Physical Chemistry , 1st ed.

The expression is rearranged to solve for k, and the result is 197 N/m. There

are 10^5 dynes per newton, 1000 mdyn per dyne, and 10^10 Å per meter, so it

is easy to show that this is equal to 1.97 mdyn/Å.

b.In the second case, again using equation 11.3:

6.000 

1013 s^1 

2

1





1.673

k

 10 ^27 kg



Evaluating this, one gets 237.8 N/m, which equals 2.378 mdyn/Å.

11.3 The Quantum-Mechanical Harmonic Oscillator

Quantum mechanically, a wavefunction for a one-dimensional harmonic os-

cillator can be determined using the (time-independent) Schrödinger equation

 2



m

2



d

d

x

2

 2 Vˆ(x)E

The potential energy for the quantum-mechanical system has the same form

as the potential energy for the classical system. (Generally speaking, since po-

tential energies are energies ofposition,the quantum-mechanical form of the

potential energy is the same as the classical form. But now because of the form

of the Schrödinger equation, the potential energy operatoris multiplied by the

wavefunction .) The Schrödinger equation for the harmonic oscillator is

 2



m

2



d

d

x

2

 2 

1

2

kx^2 E (11.4)

and acceptable wavefunctions for this one-dimensional system must satisfy this

eigenvalue equation.

This differential equation does have an analytic solution. The method we use

here is one general technique for solving differential equations: we define the

wavefunction as a power series. What we will ultimately find is that in order to

solve the Schrödinger equation, the power series must have a special form.

First, the Schrödinger equation 11.4 will be rewritten using equation 11.3 to

substitute for k. Rearranging equation 11.3, one finds that the force constant kis

k 4 ^2

2 m (11.5)

and so the Schrödinger equation for a one-dimensional harmonic oscillator

becomes

 2



m

2



d

d

x

2

 2  2 

(^2
2) mx 2
E
Now we will do three things. First, we define
 

2 



m



Second, we divide both sides of the equation by the term ^2 /2m. Third, we

bring all terms over to one side of the equation so that we have an expression

equaling zero. The Schrödinger equation becomes

d

d

x

2

 2 

(^2) x 2


2



m

2

E

 0

318 CHAPTER 11 Quantum Mechanics: Model Systems and the Hydrogen Atom