# The Function of the Free Will Assumption in No-Go Theorems [closed]

I'm interested in the function played by the free will assumption made by any number of no-go theorems in quantum mechanics. While searching the archives for prior questions, I found this one from 5 years ago.

The most highly rated answer includes a claim, "To assert that the indeterminacy of measurement results is, in some way, equivalent to a notion of "free will" ... is a proposition that is not grounded upon any physical principle", that I find intriguing but dubious. Could we not work our way backwards to infer the physical principle involved?

It seems that the function of the free will assumption in Bell-type no-go theorems is to secure or underwrite the claim that measurement settings are free variables; meaning, in the words of J. S. Bell, that "the values of such variables have implications only in their future light cones".

This seems to presuppose the assumption that may be taken as a principle of physics (applicable, at least, to experiments in which Alice and Bob measure the spin of entangled particles at a space-like separation); namely, that physicists are capable of designing and performing physics experiments in which measurement device settings are free variables.

This assumption seems uncontroversial; but, if made, it naturally invites the question as to what gives physicists this capability; and, it would seem that we could legitimately call the source of that capability the free will of the physicist.

True or false?

## closed as primarily opinion-based by safesphere, Jon Custer, ZeroTheHero, Kyle Kanos, tomAug 22 at 16:04

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

• I recommend not using the term free will on this site, because it does not have generally accepted meaning in physics, and in philosophy there are very different views on the concept of free will, some consistent with deterministic universe, some rejecting it, so using it in some specific sense that half people won't agree with to discuss Bell's theorem is confusing and unnecessary. – Ján Lalinský Aug 20 at 21:48

NanoBotic,

The relevant assumption that is used in Bell's theorem is statistical independence. The theorem requires that the states of the detectors and the hidden variables are independent on each other. Free will is not an absolute requirement. For example one can still have independence if detectors and the hidden variables are set according to decimals of Pi, sqrt(2) and sqrt (7) respectively.

It is simple to ensure that the detectors' settings are independent of each other because we can control them directly. The hidden variables, on the other hand are uncontrollable and it is not that easy to make sure that they are independent of the detectors. So, in the absence of a good argument based on physical principles some physicists try to play the emotional card of human free will. What they want is to say that there is no way the hidden variables could be correlated with the measurement settings because those are "freely chosen" just before detection, when it is to late for the hidden variables to be influenced by some local mechanism. Needless to say, this is question-begging, because this type of free will is incompatible with determinism and Bell's theorem is supposed to rule out deterministic hidden variable theories.

• Another alternative would be the "no conspiracies" assumption favored by some theorists. As long as the measurement device settings ended up as free variables, I doubt that the reliability of the theorem results would be affected. – NanoBotic Aug 15 at 17:28
• I think that statistical independence has nothing to do with conspiracies. There are physical systems that are independent in regards to some property (like the outcomes of two coins flips) and some others that are not independent (like the positions of the planets relative to the star they orbit). The assumption that the hidden variables and the detectors' settings are like the coins and not like the planets is just begging the question. – Andrei Aug 16 at 13:38

Nope. Pretentious diction of that answer aside, there are no assumptions of "free will" in any no-go theorem of quantum mechanics.

For 3 directions of measurement (a, b, and c) in a plane with the average product of measured spins $$P(\mathbf{a},\mathbf{b})$$, Bell's theorem states that the quantum prediction $$P(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot\mathbf{b}$$ is incompatible with the prediction of any local hidden variable theory $$|P(\mathbf{a},\mathbf{b})-P(\mathbf{a},\mathbf{c})|\le 1 + P(\mathbf{b},\mathbf{c})$$. Free will is not assumed, just apparent wavefunction collapse (which is now understood to arise from quantum decoherence).

The "no-clone" no-go theorem requires even less. If some unitary operator clones any state $$|\psi\rangle$$ onto blank state $$|X\rangle$$, $$|\psi\rangle|X\rangle\to|\psi\rangle|\psi\rangle$$, then it linearly distributes over the superposition $$|\psi\rangle=\alpha|\psi_1\rangle+\beta|\psi_2\rangle$$ as $$|\psi\rangle|X\rangle\to\alpha|\psi_1\rangle|\psi_1\rangle+\beta|\psi_2\rangle|\psi_2\rangle$$, thus failing to yield the desired clone (which was $$|\psi\rangle|X\rangle\to(\alpha|\psi_1\rangle+\beta|\psi_2\rangle)(\alpha|\psi_1\rangle+\beta|\psi_2\rangle)$$).