The Schrödinger Equation for one particle, mass m, in 3D:
H ( x, y, z ) ( x, y, z )  E ( x, y, z )
 
 2m
 
 2m

2
2

  V ( x, y, z )  ( x, y, z )  E ( x, y, z )


d
d 
d



  V ( x, y, z )  ( x, y, z )  E ( x, y, z )
 dx dy dz 

2
2
2
2
2
2
2
The Schrödinger Equation for two particles, masses m1 and m2, in 1D (x), coordinates x1
and x2:
H ( x , x ) ( x , x )  E ( x , x )
1
2
1
2
1
2
 







V
(
x
,
x
)
 2m
  ( x , x )  E ( x , x )
2
m


  d

 d



V
(
x
,
x
)
 2m dx 2m dx
  ( x , x )  E ( x , x )


2
2
2
2
1
2
1
1
2
1
2
1
2
2
2
2
2
2
2
1
1
2
1
2
2
1
2
1
2
2
This can be easily extended. The most complicated case we consider is the hydrogen
atom:
H ( x , y , z , x , y , z ) ( x , y , z , x , y , z ) 
1
1
1
2
2
2
1
1
1
2
2
E ( x , y , z , x , y , z )
1
1
1
2
2
 







V
(
r
)
  E
 2m

2m


 

e 
 2m   2m   4 r    E


2
m2, x2, y2 z2, -e
2
2
z
2
2
2
1
2
1
m1, x1 , y1, z1, +e
2
2
2
2
2
2
1
1
r
x
2
2
O
y
 
  2m

r  x
2
1
2
d
d
d  

 


dx
dy
dz

 2m
y z 
2
2
2
2
2
2
1
1
1
2
2
2
2
d
d
d 
e

 


 dx dy dz  4
2
2
2
2
2
2
2
2
2
2
O

  E
r 
1/ 2
Normalization in 3D:
 ( x, y, z ) d   ( x, y, z ) * ( x, y, z ) dx dy dz  1
2
all space
d  dx dy dz
For the 3 types of molecular motion the energy equations obtained from solving these
higher dimensional problems are:
Translation in 3D – ‘Particle in the 3D box’ – energy equation above.
Rotation. Rotation of a rigid 3D shape.
Linear Molecules have the same energy equation as diatomic molecules. ‘Rigid Rotator.’
Energy expression for rotation of a non-linear molecule is complicated but is exactly
known. The energy equation depends on the shape of the molecule. Generally there are 3
axes of rotation through the centre of mass (X, Y, Z), 3 moments of inertia about these
axes, IA, IB, IC , 3 rotational constants A, B, C and 3 quantum numbers J, KA, KC .
Vibrations.
Polyatomic molecules have many vibrations. Many of these behave like SHO, stretches
and angle bends. (Sometimes they are degenerate, SHO in 2D, e.g. the bend of CO2.)
Others require different models, e.g. internal rotation of a CH3- group which behaves like
a 2D rotator.
Many of these problems involve changing the coordinates from Cartesian (x, y, z) to some
other set of 3D coordinates. This is in order to ‘separate’ the coordinates into Schrödinger
Equations in 1 dimension.
Example of separating the coordinates:
Particle in 3D box
Particle (molecule) mass m in a 3 dimensional volume (room) of lengths A, B and C in
the x, y, z directions.
V(x, y, z) = ∞ if
x < 0 or x >A, y < 0 or y > B or z < 0 or z > C
V(x, y, z) = 0 if
0 ≤ x ≤ A and 0≤y ≤ B and 0 ≤ z ≤ C
z
C
A
B
x
y
Now show that this 1 particle in 3D problem separates into 3, 1 particle 1D problems.
The fundamental point in this example is:
If the Hamiltonian can be separated:
H ( x, y, z )  H ( x)  H ( y)  H ( z )
1
2
3
then the one dimensional Schrödinger Equations can be solved the energies added and the
wavefunctions multiplied.
EE E E
x
y
*see demonstration in 2D below
z
 ( x, y, z )   ( x) ( y ) ( z )
1
2
3
In this case
   


 



  V ( x, y, z )
y
z 
 2m  x

H ( x, y, z )  
2
2
2
2
2
2
2
and clearly: V ( x, y, z )  V ( x)  V ( y )  V ( z )
  


 x
 2m 
2
H ( x, y, z )  
therefore:
  
 
 2m
2

2



y
z
2



  V ( x)  V ( y )  V ( z ) 



2
2
2
   

   

 V ( x)  
 V ( y )  
 V ( z ) 

x
 

 2m y
  2m z
 H ( x)  H ( y )  H ( z )
2
2
2
2
2
2
2
2
2
  d

H ( x)  
 V ( x) etc., which is the 1D particle in the box in
 2m dx

the x direction.
2
2
where:
1
2
nh
2
n x 
E 
, ( x)    sin 
, n  1,2,3
 A
 A 
8mA
2
Therefore:
1/ 2
2
x
n
2
ph
2
p y 
E 
, ( y )    sin 
, p  1,2,3
B
 B 
8mB
qh
2
q z 
E 
, ( z )    sin 
, q  1,2,3
C 
 C 
8mC
2
and similarly:
1/ 2
2
y
p
2
2
1/ 2
2
z
q
2
nh
ph
qh
E


8mA 8mB 8mC
2
2
2
2
Therefore:
2
2
2
1/ 2
*demonstration for 2 D
H ( x, y ) ( x, y )  E ( x, y )
if : H ( x, y )  H ( x)  H ( y )
let : ( x, y )   ( x) ( y ) and E  E  E
1
2
1
2
 ( y ) H ( x) ( x)   ( x) H ( y ) ( y )  E ( x) ( y )
 throughout by ( x) ( y ) on the left
H ( x) ( x) H ( y ) ( y )

EE E
 ( x)
 ( y)
 H ( x) ( x)  E  ( x) and H ( y ) ( y )  E  ( y )
1
1
1
2
1
2
2
1
2
1
1
2
2
1
2
8 
 n x   p x   q x 
, ( x, y, z )  
 sin 
 sin 
 sin 

 ABC 
 A   B   C 
n  1,2,3 , p  1,2,3 , q  1,2,3
NOTE: 1 quantum number for each coordinate!
2
2
1
1
2
2
2