# 3.3. Examples of non-linear variational calculus¶

Linear variational calculus leads to a (generalised) matrix eigenvalue
problem to be solved. Variational methods form a much wider class,
including trial functions which depend non-linearly on the variational
parameters. Numerically this is quite complicated to solve, but several
analytic non-linear variational calculations exist which give quite good
results even if only one or two variational parameters are used. A nice
example is the hydrogen atom, which we try to solve using a variational
wave function ('trial function')
of the form
You may note that this is the form of the *exact* ground state of the
hydrogen atom. However, to illustrate the variational method, we first
*relax* the value of the parameter and then vary it to minimise the
expectation value of the energy. We should then find the exact ground
state wave function and energy.

The Schrödinger equation was already given in atomic units in the previous section:

It is useful to first normalise the trial wave function: so that we have

Its is now easy to calculate the expectation value of the kinetic energy. Using the fact that for a function in 3D which only depends on , we have we find, after some calculation that For the potential energy, we find Therefore, the expectation value of the energy for the trial wave function is given by The minimum of this expression is found at and yields an energy of in units of eV, which is the correct ground state energy of eV.

Now we turn to a more complicated problem: the helium atom, which (when
the nucleus is considered not to move because of its large mass) is
described by the Hamiltonian
In atomic
units, this becomes, with :
For the trial wave function we use the form
This function is chosen such that it yields the ground state of two
*noninteracting* electrons moving in the field of the helium nucleus
(the nuclear charge leads to a scaling of 2 in the exponent). So
the trial wave function is simply the successful trial wave function (in
the sense that it contains the exact solution) for the
independent-electron case. In particular, when an electron approaches
the nucleus, its behaviour is properly described by this wave function
as the electron-nucleus interaction largely dominates the
electron-electron interaction in that case.

We have taken the wave function to be symmetric in the coordinates and . The two electrons should however form an antisymmetric wave function as they are fermions. The antisymmetry is taken care of by the spin wave function Later we shall go much deeper into the structure of many-body wave functions -- here we just mention that, as the Hamiltonian does not contain any spin dependence, we can forget about the spin part of the wave function and can safely assume that the ground state wave function is symmetric in and .

We first must normalise this solution. This can be done for and independently and we obtain so that the normalised solution reads In order to find the expectation value of the Hamiltonian for this wave function, it is convenient to write it in the form where This is the Hamiltonian for an electron moving in the helium potential.

We calculate the kinetic energy following the method used for the hydrogen atom. The result is (for the two electrons together) A quick way to arrive at this result is by taking the kinetic energy for the hydrogen atom, replacing and multiplying by 2 because we now have two electrons. The potential energy due to the attraction between the nucleus and the electrons is found to be as can be verified by taking the hydrogen result , replacing and multiplying by 2 because the nuclear charge is twice as large as for the hydrogen atom and again by 2 because we now have two electrons.

Now we must add to this the contribution from the electron repulsion: If we fix , we can evaluate the integral over . Choosing the -axis to be the direction of , we have (without the prefactor): We first evaluate the integral over . Choosing , this is of the form and we are left with The term in square brackets equals where the function returns the largest of the two numbers and . Therefore, we need to split the integral over into a part running from to and a part running from to :

Now it remains to multiply this by the normalisation factor and by , and then integrate over (there is no dependence on the two angular variables for ). All the integrals are straightforward and the final result is in units of Hartree = eV. So the total energy is given by Taking , i.e. assuming that both electrons are in the ground state of the atom with the electron-electron interaction switched off, yields an energy of The experimental value is eV, so although our result is not extremely bad, it is not impressively accurate either.

Now let us relax and find the minimum of the trial energy. This is found at Bohr radii. The energy is then found at Hartree eV.