Roothaan Hall Equations:

The Hartree-Fock equations are a set of modified Schrodinger equations:



  • The Fock operator ($f_{i}$) is given by (restricted case):

    $f_{i}= \hat{h}_{i}+\sum_{j=1}^{n/2}[2\hat{J}_{j}(i)-\hat{K}_{j}(i)]$

  • and the molecular orbitals are expressed as a linear combination of ($N_{b}$) atomic orbitals ($\chi_{o}$):

    $\psi_{m} =\sum_{o=1}^{N_{b}}c_{om}\chi_{o}$

By substituting $\psi_{m}$, one obtains:

$f_{i}\sum_{o=1}^{N_{b}}c_{om}\chi_{o}= \epsilon_{m}\sum_{o=1}^{N_{b}}c_{om}\chi_{o}$

If one now multiplies from the left by $\chi_{o'}$, and integrates over the coordinates of particle i:

$\sum_{o=1}^{N_{b}}c_{om}\int\chi_{o'}f_{i}\chi_{o}dr_{1}= \epsilon_{m}\sum_{o=1}^{N_{b}}c_{om}\int\chi_{o'}\chi_{o}dr_{1}$

$\sum_{o=1}^{N_{b}}F_{o'o}c_{om}= \epsilon_{m}\sum_{o=1}^{N_{b}}S_{o'o}c_{om}$


  • $F_{o'o}=\int\chi_{o'}f_{i}\chi_{o}dr_{1}$

  • $S_{o'o}=\int\chi_{o'}\chi_{o}dr_{1}$

The expression has the form of a relation between matrix elements of the product matrices FC and SC. If one introduces the diagonal matrix $\epsilon$ along the diagonal, the expression can be written as the matrix equality:

$FC = SC\epsilon$

An Example:

To set up the Roothann equations for the HF molecule using the $N_{b} = 2$ basis set H1s ($\chi_{a}$) and F2p$_{z}$ ($\chi_{a}$) one can write the two molecular orbitals (m = a, b) as:

$\psi_{a}=c_{Aa}\chi_{A} + c_{Ba}\chi_{B}$

$\psi_{b}=c_{Ab}\chi_{A} + c_{Bb}\chi_{B}$

The following matrices are obtained:

$F = \begin{bmatrix}F_{A}(A)&F_{A}(B)\\ F_{B}(A)&F_{B}(B)\end{bmatrix}$

$S = \begin{bmatrix} 1 & S \\ S & 1 \end{bmatrix}$

$C = \begin{bmatrix} c_{Aa} & c_{Ab} \\ c_{Ba} & c_{Bb} \end{bmatrix}$

Then the Roothan equations ($FC=SC\epsilon$) are:

$\begin{bmatrix}F_{AA}&F_{AB}\\ F_{BA}&F_{BB}\end{bmatrix} \begin{bmatrix} c_{Aa} & c_{Ab} \\ c_{Ba} & c_{Bb} \end{bmatrix} = \begin{bmatrix} 1 & S \\ S & 1 \end{bmatrix} \begin{bmatrix} c_{Aa} & c_{Ab} \\ c_{Ba} & c_{Bb} \end{bmatrix} \begin{bmatrix} \epsilon_{a} & 0 \\ 0 & \epsilon_{b} \end{bmatrix}$


In many textbooks and lectures, the Roothan equations are often described after an introduction to the Slater determinant.

If the total wavefunction from the HF example can be written in the form of a Slater determinant:

$\Psi = \frac{1}{\sqrt{N!}}\begin{bmatrix} \psi_{a}(i)&\psi_{b}(i)\\ \psi_{a}(j)&\psi_{b}(j)\end{bmatrix}$

How is the normalization constant used in the Slater determinant built into these equations?

Note: The derivation has come from Atkins' Physical Chemistry 9th Edition


2 Answers 2


The normalization constant is built into the Roothaan-Hall equation. Namely, when you derive the Hartree-Fock energy expression from the wave function, you integrate out all the orbitals that don't "touch" the Hamiltonian, and the permutations of the orbitals "kill off" the normalization constant.

Now, when you have the energy expression, you can substitute the expansion of the orbitals in the basis set, and vary the energy with respect to the expansion coefficients; you get the Roothaan-Hall equations as a result.


The normalization constant for the Slater determinant is not incorporated into these equations. The HF equations give you the molecular orbitals while the Slater determinant is just a way of putting those orbitals together to make the full electronic wavefunction. You can find the orbitals without worrying about the wavefunction normalization.

  • $\begingroup$ Is this in contradiction with the answer by @SusiLehtola? $\endgroup$
    – Camps
    Commented Aug 31, 2020 at 14:30
  • 3
    $\begingroup$ @Camps I believe I misinterpreted the question. I took it to mean "does Roothaan-Hall incorporate the wavefunction normalization into the orbital?" that is, could I evaluate the wavefunction as just the determinant (not the Slater determinant) of the orbitals. Susi has the right idea that they are asking about the incorporation of this normalization condition into the derivation of the Roothaan-Hall equations. $\endgroup$
    – Tyberius
    Commented Aug 31, 2020 at 14:55

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