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In some chemistry classes I was taught the (seemingly usual) 'tale of exactly two atoms' that form bonding and anti-bonding states in the LCAO-theory (similar to this question). I've not seen the molecular orbitals mentioned since that time until now. I'm trying to learn to use pyscf, and there the author talks about molecular orbitals all the time. Even for things like full-ci. I don't understand that, but it has convinced me that there is a link between state functions of the system and the molecular orbitals (which is more general than the tale of exactly 2 atoms as linked to above).

Is the idea that a molecular orbital is a state function (or at least within the approximation)? I assume the chemists would have said that if it was true. So assuming that it is false, could you please explain how the molecular orbitals relate to the state function of the $n$-body problem in as much generality as possible?


EDIT: The question has been closed because it 'needs details or clarity'. However, it seems the person that provided the answer had no problems understanding it. That is not to say that the question can't be improved. So could the person(s) that closed the question kindly say where the question is unclear? It will help me to understand if I can satisfy the critics or if that will be impossible.

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Molecular orbitals are simply used to construct $N$-electron basis states (Slater determinants).

First, let us have a one-particle Hilbert space $\cal{H}$ of dimension $M$, with $M$ linearly independent vectors $\{\phi_p\}_{p=1}^M$ forming a basis on it. In non-relativistic one-electron QM, $\cal{H}=L^2(\mathbb{R}^3)\otimes\mathbb{C}^2$ (scattering states now not considered for simplicity), so in principle, $M\rightarrow\infty$ should be taken; we nevertheless keep $M$ finite to properly see the dimensions (also, $M$ is obviously finite in all numerical calculations). For the sake of simplicity, let us further take the basis to be orthonormal: $$\langle\phi_p|\phi_q\rangle=\delta_{pq} \ , $$ although this is not necessary. Since this is a basis, any one-particle function can be expanded in it as $$ |\psi\rangle=\sum_{p=1}^Mc_p|\phi_p\rangle \ . $$

The important thing to realize is that constructing the antisymmetrized products of $N$ such vectors in all ${M}\choose{N}$ possible ways will leave you with an orthonormal basis on the antisymmetric subspace of the $N$-particle Hilbert space. For example, for $N=2$, the antisymmetrized products $$ |\Phi_{pq}\rangle=\frac{1}{\sqrt{2}}\left[ |\phi_p\rangle\otimes|\phi_q\rangle-|\phi_q\rangle\otimes|\phi_p\rangle \right] $$ for $p,q=1,...,M$, $p<q$, form a basis in the antisymmetric subspace of the two-particle Hilbert space $\cal{H}\otimes\cal{H}$ (which is the relevant subspace for two electrons). Any two-electron function can be expanded in this ${{M}\choose{2}}=M(M-1)/2$ -dimensional basis: $$ |\Psi\rangle=\sum_{\substack{p,q=1 \\ p<q}}^Mc_{pq}|\Phi_{pq}\rangle \ , $$ orthonormality being understood as $$ \langle\Phi_{rs}|\Phi_{pq}\rangle=\delta_{pr}\delta_{qs}-\delta_{ps}\delta_{qr} \ . $$ The idea is the same for the general $N$-electron case, the antisymmetrized products (Slater determinants) $$ |\Phi_{p_1p_2...p_N}\rangle= \frac{1}{\sqrt{N!}}\sum_{{\cal{P}}\in S_N}(-1)^{\pi_{\cal{P}}}{\cal{P}} \left[|\phi_{p_1}\rangle\otimes|\phi_{p_2}\rangle\otimes...\otimes|\phi_{p_N}\rangle\right] $$ can be used in $$ |\Psi\rangle=\sum_{\substack{p_1,p_2,...,p_N=1 \\ p_1<p_2<...<p_N}}^Mc_{p_1p_2...p_N}|\Phi_{p_1p_2...p_N}\rangle \ . $$ Such expansions are used extensively in the many-body theory. In the most basic version of Full-CI, these ${M}\choose{N}$ Slater determinants are used to construct the matrix representation of the Hamiltonian, and the energies and coefficients $c_{p_1p_2...p_N}$ of the eigenstates are obtained from the diagonalization of this matrix.

Note that, apart from orthonormality, I assumed nothing about the one-particle basis. These orbitals are usually found from e.g. Hartree-Fock or Kohn-Sham calculations. On the level of theory, LCAO does not have much to do with any of this, it is just often (almost always) useful to expand molecular orbitals over atom-centered functions: $$ |\phi_p\rangle=\sum_{\mu=1}^Md_{\mu p}|\chi_\mu\rangle \ . $$ This is practical from a computational/interpretational point of view, as it is often easier to see e.g. bonding/antibonding orbitals in this way. But the expansion over atomic functions is by no means necessary, you can do quantum chemistry in e.g. a plane wave basis, too.

