**Spring 2015**

Large-scale diagonalization (Iterative methods, Lanczo's method, dimensionalities $10^{10}$ states)

Coupled cluster theory, favoured method in quantum chemistry, molecular and atomic physics. Applications to ab initio calculations in nuclear physics as well for large nuclei.

Perturbative many-body methods

Density functional theories/Mean-field theory and Hartree-Fock theory (covered partly also in FYS-MENA4111)

Monte-Carlo methods (Only in FYS4411, Computational quantum mechanics)

Green's function theories (depending on interest)

and other. The physics of the system hints at which many-body methods to use.

Blaizot and Ripka,

*Quantum Theory of Finite systems*, MIT press 1986Negele and Orland,

*Quantum Many-Particle Systems*, Addison-Wesley, 1987.Fetter and Walecka,

*Quantum Theory of Many-Particle Systems*, McGraw-Hill, 1971.Helgaker, Jorgensen and Olsen,

*Molecular Electronic Structure Theory*, Wiley, 2001.Mattuck,

*Guide to Feynman Diagrams in the Many-Body Problem*, Dover, 1971.Dickhoff and Van Neck,

*Many-Body Theory Exposed*, World Scientific, 2006.

An operator is defined as $\hat{O}$ throughout. Unless otherwise specified the number of particles is always $N$ and $d$ is the dimension of the system. In nuclear physics we normally define the total number of particles to be $A=N+Z$, where $N$ is total number of neutrons and $Z$ the total number of protons. In case of other baryons such isobars $\Delta$ or various hyperons such as $\Lambda$ or $\Sigma$, one needs to add their definitions. Hereafter, $N$ is reserved for the total number of particles, unless otherwise specificied.

The quantum numbers of a single-particle state in coordinate space are defined by the variable

$$
x=({\bf r},\sigma),
$$

where

$$
{\bf r}\in {\mathbb{R}}^{d},
$$

$$
x\in {\mathbb{R}}^{d}\oplus (\frac{1}{2}),
$$

and the integral $\int dx = \sum_{\sigma}\int d^dr = \sum_{\sigma}\int d{\bf r}$, and

$$
\int d^Nx= \int dx_1\int dx_2\dots\int dx_N.
$$

$$
\Psi_{\lambda}=\Psi_{\lambda}(x_1,x_2,\dots,x_N),
$$

$$
\Psi_0=\Psi_0(x_1,x_2,\dots,x_N).
$$

$$
\Phi_{\lambda}=\Phi_{\lambda}(x_1,x_2,\dots,x_N),
$$

with

$$
\Phi_0=\Phi_0(x_1,x_2,\dots,x_N),
$$

$$
\Psi_{\lambda}\in {\cal H}_N:= {\cal H}_1\oplus{\cal H}_1\oplus\dots\oplus{\cal H}_1,
$$

$$
{\cal H}_1:= L^2(\mathbb{R}^{d}\oplus (\sigma)).
$$

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,

$$
[\hat{H},\hat{P}] = 0,
$$

$$
\hat{P}_{ij}\Psi_{\lambda}(x_1, x_2, \dots,x_i,\dots,x_j,\dots,x_N)=
\beta\Psi_{\lambda}(x_1, x_2, \dots,x_j,\dots,x_i,\dots,x_N),
$$

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 $\beta = -1$.

The Schrodinger equation reads

$$
\begin{equation}
\hat{H}(x_1, x_2, \dots , x_N) \Psi_{\lambda}(x_1, x_2, \dots , x_N) =
E_\lambda \Psi_\lambda(x_1, x_2, \dots , x_N), \label{eq:basicSE1} \tag{1}
\end{equation}
$$

where the vector $x_i$ represents the coordinates (spatial and spin) of particle $i$, $\lambda$ stands for all the quantum numbers needed to classify a given $N$-particle state and $\Psi_{\lambda}$ is the pertaining eigenfunction. Throughout this course, $\Psi$ refers to the exact eigenfunction, unless otherwise stated.

