d \lambda = 1/2

Same for P(b)

and for P(a,b):
P(a,b) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon^2 \cos^2(a - \lambda)\cos^2(b - \lambda) d\lambda

and this leads to:
P(a,b) = \frac{\epsilon^2 }{8} (2 + \cos 2 (a - b))

Of course we don't know the absolute number of trials (the true value of epsilon) if we don't consider the photon model, but we can do away with that by calculating:

\frac{P(a,b)}{P(a) P(b)} = \frac{1}{2} (2 + \cos 2(a-b))

Note that this is NOT equal to the quantum prediction, especially for the fact that the above correlation function doesn't go down to 0, when the two polarizers are perpendicular.

cheers,
Patrick.

 

[vanesch]

My previous message may give the impression that one cannot obtain a cos^2(a-b) curve with a hidden variable model. This is not true, and I just made one. It goes as follows:
Imagine that each pair of light pulses that is sent out has two hidden variables: one is lambda, the polarization direction (which will be uniformly distributed in 0-2Pi) and the other one is a one-bit random variable: if the Alice pulse receives the 1 bit, then the Bob pulse receives the 0 bit, and vice versa. This random variable is distributed 50/50, and we call it the tau variable.

Now, a polarizer could work in the following way, upon reception of a light pulse with hidden variables lambda and tau:
If tau = 1, then the intensity that gets through the polarizer under angle a equals the incoming intensity if |a - lambda|< delta (a small angle) and is blocked completely if not.
However, if tau = 0, then the intensity that gets through the polarizer is equal to the incoming intensity times cos^2(a - lambda).

Next, the probability of a detector click is proportional to the incoming intensity.

Applying this model yields a correlation about proportional to cos^2(a-b). Mind you, I say: proportional !

Indeed, for the individual polarizers, we get essentially 1/4 of the total number of trials (half of the time we get a bit 1, so then the probability of letting any intensity through is very rare, because lambda needs to be close to a, and the other half of the time, we get the cos^2 curve, which gives us 1/2 on average).
However, for the coincidence, in order for both to click, one of both will have a bit 1. So we KNOW that we are in one of the rare cases when lambda is close, or to a, or to b. In that case, the OTHER polarizer receives the 0 bit, and hence the probability for the OTHER one of clicking is given by the cos^2 rule. Only, we suppressed seriously the entire population and a very small fraction of the trials do give rise to a correlation. But if we add in arbitrary "efficiency" coefficients, we can say that we have a cos^2 relation.

This ad hoc model suffers of course from a lot of difficulties and is made up for the purpose. First of all, this is not classical optics either. The bit left or right mechanism is totally taken out of thin air. Next, although this model can explain certain aspects of the cos^2 curve, it would fail miserably on energy balances: we wouldn't have conservation of total radiant energy when taking the flux that gets through a polarizer, and that gets to the perpendicular polarizer.
But it is a technique to show that a curve, proportional to cos^2 can eventually be constructed.

cheers,
Patrick.

 

[cartuz]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true. But as many people stated that, I believed it for a while myself. However, as I tried to show, if you assume the following:

1) Intensity of (classical) radiation when incident radiation has polarization direction lambda and the polarizer has direction a is given by the original intensity multiplied by cos^2(lambda-a).

2) Identical incident radiation (during one "pulse") at Alice's and Bob's, with identical (or opposite, pick your choice) polarization directions

3) photodetection probability (clicking probability) proportional to intensity, and, for a given intensity, statistically independent of any other photodetection somewhere else.

If we assume that the source sends out pulse trains, each with an (unknown) value of polarization, and equal intensity, then:

The probability at Alice (polarizer at angle a) of clicking is:
P(a,lambda) = eps cos^2(a - lambda)

The probability at Bob (polarizer at angle b) of clicking is:
P(b,lambda) = eps cos^2(b - lambda)

and the probability of clicking together, these probabilities being considered independent, is given by:

P(a,b,lambda) = eps^2 cos^2(a-lambda) cos^2(b-lambda)

From a symmetry argument, one can deduce that lambda must be drawn from a uniform distribution between 0 and 2 Pi, and so, the observed overall probabilities of clicking on N trials are:

P(a) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon \cos^2(a - \lambda) d \lambda = 1/2

Same for P(b)

and for P(a,b):
P(a,b) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon^2 \cos^2(a - \lambda)\cos^2(b - \lambda) d\lambda

and this leads to:
P(a,b) = \frac{\epsilon^2 }{8} (2 + \cos 2 (a - b))

Of course we don't know the absolute number of trials (the true value of epsilon) if we don't consider the photon model, but we can do away with that by calculating:

\frac{P(a,b)}{P(a) P(b)} = \frac{1}{2} (2 + \cos 2(a-b))

Note that this is NOT equal to the quantum prediction, especially for the fact that the above correlation function doesn't go down to 0, when the two polarizers are perpendicular.

cheers,
Patrick.

