How to obtain excited state energies from configuration interaction? (Helium example)

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:



```
-----------------------------
CIS-EXCITED STATES (SINGLETS)
-----------------------------

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)
-----------------------
CIS/TD-DFT TOTAL ENERGY
-----------------------

    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*.


