If we ignore the interaction potential (i.e., the term proportional to
1/|**r**_{2}-**r**_{1}| ), the
Hamiltonian is the sum of two terms, each identical in form to the
H-atom Hamiltonian we've already solved, each a function
of just one electron's coordinates. Just as in the
separation of variables: if the variables in the Hamiltonian can be separated
into different terms, we use a product wavefunction to solve the
Schrödinger's equation:

The eigenenergies of this simplified system are just the sum of the 1-electron eigenenergies:

We now include the interaction potential as a perturbation. The first order energy shift is just the expectation value of the perturbation using the unperturbed (simple product) wavefunctions:

If we slip back into normal (dimensioned) coordinates
we can provide a simple explanation of this term. Electron
1 is distributed in space as given by its wavefunction. Thus there
is a negative charge density surrounding the nucleus:
=-*e*||^{2}
where *e* is the unit charge (i.e., -*e* is the charge
on an electron). This charge density produces a electrostatic
potential we can calculate (we use here cgs units: no
4_{0}):

Electron 2 therefore feels, in addition to the the electrostatic potential of the nucleus, the above electrostatic potential. We can calculate this additional energy:

which is exactly the above term (but in dimensioned coordinates).

The fact that the electrons in a multi-electron atom feel the total electrostatic potential due to the total charge: both the attractive nucleus and repulsive electron-cloud, suggests an approximation scheme. As a first approximation solve Schrödinger's equation with just the nuclear potential. Calculate the charge density of the resulting electron clouds and now use the total charge density (nucleus and electron-cloud) to calculate the total electrostatic potential. Use this improved electrostatic potential in Schrödinger's equation; calculate new wavefunctions and electron cloud densities; again calculate an improved electrostatic potential. Continue this process until it become self-consistent, i.e., a new iteration has no noticeable effect of the wavefunctions/eigenenergies. This method is called the self consistent field method or the Hartree method (or Hartree-Fock method, for an improved version accounting for wavefunction symmetry [see below]).

The above interaction integral looks like a mess to
evaluate. We proceed by expanding 1/|**r**_{2}-**r**_{1}|
in terms of spherical harmonics:

where, *r*_{>} is the larger
of *r*_{1} and *r*_{2} and
*r*_{<} is the smaller
of *r*_{1} and *r*_{2}.
With this substitution, the interaction integral become a
sum of terms like this:

Almost all of these terms are zero because most of the
3-*Y*_{lm} angular integrals are zero.

*m* must be zero for 3-Js to be non-zero.
(This is basically the statement that if the charge density
is -symmetric, the resulting potential
must be -symmetric). And
0<=*l*<=2*l*_{i}. (This is basically
the statement that a charge density going like
|*Y*_{l0}|^{2} *P*_{l}^{2}
has at most a 2*l*-multipole.) Further *l* must be even.
(This is basically the statement that if the charge density is inversion-symmetric,
the potential must be also.)

We pick a particular two electron problem: the first excited state
of helium. Thus electron 1 is in the 1*s* state
(*n*_{1}=1, *l*_{1}=0, *m*_{1}=0)
and electron 2 is in one of the *n*_{2}=2 states:
2*s* or 2*p*. So *l*=0 and the sum is just
one term with trivial angular integrals since *Y*_{00}
is just a constant: 1/(4)^{½}.
Thus in both 3-*Y*_{lm} integrals,
*Y*_{lm}=*Y*_{00} is constant and can be pulled
from the integral. Since 4|*Y*_{00}|^{2}=1,
the first-order energy correction is just 1/*Z*=½ the
radial integral.

We are left doing the radial integral using *Mathematica*:

Thus the energy of the 1*s*2*s* state is
approximately:

-½(1+¼)+½(0.209877)= -.520

Thus the energy of the 1*s*2*p* state is
approximately:

-½(1+¼)+½(0.242798)= -.504

i.e., the 2*p* state lies .016×(*Z*^{2}27.2 eV)=1.8 eV
above the 2*s* state. Our 2*s*-2*p* splitting is about a
factor of two too large. The over all binding energy is fairly close to the
approximately (-58 eV)=-.533 seen experimentally--perturbation theory has
much improved the initial energy estimate of -.625. The problems at the end
of this section describe yet more accurate approximation methods.

The main point here is that the *l* degeneracy seen in the H-atom [i.e., that
all the states with the same principal quantum number *n* but
different values of *l* (0<=*l*<=*n*-1) have the
same energy: *E'*=-½(1/*n*^{2})] is
broken in multi-electron atoms. The repulsive electrostatic potential
of the orbiting electrons changes the potential from an exactly
1/*r* form to something more complex. As a result the
*ns* state ends up below the *np* states, which are in turn
below the *nd* states. We call this phenomenon *l-tilting*.
We will see that in many-electron atoms, *l*-tilting can be so
large that the 4*s* state can be below the 3*d* state.

Remember that classically the maximum *l* states (*l*=*n*-1)
are circular orbits where *r*=*a* the semimajor axis, whereas
the *l*=0 orbits range from *r*=0 to *r*=2*a*.
Thus the low-*l* states are both on average are further away from the core
electrons (and hence feel a reduced interaction) and occasionally very close
to the nucleus (and hence occasionally feel the full nuclear attraction).