Excuse me for it is hard to read a long English texts for me. Because it is ordinary I read texts with mathematics only.
We can clear see that Bell's Inequalities it is possible to complete by curved and stationary geometry, which play role the non-local hidden variables. Why the geometry must be non-curved? Curved geometry is more suitable because it is the general case.
In that case Bell's Inequalities will be violet!
It is clear from
Correlation factor M of random variables \lambda ^{i} are projections
onto directions A^{\nu } and B^{n} defined by polarizers (all these
vectors being unit) is

\left| M_{AB}\right| =\left| \left\langle AB\right\rangle \right| =\left|<br /> \langle \lambda ^{i}A^{k}g_{ik}\lambda ^{m}B^{n}g_{mn}\rangle \right|

The deferential geometry gives

\cos \Phi =\frac{g_{ik}\lambda ^{i}A^{k}}{\sqrt{\lambda ^{i}\lambda _{i}}\sqrt{A^{k}A_{k}}},

\cos (\Phi +\theta )=\frac{g_{mn}\lambda ^{m}B^{n}}{\sqrt{\lambda<br /> ^{m}\lambda _{m}}\sqrt{B^{n}B_{n}}}.

Here i,k,m,n possesses 0,1,2,3; \theta is angle between polarizers, then
\left| M_{AB}\right| =\left| \frac{1}{2\pi }\int_{0}^{2\pi }\rho (\Phi<br /> )\cos \Phi \cos \left( \Phi +\theta \right) d\Phi \right|

This the case when entanglement explained be by stationary gravitational fields (curved geometry) is perspective, I hope.

 

[DrChinese]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true. But as many people stated that, I believed it for a while myself. However, as I tried to show, if you assume the following:

Patrick,

I don't mean to pick apart words. But there are two ways to interpret the situation you describe. You are of course correct that the application of Malus' Law in the "classical" manner you describe yields a different prediction for joint detection.

But that is not the only way to apply Malus' Law to this case. Since it describes the results only when initial polarizeration is KNOWN, you should wait to apply it until an observation is performed on one or the other of the entangled photons...similar to how it is done in classical application (in which the first polarizer tells us the initial polarization). Then the joint results will match QM exactly.

In other words, if you apply Malus' Law ASSUMING hidden variables, you get a different result (for joint detection) than if you apply it using (what I think of as) a "traditional" application which does not specifically assume HV. Presumably, Malus never thought of hidden variables one way or the other.

So to summarize: if you push a "classical" application of Malus' Law as you describe, you immediately run into problems because the results disagree with experiment (as you pointed out). If you push the hidden variable version and apply as I described, you immediately run into problems with Bell's Theorem. So you know that there is something wrong with the "classical" or "traditional" views (which are just labels as I have used them here) either way you choose to look at it.

 

[vanesch]

But that is not the only way to apply Malus' Law to this case. Since it describes the results only when initial polarizeration is KNOWN, you should wait to apply it until an observation is performed on one or the other of the entangled photons...similar to how it is done in classical application (in which the first polarizer tells us the initial polarization). Then the joint results will match QM exactly.

Because people told this already a few times, intuitively I thought that that was very acceptable, and that's why for a long time I thought that "classical optics" predicted the same correlations as quantum theory.
When you apply it as you state, when you say: ah, one detector clicked, so the light MUST BE in the same polarization direction as that polarizer, YOU ARE IN FACT APPLYING QUANTUM THEORY ! You projected and normalized the ENTIRE state on the direction of the first polarizer, as is typically done in QM (in the Copenhagen view, let's be clear).
But that is NOT what is done in classical optics. In classical optics a polarizer ONLY SELECTS THE COMPONENT of the light in the direction of the polarizer FOR THE LIGHT AT THAT POLARIZER. When you send light polarized under 45 degrees with the polarizer, only HALF of the intensity gets through ; and if you have two "identical copies of light" (as entanglement is seen in classical optics), it is not because on one side, you selected only half of the intensity (because your polarizer made an angle of 45 degrees) that suddenly the light will jump over 45 degrees on the other side to match the detection. That is a typical pure quantum phenomenon.

And that's why I wanted to point out, that, in the case Alice and Bob put their polarizers at 90 degrees, in the classical picture, THERE IS STILL LIGHT COMING THROUGH at both sides: namely all that light that is not exactly polarized at 0 or at 90 degrees. If light is incident under 45 degrees, at both sides, they get half the intensity, so there is a real chance of having coincident clicks.

In other words, if you apply Malus' Law ASSUMING hidden variables, you get a different result (for joint detection) than if you apply it using (what I think of as) a "traditional" application which does not specifically assume HV. Presumably, Malus never thought of hidden variables one way or the other.

I was not really using "hidden variables" ; the "hidden variable" was the random polarization of classical light. We know that the source sent out light pulses which have identical polarization for bob and alice, wavetrain per wavetrain. But this common polarization can fluctuate randomly (as the phase of light can fluctuate randomly outside of the coherence time). There is nothing surprising about that in classical optics.
Then I calculated the light intensity that got through each polarizer individually, using Malus' law, and then I applied a probability law for each photodetector, that gives us a probability of clicking per unit of time which is proportional to the incident intensity (after the polarizer).
I would think that that is exactly what one is supposed to do in classical optics, no ? I didn't "push" anything.

So to summarize: if you push a "classical" application of Malus' Law as you describe, you immediately run into problems because the results disagree with experiment (as you pointed out). If you push the hidden variable version and apply as I described, you immediately run into problems with Bell's Theorem. So you know that there is something wrong with the "classical" or "traditional" views (which are just labels as I have used them here) either way you choose to look at it.

