\chapter{Traditional coupled-cluster theory}

\begin{dfn}\label{dfn:cc-effective-hamiltonian}
\thmtitle{Traditional coupled-cluster theory}
A \textit{wave operator} maps a determinant into a correlated wavefunction, $\Y=\W\F$.
The \textit{coupled-cluster Ansatz} is characterized by an exponential parametrization of the wave operator.
\begin{align}
\label{eq:cc-effective-hamiltonian}
  H_{\text{c}}\Y_{\text{CC}}
=
  E_{\text{c}}\Y_{\text{CC}}
&&
  \Y_{\text{CC}}
\equiv
  \text{exp}(T)
  \F
&&
  T
\equiv
  T_1
+
  T_2
+
  \cd
+
  T_n
&&
  T_k
\equiv
  (\tfr{1}{k!})^2
  t_{a_1\cd a_k}^{i_1\cd i_k}
  \tl{a}^{a_1\cd a_k}_{i_1\cd i_k}
\end{align}
The coupled-cluster Schr\"odinger equation can be projected onto the determinant basis to arrive at a series of equations
\begin{align}
\label{eq:projected-cc-equations}
  \ip{\F|H_{\text{c}}|\Y_{\text{CC}}}
=
  E_{\text{c}}
&&
  \ip{\F_{ij\cd}^{ab\cd}|H_{\text{c}}|\Y_{\text{CC}}}
=
  E_{\text{c}}
  \ip{\F_{ij\cd}^{ab\cd}|\Y_{\text{CC}}}
\end{align}
which specify the coupled-cluster energy and the \textit{amplitudes}, $t_{ab\cd}^{ij\cd}$.
A different approach, known as \textit{traditional coupled-cluster (TCC) theory}, first multiplies the Schr\"odinger equation on the left by the inverse of the wave operator
\begin{align}
  \ol{H}_{\text{c}}
  \F
=
  E_{\text{c}}
  \F
&&
  \ol{H}_{\text{c}}
\equiv
  \text{exp}(-T)
  H_{\text{c}}\,
  \text{exp}(T)
\end{align}
to define an \textit{effective Hamiltonian}, $\ol{H}_{\text{c}}$.
The eigenvalue of this similarity-transformed\footnote{See \url{https://en.wikipedia.org/wiki/Matrix_similarity}.} Hamiltonian is the exact correlation energy, $E_{\text{c}}$, but its eigenstate is the reference determinant, $\F$, rather than the correlated wavefunction.
Note that, unlike the true Hamiltonian, $\ol{H}_{\text{c}}$ is non-Hermitian.
Projection onto the determinant basis yields energy and amplitude equations
\begin{align}
\label{eq:traditional-cc-equations}
  \ip{\F|\ol{H}_{\text{c}}|\F}
=
  E_{\text{c}}
&&
  \ip{\F_{ij\cd}^{ab\cd}|\ol{H}_{\text{c}}|\F}
=
  0
\end{align}
which look similar to eq~\ref{eq:projected-cc-equations}, except that the right-hand side of the amplitude equations is now zero.
The next few results show that the TCC similarity transformation removes disconnected terms in eq~\ref{eq:projected-cc-equations}, of which $E_{\text{c}}t_{ab\cd}^{ij\cd}$ is an example.
\end{dfn}

\begin{thm}\label{thm:hausdorff}
\thmtitle{The Hausdorff Expansion}
\thmstatement{
$\ds{
  e^{- X}Ye^{X}
=
  \sum_{n=0}^\infty
  \fr{1}{n!}
  [\,\cdot\,, X]^n(Y)
}$
}\,\footnote{
$
  [\,\cdot\,, X]^n(Y)
$
denotes a nested commutator,
$
  [\cd[[Y,X],X]\cd,X]
$.
For $n=0$, we define
$
  [\,\cdot\,, X]^0(Y)
\equiv
  Y
$.
}\vspace{5pt}
\thmproof{
  This follows from
  $
    \pd{^n}{\la^n}
    e^{-\la X}Ye^{\la X}
  =
    [\,\cdot\,, X]^n(
    e^{-\la X}Ye^{\la X}
    )
  $,
  which we will prove by induction.
  This is trivially true for $n=0$.
  Assuming it holds for $n$, the following shows that it also holds for $n+1$, completing the induction.
\begin{align*}
  \tpd{^{n+1}}{\la^{n+1}}
  e^{-\la X}Ye^{\la X}
=
  [\,\cdot\,, X]^n(
  \tpd{}{\la}
  e^{-\la X}Ye^{\la X}
  )
=
  [\,\cdot\,, X]^n(
    e^{-\la X}Ye^{\la X} X
  -
    X e^{-\la X}Ye^{\la X}
  )
=
  [\,\cdot\,, X]^{n+1}(
    e^{-\la X}Ye^{\la X}
  )
\end{align*}
  Substituting this result into a Taylor expansion of
  $
    e^{-\la X}Ye^{\la X}
  $
  about
  $\la=0$
  evaluated at
  $\la=1$
  completes the proof.
}
\end{thm}

\begin{ex}
The Hausdorff expansion can be used to express the TCC effective Hamiltonian in powers of $T$.
\begin{align*}
  \ol{H}_{\text{c}}
=
  \text{exp}(-T)
  H_{\text{c}}\,
  \text{exp}(T)
=
  H_{\text{c}}
+
  [H_{\text{c}}, T]
+
  \tfr{1}{2!}
  [[H_{\text{c}}, T], T]
+
  \tfr{1}{3!}
  [[[H_{\text{c}}, T], T], T]
+
  \tfr{1}{4!}
  [[[[H_{\text{c}}, T], T], T], T]
+
  \cd
\end{align*}
This expansion can be further simplified by analyzing the commutators with $T$ using Wick's theorem.
\end{ex}

\begin{prop}\label{prop:wicks-theorem-for-commutators}
\thmstatement{
If $Q$ and $Q'$ are normal ordered and one of them has an even operator count,
$
  [Q,Q']
=
  \gno{
    \ctr{}{Q}{}{Q}
    QQ'
  }
-
  \gno{
    \ctr{}{Q}{'}{Q}
    Q'Q
  }
$.%
}\vspace{2pt}%
\thmproof{
By Wick's theorem,
$
  QQ'
-
  Q'Q
=
  \gno{QQ'}
+
  \gno{
    \ctr{}{Q}{}{Q}
    QQ'
  }
-
  \gno{Q'Q}
-
  \gno{
    \ctr{}{Q}{'}{Q}
    Q'Q
  }
$.
The proposition follows from the fact that
$
  \gno{QQ'}
=
  \gno{Q'Q}
$
when one of the strings contains an even number of operators.
}
\end{prop}

\begin{cor}
\label{cor:tcc-similarity-transformation-connected}
\thmstatement{
  TCC similarity-transformed operators,
  $
    \ol{W}
  \equiv
    \text{exp}(-T)
    W
    \text{exp}(T)
  $,
  can be evaluated as
  $
    \ol{W}
  =
    (
      W\,
      \text{exp}(T)
    )_{\text{C}}
  $,
  where the subscript $\text{C}$ denotes a restriction to connected diagrams.
}
\thmproof{
  \Cref{prop:wicks-theorem-for-commutators} implies
$
  [W,T]
=
\gno{
  \ctr{}{W}{}{T}
  WT
}
$
and, by straightforward induction,
$
  [\,\cdot\,,T]^n(W)
=
\gno{
  \ctr[1.4]{}{W}{TT\cd}{T}
  \ctr[0.7]{}{W}{T}{T}
  \ctr[0.0]{}{W}{}{T}
  WTT\cd T
}
=
  (WT^n)_{\text{C}}
$,
since
$T$
has no contractions with operators to its right.\footnote{This is easily seen from the diagram.  It comes from the fact that $T$ is composed entirely of quasi-particle creation operators.}
  Applying \cref{thm:hausdorff} to $\ol{W}$ and using this result completes the proof.
}
\end{cor}

\begin{rmk}\label{rmk:connected-expansion}
Applying \cref{cor:tcc-similarity-transformation-connected} to the coupled-cluster effective Hamiltonian gives the following expansion
\begin{align}
  \ol{H}_{\text{c}}
=
  (H_{\text{c}}\,
   \text{exp}(T))_{\text{C}}
=
  (
    H_{\text{c}}
  +
    H_{\text{c}}
    T
  +
    \tfr{1}{2!}
    H_{\text{c}}
    T^2
  +
    \tfr{1}{3!}
    H_{\text{c}}
    T^3
  +
    \tfr{1}{4!}
    H_{\text{c}}
    T^4
  )_{\text{C}}
\end{align}
which ends at the fourth power because $H_{\text{c}}$ is a linear combination of one- and two-particle operators, which can contract at most two and four $T$'s, respectively.
Using this result, the energy and amplitude equations are often written as
\begin{align}
\label{eq:traditional-cc-equations-2}
  \ip{\F|H_{\text{c}}\,\text{exp}(T)|\F}_{\text{C}}
=
  E_{\text{c}}
&&
  \ip{\F_{ij\cd}^{ab\cd}|H_{\text{c}}\,\text{exp}(T)|\F}_{\text{C}}
=
  0
\end{align}
where the subscript $\text{C}$ on the expectation value ket is shorthand for
$
  \ip{\F_{ij\cd}^{ab\cd}|(H_{\text{c}}\,\text{exp}(T))_{\text{C}}|\F}
$.
\end{rmk}

\begin{ntt}
The following is suggested notation for the diagonal and off-diagonal contributions to the Fock operator.
\begin{align}
\diagram{
  \draw (-0.5,0) node[circlex] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
=
\diagram{
  \draw (-0.5,0) node[circlez] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
&&
\diagram{
  \draw (-0.5,0) node[circlez] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
\equiv
  H_0
&&
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
\equiv
  f_p^q(1-\delta_p^q)
  \tl{a}^p_q
\end{align}
so that
$
  H
=
  E_{\text{ref}}
+
\diagram{
  \draw (-0.5,0) node[circlez] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.35);
  \draw[-<-] (h1) to ++(0,-0.35);
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.35);
  \draw[-<-] (h1) to ++(0,-0.35);
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot=white}{sawtooth};
  \draw[->-] (g1) to ++(0,+0.35);
  \draw[-<-] (g1) to ++(0,-0.35);
  \draw[->-] (g2) to ++(0,+0.35);
  \draw[-<-] (g2) to ++(0,-0.35);
}
$\
is the full electronic Hamiltonian
and
$
  V_{\text{c}}
=
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.35);
  \draw[-<-] (h1) to ++(0,-0.35);
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot=white}{sawtooth};
  \draw[->-] (g1) to ++(0,+0.35);
  \draw[-<-] (g1) to ++(0,-0.35);
  \draw[->-] (g2) to ++(0,+0.35);
  \draw[-<-] (g2) to ++(0,-0.35);
}
$.
Note that
\begin{align*}
\diagram{
  \draw (-0.5,0) node[circlez] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
=
\diagram{
  \draw (-0.5,0) node[circlez] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
+
\diagram{
  \draw (-0.5,0) node[circlez] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[-<-] (h1) to ++(0,+0.5);
  \draw[->-] (h1) to ++(0,-0.5);
}
&&
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
=
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(-0.25,+0.5);
  \draw[-<-] (h1) to ++(+0.25,+0.5);
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(-0.25,-0.5);
  \draw[-<-] (h1) to ++(+0.25,-0.5);
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[-<-] (h1) to ++(0,+0.5);
  \draw[->-] (h1) to ++(0,-0.5);
}
\end{align*}
where the excitation level $\pm1$ contributions to $H_0$ have been omitted because its interaction tensor is diagonal.
\end{ntt}

