The Schroedinger Equation
When twice the mass of the electron is set equal to 1, h bar is set equal to 1, and r is measured
in Bohr radii, e2 comes out 2. Thus the Schrödinger equation for the hydrogen atom is
2
 2    ER
r
The value of ∇2 in 3 dimensions is derived in Del2.doc
The ground state solution is
  exp  r 
This yields E = -1 Rydberg. The potential at r = 1 Bohr radius is -2 Rydbergs.
If m rather than 2m is set equal to 1, e2 becomes 1.
 2 1

   EH 
2
r
The result is to simply divide everything by 2 so that the energy of the Hydrogen atom in Hartrees is 1/2.
The wave function is the same, but can be written as

  exp  2 EH r

Variance
The variance in the Schrödinger equation (in Hartrees) is given by
 2 c, E  

 2  r , c 
2
2
 V  r    r , c   E  r , c  dr
   r , c  dr
2
The ½ indicates that the value of e2 in V is 1 and E is measured in Hartrees.
The Schrödinger equation can be approximately “solved” by minimizing 
2
c, E  .