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I seem to be having a conceptual problem with HF theory as to where the Fock operator comes from. In order to explain the context of my conceptual problem, I will include how HF was derived according to the resources I have used:

Starting from the electronic time-independent Schrodinger equation (assume Born-Oppenheimer approximation), the electronic Hamiltonian containing the correlated motion of the set of all electrons $\boldsymbol{\vec{r}}$: $$ \hat{H}_{elec}(\boldsymbol{\vec{r}})\Psi_{elec}(\boldsymbol{\vec{r}})=E_{elec}\Psi_{elec}(\boldsymbol{\vec{r}}) $$ $$ \hat{H}_{elec}(\boldsymbol{\vec{r}})=-\sum_{i=1}^{N}\frac{1}{2}\nabla_{i}^{2}-\sum_{i=1}^{N}\sum_{k=1}^{M}\frac{Z_{k}}{\vec{r}_{ik}}+\sum_{i<j}^{N}\frac{1}{\vec{r}_{ij}} $$ Step 1: Assume (incorrectly) no electron correlation. The many-electron electronic Hamiltonian operator $\hat{H}_{elec}$ becomes separable into one-electron Hamiltonian operators $\hat{h}_{i}$: $$ \hat{H}_{elec}(\boldsymbol{\vec{r}})=\sum_{i=1}^{N}\hat{h}_{i}(\vec{r}_{i}) $$ $$ \hat{h}_{i}(\vec{r}_{i})=\frac{1}{2}\nabla_{i}^{2}-\sum_{k=1}^{M}\frac{Z_{k}}{\vec{r}_{ik}} $$ As a result, the many-electron wavefunction $\Psi_{elec}$ is expressed as the product of one-electron wavefunctions $\psi_{i}$, and the total electronic energy $E_{elec}$ is the sum of one-electron energies $\epsilon_{i}$: $$ \Psi_{elec}(\boldsymbol{\vec{r}})=\prod_{i=1}^{N}\psi_{i}(r_{i}) $$ $$ E_{elec}=\sum_{i=1}^{N}\epsilon_{i} $$ Step 2: The assumption that electrons are not correlated is severe; however, use the product of one-electron wavefunctions as a first order approximation for which is expanded upon in the following steps. $$ \Psi_{elec}(\boldsymbol{\vec{r}})\approx\prod_{i=1}^{N}\psi_{i}(r_{i}) $$ The one-electron product wavefunction must be modified to include spin and the antisymmetry of electrons: $$ \Psi(\vec{\boldsymbol{x}})=\frac{1}{\sqrt{N!}}\begin{vmatrix} \chi_{i}(\vec{x}_{i})&\chi_{j}(\vec{x}_{i})&\cdots&\chi_{N}(\vec{x}_{i})\\ \chi_{i}(\vec{x}_{j})&\chi_{j}(\vec{x}_{j})&\cdots&\chi_{N}(\vec{x}_{j})\\ \vdots&\vdots&\ddots&\vdots\\ \chi_{i}(\vec{x}_{N})&\chi_{j}(\vec{x}_{N})&\cdots&\chi_{N}(\vec{x}_{N}) \end{vmatrix} $$ where $\chi_{i}(\vec{x}_{i})=\psi_{i}(\vec{r}_{i})\sigma_{i}(\omega_{i})$

Step 3: Generate an energy expression using the exact many-electron electronic Hamiltonian operator and antisymmetric wavefunction: $$ E_{elec}=\left \langle \Psi(\vec{\boldsymbol{x}})|\hat{H}_{elec}|\Psi(\vec{\boldsymbol{x}}) \right \rangle=\sum_{i=1}^{N}h_{i}+\sum_{i<j}^{N}(J_{ij}-K_{ij}) $$ where the one-electron energies are: $$ h_{i}=\left \langle i|\hat{h}_{i}|i \right \rangle=\int\chi_{i}(\vec{x}_{i})\left [\sum_{i=1}^{N}\left(-\frac{1}{2}\nabla_{i}^{2}-\sum_{k=1}^{M}\frac{Z_{k}}{\vec{r}_{ik}}\right)\right]\chi_{i}(\vec{x}_{i})d\vec{x}_{i} $$ and the two-electron energies are: $$ J_{ij}=\left [ ii|jj \right ]=\iint\chi_{i}^{\ast}(\vec{x}_{i})\chi_{i}(\vec{x}_{i})\left [\frac{1}{r_{ij}} \right ]\chi_{j}^{\ast}(\vec{x}_{j})\chi_{j}(\vec{x}_{j})d\vec{x}_{j}d\vec{x}_{i} $$ $$ K_{ij}=\left [ ij|ji \right ]=\iint\chi_{i}^{\ast}(\vec{x}_{i})\chi_{j}(\vec{x}_{i})\left [ \frac{1}{r_{ij}} \right ]\chi_{j}^{\ast}(\vec{x}_{j})\chi_{i}(\vec{x}_{j})d\vec{x}_{j}d\vec{x}_{i} $$ Step 4: Minimize energy of antisymmetric wavefunction with respect to all spin orbitals $\chi_{a}(\vec{x}_{i})$, with the constraint that all spin orbitals remain orthonormal during energy minimization. Lagrange's method of undetermined multipliers is implemented to find an energy minimum under the orthonormality constraint, where the Lagrange function $L\left[\left\{\chi_{a}\right\}\right]$ is expressed as follows: $$ L\left[\left\{\chi_{a}\right\}\right]=E_{0}\left[\left\{\chi_{a}\right\}\right]-\sum_{a=1}^{N}\sum_{b=1}^{N}\epsilon_{ab}(\int\chi_{a}^{\ast}\chi_{b}-\delta_{ab}) $$ The Lagrange function is then differentiated with respect to the set of spin orbitals and the following expression results: $$ \hat{f}_{i}(x_{i})\chi_{a}(x_{i})=\sum_{b=1}^{N}\epsilon_{ba}\chi_{b}(x_{i}) $$

