This question is a continuation of my previous question https://mathematica.stackexchange.com/q/281856/

I have the Gaussian basis set, this basis is non-orthogonal. In the case of non-orthogonal basis set to find min Energy I have to solve the next equation:

$$H\vec{\psi}=EB\vec{\psi}$$ where $H$ is a Hamiltonian matrix $B$ is a matrix of overlap integrals: $$B_{i,j}=\langle\psi_i|\psi_j\rangle.$$


  1. Do I understand correctly that, as in the case of orthogonal basis set, the wave function of the ground state has the following form? $$\psi_{gs}=\sum_i {c_i\psi_i}$$ where $\psi_i$ are the basis functions, $c_i$ are the ground state eigenvector components

  2. The ground state energy ($E_{min}$) can now be calculated as follows? $$E_{gs}=\langle\psi_{gs}|H|\psi_{gs}\rangle$$ Those does the overlap integrals no longer participate in obtaining the ground state?

  3. I would like to get the average of some operator, for example the square of the coordinate on the ground state functions:

$$\langle r^2\rangle=\langle\psi_{gs}|r^2|\psi_{gs}\rangle$$In this case, the overlap integrals also do not participate in this expression in any way?

Following the AXensen's answer:

After the equation is solved $H\vec{\psi}=EB\vec{\psi}$ there is a set of the eigenvalues and the set of the eigenvectors components, but these eigenvectors are not normalized yet. Let's take a look at eigenvector normalization with a particular example:

Let the basis consist of two functions $\psi_1$ and $\psi_2$, then $\begin{pmatrix} B_{11} & B_{12}\\ B_{21}& B_{22} \end{pmatrix}$ is the overlap integrals matrix. $c_1$, $c_2$ are the eigenvector components.
Then, in order to find the normalization coefficient, it is necessary to solve the following equation:

$$\begin{pmatrix} N*c_{1}^* & N*c_{2}^* \end{pmatrix} \begin{pmatrix} B_{11} & B_{12}\\ B_{21}& B_{22} \end{pmatrix} \begin{pmatrix} N*c_{1} \\ N*c_{2} \end{pmatrix}=1$$

After that it will be possible to write down normalized wave function (for example the wave function of ground state): $\psi_{gs}=N(c_{1}\psi_1+c_{2}\psi_2)$

  • $\begingroup$ (1.) is actually answered in the thread linked (it serves for deriving the first equation in your question) $\endgroup$
    – Roger V.
    Apr 18, 2023 at 16:22
  • $\begingroup$ @AXensen, thanks! I have asked question below your answer, please have a look at it. $\endgroup$
    – Mam Mam
    Apr 18, 2023 at 23:16

1 Answer 1


As long as your groundstate is normalized, which is not the same thing as $\sum |c_i|^2=1$, then everything you said is correct.

The condition is instead:

$$ \langle \psi_{gs}|\psi_{gs}\rangle=1 $$ Which can be written conveniently as $$ \vec{c}_i^TB\vec{c}_i=1 $$ or alternatively as $$ \sum_{ij}c_j^*c_i\langle \psi_i|\psi_j\rangle $$ Which is the same thing - almost by the definition of the matrix $B$. Also be aware that for similar reasons, $$ \langle \psi_{gs}|H|\psi_{gs}\rangle\neq|c_i|^2\langle\psi_i|H|\psi_i\rangle $$

  • $\begingroup$ Thanks! Could you write how the ground state wavefunction needs to be normalized in this case? Since it's not the same as $∑|c_i|^2=1$ $\endgroup$
    – Mam Mam
    Apr 18, 2023 at 15:59
  • 1
    $\begingroup$ @MamMam done. edited my answer $\endgroup$
    – AXensen
    Apr 18, 2023 at 16:11
  • $\begingroup$ Thank you very much for the details! I have edited my question and added part with the particular example, which describes wave function normalization. Could you please look at this part. Did I understand your explanation correctly? $\endgroup$
    – Mam Mam
    Apr 18, 2023 at 23:05
  • 1
    $\begingroup$ @MamMam looks good to me $\endgroup$
    – AXensen
    Apr 19, 2023 at 9:15

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