We write the Hamilton operator, or Hamiltonian, in a generic way

$$
\hat{H} = \hat{T} + \hat{V}
$$

where $\hat{T}$ represents the kinetic energy of the system

$$
\hat{T} = \sum_{i=1}^N \frac{\mathbf{p}_i^2}{2m_i} = \sum_{i=1}^N \left( -\frac{\hbar^2}{2m_i} \mathbf{\nabla_i}^2 \right) =
\sum_{i=1}^N t(x_i)
$$

while the operator $\hat{V}$ for the potential energy is given by

$$
\begin{equation}
\hat{V} = \sum_{i=1}^N \hat{u}_{\mathrm{ext}}(x_i) + \sum_{ji=1}^N v(x_i,x_j)+\sum_{ijk=1}^Nv(x_i,x_j,x_k)+\dots
\label{eq:firstv} \tag{2}
\end{equation}
$$

Hereafter we use natural units, viz. $\hbar=c=e=1$, with $e$ the elementary charge and $c$ the speed of light. This means that momenta and masses have dimension energy.

If one does quantum chemistry, after having introduced the Born-Oppenheimer approximation which effectively freezes out the nucleonic degrees of freedom, the Hamiltonian for $N=n_e$ electrons takes the following form

$$
\hat{H} = \sum_{i=1}^{n_e} t(x_i) - \sum_{i=1}^{n_e} k\frac{Z}{r_i} + \sum_{i < j}^{n_e} \frac{k}{r_{ij}},
$$

$$
\begin{equation}
\hat{H} = \hat{H}_0 + \hat{H}_I
= \sum_{i=1}^{n_e}\hat{h}_0(x_i) + \sum_{i < j}^{n_e}\frac{1}{r_{ij}},
\label{H1H2} \tag{3}
\end{equation}
$$

where we have defined

$$
r_{ij}=| {\bf r}_i-{\bf r}_j|,
$$

and

$$
\begin{equation}
\hat{h}_0(x_i) = \hat{t}(x_i) - \frac{Z}{x_i}.
\label{hi} \tag{4}
\end{equation}
$$

The first term of Eq. (3), $H_0$, is the sum of the $N$
*one-body* Hamiltonians $\hat{h}_0$. Each individual
Hamiltonian $\hat{h}_0$ contains the kinetic energy operator of an
electron and its potential energy due to the attraction of the
nucleus. The second term, $H_I$, is the sum of the $n_e(n_e-1)/2$
two-body interactions between each pair of electrons. Note that the double sum carries a restriction $i < j$.

The potential energy term due to the attraction of the nucleus defines the onebody field $u_i=u_{\mathrm{ext}}(x_i)$ of Eq. (2).
We have moved this term into the $\hat{H}_0$ part of the Hamiltonian, instead of keeping it in $\hat{V}$ as in Eq. (2).
The reason is that we will hereafter treat $\hat{H}_0$ as our non-interacting Hamiltonian. For a many-body wavefunction $\Phi_{\lambda}$ defined by an

appropriate single-particle basis, we may solve exactly the non-interacting eigenvalue problem

$$
\hat{H}_0\Phi_{\lambda}= w_{\lambda}\Phi_{\lambda},
$$

with $w_{\lambda}$ being the non-interacting energy. This energy is defined by the sum over single-particle energies to be defined below. For atoms the single-particle energies could be the hydrogen-like single-particle energies corrected for the charge $Z$. For nuclei and quantum dots, these energies could be given by the harmonic oscillator in three and two dimensions, respectively.

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

$$
\begin{equation}
\hat{H} = \hat{H}_0 + \hat{H}_I
= \sum_{i=1}^N \hat{h}_0(x_i) + \sum_{i < j}^N V(r_{ij}),
\label{Hnuclei} \tag{5}
\end{equation}
$$

with

$$
\begin{equation}
H_0=\sum_{i=1}^N \hat{h}_0(x_i) = \sum_{i=1}^N\left(\hat{t}(x_i) + \hat{u}_{\mathrm{ext}}(x_i)\right).
\label{hinuclei} \tag{6}
\end{equation}
$$

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.

