From f4096d4ddffd86b25183ba803df8dae953f0b249 Mon Sep 17 00:00:00 2001 From: Morten Hjorth-Jensen Date: Fri, 29 Dec 2023 09:37:02 +0100 Subject: [PATCH] update text --- .../BookProject/{chapter3.tex => fci.tex} | 0 .../BookProject/{chapter4.tex => hfock.tex} | 0 doc/BookChapters/BookProject/introduction.tex | 427 ++++++++ doc/BookChapters/BookProject/secondquant.tex | 987 ++++++++++++++++++ doc/BookChapters/BookProject/systems.tex | 427 ++++++++ 5 files changed, 1841 insertions(+) rename doc/BookChapters/BookProject/{chapter3.tex => fci.tex} (100%) rename doc/BookChapters/BookProject/{chapter4.tex => hfock.tex} (100%) create mode 100644 doc/BookChapters/BookProject/introduction.tex create mode 100644 doc/BookChapters/BookProject/secondquant.tex create mode 100644 doc/BookChapters/BookProject/systems.tex diff --git a/doc/BookChapters/BookProject/chapter3.tex b/doc/BookChapters/BookProject/fci.tex similarity index 100% rename from doc/BookChapters/BookProject/chapter3.tex rename to doc/BookChapters/BookProject/fci.tex diff --git a/doc/BookChapters/BookProject/chapter4.tex b/doc/BookChapters/BookProject/hfock.tex similarity index 100% rename from doc/BookChapters/BookProject/chapter4.tex rename to doc/BookChapters/BookProject/hfock.tex diff --git a/doc/BookChapters/BookProject/introduction.tex b/doc/BookChapters/BookProject/introduction.tex new file mode 100644 index 00000000..1e2d0367 --- /dev/null +++ b/doc/BookChapters/BookProject/introduction.tex @@ -0,0 +1,427 @@ + +% ------------------- main content ---------------------- + +\chapter{Many-body Hamiltonians, basic linear algebra and Second Quantization} + +\subsection*{Definitions and notations} + +Before we proceed we need some definitions. +We will assume that the interacting part of the Hamiltonian +can be approximated by a two-body interaction. +This means that our Hamiltonian is written as the sum of some onebody part and a twobody part +\begin{equation} + \hat{H} = \hat{H}_0 + \hat{H}_I + = \sum_{i=1}^A \hat{h}_0(x_i) + \sum_{i < j}^A \hat{v}(r_{ij}), +\label{Hnuclei} +\end{equation} +with +\begin{equation} + H_0=\sum_{i=1}^A \hat{h}_0(x_i). +\label{hinuclei} +\end{equation} +The onebody part $u_{\mathrm{ext}}(x_i)$ is normally approximated by a harmonic oscillator potential or the Coulomb interaction an electron feels from the nucleus. However, other potentials are fully possible, such as +one derived from the self-consistent solution of the Hartree-Fock equations to be discussed here. + +Our Hamiltonian is invariant under the permutation (interchange) of two particles. +Since we deal with fermions however, the total wave function is antisymmetric. +Let $\hat{P}$ be an operator which interchanges two particles. +Due to the symmetries we have ascribed to our Hamiltonian, this operator commutes with the total Hamiltonian, +\[ +[\hat{H},\hat{P}] = 0, + \] +meaning that $\Psi_{\lambda}(x_1, x_2, \dots , x_A)$ is an eigenfunction of +$\hat{P}$ as well, that is +\[ +\hat{P}_{ij}\Psi_{\lambda}(x_1, x_2, \dots,x_i,\dots,x_j,\dots,x_A)= +\beta\Psi_{\lambda}(x_1, x_2, \dots,x_i,\dots,x_j,\dots,x_A), +\] +where $\beta$ is the eigenvalue of $\hat{P}$. We have introduced the suffix $ij$ in order to indicate that we permute particles $i$ and $j$. +The Pauli principle tells us that the total wave function for a system of fermions +has to be antisymmetric, resulting in the eigenvalue $\beta = -1$. + +In our case we assume that we can approximate the exact eigenfunction with a Slater determinant +\begin{equation} + \Phi(x_1, x_2,\dots ,x_A,\alpha,\beta,\dots, \sigma)=\frac{1}{\sqrt{A!}} +\left| \begin{array}{ccccc} \psi_{\alpha}(x_1)& \psi_{\alpha}(x_2)& \dots & \dots & \psi_{\alpha}(x_A)\\ + \psi_{\beta}(x_1)&\psi_{\beta}(x_2)& \dots & \dots & \psi_{\beta}(x_A)\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + \psi_{\sigma}(x_1)&\psi_{\sigma}(x_2)& \dots & \dots & \psi_{\sigma}(x_A)\end{array} \right|, \label{eq:HartreeFockDet} +\end{equation} +where $x_i$ stand for the coordinates and spin values of a particle $i$ and $\alpha,\beta,\dots, \gamma$ +are quantum numbers needed to describe remaining quantum numbers. + +\paragraph{Brief reminder on some linear algebra properties.} +Before we proceed with a more compact representation of a Slater determinant, we would like to repeat some linear algebra properties which will be useful for our derivations of the energy as function of a Slater determinant, Hartree-Fock theory and later the nuclear shell model. + +The inverse of a matrix is defined by + +\[ +\mathbf{A}^{-1} \cdot \mathbf{A} = I +\] +A unitary matrix $\mathbf{A}$ is one whose inverse is its adjoint +\[ +\mathbf{A}^{-1}=\mathbf{A}^{\dagger} +\] +A real unitary matrix is called orthogonal and its inverse is equal to its transpose. +A hermitian matrix is its own self-adjoint, that is +\[ +\mathbf{A}=\mathbf{A}^{\dagger}. +\] + + +\begin{quote} +\begin{tabular}{ccc} +\hline +\multicolumn{1}{c}{ Relations } & \multicolumn{1}{c}{ Name } & \multicolumn{1}{c}{ matrix elements } \\ +\hline +$A = A^{T}$ & symmetric & $a_{ij} = a_{ji}$ \\ +$A = \left (A^{T} \right )^{-1}$ & real orthogonal & $\sum_k a_{ik} a_{jk} = \sum_k a_{ki} a_{kj} = \delta_{ij}$ \\ +$A = A^{ * }$ & real matrix & $a_{ij} = a_{ij}^{ * }$ \\ +$A = A^{\dagger}$ & hermitian & $a_{ij} = a_{ji}^{ * }$ \\ +$A = \left (A^{\dagger} \right )^{-1}$ & unitary & $\sum_k a_{ik} a_{jk}^{ * } = \sum_k a_{ki}^{ * } a_{kj} = \delta_{ij}$ \\ +\hline +\end{tabular} +\end{quote} + +\noindent +Since we will deal with Fermions (identical and indistinguishable particles) we will +form an ansatz for a given state in terms of so-called Slater determinants determined +by a chosen basis of single-particle functions. + +For a given $n\times n$ matrix $\mathbf{A}$ we can write its determinant +\[ + det(\mathbf{A})=|\mathbf{A}|= +\left| \begin{array}{ccccc} a_{11}& a_{12}& \dots & \dots & a_{1n}\\ + a_{21}&a_{22}& \dots & \dots & a_{2n}\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + a_{n1}& a_{n2}& \dots & \dots & a_{nn}\end{array} \right|, +\] +in a more compact form as +\[ +|\mathbf{A}|= \sum_{i=1}^{n!}(-1)^{p_i}\hat{P}_i a_{11}a_{22}\dots a_{nn}, +\] +where $\hat{P}_i$ is a permutation operator which permutes the column indices $1,2,3,\dots,n$ +and the sum runs over all $n!$ permutations. The quantity $p_i$ represents the number of transpositions of column indices that are needed in order to bring a given permutation back to its initial ordering, in our case given by $a_{11}a_{22}\dots a_{nn}$ here. + +A simple $2\times 2$ determinant illustrates this. We have +\[ + det(\mathbf{A})= +\left| \begin{array}{cc} a_{11}& a_{12}\\ + a_{21}&a_{22}\end{array} \right|= (-1)^0a_{11}a_{22}+(-1)^1a_{12}a_{21}, +\] +where in the last term we have interchanged the column indices $1$ and $2$. The natural ordering we have chosen is $a_{11}a_{22}$. + +\paragraph{Back to the derivation of the energy.} +The single-particle function $\psi_{\alpha}(x_i)$ are eigenfunctions of the onebody +Hamiltonian $h_i$, that is +\[ +\hat{h}_0(x_i)=\hat{t}(x_i) + \hat{u}_{\mathrm{ext}}(x_i), +\] +with eigenvalues +\[ +\hat{h}_0(x_i) \psi_{\alpha}(x_i)=\left(\hat{t}(x_i) + \hat{u}_{\mathrm{ext}}(x_i)\right)\psi_{\alpha}(x_i)=\varepsilon_{\alpha}\psi_{\alpha}(x_i). +\] +The energies $\varepsilon_{\alpha}$ are the so-called non-interacting single-particle energies, or unperturbed energies. +The total energy is in this case the sum over all single-particle energies, if no two-body or more complicated +many-body interactions are present. + +Let us denote the ground state energy by $E_0$. According to the +variational principle we have +\[ + E_0 \le E[\Phi] = \int \Phi^*\hat{H}\Phi d\mathbf{\tau} +\] +where $\Phi$ is a trial function which we assume to be normalized +\[ + \int \Phi^*\Phi d\mathbf{\tau} = 1, +\] +where we have used the shorthand $d\mathbf{\tau}=dx_1dr_2\dots dr_A$. + +In the Hartree-Fock method the trial function is the Slater +determinant of Eq.~(\ref{eq:HartreeFockDet}) which can be rewritten as +\[ + \Phi(x_1,x_2,\dots,x_A,\alpha,\beta,\dots,\nu) = \frac{1}{\sqrt{A!}}\sum_{P} (-)^P\hat{P}\psi_{\alpha}(x_1) + \psi_{\beta}(x_2)\dots\psi_{\nu}(x_A)=\sqrt{A!}\hat{A}\Phi_H, +\] +where we have introduced the antisymmetrization operator $\hat{A}$ defined by the +summation over all possible permutations of two particles. + +It is defined as +\begin{equation} + \hat{A} = \frac{1}{A!}\sum_{p} (-)^p\hat{P}, +\label{antiSymmetryOperator} +\end{equation} +with $p$ standing for the number of permutations. We have introduced for later use the so-called +Hartree-function, defined by the simple product of all possible single-particle functions +\[ + \Phi_H(x_1,x_2,\dots,x_A,\alpha,\beta,\dots,\nu) = + \psi_{\alpha}(x_1) + \psi_{\beta}(x_2)\dots\psi_{\nu}(x_A). +\] + +Both $\hat{H}_0$ and $\hat{H}_I$ are invariant under all possible permutations of any two particles +and hence commute with $\hat{A}$ +\begin{equation} + [H_0,\hat{A}] = [H_I,\hat{A}] = 0. \label{commutionAntiSym} +\end{equation} +Furthermore, $\hat{A}$ satisfies +\begin{equation} + \hat{A}^2 = \hat{A}, \label{AntiSymSquared} +\end{equation} +since every permutation of the Slater +determinant reproduces it. + +The expectation value of $\hat{H}_0$ +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = A! \int \Phi_H^*\hat{A}\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau} +\] +is readily reduced to +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = A! \int \Phi_H^*\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau}, +\] +where we have used Eqs.~(\ref{commutionAntiSym}) and +(\ref{AntiSymSquared}). The next step is to replace the antisymmetrization +operator by its definition and to +replace $\hat{H}_0$ with the sum of one-body operators +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = \sum_{i=1}^A \sum_{p} (-)^p\int + \Phi_H^*\hat{h}_0\hat{P}\Phi_H d\mathbf{\tau}. +\] + +The integral vanishes if two or more particles are permuted in only one +of the Hartree-functions $\Phi_H$ because the individual single-particle wave functions are +orthogonal. We obtain then +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}= \sum_{i=1}^A \int \Phi_H^*\hat{h}_0\Phi_H d\mathbf{\tau}. +\] +Orthogonality of the single-particle functions allows us to further simplify the integral, and we +arrive at the following expression for the expectation values of the +sum of one-body Hamiltonians +\begin{equation} + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = \sum_{\mu=1}^A \int \psi_{\mu}^*(x)\hat{h}_0\psi_{\mu}(x)dx + d\mathbf{r}. + \label{H1Expectation} +\end{equation} + +We introduce the following shorthand for the above integral +\[ +\langle \mu | \hat{h}_0 | \mu \rangle = \int \psi_{\mu}^*(x)\hat{h}_0\psi_{\mu}(x)dx, +\] +and rewrite Eq.~(\ref{H1Expectation}) as +\begin{equation} + \int \Phi^*\hat{H}_0\Phi d\tau + = \sum_{\mu=1}^A \langle \mu | \hat{h}_0 | \mu \rangle. + \label{H1Expectation1} +\end{equation} + +The expectation value of the two-body part of the Hamiltonian is obtained in a +similar manner. We have +\[ + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = A! \int \Phi_H^*\hat{A}\hat{H}_I\hat{A}\Phi_H d\mathbf{\tau}, +\] +which reduces to +\[ + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = \sum_{i\le j=1}^A \sum_{p} (-)^p\int + \Phi_H^*\hat{v}(r_{ij})\hat{P}\Phi_H d\mathbf{\tau}, +\] +by following the same arguments as for the one-body +Hamiltonian. + +Because of the dependence on the inter-particle distance $r_{ij}$, permutations of +any two particles no longer vanish, and we get +\[ + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = \sum_{i < j=1}^A \int + \Phi_H^*\hat{v}(r_{ij})(1-P_{ij})\Phi_H d\mathbf{\tau}. +\] +where $P_{ij}$ is the permutation operator that interchanges +particle $i$ and particle $j$. Again we use the assumption that the single-particle wave functions +are orthogonal. + +We obtain +\begin{align} + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = \frac{1}{2}\sum_{\mu=1}^A\sum_{\nu=1}^A + &\left[ \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)\hat{v}(r_{ij})\psi_{\mu}(x_i)\psi_{\nu}(x_j) + dx_idx_j \right.