We have two particles, A and B, that are entangled to have spin 0, which means that if we measure A’s spin in one direction (resulting in up/down) and B’s spin in the opposite direction, we’ll always always always get equal answers (i.e. always get up-up or down-down).

How does B know to give the same result as A? It must be that either:

  1. Some information was transmitted to B when we measured A, possibly faster than light (which would be crazy). Or,

  2. When the particles got entangled, they agreed, for every direction, which of them would measure up in that direction and which would measure down. So they agreed on a globe where every point is colored either Amber (A measures up in that direction, and B down) or Blue (B measures up, and A down), and each particle carries around that globe so it knows how to respond when we ask its spin.

Since those are the only two possibilities, and (1) is crazy, (2) must be true. Now let’s prove that (2) is false.

We can think of our entanglement process as being a random variable from which we sample colored globes. For each globe, we can measure the colors of two points: one point by running Particle A through a detector, and one point by running Particle B through a detector.

Let’s consider three directions:

  • \(N-\theta\),
  • \(N\),
  • \(N+\theta\),

for some tiny \(\theta\). By pointing a detector in one of these directions and running a particle through it, we can measure the color of that particle’s globe in that direction. So each of these directions gives rise to a random variable from which we sample colors. Call these three random variables X, Y, and Z, respectively.

Let’s point one detector \(N-\theta\), and one detector \(N\), and run a bunch of particle-pairs through them, sampling from X and Y simultaneously. We can measure \(p := P(X \ne Y)\).

Likewise, we can confirm that \(p = P(Y \ne Z)\), because theta is small enough that everything is locally linear.

Lemma: for any random variables X,Y,Z, \(P(X \ne Z)\) can’t be more than \(P(X \ne Y) + P(X \ne Z)\). Proof left to reader.

So, if we point one detector \(N-\theta\) and the other one \(N+\theta\), they can’t disagree more than \(2p\) of the time.

Except, wait. Quantum mechanics, and experiment, both indicate that P(disagree) goes up like the square of the angle between the detectors, i.e. in our final configuration the detectors disagree \(4p\) of the time. This is completely impossible using the predetermined-colored-globe model.

So, our options are: (1) FTL information transmission, or (2) something that experiment shows to be false. Shoot.