Physical Chemistry , 1st ed.

were, then would be zero between 0 and a, and then it would be zero every-
where and the particle would not exist anywhere. We reject that possibility,
since the particle’s existence is unquestioned. In order for the wavefunction to
equal zero at xa, the value of sin ka mustbe zero:

sin ka 0

When is sin kaequal to 0? In terms of radians, this occurs when kaequals 0,
,2 ,3 ,4 ,...or at all integral values of. We reject the value 0 because
sin 0 equals 0 and so the wavefunction would not exist anywhere. We thus have
the following restriction on the argument of the sine function:

kan n1,2,3,...

Solving for k,

k

n

a

where nis a positive integer. Although there is no mathematical reason ncan’t
be a negative integer, use of negative integers adds nothing new to the solution,
so they are ignored. This will not always be the case.
Having an expression for kallows us to rewrite both the wavefunction and
the expression for the energies:

(x) Bsin

n

a

x

E

n

2

2

m

2

a



2

2



8

n

m

(^2) h
a
2
(^2)
where the definition for has been substituted in the last expression for en-
ergy. The energy values depend on some constants and on n, which is restricted
to positive integer values. This means that the energy cannot have just any
value; it can have only values determined by h,m,a, and—most importantly—
n. The energy of the particle in the box is quantized,since the energy value is
restricted to having only certain values. The integer nis called a quantum
number.
The determination of the wavefunction is not complete. It must be nor-
malized. It is assumed to be multiplied by some constant Nsuch that

a
0
(N)(N) dx 1
The limits on the integral are 0 to abecause the only region of interest for
the nonzero wavefunction is from x0 to xa. The infinitesimal dis sim-
ply dx.
We will assume that the normalization constant is part of the constant Bthat
multiplies the sine part of the wavefunction. The integral to be evaluated is

a
0 
Nsin
n
a
x
(^) Nsin
n
a
x
(^) dx 1
The complex conjugate does not change anything inside the parentheses, since
everything is a real number or real function. A function similar to this was
evaluated in Example 10.7. By following the same procedure as in that exam-
ple (and you should verify that the procedure has the following result), we find

that N2/a. Since both the wavefunction and the energy are dependent on

some quantum number n, they are usually given a subscript n, like nand En,

10.8 An Analytic Solution: The Particle-in-a-Box 291