Yes, of course. But it just appeared to me that what I had been taking for granted because so many people said it, namely that in PURELY CLASSICAL OPTICS, you get out the same correlations as in quantum theory, is absolutely not true ! And the most striking aspect is again the pure ANTI correlation when Alice and bob have perpendicular polarizers, which is impossible to obtain in classical optics. (but for which you can build an ad hoc hidden variable model, as I did - without physical plausibility).

cheers,
Patrick.

 

[kleinwolf]

Excuse me for it is hard to read a long English texts for me. Because it is ordinary I read texts with mathematics only.
We can clear see that Bell's Inequalities it is possible to complete by curved and stationary geometry, which play role the non-local hidden variables. Why the geometry must be non-curved? Curved geometry is more suitable because it is the general case.
In that case Bell's Inequalities will be violet!
It is clear from
Correlation factor M of random variables \lambda ^{i} are projections
onto directions A^{\nu } and B^{n} defined by polarizers (all these
vectors being unit) is

\left| M_{AB}\right| =\left| \left\langle AB\right\rangle \right| =\left|<br /> \langle \lambda ^{i}A^{k}g_{ik}\lambda ^{m}B^{n}g_{mn}\rangle \right|

The deferential geometry gives

\cos \Phi =\frac{g_{ik}\lambda ^{i}A^{k}}{\sqrt{\lambda ^{i}\lambda _{i}}\sqrt{A^{k}A_{k}}},

\cos (\Phi +\theta )=\frac{g_{mn}\lambda ^{m}B^{n}}{\sqrt{\lambda<br /> ^{m}\lambda _{m}}\sqrt{B^{n}B_{n}}}.

Here i,k,m,n possesses 0,1,2,3; \theta is angle between polarizers, then
\left| M_{AB}\right| =\left| \frac{1}{2\pi }\int_{0}^{2\pi }\rho (\Phi<br /> )\cos \Phi \cos \left( \Phi +\theta \right) d\Phi \right|

This the case when entanglement explained be by stationary gravitational fields (curved geometry) is perspective, I hope.

Well, in fact in this case, then Bell's idea is not really reached because you have something like :

M_{AB}=\frac{1}{2\pi}\int_0^{2\pi}\cos(\theta_A-\phi)\cos(\theta_B-\phi)\rho(\phi)d\phi
=\int_0^{2\pi}A(\theta_A,\phi)B(\theta_B,\phi)\rho(\phi)d\phi

But your functions A and B ARE NOT THE RESULTS OF MEASUREMENT (because the results can be only 1 and -1)...That's why Bell could deduce a difference between hidden variable and QM...Remind that the results of measurement of the quantum-mechanical operator (\sigma\cdot n_A)\otimes(\sigma\cdot n_B) are +1 or -1 (the eigenvalues)...but nothing is allowed inbetween.

What you could say is that in your case :A(\theta_A,\phi)=\int a(\theta_A,\phi,\lambda)u(\lamda)d\lambda

where a(\theta_A,\phi,\lambda)=+1,-1

So that then you can apply Bell's theorem on a,b...so that your proposition doesn't violate the inequqality (CHSH for example).

 

[kleinwolf]

Excuse me for it is hard to read a long English texts for me. Because it is ordinary I read texts with mathematics only.
We can clear see that Bell's Inequalities it is possible to complete by curved and stationary geometry, which play role the non-local hidden variables. Why the geometry must be non-curved? Curved geometry is more suitable because it is the general case.
In that case Bell's Inequalities will be violet!
It is clear from
Correlation factor M of random variables \lambda ^{i} are projections
onto directions A^{\nu } and B^{n} defined by polarizers (all these
vectors being unit) is

\left| M_{AB}\right| =\left| \left\langle AB\right\rangle \right| =\left|<br /> \langle \lambda ^{i}A^{k}g_{ik}\lambda ^{m}B^{n}g_{mn}\rangle \right|

The deferential geometry gives

\cos \Phi =\frac{g_{ik}\lambda ^{i}A^{k}}{\sqrt{\lambda ^{i}\lambda _{i}}\sqrt{A^{k}A_{k}}},

\cos (\Phi +\theta )=\frac{g_{mn}\lambda ^{m}B^{n}}{\sqrt{\lambda<br /> ^{m}\lambda _{m}}\sqrt{B^{n}B_{n}}}.

Here i,k,m,n possesses 0,1,2,3; \theta is angle between polarizers, then
\left| M_{AB}\right| =\left| \frac{1}{2\pi }\int_{0}^{2\pi }\rho (\Phi<br /> )\cos \Phi \cos \left( \Phi +\theta \right) d\Phi \right|

This the case when entanglement explained be by stationary gravitational fields (curved geometry) is perspective, I hope.

Well, in fact in this case, then Bell's idea is not really reached because you have something like :

M_{AB}=\frac{1}{2\pi}\int_0^{2\pi}\cos(\theta_A-\phi)\cos(\theta_B-\phi)\rho(\phi)d\phi
=\int_0^{2\pi}A(\theta_A,\phi)B(\theta_B,\phi)\rho(\phi)d\phi

But your functions A and B are not the results of measurement (because the results can be only 1 and -1)...That's why Bell could deduce a difference between hidden variable and QM...Remind that the results of measurement of the quantum-mechanical operator (\sigma\cdot n_A)\otimes(\sigma\cdot n_B) are +1 or -1 (the eigenvalues)...but nothing is allowed inbetween.

