The Free Will Theorem

Today we prove Conway-Kochen free will theorem:

SPIN + TWIN + MIN + (very limited) free choice of the experimenters 

= free choice of the particles

SPIN is just a basic statement about the behavior of a spin-1 particle. It states that angular momentum operators exist and they behave in such a way to make the 101 property be true (see below).

TWIN states that it's possible to entangle two particles, such that their spins, when measured in the same direction, are opposite.

MIN is stated rather obscurely. I believe it means that it is impossible for an event to depend on another event outside of its past light cone. This is basically what special relativity states.

Free choice is defined as "non-functional", that is, not described by a function. In this interpretation, to say I have no free choice in going left or right, is to say that there is a function $f$, such that the direction I am going is $f($everything in my past lightcone$)$.

This proof goes in two steps. The first step is the Kochen-Specker Theorme uses only SPIN. The second step uses TWIN and MIN to construct an entanglement separated by a very long distance (like all those Bell-inequality experiments), and then assume (a very limited amount of) free choice of the experimenters, but not the particles, to get a contradiction.

Step 1: Kochen-Specker Theorem (1966)

This is well-known and I will direct you to plus magazine's proof. First read this, then read this. For those who know a bit more quantum mechanics, here's what 101 property means: Consider a spin-1 particle. Let $S_x$ be the operator of the angular momentum along vector $x$ for the particle, then $S_x$ has three possible eigenvalues: $\hbar, 0, -\hbar$. Normalize by setting $\hbar = 1$, we find that $S_x^2$ has two possible eigenvalues: $0, 1$. Then, it can be shown that for any triple of orthogonal vectors $x, y, z$, we have $S_x^2+S_y^2+S_z^2 = 2$, and so the measurement results must be one of $(1, 1, 0), (1, 0, 1), (0, 1, 1)$.

Notice that since $S_x^2 = S_{-x}^2$, we can safely consider a direction as defined by a line through the origin, rather than a vector.

Another note: sometimes, the configuration of 33 lines is called the Peres configuration. It's easy to verify that, if we represent each line as a vertex, and connect two vertices iff they represent orthogonal lines, then we obtain a graph with 72 edges, making up 16 triangles (corresponding to triple-orthogonal-lines) and 24 edges that do not make up any triangle.

The Stanford Encyclopedia contains more variations and ways to escape the conclusion of the Kochen-Specker theorem.

The best proof of Clairaut's theorem on equality of mixed partials

This is the best proof of Clairaut's theorem on equality of mixed partials that I have ever thought up. And I bet it's the best proof of the theorem you'll ever see!

I'll prove the 2-variable version, but the generalization to $n$-variables is obvious.

Clairaut's Theorem: Suppose $f$ is a real-valued function of two variables $x,y$ and $f(x,y)$ is defined on an open subset $U$ of $\mathbb{R}^2$. Suppose further that both the second-order mixed partial derivatives $\partial_x\partial_y f(x,y)$ and $\partial_y\partial_x f(x,y)$ exist and are continuous on $U$. Then, we have: $$\partial_x\partial_y f = \partial_y\partial_x f$$ on all of $U$.