\begin{rmk}
It can be shown that the determinant basis forms an eigenbasis for the diagonal part of the Fock operator.\footnotemark
\begin{align}
  H_0\F_{i_1\cd i_k}^{a_1\cd a_k}
=
  \mc{E}_{i_1\cd i_k}^{a_1\cd a_k}
  \F_{i_1\cd i_k}^{a_1\cd a_k}
&&
  H_0
\equiv
  f_p^p\tl{a}^p_p
&&
  \mc{E}_{q_1\cd q_k}^{p_1\cd p_k}
\equiv
  \sum_{r=1}^k
  f_{p_r}^{p_r}
-
  \sum_{r=1}^k
  f_{q_r}^{q_r}
\end{align}
Noting that $H_0$ is Hermitian, this implies
$\ip{\F_{ij\cd}^{ab\cd}|H_0\,T|\F}=\mc{E}_{ij\cd}^{ab\cd}\ip{\F_{ij\cd}^{ab\cd}|T|\F}=\mc{E}_{ij\cd}^{ab\cd}t_{ab\cd}^{ij\cd}$.
This can be used to rearrange the amplitude equation in
(\ref{eq:traditional-cc-equations-2}) as follows, which defines the working
equations used to iteratively solve TCC.%
\footnote{%
    Note that $\ip{\F_{ij\cd}^{ab\cd}|H_0\,\exp(T)|\F}_{\text{C}}=\ip{\F_{ij\cd}^{ab\cd}|H_0\,T|\F}$.
}
\begin{align}
  t_{ab\cd}^{ij\cd}
=
  (\mc{E}_{ab\cd}^{ij\cd})^{-1}
  \ip{\F_{ab\cd}^{ij\cd}|V_{\text{c}}\,\text{exp}(T)|\F}_{\text{C}}
&&
  V_{\text{c}}
\equiv
  H_{\text{c}}
-
  H_0
=
  f_p^q(1-\delta_p^q)
  \tl{a}^p_q
+
  \tfr{1}{4}
  \ol{g}_{pq}^{rs}
  \tl{a}^{pq}_{rs}
\end{align}
In M\o ller-Plesset perturbation theory, $H_0$ is known the \textit{zeroth order Hamiltonian} and $V_{\text{c}}$ is the \textit{perturbation}.
These operators are also known as the \textit{model Hamiltonian} and \textit{fluctuation potential}, respectively.
\end{rmk}
\footnotetext{
The proof is as follows.
First, note that $a^p_p\F_\si = n_p^\si \F_\si$, where $n_p^\si$ is the occupation of $\y_p$ in $\F_\si$.
By Wick's theorem, $a^p_p=\tl{a}^p_p+n_p^{\text{ref}}$, where $n_p^{\text{ref}}$ denotes the occupation of $\y_p$ in $\F$.
Therefore, $\tl{a}^p_p\F_\si = (n_p^\si - n_p^{\text{ref}})\F_\si$ and $H_0\F_\si = \pr{\sum_{p\in\F_\si}f_p^p - \sum_{p\in\F}f_p^p}\F_\si$.
}

\begin{dfn}
\thmtitle{Excitation level}
The \textit{excitation level} of a graph equals the net number of particles or quasi-particles it creates, divided by two.
For example, the quasi-particle excitation levels of the $T_1$, $T_2$ and $T_3$ operators are $1$, $2$, and $3$, respectively, and that of $\tl{a}_{abcd}^{ijkl}$ is $-4$.
A convenient rule for evaluating reference expectation values is that the total excitation level of a closed graph must balance out to zero.
\end{dfn}

\begin{ex}
The excitation levels in the quasi-particle expansions of one- and two-particle operators are as follows.
\begin{align*}
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot=white] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
  \node at (0,-0.8) {$(0)$};
}
=&\
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(0,+0.5);
  \draw[-<-] (h1) to ++(0,-0.5);
  \node at (0,-0.8) {$(0)$};
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(-0.25,+0.5);
  \draw[-<-] (h1) to ++(+0.25,+0.5);
  \node at (0,-0.8) {$(+1)$};
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(-0.25,-0.5);
  \draw[-<-] (h1) to ++(+0.25,-0.5);
  \node at (0,-0.8) {$(-1)$};
}
+
\diagram{
  \draw (-0.5,0) node[circlep] (h) {} -- (0,0) node[ddot] (h1) {};
  \draw[->-] (h1) to ++(0,-0.5);
  \draw[-<-] (h1) to ++(0,+0.5);
  \node at (0,-0.8) {$(0)$};
}
\\[10pt]
\diagram{
  \interaction{2}{g}{(0,0)}{ddot=white}{sawtooth};
  \draw[->-] (g1) to ++(0,+0.5);
  \draw[-<-] (g1) to ++(0,-0.5);
  \draw[->-] (g2) to ++(0,+0.5);
  \draw[-<-] (g2) to ++(0,-0.5);
  \node at (0.5,-0.8) {$(0)$};
}
=&\
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(0,+0.5);
  \draw[-<-] (g1) to ++(0,-0.5);
  \draw[->-] (g2) to ++(0,+0.5);
  \draw[-<-] (g2) to ++(0,-0.5);
  \node at (0.5,-0.8) {$(0)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(0,+0.5);
  \draw[-<-] (g1) to ++(0,-0.5);
  \draw[->-] (g2) to ++(-0.25,+0.5);
  \draw[-<-] (g2) to ++(+0.25,+0.5);
  \node at (0.5,-0.8) {$(+1)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(0,+0.5);
  \draw[-<-] (g1) to ++(0,-0.5);
  \draw[->-] (g2) to ++(-0.25,-0.5);
  \draw[-<-] (g2) to ++(+0.25,-0.5);
  \node at (0.5,-0.8) {$(-1)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(-0.25,+0.5);
  \draw[-<-] (g1) to ++(+0.25,+0.5);
  \draw[->-] (g2) to ++(-0.25,+0.5);
  \draw[-<-] (g2) to ++(+0.25,+0.5);
  \node at (0.5,-0.8) {$(+2)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(-0.25,-0.5);
  \draw[-<-] (g1) to ++(+0.25,-0.5);
  \draw[->-] (g2) to ++(-0.25,+0.5);
  \draw[-<-] (g2) to ++(+0.25,+0.5);
  \node at (0.5,-0.8) {$(0)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(-0.25,-0.5);
  \draw[-<-] (g1) to ++(+0.25,-0.5);
  \draw[->-] (g2) to ++(-0.25,-0.5);
  \draw[-<-] (g2) to ++(+0.25,-0.5);
  \node at (0.5,-0.8) {$(-2)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(0,-0.5);
  \draw[-<-] (g1) to ++(0,+0.5);
  \draw[->-] (g2) to ++(-0.25,+0.5);
  \draw[-<-] (g2) to ++(+0.25,+0.5);
  \node at (0.5,-0.8) {$(+1)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(0,-0.5);
  \draw[-<-] (g1) to ++(0,+0.5);
  \draw[->-] (g2) to ++(-0.25,-0.5);
  \draw[-<-] (g2) to ++(+0.25,-0.5);
  \node at (0.5,-0.8) {$(-1)$};
}
+
\diagram{
  \interaction{2}{g}{(0,0)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(0,-0.5);
  \draw[-<-] (g1) to ++(0,+0.5);
  \draw[->-] (g2) to ++(0,-0.5);
  \draw[-<-] (g2) to ++(0,+0.5);
  \node at (0.5,-0.8) {$(0)$};
}\,\,.
\end{align*}
\end{ex}

\begin{ex}
\thmtitle{The CCSDTQ equations}
Truncating the cluster operator at quadruples, $T\approx T_1 + T_2 + T_3 + T_4$, gives the CCSDTQ approximation.
The resulting singles, doubles, triples, and quadruples amplitude equations are given by\begin{align}
\nonumber
  t_a^i
=&\
  (\mc{E}_a^i)^{-1}
  \ip{\F_i^a|
    V_{\text{c}}
    (
      1
    +
      T_1
    +
      T_2
    +
      T_3
    +
      \tfr{1}{2}
      T_1^2
    +
      T_1T_2
    +
      \tfr{1}{3!}
      T_1^3
    )
  |\F}_{\text{C}}
\\[3pt]
\nonumber
  t_{ab}^{ij}
=&\
  (\mc{E}_{ab}^{ij})^{-1}
  \br{\F_{ij}^{ab}}
    V_{\text{c}}
    (
      1
    +
      T_1
    +
      T_2
    +
      T_3
    +
      T_4
    +
      \tfr{1}{2}
      T_1^2
    +
      T_1T_2
    +
      T_1T_3
    +
      \tfr{1}{2}
      T_2^2
    +
      \tfr{1}{3!}
      T_1^3
    +
      \tfr{1}{2}
      T_1^2T_2
    +
      \tfr{1}{4!}
      T_1^4
    )
  \kt{\F}_{\text{C}}
\\[3pt]
\nonumber
  t_{abc}^{ijk}
=&\
  (\mc{E}_{abc}^{ijk})^{-1}
  \br{\F_{ijk}^{abc}}
    V_{\text{c}}
    (
      T_2
    +
      T_3
    +
      T_4
    +
      T_1T_2
    +
      T_1T_3
    +
      \tfr{1}{2}
      T_2^2
    +
      T_1T_4
    +
      T_2T_3
    +
      \tfr{1}{2}
      T_1^2T_2
    +
      \tfr{1}{2}
      T_1^2T_3
    +
      \tfr{1}{2}
      T_1T_2^2
    +
      \tfr{1}{3!}
      T_1^3T_2
    )
  \kt{\F}_{\text{C}}
\\[3pt]
\nonumber
  t_{abcd}^{ijkl}
=&\
  (\mc{E}_{abcd}^{ijkl})^{-1}
  \br{\F_{ijkl}^{abcd}}
    V_{\text{c}}
    (
      T_3
    +
      T_4
    +
      T_1T_3
    +
      \tfr{1}{2}
      T_2^2
    +
      T_1T_4
    +
      T_2T_3
    +
      T_2T_4
    +
      \tfr{1}{2}
      T_3^2
    +
      \tfr{1}{2}
      T_1^2T_3
    +
      \tfr{1}{2}
      T_1T_2^2
    +
      \tfr{1}{2}
      T_1^2T_4
\\
\nonumber
&
\hphantom{
  (\mc{E}_{abcd}^{ijkl})^{-1}
  \br{\F_{ijkl}^{abcd}}
    V_{\text{c}}
    (
      T_3\
 }
    +
      T_1T_2T_3
    +
      \tfr{1}{3!}
      T_2^3
    +
      \tfr{1}{3!}
      T_1^3T_3
    +
      \tfr{1}{2!2!}
      T_1^2T_2^2
    )
  \kt{\F}_{\text{C}}
\end{align}
where several contributions to $\text{exp}(T_1+T_2+T_3+T_4)$ have been omitted either because the excitation levels do not balance or because they require one of the cluster operators to be disconnected from the Hamiltonian.
\end{ex}

\begin{dfn}
\thmtitle{Isomorphism}
For any invertible map $S:\ol{V}\rightarrow V$, we can express operators and vectors on $V$ as
\begin{align}
  A
=
  S
  \ol{A}
  S^{-1}
&&
  \kt{v}
=
  S
  \kt{\ol{v}}
&&
  \br{v}
=
  \br{\ol{v}}
  S^{-1}
\end{align}
in terms of operators and vectors on $\ol{V}$.
Note that the transformed bra and ket, $\br{\ol{v}}$ and $\kt{\ol{v}}$, corresponding to $v$ are not adjoints unless the transformation is unitary, $S^{-1}=S\dg$.
The similarity-transformed operator $\ol{A}$ retains all of the basis-independent properties of $A$, such as its trace, determinant, and eigenvalue spectrum, and its matrix elements satisfy $\ip{\ol{v}|\ol{A}|\ol{v}'}=\ip{v|A|v'}$.
More broadly, the invertibility of $S$ implies an \textit{isomorphism} between $V$ and $\ol{V}$, such that all statements about $V$ are in one-to-one correspondence with statements about $\ol{V}$ under this transformation.
\end{dfn}