Our Hamiltonian is invariant under the permutation (interchange) of two particles. % (exercise here, prove it) 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,

$$
[\hat{H},\hat{P}] = 0,
$$

$$
\hat{P}_{ij}\Psi_{\lambda}(x_1, x_2, \dots,x_i,\dots,x_j,\dots,x_N)=
\beta\Psi_{\lambda}(x_1, x_2, \dots,x_i,\dots,x_j,\dots,x_N),
$$

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

$$
\begin{equation}
\Phi(x_1, x_2,\dots ,x_N,\alpha,\beta,\dots, \sigma)=\frac{1}{\sqrt{N!}}
\left| \begin{array}{ccccc} \psi_{\alpha}(x_1)& \psi_{\alpha}(x_2)& \dots & \dots & \psi_{\alpha}(x_N)\\
\psi_{\beta}(x_1)&\psi_{\beta}(x_2)& \dots & \dots & \psi_{\beta}(x_N)\\
\dots & \dots & \dots & \dots & \dots \\
\dots & \dots & \dots & \dots & \dots \\
\psi_{\sigma}(x_1)&\psi_{\sigma}(x_2)& \dots & \dots & \psi_{\sigma}(x_N)\end{array} \right|, \label{eq:HartreeFockDet} \tag{7}
\end{equation}
$$

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.

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 $d\mathbf{\tau}=d\mathbf{r}_1d\mathbf{r}_2\dots d\mathbf{r}_N$.

In the Hartree-Fock method the trial function is the Slater determinant of Eq. (7) which can be rewritten as

$$
\Phi(x_1,x_2,\dots,x_N,\alpha,\beta,\dots,\nu) = \frac{1}{\sqrt{N!}}\sum_{P} (-)^P\hat{P}\psi_{\alpha}(x_1)
\psi_{\beta}(x_2)\dots\psi_{\nu}(x_N)=\sqrt{N!}\hat{A}\Phi_H,
$$

$$
\begin{equation}
\hat{A} = \frac{1}{N!}\sum_{p} (-)^p\hat{P},
\label{antiSymmetryOperator} \tag{8}
\end{equation}
$$

$$
\Phi_H(x_1,x_2,\dots,x_N,\alpha,\beta,\dots,\nu) =
\psi_{\alpha}(x_1)
\psi_{\beta}(x_2)\dots\psi_{\nu}(x_N).
$$

$$
\begin{equation}
[H_0,\hat{A}] = [H_I,\hat{A}] = 0. \label{commutionAntiSym} \tag{9}
\end{equation}
$$

Furthermore, $\hat{A}$ satisfies

$$
\begin{equation}
\hat{A}^2 = \hat{A}, \label{AntiSymSquared} \tag{10}
\end{equation}
$$

$$
\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}
= N! \int \Phi_H^*\hat{A}\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau}
$$

is readily reduced to

$$
\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}
= N! \int \Phi_H^*\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau},
$$

$$
\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}
= \sum_{i=1}^N \sum_{p} (-)^p\int
\Phi_H^*\hat{h}_0\hat{P}\Phi_H d\mathbf{\tau}.
$$

$$
\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}= \sum_{i=1}^N \int \Phi_H^*\hat{h}_0\Phi_H d\mathbf{\tau}.
$$

$$
\begin{equation}
\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}
= \sum_{\mu=1}^N \int \psi_{\mu}^*(\mathbf{r})\hat{h}_0\psi_{\mu}(\mathbf{r})
d\mathbf{r}.
\label{H1Expectation} \tag{11}
\end{equation}
$$

$$
\langle \mu | \hat{h}_0 | \mu \rangle = \int \psi_{\mu}^*(\mathbf{r})\hat{h}_0\psi_{\mu}(\mathbf{r}),
$$

and rewrite Eq. (11) as

$$
\begin{equation}
\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}
= \sum_{\mu=1}^N \langle \mu | \hat{h}_0 | \mu \rangle.
\label{H1Expectation1} \tag{12}
\end{equation}
$$

