225 lines
9.3 KiB
TeX
225 lines
9.3 KiB
TeX
\documentclass[11pt]{article}
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\begin{document}
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\section{Introduction}
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--Review Hartree-Fock in the wedge-product notation.
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We will introduce a framework for calculating approximations to
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the ground state many-electron wavefunction for a given molecule,
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assuming the positions of the nuclei are given. This framework
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reproduces the Hartree-Fock solution as the lowest order, and in
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principle gives the exact wavefunction as the highest order
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solution. The idea is to use some intermediate order approximation
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that is computationally practical, and sufficiently accurate.
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We will start by reviewing the Hartree-Fock method in a new notation.
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\subsection{Hartree-Fock theory}
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The problem is to find an antisymmetric, $n$-electron wavefunction (actually, a spin orbital)
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$\Psi =
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\Psi(x_1, \cdots, x_n)$ that minimizes,
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\begin{equation}
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E_\Psi = \langle \Psi | H | \Psi \rangle \,,
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\end{equation}
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where $H = H_{core} + H_{ee}$ is the Hamiltonian, with $H_{core}$
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being the 1-electron core Hamiltonian representing electron kinetic energy
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and the electron-nucleus interactions, and $H_{ee}$ representing the
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electron-electron interactions,
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\begin{equation}
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H_{ee} = \sum_{k=1}^{n-1} \sum_{l=k+1}^n \frac{e^2}{|\mathbf{r}_k - \mathbf{r}_l|}\,.
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\end{equation}
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The Hartree-Fock method restricts the search for the
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minimizing $\Psi$ to the set of Slater determinants of 1-electron spin
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orbitals. This automatically gives $n$-electron wavefunctions that are
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antisymmetric. Let us introduce a new notation to represent these
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Slater determinants.
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\subsection{Antisymmetric products}
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Let $\psi_A(x)$ and $\psi_B(x)$ be two single-electron wavefunctions
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(let us ignore spin for the moment; everything we say below can (and
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in fact, should) be written in terms of spin orbitals). We define
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their antisymmetric product, or the {\it wedge} product $\psi_A
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\wedge \psi_B$ as an antisymmetric function for two variables,
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\begin{equation}
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(\psi_A \wedge \psi_B)(x_1, x_2) = \frac{1}{\sqrt{2!}} \big(\psi_A(x_1)
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\psi_B(x_2) - \psi_A(x_2)\psi_B(x_1)\big)\,.
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\end{equation}
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Note that this implies $\psi_A\wedge \psi_B=-\psi_B\wedge\psi_A$.
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This is just a fancy notation for the Slater determinant of two
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variables, but it will turn out to be convenient for the
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generalization we will describe. As in the Slater determinant, the
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factor $\frac{1}{\sqrt{2!}}$ ensures that the two-particle wavefunction
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$\psi_A\wedge \psi_B$ is normalized when $\psi_A$ and $\psi_B$ are
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orthogonal to each other\footnote{A note here on the fact that they
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can always be made orthogonal to each other without changing the
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determinant.}
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and are normalized as single-electron
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wavefunctions. Similarly, we can represent a
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$3\times 3$ Slater determinant by a triple wedge product of
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single-electron wavefunctions, which gives a function of three varibles,
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\begin{eqnarray}
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(\psi_A \wedge \psi_B \wedge \psi_C)(x_1,x_2,x3) =
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\frac{1}{\sqrt{3!}}\Big[
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\psi_A(x_1) \psi_B(x_2) \psi_C(x_3) + \cr
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\psi_A(x_2) \psi_B(x_3) \psi_C(x_1) + \cr
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\psi_A(x_3) \psi_B(x_1) \psi_C(x_2) - \cr
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\psi_A(x_1) \psi_B(x_3) \psi_C(x_2) - \cr
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\psi_A(x_3) \psi_B(x_2) \psi_C(x_1) - \cr
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\psi_A(x_2) \psi_B(x_1) \psi_C(x_3)\Big] \,.
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\end{eqnarray}
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The $n$-fold product $\psi_1\wedge\cdots\wedge\psi_n$ is defined similarly,
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by summing over all permutations of the arguments, and introducing the
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appropriate signs just as in the definition of the determinant,
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\begin{equation}
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(\psi_1\wedge\cdots\wedge\psi_n)(x_1,\cdots,x_n) = \frac{1}{\sqrt{n!}} \sum_{\sigma}
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sgn(\sigma)\psi_1(x_{\sigma_1}) \psi_2(x_{\sigma_2})\cdots \psi_n(x_{\sigma_n})\,,
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\end{equation}
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where $\sigma$ denotes a permutation of $1\cdots n$, so the sum over
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$\sigma$ denotes a sum over all such permutations. $sgn(\sigma)$
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denotes the sign of the permutation, being equal to $1$ if $\sigma$
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can be written as an even number of successive pair exchanges, and
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$-1$ otherwise.
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Although these definitions are enough for rewriting Hartree-Fock
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theory using the notation of wedge products, we will also
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define the wedge products of multiparticle wavefunctions, since these
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will be crucial to the generalization we will describe.
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Let $\omega$ be a two-particle, antisymmetric wavefunction, and $\psi($
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be a single-particle wavefunction. We define their wedge product
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$\omega \wedge \psi$ to be a totally antisymmetric three-particle
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wavefunction given by,
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\begin{eqnarray}
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(\omega \wedge \psi)(x_1, x_2, x_3) = \frac{1}{N_{2,1}} \Big[
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\omega(x_1,x_2)\psi(x_3)+\cr
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\omega(x_2,x_3)\psi(x_1)+\cr
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\omega(x_3,x_1)\psi(x_2)-\cr
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\omega(x_1,x_3)\psi(x_2)-\cr
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\omega(x_3,x_2)\psi(x_1)-\cr
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\omega(x_2,x_1)\psi(x_3)\Big] \,,
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\end{eqnarray}
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where $N_{2,1}$ is a normalization factor to be discussed shortly.
