I recently watched a video from Brian Greene that goes over an example of Bell’s inequality. Video here. In this example, he imagines the generation of a pair of entangled particles, each of which is correlated in such a way that the spin of one is the opposite to the other when measured. He says that if there is a local hidden variable, then the spin of each particle is determined and the measurement process simply reveals the outcome of what is determined. This allows us to compute all the different kinds of combinations of spins that can be measured in different axes. This in turn allows us to come up with Bell’s inequality, a condition that must not be violated if this variable indeed exists.

Borrowing words from another answer on this stack exchange,

If we have local hidden variables, then there's no real "collapse". When the particles were created, they had definite "plans" for what they would do if they encountered a detector aligned along any conceivable axis. What we call a "singlet state" is just a conspiracy that the two particles entered into when they were created, comparing each other's "plans" and making sure that if particle A came out of a detector spin-up, then particle B would come out of the same kind of detector spin-down.”

My question is the following: What if these hidden variables don’t do this in a definite way? That is to say, is it possible for there to be a local hidden variable for each electron such that through the course of its path until the measurement device, its spin is not definite. In other words, the spin may change throughout that path, but at any particular moment, the spin is definite.

Secondly, let’s assume that we are measuring the spin on a particular axis. Is it possible for the initially prepared entangled pair to function in a way where once each electron heads towards its detector, there is some variable that changes the spin of either one or both of the electrons such that they are not inverse correlated anymore?

In other words, let’s assume that on one particular axis, one electron in a pair that is generated has spin up and the other spin down. Can it now be the case that by the time each electron is measured, on a particular axis, each electron’s spin is measured up in each measurement device?

In reference to the quote mentioned above, could the “conspiracy plan” simply change in some sort of local hidden way?

  • 2
    $\begingroup$ The observed results violate Bell's theorem. No theory---including yours---can change that fact. So I'm unclear on what problem your theory is supposed to solve. $\endgroup$
    – WillO
    Nov 9, 2023 at 15:49
  • $\begingroup$ @WillO My question is that if there is some local hidden “plan” within each particle of this initially entangled pair that allows spins to change, independently for each electron, must the experimental results due to this still not violate Bell’s inequality? $\endgroup$
    – user383648
    Nov 9, 2023 at 15:56
  • $\begingroup$ @user383648 Your asking the right questions but don't get sidetracked by the term entangled. You are correct to describe this as correlating the two particles. Two photons perfectly correlated, have all their variables correlated. Polarization is the obvious variable but you also have speed, frequency, phase, direction and linear dependency (or timing) from the source to the analyzer, to the detector. One additional variable never mentioned (hidden) is related to frequency and timing. If the photons energy oscillated up and down then transiting a polarizer would be completely linear dependent. $\endgroup$ Nov 14, 2023 at 19:05

3 Answers 3


You’re right that Greene oversimplified the potential role of hidden variables, but this oversimplification doesn’t change the general conclusion of Bell’s Theorem.

Greene was imagining that if you knew the hidden variable $\lambda$ on one side of the experiment, that you could predict exactly what outcome would occur for a given measurement on that side, with certainty. This is overly restrictive. Since quantum measurements are generally probabilistic anyways, you seem to be asking whether you could instead devise a hidden variable theory where knowing $\lambda$ exactly still only allowed you to make a probabilistic calculation of the outcomes. The answer is yes, but it doesn't help.

Mathematically, if $a$ is the measurement setting on that side of the experiment, and $A$ is the outcome, it’s certainly logically possible to build a hidden variable model that utilized some probability distribution $P_a(A|\lambda)$, some probabilistic measurement-outcome rule. (Here I’ve put the "$a$" as a subscript indicating that the setting is an externally-controlled input, not derived by the model itself, a notation advocated in this Rev Mod Phys review of the topic.)

Similarly, on the other side, you would then have some probabilistic measurement rule $P_b(B|\lambda)$.

You are also asking whether $\lambda$ itself might change on each wing of the experiment. And certainly, that would be reasonable to allow; any serious attempt at a hidden variable model would have to do this to have any hope of matching actual experiments. (After all, I could take one of the two particles and rotate it by 90 degrees; certainly the corresponding hidden variables that pointed in any particular direction would then presumably want to also rotate by the same amount.)

