Orbitals and Pauli Exclusion Principle without Slater determinants

I'm trying to understand how to apply the pauli exclusion principle to the Born-Oppenheimer approximation of molecular structure.


Let me preface this by saying I'm not a chemist nor physicist. I'm a mathematician by trade which has the side affect of understanding electronic structure as solutions to a PDE.


So given some set of fixed nuclei and a given number of electrons I can construct the Schroedinger equation:


$$ \left(-\frac12\sum\_{i=1}^N\Delta\_i^2 - \sum\_{i=1}^N\sum\_{A=1}^M\frac{Z\_A}{|r\_i - r\_A|} + \sum\_{i=1}^N\sum\_{j>i}^N \frac{1}{|r\_i - r\_j|} \right) \Psi = \varepsilon \Psi $$


I've read in Szabo and Ostlund that the pauli exclusion principle manifests itself by saying the wavefunction $\Psi$ must be antisymmetric in spin+location coordinates.


The next thing most people do is to introduce Slater determinants from 1-electron orbitals. This is an approximation to the solution space of wave functions, evidenced by the fact that FCI procedures must assume a sum of slater determinants for more accurate solutions. I want to keep this theoretical, so for now I don't care much for computational considerations.


The problem becomes as such, when talking about slater determinants it makes sense to append a spin variable and to talk about spin-orbitals of single electrons. Without slater determinants, single electron orbitals don't make sense (we only have the N-electron Schroedinger equation). Appending a spin coordinate also doesn't make sense because anti-symmetry of $\Psi$ and the Schroedinger equation don't see it.


So please help me out here, assuming I can solve the electronic Schroedinger equation exactly, what conditions do I need to place on the wavefunction so that the Pauli Exclusion Principle is satisfied (giving me a physical solution)?




I think what you are asking is "What form does $\Psi$ need to take so that the exchange interaction exists when the electronic Hamiltonian given acts upon it?". (Noting that the exchange interaction is due to the fact that electrons are *indistinguishable* and *antisymmetric* with respect to their positions). 


From what I understand the Hamiltonian you have given acts upon electrons and so does not include the exchange interactions. As far as I'm aware the analytical form of the electronic Hamiltonian that includes exchange is not known. If it were, then density functional theory would not have so many approximate exchange-correlation functionals.


I will try to show the problem by presenting the form of the Schrodinger Equation in a slightly different way than is presented in Szabo and Ostlund's book. 


Firstly, it is understood that $N$ in the Hamiltonian you give in the question represents *electrons* - not *spin orbitals* and we know that the electron density $\rho$ at $\textbf{r}$ is represented: \begin{equation} \label{eq1} \rho(\textbf{r}) = |\Psi(\textbf{r})|^2 = \langle \Psi|\textbf{r}\rangle\langle \textbf{r}|\Psi \rangle = \Psi^\*(\textbf{r})\Psi(\textbf{r}) \end{equation}
The Hamiltonian you give acts upon the electron density by using spin orbitals to represent it.


Note, a physical electron density must therefore arise from a wavefunction that is anti-symmetric. 


The Hamiltonian you have given can be seen to act upon the wavefunction as follows:
\begin{align}
E = &\int \Psi^\*(\textbf{r}\_1)(-\frac{1}{2}\nabla^2-\sum\_{A=1}^{M}\frac{Z\_A}{|\textbf{r}\_1-\textbf{r}\_A|}) \Psi(\textbf{r}\_1) d\textbf{r}\_1 \\ &+\frac{1}{2}\int\int \Psi^\*(\textbf{r}\_1) \Psi^\*(\textbf{r}\_2) (\frac{1}{|\textbf{r}\_1-\textbf{r}\_2|}) \Psi(\textbf{r}\_1)\Psi(\textbf{r}\_2)d\textbf{r}\_1d\textbf{r}\_2
\end{align}  

Note that the second term - concerning two-electron integrals shows the electron density at $\textbf{r}\_1$ interacting with the electron density of $\textbf{r}\_2$; this is clear when rearranging the final term (knowing that the $r^{-1}\_{12}$ operator is a Hermitian operator) and substituting in the first equation: 
\begin{align}
\langle 12|12\rangle &=\int\int \Psi^\*(\textbf{r}\_1) \Psi(\textbf{r}\_1) (\frac{1}{|\textbf{r}\_1-\textbf{r}\_2|}) \Psi^\*(\textbf{r}\_2)\Psi(\textbf{r}\_2)d\textbf{r}\_1d\textbf{r}\_2 \\ &= \int\int \rho(\textbf{r}\_1) (\frac{1}{|\textbf{r}\_1-\textbf{r}\_2|}) \rho(\textbf{r}\_2)d\textbf{r}\_1d\textbf{r}\_2 
\end{align}
Now, it is well known that electrons are *indistinguishable*, (i.e. they exist with a given probability density over all space and an electron cannot be labelled) such that: $$ |\Psi(1,2)|^2 = |\Psi(2,1)|^2 $$ Where $1=\textbf{r}\_1$ and $2=\textbf{r}\_2$.