\\ + &\left. + - \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j) + \hat{v}(r_{ij})\psi_{\nu}(x_i)\psi_{\mu}(x_j) + dx_idx_j + \right]. \label{H2Expectation} +\end{align} +The first term is the so-called direct term. It is frequently also called the Hartree term, +while the second is due to the Pauli principle and is called +the exchange term or just the Fock term. +The factor $1/2$ is introduced because we now run over +all pairs twice. + +The last equation allows us to introduce some further definitions. +The single-particle wave functions $\psi_{\mu}(x)$, defined by the quantum numbers $\mu$ and $x$ +are defined as the overlap +\[ + \psi_{\alpha}(x) = \langle x | \alpha \rangle . +\] + +We introduce the following shorthands for the above two integrals +\[ +\langle \mu\nu|\hat{v}|\mu\nu\rangle = \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)\hat{v}(r_{ij})\psi_{\mu}(x_i)\psi_{\nu}(x_j) + dx_idx_j, +\] +and +\[ +\langle \mu\nu|\hat{v}|\nu\mu\rangle = \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j) + \hat{v}(r_{ij})\psi_{\nu}(x_i)\psi_{\mu}(x_j) + dx_idx_j. +\] + +\subsection*{Preparing for later studies: varying the coefficients of a wave function expansion and orthogonal transformations} + +It is common to expand the single-particle functions in a known basis and vary the coefficients, +that is, the new single-particle wave function is written as a linear expansion +in terms of a fixed chosen orthogonal basis (for example the well-known harmonic oscillator functions or the hydrogen-like functions etc). +We define our new single-particle basis (this is a normal approach for Hartree-Fock theory) by performing a unitary transformation +on our previous basis (labelled with greek indices) as +\begin{equation} +\psi_p^{new} = \sum_{\lambda} C_{p\lambda}\phi_{\lambda}. \label{eq:newbasis} +\end{equation} +In this case we vary the coefficients $C_{p\lambda}$. If the basis has infinitely many solutions, we need +to truncate the above sum. We assume that the basis $\phi_{\lambda}$ is orthogonal. + +It is normal to choose a single-particle basis defined as the eigenfunctions +of parts of the full Hamiltonian. The typical situation consists of the solutions of the one-body part of the Hamiltonian, that is we have +\[ +\hat{h}_0\phi_{\lambda}=\epsilon_{\lambda}\phi_{\lambda}. +\] +The single-particle wave functions $\phi_{\lambda}(\mathbf{r})$, defined by the quantum numbers $\lambda$ and $\mathbf{r}$ +are defined as the overlap +\[ + \phi_{\lambda}(\mathbf{r}) = \langle \mathbf{r} | \lambda \rangle . +\] + +In deriving the Hartree-Fock equations, we will expand the single-particle functions in a known basis and vary the coefficients, +that is, the new single-particle wave function is written as a linear expansion +in terms of a fixed chosen orthogonal basis (for example the well-known harmonic oscillator functions or the hydrogen-like functions etc). + +We stated that a unitary transformation keeps the orthogonality. To see this consider first a basis of vectors $\mathbf{v}_i$, +\[ +\mathbf{v}_i = \begin{bmatrix} v_{i1} \\ \dots \\ \dots \\v_{in} \end{bmatrix} +\] +We assume that the basis is orthogonal, that is +\[ +\mathbf{v}_j^T\mathbf{v}_i = \delta_{ij}. +\] +An orthogonal or unitary transformation +\[ +\mathbf{w}_i=\mathbf{U}\mathbf{v}_i, +\] +preserves the dot product and orthogonality since +\[ +\mathbf{w}_j^T\mathbf{w}_i=(\mathbf{U}\mathbf{v}_j)^T\mathbf{U}\mathbf{v}_i=\mathbf{v}_j^T\mathbf{U}^T\mathbf{U}\mathbf{v}_i= \mathbf{v}_j^T\mathbf{v}_i = \delta_{ij}. +\] + +This means that if the coefficients $C_{p\lambda}$ belong to a unitary or orthogonal trasformation (using the Dirac bra-ket notation) +\[ +\vert p\rangle = \sum_{\lambda} C_{p\lambda}\vert\lambda\rangle, +\] +orthogonality is preserved, that is $\langle \alpha \vert \beta\rangle = \delta_{\alpha\beta}$ +and $\langle p \vert q\rangle = \delta_{pq}$. + +This propertry is extremely useful when we build up a basis of many-body Stater determinant based states. + +\textbf{Note also that although a basis $\vert \alpha\rangle$ contains an infinity of states, for practical calculations we have always to make some truncations.} + +Before we develop for example the Hartree-Fock equations, there is another very useful property of determinants that we will use both in connection with Hartree-Fock calculations and later shell-model calculations. + +Consider the following determinant +\[ +\left| \begin{array}{cc} \alpha_1b_{11}+\alpha_2sb_{12}& a_{12}\\ + \alpha_1b_{21}+\alpha_2b_{22}&a_{22}\end{array} \right|=\alpha_1\left|\begin{array}{cc} b_{11}& a_{12}\\ + b_{21}&a_{22}\end{array} \right|+\alpha_2\left| \begin{array}{cc} b_{12}& a_{12}\\b_{22}&a_{22}\end{array} \right| +\] + +We can generalize this to an $n\times n$ matrix and have +\[ +\left| \begin{array}{cccccc} a_{11}& a_{12} & \dots & \sum_{k=1}^n c_k b_{1k} &\dots & a_{1n}\\ +a_{21}& a_{22} & \dots & \sum_{k=1}^n c_k b_{2k} &\dots & a_{2n}\\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +a_{n1}& a_{n2} & \dots & \sum_{k=1}^n c_k b_{nk} &\dots & a_{nn}\end{array} \right|= +\sum_{k=1}^n c_k\left| \begin{array}{cccccc} a_{11}& a_{12} & \dots & b_{1k} &\dots & a_{1n}\\ +a_{21}& a_{22} & \dots & b_{2k} &\dots & a_{2n}\\ +\dots & \dots & \dots & \dots & \dots & \dots\\ +\dots & \dots & \dots & \dots & \dots & \dots\\ +a_{n1}& a_{n2} & \dots & b_{nk} &\dots & a_{nn}\end{array} \right| . +\] +This is a property we will use in our Hartree-Fock discussions. + +We can generalize the previous results, now +with all elements $a_{ij}$ being given as functions of +linear combinations of various coefficients $c$ and elements $b_{ij}$, +\[ +\left| \begin{array}{cccccc} \sum_{k=1}^n b_{1k}c_{k1}& \sum_{k=1}^n b_{1k}c_{k2} & \dots & \sum_{k=1}^n b_{1k}c_{kj} &\dots & \sum_{k=1}^n b_{1k}c_{kn}\\ +\sum_{k=1}^n b_{2k}c_{k1}& \sum_{k=1}^n b_{2k}c_{k2} & \dots & \sum_{k=1}^n b_{2k}c_{kj} &\dots & \sum_{k=1}^n b_{2k}c_{kn}\\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots &\dots \\ +\sum_{k=1}^n b_{nk}c_{k1}& \sum_{k=1}^n b_{nk}c_{k2} & \dots & \sum_{k=1}^n b_{nk}c_{kj} &\dots & \sum_{k=1}^n b_{nk}c_{kn}\end{array} \right|=det(\mathbf{C})det(\mathbf{B}), +\] +where $det(\mathbf{C})$ and $det(\mathbf{B})$ are the determinants of $n\times n$ matrices +with elements $c_{ij}$ and $b_{ij}$ respectively. +This is a property we will use in our Hartree-Fock discussions. Convince yourself about the correctness of the above expression by setting $n=2$. + +With our definition of the new basis in terms of an orthogonal basis we have +\[ +\psi_p(x) = \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x). +\] +If the coefficients $C_{p\lambda}$ belong to an orthogonal or unitary matrix, the new basis +is also orthogonal. +Our Slater determinant in the new basis $\psi_p(x)$ is written as +\[ +\frac{1}{\sqrt{A!}} +\left| \begin{array}{ccccc} \psi_{p}(x_1)& \psi_{p}(x_2)& \dots & \dots & \psi_{p}(x_A)\\ + \psi_{q}(x_1)&\psi_{q}(x_2)& \dots & \dots & \psi_{q}(x_A)\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + \psi_{t}(x_1)&\psi_{t}(x_2)& \dots & \dots & \psi_{t}(x_A)\end{array} \right|=\frac{1}{\sqrt{A!}} +\left| \begin{array}{ccccc} \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_1)& \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_A)\\ + \sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_1)&\sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_A)\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + \sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_1)&\sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_A)\end{array} \right|, +\] +which is nothing but $det(\mathbf{C})det(\Phi)$, with $det(\Phi)$ being the determinant given by the basis functions $\phi_{\lambda}(x)$. + +In our discussions hereafter we will use our definitions of single-particle states above and below the Fermi ($F$) level given by the labels +$ijkl\dots \le F$ for so-called single-hole states and $abcd\dots > F$ for so-called particle states. +For general single-particle states we employ the labels $pqrs\dots$. + +The energy functional is +\[ + E[\Phi] + = \sum_{\mu=1}^A \langle \mu | h | \mu \rangle + + \frac{1}{2}\sum_{{\mu}=1}^A\sum_{{\nu}=1}^A \langle \mu\nu|\hat{v}|\mu\nu\rangle_{AS}, +\] +we found the expression for the energy functional in terms of the basis function $\phi_{\lambda}(\mathbf{r})$. We then varied the above energy functional with respect to the basis functions $|\mu \rangle$. +Now we are interested in defining a new basis defined in terms of +a chosen basis as defined in Eq.~(\ref{eq:newbasis}). We can then rewrite the energy functional as +\begin{equation} + E[\Phi^{New}] + = \sum_{i=1}^A \langle i | h | i \rangle + + \frac{1}{2}\sum_{ij=1}^A\langle ij|\hat{v}|ij\rangle_{AS}, \label{FunctionalEPhi2} +\end{equation} +where $\Phi^{New}$ is the new Slater determinant defined by the new basis of Eq.~(\ref{eq:newbasis}). + +Using Eq.~(\ref{eq:newbasis}) we can rewrite Eq.~(\ref{FunctionalEPhi2}) as +\begin{equation} + E[\Psi] + = \sum_{i=1}^A \sum_{\alpha\beta} C^*_{i\alpha}C_{i\beta}\langle \alpha | h | \beta \rangle + + \frac{1}{2}\sum_{ij=1}^A\sum_{{\alpha\beta\gamma\delta}} C^*_{i\alpha}C^*_{j\beta}C_{i\gamma}C_{j\delta}\langle \alpha\beta|\hat{v}|\gamma\delta\rangle_{AS}. \label{FunctionalEPhi3} +\end{equation} + diff --git a/doc/BookChapters/BookProject/secondquant.tex b/doc/BookChapters/BookProject/secondquant.tex new file mode 100644 index 00000000..04521412 --- /dev/null +++ b/doc/BookChapters/BookProject/secondquant.tex @@ -0,0 +1,987 @@ +\chapter{Second quantization} + +We introduce the time-independent operators +$a_\alpha^{\dagger}$ and $a_\alpha$ which create and annihilate, respectively, a particle +in the single-particle state +$\varphi_\alpha$. +We define the fermion creation operator +$a_\alpha^{\dagger}$ +\begin{equation} + a_\alpha^{\dagger}|0\rangle \equiv |\alpha\rangle \label{eq:2-1a}, +\end{equation} +and +\begin{equation} + a_\alpha^{\dagger}|\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} \equiv |\alpha\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} \label{eq:2-1b} +\end{equation} + +In Eq.~(\ref{eq:2-1a}) +the operator $a_\alpha^{\dagger}$ acts on the vacuum state +$|0\rangle$, which does not contain any particles. Alternatively, we could define a closed-shell nucleus or atom as our new vacuum, but then +we need to introduce the particle-hole formalism, see the discussion to come. + +In Eq.~(\ref{eq:2-1b}) $a_\alpha^{\dagger}$ acts on an antisymmetric $n$-particle state and +creates an antisymmetric $(n+1)$-particle state, where the one-body state +$\varphi_\alpha$ is occupied, under the condition that +$\alpha \ne \alpha_1, \alpha_2, \dots, \alpha_n$. +It follows that we can express an antisymmetric state as the product of the creation +operators acting on the vacuum state. +\begin{equation} + |\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} = a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle \label{eq:2-2} +\end{equation} + +It is easy to derive the commutation and anticommutation rules for the fermionic creation operators +$a_\alpha^{\dagger}$. Using the antisymmetry of the states +(\ref{eq:2-2}) +\begin{equation} + |\alpha_1\dots \alpha_i\dots \alpha_k\dots \alpha_n\rangle_{\mathrm{AS}} = + - |\alpha_1\dots \alpha_k\dots \alpha_i\dots \alpha_n\rangle_{\mathrm{AS}} \label{eq:2-3a} +\end{equation} +we obtain +\begin{equation} + a_{\alpha_i}^{\dagger} a_{\alpha_k}^{\dagger} = - a_{\alpha_k}^{\dagger} a_{\alpha_i}^{\dagger} \label{eq:2-3b} +\end{equation} + +Using the Pauli principle +\begin{equation} + |\alpha_1\dots \alpha_i\dots \alpha_i\dots \alpha_n\rangle_{\mathrm{AS}} = 0 \label{eq:2-4a} +\end{equation} +it follows that +\begin{equation} + a_{\alpha_i}^{\dagger} a_{\alpha_i}^{\dagger} = 0. \label{eq:2-4b} +\end{equation} +If we combine Eqs.