What you could say is that in your case :A(\theta_A,\phi)=\int a(\theta_A,\phi,\lambda)u(\lambda)d\lambda

where a(\theta_A,\phi,\lambda)=+1,-1

So that then you can apply Bell's theorem on a,b...so that it seems your proposition doesn't violate the inequality (CHSH for example).

The point is that if the correlation is expressed as hidden variables only :

C(A,B)=\int_a^b a(\vec{n}_A,\vec{\lambda})b(\vec{n}_A,\vec{\lambda})\rho(\vec{\lambda})d\lambda_1...d\lambda_n

Then Bell's inequality is respected..(where a,b are constant (e.g. infinity or 0->2pi)...

however, if the correlation contains a visible variable :

C(A,B,\mu)=\int_{f(\mu)}^{g(\mu)}A(n_A,\lambda,\mu)B(n_B,\lambda,\mu)\rho(\lambda,\mu)d\lambda

Then you can violate the inequality, in the sense you choose \mu so that you cannot factorize the results like in Bell's theorem.

 

[cartuz]

Well, in fact in this case, then Bell's idea is not really reached because you have something like :

M_{AB}=\frac{1}{2\pi}\int_0^{2\pi}\cos(\theta_A-\phi)\cos(\theta_B-\phi)\rho(\phi)d\phi
=\int_0^{2\pi}A(\theta_A,\phi)B(\theta_B,\phi)\rho(\phi)d\phi

But your functions A and B are not the results of measurement (because the results can be only 1 and -1)...That's why Bell could deduce a difference between hidden variable and QM...Remind that the results of measurement of the quantum-mechanical operator (\sigma\cdot n_A)\otimes(\sigma\cdot n_B) are +1 or -1 (the eigenvalues)...but nothing is allowed inbetween.

What you could say is that in your case :A(\theta_A,\phi)=\int a(\theta_A,\phi,\lambda)u(\lambda)d\lambda

where a(\theta_A,\phi,\lambda)=+1,-1

So that then you can apply Bell's theorem on a,b...so that it seems your proposition doesn't violate the inequality (CHSH for example).

The point is that if the correlation is expressed as hidden variables only :

C(A,B)=\int_a^b a(\vec{n}_A,\vec{\lambda})b(\vec{n}_A,\vec{\lambda})\rho(\vec{\lambda})d\lambda_1...d\lambda_n

Then Bell's inequality is respected..(where a,b are constant (e.g. infinity or 0->2pi)...

however, if the correlation contains a visible variable :

C(A,B,\mu)=\int_{f(\mu)}^{g(\mu)}A(n_A,\lambda,\mu)B(n_B,\lambda,\mu)\rho(\lambda,\mu)d\lambda

Then you can violate the inequality, in the sense you choose \mu so that you cannot factorize the results like in Bell's theorem.

Thank you for reply and excuse me for delay.Are this mean that I
must to add here \mu which have the sense of distribution of metric g_{ik}?
Yes, I understand that metric allow correlated different quantatives in the space.

 

[Sherlock]

The statement in question was that the cos^2 theta formula
is incompatible with hidden variables. It isn't. If you consider just
the individual results, then you can write the probability of
detection as P = cos^2 |a - lambda|, where a is the setting
of an individual polarizer and lambda is the emission polarization.

Well, yes, that's exactly what I did. But apparently now you
assume EQUAL polarizations (lambda) at both sides, and not
OPPOSITE polarizations. So be it.

For a GIVEN lambda (unknown, I agree), you say that, if we put up a polarizer at Alice in direction a, it has a probability equal to cos^2(a-lambda) of clicking (assuming it "100% efficient" ; we'll come to that later). This means then also, I would think, that we have a probability cos^2(a-lambda) of clicking at Bob's place if he also puts his polarizer in direction a, right ?
And if Bob puts his polarizer in direction b, I assume that his probability of clicking for the same lambda is cos^2(b-lambda), right ?
Of course, specifying individual probabilities doesn't give us the joint distribution, except if you say that these probabilities are independent. But normally, what happens at Bob is independent of what happens at Alice, once lambda is given. So the JOINT PROBABILITY that the detector (in direction a) at Alice clicks, and that within the same time interval, the detector (in direction b at Bob) clicks is then given by

P(a,b,lambda) = cos^2(a-lambda) cos^2(b-lambda)

If that is not the case, then give me your joint probability for a given lambda.

First, the statement I was replying to was that local hidden
variables don't *exist*. I disagreed. They do exist, and
it can be demonstrated by looking at *individual* results.
Lambda is what's being analyzed in the individual context.

Regarding the *joint* probability, it doesn't depend
on a given lambda. That is, lambda isn't what's being
analyzed. What's being analyzed in the combined
context is a global constant. Lambda isn't the
global *constant*, so a description of joint
probability (such as what you evaluated) based on
lambda might give the correct functional form, but a
reduced spread (as you pointed out).

Ok, so, in the joint context, the polarizers, taken
together, are analyzing the degree to which photon_1
and photon_2 of any given pair are polarized identically.

So, if A records a photon detection, then the
probability of detection at B with the polarizers
aligned is 1 -- and the probability of detection
at B varies as cos^2 |a - b|. Isn't this
just standard quantum optics?

But where is the identical polarization produced?
As far as we *know* it *could* be produced at
the polarizers or detectors. But there's
an explanation that fits the observations,
and doesn't require the existence of undetectable
superluminal 'influences'.