\begin{rmk}
Since exponential operators are automatically invertible, the TCC wave operator defines a similarity transformation of Fock space into itself.
The image of the Schr\"odinger equation under this transformation is as follows
\begin{align}
\label{eq:similarity-transformed-schrodinger-equation}
  \ol{H}
  \kt{\ol{\Y}_k}
=
  E_k
  \kt{\ol{\Y}_k}
&&
  \br{\ol{\Y}_k}
  \ol{H}
=
  \br{\ol{\Y}_k}
  E_k
&&
\begin{array}{c@{\ }l}
  \ol{H}
&=
  \text{exp}(-T)
  H\,
  \text{exp}(T)
\\
  \kt{\ol{\Y}_k}
&=
  \text{exp}(-T)
  \kt{\Y_k}
\\
  \br{\ol{\Y}_k}
&=
  \br{\Y_k}\,
  \text{exp}(T)
\end{array}
\end{align}
where the $k\eth$ left and right eigenstates are not adjoints because $\text{exp}(T)$ is inherently non-unitary.
In TCC, $T$ is determined by the requirement that the ground-state right eigenvector of $\ol{H}$ be the reference determinant,
$
  \kt{\ol{\Y}_0}
\overset{!}{=}
  \kt{\F}
$.
\end{rmk}

\begin{dfn}
\thmtitle{Equation-of-motion coupled-cluster theory}
Expanding the left and right eigenstates of $\ol{H}$ in the determinant basis leads to the \textit{equation-of-motion (EOM) coupled-cluster equations}
\begin{align}
\begin{array}{r@{\ }lr@{\ }l}
  \ol{H}
  \,{}^kR
  \kt{\F}
&
=
  E_k
  \,{}^kR
  \kt{\F}
&
  R
&=
  R_0
+
  R_1
+
  \cd
+
  R_n
\\[7pt]
  \br{\F}
  \,{}^kL\,
  \ol{H}
&=
  \br{\F}
  \,{}^kL\,
  E_k
&
  L
&=
  L_0
+
  L_1
+
  \cd
+
  L_n
\end{array}
&&
  \ip{\F|
    {}^kL\,
    {}^lR
  |\F}
\overset{!}{=}
  \delta_{kl}
\end{align}
which are analogous to the configuration interaction eigenvalue equation, ${}^kR$ and ${}^kL$ being linear excitation and de-excitation operators analogous to $C$ and $C\dg$.
The condition on the right indicates that left and right eigenstates are chosen to form a \textit{biorthonormal system} by normalizing the overlap of the $k\eth$ left and right eigenfunctions to equal one.
Observable expectation values are given by
$
  \ip{\Y_k|W|\Y_k}
=
  \ip{\F|{}^kL\,\ol{W}\,{}^kR|\F}
$
in terms of left and right EOM coefficients.
Transition matrix elements are given by
$
  \ip{\Y_k|W|\Y_l}
=
  \ip{\F|{}^kL\,\ol{W}\,{}^lR|\F}
$.
\end{dfn}

\begin{dfn}
\thmtitle{The coupled-cluster Lagrangian}
The ground-state TCC equations are equivalent to requiring ${}^0R=1$, which implies ${}^0L_0=1$ from the biorthonormality condition.
Therefore, the ground-state left eigenvector has the form
\begin{align}
\label{eq:cc-ground-state-wave-operators}
  {}^0L
=
  1
+
  \La
&&
  \La
=
  \La_1
+
  \cd
+
  \La_n
&&
  \La_k
\equiv
  (\tfr{1}{k!})^2
  \la_{i_1\cd i_k}^{a_1\cd a_k}
  \tl{a}^{i_1\cd i_k}_{a_1\cd a_k}
\end{align}
and the ground-state energy can be written as follows, in an expression known as the \textit{coupled-cluster Lagrangian}.
\begin{align}
  \ip{\Y_0|H|\Y_0}
=
  \ip{\F|
    (
      1
    +
      \La
    )
    \ol{H}
  |\F}
\equiv
  \mc{L}(\bo{t},\bm{\la})
\end{align}
To see why this constitutes a Lagrangian, note that setting its gradient with respect to $\la_{i_1\cd i_k}^{a_1\cd a_k}$ equal to zero yields the TCC amplitude equations of eq~\ref{eq:traditional-cc-equations}.
If these are satisfied, the $\la$-dependent part of the equation vanishes and the Lagrangian returns the coupled-cluster energy:
$
  \ip{\F|\ol{H}|\F}
=
  E
$.
Therefore, the $\la$ coefficients can be viewed as Lagrange multipliers enforcing the TCC amplitude equations as a constraint.
\end{dfn}

\begin{dfn}
\thmtitle{The coupled-cluster lambda equations}
Setting the gradient of $\mc{L}$ with respect to $t_{a_1\cd a_k}^{i_1\cd i_k}$ equal to zero gives the \textit{coupled-cluster lambda equations}, which determine the Lagrange multpliers.
\begin{align}
  \br{\F}
  (
    1
  +
    \La
  )
  H_{\text{c}}\,
  \text{exp}(T)
  \kt{\F_{i_1\cd i_k}^{a_1\cd a_k}}_{\text{C}}
\overset{!}{=}
  0
\end{align}
The subscript $\text{C}$ now denotes that $H_{\text{c}}$ is connected both to the ket and to the $T$ operators.
This can be rearranged as\footnote{Note that $\ip{\F|\La H_0 T^p|\F_{i_1\cd i_k}^{a_1\cd a_k}}_{\text{C}}=0$ for $p\geq 1$, since the model Hamiltonian can only connect to one operator on either side.}
\begin{align}
  \la_{i_1\cd i_k}^{a_1\cd a_k}
=
  (\mc{E}_{a_1\cd a_k}^{i_1\cd i_k})^{-1}
  \br{\F}
  (
    1
  +
    \La
  )
  V_{\text{c}}\,
  \text{exp}(T)
  \kt{\F_{i_1\cd i_k}^{a_1\cd a_k}}_{\text{C}}
\end{align}
which sets up the iterative procedure for determining $\la_{i_1\cd i_k}^{a_1\cd a_k}$ from a given set of amplitudes.
\end{dfn}

\begin{ex}
\thmtitle{The CCSD lambda equations}
Truncating the wave operator at doubles, $T\approx T_1 + T_2$, gives the CCSD approximation.
The resulting singles and doubles lambda equations are given by the following
\begin{align*}
  \la_i^a
=&\
  (\mc{E}_a^i)^{-1}
  \br{\F}
    V_{\text{c}}\,
    (
      1
    +
      T_1
    )
  +
    \La_1
    V_{\text{c}}\,
    (
      1
    +
      T_1
    +
      T_2
    +
      \tfr{1}{2}
      T_1^2
    )
  +
    \La_2
    V_{\text{c}}\,
    (
      1
    +
      T_1
    +
      T_2
    +
      \tfr{1}{2}
      T_1^2
    +
      T_1
      T_2
    )
  \kt{\F_i^a}_{\text{C}}
\\
  \la_{ij}^{ab}
=&\
  (\mc{E}_{ab}^{ij})^{-1}
  \br{\F}
    V_{\text{c}}\,
  +
    \La_1
    V_{\text{c}}\,
    (
      1
    +
      T_1
    )
  +
    \La_2
    V_{\text{c}}\,
    (
      1
    +
      T_1
    +
      T_2
    +
      \tfr{1}{2}
      T_1^2
    )
  \kt{\F_{ij}^{ab}}_{\text{C}}
\end{align*}
where we have omitted any contributions to $\text{exp}(T_1 + T_2)$ that vanish.
\end{ex}

