Suppose that a maximal test performed at the time t1 yields the outcome u, and that we want to calculate the probability with which a maximal test performed at the later time t2 yields the outcome w. Further suppose that at some intermediate time t another maximal test is made, and that its possible outcomes are v1, v2, v3,… Let v be one of these values. Because a maximal test renders the outcomes of earlier measurements irrelevant, the joint probability p(w,v|u) with which the intermediate and final tests yield v and w, respectively, given the initial outcome u, is the product of two probabilities: the probability p(v|u) of v given u, and the probability p(w|v) of w given v. By Born’s rule, this is
p(w,v|u) = |<w|v> <v|u>|2,
where u, v, and w are unit vectors in the subspaces representing u, v, and w, respectively. To obtain the probability of w given u, regardless of the intermediate outcome, we must calculate this probability for all possible intermediate outcomes and add the results:
pA(w|u) = |<w|v1> <v1|u>|2 + |<w|v2> <v2|u>|2 + |<w|v3> <v3|u>|2 + ···
In (other) words, first square the magnitudes of the amplitudes <w|v1> <v1|u>, etc., then add the results. This is Rule A. (Hence the subscript A.)
Rule B next. Since the vectors v1, v2, v3,… form a basis, any vector u can be written as
u = u1v1 + u2v2 + u3v3 + ···,
where u1, u2, u3, … are the components of u with respect to this basis. The scalar product of v1 and u is
<v1|u> = u1<v1|v1> + u2<v1|v2> + ···.
Since basis vectors are unit vectors, we have that <v1|v1> = 1, and since they are mutually orthogonal, we have that <v1|v2> = 0. Hence <v1|u> = u1, <v2|u> = u2, etc. Consequently,
u = v1 <v1|u> + v2 <v2|u> + v3 <v3|u> + ···
<w|u> = <w|v1> <v1|u> + <w|v2> <v2|u> + <w|v3> <v3|u> + ···.
If no intermediate measurement is made, the probability of w given u is
pB(w|u) = |<w|u>|2 = |<w|v1> <v1|u> + <w|v2> <v2|u> + <w|v3> <v3|u> + ···|2
In (other) words, first add the amplitudes <w|v1> <v1|u>, etc., then square the magnitude of the result. This is Rule B. (Hence the subscript B.)