Does quantum theory have an answer for the
question of where the identical polarization
is produced?

As far as I can tell, you could
interpret the projection as being due to local
or nonlocal transitions. That is, you *can*
assume that the identical polarization of
photon_1 and photon_2 is produced via the
emission process, and this assumption isn't
contradicted by the data. It also isn't, imo,
contradicted by Bell's analysis which deals
with the variable lambda. It also fits the
original idea of entanglement being due
to past interaction or common source.

Since interpretations of Bell's analysis
seem to be the only thing that superluminality
(wrt to associated experiments) has going
for it, it doesn't seem to me to be the
most reasonable option.

Regarding your statement that, using a wave
picture for light, a detector sitting behind
perpendicularly crossed linear polarizers has a >0
probability of registering a detection:
I don't think that's so. Anyway, I'm not
sure what that has to do with what we're
talking about -- which is, I thought, the
nature of entanglement in general, and in Bell/EPR
experiments in particular.

 

[vanesch]

Regarding the *joint* probability, it doesn't depend
on a given lambda. That is, lambda isn't what's being
analyzed. What's being analyzed in the combined
context is a global constant. Lambda isn't the
global *constant*, so a description of joint
probability (such as what you evaluated) based on
lambda might give the correct functional form, but a
reduced spread (as you pointed out).

But it isn't doing that ! It is doing that only in my second,
ad hoc, model.

Ok, so, in the joint context, the polarizers, taken
together, are analyzing the degree to which photon_1
and photon_2 of any given pair are polarized identically.

So, if A records a photon detection, then the
probability of detection at B with the polarizers
aligned is 1 -- and the probability of detection
at B varies as cos^2 |a - b|. Isn't this
just standard quantum optics?

No, not at all. That's standard optics IF IT IS THE SAME BEAM:

beam in ==> (pol 1) =(beam1)==> (pol2) ==> beam 2.

If pol1 and pol2 are aligned, of course beam 1 is reduced in intensity from beam in, but beam 2 is not reduced anymore, it is equal to beam1. And if pol1 and pol2 are perpendicular, beam2 equals 0. BUT THAT IS BECAUSE POL2 WORKS ON THE BEAM THAT GOT THROUGH POL1, which selected the component of beamin in the pol1 direction. If you then select AGAIN that component, of course everything gets through. It is as if you have a vector in the xy plane, and you project on the x-axis, and then you project the projection AGAIN on the x-axis. You work the second time with the projection.

But if you have:
beam2 <== (pol2) == (beamin) ====(pol1)==> beam1

then beam1 is simply the component of beamin in the pol1 direction, and beam2 is simply the component of beamin in the pol2 direction, and in classical optics these projections are done on the original beam. In the xy analogy, if pol1 and pol2 are resp. on the x and the y axis, beam1 gives you the x-component, and beam2 gives you the y component of the arbitrary vector coming from beamin. It is the special quantum context which makes that the measurement of the component x at pol1 makes that suddenly the beam at pol2 is projected out together in that x-direction (to take up the Copenhagen picture). But that's not what is done in classical optics.

In fact, in classical optics, there is no difference between this setup, and simply an incoming beam that is split on a (non-polarizing) beamsplitter, the transmitted beam arriving at pol1 and the reflected beam arriving at pol2.

That is, you *can*
assume that the identical polarization of
photon_1 and photon_2 is produced via the
emission process, and this assumption isn't
contradicted by the data. It also isn't, imo,
contradicted by Bell's analysis which deals
with the variable lambda. It also fits the
original idea of entanglement being due
to past interaction or common source.

Well, of course I assume that the two beams have identical polarization (lambda)!

Let us assume, for fun, that the source is polarized: it is sending out light always under 45 degrees, at both sides. Now assume that pol1 is set parallel to the x-axis, and pol2 is set parallel to the y axis. What do you think are the intensities at both sides in classical optics ?

Regarding your statement that, using a wave
picture for light, a detector sitting behind
perpendicularly crossed linear polarizers has a >0
probability of registering a detection:
I don't think that's so.

Eh, a detector behind CROSSED linear polarizers doesn't see any light, of course ! I never said that. I said that a detector behind ONE polarizer sees some light, and if ANOTHER BEAM (with identical polarization) falls upon ANOTHER POLARIZER (which is crossed to the first), it will ALSO SEE SOME LIGHT (in classical optics).

Anyway, I'm not
sure what that has to do with what we're
talking about -- which is, I thought, the
nature of entanglement in general, and in Bell/EPR
experiments in particular.

?

cheers,
Patrick.

 

[Sherlock]

But it isn't doing that ! It is doing
that only in my second, ad hoc, model.

I'm not sure what you're referring to here.

 

[Sherlock]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true.

Actually, I wasn't arguing that -- but
rather just offering a perspective on why the
qm formulation works. A view of some possible
physical reasons for the correlations in optical Bell/EPR
experiments. These setups *do* have some
important features in common with the classical
setups that first produced the cos^2 theta formula.

The thing about photon detections is that there's
no way to tell whether all or some portion of the
light incident on a polarizer has been transmitted
by the polarizer when a detection occurs.

I suspect that you and I might have somewhat different
conceptions of what the word "photon" might refer to,
aside from it's existence as a theoretical entity and
a recorded detection.