~(\ref{eq:2-3b}) and (\ref{eq:2-4b}), we obtain the well-known anti-commutation rule +\begin{equation} + a_{\alpha}^{\dagger} a_{\beta}^{\dagger} + a_{\beta}^{\dagger} a_{\alpha}^{\dagger} \equiv + \{a_{\alpha}^{\dagger},a_{\beta}^{\dagger}\} = 0 \label{eq:2-5} +\end{equation} + +The hermitian conjugate of $a_\alpha^{\dagger}$ is +\begin{equation} + a_{\alpha} = ( a_{\alpha}^{\dagger} )^{\dagger} \label{eq:2-6} +\end{equation} +If we take the hermitian conjugate of Eq.~(\ref{eq:2-5}), we arrive at +\begin{equation} + \{a_{\alpha},a_{\beta}\} = 0 \label{eq:2-7} +\end{equation} + +What is the physical interpretation of the operator $a_\alpha$ and what is the effect of +$a_\alpha$ on a given state $|\alpha_1\alpha_2\dots\alpha_n\rangle_{\mathrm{AS}}$? +Consider the following matrix element +\begin{equation} + \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha_1'\alpha_2' \dots \alpha_m'\rangle \label{eq:2-8} +\end{equation} +where both sides are antisymmetric. We distinguish between two cases. The first (1) is when +$\alpha \in \{\alpha_i\}$. Using the Pauli principle of Eq.~(\ref{eq:2-4a}) it follows +\begin{equation} + \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha = 0 \label{eq:2-9a} +\end{equation} +The second (2) case is when $\alpha \notin \{\alpha_i\}$. It follows that an hermitian conjugation +\begin{equation} + \langle \alpha_1\alpha_2 \dots \alpha_n|a_\alpha = \langle\alpha\alpha_1\alpha_2 \dots \alpha_n| \label{eq:2-9b} +\end{equation} + +Eq.~(\ref{eq:2-9b}) holds for case (1) since the lefthand side is zero due to the Pauli principle. We write +Eq.~(\ref{eq:2-8}) as +\begin{equation} + \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha_1'\alpha_2' \dots \alpha_m'\rangle = + \langle \alpha_1\alpha_2 \dots \alpha_n|\alpha\alpha_1'\alpha_2' \dots \alpha_m'\rangle \label{eq:2-10} +\end{equation} +Here we must have $m = n+1$ if Eq.~(\ref{eq:2-10}) has to be trivially different from zero. + +For the last case, the minus and plus signs apply when the sequence +$\alpha ,\alpha_1, \alpha_2, \dots, \alpha_n$ and +$\alpha_1', \alpha_2', \dots, \alpha_{n+1}'$ are related to each other via even and odd permutations. +If we assume that $\alpha \notin \{\alpha_i\}$ we obtain +\begin{equation} + \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha_1'\alpha_2' \dots \alpha_{n+1}'\rangle = 0 \label{eq:2-12} +\end{equation} +when $\alpha \in \{\alpha_i'\}$. If $\alpha \notin \{\alpha_i'\}$, we obtain +\begin{equation} + a_\alpha\underbrace{|\alpha_1'\alpha_2' \dots \alpha_{n+1}'}\rangle_{\neq \alpha} = 0 \label{eq:2-13a} +\end{equation} +and in particular +\begin{equation} + a_\alpha |0\rangle = 0 \label{eq:2-13b} +\end{equation} + +If $\{\alpha\alpha_i\} = \{\alpha_i'\}$, performing the right permutations, the sequence +$\alpha ,\alpha_1, \alpha_2, \dots, \alpha_n$ is identical with the sequence +$\alpha_1', \alpha_2', \dots, \alpha_{n+1}'$. This results in +\begin{equation} + \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha\alpha_1\alpha_2 \dots \alpha_{n}\rangle = 1 \label{eq:2-14} +\end{equation} +and thus +\begin{equation} + a_\alpha |\alpha\alpha_1\alpha_2 \dots \alpha_{n}\rangle = |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \label{eq:2-15} +\end{equation} + +The action of the operator +$a_\alpha$ from the left on a state vector is to to remove one particle in the state +$\alpha$. +If the state vector does not contain the single-particle state $\alpha$, the outcome of the operation is zero. +The operator $a_\alpha$ is normally called for a destruction or annihilation operator. + +The next step is to establish the commutator algebra of $a_\alpha^{\dagger}$ and +$a_\beta$. + +The action of the anti-commutator +$\{a_\alpha^{\dagger}$,$a_\alpha\}$ on a given $n$-particle state is +\begin{align} + a_\alpha^{\dagger} a_\alpha \underbrace{|\alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} &= 0 \nonumber \\ + a_\alpha a_\alpha^{\dagger} \underbrace{|\alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} &= + a_\alpha \underbrace{|\alpha \alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} = + \underbrace{|\alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} \label{eq:2-16a} +\end{align} +if the single-particle state $\alpha$ is not contained in the state. + + If it is present +we arrive at +\begin{align} + a_\alpha^{\dagger} a_\alpha |\alpha_1\alpha_2 \dots \alpha_{k}\alpha \alpha_{k+1} \dots \alpha_{n-1}\rangle &= + a_\alpha^{\dagger} a_\alpha (-1)^k |\alpha \alpha_1\alpha_2 \dots \alpha_{n-1}\rangle \nonumber \\ + = (-1)^k |\alpha \alpha_1\alpha_2 \dots \alpha_{n-1}\rangle &= + |\alpha_1\alpha_2 \dots \alpha_{k}\alpha \alpha_{k+1} \dots \alpha_{n-1}\rangle \nonumber \\ + a_\alpha a_\alpha^{\dagger}|\alpha_1\alpha_2 \dots \alpha_{k}\alpha \alpha_{k+1} \dots \alpha_{n-1}\rangle &= 0 \label{eq:2-16b} +\end{align} +From Eqs.~(\ref{eq:2-16a}) and (\ref{eq:2-16b}) we arrive at +\begin{equation} + \{a_\alpha^{\dagger} , a_\alpha \} = a_\alpha^{\dagger} a_\alpha + a_\alpha a_\alpha^{\dagger} = 1 \label{eq:2-17} +\end{equation} + +The action of $\left\{a_\alpha^{\dagger}, a_\beta\right\}$, with +$\alpha \ne \beta$ on a given state yields three possibilities. +The first case is a state vector which contains both $\alpha$ and $\beta$, then either +$\alpha$ or $\beta$ and finally none of them. + +The first case results in +\begin{align} + a_\alpha^{\dagger} a_\beta |\alpha\beta\alpha_1\alpha_2 \dots \alpha_{n-2}\rangle = 0 \nonumber \\ + a_\beta a_\alpha^{\dagger} |\alpha\beta\alpha_1\alpha_2 \dots \alpha_{n-2}\rangle = 0 \label{eq:2-18a} +\end{align} +while the second case gives +\begin{align} + a_\alpha^{\dagger} a_\beta |\beta \underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle =& + |\alpha \underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle \nonumber \\ + a_\beta a_\alpha^{\dagger} |\beta \underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle =& + a_\beta |\alpha\beta\underbrace{\beta \alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle \nonumber \\ + =& - |\alpha\underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle \label{eq:2-18b} +\end{align} + +Finally if the state vector does not contain $\alpha$ and $\beta$ +\begin{align} + a_\alpha^{\dagger} a_\beta |\underbrace{\alpha_1\alpha_2 \dots \alpha_{n}}_{\neq \alpha,\beta}\rangle &=& 0 \nonumber \\ + a_\beta a_\alpha^{\dagger} |\underbrace{\alpha_1\alpha_2 \dots \alpha_{n}}_{\neq \alpha,\beta}\rangle &=& + a_\beta |\alpha \underbrace{\alpha_1\alpha_2 \dots \alpha_{n}}_{\neq \alpha,\beta}\rangle = 0 \label{eq:2-18c} +\end{align} +For all three cases we have +\begin{equation} + \{a_\alpha^{\dagger},a_\beta \} = a_\alpha^{\dagger} a_\beta + a_\beta a_\alpha^{\dagger} = 0, \quad \alpha \neq \beta \label{eq:2-19} +\end{equation} + +We can summarize our findings in Eqs.~(\ref{eq:2-17}) and (\ref{eq:2-19}) as +\begin{equation} + \{a_\alpha^{\dagger},a_\beta \} = \delta_{\alpha\beta} \label{eq:2-20} +\end{equation} +with $\delta_{\alpha\beta}$ is the Kroenecker $\delta$-symbol. + +The properties of the creation and annihilation operators can be summarized as (for fermions) +\[ + a_\alpha^{\dagger}|0\rangle \equiv |\alpha\rangle, +\] +and +\[ + a_\alpha^{\dagger}|\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} \equiv |\alpha\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}}. +\] +from which follows +\[ + |\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} = a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle. +\] + +The hermitian conjugate has the folowing properties +\[ + a_{\alpha} = ( a_{\alpha}^{\dagger} )^{\dagger}. +\] +Finally we found +\[ + a_\alpha\underbrace{|\alpha_1'\alpha_2' \dots \alpha_{n+1}'}\rangle_{\neq \alpha} = 0, \quad + \textrm{in particular } a_\alpha |0\rangle = 0, +\] +and +\[ + a_\alpha |\alpha\alpha_1\alpha_2 \dots \alpha_{n}\rangle = |\alpha_1\alpha_2 \dots \alpha_{n}\rangle, +\] +and the corresponding commutator algebra +\[ + \{a_{\alpha}^{\dagger},a_{\beta}^{\dagger}\} = \{a_{\alpha},a_{\beta}\} = 0 \hspace{0.5cm} +\{a_\alpha^{\dagger},a_\beta \} = \delta_{\alpha\beta}. +\] + +\subsection*{One-body operators in second quantization} + +A very useful operator is the so-called number-operator. Most physics cases we will +study in this text conserve the total number of particles. The number operator is therefore +a useful quantity which allows us to test that our many-body formalism conserves the number of particles. +In for example $(d,p)$ or $(p,d)$ reactions it is important to be able to describe quantum mechanical states +where particles get added or removed. +A creation operator $a_\alpha^{\dagger}$ adds one particle to the single-particle state +$\alpha$ of a give many-body state vector, while an annihilation operator $a_\alpha$ +removes a particle from a single-particle +state $\alpha$. + +Let us consider an operator proportional with $a_\alpha^{\dagger} a_\beta$ and +$\alpha=\beta$. It acts on an $n$-particle state +resulting in +\begin{equation} + a_\alpha^{\dagger} a_\alpha |\alpha_1\alpha_2 \dots \alpha_{n}\rangle = + \begin{cases} + 0 &\alpha \notin \{\alpha_i\} \\ + \\ + |\alpha_1\alpha_2 \dots \alpha_{n}\rangle & \alpha \in \{\alpha_i\} + \end{cases} +\end{equation} +Summing over all possible one-particle states we arrive at +\begin{equation} + \left( \sum_\alpha a_\alpha^{\dagger} a_\alpha \right) |\alpha_1\alpha_2 \dots \alpha_{n}\rangle = + n |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \label{eq:2-21} +\end{equation} + +The operator +\begin{equation} + \hat{N} = \sum_\alpha a_\alpha^{\dagger} a_\alpha \label{eq:2-22} +\end{equation} +is called the number operator since it counts the number of particles in a give state vector when it acts +on the different single-particle states. It acts on one single-particle state at the time and falls +therefore under category one-body operators. +Next we look at another important one-body operator, namely $\hat{H}_0$ and study its operator form in the +occupation number representation. + +We want to obtain an expression for a one-body operator which conserves the number of particles. +Here we study the one-body operator for the kinetic energy plus an eventual external one-body potential. +The action of this operator on a particular $n$-body state with its pertinent expectation value has already +been studied in coordinate space. +In coordinate space the operator reads +\begin{equation} + \hat{H}_0 = \sum_i \hat{h}_0(x_i) \label{eq:2-23} +\end{equation} +and the anti-symmetric $n$-particle Slater determinant is defined as +\[ +\Phi(x_1, x_2,\dots ,x_n,\alpha_1,\alpha_2,\dots, \alpha_n)= \frac{1}{\sqrt{n!}} \sum_p (-1)^p\hat{P}\psi_{\alpha_1}(x_1)\psi_{\alpha_2}(x_2) \dots \psi_{\alpha_n}(x_n). +\] + +Defining +\begin{equation} + \hat{h}_0(x_i) \psi_{\alpha_i}(x_i) = \sum_{\alpha_k'} \psi_{\alpha_k'}(x_i) \langle\alpha_k'|\hat{h}_0|\alpha_k\rangle \label{eq:2-25} +\end{equation} +we can easily evaluate the action of $\hat{H}_0$ on each product of one-particle functions in Slater determinant. +From Eq.