Advanced Chemistry - Mono Mole

December 7, 2022February 28, 2025

Hartree-Fock-Roothaan method

The Hartree-Fock-Roothaan method, an extension of the Hartree-Fock method, uses spin orbitals that are linear combinations of a set of basis wavefunctions $\phi_i(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}\chi_{\alpha}(x)$ .

Unlike the Hartree-Fock method, in which the parameters $\zeta_i$ of the wavefunctions are varied, the best basis wavefunctions $\chi_{\alpha}$ are chosen in the Hartree-Fock-Roothaan method and the coefficients $c_{i\alpha}$ are instead varied in an iterative process to determine the state of the system.

To derive the Hartree-Fock-Roothaan equations, we begin by substituting $\phi_i^*(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}^*\chi_{\alpha}^*(x)$ and $\phi_i(x)=\sum_{\beta=1}^{n'}c_{i\beta}\chi_{\beta}(x)$ in eq88, where we have relabelled $\phi_r^{'*}(x)$ and $\phi_r^{'}(x)$ as $\phi_i^*(x)$ and $\phi_i(x)$ respectively, to give

$\sum_{i=1}^{n}H_{i}=\sum_{i=1}^{n}\sum_{\alpha,\beta=1}^{n'}c_{i\alpha}^*c_{i\beta}[\alpha\vert\beta]\;\;\;\;\;\;\;\;(152)$

where $[\alpha\vert\beta]=\int\chi_{\alpha}^*(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\chi_{\beta}(x)dx$ .

Next we substitute $\phi_i^*(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}^*\chi_{\alpha}^*(x)$ , $\phi_i(x)=\sum_{\beta=1}^{n'}c_{i\beta}\chi_{\beta}(x)$ , $\phi_j^*(x)=\sum_{\gamma=1}^{n'}c_{j\gamma}^*\chi_{\gamma}^*(x)$ and $\phi_j(x)=\sum_{\delta=1}^{n'}c_{j\delta}\chi_{\delta}(x)$ in eq92 to give

$\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^{n}(J_{ij}-K_{ij})=\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^{n}\sum_{\alpha,\beta,\gamma,\delta=1}^{n'}c_{i\alpha}^*c_{i\beta}c_{j\gamma}^*c_{j\delta}\{[\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]\}\;\;\;\;\;\;\;\;(153)$

where $[\alpha\beta\vert\gamma\delta]=\int\int\chi_{\alpha}^*(x)\chi_{\beta}(x)\frac{\chi_{\gamma}^*(x')\chi_{\delta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$ and $[\alpha\delta\vert\gamma\beta]=\int\int\chi_{\alpha}^*(x)\chi_{\delta}(x)\frac{\chi_{\gamma}^*(x')\chi_{\beta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$ .

Substituting eq152 and eq153 in eq94 gives

$E=\sum_{i=1}^{n}\sum_{\alpha,\beta=1}^{n'}c_{i\alpha}^*\left\{[\alpha\vert\beta]+\frac{1}{2}\sum_{j\neq i}\sum_{\gamma,\delta=1}^{n'}c_{j\gamma}^*c_{j\delta}\left([\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]\right)\right\}c_{i\beta}\;\;\;\;\;\;\;\;(154)$

The constraint for the Lagrange method is found by substituting $\phi_i^*(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}^*\chi_{\alpha}^*(x)$ and $\phi_j(x)=\sum_{\beta=1}^{n'}c_{j\beta}\chi_{\beta}(x)$ in $\int\phi_i^*(x)\phi_j(x)dx=\delta_{ij}$ to give:

$\sum_{\alpha,\beta=1}^{n'}c_{i\alpha}^*c_{j\beta}S_{\alpha\beta}=\delta_{ij}$

where $S_{\alpha\beta}=\int\chi_{\alpha}^*(x)\chi_{\beta}(x)dx$ .