Regarding anticoincidence experiments using
beamsplitters -- it's the same problem. There's
no way to tell if the light incident on a beamsplitter
and subsequently producing a photon detection at
one detector or the other (but never both)
was unevenly split or traveled only one path or
the other.

Do the uncertainty relations associated with these
types of experiments forbid ever knowing the
answers to these questions (given the current
fundamental quantum of action)?

 

[vanesch]

The thing about photon detections is that there's
no way to tell whether all or some portion of the
light incident on a polarizer has been transmitted
by the polarizer when a detection occurs.

I suspect that you and I might have somewhat different
conceptions of what the word "photon" might refer to,
aside from it's existence as a theoretical entity and
a recorded detection.

Well, there is of course one true concept of a photon, and that's the theory that defines it, namely QED. But if you insist on classical optics (which is Maxwell's equations), then a way to try to explain photo-electric clicking is by assuming that the EM wave amplitude is strongly pulsed: you don't have the intro textbook sine wave, but you have essentially most of the time, very low amplitudes and then you have sudden pulses (wave packets). When you look at the monochromaticity required (the delta lambda / lambda) and the time scale of detection (a few ns) versus the period of EM field oscillation (order fs), then there is all the room in the world to make these wave packets which are peaked in amplitude on the ns scale and appear still essentially monochromatic. Adding a semiclassical model of the source (where atoms radiate pulses of light during short time intervals) you have a natural setting for claiming that the EM wave is pulsed that way.
So *IF YOU INSIST ON THIS SEMICLASSICAL MODEL* (which, I recall, can explain quite a lot of optical phenomena), with individual light pulses which are EM wave trains according to Maxwell, then I don't see how you can arrive at any other prediction for the correlations than what I calculated, namely eps^2/8 (2 - cos 2(a-b) ).

Mind you that the workings of a beam splitter, a polarizing filter and a photodetector, in this semi-classical model do not have many liberties. Especially the beam splitter: if you ever hope to get interference using this pulsed light, a beam splitter has to send HALF of the EM energy (1/sqrt(2) of the E-field amplitude) very accurately to both sides. If it sends a whole pulse to the left, and then a whole pulse to the right, upon recombination, you wouldn't have any interference. Now, beamsplitters do give rise to interference. So that limits strongly how they can handle the classical EM wave.
In the same way, a photodetector can be checked against bolometric energy flux measurements: there is a very strict relation between the total number of counts during a certain time, and the total incident EM radiation. If you assume that the photodetector doesn't have any memory mechanism beyond the few ns scale then the probability of detection can only depend upon the incident EM energy (the flux of the Pointing vector). You can then also check its dependence, or not, of any polarization state.

Again, interference experiments with light getting through two polarizers show, in a similar way as done with a beam splitter, that classical EM wave pulses do not sometimes get through entirely, and sometimes don't get through, but that their intensities are lowered according to Malus's law, per pulse.

All this in the hypothesis of *classical EM radiation*.

You can think of many experiments that way, people have done them for more than a century, the classical behaviour of these components is completely constrained, and allows one to make precise predictions, based upon classical optics.

And for certain experiments, these predictions are in contradiction:
a) with QED predictions
b) with experimental results
but this only happens in the case of non-classical states of light (according to QED), such as 1-photon and 2-photon states in superposition (entangled photons).

Regarding anticoincidence experiments using
beamsplitters -- it's the same problem. There's
no way to tell if the light incident on a beamsplitter
and subsequently producing a photon detection at
one detector or the other (but never both)
was unevenly split or traveled only one path or
the other.

There is a way: interference of the resulting beams. If they interfere, they have to be present at the same time, and not one after the other.
You have to be able to do E1(t) + E2(t) at the screen. If at one time, you have a full E1 but no E2, and at another time, you have a full E2 but no E1, then you won't see interference.

cheers,
Patrick.

 

[Sherlock]

Well, there is of course one true concept of a photon, and that's the theory that defines it, namely QED. But if you insist on classical optics (which is Maxwell's equations), then a way to try to explain photo-electric clicking is by assuming that the EM wave amplitude is strongly pulsed: you don't have the intro textbook sine wave, but you have essentially most of the time, very low amplitudes and then you have sudden pulses (wave packets). When you look at the monochromaticity required (the delta lambda / lambda) and the time scale of detection (a few ns) versus the period of EM field oscillation (order fs), then there is all the room in the world to make these wave packets which are peaked in amplitude on the ns scale and appear still essentially monochromatic. Adding a semiclassical model of the source (where atoms radiate pulses of light during short time intervals) you have a natural setting for claiming that the EM wave is pulsed that way.
So *IF YOU INSIST ON THIS SEMICLASSICAL MODEL* (which, I recall, can explain quite a lot of optical phenomena), with individual light pulses which are EM wave trains according to Maxwell, then I don't see how you can arrive at any other prediction for the correlations than what I calculated, namely eps^2/8 (2 - cos 2(a-b) ).