\begin{ex}
Assuming Brillouin's theorem holds, the CCD lambda equations are as follows.
\begin{align*}
  \la_{ij}^{ab}
  \mc{E}_{ab}^{ij}
=&\
  \ip{\F|
    V_{\text{c}}
  +
    \La_2
    V_{\text{c}}
  +
    \La_2 V_{\text{c}}T_2
  |\F_{ij}^{ab}}_{\text{C}}
\\=&\
\diagram{
  \interaction{2}{g}{(0,0.5)}{ddot}{sawtooth};
  \draw[->-] (g1) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-] (g1) to ++(+0.25,-1) node[smalldot] {};
  \draw[->-] (g2) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-] (g2) to ++(+0.25,-1) node[smalldot] {};
}
+
\diagram{
  \interaction{2}{t}{(0,0.5)}{ddot}{overhang};
  \draw[->-=0.65] (t1) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-=0.25,-<-=0.75]
      (t1)
    to
      node[ddot,midway] (g1) {}
    ++(+0.25,-1)
      node[smalldot] {};
  \draw[->-=0.65] (t2) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-=0.25,-<-=0.75]
      (t2)
    to
      node[ddot,midway] (g2) {}
    ++(+0.25,-1)
      node[smalldot] {};
   \draw[sawtooth] (g1) to (g2);
}
+
\diagram{
  \interaction{2}{t}{(0,0.5)}{ddot}{overhang};
  \draw[-<-=0.65] (t1) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-=0.25,->-=0.75]
      (t1)
    to
      node[ddot,midway] (g1) {}
    ++(+0.25,-1)
      node[smalldot] {};
  \draw[-<-=0.65] (t2) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-=0.25,->-=0.75]
      (t2)
    to
      node[ddot,midway] (g2) {}
    ++(+0.25,-1)
      node[smalldot] {};
   \draw[sawtooth] (g1) to (g2);
}
+
\diagram{
  \interaction{2}{t}{(0,0.5)}{ddot}{overhang};
  \interaction{2}{g}{(1,0.)}{ddot}{sawtooth};
  \draw[-<-] (t1) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-] (t1) to ++(+0.25,-1) node[smalldot] {};
  \draw[->-,bend left=40 ] (g1) to (t2);
  \draw[-<-,bend right=40] (g1) to (t2);
  \draw[-<-] (g2) to ++(-0.25,-0.5) node[smalldot] {};
  \draw[->-] (g2) to ++(+0.25,-0.5) node[smalldot] {};
}
+
\diagram{
  \interaction{2}{l}{(0,0.5)}{ddot}{overhang};
  \interaction{2}{g}{(2,0.5)}{ddot}{sawtooth};
  \draw[overhang] (1.5,0) node[ddot] (t1) {} to ++(1.5,0) node[ddot] (t2) {};
  \draw[-<-=0.65] (l1) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-=0.65] (l1) to ++(+0.25,-1) node[smalldot] {};
  \draw[-<-=0.65] (l2) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-] (t1) to (l2);
  \draw[->-] (t1) to (g1);
  \draw[->-=0.65] (g1) to ++(+0.25,-1) node[smalldot] {};
  \draw[-<-,bend left=40 ] (t2) to (g2);
  \draw[->-,bend right=40] (t2) to (g2);
}
+
\diagram{
  \interaction{2}{l}{(0,0.5)}{ddot}{overhang};
  \interaction{2}{g}{(2,0.5)}{ddot}{sawtooth};
  \draw[overhang] (1.5,0) node[ddot] (t1) {} to ++(1.5,0) node[ddot] (t2) {};
  \draw[->-=0.65] (l1) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-=0.65] (l1) to ++(+0.25,-1) node[smalldot] {};
  \draw[->-=0.65] (l2) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-] (t1) to (l2);
  \draw[-<-] (t1) to (g1);
  \draw[-<-=0.65] (g1) to ++(+0.25,-1) node[smalldot] {};
  \draw[->-,bend left=40 ] (t2) to (g2);
  \draw[-<-,bend right=40] (t2) to (g2);
}
\\&\
+
\diagram{
  \interaction{2}{l}{(0,0.5)}{ddot}{overhang};
  \interaction{2}{g}{(2.2,0.5)}{ddot}{sawtooth};
  \interaction{2}{t}{(1.1,0.0)}{ddot}{overhang};
  \draw[->-=0.7] (l1) to ++(-0.4,-1) node[smalldot] {};
  \draw[-<-=0.3] (l1) to (t1);
  \draw[->-=0.7] (l2) to ++(-0.4,-1) node[smalldot] {};
  \draw[-<-=0.3] (l2) to (t2);
  \draw[->-=0.3] (g1) to (t1);
  \draw[-<-=0.7] (g1) to ++(+0.4,-1) node[smalldot] {};
  \draw[->-=0.3] (g2) to (t2);
  \draw[-<-=0.7] (g2) to ++(+0.4,-1) node[smalldot] {};
}
+
\diagram{
  \interaction{2}{l}{(0,0.5)}{ddot}{overhang};
  \interaction{2}{g}{(2,0.5)}{ddot}{sawtooth};
  \interaction{2}{t}{(1,0.0)}{ddot}{overhang};
  \draw[-<-=0.65] (l1) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-=0.65] (l1) to ++(+0.25,-1) node[smalldot] {};
  \draw[->-,bend left=40 ] (t1) to (l2);
  \draw[-<-,bend right=40] (t1) to (l2);
  \draw[->-,bend left=40 ] (t2) to (g1);
  \draw[-<-,bend right=40] (t2) to (g1);
  \draw[-<-=0.65] (g2) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-=0.65] (g2) to ++(+0.25,-1) node[smalldot] {};
}
+
\diagram{
  \interaction{2}{g}{(0,0.5)}{ddot}{sawtooth};
  \interaction{2}{l}{(2.2,0.5)}{ddot}{overhang};
  \interaction{2}{t}{(1.1,0.0)}{ddot}{overhang};
  \draw[->-=0.7] (g1) to ++(-0.4,-1) node[smalldot] {};
  \draw[-<-=0.3] (g1) to (t1);
  \draw[->-=0.7] (g2) to ++(-0.4,-1) node[smalldot] {};
  \draw[-<-=0.3] (g2) to (t2);
  \draw[->-=0.3] (l1) to (t1);
  \draw[-<-=0.7] (l1) to ++(+0.4,-1) node[smalldot] {};
  \draw[->-=0.3] (l2) to (t2);
  \draw[-<-=0.7] (l2) to ++(+0.4,-1) node[smalldot] {};
}
\\&\
+
\diagram{
  \interaction{2}{g}{(0,0.5)}{ddot}{sawtooth};
  \interaction{2}{l}{(2,0.5)}{ddot}{overhang};
  \draw[overhang] (1.5,0) node[ddot] (t1) {} to ++(1.5,0) node[ddot] (t2) {};
  \draw[-<-=0.65] (g1) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-=0.65] (g1) to ++(+0.25,-1) node[smalldot] {};
  \draw[-<-=0.65] (g2) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-] (t1) to (g2);
  \draw[->-] (t1) to (l1);
  \draw[->-=0.65] (l1) to ++(+0.25,-1) node[smalldot] {};
  \draw[-<-,bend left=40 ] (t2) to (l2);
  \draw[->-,bend right=40] (t2) to (l2);
}
+
\diagram{
  \interaction{2}{g}{(0,0.5)}{ddot}{sawtooth};
  \interaction{2}{l}{(2,0.5)}{ddot}{overhang};
  \draw[overhang] (1.5,0) node[ddot] (t1) {} to ++(1.5,0) node[ddot] (t2) {};
  \draw[->-=0.65] (g1) to ++(-0.25,-1) node[smalldot] {};
  \draw[-<-=0.65] (g1) to ++(+0.25,-1) node[smalldot] {};
  \draw[->-=0.65] (g2) to ++(-0.25,-1) node[smalldot] {};
  \draw[->-] (t1) to (g2);
  \draw[-<-] (t1) to (l1);
  \draw[-<-=0.65] (l1) to ++(+0.25,-1) node[smalldot] {};
  \draw[->-,bend left=40 ] (t2) to (l2);
  \draw[-<-,bend right=40] (t2) to (l2);
}
\\=&\
  \ol{g}_{ij}^{ab}
+
  \tfr{1}{2}
  \la_{ij}^{cd}
  \ol{g}_{cd}^{ab}
+
  \tfr{1}{2}
  \la_{kl}^{ab}
  \ol{g}_{ij}^{kl}
+
  P_{(i/j)}^{(a/b)}
  \la_{ik}^{ac}
  \ol{g}_{cj}^{kb}
-
  \tfr{1}{2}
  P_{(i/j)}
  \la_{ik}^{ab}
  t_{cd}^{kl}
  \ol{g}_{jl}^{cd}
-
  \tfr{1}{2}
  P^{(a/b)}
  \la_{ij}^{ac}
  t_{cd}^{kl}
  \ol{g}_{kl}^{bd}
\\&\
+
  \tfr{1}{2^2}
  \la_{ij}^{cd}
  t_{cd}^{kl}
  \ol{g}_{kl}^{ab}
+
  P_{(i/j)}^{(a/b)}
  \la_{ik}^{ac}
  t_{cd}^{kl}
  \ol{g}_{lj}^{db}
+
  \tfr{1}{2^2}
  \ol{g}_{ij}^{cd}
  t_{cd}^{kl}
  \la_{kl}^{ab}
-
  \tfr{1}{2}
  P_{(i/j)}
  \ol{g}_{ik}^{ab}
  t_{cd}^{kl}
  \la_{jl}^{cd}
-
  \tfr{1}{2}
  P^{(a/b)}
  \ol{g}_{ij}^{ac}
  t_{cd}^{kl}
  \la_{kl}^{bd}
\end{align*}
\end{ex}

\begin{samepage}
\begin{rmk}
\thmtitle{The Hellmann-Feynman theorem}
If the Hamiltonian depends on a parameter $\xi$, such as a nuclear coordinate or an electric field strength, we can express the Schr\"odinger equation as a function of that parameter
\begin{align}
\label{eq:xi-dependent-schrodinger-equation}
  H(\xi)
  \Y(\xi)
=
  E(\xi)
  \Y(\xi)
&&
  \ip{\Y(\xi)|\Y(\xi)}
\overset{!}{=}
  1
\end{align}
Then the total derivative of the energy $E(\xi)=\ip{\Y(\xi)|H(\xi)|\Y(\xi)}$ with respect to $\xi$ is given by the following.
\begin{align}
  \fd{E(\xi)}{\xi}
=
  \ip{\Y(\xi)|
  \pd{H(\xi)}{\xi}
  |\Y(\xi)}
+
\cancel{
  \ip{\pd{\Y(\xi)}{\xi}|
  H(\xi)
  |\Y(\xi)}
+
  \ip{\Y(\xi)|
  H(\xi)
  |\pd{\Y(\xi)}{\xi}}
}
\end{align}
The second and third terms cancel by the \textit{Hellmann-Feynman theorem}, which one can prove by substituting the Schr\"odinger equation into both terms and employing the derivative of the normalization condition with respect to $\xi$.
In words, it says that the first derivative of the energy does not depend on the  ``response'' of the wavefunction.
More generally, for approximate methods, the wavefunction may be parametrized by a set of coefficients $\bo{c}$ which are \textit{stationary} in the sense that their energy gradient equals zero.
Denoting the remaining non-stationary coordinates by $\bo{p}$, we have
\begin{align}
  \fd{E(\xi)}{\xi}
=
  \pd{E(\xi)}{\xi}
+
  \cancel{
  \pd{E}{\bo{c}}
  }
  \cdot
  \fd{\bo{c}}{\xi}
+
  \pd{E}{\bo{p}}
  \cdot
  \fd{\bo{p}}{\xi}
&&
  \pd{E}{\xi}
=
  \ip{\Y(\bo{c},\bo{p})|
  \pd{H(\xi)}{\xi}
  |\Y(\bo{c},\bo{p})}
\end{align}
which applies to configuration interaction and other variational methods.
For most methods $\bo{p}$ will include parameters which determine the molecular orbital coefficients, since the shape of the Hartree-Fock orbitals depends on $\xi$.
When $\xi$ is a nuclear coordinate, the atomic-orbital basis functions themselves change\footnote{The parameters defining the basis functions are constant, but their centers of origin move with the nuclei.}
and we must include parameters to account for this as well.
The Hellmann-Feynman theorem does not apply to TCC energy, which is not stationary in any of its parameters.
However, it does apply to the coupled-cluster Lagrangian as follows.
\begin{align}
  \fd{\mc{L}(\xi)}{\xi}
=
  \pd{\mc{L}(\xi)}{\xi}
+
  \cancel{
  \pd{\mc{L}}{\bo{t}}
  }
  \cdot
  \fd{\bo{t}}{\xi}
+
  \cancel{
  \pd{\mc{L}}{\bm{\la}}
  }
  \cdot
  \fd{\bm{\la}}{\xi}
+
  \pd{\mc{L}}{\bo{p}}
  \cdot
  \fd{\bo{p}}{\xi}
&&
  \pd{\mc{L}(\xi)}{\xi}
=
  \ip{\F(\bo{p})|
    (
      1
    +
      \La
    )\,
    \pd{H(\xi)}{\xi}\,
    \text{exp}(T)
  |\F(\bo{p})}_{\text{C}}
\end{align}
This is known as the \textit{generalized Hellmann-Feynman theorem} for coupled-cluster theory.
Note that one must solve both the amplitude equations and the lambda equations in order to evaluate the equation on the right.
\end{rmk}
\end{samepage}


%auto-ignore
\providecommand{\MainFolder}{..}
\documentclass[\MainFolder/Text.tex]{subfiles}

\begin{document}
\section{Homology of twisted IBL-infinity structure for complex projective space}
%Computation of \texorpdfstring{$\OPQ_{210}$ and $\OPQ_{120}$}{q210 and q120}
\label{Section:CPn1}
Let $\Kaehler\in \DR^2(\CP^n)$ be the Fubini--Study K\"ahler form on $\CP^n$ (see \cite[Examples 3.1.9]{Huybrechts2004}). The powers of $\Kaehler$ are harmonic,\footnote{This follows by induction on the power of $\Kaehler$ using the fact that, on a general K\"ahler manifold $M$, the Lefschetz operator $\Lef: \DR(M)\rightarrow \DR(M)$ defined by $\Lef(\eta)\coloneqq \eta \wedge \Kaehler$ for all $\eta\in \DR(M)$ commutes with the Hodge--de Rham Laplacian~$\Delta$ (see \cite[Chapter 3]{Huybrechts2004}). } and we get easily
\[ \Harm(\CP^n) = \langle 1,\Kaehler, \dotsc, \Kaehler^n \rangle. \]
We denote the Riemannian volume of $\CP^n$ by
\[ V\coloneqq \int_{\CP^n} \frac{1}{n!}\Kaehler^n. \]
Consider the basis $e_0$, $\dotsc$, $e_n$ of $\Harm(\CP^n)[1]$ defined for all $i=0$, $\dotsc$, $n$ by
\[ e_i\coloneqq \frac{\NK^i}{(n! V)^{\frac{i}{n}}}, \quad \text{where}\quad\NK^i\coloneqq \SuspU \Kaehler^i. \]
The matrix of the pairing $\Pair$ from~\eqref{Eq:DeRhamDGA} with respect to the basis $e_0$,~$\dotsc$,~$e_{n}$ reads:
\[ (\Pair^{ij}) = \begin{pmatrix}
0 & \dotsb & 1 \\
\vdots & {\displaystyle\, .^{{\displaystyle \, .^{\displaystyle\,.}}}} & \vdots \\
1 & \dotsb & 0
\end{pmatrix}. \]
The basis $e^0$, $\dotsc$, $e^n$ dual to $e_0$, $\dotsc$, $e_n$ with respect to $\Pair$ thus satisfies
\[ e^i = e_{n-i}\quad\text{for all }i=0,\dotsc,n. \]
Therefore, the following holds for the matrix $(T^{ij})$ from~\eqref{Eq:PropagatorT}:
\begin{equation*}
(T^{ij}) = - (\Pair^{ij}).
\end{equation*}
For all $1\le i, j, k \le n$, we have
\[ \mu_2(e_i, e_j) = e_{i+j}\quad \text{and}\quad
\MC_{10}(\Susp e_{i}e_j e_k) = \delta_{i+j+k,n}. \]
For $\psi$, $\psi_1$, $\psi_2 \in \CDBCyc \Harm$ and generating words $\omega$, $\omega_1$, $\omega_2 \in \BCyc\Harm$, we chave
\begin{equation*}
\begin{aligned}
\OPQ_{210}(\Susp^2 \psi_1 \otimes \psi_2)(\Susp \omega) &= -\sum_{i=0}^n \sum \varepsilon(\omega\mapsto \omega^1\omega^2)(-1)^{\Abs{\omega^1}} \psi_1(e_i \omega^1)\psi_2(e_{n-i}\omega^2), \\
\OPQ_{120}(\Susp \psi)(\Susp^2 \omega_1 \otimes \omega_2) & = - \sum_{i=0}^n \sum \varepsilon(\omega_1 \mapsto \omega_1^1)\varepsilon(\omega_2\mapsto \omega_2^1) (-1)^{\Abs{\omega_1}} \psi(e_i \omega_1^1 e_{n-i} \omega_2^1).
\end{aligned}
\end{equation*}