Therefore, the Lagrangian is:

$F(c_{ix}^*c_{ix})=E(c_{ix}^*c_{ix})-\sum_{ij}^n\lambda_{ij}\left(\sum_{\alpha,\beta}^{n'}c_{i\alpha}^*c_{j\beta}S_{\alpha\beta}-\delta_{ij}\right)$

where $x=\alpha,\beta,\gamma,\delta$ and $\lambda_{ij}$ are the undetermined multipliers.

As shown in the derivation of the canonical Hartree-Fock equations, we can select a set of coefficients $c_{i\alpha}^*$ and $c_{j\beta}$ that diagonalises the Hermitian matrix with elements $\lambda_{ij}$ . The Lagrangian becomes

$F(c_{ix}^*c_{ix})=E(c_{ix}^*c_{ix})-\sum_{i}^n\lambda_{i}\left(\sum_{\alpha,\beta}^{n'}c_{i\alpha}^*c_{i\beta}S_{\alpha\beta}-1\right)$

The total differential of the Lagrangian is

$dF=\sum_{k=1}^n\left[\frac{\partial E}{\partial\mu}-\frac{\partial \sum_{i}^n\lambda_{i}\sum_{\alpha,\beta}^{n'}c_{i\alpha}^*c_{i\beta}S_{\alpha\beta}}{\partial\mu}\right]d\mu=0\;\;\;\;\;\;\;\;(155)$

where $\mu=c_{kx}^*,c_{kx}$ .

Eq155 has the same form as eq12. If we can find a set of values of $\lambda_{i}$ that renders the dependent variable terms of $d\mu$ zero, we are left with the independent variable terms. Consequently, all the coefficients of $d\mu$ are equal to zero and they form a set of equations that can be solved simultaneously. To simplify eq155, substitute eq154 in it to give

$\sum_{k=1}^n\biggr[\frac{\sum_{i=1}^n\sum_{\alpha,\beta}^{n'}c_{i\alpha}^*\partial c_{i\beta}[\alpha\vert\beta]}{\partial c_{k\beta}}+\frac{\sum_{i=1}^n\sum_{\alpha,\beta}^{n'}\partial c_{i\alpha}^*c_{i\beta}[\alpha\vert\beta]}{\partial c_{k\alpha}^*}$ $+\frac{1}{2}\frac{\sum_{j\neq i}\sum_{i=1}^n\sum_{\alpha,\beta,\gamma,\delta}^{n'}\partial c_{i\alpha}^*c_{i\beta}c_{j\gamma}^*c_{j\delta}[\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]}{\partial c_{k\alpha}^*}$ $+\frac{1}{2}\frac{\sum_{j\neq i}\sum_{i=1}^n\sum_{\alpha,\beta,\gamma,\delta}^{n'}c_{i\alpha}^*\partial c_{i\beta}c_{j\gamma}^*c_{j\delta}[\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]}{\partial c_{k\beta}}$ $+\frac{1}{2}\frac{\sum_{j\neq i}\sum_{i=1}^n\sum_{\alpha,\beta,\gamma,\delta}^{n'}c_{i\alpha}^*c_{i\beta}\partial c_{j\gamma}^*c_{j\delta}[\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]}{\partial c_{k\gamma}^*}$ $+\frac{1}{2}\frac{\sum_{j\neq i}\sum_{i=1}^n\sum_{\alpha,\beta,\gamma,\delta}^{n'}c_{i\alpha}^*c_{i\beta}c_{j\gamma}^*\partial c_{j\delta}[\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]}{\partial c_{k\delta}}$ $-\frac{\sum_{i=1}^n\lambda_i\sum_{\alpha,\beta}^{n'}\partial c_{i\alpha}^*c_{i\beta}S_{\alpha\beta}}{\partial c_{k\alpha}^*}-\frac{\sum_{i=1}^n\lambda_i\sum_{\alpha,\beta}^{n'}c_{i\alpha}^*\partial c_{i\beta}S_{\alpha\beta}}{\partial c_{k\beta}}\biggr]dc_{kx}=0$

The next step involves the following:

1. Renaming the dummy indices of the 5^th and 6^th terms by swapping i and j, $\alpha$ and $\gamma$ , $\beta$ and $\delta$ , and the dummy coordinates $x$ and $x'$ .
2. Noting that $[\gamma\delta\vert\alpha\beta]=[\alpha\beta\vert\gamma\delta]$ and $[\gamma\beta\vert\alpha\delta]=[\alpha\delta\vert\gamma\beta]$ .
3. Noting that $\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n(\cdots)=\frac{1}{2}\sum_{j=1}^n\sum_{i=1}^n(\cdots)=\frac{1}{2}\sum_{i\neq j}\sum_{j=1}^n(\cdots)$ (see this article for explanation).
4. Expanding each term of the equation and carrying out the partial differentiation.

