# Variational (Rayleigh-Ritz) method and multielectron wavefunction

I am studying Rayleigh-Ritz method and I have some problem to understand what the final wavefunction looks like. Suppose that we have a system with $$N$$ electrons, atom or molecule. We are interested in solving the time indepedent Schrodinger equation:

$$H\Psi = E\Psi$$

We assume a trial wavefunction $$\Phi = \sum_{i=1}^k c_i\chi_i$$

and we are trying to find the parameters $$c_i$$ that correspond to critical points (I am omitting the mathematics and the general procedure).

After the procedure we obtain different sets of parameters, $$\{c_i \}_0$$, $$\{c_i\}_1$$, $$\{c_i\}_2$$, $$\ldots$$, $$\{c_i\}_k$$. When substituted to second equation these parameters give back $$\Phi_0$$, $$\Phi_1$$, $$\Phi_2$$, $$\ldots$$, $$\Phi_k$$. I interpret these wavefunctions as the ground state, first excited state etc.

Is this interpretation correct? What troubles me is that if $$\Phi$$ is just a linear combination of single-electron orbitals then how can we use it to represent a multi-electron wavefunction? I mean we have $$N$$ electrons, so $$\Phi$$ must have the form $$\Phi (\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N)$$ which I can't understand how it can be obtained by just taking a sum.

• In general this interpretation is not correct. A Hamiltonian set up to give you the ground state will only ever give you the ground state. If all, these are excited configurations or expansions of the ground state; they don't have a physical meaning of interpretation. Commented Apr 23, 2022 at 22:56

## One Electron

Starting with your trial wavefunction: $$\Phi = \sum^k_{i=1} c_i \chi_i,$$ you seem to be thinking about one-electron orbitals $$\chi_i(r)$$ which will add up to a one-electron molecular orbital (MO) $$\Phi$$.

In that case, when you find $$c_i$$ by minimising the energy $$E = \frac{\langle\Phi \vert \hat{H} \vert \Phi\rangle}{\langle \Phi \vert \Phi \rangle}$$ that vector of coefficients will give you the ground state one-electron wave function.

For one-electron wave functions, E can be written as a matrix equation with the one-particle operator $$\hat{H}$$, where the eigenvectors are the $$c_i$$ vectors. In that case (exact factorisation) the eigenvectors corresponding to higher eigenvalues correspond to other eigenfunctions of the system which are excited one-electron states.

If your basis of input functions $$\chi_i$$ isn't complete for the relevant space, your final ground state wave function $$\Phi$$ will only be an approximation, but the "best" approximation possible (lowest energy) in a given basis.

For example, the Hartree-Fock (HF) approximation to the one-electron hydrogen atom is analogous to the Rayleigh-Ritz (RR) approximation to the ground state. When using a standard gaussian basis, used in most quantum chemistry programs, the quality of that answer and wave function will improve as you increase the number on input basis functions $$\chi_i$$.

## Many Electrons

The simplicity of the RR approximation/ansatz relies on the fact that the equation for $$E$$ is a matrix equation that can be solved for eigenvectors $$c_i$$ which describe $$\Phi$$ in terms of $$\chi_i$$.

Many electron wave functions are anti-symmetric with respect to exchange of electrons. One method for creating an anti-symmetric ansatz from one particle wave functions is the Slater determinant $$\Psi$$. Modern approximations of many electron wave functions rely on composing wave functions of Slater determinants, which are themselves built up of one electron wave functions.

If we want to recover the linear form of the original expression for $$\Phi$$ we would need to use a basis of Slater determinants for our trial basis $$\chi_i$$, $$\Phi\left(r_1,r_2,...,r_n\right) = \sum^k_{i=1} c_i \Psi_i\left(r_1,r_2,...,r_n\right)$$

The question this begs, is how to choose a suitable many electron basis $$\Psi_i$$.

The Configuration Interaction method uses this approach - a Rayleigh-Ritz approximation using a basis of Slater determinants. It selects the Slater determinants by first solving the system with the HF approximation. This gives a set of one electron MOs, some of which are occupied and some unoccupied. By building Slater determinants with the occupied MO, then all but one occupied MO and one unoccupied MO (single excited Slater determinants), then all but two (double excited Slater determinants) etc, you can build an arbitrarily large basis of Slater determinants.

For a system containing $$N$$ electrons, $$O$$ occupied MOs and $$V$$ unoccupied MOs and $$M=O+V$$ total MOs, the number of Slater determinants grows exponentially as higher excitations are added.

• The ground state has 1 HF Slater determinant,
• There are $$VO \sim M^2$$ single excited Slater determinants,
• There are $$\frac{V(V-1)O(O-1)}{4} \sim M^4$$ double excited Slater determinants

and so on.

As the size of the basis rapidly explodes, often on the ground state and double excitations are included, giving the CID or configuration interaction doubles method, the simplest possible CI method.

Note: For symmetry reasons, the singly excited determinants aren't coupled to the HF ground state by the Hamiltonian, and would have no contribution ($$c_i=0$$) as the only excited determinant are included, the simplest CI method is CID, singly excited determinants can then be added to give CISD. See Brillouin's theorem.

### Excited states

If the whole CI matrix is diagonalised, rather than relying on a iterative method to find the lowest eigenvalue, you can get excited states represented as sum of determinants. However by constraining them to be within the basis of determinants made with HF MO they are generally considered poor approximations. The exponetially increasing size of the CI matrix typically makes calculation such eigenvectors prohibitive.

The Configuration Interaction Singles (CIS) method is essentially doing this with only the singly excited determinants to calculate approximate one-particle excitation energies.

### Problems with CI

One notable problem with CI methods is size extensivity - the property that N non-interacting copies of the same systems, e.g. 10 water molecules 1 m away from each other, has the energy of $$N$$ times that of one copy of the system. This property is true for HF for example.

Obviously this is physically true, but by limiting the total number of excitations in CI even as the system size increases, CI is not size extensive. Methods such as Couple Cluster, which approximates $$Phi$$ as a product of Slater determinants rather than a sum, have since been developed to build on the deficencies of CI.

• If I understand it correctly the procedure is still the same (when moving to many-electron problems) but we change the form of basis functions? Now the basis functions are not one electron wavefunctions but Slater-determinants, which means that the $$\sum_i c_i\chi_i$$ leads to a function of $n$ variables if the system is composed of $n$ electrons? Also the elements of the Slater matrix are what we call molecular orbitals? Commented Apr 26, 2022 at 20:21
• Yes, if you do a one electron calculation with an basis of atomic orbitals, your solutions will be molecular orbitals - linear combinations of atomic orbitals. If you take $n$ molecular orbitals you can make an $n$ electron Slater determinant. You will take $m$ Slater determinants, where $m$ is typically much much larger than $n$, and solve for $m$ $c_i$ variables. Commented Apr 26, 2022 at 21:05