~(\ref{eq:2-25}) we obtain the following result without permuting any particle pair +\begin{align} + && \left( \sum_i \hat{h}_0(x_i) \right) \psi_{\alpha_1}(x_1)\psi_{\alpha_2}(x_2) \dots \psi_{\alpha_n}(x_n) \nonumber \\ + & =&\sum_{\alpha_1'} \langle \alpha_1'|\hat{h}_0|\alpha_1\rangle + \psi_{\alpha_1'}(x_1)\psi_{\alpha_2}(x_2) \dots \psi_{\alpha_n}(x_n) \nonumber \\ + &+&\sum_{\alpha_2'} \langle \alpha_2'|\hat{h}_0|\alpha_2\rangle + \psi_{\alpha_1}(x_1)\psi_{\alpha_2'}(x_2) \dots \psi_{\alpha_n}(x_n) \nonumber \\ + &+& \dots \nonumber \\ + &+&\sum_{\alpha_n'} \langle \alpha_n'|\hat{h}_0|\alpha_n\rangle + \psi_{\alpha_1}(x_1)\psi_{\alpha_2}(x_2) \dots \psi_{\alpha_n'}(x_n) \label{eq:2-26} +\end{align} + +If we interchange particles $1$ and $2$ we obtain +\begin{align} + && \left( \sum_i \hat{h}_0(x_i) \right) \psi_{\alpha_1}(x_2)\psi_{\alpha_1}(x_2) \dots \psi_{\alpha_n}(x_n) \nonumber \\ + & =&\sum_{\alpha_2'} \langle \alpha_2'|\hat{h}_0|\alpha_2\rangle + \psi_{\alpha_1}(x_2)\psi_{\alpha_2'}(x_1) \dots \psi_{\alpha_n}(x_n) \nonumber \\ + &+&\sum_{\alpha_1'} \langle \alpha_1'|\hat{h}_0|\alpha_1\rangle + \psi_{\alpha_1'}(x_2)\psi_{\alpha_2}(x_1) \dots \psi_{\alpha_n}(x_n) \nonumber \\ + &+& \dots \nonumber \\ + &+&\sum_{\alpha_n'} \langle \alpha_n'|\hat{h}_0|\alpha_n\rangle + \psi_{\alpha_1}(x_2)\psi_{\alpha_1}(x_2) \dots \psi_{\alpha_n'}(x_n) \label{eq:2-27} +\end{align} + +We can continue by computing all possible permutations. +We rewrite also our Slater determinant in its second quantized form and skip the dependence on the quantum numbers $x_i.$ +Summing up all contributions and taking care of all phases +$(-1)^p$ we arrive at +\begin{align} + \hat{H}_0|\alpha_1,\alpha_2,\dots, \alpha_n\rangle &=& \sum_{\alpha_1'}\langle \alpha_1'|\hat{h}_0|\alpha_1\rangle + |\alpha_1'\alpha_2 \dots \alpha_{n}\rangle \nonumber \\ + &+& \sum_{\alpha_2'} \langle \alpha_2'|\hat{h}_0|\alpha_2\rangle + |\alpha_1\alpha_2' \dots \alpha_{n}\rangle \nonumber \\ + &+& \dots \nonumber \\ + &+& \sum_{\alpha_n'} \langle \alpha_n'|\hat{h}_0|\alpha_n\rangle + |\alpha_1\alpha_2 \dots \alpha_{n}'\rangle \label{eq:2-28} +\end{align} + +In Eq.~(\ref{eq:2-28}) +we have expressed the action of the one-body operator +of Eq.~(\ref{eq:2-23}) on the $n$-body state in its second quantized form. +This equation can be further manipulated if we use the properties of the creation and annihilation operator +on each primed quantum number, that is +\begin{equation} + |\alpha_1\alpha_2 \dots \alpha_k' \dots \alpha_{n}\rangle = + a_{\alpha_k'}^{\dagger} a_{\alpha_k} |\alpha_1\alpha_2 \dots \alpha_k \dots \alpha_{n}\rangle \label{eq:2-29} +\end{equation} +Inserting this in the right-hand side of Eq.~(\ref{eq:2-28}) results in +\begin{align} + \hat{H}_0|\alpha_1\alpha_2 \dots \alpha_{n}\rangle &=& \sum_{\alpha_1'}\langle \alpha_1'|\hat{h}_0|\alpha_1\rangle + a_{\alpha_1'}^{\dagger} a_{\alpha_1} |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \nonumber \\ + &+& \sum_{\alpha_2'} \langle \alpha_2'|\hat{h}_0|\alpha_2\rangle + a_{\alpha_2'}^{\dagger} a_{\alpha_2} |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \nonumber \\ + &+& \dots \nonumber \\ + &+& \sum_{\alpha_n'} \langle \alpha_n'|\hat{h}_0|\alpha_n\rangle + a_{\alpha_n'}^{\dagger} a_{\alpha_n} |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \nonumber \\ + &=& \sum_{\alpha, \beta} \langle \alpha|\hat{h}_0|\beta\rangle a_\alpha^{\dagger} a_\beta + |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \label{eq:2-30a} +\end{align} + +In the number occupation representation or second quantization we get the following expression for a one-body +operator which conserves the number of particles +\begin{equation} + \hat{H}_0 = \sum_{\alpha\beta} \langle \alpha|\hat{h}_0|\beta\rangle a_\alpha^{\dagger} a_\beta \label{eq:2-30b} +\end{equation} +Obviously, $\hat{H}_0$ can be replaced by any other one-body operator which preserved the number +of particles. The stucture of the operator is therefore not limited to say the kinetic or single-particle energy only. + +The opearator $\hat{H}_0$ takes a particle from the single-particle state $\beta$ to the single-particle state $\alpha$ +with a probability for the transition given by the expectation value $\langle \alpha|\hat{h}_0|\beta\rangle$. + +It is instructive to verify Eq.~(\ref{eq:2-30b}) by computing the expectation value of $\hat{H}_0$ +between two single-particle states +\begin{equation} + \langle \alpha_1|\hat{h}_0|\alpha_2\rangle = \sum_{\alpha\beta} \langle \alpha|\hat{h}_0|\beta\rangle + \langle 0|a_{\alpha_1}a_\alpha^{\dagger} a_\beta a_{\alpha_2}^{\dagger}|0\rangle \label{eq:2-30c} +\end{equation} + +Using the commutation relations for the creation and annihilation operators we have +\begin{equation} +a_{\alpha_1}a_\alpha^{\dagger} a_\beta a_{\alpha_2}^{\dagger} = (\delta_{\alpha \alpha_1} - a_\alpha^{\dagger} a_{\alpha_1} )(\delta_{\beta \alpha_2} - a_{\alpha_2}^{\dagger} a_{\beta} ), \label{eq:2-30d} +\end{equation} +which results in +\begin{equation} +\langle 0|a_{\alpha_1}a_\alpha^{\dagger} a_\beta a_{\alpha_2}^{\dagger}|0\rangle = \delta_{\alpha \alpha_1} \delta_{\beta \alpha_2} \label{eq:2-30e} +\end{equation} +and +\begin{equation} +\langle \alpha_1|\hat{h}_0|\alpha_2\rangle = \sum_{\alpha\beta} \langle \alpha|\hat{h}_0|\beta\rangle\delta_{\alpha \alpha_1} \delta_{\beta \alpha_2} = \langle \alpha_1|\hat{h}_0|\alpha_2\rangle \label{eq:2-30f} +\end{equation} + +\subsection*{Two-body operators in second quantization} + +Let us now derive the expression for our two-body interaction part, which also conserves the number of particles. +We can proceed in exactly the same way as for the one-body operator. In the coordinate representation our +two-body interaction part takes the following expression +\begin{equation} + \hat{H}_I = \sum_{i < j} V(x_i,x_j) \label{eq:2-31} +\end{equation} +where the summation runs over distinct pairs. The term $V$ can be an interaction model for the nucleon-nucleon interaction +or the interaction between two electrons. It can also include additional two-body interaction terms. + +The action of this operator on a product of +two single-particle functions is defined as +\begin{equation} + V(x_i,x_j) \psi_{\alpha_k}(x_i) \psi_{\alpha_l}(x_j) = \sum_{\alpha_k'\alpha_l'} + \psi_{\alpha_k}'(x_i)\psi_{\alpha_l}'(x_j) + \langle \alpha_k'\alpha_l'|\hat{v}|\alpha_k\alpha_l\rangle \label{eq:2-32} +\end{equation} + +We can now let $\hat{H}_I$ act on all terms in the linear combination for $|\alpha_1\alpha_2\dots\alpha_n\rangle$. Without any permutations we have +\begin{align} + && \left( \sum_{i < j} V(x_i,x_j) \right) \psi_{\alpha_1}(x_1)\psi_{\alpha_2}(x_2)\dots \psi_{\alpha_n}(x_n) \nonumber \\ + &=& \sum_{\alpha_1'\alpha_2'} \langle \alpha_1'\alpha_2'|\hat{v}|\alpha_1\alpha_2\rangle + \psi_{\alpha_1}'(x_1)\psi_{\alpha_2}'(x_2)\dots \psi_{\alpha_n}(x_n) \nonumber \\ + & +& \dots \nonumber \\ + &+& \sum_{\alpha_1'\alpha_n'} \langle \alpha_1'\alpha_n'|\hat{v}|\alpha_1\alpha_n\rangle + \psi_{\alpha_1}'(x_1)\psi_{\alpha_2}(x_2)\dots \psi_{\alpha_n}'(x_n) \nonumber \\ + & +& \dots \nonumber \\ + &+& \sum_{\alpha_2'\alpha_n'} \langle \alpha_2'\alpha_n'|\hat{v}|\alpha_2\alpha_n\rangle + \psi_{\alpha_1}(x_1)\psi_{\alpha_2}'(x_2)\dots \psi_{\alpha_n}'(x_n) \nonumber \\ + & +& \dots \label{eq:2-33} +\end{align} +where on the rhs we have a term for each distinct pairs. + +For the other terms on the rhs we obtain similar expressions and summing over all terms we obtain +\begin{align} + H_I |\alpha_1\alpha_2\dots\alpha_n\rangle &=& \sum_{\alpha_1', \alpha_2'} \langle \alpha_1'\alpha_2'|\hat{v}|\alpha_1\alpha_2\rangle + |\alpha_1'\alpha_2'\dots\alpha_n\rangle \nonumber \\ + &+& \dots \nonumber \\ + &+& \sum_{\alpha_1', \alpha_n'} \langle \alpha_1'\alpha_n'|\hat{v}|\alpha_1\alpha_n\rangle + |\alpha_1'\alpha_2\dots\alpha_n'\rangle \nonumber \\ + &+& \dots \nonumber \\ + &+& \sum_{\alpha_2', \alpha_n'} \langle \alpha_2'\alpha_n'|\hat{v}|\alpha_2\alpha_n\rangle + |\alpha_1\alpha_2'\dots\alpha_n'\rangle \nonumber \\ + &+& \dots \label{eq:2-34} +\end{align} + +We introduce second quantization via the relation +\begin{align} + && a_{\alpha_k'}^{\dagger} a_{\alpha_l'}^{\dagger} a_{\alpha_l} a_{\alpha_k} + |\alpha_1\alpha_2\dots\alpha_k\dots\alpha_l\dots\alpha_n\rangle \nonumber \\ + &=& (-1)^{k-1} (-1)^{l-2} a_{\alpha_k'}^{\dagger} a_{\alpha_l'}^{\dagger} a_{\alpha_l} a_{\alpha_k} + |\alpha_k\alpha_l \underbrace{\alpha_1\alpha_2\dots\alpha_n}_{\neq \alpha_k,\alpha_l}\rangle \nonumber \\ + &=& (-1)^{k-1} (-1)^{l-2} + |\alpha_k'\alpha_l' \underbrace{\alpha_1\alpha_2\dots\alpha_n}_{\neq \alpha_k',\alpha_l'}\rangle \nonumber \\ + &=& |\alpha_1\alpha_2\dots\alpha_k'\dots\alpha_l'\dots\alpha_n\rangle \label{eq:2-35} +\end{align} + +Inserting this in (\ref{eq:2-34}) gives +\begin{align} + H_I |\alpha_1\alpha_2\dots\alpha_n\rangle + &=& \sum_{\alpha_1', \alpha_2'} \langle \alpha_1'\alpha_2'|\hat{v}|\alpha_1\alpha_2\rangle + a_{\alpha_1'}^{\dagger} a_{\alpha_2'}^{\dagger} a_{\alpha_2} a_{\alpha_1} + |\alpha_1\alpha_2\dots\alpha_n\rangle \nonumber \\ + &+& \dots \nonumber \\ + &=& \sum_{\alpha_1', \alpha_n'} \langle \alpha_1'\alpha_n'|\hat{v}|\alpha_1\alpha_n\rangle + a_{\alpha_1'}^{\dagger} a_{\alpha_n'}^{\dagger} a_{\alpha_n} a_{\alpha_1} + |\alpha_1\alpha_2\dots\alpha_n\rangle \nonumber \\ + &+& \dots \nonumber \\ + &=& \sum_{\alpha_2', \alpha_n'} \langle \alpha_2'\alpha_n'|\hat{v}|\alpha_2\alpha_n\rangle + a_{\alpha_2'}^{\dagger} a_{\alpha_n'}^{\dagger} a_{\alpha_n} a_{\alpha_2} + |\alpha_1\alpha_2\dots\alpha_n\rangle \nonumber \\ + &+& \dots \nonumber \\ + &=& \sum_{\alpha, \beta, \gamma, \delta} ' \langle \alpha\beta|\hat{v}|\gamma\delta\rangle + a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma + |\alpha_1\alpha_2\dots\alpha_n\rangle \label{eq:2-36} +\end{align} + +Here we let $\sum'$ indicate that the sums running over $\alpha$ and $\beta$ run over all +single-particle states, while the summations $\gamma$ and $\delta$ +run over all pairs of single-particle states. We wish to remove this restriction and since +\begin{equation} + \langle \alpha\beta|\hat{v}|\gamma\delta\rangle = \langle \beta\alpha|\hat{v}|\delta\gamma\rangle \label{eq:2-37} +\end{equation} +we get +\begin{align} + \sum_{\alpha\beta} \langle \alpha\beta|\hat{v}|\gamma\delta\rangle a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma &=& + \sum_{\alpha\beta} \langle \beta\alpha|\hat{v}|\delta\gamma\rangle + a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma \label{eq:2-38a} \\ + &=& \sum_{\alpha\beta}\langle \beta\alpha|\hat{v}|\delta\gamma\rangle + a^{\dagger}_\beta a^{\dagger}_\alpha a_\gamma a_\delta \label{eq:2-38b} +\end{align} +where we have used the anti-commutation rules. + +Changing the summation indices +$\alpha$ and $\beta$ in (\ref{eq:2-38b}) we obtain +\begin{equation} + \sum_{\alpha\beta} \langle \alpha\beta|\hat{v}|\gamma\delta\rangle a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma = + \sum_{\alpha\beta} \langle \alpha\beta|\hat{v}|\delta\gamma\rangle + a^{\dagger}_\alpha a^{\dagger}_\beta a_\gamma a_\delta \label{eq:2-38c} +\end{equation} +From this it follows that the restriction on the summation over $\gamma$ and $\delta$ can be removed if we multiply with a factor $\frac{1}{2}$, resulting in +\begin{equation} + \hat{H}_I = \frac{1}{2} \sum_{\alpha\beta\gamma\delta} \langle \alpha\beta|\hat{v}|\gamma\delta\rangle + a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma \label{eq:2-39} +\end{equation} +where we sum freely over all single-particle states $\alpha$, +$\beta$, $\gamma$ og $\delta$. + +With this expression we can now verify that the second quantization form of $\hat{H}_I$ in Eq.