The cyclic homology of $\Harm(\CP^n)$ is that of the truncated polynomial algebra
\[ A \coloneqq\R[x]/(x^{n+1})\quad\text{with }\Deg(x)=2. \]
The following lemma computes its cyclic homology.
\begin{Lemma}[Cyclic homology of truncated graded polynomial algebra]
Consider $A\coloneqq\R[x]/(x^{n+1})$ with $\Deg(x)=d\in \Z$. For all $i=1$, $\dotsc$, $n$ and $k\in \N_0$, there are cycles $\tilde{t}_{2k+1,i}\in  \tilde{D}_q(A)$ of weights $w(\tilde{t}_{2k+1,i}) = 2k+1$ and degrees $\Abs{\tilde{t}_{2k+1,i}} = d(i+(n+1)k)$, where $q = w(\tilde{t}_{2k+1,i})-\Abs{\tilde{t}_{2k+1,i}}-1$, which form a basis of $\ClasCycH(A)$ (the non-degree shifted cyclic homology defined on page~\pageref{Eq:NDSComplex}).
\end{Lemma}
\begin{proof}
A computation of the cyclic homology of $A$ for $\Abs{x}=0$ is the goal of \cite[Exercise 4.1.8.]{LodayCyclic} or \cite[Exercise 9.1.1]{Weibel1994}.
The hint is to compute the Hochschild homology $\H\H_n(A) = \Tor_n^{A_e}(A,A)$, where $A_e$ is the enveloping algebra of $A$, using a non-canonical (i.e., not the bar complex) projective resolution of the $A_e$-module $A$ given by
\begin{equation}\label{Eq:ProjRes}
\begin{tikzcd}
\dotsb \arrow[r] & A_e \arrow{r}{\cdot v} & A_e \arrow{r}{\cdot u} & A_e \arrow{r}{\mu} & A \arrow{r} & 0,
\end{tikzcd}
\end{equation}
where $u = x \otimes 1 - 1 \otimes x$ and $v = \sum_{i=0}^{n} x^i \otimes x^{n-i}\in A_e$.
The resolution continues to the left with $\cdot u$ und $\cdot v$ periodically.
Clearly, $\mu$ composed with $\cdot u$ vanishes and $u\cdot v = v\cdot u = x^{n+1}\otimes 1 - 1 \otimes x^{n+1} = 0$; one can check that \eqref{Eq:ProjRes} is indeed a resolution.
If $\Deg(x)=d$, then $\Deg(u) = d$ and $\Deg(v) = n d$.

We lift the resolution \eqref{Eq:ProjRes} to the graded category (i.e., we require that the maps are homogenous) by taking the degree shifts
\[
\begin{tikzcd}
\dotsb \arrow{r} &
A_e[-(n+1)di] \arrow{r}{d_{2i}} \ar[draw=none]{d}[name=X, anchor=center]{}&
A_e[-(n+1)di + nd] \arrow{r}{d_{2i-1}} &
A_e[-d(n+1)(i-1)]  \ar[rounded corners,
            to path={ -- ([xshift=2ex]\tikztostart.east)
                      |- (X.center) \tikztonodes
                      -| ([xshift=-2ex]\tikztotarget.west)
                      -- (\tikztotarget)}]{dlll}[at end]{}\\
\dotsb\arrow{r} &
A_e[-(n+1)d] \arrow{r}{d_{2}} &
A_e[-d] \arrow{r}{d_1}\ar[draw=none]{d}[name=Y, anchor=center]{} &
A_e  \ar[rounded corners,
            to path={ -- ([xshift=2ex]\tikztostart.east)
                      |- (Y.center) \tikztonodes
                      -| ([xshift=-2ex]\tikztotarget.west)
                      -- (\tikztotarget)}]{dl}{\mu} \\
&  & A \arrow{r} & 0,
\end{tikzcd}
\]
where we denoted by $d_i$ the maps from \eqref{Eq:ProjRes}.
Tensoring with $A$, we get the graded vector spaces $A \otimes_{A_e} A_e[\cdot] \simeq A[\cdot]$, and the maps $d_j$ become multiplications with $u$, $v \in A_e$ in $A$ as a right $A_e$-module.
For all polynomials $p\in A$, we have
$$ \begin{aligned}
    d_{2i-1}(p) = p\cdot (x\otimes 1 - 1 \otimes x) &= px - (-1)^{\Abs{x} \Abs{p}}xp = 0, \\
    d_{2i}(p) = p\cdot (\sum_{i=0}^n x^i \otimes x^{n-i}) & = (-1)^{n \Abs{x}\Abs{p}}(n+1) x^n p.
\end{aligned}$$
The homology of this chain complex consists of graded vector spaces $\H\H_{(l)}$ for $l\ge 0$ which correspond to the homology of the bar complex graded by weights, i.e., $\H\H_{(l)}$ would be represented by cycles in $A^{\otimes l+1}$ if the bar resolution was taken. We compute
\begin{equation}\label{Eq:HochCPn}
\H\H_{(l)} = \begin{cases}
 (x A)[-(n+1)di] & \text{for }l=2i, \\
 \R[x]/(x^n)[-(n+1)di + nd] & \text{for }l = 2i - 1, \\
 A & \text{for }l = 0.
\end{cases}
\end{equation}
Now, because the differential $\tilde{\delta}$ is zero and $\tilde{\Hd}$ is degree preserving, it holds (c.f., \eqref{Eq:TotComplNDS})
\[\ClasHH(A) = \bigoplus_{l\in \N_0} \H\H_{(l)}.\]
Clearly, the same will hold for $\ClasCycH(A)$, and hence we can ignore the gradation by degree and just use the gradation by weights in the bar complex, i.e., the non-graded theory.

In order to compute $\CycH_{(l)}(A)$, we consider the Connes' exact sequence in homology, or $\mathrm{ISB}$-sequence, see \cite[Theorem~2.2.1]{LodayCyclic}.
It arises from the exact sequence of bicomplexes $0 \rightarrow \LodCycBi^{\{2\}} \hookrightarrow \LodCycBi \twoheadrightarrow \LodCycBi[2,0] \rightarrow 0$, where~$\LodCycBi$ is the Loday's cyclic bicomplex from~\cite[Paragraph~2.1.2]{LodayCyclic} (bar complexes in columns), $\LodCycBi^{\{2\}}$ is the sub-bicomplex consisting of the first two columns of $\LodCycBi$ and $\LodCycBi[2,0]$ is the part of $\LodCycBi$ starting with the third column.
Because $A$ is augmented and unital, \cite[Theorem~4.1.13]{LodayCyclic} guarantees a splitting of the ISB-sequence into
\begin{equation}\label{Eq:ConnexIBS}
0 \longrightarrow \widebar{\H}^\lambda_{(l-1)} \longrightarrow \widebar{\H\H}_{(l)} \longrightarrow \widebar{\H}^\lambda_{(l)} \longrightarrow 0\quad\text{for }l\ge 1,
\end{equation}
where the bar $\bar{\cdot}$ denotes the reduced homology. It holds $\CycH_{(0)}= \H\H_{(0)}= A$, and hence $\widebar{\H}^\lambda_{(0)} = \langle x, \dotsc, x^n \rangle$.
Using \eqref{Eq:HochCPn}, the first map for $l=1$ in \eqref{Eq:ConnexIBS} reads $\langle x,\dotsc, x^n \rangle \hookrightarrow \langle 1, x, \dotsc, x^{n-1}\rangle[-d]$, and hence it is an isomorphism.
It follows that $\widebar{\H}^\lambda_{(1)} = 0$.
For $k\ge 1$, we obtain inductively $\widebar{\H}^\lambda_{(2k)} \simeq \widebar{\H\H}_{(2k)} = \langle x,\dotsc,x^n\rangle[-(n+1)dk] \hookrightarrow \widebar{\H\H}_{(2k+1)} = \langle 1, x, \dotsc, x^{n-1} \rangle[-(n+1)dk - d]$. This again has to be an isomorphism, and hence $\widebar{\H}^\lambda_{(2k+1)} = 0$.
\end{proof}
%The case of $\Abs{x} = d$ can be solved by taking suitable degree shifts in the proposed projective resolution which is used to compute $\H\H(A)$. Unfortunately, using a non-canonical projective resolution, we lose the concrete form of the cyclic cycles and obtain just the following result:
%For all $i=1$, $\dotsc$, $n$ and $k\in \N_0$, there are cycles $\tilde{t}_{2k+1,i}\in  \tilde{D}_q(A)$ of weights $2k+1$ and degrees $d(i+(n+1)k)$ which form a basis of $\ClasCycH(A)$.
We apply the degree shift $U: \tilde{D}(A) \rightarrow D(A)$ from Proposition~\ref{Prop:DGA} to get the generators
\[ t_{w,i} \coloneqq U(\tilde{t}_{w,i}) \in D^\lambda( \Harm(\CP^n)) \]
of weights $w$ and degrees $2i+ (w-1)n -1$, so that
\[ \H^\lambda(\Harm(\CP^n)) = \langle t_{w,i}, \NOne^{w} \mid w\in \N \text{ odd}, i=1,\dotsc, n\rangle. \]
By the universal coefficient theorem, we have $\H_\lambda^* = (\H^\lambda)^{\GD}$ with respect to the grading by the degree. Given $d\in \Z$, the equation $d= 2i + (w-1)n - 1$ has only finitely many solution $(w,i) \in \N \times \{1,\dotsc,n\}$, and hence we get
\begin{equation}\label{Eq:CPnHom}
\HIBL^\MC(\CycC(\Harm(\CP^n))) = \langle \Susp t_{w,i}^*, \Susp\NOne^{w*} \mid w\in \N \text{ odd}, i=1,\dotsc, n \rangle,
\end{equation}
where $t_{w,i}^*$ and $\NOne^{w*} \in \DBCyc \Harm$ are the duals to $t_{w,i}$ and $\NOne^{w}$, respectively (see Remark~\ref{Rem:UCT}). Notice that both $\Abs{\Susp t_{w,i}^*}$ and $\Abs{\Susp \NOne^{w*}}$ are even since $\Abs{\Susp} = 2n-3$.

Because $\CP^n$ is geometrically formal, Proposition~\ref{Prop:GeomForm} implies that $\PMC_{10} = \MC_{10}$. Because $\HIBL^\MC(\CycC)$ is concentrated in even degrees and because a general $\IBLInfty$-operation $\OPQ_{klg}$ is odd, all operations vanish on the homology. Therefore, for the \emph{twisted $\IBL$-algebras} we have
\begin{equation*}
\IBL(\HIBL^\PMC(\CycC)) = \IBL(\HIBL^\MC(\CycC)) = (\HIBL^\MC(\CycC), \OPQ_{210} \equiv 0, \OPQ_{120}\equiv 0),
\end{equation*}
where $\HIBL^\MC(\CycC)$ is given by \eqref{Eq:CPnHom}.

