observable, we can prove, as claimed above, that if two states are
distinguishable because they have definite, different values of some
observable, then they are orthogonal.^4a/Some examples of interpreta-
tion of the inner product. The
first three examples are explained
immediately below. The fourth,
about averages, is justified on
p. 985.Suppose thatuandvare both properly normalized wavefunc-
tions. If|〈u|v〉|= 1, then the states are identical.^5 If〈u|v〉= 0, then
uandvare completely distinguishable from one another. There is
also the intermediate case where〈u|v〉has a magnitude greater than
0 but less than 1. In this case, we could say thatuis a mixture of
vplus some other statewthat is distinguishable fromv, i.e., that|u〉=α|v〉+β|w〉.where〈v|w〉= 0. We then have
〈u|v〉= (α〈v|+β〈w|)|v〉=α.Now suppose that we make measurements capable of determining
whether or not the system is in the statev. If the system is pre-
pared in stateu, and we make these measurements on it, then by
(^4) Proof: Consider statesuandvwithOu=e 1 uandOv=e 2 v. IfOis
Hermitian, we have〈Ou|v〉=〈u|Ov〉, soe∗ 1 〈u|v〉=e 2 〈u|v〉. But sincee 1 ande 2
are real and unequal, we must have〈u|v〉= 0.
(^5) If the inner product is, for example,−1, then the wavefunctions differ only
by an unobservable difference in phase, so they really describe the same state.
Section 14.6 The underlying structure of quantum mechanics, part 2 983