In a previous post, I talked about the uncertainty relations. The theme there was the way that geometry undergirds quantum probability. This post and the one that follows introduce a related perspective. Even though I don’t share his Bayesian outlook, the discussion owes much to a paper by the late and much-missed Itamar Pitowsky.*

Imagine a set-up like the one in the picture:

The green circles at the vertices of the triangles are lights. The one labeled “l” is on the others are off. Below the seven linked triangles you see a screen (on the left) a dial with seven positions, and a button below the dial. Here’s the idea.

You type a string of digits and letters; they appear on the screen. Then you choose one of the 7 triangles and set the dial so that it points to the corresponding number. When you push the button, exactly of the lights at the vertices of the triangle lights up. In the picture, the triangle was number 7 and light l came on. For a given input string and triangle, it’s generally not possible to predict which light will light up, but given the input and the choice of triangle, there’s a definite probability for each of the relevant lights to light up. (We can ignore the rule that fixes the probabilities.) For example, it might be that for input B35D68, the probabilities for lights j, k and l are p(j) = .2, p(k) = .35, p(l) = .45.

We need one more stipulation: given a choice of input, the probability for a light to come on doesn’t depend on which triangle pick. For example: we supposed that for input B35D68 and dial setting 7, p(k) = .35. The stipulation says that p(k) would also have been .35 if the dial had been set to 5.

What does this have to do with quantum mechanics?

For quantum spin-one systems (don’t worry about exactly what those are), experiments have three possible outcomes. To connect this post with the previous one, the possible outcomes of any experiment could be represented by one of three mutually perpendicular lines. But in a three-dimensional space, a single line belongs to many different sets of three mutually perpendicular lines. The physical correlate: there are distinct experiments on spin-one systems that “overlap”: there’s an outcome of one experiment represented by the same line as one of the outcomes of the other experiment. This means that no matter what state the system starts in, these two outcomes will have the same probability.

In our set-up, each triangle each can be paired with a three-outcome quantum experiment. Vertices of a triangle correspond to triples of perpendicular lines (“eigenvectors,” for the cogniscenti.) Each code number picks out a quantum state, and the cosine-squared rule gives the probabilities. But we can leave those details in the background. What matters are the relations of overlap among the triangles.

Our set-up could be powered by a quantum system. Each run of an “experiment” could mirror what happens when we pick one of 7 appropriately-chosen possible experiments, on a spin-one quantum system. And what follows rests on these two rules:

(1) If three vertices form a triangle, their associated probabilities sum to 1

(2) If a vertex belongs to more than one triangle, its probability is same no matter which triangle we associate it with.

Applying (1) and (2) to our diagram, we get this conclusion:

For any input,

3) p(a) + p(m) ≤ 3/2.

Here’s why. Think about the triangles (a,f,g) and (i,h,m). (1) tells us that A) p(a) + p(f) + p(g) + p(i) + p(h) + p(m) = 2. Now try to suppose p(a) + p(m) > 3/2. We’ll see that this assumption is impossible. Notice first that if p(a) + p(m) > 3/2, then p(f) + p(g) + p(i) + p(h) < 1/2. Otherwise, the total in A) will be bigger than 2. But if

p(f) + p(g) + p(i) + p(h) ≤ 1/2

then we automatically have

p(g) + p(h) ≤ 1/2.

An exactly similar argument shows that

p(c) + p(d) ≤ 1/2.

Now consider triangles (g,h,k) and (c,d,j). Since p(g) and p(h) don’t depend on the triangle we associate with them, and since they add to less than 1/2, we must have p(k) > 1/2. By an exactly similar argument, p(j) > 1/2. But now the triangle (j,k,l) violates (1):the probabilities already add to more than 1. So our assumption that p(a) + p(m) > 3/2 has to be wrong. We have

3) p(a) + p(m) ≤ 3/2.

As Pitowsky points out, this is a kind of uncertainty relation. Being certain of *a* constraints the probability of *m*, and vice-versa. We can’t be certain of *a* and *m* at the same time. Notice, however: being certain that a would come on if we picked triangle 1 or 4 doesn’t rule out the possibility that light *m* would come on if we picked triangle 3 or 6 instead — at least, not as far as 3) itself is concerned.

What’s fascinating, as Pitowsky stresses, is that the probability constraints emerge from the combinatorics. It’s a matter of how outcome sets for different possible experiments fit together. This fits with the theme of the previous post: quantum uncertainty isn’t about measurement disturbance. It’s a classically-unexpected dependence relation among possible experiments. In the next post we’ll see just how surprising this is from a classical point of view.

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*”Betting on the Outcomes of Measurements: a Bayesian Theory of Quantum Probability,” Studies in the History and Philosophy of Modern Physics, 34 (2003) 395 ff.