According to \cite[Section 3.1.2]{Basu2011}, the minimal model for the Borel construction $\LoopBorel \CP^n$ is the cdga $\Lambda^{\mathrlap{\Sph{1}}\hphantom{S}}(n+1,1)$, which is freely generated (over $\R$) by the homogenous vectors $x_1$, $x_2$, $y_1$, $y_2$, $u$ of degrees
\[ \Abs{x_1} = 2,\quad \Abs{x_2} = 2n+1,\quad \Abs{y_1} = 1,\quad \Abs{y_2} = 2n, \quad \Abs{u} = 2, \]
whose differential $\Dd$ satisfies
\[ \Dd y_1 = 0,\quad \Dd x_1 = y_1 u,\quad \Dd y_2 = -(n+1) x_1^n y_1,\quad \Dd x_2 = x_1^{n+1} + y_2 u. \]
By \cite[Theorem 3.6]{Basu2011}, the string cohomology $\StringCoH^*(\Loop \CP^n; \R)\simeq \H(\Lambda^{\mathrlap{\Sph{1}}\hphantom{S}}(n+1,1),\Dd)$ satisfies for all $m\in \N_0$ the following:
\begin{align*}
\StringCoH^m(\Loop \CP^n; \R) = \begin{cases}
\langle u^j \rangle & \text{if }m=2j, \\
\langle y_1 y_2^p x_1^q \mid 0\le q \le n-1, p\ge 0; q + n p = j\rangle & \text{if }m=2j+1.
\end{cases}
\end{align*}
The right-hand side can be identified with $\StringH(\Loop \CP^n; \R)$ by the universal coefficient theorem. According to \cite[Proposition 3.7]{Basu2011}, we have $\StringOp_2 = 0$ and $\StringCoOp_2 = 0$. We conclude that the map
\[\begin{aligned}
 \HIBL^\PMC(\RedCycC(\Harm(\CP^n)))[1] & \longrightarrow \RedStringH(\Loop \CP^n; \R)[3-n] \\
\Susp t^*_{2k+1,l} & \longmapsto \Susp y_1 y_2^k x_1^{l-1}\qquad\text{for }k\ge 0\text{ and }l=1,\dotsc, n
\end{aligned} \]
induces an isomorphism of $\IBL$-algebras
\[ \IBL(\HIBL^\PMC(\RedCycC(\Harm(\CP^n)))) \simeq \IBL(\RedStringH(\Loop \CP^n; \R)[3-n]). \]
\end{document}


\subsection{Echo State Networks}

\begin{figure}[htp]
  \centering
  \subfloat[Training Flow]{
    \includegraphics[clip, width=0.85\columnwidth]{reservoir_training.png}
  }

  \subfloat[Predicting Flow]{
    \includegraphics[clip, width=0.85\columnwidth]{reservoir_predicting.png}
  }
  \caption{(a) Shows the behavior of an \gls{esn} during the training phase. (b) Shows \gls{esn} behavior during the predicting phase. The output $U(t+\Delta t)$ is used as the next input value. }
  \label{fig:reservoir_graph}
\end{figure}

An \gls{esn}, sometimes called a ``reservoir
computer,''\cite{pathak_using_2017, pathak_model-free_2018, vlachas_backpropagation_2020} is a type of recurrent
neural network \cite{lukosevicius_reservoir_2009}
 that replaces the many hidden layers of a conventional feed-forward
neural network with a reservoir that is:
\begin{enumerate}
  \item sparse,
  \item connected by uniformly random weights, centered at zero,
  \item and large (i.e. has many neurons).
\end{enumerate}

The reservoir is therefore a randomly instantiated adjacency matrix,
\textit{\textbf{W}}, of size $N \times N$. An input matrix  $W^{in}$, of size
$N \times K$, maps the input vector, $U(t)$ with
$K$ units, onto the reservoir. The activation states of the reservoir are
calculated by \cite{shi_energy_2016, pathak_model-free_2018, lukosevicius_practical_2012}
 \begin{align}
   x(t) &= \tanh \left(W^{in}\cdot U(t) + \mathbf{W}x(t-1)\right)
   \intertext{where}
   x(t) &= \text{the collection of reservoir activations}\nonumber.
 \end{align}
 The output, $U(t+\Delta t)$, is read by an output weight matrix,
 $W^{out}$, thus:
 \begin{align}
   U(t+\Delta t) &= \left(W^{out}\right)^T\cdot x(t).
 \end{align}
 In the training phase, we
 discard the output and the next training input is passed to
 the network. During the prediction phase, we keep the output and use it as the
 next input. Figure \ref{fig:reservoir_graph} illustrates this behavior. The
 speed of \glspl{esn} is owed
 to this structure -- only $W^{out}$ has tunable weights. Everything else is
 fixed. In this work, we adapted the open source Python package \texttt{pyESN} \cite{korndorfer_pyesn_2015} to construct and train the network.

 \subsection{Hyper-Parameter Optimization}

 \glspl{esn} are fast because a large reservoir, that does not require
 training, replaces the hidden layers in a conventional feed-forward neural
 network.
 The trade-off is that \glspl{esn} are sensitive to various hyper-parameters
 that must be optimized \cite{lukosevicius_practical_2012}. Table \ref{tab:parameters} summarizes these hyper-parameters. The spectral radius ($\rho$) should satisfy the ``echo state property'' which means that
 previous reservoir activations have a decaying influence on future states. This
 is usually guaranteed for $\rho < 1$, but is not a requirement
 \cite{lukosevicius_practical_2012}.
 \begin{table*}[ht]
   \centering
   \caption{Description of Model Hyper-Parameters}
   \resizebox{\textwidth}{!}{
   \begin{tabular}{l l l}
     \hline
     Hyper-parameter & Purpose & Tested Values\\
     \hline
     \texttt{noise} & Neuron regularization & [0.0001, 0.0003, 0.0007, 0.001, \\
     &&0.003, 0.005, 0.007, 0.01]\\
     $\rho$ & Spectral radius & [0.5, 0.7, 0.9, 1, 1.1, \\
     &&1.2, 1.3, 1.5]\\
     $N$ & Size of reservoir, \textbf{W} & [600, 800, 1000, 1500, 2000, \\
     &&2500, 3000, 4000]\\
     \texttt{sparsity} & The density of connections in \textbf{W}& [0.005, 0.01, 0.03, 0.05, \\
     &&0.1, 0.12, 0.15, 0.2]\\
     Training Length & Training set size & $L \in$ [5000, 25000], step size = 300\\
     \hline
   \end{tabular}
   } % end resizebox
   \label{tab:parameters}
 \end{table*}
% \FloatBarrier
 We optimize the hyper-parameters by performing a grid search over the test
 values specified in Table \ref{tab:parameters}. We took the following
 optimization steps for each prediction task:
 \begin{enumerate}
   \item Select a hyper-parameter or pair of parameters.
   \item Generate \gls{esn} prediction with the specified parameters.
   \item Calculate and record the root mean squared error (RMSE).
   \item Continue until last entry in the parameter set is reached.
   \item Set the network parameters to the hyper-parameter value that minimizes the
   RMSE.
 \end{enumerate}
 Figure \ref{fig:rhoxnoise-demand04} shows an example
 heatmap that optimized the spectral radius and noise hyper-parameters for the 4-hour
 ahead demand forecast and illustrates the sensitivity of \glspl{esn} to
 hyperparameter values.

 \begin{figure}[h]
   \includegraphics[width=\columnwidth]{./images/04_demand_rho_noise_loss.png}
   % \input{./images/04_wind_elevation_rho_noise_loss-img0.png}
   \caption{An example heatmap of the RMSE for 4-hour ahead demand prediction with different combinations of spectral radius, $\rho$, and noise.}
   \label{fig:rhoxnoise-demand04}
 \end{figure}

 \subsection{Prediction Tasks}
In order to verify that the performance of our \gls{esn} implementation was
consistent with the literature \cite{pathak_using_2017} we first
performed a benchmarking task by making a prediction for the Lorenz
1963 model. The Lorenz 1963 model is described by a system of coupled differential equations:
\begin{align}
  \frac{\text{d}x}{\text{d}t} &= \sigma (y - x),\\
  \frac{\text{d}y}{\text{d}t} &= x(\rho - z)- y,\\
  \frac{\text{d}z}{\text{d}t} &= xy - \beta z,
  \intertext{where}
  \sigma &= 10,\nonumber\\
  \beta &= 8/3,\nonumber\\
  \rho &=28, \nonumber
\end{align}
since the system demonstrates chaotic behavior for these parameter values
\cite{lorenz_deterministic_1963}.

Next, we optimized predictions for
univariate time-series
representing total demand, solar energy, and wind energy 4-hours ahead and
48-hours ahead. Finally, we repeated those same six tasks with an additional
predictor. One value from each column in Table \ref{tab:tasks} was selected for
each task for a total of 42 predictions.

\begin{table*}[ht]
  \centering
  \caption{Summary of prediction tasks.}
  \label{tab:tasks}
  \resizebox{\columnwidth}{!}{
  \begin{tabular}{c c c}
    \hline
    Target & Future & Additional Predictor\\
    \hline
    && None \\
    Total Demand && Solar Elevation\\
    &4-hours ahead& Humidity\\
    Solar Energy && Pressure\\
    &48-hours ahead& Wet Bulb Temp.\\
    Wind Energy && Dry Bulb Temp.\\
    && Wind Speed\\
    \hline
  \end{tabular}%
  } % end resizebox
\end{table*}
% \FloatBarrier


