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Related to my previous question (Why would classical correlation in Bell's experiment be a linear function of angle?), as a newbie in quantum mechanics, I am also unable to find the reason to why are the results of the Bell's experiment need to be attributed to "losing classical realism"?

I am ignoring the non-locality argument as a rather poor (FTL) explanation of the phenomena, but I don't see why the following couldn't be true (and I am not a physicist and am probably lacking some fundamental knowledge, so please bear with my nonsense):

What if the the discrepancies in Bell's experiments would simply be attributed to uncertainties in the measurement process, instead of insisting that we must give up realism?

In other words:

  1. Two entangled particles always leave the source differently polarized relative to each other,
  2. Due to measurement uncertainties at that scales, the probability of us measuring the correct polarization for both particles is a function of the plate angle?

In other words:

  1. If the angle of the deflector matches at both measurement points, we will measure both particles "the same way", and get 100% correlation, even if this means that we actually measured the opposite spin on both of them in all cases.
  2. Similarly, if the angle slightly differs, we will still measure spins in a "pretty similar" way, leading to a large correlation between our measurements, again without knowing if we actually measured them properly.

For particle experiments, all these measurements seem to be performed by interacting with the particle, and then interpreting the output, which may or may not represent the actual state of the particle, in my opinion.

Lou
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  • But we can remeasure the photon. If a photon passes once through a filter at angle theta, it will forever pass through such a filter. It is difficult to reconcile that effect with a measurement error. – BowlOfRed Jul 31 '14 at 08:30
  • @bowlofred: but I didn't mean a measurement error in the sense of accidentally reading the wrong value once, but instead interacting with the particle in a way that "collapses" its wave function into a certain state which might not be a true representation of its state while exiting the source. I guess it's a bit vague, now that I read this to my self. – Lou Jul 31 '14 at 09:26

2 Answers2

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For now I will only give you an overview of the ideas involved and show you how you should interpret the idea of a "local realistic theory" that cannot exist at the microscopic scale. Once you've read it, and if you feel you need more mathematical rigor to be convinced, then I will draw you step by step the proof of Bell's inequality (it is not the only one leading to the same claims, just one of the first ones that did so), as it is rather neat.

Einstein reality: Property of system already determined before the measurement. Meaning the system "has them".

Einstein locality: Physical reality described in a local manner. Independently of measurements that are carried out on spatially separated systems: "no action at a distance".

Now Bell's inequality showed that for entangled system, Einstein's description (or maybe expectation) of reality/locality of physical systems are both undermined. Bell's approach:

Assuming that each photon carries a hidden variable $\lambda$ that determines the outcome of polarization experiments at A and B for any angles of the polarimeters $\delta_1$ and $\delta_2$: $$\begin{align} S_A^{\lambda}(\delta_1) = {+1,-1}\\ S_B^{\lambda}(\delta_2) = {+1,-1} \end{align} $$ The two $S$ functions contain the possible outcomes of a polarization measurement (for each system as detailed in the equation), and the result already defined (either -1 or +1) because $S$ is dependent on a hidden variable $\lambda$ providing the measurement outcome before it has taken place.

The variable $\lambda$ itself has a probability density distribution as follows: $$ \rho (\lambda) \ge 0, \int \rho(\lambda)d\lambda=1$$

Now using the classical correlation coefficient (product of $S_A$ and $S_B$ expresses locality): $$\epsilon^{cl}(\delta_1,\delta_2)= \int \rho(\lambda)S_A^{\lambda}(\delta_1)S_B^{\lambda}(\delta_2)d\lambda$$

From this equation, Bell derived his famous inequality (proof of which I was referring to at the start): $$\left|\epsilon^{cl}(\delta_1,\delta_2) - \epsilon^{cl}(\delta_1,\delta_3)\right| \leq 1-\epsilon^{cl}(\delta_2,\delta_3) $$

Having now all the necessary ingredients, the next step is to measure correlation coefficients at different angles $\delta_1, \delta_2, \delta_3$, and see if Bell's inequation holds or not: (if it holds then Einstein's views would have been plausible) Now by choosing: $\delta_1=30°, \delta_2=60°, \delta_3=90°$ Calculating the correlation coefficients in quantum mechanics and then compared.