We have,

$\sum_{i=1}^n\sum_{\alpha,\beta}^{n'}c_{i\alpha}^*\{[\alpha\vert\beta]+\sum_{j\neq i}^n\sum_{\gamma,\delta}^{n'}c_{j\gamma}^*c_{j\delta}\left([\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]\right)-\lambda_iS_{\alpha\beta}\}dc_{i\beta}$ $+\sum_{i=1}^n\sum_{\alpha,\beta}^{n'}c_{i\beta}\{[\alpha\vert\beta]+\sum_{j\neq i}^n\sum_{\gamma,\delta}^{n'}c_{j\gamma}^*c_{j\delta}\left([\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]\right)-\lambda_iS_{\alpha\beta}\}dc_{i\alpha}^*=0$

Switching the dummy labels $\alpha$ , $\beta$ and $\gamma$ , $\delta$ , and the dummy coordinates $x$ and $x'$ for the 1^st term on the RHS of the above equation,

$\sum_{i=1}^n\sum_{\alpha,\beta}^{n'}c_{i\beta}^*\{[\beta\vert\alpha]+\sum_{j\neq i}^n\sum_{\gamma,\delta}^{n'}c_{j\delta}^*c_{j\gamma}\left([\beta\alpha\vert\delta\gamma]-[\beta\gamma\vert\delta\alpha]\right)-\lambda_iS_{\beta\alpha}\}dc_{i\alpha}$ $+\sum_{i=1}^n\sum_{\alpha,\beta}^{n'}c_{i\beta}\{[\alpha\vert\beta]+\sum_{j\neq i}^n\sum_{\gamma,\delta}^{n'}c_{j\gamma}^*c_{j\delta}\left([\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta]\right)-\lambda_iS_{\alpha\beta}\}dc_{i\alpha}^*=0\;\;\;\;(156)$

where the 1^st term of the above equation is the complex conjugate of the 2^nd term.

Since all coefficients of $dc_{i\alpha}$ and $dc_{i\alpha}^*$ are equal to zero, we have

$\sum_{\beta}^{n'}(H_{\alpha\beta}-\lambda_iS_{\alpha\beta})c_{i\beta}=0\;\;\;\;\;\;\;\;(157)$

where $H_{\alpha\beta}=[\alpha\vert\beta]+\sum_{j\neq i}^n\sum_{\gamma,\delta}^{n'}c_{j\gamma}^*c_{j\delta}([\alpha\beta\vert\gamma\delta]-[\alpha\delta\vert\gamma\beta])$ .

Similarly, the complex conjugate in eq156 gives:

$\sum_{\alpha}^{n'}(H_{\beta\alpha}-\lambda_iS_{\beta\alpha})c_{i\alpha}=0\;\;\;\;\;\;\;\;(158)$

Eq157 and eq158 are known as the Hartree-Fock-Roothaan equations.

Eq157 has non-trivial solutions if the determinant $\vert H_{\alpha\beta}-\lambda_iS_{\alpha\beta}\vert$ equals to zero. The computation process involves:

Evaluating the integrals $H_{\alpha\beta}$ and $S_{\alpha\beta}$ either analytically or numerically using initial guess values of $c_{i\beta}$ , and with basis wavefunctions where the parameters are fixed.
Substituting the evaluated integrals in the characteristic equation and solving for $\lambda_i$ , which is then used to obtain improved values of $c_{i\beta}$ for the next iteration.
Repeating steps 1 and 2 with the improved values of $c_{i\beta}$ from the previous iteration until self-consistency is attained.

