A theory of CI method

For going beyond the Hartree–Fock (one determinant) approximation one introduce configuration interaction (CI). Interaction means the mixing (interaction) of different electronic configurations (states).

Algorithm of usage CI method:

  1. The solution of the Hartree-Fock equations gives a set of $ M $ spin orbitals, and to construct Slater's determinant use only $ N $, which correspond to the minimum orbital energies (Aufbau principle).

  2. Some of the remaining $ M-N $ functions (corresponding to virtual orbitals) are used to construct additional Slater determinants.

    These determinants are obtained by substituting a certain number of spin orbitals of the original determinant $ \Phi^{0} $ for the corresponding number of virtual spin orbitals. The obtained determinants are called excited and are denoted as $\Phi_K $.

    The wave function has the form: \begin{equation} \Phi = \Phi^{0} + \sum_{K = 1}^{L} C_K\Phi_K. \end{equation}

  3. The search for $ \Phi $ is reduced to the variational problem of minimizing electronic energy by varying the coefficients $ C_K $.

  4. When forming the set $ \{\Phi_K\}_ {1, \ldots, L} $ often are limited to determinants, single- and double-excited with respect to $ \Phi^{0} $.

Heluim example

SCF for Helium ground state singlet gives $$ \phi_1 = 0.842 \chi_1 + 0.183 \chi_2, \quad \text{with orbital energy}\quad \epsilon_1= -0.918 $$ and $$ \phi_2 = -1.620 \chi_1 + 1.816 \chi_2, \quad \text{with orbital energy}\quad \epsilon_2= 2.810. $$ The total HF ground state energy $E_\text{total} = -2.862~E_\mathrm{h}$.

The ground state Slater determinant is $$ \Phi^{0} = |\phi_1 \overline{\phi_1}|, $$ and excited one $$ \Phi^{1} = |\phi_2 \overline{\phi_2}|. $$

So I can construct a secular equation: $$ \begin{vmatrix} \langle \Phi_0| \hat{H} | \Phi_0 \rangle - E & \langle \Phi_0| \hat{H} | \Phi_1 \rangle \\ \langle \Phi_0| \hat{H} | \Phi_1 \rangle & \langle \Phi_1| \hat{H} | \Phi_1 \rangle - E \end{vmatrix} = \begin{vmatrix} -2.862 - E & 0.2895 \\ 0.2895 & 3.226 - E \end{vmatrix} =0 $$ and get two $E$'s: $E_{1,2} = -2.876, 3.24~E_\mathrm{h}$.

So, using CI method I get a profit in ground state energy: $-2.876$ instead $-2.862$.

Using a CI method in Quantum Chemistry Computational soft (for example, ORCA) I also can get a correct excited state energies of Helium atom.

Now a question. The energy $E_2 =3.24~E_\mathrm{h}$ is not at all like the energy of an excited state, because it is huge and positive. Since ORCA calculation gives:


the weight of the individual excitations are printed if larger than 1.0e-02

STATE  1:  E=   1.908913 au     51.944 eV   418958.0 cm**-1 <S**2> =   0.000000
     0a ->   1a  :     1.000000 (c=  1.00000000)

    E(SCF)  =     -2.855160479 Eh
    DE(CIS) =      1.908912949 Eh (Root  1)
    ----------------------------- ---------
    E(tot)  =     -0.946247530 Eh

How can I obtain correct result for energy of a first excited state doing by hand?

The examples are taken from Quantum Chemistry 7th Edition by Ira Levine.



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