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The idea of the definition is to permute the variables plugged in the three
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``slots'' in $\omega(\cdot, \cdot)\psi(\cdot)$, and introduce the
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appropriate signs for each permutation.\footnote{$\psi\wedge\omega$ is
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defined in a similar manner, and turns out to be equal to
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$\omega\wedge\psi$. Note that this commutative property does not
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hold in the more general case of a wedge product between a
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$p$-particle wavefunction and and a $q$-particle wavefunction.}
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Note that some of the terms
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are identical to each other due to the antisymmetry of $\omega$, e.g.,
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$-\omega(x_1,x_3)\psi(x_2) = \omega(x_3,x_1)\psi(x_2)$. The
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normalization factor $N_{2,1} = \frac{1}{2\sqrt{3}}$ is
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chosen\footnote{The generalized version which will
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perhaps be discussed in a future version is given by
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$N_{p,q}=\left(p!q!{p+q\choose p}\right)^{-1}$} so that
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the wedge product $\omega\wedge\psi$ is
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normalized when $\omega$ and
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$\psi$ are normalized as two-particle and single-particle
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wavefunctions, respectively, and are orthogonal to each other in the
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sense that,
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\begin{equation}
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\int \omega(x,y)\psi(y)dy = 0\,.
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\end{equation}
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Due to the antisymmetry of $\omega$, this implies $\int
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\omega(y,x)\psi(y)dy=0$ also. We will come back to this notion of
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orthogonality in the following sections. {\it Remark on the
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overcounting due to the antisymmetry of $\omega$}
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This definition of the wedge product of a two-particle
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wavefunction with a one-particle wavefunction is compatible with the
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triple wedge product of single particle wavefunctions introduced above.
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% give equation number
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If one defines $\omega = \psi_A\wedge\psi_B$, then
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$\omega \wedge \psi_C = (\psi_A\wedge\psi_B)\wedge\psi_C$
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turns out to be equal to $\psi_A\wedge\psi_B\wedge\psi_C$ as given above.
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% give equation number
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Likewise, defining $\omega =
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\psi_B\wedge\psi_C$ and using a definition of
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$\psi\wedge\omega$ similar to that of $\omega\wedge\psi$,
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% give eq no
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one can see that $\psi_A\wedge(\psi_B\wedge\psi_C) =
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\psi_A\wedge\psi_B\wedge\psi_C$.
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It is possible to generalize these definitions and
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--Orthogonality in one pair of slots implies that in any other pair.
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--Remark on the fact that the orthogonalization of a two-part wavefn
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with a one-part. wavefne is always possible without changing their wedge
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product. (Generalize this to p-part and q-part.)
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--Include remarks about the normalization factors not being standard.
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\subsection{Orthogonality}
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\subsection{The setup}
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--Here, describe the general setup: we basically have a partially
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ordered set of spaces of wavefunctions, the maximal element being the
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set of all antisym n-particle wavefunctions, and the minimal one being
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the set of Hartree-Fock wavefunctions.
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\section{The first improvement}
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Let us start by exploring the lowest order improvement over
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Hartree-Fock theory in this framework, namely, minimizing the expectation
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value of energy over the space of wavefunctions of the form,
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\begin{equation}
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\Psi(x_1\cdots x_n) =
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(\omega\wedge\psi_3\wedge\psi_4\cdots\wedge\psi_n)(x_1\cdots x_n) \,,
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\end{equation}
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where $\omega$ is a normalized, two-particle, antisymmetric wavefunction,
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$\{\psi_i: i=3\cdots n\}$ is an orthonormal set of single-particle
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wavefunctions\footnote{Indexing of the $\psi$s starts from 3 for
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notational convenience.},
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and $\omega$ is orthogonal to the $\psi_i$s in the sense
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described above. % give section number
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\subsection{Core terms}
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Let's first calculate the expectation value of $H_{core}$ for this
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wavefunction. Using $H_{core}(x_1,\cdots, x_n)=\sum_k
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H_{core}^{(k)}(x_k)$,
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\begin{equation}
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E_\Psi = \langle\Psi|H_{core}|\Psi\rangle = \sum_k
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\langle\Psi|H_{core}^{(k)}|\Psi\rangle\,,
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\end{equation}
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and expanding the wavefunction as a sum of
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products with permuted arguments, we get,
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\begin{eqnarray*}
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\lefteqn{E_\Psi=\int dx_1\cdots dx_n
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\sum_{k,\sigma,\tau}(-)^\sigma (-)^\tau}\\& &
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\Big[\omega(x_{\sigma_1},x_{\sigma_2})
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\psi_3(x_{\sigma_3})\cdots\psi_n(x_{\sigma_n})\Big]H_{core}^{(k)}\Big[\omega(x_{\tau_1},x_{\tau_2})
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\psi_3(x_{\tau_3})\cdots\psi_n(x_{\tau_n})\Big]\,,
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\end{eqnarray*}
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where $\sigma$ and $\tau$ are permutations of $1\cdots n$, and
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$(-)^\sigma$ and $(-)^\tau$ are shorthand for $sgn(\sigma)$ and
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$sgn(\tau)$, respectively.
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\subsubsection{Utilizing orthogonality}
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\subsection{e-e terms}
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Evaluate the e-e terms.
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\subsection{Total energy}
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Give the final formula.
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\section{Introducing a basis}
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Extract the analogue of Roothaan-Hall equations by defining a basis.
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\section{Computational considerations}
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How to implement a preliminary version of the algorithm.
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\end{document}
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