Unlike for hydrogen, a closed-form solution to the Schrödinger equation for the helium atom has not been found.

However, various approximations, such as the Hartree–Fock method, can be used to estimate the ground state energy and wavefunction of the atom.

[2] Its success was considered to be one of the earliest signs of validity of Schrödinger's wave mechanics.

The Hamiltonian of helium, considered as a three-body system of two electrons and a nucleus and after separating out the centre-of-mass motion, can be written as[citation needed]

The Hamiltonian operator, since it only acts on the spatial component, gives the eigenvector equation:

This energy, however, is not degenerate with multiplicity given by the dimension of the space of combined spin states because of a symmetrization postulate, which requires that physical solutions for identical fermions should be totally antisymmetric, imposing a restriction on the choice of

Since it is also rotationally invariant, the total x, y or z component of angular momentum operator also commutes with the Hamiltonian.

Alternatively, a more generalized representation of the above can be provided without considering the spatial and spin parts separately.

A convenient basis consists of one anti-symmetric matrix (with total spin

in the above (scalar) Hamiltonian are neglected (e.g. an external magnetic field, or relativistic effects, like angular momentum coupling), the four Schrödinger equations can be solved independently.

[6][4] This is identical to the previously discussed method of finding spatial wavefunction eigenstates independently of the spin states, here spatial wavefunctions of different spin states correspond to the different components of the matrix.

need not correspond to physical states of identical electrons as per the symmetrization postulate.

The proper wave function then must be composed of the symmetric (+) and antisymmetric(−) linear combinations:

A good theoretical descriptions of helium including the perturbation term can be obtained within the Hartree–Fock and Thomas–Fermi approximations (see below).

However it is just an approximation, and there are more accurate and efficient methods used today to solve atomic systems.

The "many-body problem" for helium and other few electron systems can be solved quite accurately.

We can find the first order correction in energy due to electron repulsion in Hamiltonian

[8] A better approximation for ground state energy is obtained by choosing better trial wavefunction in variational method.

As Z gets bigger, our approach should yield better results, since the electron-electron repulsion term will get smaller.

The net effect of each electron on the motion of the other one is to screen somewhat the charge of the nucleus, so a simple guess for V(r) is

If Ze was 1.70, that would make the expression above for the ground state energy agree with the experimental value E0 = −2.903 a.u.

For the ground state of helium, for the average shielding approximation, the screening effect of each electron on the other one is equivalent to about

[9] To obtain a more accurate energy the variational principle can be applied to the electron-electron potential Vee using the wave function

This is closer to the experimental value, but if a better trial wave function is used, an even more accurate answer could be obtained.

An ideal wave function would be one that doesn't ignore the influence of the other electron.

Treating Z as a variational parameter to minimize H. The Hamiltonian using the wave function above is given by:

This shows that the other electron somewhat shields the nucleus reducing the effective charge from 2 to 1.69.

Since due to the symmetrization postulate, the combined spatial wavefunctions differ on symmetric or antisymmetric nature, the J term is responsible for the splitting of energy levels between ortho and para helium states.

The first integral is said to be analogous to classical potential due to Coulomb interaction, where the squares of wavefunctions are interpreted as electron density.

The Schrodinger equation for helium, like that of hydrogen, can be solved to accuracies equivalent to the most precise experimental values.