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As well-known, John S. Bell proved in 1964 that local realism (as espoused e.g. by Einstein) was incompatible with the predictions of quantum mechanics. His proof involves the famous Bell inequality. There are later proofs which do not use inequalities but none of them is as simple as the one I'm going to put forward (though I guess my proof is wrong and hope you'll help me see why).

I assume quantum mechanics to reveal its incompatibility with local realism.

Consider a pair of non-maximally entangled spins

$$\sqrt{2/3} |+ -\rangle + \sqrt{1/3} |- +\rangle\,.$$

Consider experiment A. We produce many so entangled pairs and only the value of the x component of the first spin is measured in each case. The prediction is that the value +1 will be found about 2/3 of the time.

Consider now experiment B. Again we produce many such pairs. On each occasion, first the value of the y component of the second spin is measured; then the value of the x component of the first spin is measured. The measurements are separated by a space-like interval. As the values of the spin can never coincide, once we measure the y component of the second one, we know with certainty the value of the y component of the first one; and as different components of the spin do not commute, once the value of the y component of the first spin is known, the value of its x component becomes absolutely uncertain; hence, its measurement will now yield value +1 about 1/2 the time, not 2/3 as in experiment A.

Comparing both experiments, we see that in experiment B we have managed to alter the probability distribution for the values of the x component of the first spin and, if locality holds, we've achieved this without in any way disturbing the first spin. This is incompatible with those values' being an intrinsic element of reality of the spin; that is, it is incompatible with realism.

Thus, locality-plus-realism is incompatible with quantum mechanics.

As the reasoning is extremely simple and I haven't seen it published anywhere, it is surely flawed. However, I can't find the flaw.

Can someone find it?

Thanks.

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    $\begingroup$ Do $+$ and $-$ refer to spins in the $x$ direction? Then your first mistake is in the paragraph beginning "Consider now experiment B" where you write "Once we measure the $y$ componenet of the second one, we know with certainty the value of the $y$ component of the first one". $\endgroup$ – WillO Oct 11 '16 at 12:01
  • $\begingroup$ There's no flaw, Bell's theorem rules out local hidden variable theories. $\endgroup$ – Yogi DMT Oct 11 '16 at 13:58
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First a quick check:

Comparing both experiments, we see that in experiment B we have managed to alter the probability distribution for the values of the x component of the first spin

This would imply not only Bell's theorem but also faster-than-light communication: The one measuring the first spin would know whether the second spin was measured if given a large enough sample. Therefore, it is safe to assume your answer must be wrong and what remains is to find the error.


Then in detail: As WillO pointed out in the comments, the error is here:

As the values of the spin can never coincide, once we measure the y component of the second one, we know with certainty the value of the y component of the first one

I'll use $|x_\pm\rangle$ and $|y_\pm\rangle$ to denote the spin-eigenstates in the $x$ and $y$ direction.

Your initial state is $$ |\psi_0\rangle = \sqrt{\frac{2}{3}} |x_+ x_-\rangle + \sqrt{\frac{1}{3}} |x_- x_+\rangle. $$

Using $|x_+\rangle = \frac{1-i}{2} |y_+\rangle + \frac{1+i}{2} |y_-\rangle$ and $|x_-\rangle = \frac{1+i}{2} |y_+\rangle + \frac{1-i}{2} |y_-\rangle$ (as can bee see e.g. from these representations), we get

$$ \begin{align*} |x_+ x_-\rangle &= \left(\frac{1-i}{2} |y_+\rangle + \frac{1+i}{2} |y_-\rangle\right) \left(\frac{1+i}{2} |y_+\rangle + \frac{1-i}{2} |y_-\rangle\right) \\ &= \frac{1}{2} |y_+ y_+\rangle + \frac{-i}{2} |y_+ y_-\rangle + \frac{i}{2}|y_- y_+\rangle + \frac{1}{2}|y_- y_-\rangle, \\ |x_- x_+\rangle &= \frac{1}{2} |y_+ y_+\rangle + \frac{-i}{2} |y_- y_+\rangle + \frac{i}{2}|y_+ y_-\rangle + \frac{1}{2}|y_- y_-\rangle. \end{align*} $$ So you'll see that $|\psi_0\rangle$ does have nonvanishing factors for components $|y_+y_+\rangle$ and $|y_-y_-\rangle$. Therefore your claim that the the measured spins in the $y$ direction can't be the same is false.

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  • $\begingroup$ JiK has elaborated on WillO's answer and has completely clarified the problem. Thanks. $\endgroup$ – Laureano Luna Oct 11 '16 at 15:24
  • $\begingroup$ it seems to me you have used the Pauli matrices to achieve a change of basis. Can you explain the logic behind that operation, please? $\endgroup$ – Laureano Luna Oct 13 '16 at 12:21
  • $\begingroup$ Upon reflection, I wonder whether the explanation by JiK holds for pairs of entangled electrons whose total spin is required to be zero by conservation of angular momentum. I quote from the page linked to below: "Assume that angular momentum conservation dictates that the total spin of the two electrons is zero. The magnitude of the total spin is zero and the component of the total spin measured along any axis is zero". electron6.phys.utk.edu/phys250/modules/module%203/… $\endgroup$ – Laureano Luna Dec 3 '16 at 1:03

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