The Function of the Free Will Assumption in No-Go Theorems 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?
 A: 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.
A: 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 $|\psi\rangle|X\rangle\to(\alpha|\psi_1\rangle+\beta|\psi_2\rangle)(\alpha|\psi_1\rangle+\beta|\psi_2\rangle)$).