\section{Integral equation derivation} \label{sec:anapp}

%\subsection{Adapting the completed double layer representation
%to solve the Dirichlet problem}
We would like to adapt the completed double layer representation
for solutions of Stokes flow \cref{eq:helmstokes,eq:helmstokescty,eq:helmstokesnoslip} to solve
the clamped plate problem \cref{eq:buck1,eq:buck2,eq:buck3}.
Let $f$ and $g$ be the boundary data as in \cref{eq:biharmD1,eq:biharmD2,eq:biharmD3}.
By computing tangential derivatives of $f$ on each
boundary component, we get the following related oscillatory Stokes
problem:
\begin{align}
 \Delta (\Delta + \lambda^2) \tilde{w}  = 0 &\quad \bx \in D \, ,\nonumber\\
 \frac{\partial \tilde{w}}{\partial \tau} =
 \frac{\partial f}{\partial \tau}
&\quad \bx \in \Gamma \label{eq:biharm5} \, ,\\
 \frac{\partial \tilde{w}}{\partial n} = g &\quad \bx
\in \Gamma
\, .
\nonumber
\end{align}
There are two main issues to be addressed in using the completed
double layer representation in this context. First,
the representation is designed for
oscillatory Stokes flow, in which the quantities of interest are derivatives
of the potential $\tilde{w}$ and not $\tilde{w}$ itself; the
representation for $\tilde{w}$ may not be single-valued.
We will establish that, in the context of \cref{eq:biharm5},
the stream function is necessarily single-valued. We also discuss
some numerical issues related to evaluating the stream function.
The second issue to address
is that the solution $\tilde{w}$ only satisfies the original boundary
condition for the value of $\tilde{w}$ up to a constant on
each boundary component.
In fact, for multiply connected domains, the completed double layer
representation is incomplete for the Dirichlet
problem \cref{eq:biharmD1,eq:biharmD2,eq:biharmD3}. We present a remedy for this
issue and provide some physical intuition.
%
%\subsection{Single-valued stream functions}
%To solve the Dirichlet problem \cref{eq:biharmD1,eq:biharmD2,eq:biharmD3},
%it is necessary to compute
%a single-valued biharmonic potential.
%In the case of a multiply connected domain,
%there is no guarantee that a single-valued stream
%function exists for a given velocity field.
%
%Consider the following example.
%Let $(r,\theta)$ denote standard polar coordinates. It is
%easy to verify that the velocity field
%$\bu=\frac{1}{r}\hat{r}$
%solves the equations of Stokes flow in an annulus centered at the
%origin. A stream function for this flow is $w=\theta$, which
%is not single-valued; indeed, there are no single-valued
%stream functions which generate this flow.
%
%Let $D$ be a multiply connected domain with boundary
%$\Gamma = \cup_{i=0}^N \Gamma_i$, as in the previous section.
%We note that the gradient of a stream function is determined
%by the velocity field, i.e.
%
%\begin{equation}
%\nabla w = -\bu^\perp := \begin{pmatrix}
%  - u_2 \\ u_1
%\end{pmatrix} \; .
%\end{equation}
%Therefore, a velocity field has single-valued stream
%functions if and only if $\bu^\perp$ is conservative.
%Using standard results from multivariable calculus,
%we can  characterize such flows.
%
%\begin{proposition}
%Suppose that $\bu$ is a divergence-free velocity
%field which is $C^1$ on $D$ and continuous on
%$D\cup \Gamma$. The field $\bu^\perp$
%is conservative if and only if
%
%\begin{equation}
% \int_{\Gamma_i} \bu\cdot \bn \, dS
%= 0 \quad i=0,1,\ldots N \, .
%\label{eq:NecSuffStreamFuncExistence}
%\end{equation}
%
%\end{proposition}
%
%The equalities \cref{eq:NecSuffStreamFuncExistence}
%constitute $N$ linearly independent constraints on the
%boundary data because the divergence-free condition
%\cref{eq:MassConservation} implies that
%$\int_{\Gamma} \bu \cdot \bn \, dS = 0$.
%It turns out that
%these conditions are satisfied when the Dirichlet
%problem is recast as a Stokes flow \cref{eq:biharm5}, as
%it is easily verified that
%\begin{equation}
% \int_{\Gamma_i} \bu\cdot \bn \, dS
%= \int_{\Gamma_i} \frac{\partial f}{\partial \tau} \, dS =0 \, .
%\end{equation}
%Thus, any stream function $\tilde{w}$ obtained for the Stokes
%flow \cref{eq:biharm5} is necessarily single-valued.
%% NOT GOING THIS WAY ANYMORE
%%It can be shown using the Helmholtz Hodge decomposition that the
%%conditions \cref{eq:NecSuffStreamFuncExistence} are sufficient for
%%the existence of a single-valued stream function.
%
%\subsection{Evaluating the stream function} \label{subsec:stream}
%
%Given compatible boundary data for the velocity field $\bu$,
%the completed double layer representation for Stokes flow
%\cref{eq:IntRepStokes} guarantees the existence of
%a solution density $\boldsymbol\mu$
%and a corresponding stream function $\tilde{w}$. The Goursat
%function formula
%for $\tilde{w}$, see \cref{sec:stokeslayer},
%is necessarily single-valued, as explained above,
%but it is not immediately obvious from the formula
%that this should be true.
%
%The difficulty in the representation of $\tilde{w}$
%comes from the part
%of the stream function corresponding to the double layer potential
% \cref{eq:wDL}. The
%second term in the expression for the double layer potential is
%\begin{equation}
% v_1(z) = \mbox{Re}\left[\frac{1}{4\pi i}\int_{\Gamma}\left(
%\overline{\rho\left(\xi\right)}d\xi+\rho\left(\xi\right)
%\overline{d\xi}\right)\log\left(\xi-z\right)\right] \, .
%\end{equation}
%To compute this term, in a na\"{i}ve numerical implementation,
%the question of which is the appropriate branch of the
%logarithm to use would arise at many steps.
%To avoid this complication, it is possible instead to compute
%$v_1$, up to a constant, as the harmonic conjugate of the
%function
%\begin{equation}
% v_2 = \frac{1}{4\pi}\int_{\Gamma}\left(\overline{\rho\left(\xi\right)}
%d\xi+\rho\left(\xi\right)\overline{d\xi}\right)\log
%\left(\left|\xi-z\right|\right) \, . \label{eq:harmconjg}
%\end{equation}
%
%We will use this approach to evalute $v_1$ numerically.
%As a result of the Cauchy-Riemann equations, the
%harmonic conjugate of $v_2$, satisfies the following
%Neumann problem for the Laplace equation:
%\begin{align}
% \Delta v_1 &= 0 &\quad x\in D \, ,\\
% \frac{\partial v_1}{\partial n} &= -\frac{\partial v_2}{\partial \tau}
%&\quad x\in\Gamma \, .
%\end{align}
%It is possible then to use standard integral equation
%methods to compute $v_1$.
%
%Let $v_1 = S^L_\Gamma \sigma$, where $S^L_\Gamma \sigma$
%is the single layer potential for Laplace's equation, given by
%\begin{equation}
%S^L_\Gamma\sigma (\bx) = -\frac{1}{2\pi} \int_{\Gamma} \log
%\left|\bx - \by \right|\sigma(\by)\, dS \left(\by \right) \, ,
%\end{equation}
%where $\sigma\in \mathcal{X}= C^{0,\alpha}\left(\Gamma\right)$, for
%some $\alpha \in (0,1)$, is an unknown density
%(see \cite{kress1999linear, guenther1988partial}).
%Imposing the Neumann boundary conditions results in the
%following boundary integral equation for $\sigma$:
%\begin{align}
%-\frac{\partial v_2}{\partial \tau} (\bx) &= \frac{1}{2} \sigma \left(\bx\right) - \frac{1}{2\pi}\oint_{\Gamma}
%\frac{\partial}{\partial n_x}
%\log \left| \bx - \by \right |\sigma(\by)\, dS \left(\by \right)
%\, , \\
%-\frac{\partial v_2}{\partial \tau} &= \left( \frac{1}{2}I_{\mathcal{X}} + K^L_\Gamma  \right) \sigma
%\, ,
%\label{eq:BlockSystemRow2P1tmp}
%\end{align}
%where the operator $K^L_\Gamma$ is compact, so that the integral equation
%is second kind.
%For a derivation of this result, see \cite{kress1999linear}.
%
%It is well known that the operator $\frac{1}{2}I_{\mathcal{X}} + K^L_\Gamma$
%has a one dimensional null space. Thus,
%we choose to solve the above integral equation subject to
%the constraint $\int_{\Gamma} \sigma \, dS = 0$.
%Furthermore, it is known that solving the Neumann problem subject
%to the above constraint is equivalent
%to solving
%\begin{align}
%\left( \frac{1}{2}I_{\mathcal{X}} + K^L_\Gamma  + W_\Gamma \right) \sigma
%= -\frac{\partial v_2}{\partial \tau}
%\label{eq:BlockSystemRow2P1}
%\end{align}
%where $W_\Gamma\sigma = \int_{\Gamma} \sigma \, dS$.
%To prove this, we need
%the following property of
%the Green's function for Laplace's equation.
%\begin{align}
%-\frac{1}{2\pi}\oint_{\gamma} \frac{\partial}{\partial n_{\by}}
%\log \left|\bx - \by\right|\, dS \left(\by \right)
%&=-\frac{1}{2}\quad\bx\in\gamma \, , \label{eq:DLP2}
%\end{align}
%Using the above property, it follows that
%\begin{align}
% \int_{\Gamma} K^L_\Gamma \sigma \, dS(\x)
%= -\frac{1}{2}\int_{\Gamma} \sigma \left(\bx\right) \, dS(\x)
%\label{eq:KLint}
%\end{align}
%Integrating equation \cref{eq:BlockSystemRow2P1} and using
%equation \cref{eq:KLint}, we get
%\begin{align}
%\int_{\Gamma} \left( \frac{1}{2}I_{\mathcal{X}} +
%K^L_\Gamma  + L \right) \sigma \, dS(\x)
%&= -\int_{\Gamma} \frac{\partial v_2}{\partial \tau} \, dS(\x) \\
%\left|\Gamma\right| L\sigma  &= 0
%\end{align}
%which proves the result.

\subsection{Making the representation complete}
As mentioned above, the solution $\tilde{w}$ of
the auxiliary oscillatory Stokes problem \cref{eq:biharm5}
only satisfies the boundary conditions of the
original Dirichlet problem \cref{eq:biharmD1,eq:biharmD2,eq:biharmD3}
up to a constant on each boundary component. For a simply
connected domain, this constant can be recovered from the
fact that adding an arbitrary constant to a stream function
does not change the velocity field.
Thus, in simply connected domains, there is an equivalence in the
solutions of \cref{eq:biharm5} and \cref{eq:biharmD1,eq:biharmD2,eq:biharmD3}.

To analyze the case of a multiply connected domain,
we first consider
radially symmetric solutions on an annulus centered at the
origin. Let $w\left(r\right)$ be a
radially symmetric biharmonic-Stokes
potential. Then $w(r)$ solves the ordinary differential
equation (ODE)
\begin{equation}
\Delta_{r} (\Delta_{r} + k^2)w = 0 \, ,
\end{equation}
where $\Delta_{r}$ is the radial component of the Laplacian in polar
coordinates given by
$$
\Delta_{r} = \frac{1}{r}\frac{d}{dr} r \frac{d}{dr} \, .
$$
Four linearly independent solutions of this ODE are $1$, $\log{(r)}$,
$J_{0}(kr)$, and $H_{0}(kr)$,
where $J_{0}$ is the order zero Bessel function of the first kind, and $H_{0}$ is the order zero Hankel function of the first kind.
For each solution, we can compute the associated velocity field
$\bu = \nabla^{\perp} w$. By construction, $\bu$ satisfies
the continuity condition \cref{eq:MassConservation}.
For the momentum equation \cref{eq:StokesFlowEq} to be
satisfied, we need that $\Delta\bu + k^2 \bu$ is a conservative
vector field, which
is equivalent to the condition that
$\int_{\gamma} (\Delta\bu + k^2 \bu) \cdot d\bl = 0$
for any closed
loop $\gamma$ in the annulus. For $\tilde{w}=1$, $J_{0}(kr)$, and $H_{0}(kr)$,
the corresponding velocity field satisfies
$\Delta \bu + k^2 \bu=0$ so that $\Delta \bu$
is trivially a conservative vector field.
On the other hand, for $\tilde{w}=\log{(r)}$, the corresponding velocity field
satisfies $\Delta \bu + k^2 \bu=\frac{k^2}{r} \hat{\theta}$.
By considering a curve $\gamma$ encircling the origin,
we see that $\Delta \bu + k^2 \bu$ is not a conservative vector
field and that any pressure for the velocity field
associated with $\log\left(r\right)$ is not single-valued.
%
%The above analysis
%can be extended to show that any solution of the biharmonic equation
%of the form $\log{|r-r_{j}|}$ where $r_{j}$ is located in one of the holes of a multiply connected
%domain.
%cannot be represented as a Stokes velocity field.
%In simply connected
%domains, since there are no holes, corresponding to each solution of the modified biharmonic equation with clamped plate
%boundary conditions, there exists a solution to . For multiply connected domains with genus $N$,
%the set of stream functions for Stokes velocity fields
%misses an $N$ dimensional space of solutions, corresponding
%to biharmonic charges located in the holes of the domain.
%Following this reasoning, we obtain a complete representation
%for biharmonic potentials on multiply connected domains
%by adding $N$ charges, one per each hole of the domain,
%to the representation for $w$. The details of this approach, and
%the proof that it is sufficient, is in the next section.


\section{Our problem}

Part of our problem is the detection and classification of faults in the HVAC system of the SE2 building in the Merced campus of the University of California. Such system has an approximate of 2000 sensor and we take samples each 5 minutes. It is our intention to use data from the past 2 years to train our model, this gives us a size for the training set of $n = 210240$ training samples. Thus, lets assume that our ANN will have the following topology: $m = 2000$ for the 2000 input features (this is just a rough estimate since we will apply dimensionality reduction and feature extraction techniques to reduce this number as much as we can), $k = 4$ layers for an ANN of 1 input layer, 2 hidden layers and 1 output layer. In this case we assume that the number of output layers (fault classes) $o$ is much smaller than the number of input layers, we expect a number on the order of $100$ different types of faults, as for the number of neurons on each hidden layer we will set it arbitrary to $2000$ (the number of input neurons) as it has been shown that the number of hidden layers for other classification/regression problems is usually smaller than the number of input layers, thus, setting the number of hidden layers equal to the number of input layers will help us set a boundary on the number of computations.