First definition of correlation in quantum mechanics: $$\begin{align} \epsilon^{AB}(\alpha,\beta) :&= \left<\Phi_+^{AB}\right. \left|E^{A}(\alpha)\otimes E^{B}(\beta) \right| \left. \Phi_+^{AB}\right> \\ \epsilon^{AB}(\alpha,\beta) &= P_{++}+P_{--}-P_{+-}-P_{-+} \\ \epsilon^{AB}(\alpha,\beta) &= \cos2(\beta-\alpha) \end{align}$$ The above are the generalized formulas, where $\alpha$ and $\beta$ are the polarimeter angles, $E^{A}$ and $E^{B}$ are the polarization operators of system of photon A and system of photon B respectively, $\Phi_+^{AB}$ (entangled state of choice) is one of the 4 Bell states (for 2 particle systems) and $P_{++},...$ are the probabilities to measure both polarization as horizontal, $--$ for measuring both vertical polarizations and so on. To reach the simplified formula with $\cos$, just calculate each term in the second equation (using the first equation).

Back to our measurements, now using $\epsilon^{AB}(\alpha,\beta) = \cos2(\beta-\alpha)$ we have: $$\epsilon^{AB}(\delta_1,\delta_2)=\frac{1}{2}, \epsilon^{AB}(\delta_1,\delta_3)=-\frac{1}{2}, \epsilon^{AB}(\delta_2,\delta_3)=\frac{1}{2} $$ Inserting the results back into Bell's inequality yields: $1 \leq \frac{1}{2}$

It is clear that Bell's inequality is violated using the quantum mechanical definition of the correlation coefficient, meaning the Quantum theory and local-realistic theories leads to contradictory results.

To sum up, it was shown that there cannot be a so called "hidden" variable for each measurement that would predict the outcome before it is actually performed. Which brings us to the correct assessment of entangled states which is:

"The quantum state of each particle cannot be described independently, and measurements can be correlated even if the two entangled systems are light years apart."

Ellie
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  • +1, but I think that statement about Einstein should be: "at least one of Einstein's reality/locality has to be undermined"? Correlation itself doesn't imply causation, and I think the accepted view is that "action at a distance" is not one of the feasible explanations of the phenomena. – Lou Jul 31 '14 at 10:24
  • Actually it does deny both assertions, as Bell's inequality was derived from the classical definition of correlation coefficient which itself already entailed both reality and locality of Einstein, to see why, just take a look at the expression of $\epsilon^{cl}(\delta_1,\delta_2)$, dependance of $S$ functions on $\lambda$ fulfills the reality condition and the product of $S_A$ and $S_B$ expresses the locality. – Ellie Jul 31 '14 at 11:28
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    But isn't that's exactly why if only one of them is invalid, the expression doesn't hold? – Lou Jul 31 '14 at 11:32
  • I think the point is best made clear by showing the derivation of the inequality, I will write it out for you as soon as I find the time (rather lengthy). – Ellie Jul 31 '14 at 11:34
  • Thanks, no need to waste your time writing all that, I am sure I can find detailed math proofs online, this is quite enough. – Lou Jul 31 '14 at 12:27
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You may be thinking of the particles as idealized coins that are in identical initial states, and measuring along a certain angle as flipping a coin in a certain way. As long as the world is deterministic, they will produce identical results if flipped in the same way, and probably similar results if flipped in similar ways. That's fine, but there's nothing random about it: the result is a deterministic function of the measurement angle and a value representing the initial state of the coin. That's enough to prove Bell's result.

The pioneers of quantum mechanics probably shared your intuition that when flipping the coins in similar but not identical ways, the results could be correlated enough to match the quantum prediction. Otherwise one of them would have proven this theorem long before Bell. But that intuition is wrong, provably. If you find Bell's argument too difficult to follow, take another look at my simplified three-angle version from the previous question.

benrg
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