Download Quantum Theory and Molecular Energy

What is the wavefunction?
The Born Interpretation of the Wavefunction:
It is a mathematical (sometimes imaginary) function of the coordinate(s). The square of
the wavefunction is interpreted as being proportional to the probability of the particle(s)
being a particular value of the coordinates.
In 1 dimension:
The probability of the particle described by  (x) being between x and x + dx is
proportional to:
 ( x)* ( x) dx   ( x) dx
2
Probability   dx   * dx
2
Probability   N  *  N dx
therefore:
  N  *  N dx  1


as the particle must be somewhere!
N is called the Normalised Wavefunction.
The Born Interpretation requires:
ψ is continuous everywhere:
ψ must have a continuous slope:
ψ
ψ
x
x
is not allowed
is not allowed
ψ is single valued:
ψ is finite everywhere:
ψ
ψ
x
is not allowed
x
is not allowed
This restricts the possible mathematical forms of the wavefunction.
The Born Interpretation introduces the major differences between the results of Classical
and Quantum Mechanics.
To demonstrate all the above points, solve a real example.
The Particle in the Box.
Consider a particle mass m, in one dimension x, in a box defined by a potential
V(x) = 0, 0 ≤ x ≤ L
V(x) = ∞, x ≤ 0, x ≥ L
V
x
x= 0
x= L
Consider the Schrödinger Equation in the regions x ≤ 0 and x ≥ L where V(x) = ∞:
H ( x) ( x)  E ( x)
  d

 V ( x)  ( x)  E ( x)

 2m dx

2
2
2
V ( x)   for all x
  d  ( x)
 E    ( x)
2m dx
d  ( x)
  ( x)
dx
2
2
2
2
2
This is only possible if  (x) = 0 in these regions.  (x ) 2 = 0, there is no probability of
the particle being outside the box..
Consider the Schrödinger Equation inside the box, 0 ≤ x ≤ L
H ( x) ( x)  E ( x)
  d

 V ( x)  ( x)  E ( x)

 2m dx

V ( x)  0 for all x inside the box
d
 2mE
 ( x) 
 ( x)  k  ( x)
dx

2
2
2
2
2
2
2
a general solution is:
 ( x)  A sin( kx)  B cos( kx)
k  (2mE /  )
2
1/ 2
Boundary Condition. Impose requirement that  (x) = 0 at x = 0 and x = L (the
boundaries of the problem) that is already established.
at x  0
 ( x)  0  A sin( 0)  B cos( 0)  B
B  0
at x  L
 ( x)  0  A sin( kL)
 kL  n , n  1,2,3
the integer quantum number n appears purely from the mathematics
but k  (2mE /  )
 (2mE /  ) L  n
n 
nh
E 

, n  1,2,3
n
2mL
8mL
2
2
2
2
1/ 2
1/ 2
2
2
2
2
2
quantized energy levels.
The wavefunctions can be calculated:
kL  n
 ( x)  A sin(kx)  A sin(n x / L)
n
Note that the energy and wavefunction are labeled by the quantum number n.
Normalisation:

 ( x) dx  A  sin( n x / L) dx  1
L
2
2
n

 A  ( 2 / L)
2
0
1/ 2
 ( x)  (2 / L) sin( n x / L)
1/ 2
n
Solving the Schrödinger Equation results in quantized energy levels and ‘remarkable’
(compared to Classical Mechanics) probability distribution of the particle.