The Chemistry Maths Book, Second Edition

396 Chapter 14Partial differential equations

As for the one-dimensional case, the constant value ofVinside the box ensures that

no force acts on the particle in this region; settingV 1 = 10 means that Eis the kinetic

energy of the particle. The infinite value ofVat the ‘walls’ and outside the box

ensures that the particle cannot leave the box;ψ 1 = 10 at the walls and outside the box.

For the particle within the box, we have the boundary value problem

(14.16)

with conditions

ψ(0, 1 y) 1 = 1 ψ(a, 1 y) 1 = 1 0(ψ 1 = 1 0 when x 1 = 1 0 or x 1 = 1 a)

ψ(x, 1 0) 1 = 1 ψ(x, 1 b) 1 = 1 0(ψ 1 = 1 0 when y 1 = 1 0 or y 1 = 1 b)

(14.17)

The partial differential equation is reduced to two ordinary differential equations by

writing the wave function as the product

ψ(x, 1 y) 1 = 1 X(x) 1 × 1 Y(y) (14.18)

Then, as in Section 14.3, substitution of this product and its derivatives into (14.13),

and division byψ 1 = 1 XY, gives

(14.19)

The term involving xonly must be constant and the term involving yonly must also

be constant. Therefore,

(14.20)

whereC

x

andC

y

are constants, andC

x

1 + 1 C

y

1 = 1 − 2 mE2A

2

.

The two-dimensional boundary value problem has been reduced to two one-

dimensional boundary value problems:

(14.21a)

(14.21b)

Both of these are the same problem as that of the particle in a one-dimensional box

discussed in Section 12.6. Equation (14.21a) describes the motion of the particle along

the x-direction, and has normalized solutions given by (12.53), with appropriate

changes of symbols:

(14.22a)

Xx

a

px

a

p

p

()= sin













,=,,,

2

123

π

...

dY

dy

CY Y Yb

y

2

2

=, () () 00 = =

dX

dx

CX X Xa

x

2

2

=, () () 00 = =

11

2

2

2

2

X

dX

dx

C

Y

dY

dy

C

xy

=, =

11 2

2

2

2

22

X

dX

dx

Y

dY

dy

mE

+=−



−

∂

∂

• 
  

∂

∂

















=



22

2

2

2

2 m

xy

E

ψψ

ψ