Physical Chemistry Third Edition

15.3 The Particle in a Box and the Free Particle 665

where

κ^2 

2 mE

h ̄

2 (15.3-5)

Equation (15.3-4) is of the same form as Eq. (14.3-10). Its general solution is

ψ(II)(x)Bcos(κx)+Csin(κx) (15.3-6)

We must now make our solution conform to the continuity condition. In order forψto

be continuous atx0 andxa, we must require that

ψ(II)(0)ψ(I)(0) 0 ; ψ(II)(a)ψ(III)(a) 0 (15.3-7)

These conditions are similar to the boundary conditions for the vibrations of a string

in Section 14.3. In order forψ(II)(0) to vanish, the constantBmust vanish, because

cos(0)1. The coordinate wave function (energy eigenfunction) for region II is now

ψ(II)(x)Csin(κx) (15.3-8)

The continuity condition has another consequence. The condition thatψ(II)(a) 0

imposes a condition onκ, as in Eq. (14.3-14). The sine function vanishes when its

argument is an integral multiple ofπ, so that the quantityκcan take on the values

given by

nπκa or κnπ/a (15.3-9)

wherenis a quantum number that can take on integral values. The energy eigenfunctions

inside the box are now given by:

ψn(x)Csin(nπx/a) (15.3-10)

where we omit the superscript (II). We do not allown0, since this would correspond

toψ0 and the absence of any wave. If there is no wave there is no particle. We also

ignore negative values ofn, since the sine function is an odd function and replacingn

by−nwould simply change the sign of the function, corresponding to replacement of

Cby−C. The value ofCis unimportant for our present purposes since the Schrödinger

equation is satisfied for any value ofCand since the energy eigenvalues do not depend

onC.

The energy eigenvalues are quantized, with values determined by the value ofn:

EEn

h ̄^2 κ^2

2 m



h ̄^2 n^2 π^2

2 ma^2



h^2 n^2

8 ma^2

(15.3-11)

The quantization of the energy comes from the Schrödinger equation and from the

boundary conditions.

Figure 15.4a represents the energy eigenvalues of a particle in a one-dimensional

box by horizontal line segments at heights proportional to their energy values, and

Figure 15.4b shows the energy eigenfunctions. Each wave function is plotted on an

axis that is placed at a height corresponding to its energy eigenvalue. Note that the

wave functions in Figure 15.4b resemble the standing waves in Figure 14.7.

The energy in Eq. (15.3-11) is all kinetic energy, since we set the potential energy

inside the box equal to zero. Since we do not allown0, the minimum possible kinetic

energy is equal toh^2 / 8 ma^2 , corresponding ton1. This minimum energy is positive

and is called thezero-point energy. It is not possible for the particle in a box to have

zero kinetic energy. This result is very different from that of classical mechanics, which

always allows a particle to be at rest.