The Chemistry Maths Book, Second Edition

12.6 The particle in a one-dimensional box 353

we have

(12.47)

with general solution, in trigonometric form,

ψ(x) 1 = 1 d

1

1 cos 1 ωx 1 + 1 d

2

1 sin 1 ωx (12.48)

We now apply the boundary conditions (12.45). Atx 1 = 10 ,

ψ(0) 1 = 1 d

1

1 cos 101 + 1 d

2

1 sin 101 = 1 d

1

1 = 10

and atx 1 = 1 l(having setd

1

1 = 10 in (12.48)),

ψ(l) 1 = 1 d

2

1 sin 1 ωl 1 = 10 (12.49)

It follows, because the sine function is zero only when its argument is a multiple of π,

thatωl 1 = 1 nπwhere nis an integer:

(12.50)

The solutions of the boundary value problem are therefore

(12.51)

where the permitted solutions have been labelled with the quantum numbern. The

valuen 1 = 10 has been discounted because the trivial solutionψ

0

(x) 1 = 10 is not a physical

solution, and negative values of nhave been discounted because ψ

−n

(x) 1 = 1 −ψ

n

(x)is

merelyψ

n

with a change of sign.

For each unique solutionψ

n

there exists an energy

(12.52)

(using E 1 =1A

2

ω

2

22 m 1 = 1 h

2

ω

2

28 π

2

m). We see that quantization of the energy is a

consequence of constraining the motion of the particle to a finite region of space by the

application of the boundary conditions; this is a general result in quantum mechanics.

We note that these results are not restricted to motion in a straight line; they are valid

for any continuous curve of length lif the variable xis measured along the curve.

The wave functions (12.51) are not yet fully specified because the coefficient d

2

is

not defined. To determine d

2

we invoke the quantum-mechanical interpretation of

the square of the wave function as a probability density (see also Example 10.4):

ψ

2

(x)dx 1 = 1 probability that the particle be in element dxat position x

E

nh

ml

n

=

22

2

8

ψ

n

xd

nx

l

() sin=,=,,,n

2

123

π

...

ω=, =,±,±,

n

l

n

π

012 ...

d

dx

2

2

2

0

ψ

+=ωψ