state Ψ =c +c′. If bothcandc′are nonzero, then we expect
to get a state with properties in between those of and. If we
measure the energy of such a state, then our wavefunction becomes
entangled with that of the particle, and we look like this:c
We measured the
energy to be 1. +c′We measured the
energy to be 4..Suppose we make the mixture an equal one,c=c′. Then the average
should be (1 + 4)/2 = 2.5. This turns out to be easily expressible
using an inner product:〈Ψ|OEΨ〉= 2.5.It’s a good exercise to work this out for yourself (problem 20, p. 1013).
The key point is that Ψ can be expressed as a superposition of states
of definite energy Ψ =c +c′ , and when the operatorOEworks
on Ψ, it givesOEΨ =c + 4c′. (And remember that by nor-
malization,|c|=|c′|= 1/√
2.)
This is a general rule for calculating averages: for a state Ψ, the
average value for an observableOis〈Ψ|OΨ〉. Because observables
are hermitian, this is the same as〈OΨ|Ψ〉.
Discussion Questions
A Suppose that by rotating vectors we could change the results of
dot products. Explain why this would be very naughty, first by using an
example in whichu·u= 1, and then, just to make it naughtier, one where
u·v= 0.
B Suppose that as a system evolved over time, inner products of
wavefunctions could change. As in discussion question A, give shockingly
naughty examples where initially we have〈Ψ|Ψ〉= 1 and〈Ψ|Φ〉= 0, but
later these inner products change.14.6.3 Completeness
We have used math to back up our claim that distinguishable
states are orthogonal. Going in the opposite direction, suppose that
〈u|v〉= 0. How can we then conclude that there exists some ob-
servableOthat can distinguish them? There is no straightforward
mathematical reason why this must be true, but it would not make
sense physically to talk about two states that were utterly distinct
and yet indistinguishable by any experiment. We therefore take this
as a postulate.^6(^6) Our statement of the completeness principle refers to taking a sum of wave-
functions. Because the physical motivation for the completeness postulate is so
appealing, physicists are willing to stretch the definition of the word “sum” in
order to make it true. The sum can be an infinite sum, and in certain cases
we may even need to make it an integral, which is a kind of continuous sum.
For example, consider a one-dimensional particle in a box. A complete set of
observables for this system can be found by picking the energy operator alone.
Now suppose we throw a particle in the box, in such a way that its position is
986 Chapter 14 Additional Topics in Quantum Physics