NEXT article: Solution of the Hartree-Fock-Roothaan equations for helium

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December 7, 2022May 14, 2026

Antisymmetriser

An antisymmetriser $\hat{A}$ is an operator that makes a wavefunction antisymmetric.

Consider a system of two-electrons, each of which is described by a spin-orbital.

Question

What is the difference between an orbital and a spin-orbital?

Answer

An orbital is a one-electron spatial wavefunction $\psi(\boldsymbol{\mathit{r}})$ , i.e. a function that gives the best estimated energy eigenvalue of an electron when operated on by the Hamiltonian. For a given spatial orbital $\psi(\boldsymbol{\mathit{r}})$ , we can form two spin-orbitals $\phi_1=\psi(\boldsymbol{\mathit{r}})\alpha(\omega)$ and $\phi_2=\psi(\boldsymbol{\mathit{r}})\beta(\omega)$ , where $\alpha(\omega)$ and $\beta(\omega)$ are spin functions. We say that the spatial orbital is doubly occupied by two electrons with the same energy. The abbreviated symbol of a spin-orbital is $\phi_i(x_j)$ , where $x_j$ represents the set of all 4 coordinates (3 spatial coordinates and 1 spin coordinate) associated with the electron.

The symmetric and antisymmetric forms of the total wavefunction of the system are $\Psi_s(x_1,x_2)=\phi_1(x_1)\phi_2(x_2)$ and $\Psi_a(x_1,x_2)=\frac{1}{\sqrt 2}\left[\phi_1(x_1)\phi_2(x_2)-\phi_2(x_1)\phi_1(x_2)\right]$ respectively. We have

$\hat{A}\Psi_s=\Psi_a\;\;\;\;\;\;\;\;(59)$

where

1. 1. $\hat{A}=\frac{1}{\sqrt 2}\left(\hat I-\hat P_{12}\right)$ .
  2. $\hat{I}$ is the identity operator.
  3. $\hat P_{12}$ , like the exchange operator, swaps the labels of any two identical particles, i.e. $\hat P_{12}\Psi(q_1,\cdots,q_i,\cdots,q_k,\cdots,q_n)=\Psi(q_1,\cdots,q_k,\cdots,q_i,\cdots,q_n)$

Eq59 implies that the antisymmetriser transforms the symmetric form of the wavefunction into a linear combination of states, each with a distinct permutation of spin-orbital labels.

We can also expressed the antisymmetriser as $\hat{A}=\frac{1}{\sqrt 2}\sum_{r=1}^2(-1)^r\hat P_r$ , where an even (odd) numbered r represents an even (odd) number of times the labels are permutated. When the antisymmetriser acts on the wavefunction of an n-electron system, $\Psi_s(x_1,x_2,\cdots,x_n)=\phi_1(x_1)\phi_2(x_2)\cdots\phi_n(x_n)$ , we would expect the linear combination to have $n!$ terms since there are $n!$ ways to permutate the labels. The antisymmetriser is therefore:

$\hat{A}=N\sum_{r=1}^{n!}(-1)^r\hat{P}_r\;\;\;\;\;\;\;\;(60)$

where N is the normalisation constant.

To evaluate N, we have

$N^2\int\cdots\int\Psi_a^*(x_1,x_2,\cdots,x_n)\Psi_a(x_1,x_2,\cdots,x_n)dx_1\cdots dx_n=1$

Due to the orthonormal property of $\phi_i(x_j)$ , we find, after expanding the product of the linear combinations of $\Psi_a(x_1,x_2,\cdots,x_n)$ and its complex conjugate, that the integrals result in $n!$ terms of unity. Hence,

$N=\frac{1}{\sqrt{n!}}\;\;\;\;\;\;\;\;(61)$

Substituting eq61 in eq60 yields:

$\hat{A}=\frac{1}{\sqrt{n!}}\sum_{r=1}^{n!}(-1)^r\hat{P}_r\;\;\;\;\;\;\;\;(62)$

Next article: Properties of the Antisymmetriser

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