CHSH inequality requires both locality and realism. I will equate here realism with counterfactual definiteness.

Now counterfactual definiteness tells us that given two different measurements on the same object, described by random variables $C_1$ and $C_2$, there exists a joint probability distribution for $C_1$ and $C_2$ (this is not always the case, search for the marginal problem and indeed we know that outcomes of measurements of noncommuting observables do not posses a joint probability distribution). Now if we can assume the existence of a joint probability distribution, then expectation values $E(C_1) + E(C_2)$ may be joined together to have $E(C_1 + C_2)$.

So suppose that we now have four random variables $A_1$ and $B_1$ local to Alice and $A_2$ and $B_2$ for Bob, which can take values $\pm 1$. The expression in the CHSH inequality is $$ |E(A_1 A_2) + E(A_1 B_2) + E(B_1 A_2) - E(B_1 B_2)| $$ Now if we can assume realism (counterfactual definiteness), there exists a joint probability distribution for the outcomes of all four random variables, and we can join the expectation values in the above together to get $$ \Bigl\lvert E\bigl(A_1 (A_2 + B_2) + B_1 (A_2 -B_2)\bigr)\Bigr\rvert $$ So now we do the standard trick where either $A_2$ and $B_2$ are equal or they are opposite, so that the first term is either $\pm 2$ and same for the second term.

Now to my question. I obviously used the realism assumption in the above. I assume locality means that the marginal distributions for $A_1$ and $B_1$ are independent of the choice of random variables Bob makes, $A_2$ or $B_2$ (but I otherwise allow for the correlation of outcomes, as long as it's independent of measurement choice, as there may be hidden variables that were encoded at the source of the state that produce correlations). Where did I use the locality assumption in this derivation? I would like if you either point out precisely where this assumption is needed in this calculation or argue that it is not needed with a convincing justification.

Edit: Given the answer below, it seems likely that the locality assumption may have been used at the point where we take $E(C_1) + E(C_2) = E(C_1 + C_2)$. However, it is still not understood why realism is not sufficient to reach this conclusion without the assumption of locality. As stated above, the question is why is the above reasoning not correct and locality assumption is needed, or alternatively I am also open to the idea that it might not be needed.

Edit 2: I have received a satisfactory answer to this question over at MathOverflow from @Steven Landsburg.

The existence of a joint probability distribution for all four variables is definitely sufficient to get the inequality. The existence of joint probability distributions for (A1,B1) and (A2,B2) (i.e. "realism") is definitely not sufficient to imply a joint probability distribution for all four. Locality is a sufficient additional assumption.

He also shows there that counterfactual definiteness for local outcomes is not a sufficient condition to obtain the CHSH inequality.


2 Answers 2


I give a full derivation and touch on this point in this other answer, but to give a quick summary here, the main idea is that the locality assumption allows us to write $$E_\lambda(A_x B_y)=E_\lambda(A_x)E_\lambda(B_y),$$ where $E_\lambda(A_x)$ denotes the expectation value of the output $A$ (say, Alice's output) for the measurement setting $x$, for a given value of the hidden variable $\lambda$, and $E_\lambda(A_x B_y)$ is the expectation value of the product of the outputs observed by Alice and Bob for measurement settings $x$ and $y$, respectively.

Once you can write this (which is just a rewriting of the locality assumption in terms of expectation values), it follows that $$S_\lambda= E_\lambda(A_0)[E_\lambda(B_0)+E_\lambda(B_1)] + E_\lambda(A_1)[E_\lambda(B_0)-E_\lambda(B_1)],$$ and thus you get $\lvert S_\lambda\rvert\le2$ and $\lvert S\rvert\le2$.


I didn't think about it long, but it's surely the assumption $E(C_1)+E(C_2)=E(C_1+C_2)$ is the culprit, and CHSH is just the same as Von-Neumann's argument against hidden variables. The failure is analyzed by Bell, in relation to Bohm's theory, in the book Speakable and Unspeakable in Quantum Mechanics.

The reason $E(C_1+C_2) = E(C_1) + E(C_2)$ is local is because you are assuming the measurements of 1 and 2 are independent to conclude that the expected values add up like this. Otherwise the measurement of $C_1$ could affect $C_2$. Bohm's theory is enough to see there is no argument independent of locality.

I don't like the CHSH inequality, because it is a rehash of Von-Neumann, it's not original, and it fails in the same way that was explained in detail by Bell using Bohm.

  • 1
    $\begingroup$ But if $C_1$ and $C_2$ are counterfactually definite random variables with some joint distribution $p(c_1, c_2)$ then $E(C_1 + C_2) = E(C_1) + E(C_2)$ is a basic mathematical fact, following from the linearity of the expectation operator. You can apply this here because there exists a single distribution governing both observables. $\endgroup$
    – SMeznaric
    Sep 11, 2012 at 16:56
  • $\begingroup$ Also, local measurements need not be independent - this is a very strong assumption, stronger than local realism. Bell's integral expression for $\int d\lambda p(x_1|\lambda) p(x_2|\lambda) \rho(\lambda)$ for example allows for classical correlations between measurement outcomes. States such as $\rho = \sum_k p_k |k><k| \otimes |k><k|$ for example do not violate the CHSH inequality, but do posses non-independence between some measurement outcomes of Alice and Bob. $\endgroup$
    – SMeznaric
    Sep 11, 2012 at 17:07
  • $\begingroup$ @SMeznaric: This is not so--- you need to consider measuring "C1" measuring "C2" and measuring "C1+C2", these are three different measurements. This is discussed at length by Bell. When I said "independent", I didn't mean "uncorrelated", I just meant "causally independent". $\endgroup$
    – Ron Maimon
    Sep 11, 2012 at 19:39
  • $\begingroup$ Perhaps I don't fully understand your point. If you read about the expected value of random variables here you can see that it is linear in random variables whenever the variables are functions from the same event space (actually the page does not even deal with the case where there is no joint distribution corresponding to the given marginals, they simply take this existence for granted, although it is not guaranteed). $\endgroup$
    – SMeznaric
    Sep 11, 2012 at 20:01
  • $\begingroup$ @SMeznaric: The assumption is not justified because measuring the sum of quantum operators is not measuring the sum of classical observables representing hidden variables. The correspondence does not respect addition of operators. This is discussed by Bell at length in Speakable and Unspeakable, it's the same assumption that fails in Von-Neumann. $\endgroup$
    – Ron Maimon
    Sep 11, 2012 at 20:22

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