Page 282 - Essentials of physical chemistry

244 Essentials of Physical Chemistry

Corollary If the wave function is not normalized, then the average value of the observable
corresponding to the operator A is given by

ð
þ1
C*ACdt
1 :
<a> ¼ ð
þ1
C*Cdt
1
Postulate V The wave function or state function of a system evolves in time according to the time-
dependent Schrödinger equation
qC
HC(r, t) ¼ i :
h
qt
While time-dependent processes can be treated, our emphasis will be on time-independent wave
functions at the undergraduate level.

PARTICLE ON A RING

There is another problem in quantum mechanics that is exactly solvable and can be applied to
chemistry. Let us assume there is a particle constrained to move only on a ring of fixed radius, a.We
could consider this to be a tiny glass bead with a hole riding on a lubricated ring of thin wire, but
anyone who has recently taken a course in organic chemistry should immediately see the analogy to
that delocalized ring drawn on aromatic compounds such as benzene. This is an important problem
because it is solvable and because it illustrates the general rules of quantum mechanics we have
outlined earlier. First we write the classical momentum and assume the potential energy on the ring
is zero, but here we have to use polar coordinates with a fixed radius and only allow the angle u to
vary so the coordinate of interest is dc ¼ ad u for a circumference c and a radius a. This problem is
solvable and is especially important because it is another illustration of the proper way to treat the
ffiffiffiffiffiffiffi
p
1. For the sake of simplicity, we assume that the potential
complex arithmetic involving i ¼
energy of the particle on the ring is zero so we only consider the kinetic energy in polar coordinates
2
h d 2 2
i adu h d
and form the Hamiltonian operator. H ¼ T þ V ¼ T þ 0 ¼ þ 0 ¼ 2 2 and
2m 2ma du
2
2
2
2
h d c d c 2ma Ec
then set up Hc ¼ Ec to be solved. Thus 2 ¼ Ec ) 2 þ 2 ¼ 0 )
2
2ma du du h
"
r ffiffiffiffiffiffiffiffiffiffiffiffiffiffi#" r ffiffiffiffiffiffiffiffiffiffiffiffiffiffi#
2
2
2ma E 2ma E
D þ i 2 D i 2 c ¼ 0. This is very similar to the PIB problem but there are
h h
some important differences. First the differential circumference dc ¼ adu has a constant in it and as
we will soon see, the boundary conditions are different because the wave function must be
continuous and connect smoothly as we go around the ring. However, we do know how to solve
p ffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffi
2ma 2 E
2ma 2 E
this kind of second-order differential equation as c ¼ C 1 e i h 2 u þ C 2 e þi h 2 u . We do not
know either C 1 or C 2 but we do know that we must enforce c(2p) ¼ c(0) to join the beginning and
end of the wave function smoothly as it goes around the ring (in a standing wave). If there is a
sudden jump in the wave function at c(2p), the second derivative will be undefined at the
break point and the whole equation will be meaningless! Since the two solutions only differ
in the sign of the exponent, that sign can be absorbed into the value of u and there is only
p ffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffi
i 2ma 2 E u 0 i2p 2ma 2 E
one form of the wave function as c ¼ Ce h 2 ) c(0) ¼ Ce ¼ c(2p) ¼ Ce h 2 ,
r ffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ma E 2ma E
2
2
2p 2 ¼ 0, 2p,4p, ..., n(2p) so we find that 2 ¼ n and then we have
h h