Concise Physical Chemistry

c16 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come

256 WAVE MECHANICS OF SIMPLE SYSTEMS

We simplify this equation immediately by settingV(x,y,z)=0, so the middle

term drops out. The kinetic energy operator looks complicated:

−

h ̄^2

2 m

d^2 (x,y,z)

dx^2

−

h ̄^2

2 m

d^2 (x,y,z)

dy^2

−

h ̄^2

2 m

d^2 (x,y,z)

dz^2

=E(x,y,z)

until we make the entirely reasonable observation that, in a cube, there is no reason

to prefer particle motion in any one direction over motion in any other. The energy

Exsupplied by particle motionxis the same asEysupplied byyandEzsupplied

byz. This being the case, we can split one complicated equation into three simple

ones:

−

h ̄^2

2 m

d^2 (x)

dx^2

=Ex(x)

−

h ̄^2

2 m

d^2 (y)

dy^2

=Ey(y)

−

h ̄^2

2 m

d^2 (z)

dz^2

=Ez(z)

Each of these equations has already been solved with the results

(x)=Asin

2 πx

λ

(y)=Asin

2 πy

λ

(z)=Asin

2 πz

λ

and

Ex=

n^2 xh^2

8 ml^2

Ey=

n^2 yh^2

8 ml^2

Ez=

n^2 zh^2

8 ml^2

In the energy equations, the length of one edge of the cubelis in the denominator. If the

geometry had been other than a cube, there would be different dimensions—perhaps

a,b, andcfor a parallelepiped. This would have made the solutions slightly more

complicated, but it would not have made any significant change in the result.