cn 2 cn (11.15)An equation that relates sequential coefficients like this is called a recursion re-
lation.It allows one to determine successive coefficients, knowing the previous
ones. Ultimately, only two constants need be known at the outset:c 0 ,from
which the even-powered coefficients c 2 ,c 4 ,c 6 ,...can be determined, and c 1 ,
from which the odd-powered coefficients c 3 ,c 5 ,c 7 ,...can be determined.
Now one of the requirements for proper wavefunctions can be applied: they
must be bounded.Although this derivation started by assuming an infinite se-
ries as a solution, the wavefunction cannot be infinite and still apply to reality.
Even the inclusion of the ex(^2) /2
term does not guarantee that the infinite sum
will be bounded. But the recursion relation in equation 11.15 provides a way
to get this guarantee. Because the coefficient cn 2 depends on cn, if for some n
the coefficient cnis exactly zero, all successive constants cn 2 ,cn 4 ,cn 6 ,...are
also exactly zero. Of course, this does not affect the other coefficients cn 1 ,
cn 3 ,....So,in order to guarantee a bounded wavefunction, we must first sep-
arate the odd and even terms into two separate power series:
feven
,even
n 0
cnxn
fodd
,odd
n 1
cnxn
We will require that the wavefunctions themselves be composed ofex
(^2) /2
times eithera sum of only odd terms ora sum of only even terms. For each
sum it is now required that, in order for the wavefunction to not be infinite, at
some value ofnthe next coefficient cn 2 must become zero. That way, all fur-
ther coefficients will also be zero. Since the coefficient cn 2 can be calculated
from the previous constant cndue to the recursion relation, we can substitute
zero for cn 2 :
0 cn
The only way for the coefficient cn 2 to become identically zero is if the nu-
merator of the fraction in the above equation becomes zero at that value ofn:
2 n
2
m
2E
0
This expression includes the total energy Eof the harmonic oscillator. Energy
is an important observable, so let us detour to consider it. In order for the
wavefunction to be noninfinite, the energy of the harmonic oscillator, when
combined with the other terms like ,n,m, and , must have only those val-
ues that satisfy the above equation. Therefore, we can solve for what the values
of the energy must be. Substituting also for 2
m/, we find a simple
conclusion:
E(n^12 )h (11.16)
where nis the value of the index where the next coefficient of the series be-
comes zero,his Planck’s constant, and is the classical frequency of the oscil-
lator. That is, the total energy of the harmonic oscillator depends onlyon its
classical frequency (determined by its mass and force constant), Planck’s con-
stant, and some integer n. Since the energy can have values only as determined
by this equation, the total energy of the harmonic oscillator is quantized.The 2 n 2 mE/^2
(n2)(n1) 2 n 2 mE/^2
(n2)(n1)322 CHAPTER 11 Quantum Mechanics: Model Systems and the Hydrogen Atom