To generalize this you could invent rules (even probabilistic rules, if you wanted) that would explain how the situation one side of the experiment transforms $\lambda$ to become $\lambda_a$, and how the situation on the other side of the experiment transforms $\lambda$ to become $\lambda_b$. Combined with these rules, the measurement rules would then be updated to $P_a(A|\lambda_a)$ and $P_b(B|\lambda_b)$.

But Bell’s Theorem still goes through in the usual way for such a model. All one needs is the locality assumptions that the original $\lambda$ screens off $b$ from $\lambda_a$, and also screens off $a$ from $\lambda_b$. Specifically, the assumption that $P_{a,b}(\lambda_a|\lambda) = P_{a}(\lambda_a|\lambda)$ and $P_{a,b}(\lambda_b|\lambda) = P_{b}(\lambda_b|\lambda)$. In other words, fiddling with the settings over on one side of the experiment isn’t going to change how $\lambda$ is evolving on the other side.

Topping the assumptions off with the usual no-retrocausality assumption, that $P_{a,b}(\lambda) = P(\lambda)$, even with the above generalizations, you can still prove Bell’s Theorem. So you can’t explain quantum experiments without either violating one of those screening-off-conditions (allowing the setting on one side to influence what’s happening on the other), or instead violating the no-future-input-dependence condition, allowing retrocausal influences of the settings on the original $\lambda$. Nonlocality or retrocausality; take your pick. There’s no easy way out of the dilemma.

  • $\begingroup$ "Greene was imagining that if you knew the hidden variable $λ$ [...] that you could predict exactly what outcome would occur [...] with certainty. This is overly restrictive." - Greene was following Bell, who intended $λ$ to include whatever randomness the theory needs (such as a single real number in $[0,1)$, which is an unlimited supply of random bits). That's the reason for the integral over $ρ(λ)dλ$. It's not restrictive, it's just a definition. $\endgroup$
    – benrg
    Nov 14, 2023 at 18:04
  • $\begingroup$ Yes, he was following Bell (1964), but that needs another step. When outcomes can be predicted with certainty (say, for identical settings) "it follows that the result of any such measurement must actually be predetermined", and from there Bell could define the outcome as $A(a,\lambda)$, a deterministic function of the setting $a$ and the precise value of $\lambda$. But by 1976 Bell dropped this logical sequence, allowing that the outcome $A$ might be determined by a probabilistic function $P(A|a,\lambda)$, a function distinct from $\rho(\lambda)$. This is certainly more general. $\endgroup$ Nov 14, 2023 at 19:59

There's an important difference between what Greene said:

When the particles were created, they had definite "plans" for what they would do if they encountered a detector aligned along any conceivable axis.

and what you said:

is it possible for there to be a local hidden variable for each electron such that [...] its spin is not definite.

Greene is not saying that the particles have definite spins, but that they have definite plans. The concept of a "plan" is very general. You can think of it as an algorithm each particle follows to decide whether to return "up" or "down". It can use the time of the measurement as input, so if you want to base the result on a time varying spin you can do that; but there needn't be any notion of spin in the algorithm at all. All that matters for Bell's theorem is that it returns an answer of "up" or "down" to any question.


In fact Bell's theorem is mixing up words, for example $$A(a,\lambda)B(b,\lambda)$$ is said to be local because A depends only on a and not b. But this a is an experimental parameter. Whereas lambda is present at the same time in A and B, since lambda is the hidden variable, it is nonlocal in this formula.

So that Bell's theorem judges with a local/nonlocal formula. It is like if you should choose between red and green and youcompare with blue.

One could put different lambdas in each wing.

But the cruciality is in the temporality, it assumes $$A(a,\lambda)B(b,\lambda)-A(a,\lambda)B(b',\lambda)$$ which should be read $$A(a,\lambda_a)B(b,\lambda_b)-A(a,\mu_a)B(b',\mu_{b'})$$

In words Bell's theorem tells that if the cause is always and everywhere the same, then CHSH is smaller than 2.


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