~(\ref{eq:2-39}) +results in the same matrix between two anti-symmetrized two-particle states as its corresponding coordinate +space representation. We have +\begin{equation} + \langle \alpha_1 \alpha_2|\hat{H}_I|\beta_1 \beta_2\rangle = + \frac{1}{2} \sum_{\alpha\beta\gamma\delta} + \langle \alpha\beta|\hat{v}|\gamma\delta\rangle \langle 0|a_{\alpha_2} a_{\alpha_1} + a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma + a_{\beta_1}^{\dagger} a_{\beta_2}^{\dagger}|0\rangle. \label{eq:2-40} +\end{equation} + +Using the commutation relations we get +\begin{align} + && a_{\alpha_2} a_{\alpha_1}a^{\dagger}_\alpha a^{\dagger}_\beta + a_\delta a_\gamma a_{\beta_1}^{\dagger} a_{\beta_2}^{\dagger} \nonumber \\ + &=& a_{\alpha_2} a_{\alpha_1}a^{\dagger}_\alpha a^{\dagger}_\beta + ( a_\delta \delta_{\gamma \beta_1} a_{\beta_2}^{\dagger} - + a_\delta a_{\beta_1}^{\dagger} a_\gamma a_{\beta_2}^{\dagger} ) \nonumber \\ + &=& a_{\alpha_2} a_{\alpha_1}a^{\dagger}_\alpha a^{\dagger}_\beta + (\delta_{\gamma \beta_1} \delta_{\delta \beta_2} - \delta_{\gamma \beta_1} a_{\beta_2}^{\dagger} a_\delta - + a_\delta a_{\beta_1}^{\dagger}\delta_{\gamma \beta_2} + + a_\delta a_{\beta_1}^{\dagger} a_{\beta_2}^{\dagger} a_\gamma ) \nonumber \\ + &=& a_{\alpha_2} a_{\alpha_1}a^{\dagger}_\alpha a^{\dagger}_\beta + (\delta_{\gamma \beta_1} \delta_{\delta \beta_2} - \delta_{\gamma \beta_1} a_{\beta_2}^{\dagger} a_\delta \nonumber \\ + && \qquad - \delta_{\delta \beta_1} \delta_{\gamma \beta_2} + \delta_{\gamma \beta_2} a_{\beta_1}^{\dagger} a_\delta + + a_\delta a_{\beta_1}^{\dagger} a_{\beta_2}^{\dagger} a_\gamma ) \label{eq:2-41} +\end{align} + +The vacuum expectation value of this product of operators becomes +\begin{align} + && \langle 0|a_{\alpha_2} a_{\alpha_1} a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma + a_{\beta_1}^{\dagger} a_{\beta_2}^{\dagger}|0\rangle \nonumber \\ + &=& (\delta_{\gamma \beta_1} \delta_{\delta \beta_2} - + \delta_{\delta \beta_1} \delta_{\gamma \beta_2} ) + \langle 0|a_{\alpha_2} a_{\alpha_1}a^{\dagger}_\alpha a^{\dagger}_\beta|0\rangle \nonumber \\ + &=& (\delta_{\gamma \beta_1} \delta_{\delta \beta_2} -\delta_{\delta \beta_1} \delta_{\gamma \beta_2} ) + (\delta_{\alpha \alpha_1} \delta_{\beta \alpha_2} -\delta_{\beta \alpha_1} \delta_{\alpha \alpha_2} ) \label{eq:2-42b} +\end{align} + +Insertion of +Eq.~(\ref{eq:2-42b}) in Eq.~(\ref{eq:2-40}) results in +\begin{align} + \langle \alpha_1\alpha_2|\hat{H}_I|\beta_1\beta_2\rangle &=& \frac{1}{2} \big[ + \langle \alpha_1\alpha_2|\hat{v}|\beta_1\beta_2\rangle- \langle \alpha_1\alpha_2|\hat{v}|\beta_2\beta_1\rangle \nonumber \\ + && - \langle \alpha_2\alpha_1|\hat{v}|\beta_1\beta_2\rangle + \langle \alpha_2\alpha_1|\hat{v}|\beta_2\beta_1\rangle \big] \nonumber \\ + &=& \langle \alpha_1\alpha_2|\hat{v}|\beta_1\beta_2\rangle - \langle \alpha_1\alpha_2|\hat{v}|\beta_2\beta_1\rangle \nonumber \\ + &=& \langle \alpha_1\alpha_2|\hat{v}|\beta_1\beta_2\rangle_{\mathrm{AS}}. \label{eq:2-43b} +\end{align} + +The two-body operator can also be expressed in terms of the anti-symmetrized matrix elements we discussed previously as +\begin{align} + \hat{H}_I &=& \frac{1}{2} \sum_{\alpha\beta\gamma\delta} \langle \alpha \beta|\hat{v}|\gamma \delta\rangle + a_\alpha^{\dagger} a_\beta^{\dagger} a_\delta a_\gamma \nonumber \\ + &=& \frac{1}{4} \sum_{\alpha\beta\gamma\delta} \left[ \langle \alpha \beta|\hat{v}|\gamma \delta\rangle - + \langle \alpha \beta|\hat{v}|\delta\gamma \rangle \right] + a_\alpha^{\dagger} a_\beta^{\dagger} a_\delta a_\gamma \nonumber \\ + &=& \frac{1}{4} \sum_{\alpha\beta\gamma\delta} \langle \alpha \beta|\hat{v}|\gamma \delta\rangle_{\mathrm{AS}} + a_\alpha^{\dagger} a_\beta^{\dagger} a_\delta a_\gamma \label{eq:2-45} +\end{align} + +The factors in front of the operator, either $\frac{1}{4}$ or +$\frac{1}{2}$ tells whether we use antisymmetrized matrix elements or not. + +We can now express the Hamiltonian operator for a many-fermion system in the occupation basis representation +as +\begin{equation} + H = \sum_{\alpha, \beta} \langle \alpha|\hat{t}+\hat{u}_{\mathrm{ext}}|\beta\rangle a_\alpha^{\dagger} a_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta} + \langle \alpha \beta|\hat{v}|\gamma \delta\rangle a_\alpha^{\dagger} a_\beta^{\dagger} a_\delta a_\gamma. \label{eq:2-46b} +\end{equation} +This is the form we will use in the rest of these lectures, assuming that we work with anti-symmetrized two-body matrix elements. + +\subsection*{Particle-hole formalism} + +Second quantization is a useful and elegant formalism for constructing many-body states and +quantum mechanical operators. One can express and translate many physical processes +into simple pictures such as Feynman diagrams. Expecation values of many-body states are also easily calculated. +However, although the equations are seemingly easy to set up, from a practical point of view, that is +the solution of Schroedinger's equation, there is no particular gain. +The many-body equation is equally hard to solve, irrespective of representation. +The cliche that +there is no free lunch brings us down to earth again. +Note however that a transformation to a particular +basis, for cases where the interaction obeys specific symmetries, can ease the solution of Schroedinger's equation. + +But there is at least one important case where second quantization comes to our rescue. +It is namely easy to introduce another reference state than the pure vacuum $|0\rangle $, where all single-particle states are active. +With many particles present it is often useful to introduce another reference state than the vacuum state$|0\rangle $. We will label this state $|c\rangle$ ($c$ for core) and as we will see it can reduce +considerably the complexity and thereby the dimensionality of the many-body problem. It allows us to sum up to infinite order specific many-body correlations. The particle-hole representation is one of these handy representations. + +In the original particle representation these states are products of the creation operators $a_{\alpha_i}^\dagger$ acting on the true vacuum $|0\rangle $. +Following Eq.~(\ref{eq:2-2}) we have +\begin{align} + |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\rangle &=& a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots + a_{\alpha_{n-1}}^\dagger a_{\alpha_n}^\dagger |0\rangle \label{eq:2-47a} \\ + |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\alpha_{n+1}\rangle &=& + a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots a_{\alpha_{n-1}}^\dagger a_{\alpha_n}^\dagger + a_{\alpha_{n+1}}^\dagger |0\rangle \label{eq:2-47b} \\ + |\alpha_1\alpha_2\dots\alpha_{n-1}\rangle &=& a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots + a_{\alpha_{n-1}}^\dagger |0\rangle \label{eq:2-47c} +\end{align} + +If we use Eq.~(\ref{eq:2-47a}) as our new reference state, we can simplify considerably the representation of +this state +\begin{equation} + |c\rangle \equiv |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\rangle = + a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots a_{\alpha_{n-1}}^\dagger a_{\alpha_n}^\dagger |0\rangle \label{eq:2-48a} +\end{equation} +The new reference states for the $n+1$ and $n-1$ states can then be written as +\begin{align} + |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\alpha_{n+1}\rangle &=& (-1)^n a_{\alpha_{n+1}}^\dagger |c\rangle + \equiv (-1)^n |\alpha_{n+1}\rangle_c \label{eq:2-48b} \\ + |\alpha_1\alpha_2\dots\alpha_{n-1}\rangle &=& (-1)^{n-1} a_{\alpha_n} |c\rangle + \equiv (-1)^{n-1} |\alpha_{n-1}\rangle_c \label{eq:2-48c} +\end{align} + +The first state has one additional particle with respect to the new vacuum state +$|c\rangle $ and is normally referred to as a one-particle state or one particle added to the +many-body reference state. +The second state has one particle less than the reference vacuum state $|c\rangle $ and is referred to as +a one-hole state. +When dealing with a new reference state it is often convenient to introduce +new creation and annihilation operators since we have +from Eq.~(\ref{eq:2-48c}) +\begin{equation} + a_\alpha |c\rangle \neq 0 \label{eq:2-49} +\end{equation} +since $\alpha$ is contained in $|c\rangle $, while for the true vacuum we have +$a_\alpha |0\rangle = 0$ for all $\alpha$. + +The new reference state leads to the definition of new creation and annihilation operators +which satisfy the following relations +\begin{align} + b_\alpha |c\rangle &=& 0 \label{eq:2-50a} \\ + \{b_\alpha^\dagger , b_\beta^\dagger \} = \{b_\alpha , b_\beta \} &=& 0 \nonumber \\ + \{b_\alpha^\dagger , b_\beta \} &=& \delta_{\alpha \beta} \label{eq:2-50c} +\end{align} +We assume also that the new reference state is properly normalized +\begin{equation} + \langle c | c \rangle = 1 \label{eq:2-51} +\end{equation} + +The physical interpretation of these new operators is that of so-called quasiparticle states. +This means that a state defined by the addition of one extra particle to a reference state $|c\rangle $ may not necesseraly be interpreted as one particle coupled to a core. +We define now new creation operators that act on a state $\alpha$ creating a new quasiparticle state +\begin{equation} + b_\alpha^\dagger|c\rangle = \Bigg\{ \begin{array}{ll} + a_\alpha^\dagger |c\rangle = |\alpha\rangle, & \alpha > F \\ + \\ + a_\alpha |c\rangle = |\alpha^{-1}\rangle, & \alpha \leq F + \end{array} \label{eq:2-52} +\end{equation} +where $F$ is the Fermi level representing the last occupied single-particle orbit +of the new reference state $|c\rangle $. + +The annihilation is the hermitian conjugate of the creation operator +\[ + b_\alpha = (b_\alpha^\dagger)^\dagger, +\] +resulting in +\begin{equation} + b_\alpha^\dagger = \Bigg\{ \begin{array}{ll} + a_\alpha^\dagger & \alpha > F \\ + \\ + a_\alpha & \alpha \leq F + \end{array} \qquad + b_\alpha = \Bigg\{ \begin{array}{ll} + a_\alpha & \alpha > F \\ + \\ + a_\alpha^\dagger & \alpha \leq F + \end{array} \label{eq:2-54} +\end{equation} + +With the new creation and annihilation operator we can now construct +many-body quasiparticle states, with one-particle-one-hole states, two-particle-two-hole +states etc in the same fashion as we previously constructed many-particle states. +We can write a general particle-hole state as +\begin{equation} + |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle \equiv + \underbrace{b_{\beta_1}^\dagger b_{\beta_2}^\dagger \dots b_{\beta_{n_p}}^\dagger}_{>F} + \underbrace{b_{\gamma_1}^\dagger b_{\gamma_2}^\dagger \dots b_{\gamma_{n_h}}^\dagger}_{\leq F} |c\rangle \label{eq:2-56} +\end{equation} +We can now rewrite our one-body and two-body operators in terms of the new creation and annihilation operators. +The number operator becomes +\begin{equation} + \hat{N} = \sum_\alpha a_\alpha^\dagger a_\alpha= +\sum_{\alpha > F} b_\alpha^\dagger b_\alpha + n_c - \sum_{\alpha \leq F} b_\alpha^\dagger b_\alpha \label{eq:2-57b} +\end{equation} +where $n_c$ is the number of particle in the new vacuum state $|c\rangle $. +The action of $\hat{N}$ on a many-body state results in +\begin{equation} + N |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle = (n_p + n_c - n_h) |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle \label{2-59} +\end{equation} +Here $n=n_p +n_c - n_h$ is the total number of particles in the quasi-particle state of +Eq.