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Entanglement and teleportation 
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#55
May1305, 03:30 AM

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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. 


#56
May1305, 03:57 AM

P: 343

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


#57
May1305, 04:27 AM

P: 343

Regarding photons, you wrote:
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. 


#58
May1305, 05:57 AM

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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 5050 to 0100 (namely 5050 when the incoming classical pulse is under 45 degrees with the polarizing BS orientation, and 0100 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 5050 > full correlation and you have anticorrelation in the 0100 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 setup, 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 spin1 and spin1/2 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. 


#59
May1305, 06:13 AM

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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. 


#60
May1305, 07:31 AM

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#61
May1305, 07:44 AM

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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. 


#62
May1305, 09:00 AM

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I said the following:
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 "noA+ hit" (where it is understood that the noA+ 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 anticorrelation: 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. 


#63
May1305, 09:25 AM

P: 343

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. 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. 


#64
May1305, 09:28 AM

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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(ab). Indeed, let's take your probability which is p(a+) = cos^2(lambdaa). Assuming independent probabilities, we have then that the correlation, which is given by p(a+) p(b+) = cos^2(lambdaa) sin^2(lambdab) 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(lambdaa) sin^2(lambdab) d lambda. If you do that, you find: 1/8 (2  Cos(2 (ab)) ) = 1/8 (32 Cos^2[ab]) And NOT 1/2 sin^2(ab) !!! 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(ab))) So this specific model doesn't give us the correct, measured correlations... cheers, Patrick. I attach the small Mathematica notebook with calculation... 


#65
May1305, 10:01 AM

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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. 


#66
May1305, 11:46 AM

P: 309

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 NONLOCAL hidden variable. Let see the latter : a nonlocal measurement is obtained by the operator : [tex]\sigma_z\otimes\(\sigma_z\cdot\vec{n}_b) [/tex]...hence Both side are measured, and there is no 1 operator on the other (non disturbing operator). Let consider [tex] \theta_b=0[/tex] Hence : both directions of measurement are the same. The clearly the only 2 possible endstates are : +> or +>, with [tex] p(+)=<+\Psi>^2=\frac{1}{2}=p(+)[/tex] This sounds very like more than intuitive and easy to understand. However, one can see the things in an other way, by looking that : [tex]M=\sigma_z\otimes\sigma_z=\left(\begin{array}{cccc} 1 &&&\\&1&&\\&&1&\\&&&1\end{array}\right)[/tex] 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 : [tex] same>=\left(\begin{array}{c}\cos(\chi)\\0\\\sin(\chi)\\0\end{array}\ri ght)[/tex] [tex]different>=\left(\begin{array}{c}0\\cos(\delta)\\0\\\sin(\delta)\end{a rray}\right)[/tex] [tex] \Psi>=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\1\\1\\0\end{array}\right)[/tex] So that : [tex] p(different)=<different\Psi>^2=\frac{1}{2}\cos(\delta)^2[/tex] [tex] p(same)=<same\Psi>^2=\frac{1}{2}\sin(\chi)^2 [/tex] Where [tex]\chi,\delta[/tex] 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....?? 


#67
May1305, 12:32 PM

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However, I don't understand your calculation. When you write out sigmaz x sigmaz, 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. 


#68
May1305, 02:33 PM

P: 309

Yes, you're entirely right...my mistake is unforgivable, since this will change all the afterwards calculation and interpretation of [tex]\delta[/tex].
Then the result is [tex] p(same)=0\quad p(diff)=\frac{1}{2}(1\sin(2\chi))[/tex] 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 ?? 


#69
May1305, 10:38 PM

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It's like: throw up a coin: 25% chance you have head, 30% you have tail :) cheers, Patrick. 


#70
May1405, 01:51 AM

P: 309

It's just because we don't understand QM. But QM is omnipotent for everyone, just put : [tex]\chi=\frac{\pi}{4}\Rightarrow p(diff)=1[/tex]
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 nonlocal part and a local one....here it's just the nonlocal one. Best regards. 


#71
May1405, 02:12 AM

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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. cheers, Patrick. 


#72
May1405, 04:37 AM

P: 309

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 wellknown...but this maybe explains a bit more....I learn like you. [tex] C(A,B)=<AB><A><B> [/tex] then we have in fact a correlation operator given by the superposition of nonlocal and local opertators : [tex] M_{nonlocal}=\sigma_z\otimes\sigma_z[/tex] [tex] M_{local A}=\sigma_z\otimes\mathbb{I} [/tex] [tex] M_{local B}=\mathbb{I}\otimes\sigma_z [/tex] So that the correlation operator is : [tex] C=M_{nonlocal}M_{local_A}\Psi\rangle\langle\PsiM_{local_B} [/tex] So that the correlation operator depends on the state we measure, hence this operator is nonlinear. We have also the correspondance : [tex]M_{nonlocal}=M_{local_A}M_{local_B}[/tex] Now the fact is that the eigenstate of [tex]\mathhbb{I}[/tex] 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 : [tex]\phi\rangle=\left(\begin{array}{c}\cos(\phi)\\\sin(\phi)\end{array}\ri ght)[/tex] [tex]\chi\rangle=\left(\begin{array}{c}\cos(\chi)\\\sin(\chi)\end{array}\ri ght)[/tex] Hence, this allows for nonorthogonal bases of R^2, the generalized spectral decomposition is : [tex]\mathbb{I}_{decomp}=\phi\rangle\langle\phi+\chi\rangle\langle\chi=\ left(\begin{array}{cc}\cos(\phi)^2+\cos(\chi)^2&\cos(\phi)\sin(\phi)+\c os(\chi)\sin(\chi)\\\cos(\phi)\sin(\phi)+\cos(\chi)\sin(\chi)&\sin(\phi )^2+\sin(\chi)^2\end{array}\right)[/tex] Hence we have the relationships : [tex] \mathbb{I}=\langle\mathbb{I}_{decomp}\rangle_{\phi,\chi} [/tex] and the other precise decomposition that specifies the parameters : [tex] \exists\phi_0,\chi_0\mathbb{I}_{decomp}(\phi=\phi_0,\chi=\chi_0)=\math bb{I} [/tex] 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. 


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