Physical Chemistry , 1st ed.

what the eigenvalues for the energy are, because with identically zero, by
the Born interpretation the particle has a zero probability of being in those
regions.
Consider the region where xranges from 0 to a. The potential energy is de-
fined as zero in this region and so the Schrödinger equation becomes

^2



m

2





x

2

(^2) E
which is a second-order differential equation. This differential equation has a
known analytic solution. That is, functions are known that can be substituted
into the above second-order differential equation to satisfy the equality. The
most general form of the solution to the above differential equation is
Acos kxBsin kx
where A,B, and kare constants to be determined by the conditions of the
system.*
Since we know the form of, we can determine the expression for Eby sub-
stituting into the Schrödinger equation and evaluating the second deriva-
tive. It becomes
E
k
2
2
m

^2

Example 10.10

Show that the expression for the energy of a particle in a box is Ek^2 ^2 /2m.

Solution

All that needs to be done is to substitute the wavefunction Acos kx

Bsin kxinto the Schrödinger equation, remembering that the potential en-

ergy Vis zero. We get

10.8 An Analytic Solution: The Particle-in-a-Box 289

x  0 x  a

x-axis

V   V  

V  0

Energy

Figure 10.5 The particle-in-a-box is the simplest ideal system that is treated by quantum
mechanics. It consists of a region between x0 and xa(some length) where the potential
energy is zero. Outside of this region (x0 or xa), the potential energy is , so any particle
in the box will not be present outside the box.

*Acceptable solutions can also be written in the form

AeikxBeikx

This form is related to Acos kxBsin kxvia Euler’s theorem, which states that
eicos isin .