~(\ref{eq:2-56}). Note that $\hat{N}$ counts the total number of particles present +\begin{equation} + N_{qp} = \sum_\alpha b_\alpha^\dagger b_\alpha, \label{eq:2-60} +\end{equation} +gives us the number of quasi-particles as can be seen by computing +\begin{equation} + N_{qp}= |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle + = (n_p + n_h)|\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle \label{eq:2-61} +\end{equation} +where $n_{qp} = n_p + n_h$ is the total number of quasi-particles. + +We express the one-body operator $\hat{H}_0$ in terms of the quasi-particle creation and annihilation operators, resulting in +\begin{align} + \hat{H}_0 &=& \sum_{\alpha\beta > F} \langle \alpha|\hat{h}_0|\beta\rangle b_\alpha^\dagger b_\beta + + \sum_{\alpha > F, \beta \leq F } \left[\langle \alpha|\hat{h}_0|\beta\rangle b_\alpha^\dagger b_\beta^\dagger + \langle \beta|\hat{h}_0|\alpha\rangle b_\beta b_\alpha \right] \nonumber \\ + &+& \sum_{\alpha \leq F} \langle \alpha|\hat{h}_0|\alpha\rangle - \sum_{\alpha\beta \leq F} \langle \beta|\hat{h}_0|\alpha\rangle b_\alpha^\dagger b_\beta \label{eq:2-63b} +\end{align} +The first term gives contribution only for particle states, while the last one +contributes only for holestates. The second term can create or destroy a set of +quasi-particles and +the third term is the contribution from the vacuum state $|c\rangle$. + +Before we continue with the expressions for the two-body operator, we introduce a nomenclature we will use for the rest of this +text. It is inspired by the notation used in quantum chemistry. +We reserve the labels $i,j,k,\dots$ for hole states and $a,b,c,\dots$ for states above $F$, viz.~particle states. +This means also that we will skip the constraint $\leq F$ or $> F$ in the summation symbols. +Our operator $\hat{H}_0$ reads now +\begin{align} + \hat{H}_0 &=& \sum_{ab} \langle a|\hat{h}|b\rangle b_a^\dagger b_b + + \sum_{ai} \left[ + \langle a|\hat{h}|i\rangle b_a^\dagger b_i^\dagger + + \langle i|\hat{h}|a\rangle b_i b_a \right] \nonumber \\ + &+& \sum_{i} \langle i|\hat{h}|i\rangle - + \sum_{ij} \langle j|\hat{h}|i\rangle + b_i^\dagger b_j \label{eq:2-63c} +\end{align} + +The two-particle operator in the particle-hole formalism is more complicated since we have +to translate four indices $\alpha\beta\gamma\delta$ to the possible combinations of particle and hole +states. When performing the commutator algebra we can regroup the operator in five different terms +\begin{equation} + \hat{H}_I = \hat{H}_I^{(a)} + \hat{H}_I^{(b)} + \hat{H}_I^{(c)} + \hat{H}_I^{(d)} + \hat{H}_I^{(e)} \label{eq:2-65} +\end{equation} +Using anti-symmetrized matrix elements, +bthe term $\hat{H}_I^{(a)}$ is +\begin{equation} + \hat{H}_I^{(a)} = \frac{1}{4} + \sum_{abcd} \langle ab|\hat{V}|cd\rangle + b_a^\dagger b_b^\dagger b_d b_c \label{eq:2-66} +\end{equation} + +The next term $\hat{H}_I^{(b)}$ reads +\begin{equation} + \hat{H}_I^{(b)} = \frac{1}{4} \sum_{abci}\left(\langle ab|\hat{V}|ci\rangle b_a^\dagger b_b^\dagger b_i^\dagger b_c +\langle ai|\hat{V}|cb\rangle b_a^\dagger b_i b_b b_c\right) \label{eq:2-67b} +\end{equation} +This term conserves the number of quasiparticles but creates or removes a +three-particle-one-hole state. +For $\hat{H}_I^{(c)}$ we have +\begin{align} + \hat{H}_I^{(c)}& =& \frac{1}{4} + \sum_{abij}\left(\langle ab|\hat{V}|ij\rangle b_a^\dagger b_b^\dagger b_j^\dagger b_i^\dagger + + \langle ij|\hat{V}|ab\rangle b_a b_b b_j b_i \right)+ \nonumber \\ + && \frac{1}{2}\sum_{abij}\langle ai|\hat{V}|bj\rangle b_a^\dagger b_j^\dagger b_b b_i + + \frac{1}{2}\sum_{abi}\langle ai|\hat{V}|bi\rangle b_a^\dagger b_b. \label{eq:2-68c} +\end{align} + +The first line stands for the creation of a two-particle-two-hole state, while the second line represents +the creation to two one-particle-one-hole pairs +while the last term represents a contribution to the particle single-particle energy +from the hole states, that is an interaction between the particle states and the hole states +within the new vacuum state. +The fourth term reads +\begin{align} + \hat{H}_I^{(d)}& = &\frac{1}{4} + \sum_{aijk}\left(\langle ai|\hat{V}|jk\rangle b_a^\dagger b_k^\dagger b_j^\dagger b_i+ +\langle ji|\hat{V}|ak\rangle b_k^\dagger b_j b_i b_a\right)+\nonumber \\ +&&\frac{1}{4}\sum_{aij}\left(\langle ai|\hat{V}|ji\rangle b_a^\dagger b_j^\dagger+ +\langle ji|\hat{V}|ai\rangle - \langle ji|\hat{V}|ia\rangle b_j b_a \right). \label{eq:2-69d} +\end{align} +The terms in the first line stand for the creation of a particle-hole state +interacting with hole states, we will label this as a two-hole-one-particle contribution. +The remaining terms are a particle-hole state interacting with the holes in the vacuum state. +Finally we have +\begin{equation} + \hat{H}_I^{(e)} = \frac{1}{4} + \sum_{ijkl} + \langle kl|\hat{V}|ij\rangle b_i^\dagger b_j^\dagger b_l b_k+ + \frac{1}{2}\sum_{ijk}\langle ij|\hat{V}|kj\rangle b_k^\dagger b_i + +\frac{1}{2}\sum_{ij}\langle ij|\hat{V}|ij\rangle \label{eq:2-70d} +\end{equation} +The first terms represents the +interaction between two holes while the second stands for the interaction between a hole and the remaining holes in the vacuum state. +It represents a contribution to single-hole energy to first order. +The last term collects all contributions to the energy of the ground state of a closed-shell system arising +from hole-hole correlations. + +\subsection*{Summarizing and defining a normal-ordered Hamiltonian} + +\[ + \Phi_{AS}(\alpha_1, \dots, \alpha_A; x_1, \dots x_A)= + \frac{1}{\sqrt{A}} \sum_{\hat{P}} (-1)^P \hat{P} \prod_{i=1}^A \psi_{\alpha_i}(x_i), +\] +which is equivalent with $|\alpha_1 \dots \alpha_A\rangle= a_{\alpha_1}^{\dagger} \dots a_{\alpha_A}^{\dagger} |0\rangle$. We have also + \[ + a_p^\dagger|0\rangle = |p\rangle, \quad a_p |q\rangle = \delta_{pq}|0\rangle + \] +\[ + \delta_{pq} = \left\{a_p, a_q^\dagger \right\}, +\] +and +\[ +0 = \left\{a_p^\dagger, a_q \right\} = \left\{a_p, a_q \right\} = \left\{a_p^\dagger, a_q^\dagger \right\} +\] +\[ +|\Phi_0\rangle = |\alpha_1 \dots \alpha_A\rangle, \quad \alpha_1, \dots, \alpha_A \leq \alpha_F +\] + +\[ +\left\{a_p^\dagger, a_q \right\}= \delta_{pq}, p, q \leq \alpha_F +\] +\[ +\left\{a_p, a_q^\dagger \right\} = \delta_{pq}, p, q > \alpha_F +\] +with $i,j,\ldots \leq \alpha_F, \quad a,b,\ldots > \alpha_F, \quad p,q, \ldots - \textrm{any}$ +\[ + a_i|\Phi_0\rangle = |\Phi_i\rangle, \hspace{0.5cm} a_a^\dagger|\Phi_0\rangle = |\Phi^a\rangle +\] +and +\[ +a_i^\dagger|\Phi_0\rangle = 0 \hspace{0.5cm} a_a|\Phi_0\rangle = 0 +\] + +The one-body operator is defined as +\[ + \hat{F} = \sum_{pq} \langle p|\hat{f}|q\rangle a_p^\dagger a_q +\] +while the two-body opreator is defined as +\[ +\hat{V} = \frac{1}{4} \sum_{pqrs} \langle pq|\hat{v}|rs\rangle_{AS} a_p^\dagger a_q^\dagger a_s a_r +\] +where we have defined the antisymmetric matrix elements +\[ +\langle pq|\hat{v}|rs\rangle_{AS} = \langle pq|\hat{v}|rs\rangle - \langle pq|\hat{v}|sr\rangle. +\] + +We can also define a three-body operator +\[ +\hat{V}_3 = \frac{1}{36} \sum_{pqrstu} \langle pqr|\hat{v}_3|stu\rangle_{AS} + a_p^\dagger a_q^\dagger a_r^\dagger a_u a_t a_s +\] +with the antisymmetrized matrix element +\begin{align} + \langle pqr|\hat{v}_3|stu\rangle_{AS} = \langle pqr|\hat{v}_3|stu\rangle + \langle pqr|\hat{v}_3|tus\rangle + \langle pqr|\hat{v}_3|ust\rangle- \langle pqr|\hat{v}_3|sut\rangle - \langle pqr|\hat{v}_3|tsu\rangle - \langle pqr|\hat{v}_3|uts\rangle. +\end{align} + +\subsection*{Operators in second quantization} + +In the build-up of a shell-model or FCI code that is meant to tackle large dimensionalities +is the action of the Hamiltonian $\hat{H}$ on a +Slater determinant represented in second quantization as +\[ + |\alpha_1\dots \alpha_n\rangle = a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle. +\] +The time consuming part stems from the action of the Hamiltonian +on the above determinant, +\[ +\left(\sum_{\alpha\beta} \langle \alpha|t+u|\beta\rangle a_\alpha^{\dagger} a_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta} + \langle \alpha \beta|\hat{v}|\gamma \delta\rangle a_\alpha^{\dagger} a_\beta^{\dagger} a_\delta a_\gamma\right)a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle. +\] +A practically useful way to implement this action is to encode a Slater determinant as a bit pattern. + +Assume that we have at our disposal $n$ different single-particle orbits +$\alpha_0,\alpha_2,\dots,\alpha_{n-1}$ and that we can distribute among these orbits $N\le n$ particles. + +A Slater determinant can then be coded as an integer of $n$ bits. As an example, if we have $n=16$ single-particle states +$\alpha_0,\alpha_1,\dots,\alpha_{15}$ and $N=4$ fermions occupying the states $\alpha_3$, $\alpha_6$, $\alpha_{10}$ and $\alpha_{13}$ +we could write this Slater determinant as +\[ +\Phi_{\Lambda} = a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle. +\] +The unoccupied single-particle states have bit value $0$ while the occupied ones are represented by bit state $1$. +In the binary notation we would write this 16 bits long integer as +\[ +\begin{array}{cccccccccccccccc} +{\alpha_0}&{\alpha_1}&{\alpha_2}&{\alpha_3}&{\alpha_4}&{\alpha_5}&{\alpha_6}&{\alpha_7} & {\alpha_8} &{\alpha_9} & {\alpha_{10}} &{\alpha_{11}} &{\alpha_{12}} &{\alpha_{13}} &{\alpha_{14}} & {\alpha_{15}} \\ +{0} & {0} &{0} &{1} &{0} &{0} &{1} &{0} &{0} &{0} &{1} &{0} &{0} &{1} &{0} & {0} \\ +\end{array} +\] +which translates into the decimal number +\[ +2^3+2^6+2^{10}+2^{13}=9288. +\] +We can thus encode a Slater determinant as a bit pattern. + +With $N$ particles that can be distributed over $n$ single-particle states, the total number of Slater determinats (and defining thereby the dimensionality of the system) is +\[ +\mathrm{dim}(\mathcal{H}) = \left(\begin{array}{c} n \\N\end{array}\right). +\] +The total number of bit patterns is $2^n$. + +We assume again that we have at our disposal $n$ different single-particle orbits +$\alpha_0,\alpha_2,\dots,\alpha_{n-1}$ and that we can distribute among these orbits $N\le n$ particles. +The ordering among these states is important as it defines the order of the creation operators. +We will write the determinant +\[ +\Phi_{\Lambda} = a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle, +\] +in a more compact way as +\[ +\Phi_{3,6,10,13} = |0001001000100100\rangle. +\] +The action of a creation operator is thus +\[ +a^{\dagger}_{\alpha_4}\Phi_{3,6,10,13} = a^{\dagger}_{\alpha_4}|0001001000100100\rangle=a^{\dagger}_{\alpha_4}a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle, +\] +which becomes +\[ +-a_{\alpha_3}^{\dagger} a^{\dagger}_{\alpha_4} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle=-|0001101000100100\rangle. +\] + +Similarly +\[ +a^{\dagger}_{\alpha_6}\Phi_{3,6,10,13} = a^{\dagger}_{\alpha_6}|0001001000100100\rangle=a^{\dagger}_{\alpha_6}a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle, +\] +which becomes +\[ +-a^{\dagger}_{\alpha_4} (a_{\alpha_6}^{\dagger})^ 2 a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle=0! +\] +This gives a simple recipe: +\begin{itemize} +\item If one of the bits $b_j$ is $1$ and we act with a creation operator on this bit, we return a null vector + +\item If $b_j=0$, we set it to $1$ and return a sign factor $(-1)^l$, where $l$ is the number of bits set before bit $j$. +\end{itemize} + +\noindent +Consider the action of $a^{\dagger}_{\alpha_2}$ on various slater determinants: +\[ +\begin{array}{ccc} +a^{\dagger}_{\alpha_2}\Phi_{00111}& = a^{\dagger}_{\alpha_2}|00111\rangle&=0\times |00111\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{01011}& = a^{\dagger}_{\alpha_2}|01011\rangle&=(-1)\times |01111\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{01101}& = a^{\dagger}_{\alpha_2}|01101\rangle&=0\times |01101\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{01110}& = a^{\dagger}_{\alpha_2}|01110\rangle&=0\times |01110\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{10011}& = a^{\dagger}_{\alpha_2}|10011\rangle&=(-1)\times |10111\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{10101}& = a^{\dagger}_{\alpha_2}|10101\rangle&=0\times |10101\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{10110}& = a^{\dagger}_{\alpha_2}|10110\rangle&=0\times |10110\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{11001}& = a^{\dagger}_{\alpha_2}|11001\rangle&=(+1)\times |11101\rangle\\ +a^{\dagger}_{\alpha_2}\Phi_{11010}& = a^{\dagger}_{\alpha_2}|11010\rangle&=(+1)\times |11110\rangle\\ +\end{array} +\] +What is the simplest way to obtain the phase when we act with one annihilation(creation) operator +on the given Slater determinant representation? + +We have an SD representation +\[ +\Phi_{\Lambda} = a_{\alpha_0}^{\dagger} a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle, +\] +in a more compact way as +\[ +\Phi_{0,3,6,10,13} = |1001001000100100\rangle. +\] +The action of +\[ +a^{\dagger}_{\alpha_4}a_{\alpha_0}\Phi_{0,3,6,10,13} = a^{\dagger}_{\alpha_4}|0001001000100100\rangle=a^{\dagger}_{\alpha_4}a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle, +\] +which becomes +\[ +-a_{\alpha_3}^{\dagger} a^{\dagger}_{\alpha_4} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle=-|0001101000100100\rangle. +\] + +The action +\[ +a_{\alpha_0}\Phi_{0,3,6,10,13} = |0001001000100100\rangle, +\] +can be obtained by subtracting the logical sum (AND operation) of $\Phi_{0,3,6,10,13}$ and +a word which represents only $\alpha_0$, that is +\[ +|1000000000000000\rangle, +\] +from $\Phi_{0,3,6,10,13}= |1001001000100100\rangle$. + +This operation gives $|0001001000100100\rangle$. + +Similarly, we can form $a^{\dagger}_{\alpha_4}a_{\alpha_0}\Phi_{0,3,6,10,13}$, say, by adding +$|0000100000000000\rangle$ to $a_{\alpha_0}\Phi_{0,3,6,10,13}$, first checking that their logical sum +is zero in order to make sure that orbital $\alpha_4$ is not already occupied. + +It is trickier however to get the phase $(-1)^l$. +One possibility is as follows +\begin{itemize} +\item Let $S_1$ be a word that represents the $1-$bit to be removed and all others set to zero. +\end{itemize} + +\noindent +In the previous example $S_1=|1000000000000000\rangle$ +\begin{itemize} +\item Define $S_2$ as the similar word that represents the bit to be added, that is in our case +\end{itemize} + +\noindent +$S_2=|0000100000000000\rangle$. +\begin{itemize} +\item Compute then $S=S_1-S_2$, which here becomes +\end{itemize} + +\noindent +\[ +S=|0111000000000000\rangle +\] +\begin{itemize} +\item Perform then the logical AND operation of $S$ with the word containing +\end{itemize} + +\noindent +\[ +\Phi_{0,3,6,10,13} = |1001001000100100\rangle, +\] +which results in $|0001000000000000\rangle$. Counting the number of $1-$bits gives the phase. Here you need however an algorithm for bitcounting. Several efficient ones available. + diff --git a/doc/BookChapters/BookProject/systems.tex b/doc/BookChapters/BookProject/systems.tex new file mode 100644 index 00000000..1e2d0367 --- /dev/null +++ b/doc/BookChapters/BookProject/systems.tex @@ -0,0 +1,427 @@ + +% ------------------- main content ---------------------- + +\chapter{Many-body Hamiltonians, basic linear algebra and Second Quantization} + +\subsection*{Definitions and notations} + +Before we proceed we need some definitions. +We will assume that the interacting part of the Hamiltonian +can be approximated by a two-body interaction. +This means that our Hamiltonian is written as the sum of some onebody part and a twobody part +\begin{equation} + \hat{H} = \hat{H}_0 + \hat{H}_I + = \sum_{i=1}^A \hat{h}_0(x_i) + \sum_{i < j}^A \hat{v}(r_{ij}), +\label{Hnuclei} +\end{equation} +with +\begin{equation} + H_0=\sum_{i=1}^A \hat{h}_0(x_i). +\label{hinuclei} +\end{equation} +The onebody part $u_{\mathrm{ext}}(x_i)$ is normally approximated by a harmonic oscillator potential or the Coulomb interaction an electron feels from the nucleus. However, other potentials are fully possible, such as +one derived from the self-consistent solution of the Hartree-Fock equations to be discussed here. + +Our Hamiltonian is invariant under the permutation (interchange) of two particles. +Since we deal with fermions however, the total wave function is antisymmetric. +Let $\hat{P}$ be an operator which interchanges two particles. +Due to the symmetries we have ascribed to our Hamiltonian, this operator commutes with the total Hamiltonian, +\[ +[\hat{H},\hat{P}] = 0, + \] +meaning that $\Psi_{\lambda}(x_1, x_2, \dots , x_A)$ is an eigenfunction of +$\hat{P}$ as well, that is +\[ +\hat{P}_{ij}\Psi_{\lambda}(x_1, x_2, \dots,x_i,\dots,x_j,\dots,x_A)= +\beta\Psi_{\lambda}(x_1, x_2, \dots,x_i,\dots,x_j,\dots,x_A), +\] +where $\beta$ is the eigenvalue of $\hat{P}$. We have introduced the suffix $ij$ in order to indicate that we permute particles $i$ and $j$. +The Pauli principle tells us that the total wave function for a system of fermions +has to be antisymmetric, resulting in the eigenvalue $\beta = -1$. + +In our case we assume that we can approximate the exact eigenfunction with a Slater determinant +\begin{equation} + \Phi(x_1, x_2,\dots ,x_A,\alpha,\beta,\dots, \sigma)=\frac{1}{\sqrt{A!}} +\left| \begin{array}{ccccc} \psi_{\alpha}(x_1)& \psi_{\alpha}(x_2)& \dots & \dots & \psi_{\alpha}(x_A)\\ + \psi_{\beta}(x_1)&\psi_{\beta}(x_2)& \dots & \dots & \psi_{\beta}(x_A)\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + \psi_{\sigma}(x_1)&\psi_{\sigma}(x_2)& \dots & \dots & \psi_{\sigma}(x_A)\end{array} \right|, \label{eq:HartreeFockDet} +\end{equation} +where $x_i$ stand for the coordinates and spin values of a particle $i$ and $\alpha,\beta,\dots, \gamma$ +are quantum numbers needed to describe remaining quantum numbers. + +\paragraph{Brief reminder on some linear algebra properties.} +Before we proceed with a more compact representation of a Slater determinant, we would like to repeat some linear algebra properties which will be useful for our derivations of the energy as function of a Slater determinant, Hartree-Fock theory and later the nuclear shell model. + +The inverse of a matrix is defined by + +\[ +\mathbf{A}^{-1} \cdot \mathbf{A} = I +\] +A unitary matrix $\mathbf{A}$ is one whose inverse is its adjoint +\[ +\mathbf{A}^{-1}=\mathbf{A}^{\dagger} +\] +A real unitary matrix is called orthogonal and its inverse is equal to its transpose. +A hermitian matrix is its own self-adjoint, that is +\[ +\mathbf{A}=\mathbf{A}^{\dagger}. +\] + + +\begin{quote} +\begin{tabular}{ccc} +\hline +\multicolumn{1}{c}{ Relations } & \multicolumn{1}{c}{ Name } & \multicolumn{1}{c}{ matrix elements } \\ +\hline +$A = A^{T}$ & symmetric & $a_{ij} = a_{ji}$ \\ +$A = \left (A^{T} \right )^{-1}$ & real orthogonal & $\sum_k a_{ik} a_{jk} = \sum_k a_{ki} a_{kj} = \delta_{ij}$ \\ +$A = A^{ * }$ & real matrix & $a_{ij} = a_{ij}^{ * }$ \\ +$A = A^{\dagger}$ & hermitian & $a_{ij} = a_{ji}^{ * }$ \\ +$A = \left (A^{\dagger} \right )^{-1}$ & unitary & $\sum_k a_{ik} a_{jk}^{ * } = \sum_k a_{ki}^{ * } a_{kj} = \delta_{ij}$ \\ +\hline +\end{tabular} +\end{quote} + +\noindent +Since we will deal with Fermions (identical and indistinguishable particles) we will +form an ansatz for a given state in terms of so-called Slater determinants determined +by a chosen basis of single-particle functions. + +For a given $n\times n$ matrix $\mathbf{A}$ we can write its determinant +\[ + det(\mathbf{A})=|\mathbf{A}|= +\left| \begin{array}{ccccc} a_{11}& a_{12}& \dots & \dots & a_{1n}\\ + a_{21}&a_{22}& \dots & \dots & a_{2n}\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + a_{n1}& a_{n2}& \dots & \dots & a_{nn}\end{array} \right|, +\] +in a more compact form as +\[ +|\mathbf{A}|= \sum_{i=1}^{n!}(-1)^{p_i}\hat{P}_i a_{11}a_{22}\dots a_{nn}, +\] +where $\hat{P}_i$ is a permutation operator which permutes the column indices $1,2,3,\dots,n$ +and the sum runs over all $n!$ permutations. The quantity $p_i$ represents the number of transpositions of column indices that are needed in order to bring a given permutation back to its initial ordering, in our case given by $a_{11}a_{22}\dots a_{nn}$ here. + +A simple $2\times 2$ determinant illustrates this. We have +\[ + det(\mathbf{A})= +\left| \begin{array}{cc} a_{11}& a_{12}\\ + a_{21}&a_{22}\end{array} \right|= (-1)^0a_{11}a_{22}+(-1)^1a_{12}a_{21}, +\] +where in the last term we have interchanged the column indices $1$ and $2$. The natural ordering we have chosen is $a_{11}a_{22}$. + +\paragraph{Back to the derivation of the energy.} +The single-particle function $\psi_{\alpha}(x_i)$ are eigenfunctions of the onebody +Hamiltonian $h_i$, that is +\[ +\hat{h}_0(x_i)=\hat{t}(x_i) + \hat{u}_{\mathrm{ext}}(x_i), +\] +with eigenvalues +\[ +\hat{h}_0(x_i) \psi_{\alpha}(x_i)=\left(\hat{t}(x_i) + \hat{u}_{\mathrm{ext}}(x_i)\right)\psi_{\alpha}(x_i)=\varepsilon_{\alpha}\psi_{\alpha}(x_i). +\] +The energies $\varepsilon_{\alpha}$ are the so-called non-interacting single-particle energies, or unperturbed energies. +The total energy is in this case the sum over all single-particle energies, if no two-body or more complicated +many-body interactions are present. + +Let us denote the ground state energy by $E_0$. According to the +variational principle we have +\[ + E_0 \le E[\Phi] = \int \Phi^*\hat{H}\Phi d\mathbf{\tau} +\] +where $\Phi$ is a trial function which we assume to be normalized +\[ + \int \Phi^*\Phi d\mathbf{\tau} = 1, +\] +where we have used the shorthand $d\mathbf{\tau}=dx_1dr_2\dots dr_A$. + +In the Hartree-Fock method the trial function is the Slater +determinant of Eq.~(\ref{eq:HartreeFockDet}) which can be rewritten as +\[ + \Phi(x_1,x_2,\dots,x_A,\alpha,\beta,\dots,\nu) = \frac{1}{\sqrt{A!}}\sum_{P} (-)^P\hat{P}\psi_{\alpha}(x_1) + \psi_{\beta}(x_2)\dots\psi_{\nu}(x_A)=\sqrt{A!}\hat{A}\Phi_H, +\] +where we have introduced the antisymmetrization operator $\hat{A}$ defined by the +summation over all possible permutations of two particles. + +It is defined as +\begin{equation} + \hat{A} = \frac{1}{A!}\sum_{p} (-)^p\hat{P}, +\label{antiSymmetryOperator} +\end{equation} +with $p$ standing for the number of permutations. We have introduced for later use the so-called +Hartree-function, defined by the simple product of all possible single-particle functions +\[ + \Phi_H(x_1,x_2,\dots,x_A,\alpha,\beta,\dots,\nu) = + \psi_{\alpha}(x_1) + \psi_{\beta}(x_2)\dots\psi_{\nu}(x_A). +\] + +Both $\hat{H}_0$ and $\hat{H}_I$ are invariant under all possible permutations of any two particles +and hence commute with $\hat{A}$ +\begin{equation} + [H_0,\hat{A}] = [H_I,\hat{A}] = 0. \label{commutionAntiSym} +\end{equation} +Furthermore, $\hat{A}$ satisfies +\begin{equation} + \hat{A}^2 = \hat{A}, \label{AntiSymSquared} +\end{equation} +since every permutation of the Slater +determinant reproduces it. + +The expectation value of $\hat{H}_0$ +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = A! \int \Phi_H^*\hat{A}\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau} +\] +is readily reduced to +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = A! \int \Phi_H^*\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau}, +\] +where we have used Eqs.