Mind you that the workings of a beam splitter, a polarizing filter and a photodetector, in this semi-classical model do not have many liberties. Especially the beam splitter: if you ever hope to get interference using this pulsed light, a beam splitter has to send HALF of the EM energy (1/sqrt(2) of the E-field amplitude) very accurately to both sides. If it sends a whole pulse to the left, and then a whole pulse to the right, upon recombination, you wouldn't have any interference. Now, beamsplitters do give rise to interference. So that limits strongly how they can handle the classical EM wave.
In the same way, a photodetector can be checked against bolometric energy flux measurements: there is a very strict relation between the total number of counts during a certain time, and the total incident EM radiation. If you assume that the photodetector doesn't have any memory mechanism beyond the few ns scale then the probability of detection can only depend upon the incident EM energy (the flux of the Pointing vector). You can then also check its dependence, or not, of any polarization state.

Again, interference experiments with light getting through two polarizers show, in a similar way as done with a beam splitter, that classical EM wave pulses do not sometimes get through entirely, and sometimes don't get through, but that their intensities are lowered according to Malus's law, per pulse.

All this in the hypothesis of *classical EM radiation*.

You can think of many experiments that way, people have done them for more than a century, the classical behaviour of these components is completely constrained, and allows one to make precise predictions, based upon classical optics.

And for certain experiments, these predictions are in contradiction:
a) with QED predictions
b) with experimental results
but this only happens in the case of non-classical states of light (according to QED), such as 1-photon and 2-photon states in superposition (entangled photons).

There is a way: interference of the resulting beams. If they interfere, they have to be present at the same time, and not one after the other.
You have to be able to do E1(t) + E2(t) at the screen. If at one time, you have a full E1 but no E2, and at another time, you have a full E2 but no E1, then you won't see interference.

So, my assessment of what entanglement *is* (offered many
messages ago) would seem to be incomplete. I can't argue with
the fact that the idea that it's due to common emission polarization
results in mathematical representations that are contradicted
by experiments. Yet, the common emission polarization would
seem to be a necessary condition for producing entangled results.

I suppose I should look at the details of the MWI stuff that you
seem to like. :) Thanks for the thoughtful comments from you and
DrChinese, et al.

There's a paperclip symbol by this thread -- what does that
mean? Also, what do the "warnings" mean? I couldn't find
an explanation of this in the faq.

 

[vanesch]

So, my assessment of what entanglement *is* (offered many
messages ago) would seem to be incomplete. I can't argue with
the fact that the idea that it's due to common emission polarization
results in mathematical representations that are contradicted
by experiments. Yet, the common emission polarization would
seem to be a necessary condition for producing entangled results.

Yes, entanglement is "more" than common emission polarization.

You can have, say, 4 different "polarization relations" between two photons.

One is: identical polarization, all the time the same. That's represented in QM by, say, |theta> |theta> (a pure product state), and classically by two beams with identical, fixed polarization theta.

The second is: identical polarization, but randomly distributed from event to event. That is represented in QM by a mixture: half |0>|0> and half |90>|90> (a density matrix). This is the "correlated polarization" situation. In classical EM, it is represented by two identical beams with polarization theta, but this time theta is drawn from a population.
THIS is the situation that can be described by the semiclassical model I talked about.

The third one is: uncorrelated polarizations. This is represented in QM by a statistical mixture:
1/4 |0>|0> ; 1/4 |0>|90> ; 1/4 |90>|0> and 1/4 |90>|90>.
Classically, we have uncorrelated beams with individual random polarizations.

The final one is entanglement ; a pure state |0>|0> + |90>|90>.
There is no classical equivalent here...
Of course it implies "identical polarization with random distribution" in a certain way, but it is a STRONGER form of correlation than with ONLY this link (which is perfectly well described by the mixture in our second case above).
It has in it, this "magical link at a distance". If you want to describe it "semiclassically" you have to introduce strange things, namely that upon observation of the polarization at one side in direction A+ (which could have been the result of a partial intensity of another polarisation getting through the polarizer), suddenly the polarization at the other side has to jump into exactly that direction ; at which point it can be analysed by another polarizer in another direction, and Malus' law applies then. But that's not how Maxwell tells us that EM waves behave ! They don't "jump" because at a distance, something was detected or not. Hence the puzzling aspects of entangled states when you want to force them into classical concepts.
Again, entanglement has no classical equivalence. It's a new state which exists only within the quantum framework.

cheers,
Patrick.

 

[ZapperZ]

I would like to point out a recent paper in the European Journal of Physics, especially for those who are still trying to grasp what is meant by "entanglement". This is especially true if you think it can be understood without bothering to look into the mathematics.

G.B. Roston et al., "Quantum entanglement, spin 1/2 and the Stern-Gerlach experiment", Eur. J. Phys. v.26, p.657 (2005).

You have no excuse if you say you don't have access to it. As I've pointed out many times on here and in my journals, this is one of IoP journals, and ALL articles appearing on IoP journals are available FREE (via registration) for the first 30 days that paper appears online.

Zz.

 

[DrChinese]

G.B. Roston et al., "Quantum entanglement, spin 1/2 and the Stern-Gerlach experiment", Eur. J. Phys. v.26, p.657 (2005).

You have no excuse if you say you don't have access to it. As I've pointed out many times on here and in my journals, this is one of IoP journals, and ALL articles appearing on IoP journals are available FREE (via registration) for the first 30 days that paper appears online.

Zz.

Here is the link to the page that takes you there (just to make it even easier):

http://www.iop.org/EJ/ejs_extra - Select "This Month's Papers"

The article was put out around May 22 +/- so should be there through much of June. This site is nice, you may want to bookmark this page.

