momentum is a sine wave of the formΨ=ei k x. We havekλ= 2π
andp=h/λ=~k, and the sign is a matter of convention. Taking
the derivative ofΨgives an eigenvaluei k, which has the wrong
units (easily fixed by tacking on a factor of~), but more importantly
is not real. This suggests defining the momentum operator asOp=−i~
d
dx.
A further note about the momentum operator is example 13 on
p. 985.
A nonexample example 6
Consider the one-dimensional particle in a box, and restrict our
attention to the two lowest-energy states and their superpositions.
Define an operatorOby the rule
O( ) =
O( ) =−( ).
SinceOis linear, defining its action on and suffices to
define its action on the superpositions of these states as well.
This operator has eigenvalues, one of which isi, corresponding
to the state −i. (It also has a second eigenvalue, which
is imaginary as well.) Because this operator doesn’t have real
eigenvalues, it is not a valid observable.
Note that in examples 5 and 6, it doesn’t matter whether the
operator isdefinedusing complex numbers. Our definition of the
momentum operator was stated using an equation that had aniin
it, but its eigenvalues are real, so that’s OK. The operatorOin
example 6 was defined using only real numbers, but its eigenvalues
are not real.
Position is an observable example 7
If we have a wavefunctionΨ(x) expressed as a function of po-
sitionx, then we simply take the operator for positionOx to be
multiplication by the numberx,
Ox(Ψ) =xΨ.
For example, ifΨ=ei x(ignoring units), thenOx(Ψ) =xei x. This
operator is definitely linear, because multiplication by a number is
linear, e.g., 7(a+b) = 7a+ 7b. The only question is whether it has
eigenvalues, and whether those are real. A state of definitex, say
a state withx= 0, would have to be represented by a wavefunc-
tionΨ(x) for which there was zero probability of havingx 6 = 0, and
this requires us to haveΨ(x) = 0 for nonzerox. But what would
be the value ofΨ(0)? It has to beinfiniteifΨis to be properly nor-
malized. With this motivation, the physicist P.A.M. Dirac defined
the Dirac delta function,δ(x) ={
0 forx 6 = 0
+∞ forx= 0Section 14.6 The underlying structure of quantum mechanics, part 2 977