It is also known that, as *fermions* (half-integal spin particles) they are anti-symmetric:
 $$ \Psi(1,2) = -\Psi(2,1) $$ 
Applying this antisymmetry to the two electron integrals we know that there has to be an antisymmetric electron interaction due to the interchanging of electronic positions. Therefore, the two electron integrals can be written: $$ \langle 12||12\rangle = \langle 12|12 \rangle -\langle 12|21 \rangle $$ $$ \langle 21||21\rangle = \langle 21|21 \rangle -\langle 21|12 \rangle $$ both the negative terms are the exchange interactions associated with the interchanging positions. Which are not included in the electronic Hamiltonian given in the question and their integral forms cannot be rearranged as shown before into terms of the electron density at a single given position.


This is why the Fock Operator is used instead of the electronic Hamiltonian!


I hope this all makes sense! I have probably made a mistake somewhere! So please let me know! 




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If you would like to know more about calculating the exchange from the electron density for known wavefunctions I recommend the paper: Pendás, A. M.; Blanco, M. A.; Francisco, E. Two-electron integrations in the quantum theory of atoms in molecules. *J. Chem. Phys.* **2004,** *120* (10), 4581–4592. [DOI: 10.1063/1.1645788](https://doi.org/10.1063/1.1645788).


For normailised wavefunctions, the exchange is given:
\begin{equation}
K = - \int \int r\_{12}^{-1} \vert \rho\_{1}(\textbf{r}\_{1};\textbf{r}\_{2}) \vert^{2}d\textbf{r}\_{1}d\textbf{r}\_{2}
\end{equation}


Where:
$$\rho\_{1}(\textbf{r}\_{1};\textbf{r}\_{1}^{\prime}) = N\_{\text{elec}} \int \Psi\_{\text{elec}} (\textbf{r}\_{1},...,\textbf{r}\_{N\_{\text{elec}}})\Psi\_{\text{elec}}^{\*}(\textbf{r}\_{1}^{\prime},...,\textbf{r}\_{N\_{\text{elec}}}) d\textbf{r}\_{2}...d\textbf{r}\_{N\_{\text{elec}}}$$


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> 
> So please help me out here, assuming I can solve the electronic Schroedinger equation exactly, what conditions do I need to place on the wavefunction so that the Pauli Exclusion Principle is satisfied (giving me a physical solution)?
> 
> 
> 


If the wavefunction is a totally antisymmetric tensor then Pauli Exclusion principle is satisfied. Take a look at another [post](https://chemistry.stackexchange.com/a/100724/67043) I wrote where I give a reference
that explains the relation of indistinguishability, antisymmetry and Pauli Exclusion principle.
The wavefunction has also to be normalizable, with first derivative with respect to coordinates integrable.


The problem of finding the ground state (after Born-Oppenheimer) is presented as a minimization
$$ E\_0 = \text{inf}\{ \langle \Psi\_e,H\Psi\_e \rangle, \Psi\_e \in \mathcal{H}\_e, ||\Psi\_e||^2\_{L^2} = 1 \} $$
where $\mathcal{H}\_e = \wedge^N H^1(\mathbb{R}^3 \times \{1,0\})$.
Hartree-Fock variationally restricts $\mathcal{H}\_e$ to the space of functions that can be written as single determinants in terms of functions of $\mathbb{R}^3$ (spin-orbitals).
Note that a determinant is just the simplest element of the alternating space $\mathcal{H}\_e$. That is when exchange appears.
After projecting the problem in a finite dimensional space (Galerkin) and imposing first order optimality you get to the Fock matrix, which you don't know while you also don't know the coefficients of discretized wavefunction in a one electron basis. So your linear problem becomes a non linear problem.



> 
> Let me preface this by saying I'm not a chemist nor physicist. I'm a mathematician by trade which has the side affect of understanding electronic structure as solutions to a PDE.
> 
> 
> 


Given that you are a mathematician you will have an easier time reading from a mathematician. I highly recommend you to read [Computational quantum chemistry: A primer](https://www.sciencedirect.com/science/article/pii/S1570865903100038). It is only about 250 pages (shorter than Szabo) and explains many methods.




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There are some points in the answer of CompChem that are not right and I would like to clarify. In comments there is not enough space.



> 
> From what I understand the Hamiltonian you have given acts upon electrons and so does not include the exchange interactions. As far as I'm aware the analytical form of the electronic Hamiltonian that includes exchange is not known. If it were, then density functional theory would not have so many approximate exchange-correlation functionals.
> 
> 
> 


If the Hamiltonian acts only on electrons does not mean that it does not include exchange interactions, it only means that the nuclei have been decoupled from electrons. A more technical explanation in terms of the spectrum of the Hamiltonian is given in the reference above. In fact the Hamiltonian given includes electron-electron interactions so it has its sibblings Coulomb and exchange interaction.


It is not true that exchange and correlation expressions are not now, CompChem is mixing DFT with wavefunction methods. OP never says DFT. The problem in DFT is that the main variable is the density and the exchange and correlation energies are functionals of the pair density, not the density. So there is not known way to express it in terms of the density.



> 
> The Hamiltonian you give acts upon the electron density by using spin orbitals to represent it.
> 
> 
> 


The same as above, the Hamiltonian given by OP acts on the wavefunction ($\Psi$), not the density. That would be true for DFT.


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