Editing Slater determinant (section)

== Example: Matrix elements in a many electron problem ==
{{See also|Slater-Condon rules}}
Many properties of the Slater determinant come to life with an example in a non-relativistic many electron problem.<ref name="ReferenceA">Solid State Physics - Grosso Parravicini - 2nd edition pp.140-143</ref>
* ''The one particle terms of the Hamiltonian will contribute in the same manner as for the simple Hartree product, namely the energy is summed and the states are independent''
* ''The multi-particle terms of the Hamiltonian will introduce exchange term to lower of the energy for the anti-symmetrized wave function''

Starting from a [[molecular Hamiltonian]]:
<math display="block">\hat{H}_\text{tot} = \sum_i \frac{\mathbf{p}^2_i}{2 m} + \sum_I \frac{\mathbf{P}^2_I}{2 M_I} + \sum_i V_\text{nucl}(\mathbf{r_i}) + \frac{1}{2}\sum_{i \ne j} \frac{e^2}{|\mathbf{r}_i-\mathbf{r}_j|} + \frac{1}{2}\sum_{I \ne J} \frac{Z_I Z_J e^2}{|\mathbf{R}_I-\mathbf{R}_J|}</math>
where <math>\mathbf{r}_i</math> are the electrons and <math>\mathbf{R}_I</math>are the nuclei and
: <math>V_\text{nucl}(\mathbf{r})= - \sum_I \frac{Z_I e^2}{|\mathbf{r}-\mathbf{R}_I|}</math>

For simplicity we freeze the nuclei at equilibrium in one position and we remain with a simplified Hamiltonian
: <math>\hat{H}_e = \sum^N_i \hat{h}(\mathbf{r}_i) + \frac{1}{2}\sum^N_{i \ne j} \frac{e^2}{r_{ij}} </math>
where
: <math>\hat{h}(\mathbf{r}) = \frac{\hat{\mathbf{p}}^2}{2m} + V_\text{nucl}(\mathbf{r})</math>

and where we will distinguish in the Hamiltonian between the first set of terms as <math>\hat{G}_1</math> (the "1" particle terms) 
and the last term <math>\hat{G}_2</math> (the "2" particle term) which contains exchange term for a Slater determinant.

: <math>\hat{G}_1 =\sum^N_i \hat{h}(\mathbf{r}_i) </math>
: <math>\hat{G}_2 =\frac{1}{2} \sum^N_{i \ne j} \frac{e^2}{r_{ij}}</math>

The two parts will behave differently when they have to interact with a Slater determinant wave function. We start to compute the expectation values of one-particle terms
: <math>\langle\Psi_0 |G_1 | \Psi_0\rangle = \frac{1}{N!}\langle \det\{\psi_1 ... \psi_N\}|G_1|\det\{\psi_1 ... \psi_N\}\rangle </math>
In the above expression, we can just select the identical permutation in the determinant in the left part, since all the other N! − 1 permutations would give the same result as the selected one. We can thus cancel N! at the denominator
: <math>\langle\Psi_0 |G_1 | \Psi_0\rangle = \langle\psi_1 ... \psi_N|G_1|\det\{\psi_1 ... \psi_N\}\rangle</math>
Because of the orthonormality of spin-orbitals it is also evident that only the identical permutation survives in the determinant on the right part of the above matrix element
: <math>\langle\Psi_0 |G_1 | \Psi_0\rangle = \langle\psi_1 ... \psi_N|G_1|\psi_1 ... \psi_N\rangle</math>
This result shows that the anti-symmetrization of the product does not have any effect for the one particle terms and it behaves as it would do in the case of the simple Hartree product.

And finally we remain with the trace over the one-particle Hamiltonians
: <math>\langle\Psi_0 |G_1 | \Psi_0\rangle = \sum_i \langle\psi_i|h|\psi_i\rangle</math>
Which tells us that to the extent of the one-particle terms the wave functions of the electrons are independent of each other and the expectation value of total system is given by the sum of expectation value of the single particles.

For the two-particle terms instead
: <math>\langle\Psi_0 |G_2 | \Psi_0\rangle = \frac{1}{N!}\langle\det\{\psi_1 ... \psi_N\}|G_2|\det\{\psi_1 ... \psi_N\}\rangle = \langle\psi_1 ... \psi_N|G_2|\det\{\psi_1 ... \psi_N\}\rangle</math>

If we focus on the action of one term of <math>G_2</math>, it will produce only the two terms
: <math> \langle\psi_1(r_1,\sigma_1) ... \psi_N(r_N, \sigma_N) |\frac{e^2}{r_{12}}|\mathrm{det}\{\psi_1(r_1,\sigma_1) ... \psi_N(r_N,\sigma_N)\}\rangle= \langle\psi_1\psi_2|\frac{e^2}{r_{12}}|\psi_1\psi_2\rangle - \langle\psi_1\psi_2|\frac{e^2}{r_{12}}|\psi_2\psi_1\rangle</math>

And finally
<math display="block">\langle\Psi_0 |G_2 | \Psi_0\rangle = \frac{1}{2}\sum_{i\ne j}\left[ \langle\psi_i \psi_j | \frac{e^2}{r_{ij}} |
\psi_i \psi_j \rangle - \langle\psi_i \psi_j | \frac{e^2}{r_{ij}} |
\psi_j \psi_i \rangle \right] </math>

which instead is a mixing term. The first contribution is called the "coulomb" term or "coulomb" integral and the second is the "exchange" term or exchange integral. Sometimes different range of index in the summation is used <math display="inline">\sum_{ij}</math> since the Coulomb and exchange contributions exactly cancel each other for <math>i = j</math>.

It is important to notice explicitly that the exchange term, which is always positive for local spin-orbitals,<ref>See appendix I in {{ cite journal|doi=10.1103/RevModPhys.23.69 | title=New Developments in Molecular Orbital Theory |first1= C. C. J. |last1= Roothaan | page=69 | volume=23 |year=1951 |journal= Reviews of Modern Physics | issue=69| url=http://elib.bsu.by/handle/123456789/154388 }}</ref> is absent in the simple Hartree product. Hence the electron-electron repulsive energy <math>\langle\Psi_0 |G_2 |\Psi_0\rangle</math> on the antisymmetrized product of spin-orbitals is always lower than the electron-electron repulsive energy on the simple Hartree product of the same spin-orbitals. Since exchange bielectronic integrals are different from zero only for spin-orbitals with parallel spins, we link the decrease in energy with the physical fact that electrons with parallel spin are kept apart in real space in Slater determinant states.