bei48482_FM

With anLwe have

p  sin^2 

L

0

 0

since sin^20 sin^2 n 0 n1, 2, 3,...

The expectation value p of the particle’s momentum is 0.

At first glance this conclusion seems strange. After all, Ep^2  2 m, and so we would

anticipate that

pn  2 mEn (5.47)

The sign provides the explanation: The particle is moving back and forth, and so

its averagemomentum for any value of nis

pav  0

which is the expectation value.

According to Eq. (5.47) there should be two momentum eigenfunctions for every

energy eigenfunction, corresponding to the two possible directions of motion. The gen-

eral procedure for finding the eigenvalues of a quantum-mechanical operator, here pˆ,

is to start from the eigenvalue equation

ˆp (^) n pn (^) n (5.48)
where each pnis a real number. This equation holds only when the wave functions (^) n
are eigenfunctions of the momentum operator ˆp, which here is
ˆp
We can see at once that the energy eigenfunctions
(^) nsin
are not also momentum eigenfunctions, because
sin  cos pn^ n
To find the correct momentum eigenfunctions, we note that
sin ei ei
1
2 i
1
2 i
eiei
2 i
nx
L
2
L
n
L
i
nx
L
2
L
d
dx
i
nx
L
2
L
d
dx
i
(n L)  (n L)
2
n
L
Momentum
eigenvalues for
trapped particle
nx
L
iL
182 Chapter Five
bei48482_ch05.qxd 1/17/02 12:17 AM Page 182