Helium atom

Perturbative method[edit | edit source]

The hamiltonian describing the helium atom ( : two electrons and a nuclear charge of ) is :

Let us consider as unperturbed hamiltonian
The solutions of each are

Since is the sum of and its solutions are tensor products of single- particle states, properly antisymmetrized according to the symmetrization postulate. Let us discuss the ground state: two electrons on the level, with opposite spins because of Pauli's exclusion principle. We can obtain the appropriate antisymmetric wavefunction using Slater determinant

The order energy value is
Now we consider the electron-electron repulsion as a perturbation and evalue the first order correction to the ground-state energy:
Before solving the direct integral it's important to give it a physical meaning:

  • can be seen as a charge density associated with electron 1 since it's the product of a charge and a spatial probability distribution. Similarly can be seen as the charge density associated with the other electron
  • In this perspective it's easy to see that the direct integral represents the energy associated with coulomb interaction between electrons

The result of such an integral is , which is positive and thus increases the ground state energy. In the case of helium, with , we obtain:

This result has to be compared with the experimental value .

Variational method[edit | edit source]

Theorem

We can improve the previous result by using a different approach, based on the following theorem:

 
Proof

Let be the set of eigenstates of . Since they constitute a complete orthonormal system we can expand each on this set:

Computing we have:

 

We can take advantage of this fact choosing a ground-state single particle wavefunction depending on a variational parameter:

The wavefunction of helium ground state is thus:
We have neglected both the spin part and the angular part of the wave function since they will not contribute to the following calculation. Now we apply the previous theorem:
That means we have to find the minimum of in order to get the best approximation, that is computing
Now
We need to evaluate two terms:
The latter is simple to evaluate since
so that this term gives
The former can be easily computed writing
so that
With these result we can compute
Differentiation with respect to gives
this result must equal 0 if we are looking for the minimum:
In case of helium we find , which is greater than : with this method electron screening has been taken into account. If we put this value of in the function we get the wavefunction of the ground state, and if we compute the energy we get
Comparison with experimental value and the one obtained with perturbation theory lead us to conclude that this is a better approximation

SUMMARY:

The hamiltonian of this system can be written as:

Where the last term can be seen as a perturbation of a two electrons-system hamiltonian:
Using the perturbation theory we find:
Solving the integral we get:
While the unperturbed energy is:
Let us state one important principle, called variational principle:
Where stands for the ground state energy of the system. A more precise result for the energy can be found looking for an "effective" charge and by evaluating at this . We get[1]:
And the value can be derived looking for a minimum in the energy. Hence:
For this particular hamiltonian we have that , which is just a consequence of the virial theorem:
Where is any eigenstate of the hamiltonian. Remember that the ground state wf of the He atom is given by the product of two hydrogen-like, ground state, wave functions. That is:
Hence, is an hydrogen-like wf and its expression is:
Actually, this is just the radial part of the wf but for most exercises and applications this is what we need.

  1. This expression holds for any He-like hamiltonian.
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