~(\ref{commutionAntiSym}) and +(\ref{AntiSymSquared}). The next step is to replace the antisymmetrization +operator by its definition and to +replace $\hat{H}_0$ with the sum of one-body operators +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = \sum_{i=1}^A \sum_{p} (-)^p\int + \Phi_H^*\hat{h}_0\hat{P}\Phi_H d\mathbf{\tau}. +\] + +The integral vanishes if two or more particles are permuted in only one +of the Hartree-functions $\Phi_H$ because the individual single-particle wave functions are +orthogonal. We obtain then +\[ + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}= \sum_{i=1}^A \int \Phi_H^*\hat{h}_0\Phi_H d\mathbf{\tau}. +\] +Orthogonality of the single-particle functions allows us to further simplify the integral, and we +arrive at the following expression for the expectation values of the +sum of one-body Hamiltonians +\begin{equation} + \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} + = \sum_{\mu=1}^A \int \psi_{\mu}^*(x)\hat{h}_0\psi_{\mu}(x)dx + d\mathbf{r}. + \label{H1Expectation} +\end{equation} + +We introduce the following shorthand for the above integral +\[ +\langle \mu | \hat{h}_0 | \mu \rangle = \int \psi_{\mu}^*(x)\hat{h}_0\psi_{\mu}(x)dx, +\] +and rewrite Eq.~(\ref{H1Expectation}) as +\begin{equation} + \int \Phi^*\hat{H}_0\Phi d\tau + = \sum_{\mu=1}^A \langle \mu | \hat{h}_0 | \mu \rangle. + \label{H1Expectation1} +\end{equation} + +The expectation value of the two-body part of the Hamiltonian is obtained in a +similar manner. We have +\[ + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = A! \int \Phi_H^*\hat{A}\hat{H}_I\hat{A}\Phi_H d\mathbf{\tau}, +\] +which reduces to +\[ + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = \sum_{i\le j=1}^A \sum_{p} (-)^p\int + \Phi_H^*\hat{v}(r_{ij})\hat{P}\Phi_H d\mathbf{\tau}, +\] +by following the same arguments as for the one-body +Hamiltonian. + +Because of the dependence on the inter-particle distance $r_{ij}$, permutations of +any two particles no longer vanish, and we get +\[ + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = \sum_{i < j=1}^A \int + \Phi_H^*\hat{v}(r_{ij})(1-P_{ij})\Phi_H d\mathbf{\tau}. +\] +where $P_{ij}$ is the permutation operator that interchanges +particle $i$ and particle $j$. Again we use the assumption that the single-particle wave functions +are orthogonal. + +We obtain +\begin{align} + \int \Phi^*\hat{H}_I\Phi d\mathbf{\tau} + = \frac{1}{2}\sum_{\mu=1}^A\sum_{\nu=1}^A + &\left[ \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)\hat{v}(r_{ij})\psi_{\mu}(x_i)\psi_{\nu}(x_j) + dx_idx_j \right.\\ + &\left. + - \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j) + \hat{v}(r_{ij})\psi_{\nu}(x_i)\psi_{\mu}(x_j) + dx_idx_j + \right]. \label{H2Expectation} +\end{align} +The first term is the so-called direct term. It is frequently also called the Hartree term, +while the second is due to the Pauli principle and is called +the exchange term or just the Fock term. +The factor $1/2$ is introduced because we now run over +all pairs twice. + +The last equation allows us to introduce some further definitions. +The single-particle wave functions $\psi_{\mu}(x)$, defined by the quantum numbers $\mu$ and $x$ +are defined as the overlap +\[ + \psi_{\alpha}(x) = \langle x | \alpha \rangle . +\] + +We introduce the following shorthands for the above two integrals +\[ +\langle \mu\nu|\hat{v}|\mu\nu\rangle = \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)\hat{v}(r_{ij})\psi_{\mu}(x_i)\psi_{\nu}(x_j) + dx_idx_j, +\] +and +\[ +\langle \mu\nu|\hat{v}|\nu\mu\rangle = \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j) + \hat{v}(r_{ij})\psi_{\nu}(x_i)\psi_{\mu}(x_j) + dx_idx_j. +\] + +\subsection*{Preparing for later studies: varying the coefficients of a wave function expansion and orthogonal transformations} + +It is common to expand the single-particle functions in a known basis and vary the coefficients, +that is, the new single-particle wave function is written as a linear expansion +in terms of a fixed chosen orthogonal basis (for example the well-known harmonic oscillator functions or the hydrogen-like functions etc). +We define our new single-particle basis (this is a normal approach for Hartree-Fock theory) by performing a unitary transformation +on our previous basis (labelled with greek indices) as +\begin{equation} +\psi_p^{new} = \sum_{\lambda} C_{p\lambda}\phi_{\lambda}. \label{eq:newbasis} +\end{equation} +In this case we vary the coefficients $C_{p\lambda}$. If the basis has infinitely many solutions, we need +to truncate the above sum. We assume that the basis $\phi_{\lambda}$ is orthogonal. + +It is normal to choose a single-particle basis defined as the eigenfunctions +of parts of the full Hamiltonian. The typical situation consists of the solutions of the one-body part of the Hamiltonian, that is we have +\[ +\hat{h}_0\phi_{\lambda}=\epsilon_{\lambda}\phi_{\lambda}. +\] +The single-particle wave functions $\phi_{\lambda}(\mathbf{r})$, defined by the quantum numbers $\lambda$ and $\mathbf{r}$ +are defined as the overlap +\[ + \phi_{\lambda}(\mathbf{r}) = \langle \mathbf{r} | \lambda \rangle . +\] + +In deriving the Hartree-Fock equations, we will expand the single-particle functions in a known basis and vary the coefficients, +that is, the new single-particle wave function is written as a linear expansion +in terms of a fixed chosen orthogonal basis (for example the well-known harmonic oscillator functions or the hydrogen-like functions etc). + +We stated that a unitary transformation keeps the orthogonality. To see this consider first a basis of vectors $\mathbf{v}_i$, +\[ +\mathbf{v}_i = \begin{bmatrix} v_{i1} \\ \dots \\ \dots \\v_{in} \end{bmatrix} +\] +We assume that the basis is orthogonal, that is +\[ +\mathbf{v}_j^T\mathbf{v}_i = \delta_{ij}. +\] +An orthogonal or unitary transformation +\[ +\mathbf{w}_i=\mathbf{U}\mathbf{v}_i, +\] +preserves the dot product and orthogonality since +\[ +\mathbf{w}_j^T\mathbf{w}_i=(\mathbf{U}\mathbf{v}_j)^T\mathbf{U}\mathbf{v}_i=\mathbf{v}_j^T\mathbf{U}^T\mathbf{U}\mathbf{v}_i= \mathbf{v}_j^T\mathbf{v}_i = \delta_{ij}. +\] + +This means that if the coefficients $C_{p\lambda}$ belong to a unitary or orthogonal trasformation (using the Dirac bra-ket notation) +\[ +\vert p\rangle = \sum_{\lambda} C_{p\lambda}\vert\lambda\rangle, +\] +orthogonality is preserved, that is $\langle \alpha \vert \beta\rangle = \delta_{\alpha\beta}$ +and $\langle p \vert q\rangle = \delta_{pq}$. + +This propertry is extremely useful when we build up a basis of many-body Stater determinant based states. + +\textbf{Note also that although a basis $\vert \alpha\rangle$ contains an infinity of states, for practical calculations we have always to make some truncations.} + +Before we develop for example the Hartree-Fock equations, there is another very useful property of determinants that we will use both in connection with Hartree-Fock calculations and later shell-model calculations. + +Consider the following determinant +\[ +\left| \begin{array}{cc} \alpha_1b_{11}+\alpha_2sb_{12}& a_{12}\\ + \alpha_1b_{21}+\alpha_2b_{22}&a_{22}\end{array} \right|=\alpha_1\left|\begin{array}{cc} b_{11}& a_{12}\\ + b_{21}&a_{22}\end{array} \right|+\alpha_2\left| \begin{array}{cc} b_{12}& a_{12}\\b_{22}&a_{22}\end{array} \right| +\] + +We can generalize this to an $n\times n$ matrix and have +\[ +\left| \begin{array}{cccccc} a_{11}& a_{12} & \dots & \sum_{k=1}^n c_k b_{1k} &\dots & a_{1n}\\ +a_{21}& a_{22} & \dots & \sum_{k=1}^n c_k b_{2k} &\dots & a_{2n}\\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +a_{n1}& a_{n2} & \dots & \sum_{k=1}^n c_k b_{nk} &\dots & a_{nn}\end{array} \right|= +\sum_{k=1}^n c_k\left| \begin{array}{cccccc} a_{11}& a_{12} & \dots & b_{1k} &\dots & a_{1n}\\ +a_{21}& a_{22} & \dots & b_{2k} &\dots & a_{2n}\\ +\dots & \dots & \dots & \dots & \dots & \dots\\ +\dots & \dots & \dots & \dots & \dots & \dots\\ +a_{n1}& a_{n2} & \dots & b_{nk} &\dots & a_{nn}\end{array} \right| . +\] +This is a property we will use in our Hartree-Fock discussions. + +We can generalize the previous results, now +with all elements $a_{ij}$ being given as functions of +linear combinations of various coefficients $c$ and elements $b_{ij}$, +\[ +\left| \begin{array}{cccccc} \sum_{k=1}^n b_{1k}c_{k1}& \sum_{k=1}^n b_{1k}c_{k2} & \dots & \sum_{k=1}^n b_{1k}c_{kj} &\dots & \sum_{k=1}^n b_{1k}c_{kn}\\ +\sum_{k=1}^n b_{2k}c_{k1}& \sum_{k=1}^n b_{2k}c_{k2} & \dots & \sum_{k=1}^n b_{2k}c_{kj} &\dots & \sum_{k=1}^n b_{2k}c_{kn}\\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots &\dots \\ +\sum_{k=1}^n b_{nk}c_{k1}& \sum_{k=1}^n b_{nk}c_{k2} & \dots & \sum_{k=1}^n b_{nk}c_{kj} &\dots & \sum_{k=1}^n b_{nk}c_{kn}\end{array} \right|=det(\mathbf{C})det(\mathbf{B}), +\] +where $det(\mathbf{C})$ and $det(\mathbf{B})$ are the determinants of $n\times n$ matrices +with elements $c_{ij}$ and $b_{ij}$ respectively. +This is a property we will use in our Hartree-Fock discussions. Convince yourself about the correctness of the above expression by setting $n=2$. + +With our definition of the new basis in terms of an orthogonal basis we have +\[ +\psi_p(x) = \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x). +\] +If the coefficients $C_{p\lambda}$ belong to an orthogonal or unitary matrix, the new basis +is also orthogonal. +Our Slater determinant in the new basis $\psi_p(x)$ is written as +\[ +\frac{1}{\sqrt{A!}} +\left| \begin{array}{ccccc} \psi_{p}(x_1)& \psi_{p}(x_2)& \dots & \dots & \psi_{p}(x_A)\\ + \psi_{q}(x_1)&\psi_{q}(x_2)& \dots & \dots & \psi_{q}(x_A)\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + \psi_{t}(x_1)&\psi_{t}(x_2)& \dots & \dots & \psi_{t}(x_A)\end{array} \right|=\frac{1}{\sqrt{A!}} +\left| \begin{array}{ccccc} \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_1)& \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_A)\\ + \sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_1)&\sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_A)\\ + \dots & \dots & \dots & \dots & \dots \\ + \dots & \dots & \dots & \dots & \dots \\ + \sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_1)&\sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_A)\end{array} \right|, +\] +which is nothing but $det(\mathbf{C})det(\Phi)$, with $det(\Phi)$ being the determinant given by the basis functions $\phi_{\lambda}(x)$. + +In our discussions hereafter we will use our definitions of single-particle states above and below the Fermi ($F$) level given by the labels +$ijkl\dots \le F$ for so-called single-hole states and $abcd\dots > F$ for so-called particle states. +For general single-particle states we employ the labels $pqrs\dots$. + +The energy functional is +\[ + E[\Phi] + = \sum_{\mu=1}^A \langle \mu | h | \mu \rangle + + \frac{1}{2}\sum_{{\mu}=1}^A\sum_{{\nu}=1}^A \langle \mu\nu|\hat{v}|\mu\nu\rangle_{AS}, +\] +we found the expression for the energy functional in terms of the basis function $\phi_{\lambda}(\mathbf{r})$. We then varied the above energy functional with respect to the basis functions $|\mu \rangle$. +Now we are interested in defining a new basis defined in terms of +a chosen basis as defined in Eq.~(\ref{eq:newbasis}). We can then rewrite the energy functional as +\begin{equation} + E[\Phi^{New}] + = \sum_{i=1}^A \langle i | h | i \rangle + + \frac{1}{2}\sum_{ij=1}^A\langle ij|\hat{v}|ij\rangle_{AS}, \label{FunctionalEPhi2} +\end{equation} +where $\Phi^{New}$ is the new Slater determinant defined by the new basis of Eq.~(\ref{eq:newbasis}). + +Using Eq.~(\ref{eq:newbasis}) we can rewrite Eq.~(\ref{FunctionalEPhi2}) as +\begin{equation} + E[\Psi] + = \sum_{i=1}^A \sum_{\alpha\beta} C^*_{i\alpha}C_{i\beta}\langle \alpha | h | \beta \rangle + + \frac{1}{2}\sum_{ij=1}^A\sum_{{\alpha\beta\gamma\delta}} C^*_{i\alpha}C^*_{j\beta}C_{i\gamma}C_{j\delta}\langle \alpha\beta|\hat{v}|\gamma\delta\rangle_{AS}. \label{FunctionalEPhi3} +\end{equation} +