Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 12
Three- Dimensional Systems
Solution of Schrodinger equation for particle move in three
dimensional in a box.
- In this case, the particle is confined to lie within a
rectangular parallelepiped with sides of lengths a, b, and
c.
- The Schrodinger equation for this system is the threedimensional
 2  2  2  2
( 2  2  2 )  E (x , y , z ) 0  x  a (1)
2m x
y
z
0  y b
0  z c
- The wave function ψ(x, y, z) satisfies the boundary
condition that it vanishes at the walls of the box, and so
 (0, y , z )   (a, y , z )  0
for all y and z
 (x , 0, z )   ( x , b , z )  0
for all x and z (2)
 (x , y , 0)   (x , y , c )  0
for all x and y
[1]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 12
- We shall use the method of separation of variables to
solve Eq. (1). We write
Ψ(x, y, z) = X(x) Y(y) Z (z)
(3)
- Substitute Eq. (3) into Eq.(1), and then divide through
ψ=XYZ to obtain
2
2
 2 X 
Y 
Z 


E
2m X
2m Y
2m Z
(4)
- Each of the three terms on the left – hand side of Eq.(4) is
a function of only x, y, or z. Therefore, each term must
equal a constant for Eq. (4) as
E x + Ey + Ez = E
(5)
where Ex, Ey, and Ez are constants and where
 2 X 
 Ex
2m X

Y 
 Ey
2m Y

Z 
 Ez
2m Z
2
2
[2]
(6)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 12
- So we can obtain from all these Eqs.
E nx n y nz
h2

8m
 n x2 n y2 n z2 
 2  2  2 
b
c 
a
n x = 1, 2,3 ,…….
n y = 1, 2,3, …….
(7)
n z = 1, 2, 3, ……..
- We should expect by symmetry that the average position
of a particle in a box at the center of the box
- We can show the average position of a particle confined
to the region shown in Figure 1 is the point (a/2, b/2, c/2)
- The position operator in three dimension is
Ȓ = X̂ i +Ŷ j +Ẑ k
and the average position is given by
r i x  j y k z
a
b
c
 r  i j k
2
2
2
(8)
Thus the average position of the particle is the center of
the box.
[3]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 12
- In a similar manner, we should expect that the average
momentum of a particle in a three dimensional box is
zero. The momentum operator in three dimension is
 

 
p  i  i
j
k

y
z 
 x
- It is a straightforward to show that
p
(9)
=0.
An interesting feature of a particle in a three dimensional
box occurs when the sides of the box are equal. In this
case, a= b =c in Eq. (7) and so,
E nx n y nz
h2

n 2  n y2  n z2 
2  x
8ma
[4]
(10)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 12
- The lowest level here, E111 , is non degenerated, but the
second level is threefold degenerated because
E 211  E 121  E 112
6h 2

8ma 2
- Figure (2) shows the distribution of the first few energy
levels of a particle in a cube.
- Note that the degeneracy occurs because of the symmetry
introduced when the general rectangular box becomes a
cube and that the degeneracy is lifted when the symmetry
is destroyed by making the sides of different lengths.
- It is a general principle of quantum mechanics that
degeneracies are the result of underlying symmetry and
that degeneracies are lifted when the symmetry is broken.
p

[5]

i