$$
\int \Phi^*\hat{H}_I\Phi d\mathbf{\tau}
= N! \int \Phi_H^*\hat{A}\hat{H}_I\hat{A}\Phi_H d\mathbf{\tau},
$$

which reduces to

$$
\int \Phi^*\hat{H}_I\Phi d\mathbf{\tau}
= \sum_{i\le j=1}^N \sum_{p} (-)^p\int
\Phi_H^*V(r_{ij})\hat{P}\Phi_H d\mathbf{\tau},
$$

$$
\int \Phi^*\hat{H}_I\Phi d\mathbf{\tau}
= \sum_{i < j=1}^N \int
\Phi_H^*V(r_{ij})(1-P_{ij})\Phi_H d\mathbf{\tau}.
$$

$$
\begin{equation}
\begin{split}
\int \Phi^*\hat{H}_I\Phi d\mathbf{\tau}
= \frac{1}{2}\sum_{\mu=1}^N\sum_{\nu=1}^N
&\left[ \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)V(r_{ij})\psi_{\mu}(x_i)\psi_{\nu}(x_j)
dx_ix_j \right.\\
&\left.
- \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)
V(r_{ij})\psi_{\nu}(x_i)\psi_{\mu}(x_j)
dx_ix_j
\right]. \label{H2Expectation} \tag{13}
\end{split}
\end{equation}
$$

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

$$
\psi_{\alpha}(x) = \langle x | \alpha \rangle .
$$

$$
\langle \mu\nu|\hat{v}|\mu\nu\rangle = \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)V(r_{ij})\psi_{\mu}(x_i)\psi_{\nu}(x_j)
dx_ix_j,
$$

and

$$
\langle \mu\nu|\hat{v}|\nu\mu\rangle = \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)
V(r_{ij})\psi_{\nu}(x_i)\psi_{\mu}(x_j)
dx_ix_j.
$$

$$
\langle \mu\nu|\hat{v}|\mu\nu\rangle_{\mathrm{AS}}= \langle \mu\nu|\hat{v}|\mu\nu\rangle-\langle \mu\nu|\hat{v}|\nu\mu\rangle,
$$

or for a general matrix element

$$
\langle \mu\nu|\hat{v}|\sigma\tau\rangle_{\mathrm{AS}}= \langle \mu\nu|\hat{v}|\sigma\tau\rangle-\langle \mu\nu|\hat{v}|\tau\sigma\rangle.
$$

It has the symmetry property

$$
\langle \mu\nu|\hat{v}|\sigma\tau\rangle_{\mathrm{AS}}= -\langle \mu\nu|\hat{v}|\tau\sigma\rangle_{\mathrm{AS}}=-\langle \nu\mu|\hat{v}|\sigma\tau\rangle_{\mathrm{AS}}.
$$

$$
\langle \mu\nu|\hat{v}|\sigma\tau\rangle_{\mathrm{AS}}= \langle \sigma\tau|\hat{v}|\mu\nu\rangle_{\mathrm{AS}}.
$$

With these notations we rewrite Eq. (13) as

$$
\begin{equation}
\int \Phi^*\hat{H}_I\Phi d\mathbf{\tau}
= \frac{1}{2}\sum_{\mu=1}^N\sum_{\nu=1}^N \langle \mu\nu|\hat{v}|\mu\nu\rangle_{\mathrm{AS}}.
\label{H2Expectation2} \tag{14}
\end{equation}
$$

$$
\begin{equation}
E[\Phi]
= \sum_{\mu=1}^N \langle \mu | \hat{h}_0 | \mu \rangle +
\frac{1}{2}\sum_{{\mu}=1}^N\sum_{{\nu}=1}^N \langle \mu\nu|\hat{v}|\mu\nu\rangle_{\mathrm{AS}}.
\label{FunctionalEPhi} \tag{15}
\end{equation}
$$

which we will use as our starting point for the Hartree-Fock calculations later in this course.