 

[Sherlock]

... entanglement has no classical equivalence. It's a new state which exists only within the quantum framework.

I feel somewhat confident in saying that an eventual qualitative understanding of quantum entanglement will be in terms of concepts developed via our ordinary sensory perception of things (ie., a wave mechanical picture) -- nothing *essentially* new or exotic, but
perhaps a lot more complicated than what's been developed so
far (eg., via quantum or classical or semiclassical descriptions).

The basic idea is that entanglement has to do with
analyzing common physical properties. There are a number
of ways that this can be produced in a universe with a
signal transmission speed limit of c. I don't think that's
been contradicted. The experimental results contradict
some simplistic ways of describing this mathematically,
that's all. If you've got a better idea wrt the essence of
entanglement, then let's hear it. :)

Now I'm going to read the Roston et al. paper referenced by
Zapperz and see if it has anything new in it.

 

[DrChinese]

The basic idea is that entanglement has to do with
analyzing common physical properties. There are a number
of ways that this can be produced in a universe with a
signal transmission speed limit of c. I don't think that's
been contradicted. The experimental results contradict
some simplistic ways of describing this mathematically,
that's all. If you've got a better idea wrt the essence of
entanglement, then let's hear it. :)

You have it backwards, as I see it. The experimental results rule out local reality. If you have a local hidden variable solution - either simplistic OR complicated - that matches experiment, let's hear it. :)

Instead of trying to restore local reality, we should try to understand local non-reality. Or non-local reality. Or non-local non-reality.

 

[Sherlock]

You have it backwards, as I see it. The experimental results rule out local reality.

That's an unwarranted conclusion. The experimental results
rule out quantitative descriptions of a certain form. As yet,
nobody's quite sure what that means as far as nature is
concerned.

If you have a local hidden variable solution - either simplistic OR complicated - that matches experiment, let's hear it. :)

I did provide, some messages back, a sort of semi-classical
approach in terms of local interactions and common source
which even vanesch allowed was ok for the usual EPR-Bell type
setups of an emitter, two polarizers and two detectors, but can't
as yet be extended to eg. beamsplitter setups.
The problem is that details of the physical characteristics of the
emitted light are lacking.

Instead of trying to restore local reality, we should try to
understand local non-reality. Or non-local reality. Or non-local
non-reality.

Local reality is still with us afaik. :) The question is whether we
need to posit superluminal signals to account for experimental
results. Some people don't think so. Some people do think so.
So, these are just two different ways to approach the
problem of explaining the correlations -- which remain
unexplained so far.

 

[vanesch]

I did provide, some messages back, a sort of semi-classical
approach in terms of local interactions and common source
which even vanesch allowed was ok for the usual EPR-Bell type
setups of an emitter, two polarizers and two detectors, but can't
as yet be extended to eg. beamsplitter setups.
The problem is that details of the physical characteristics of the
emitted light are lacking.

I first thought that indeed a semiclassical approach allowed for the reconstruction of Malus' law because that was repeated so many times here. But then I did a calculation and according to me, this semiclassical model gives you eps^2 /8 (2 - cos(2(a-b))) as a correlation function, which is NOT the prediction of QM, nor can explain the experimental results, especially in the case of perpendicular polarizers.

So could you specify again your semiclassical model ? Give us, for each "measurement interval" (a few nanoseconds):

a) what common parameters does the light have on both sides (classical polarization ; maybe also something else) which went with it thanks to a common creation ; and how these parameters are statistically distributed over the entire sample.
b) how, from these parameters, the individual detection probabilities at Alice and Bob are given if their angles of polarizers are a and b respectively
c) how you calculate from this the joint probability of detection assuming statistical independence of the probabilities cited in b).

Local reality is still with us afaik. :) The question is whether we
need to posit superluminal signals to account for experimental
results.

?

Locality implies of course the absense of superluminal signals by definition! Well, unless you are willing to sacrifice causality or special relativity...

cheers,
Patrick.

 

[DrChinese]

That's an unwarranted conclusion. The experimental results rule out quantitative descriptions of a certain form. As yet,
nobody's quite sure what that means as far as nature is
concerned.

The form that is ruled out is the one in which the photon polarization has definite values for any other angles other than the ones actually observed. If you do not choose to call that the local realistic position, that is your choice. However, that is definitely what EPR envisioned and this is what everyone else calls it.

I did provide, some messages back, a sort of semi-classical
approach in terms of local interactions and common source
which even vanesch allowed was ok for the usual EPR-Bell type
setups of an emitter, two polarizers and two detectors, but can't
as yet be extended to eg. beamsplitter setups.
The problem is that details of the physical characteristics of the
emitted light are lacking.

I don't think Vanesch said that you advanced a local realistic position he agreed with. (Of course, he can speak for himself on the matter - edit: he does in the post above.)

However, the quantum mechanical description is as physical as any theory. How about F=ma? Is that a physical description? Why would that make more sense than the HUP, for example? Just because QM uses a different mathematical language doesn't make it less of a description.

Local reality is still with us afaik. :) The question is whether we need to posit superluminal signals to account for experimental
results. Some people don't think so. Some people do think so.
So, these are just two different ways to approach the
problem of explaining the correlations -- which remain
unexplained so far.

Local reality is generally ruled out (unless you think of MWI as local reality). I agree with your question, though. Which is: are superluminal effects present?