$$ \hat{f}_{i}(\vec{x}_{i})=\hat{h}_{i}(\vec{x}_{i})+\sum_{b=1}^{N}\int\left [\chi_{b}^{\ast}(\vec{x}_{j})\frac{1}{r_{ij}}\left ( 1-\hat{P}_{ab} \right )\chi_{b}(\vec{x}_{j})d\vec{x}_{j} \right] $$

$$ \hat{H}_{elec}(\boldsymbol{\vec{x}})\approx\sum_{i=1}^{N}\hat{f}_{i}(\vec{x}_{i}) $$

Confusion: The moment I get confused occurs around around the mathematics surrounding step 4. After generating a minimum energy expression and differentiating the Lagrange function with respect to the set of spin orbitals, somehow the noncanonical form of the Hartree-Fock equations results, containing the approxmate Fock operator.

It seems to me that the exact form of the many-electron electronic Hamiltonian operator is used when the energy expression is generated (step 3), but at the end of energy minimization with the orthonormality restraint (step 4) we end up with the approximate many-electron electronic Hamiltonian.

My Question: At what point is the Fock operator introduced?

A) Does the Fock operator come in because of Step 2— as the natural result of generating a minimum energy expression by assuming the many-electron wavefunction takes the form of an antisymmetrized determinant of one-electron spin orbitals?

B) Does the Fock operator come in during Step 4— as the result of substituting an approximate expression for two-electron integrals in place of the exact expression for two-electron integrals while generating a minimum energy expression?

C) Does the Fock operator comes from somewhere else that I did not think of?

D) Am talking out my shorts, and I have no idea what is actually going on? (very possible)

Nevertheless, I appreciate anyone who is able to take the time to help me clear my confusion.

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Your steps 1 and 2 are unnecessary. The single determinant variant of Hartree-Fock is simply derived by assuming that the wave function is a single Slater determinant; after all, this is the simplest possible Ansatz for the wave function that is okay for fermionic particles like electrons.

With this wave function, one can derive the expectation value for the total energy; note that it really is just an expectation value, since the one-determinant wave function cannot be an eigenfunction of the molecular Hamiltonian. This expression contains the two-electron interactions: the Coulomb term which is also called the direct term, and the exchange term, where one has permuted the indices of the electrons. You can derive this expression in second quantization with e.g. Wick's theorem.

Now, by varying the total energy with respect to the value of the orbital with the restriction of having an orthonormal set of orbitals, you get the equation that determines the orbitals. Here, you get the Fock operator.

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  • $\begingroup$ So, I still think I am a bit confused: When the minimum energy expression is generated with the constraint that the orbitals remain orthonormal, the energy expectation term contains the two-electron integrals with respect to both x(1) and x(2). During simplification, to arrive at the Fock operator: is there a substitution made, where the instead, the exact two-electron integrals are substituted with a mean-field integral with respect to solely x(2), or is there no substitution involved, and the mean-integral arises simply as a result of using the single determinant of one-electron orbitals? $\endgroup$ Apr 21 at 19:43
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    $\begingroup$ There's no substitution, it just comes out from the single determinant approximation. Note that the Fock operator is non-local and is only defined through its action on a state; that's why we call it an operator. When you feed in a state with x1, you get out a state with x2 as the coordinate. $\endgroup$ Apr 22 at 9:02

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