Suppose that the subspaces A and B represent two possible outcomes of the same measurement. What measurement outcome is represented by the span AUB of A and B? (You will remember that the span of A and B is the smallest subspace containing both A and B.)
The following observations are relevant here. Let p(A) and p(B) be the respective probabilities of obtaining the outcomes represented by A and B. Because a line that is contained in either A or B is contained in AUB, we have that
p(AUB) = 1 whenever [p(A) = 1 or p(B) = 1].
Because a line orthogonal to both A and B is orthogonal to AUB, we also have that
p(AUB) = 0 whenever [p(A) = 0 and p(B) = 0].
This holds, in particular, if A and B represent disjoint (non-overlapping) intervals A and B in the range of a continuous variable Q. What is important here is that a line can be in AUB without being contained in either A or B. This means that the outcome AUB can be certain even if neither A nor B is certain. Obtaining the outcome AUB therefore does not imply that the value of Q is either A or B, let alone a definite number in either A or B.
Imagine two perfect — one hundred percent efficient — detectors D(A) and D(B) monitoring the two intervals. If the probabilities p(A) and p(B) are both greater than 0 (and therefore less than 1), then it isn’t certain that D(A) will click, and it isn’t certain that D(B) will click. Yet if p(AUB) = 1, then it is certain that either D(A) or D(B) will click. How come? What makes this certain?
The answer lies in the fact that quantum-mechanical probability assignments are invariably made on the (tacit) assumption that a measurement is successfully performed; there is an outcome. For instance, if A and B are disjoint regions of space, and if a measurement has indicated the presence of a particle in the union of these regions, then the tacit assumption is that a subsequent position measurement made with two detectors monitoring A and B, respectively, will yield an outcome — either of the detectors will click.
So there is no mystery here, but the implication is that quantum mechanics only gives us probabilities with which this or that outcome is obtained in a successful measurement. It does not give us the probability with which a property or value is possessed, independently from measurements, nor does it allow us to calculate the probability with which an attempted measurement will succeed. Hence it is incapable of formulating sufficient conditions for the success of a measurement.
Now consider the following two measurements. The first, M1, has three possible outcomes: A, B, and C. The second, M2, has two: AUB and C. We therefore have
p(A) + p(B) + p(C) = 1 as well as p(AUB) + p(C) = 1.
It seems reasonable to assume non-contextuality, which means that it does not make a difference what the other possible outcomes (besides C) are — only AUB or A and B. If this assumption is correct, then p(C) is the same in both equations, and
p(AUB) = p(A) + p(B).
Let us remember our objective. We are looking for a probability algorithm that is capable of accommodating nontrivial probabilities and incompatible elementary tests. If common sense in the form of non-contextuality is consistent with these requirements, we go for it. There is no need to make the world stranger than it already is. By hindsight we know that Nature concurs.
However, we also know that contextuality is an inescapable feature of situations in which probabilities are assigned either on the basis of past and future outcomes or to outcomes of measurements performed on entangled systems. We encountered two examples of such systems in Section 1.1 and 1.7. (This contextuality, which pertains to actual measurement contexts, must be distinguished from that which the proponents of pre-existent values have in mind.)