Update

I used bra-ket notations above in order to properly show the tensor product structure. The (probably) better known form of these functions is recovered by projection with the formal coordinate eigenstates $|\vec{r}\rangle$. The one-electron wave functions (orbitals) then read $$ \phi_{p}(\vec{r})= \langle\vec{r}|\phi_p\rangle= \left( \begin{matrix} \phi_{p\uparrow}(\vec{r}) \\ \phi_{p\downarrow}(\vec{r}) \end{matrix} \right) \ , $$ having two components due to their spinor character (this is the $\mathbb{C}^2$ part of ${\cal{H}}$). The components $\phi_{p\uparrow}(\vec{r})$ and $\phi_{p\downarrow}(\vec{r})$ are different in the general case; one usually wants orbitals to be eigenfunctions of $\hat{s}_z$, which forces one of the components to be zero.

The $N$-electron Slater determinants are obtained similarly after a projection with $$ |\vec{r}_1,\vec{r}_2,...,\vec{r}_N\rangle=|\vec{r}_1\rangle\otimes|\vec{r}_2\rangle\otimes...\otimes|\vec{r}_N\rangle \ , $$ leading to $$ \Phi_{p_1p_2...p_N}(\vec{r}_1,\vec{r}_2,...,\vec{r}_N)= \frac{1}{\sqrt{N!}}\sum_{{\cal{P}}\in S_N}(-1)^{\pi_{\cal{P}}}{\cal{P}} \left[\phi_{p_1}(\vec{r}_1)\otimes\phi_{p_2}(\vec{r}_2)\otimes...\otimes\phi_{p_N}(\vec{r}_N)\right] \ , $$ which shows that non-relativistic $N$-electron states have $2^N$ spinor components. Note that ${\cal{P}}$ permutes the orbital indices, not the coordinate indices.

When doing formal manipulations with wave functions, it is often useful to introduce a combined spatial+spin coordinate eigenstate $|x\rangle=|\vec{r},\sigma\rangle$ for $\sigma=\uparrow,\downarrow$, which implements a further projection onto the upper or lower spinor component: $$ \langle\vec{r},\uparrow\hspace{-0.1cm}|\phi_p\rangle=\phi_{p\uparrow}(\vec{r}) \ \ \ , \ \ \ \langle\vec{r},\downarrow\hspace{-0.1cm}|\phi_p\rangle=\phi_{p\downarrow}(\vec{r}) \ . $$ Projecting $|\Phi_{p_1p_2...p_N}\rangle$ with $$ |x_1,x_2,...,x_N\rangle=|x_1\rangle\otimes|x_2\rangle\otimes...\otimes|x_N\rangle $$ will similarly pick out a single one from the $2^N$ spinor components: $$ \Phi_{p_1p_2...p_N}(x_1,x_2,...,x_N)= \frac{1}{\sqrt{N!}}\sum_{{\cal{P}}\in S_N}(-1)^{\pi_{\cal{P}}}{\cal{P}} \left[\phi_{p_1}(x_1)\phi_{p_2}(x_2)...\phi_{p_N}(x_N)\right] \ . $$ In this way, it is often easier to talk about the antisymmetry of the electronic wave function under simultaneous interchanges in coordinate and spin space: $$ \Psi(x_1,x_2)=-\Psi(x_2,x_1) \ . $$

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    $\begingroup$ Some moderator sabotaged the question by closing it down. Nice answer though. I can see from the upvotes that it is sought after. Is the idea that the slater determinants are the molecular orbitals? $\endgroup$
    – Mikkel Rev
    Commented Jan 2 at 12:34
  • $\begingroup$ @MikkelRev Thanks for the feedback. No, molecular orbitals are always one-electron states, while Slater determinants are special $N$-electron states built from these orbitals. These determinants form a basis in which a general $N$-electron state can be expanded. I updated the answer to make some connection with the usual, textbook form of Slater determinants. $\endgroup$ Commented Jan 2 at 16:44
  • $\begingroup$ Thanks for that. I thought that $\phi_p$ in the notation above would be the atomic orbitals. But are they also the molecular orbitals? Can you say which symbol you used for the atomic orbitals, and which were used for the molecular orbitals? $\endgroup$
    – Mikkel Rev
    Commented Jan 2 at 18:34
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    $\begingroup$ @MikkelRev $\phi_p$ is a molecular orbital, $\chi_\mu$ is an atomic orbital. $\endgroup$ Commented Jan 2 at 19:41

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