Physical Chemistry , 1st ed.

or


d

d

2

x



 2 

2



m

2

E

^2 x^2  0 (11.6)

where equation 11.6 has been rearranged from the previous expression to show
in parentheses the two terms that are simply being multiplied by . The first
term is the second derivative of.
We now assume that the form of the wavefunction that satisfies this
Schrödinger equation has the form of a power series in the variable x. That is,
the wavefunction is some function f(x) that has some term containing x^0
(which is simply 1), some term containing x^1 , some term containing x^2 , ad in-
finitum, all added together. Each power ofxhas some constant called a coeffi-
cientmultiplying it, so the form off(x) (recognizing that x^0 1) is:

f(x) c 0 c 1 x^1 c 2 x^2 c 3 x^3 

The c’s are the coefficients multiplying the powers ofx. It is more concise to write
the above function using standard summation notation, as in the following:

f(x) 



n 0

cnxn (11.7)

where nis the indexof the summation. For now, the summation goes to infin-
ity. This causes a potential problem, because sums that go to an infinite number
of terms often approach infinity themselves unless there is a way for each suc-
cessive term to become smaller and smaller. A partial solution is to assume that
every term in the sum is multiplied by another term that gets much smaller as x
itself (and therefore xn) gets larger. The term that will work in this case is ex

(^2) /2
.
(Note the inclusion of the constant here. You may wonder why we use this par-
ticular exponential function. At this point, the only justification for using this
function is that it will yield an analytic solution.) This exponential is an exam-
ple of a Gaussian-type function(named after the eighteenth- to ninteenth-century
mathematician Karl Friedrich Gauss). The wavefunction for this system is now
ex
(^2) /2
f(x) (11.8)
where f(x) is the power series defined in equation 11.7.
At this point, the first and then the second derivative can be determined
with respect to x. Then the expressions for the second derivative as well as the
original function can be substituted into the proper form of the Schrödinger
equation, equation 11.6. Once we do so, the logic behind the choice of the ex-
ponential function ex
(^2) /2
will become apparent mathematically. Using the
product rule of differentiation, the first derivative is
(x)ex
(^2) /2
f(x) ex
(^2) /2
f(x)
where and f(x) refer to the first derivatives ofand f(x) with respect to
x. Using the above equation, the second derivative ofwith respect to xcan
be determined using the product rule. It is, after doing a little algebra:
ex
(^2) /2
[^2 x^2 f(x) f(x)  2 xf(x) f(x)] (11.9)
Here ex
(^2) /2
has been factored out of every term in the second derivative.
Substituting the forms ofand into the form of the Schrödinger equation
given by equation 11.6 yields
ex
(^2) /2
[^2 x^2 f(x) f(x)  2 xf(x) f(x)] 


2



m

2

E

^2 x^2 ex

(^2) /2
f(x)  0 (11.10)
11.3 The Quantum-Mechanical Harmonic Oscillator 319