Using the above information and eq. \eqref{eq:ann_complexity1} the asymptotic complexity of our ANN will be $O(210240 * 2000 * 2000^2 * 100) = O(1.6819 \times 10^{17})$ FLOPS, this neglecting the computational time spent on the minimization problem inherent to the computation of the weights for the ANN.

\section{Conclusions}

Based on the estimate of FLOPS our equipment can handle which within the order of $3 \times 10^{19}$ FLOPS/month and the complexity for the training of our modeled ANN we estimate our which is of the order of $O(1.6819 \times 10^{17}$ FLOPS we conclude our system can do the training for our model in approximately one month of computations, provided that we take the most of our equipment, such as the pipe-lining and parallel computing.


\section*{Problem 1}
\begin{problem}{A. Physics of a carbonated drink}

Carbonated bottled drinks contain dissolved carbon dioxide $CO_2$, which tends to be in the gaseous phase under normal condition. When you pour a carbonated drink into a glass, you can observe numerous small bubbles rising from the bottom. Those bubbles grow in size as they ascend and their speed changes as they travel upward.
\end{problem}

\begin{subpr}{A1. \hfill 0.5pts.}
Consider a bubble as a perfect sphere with initial radius ${r_0 = 2.0\cdot 10^{-4}m}$ located at the bottom of a glass filled with fluid of density ${\rho_f = 1.0\cdot 10^{3}kg/m^3}$ up to the height ${h = 0.2\;m}$. When the bubble reaches the surface between fluid and air, its radius increases two times from initial size near bottom of the glass.
\vspace{2mm}
\begin{center}
    \includegraphics[height=4cm]{Images/1_bbl_theory_carbonated.PNG}
\end{center}
\vspace{2mm}

Estimate amount of energy ${\Delta E}$ dissipated due to viscous effects during ascending of that bubble. For calculations use acceleration due to gravity as ${g = 9.8\;m/s^2}$. Assume that surface tension effects are negligibly small
\end{subpr}
\begin{subpr}{A2. \hfill 0.2pts.} Estimate velocity ${v_h}$ of the bubble at the distance ${h = 0.2\;m}$ from bottom of the glass
\\
For calculations use dynamic viscosity of carbonated drink as ${\eta_f = 1.0\cdot 10^{-3}Pa\cdot s}$
\end{subpr}
\begin{subpr}{A3. \hfill 1.3pts.}
Estimate acceleration ${a_h}$ of the bubble at the distance ${h = 0.2\;m}$ from bottom of the glass. Assume that surface tension effects are negligibly small.
\end{subpr}

\begin{problem}{B. Sonoluminescence}
Sonoluminescence is a wonderful phenomena where an external acoustic wave causes disturbances in the fluid, leading to the growth and rapid collapse of bubbles in the liquid. During that bubble collapse, content of the bubble become heated to extremely high temperatures, which causes excitation of the gas with emission of light.
\begin{center}
\includegraphics[height=4cm]{Images/5_bbl_theory_sonolum1.PNG}
\end{center}
\begin{center}
Photo of single bubble sonoluminescence (light blue spot in the center of glass sphere filled with fluid). Credit to Suslick K.
\end{center}
\vspace{3mm}
In this section of the problem you will have an opportunity to describe some key aspects of the sonoluminescence process
\end{problem}
\begin{subpr}{B1. \hfill 1.2pts.}
Consider a large volume of incompressible fluid of density ${\rho}$, which is initially at rest. That container with fluid has a small spherical cavity (an empty space without gas or fluid) of initial radius ${R_0}$, which starts to collapse due to an ambient fluid pressure ${P_0}$.
\begin{center}
\includegraphics[height=3cm]{Images/6_bbl_prob_empty_collapse_velocity1.PNG}
\end{center}
Evaluate velocity ${v_1}$ of the boundary of cavity when radius of the bubble would decrease to some value ${R_1}$. Assume that for such small size of the cavity surface tension and viscous effects can be neglected
\end{subpr}
\begin{problem}{}
As was shown in the previous section velocity of the boundary of an empty cavity would tend to increase to infinity during collapse of the cavity at small radius ${R\!\rightarrow\! 0}.$ Thus, modeling of an empty cavity is unrealistic, which means that bubble in fluid has to be filled with some gas or vapor. This would create a counterbalance pressure inside the bubble, allowing it to bounce back at some small non-zero radius

$$	***  $$

Sonoluminescence effect occurs when changes in ambient pressure of fluid cause a bubble with noble gas to shrink, which increases temperature of the gas significantly. Usually in experiments with sonoluminescence, fluid pressure is varied as a harmonic function of time. However, here for simplicity, will consider only a rough approximation of the process, with one step of such variations in pressure, when fluid pressure is increased very fast from initial value ${P_0}$ to some new high value  ${P_f = 100P_0}$

\begin{center}
\includegraphics[width=9cm]{Images/7_bbl_theory_velocity_too_high1.PNG}
\end{center}

For the next few questions consider a bubble, which initially has a stationary radius ${R_0 = 1.0\cdot 10^{-4}\;m},$ being filled with a noble gas xenon at temperature ${T_0 = 300.0\;K}$ and pressure ${P_0}$. Shrinkage of bubble due to increased ambient fluid pressure can be accurately modeled with adiabatic process, which can be described with adiabatic constant ${\gamma}$:
$$\gamma = \frac{C_p}{C_v} = \frac{5}{3}$$
where ${C_p}$ and ${C_v}$ are molar heat capacities of xenon at constant pressure and constant volume respectively.
\end{problem}
\begin{subpr}{B2. \hfill 1.2pts.} Determine minimal radius of the bubble ${R_{min}}$ due to increase in ambient fluid pressure from ${P_0}$ to ${P_f = 100P_0}.$ Assume that surface tension and viscous friction effects can be neglected. Also for simplicity assume that presence of water vapor inside the bubble is negligibly small\\
\textbf{Note:}
For your calculations, you are not expected to use manual iterative approach for solving transcendental equations. Instead it could be easier to use calculations in some spreadsheet such as MS Excel, or write a simple script for plotting a complicated function
\end{subpr}

\begin{subpr}{B3. \hfill 0.3pts.} Evaluate maximum temperature ${T_{max}}$ of the noble gas inside cavity during its shrinkage.
\end{subpr}
\begin{problem}{Stability}

It is interesting that a lot of processes for bubbles differ based on the initial equilibrium size of the bubble. If radius of the bubble is larger than certain critical value $R_c$ , then after small disturbances in ambient fluid pressure, the bubble will start growing uncontrollably. Otherwise if initial equilibrium radius of cavity is smaller than certain threshold, it will be stable to small variations of pressure
\end{problem}

\begin{subpr}{B4. \hfill 1.2pts.}

Consider a small bubble with radius ${R_0 = 5.0\cdot 10^{-6}m}$, which is held at pressure ${P_0 = 10^5\;Pa}$ in equilibrium with distilled water. Then fluid pressure was disturbed so that bubble expanded to radius ${R}$.
\begin{center}
\includegraphics[height=3cm]{Images/4_bbl_prob_Blake_pressure_unstable1.PNG}
\end{center}
Calculate critical radius ${R_c}$ for this system, if entire process can be treated as isothermal, with vapor pressure at that temperature equal to ${P_v = 0.5\cdot 10^{5}\;Pa}.$ Coefficient of surface tension of water is ${\sigma = 7.2\cdot 10^{-2}N/m}$
\end{subpr}

\begin{problem}{Small oscillations}

In the next questions, assume that radius of the bubble is less than critical value ${R_c}$ unless otherwise stated. For those smaller bubbles, minor disturbances in ambient fluid pressure would cause small harmonic oscillations with frequency ${\omega}.$ That frequency can be decomposed at three components as

$$\omega^2 = \omega_p^2 + \omega_{\sigma}^2 - \omega_{\eta}^2$$
where ${\omega_p}$ is natural frequency of the bubble inside liquid, neglecting effects of surface tension and attenuation due to viscosity. While ${\omega_{\sigma}}$ and ${\omega_{\eta}}$ are additional correction terms associated with surface tension effects and viscosity respectively.
\end{problem}

\begin{subpr}{B6. \hfill 1.2pts.} Evaluate frequency ${\omega_p}$ of small harmonic oscillations of the bubble of radius ${R_0 = 1.0\cdot 10^{-4}}m$ assuming that surface tension and viscosity effects can be neglected. Consider that oscillating bubble is filled with xenon, which has adiabatic constant ${\gamma = 5/3}.$ The bubble is floating inside distilled water of density ${\rho = 1.0\cdot 10^{3}kg/m^3}.$ Assume that during those oscillations  fluid pressure ${P_f}$ is always close to initial pressure ${P_0}$ inside bubble:

$$P_f\approx P_0 = 1.0\cdot 10^{5}Pa$$

For small ${z\ll 1}$ can be used following approximation
$$(1-z)^n\approx 1-nz + \frac{n(n-1)z^2}{2}$$
\end{subpr}
\begin{subpr}{B7. \hfill 0.2pts.}
It is known that for larger bubbles with initial radius ${R_0 = 1.0\cdot 10^{-4}m}$ ratio between components of frequencies related to surface tension ${\omega_{\sigma}}$ to the frequency ${\omega_p}$ calculated with neglecting surface tension and viscous effects is
$$z_{\sigma 0} = \frac{\omega_{\sigma0}}{\omega_{p0}} = 0.1$$
Determine ratio ${z_{\sigma 1} = \omega_{\sigma 1}/\omega_{p1}}$ for a smaller bubble with radius ${R_1 = R_0/10}$
\end{subpr}
\begin{subpr}{B8. \hfill 0.3pts.}
It is known that for bubble with radius ${R_0 = 1.0\cdot 10^{-4}m}$ ratio between components of frequency related to viscous forces ${\omega_{\eta}}$ to frequency ${\omega_{p}}$ calculated by neglecting surface tension and viscous effects is
$$z_{\eta 0} = \frac{\omega_{\eta 0}}{\omega_{p0}} = 4.0\cdot 10^{-4}$$
Estimate similar ratio ${z_{\eta 1} = \frac{\omega_{\eta 1}}{\omega_{p1}}}$ for bubble with smaller radius ${R_1 = R_0/10}$
\end{subpr}
\begin{problem}{Forces due to variable pressure}
In case when pressure is spatially varied, effective force applied on the object inside such pressure field would be non-zero. For example, hydro static pressure varies with depth, creating buoyancy force in the vertical direction. In this section we will describe horizontal forces exerted on bubbles in fluid in the presence of variable pressure
\end{problem}
\begin{subpr}{B9. \hfill 0.2pts.}

Consider a small cylinder with a height ${h}$ and cross-sectional area is ${S}$ placed inside a fluid, which has spatially varied pressure ${P}$ characterized with coordinate ${x}$ as
	$$P = P_0(1+c_1x^2)$$
where ${c_1}$ is some known constant
\begin{center}
\includegraphics[height=3cm]{Images/17_bbl_prob_cylinder_cx21.PNG}
\end{center}
Center of the cylinder is located at the coordinate ${x_0\gg h},$ while main axis of the cylinder is parallel to the ${X}$ axis. Evaluate effective horizontal force ${F_x}$ exerted on this cylinder from fluid
\end{subpr}

\begin{subpr}{B10. \hfill 0.5pts.}
A rigid sphere of radius ${R}$ is placed inside a fluid, with variable pressure, which changes with horizontal coordinate ${x}$ as

$$P = P_0(1+c_1x^2)$$
where ${c_1}$ is some known constant\\
Center of the sphere is located at coordinate ${x_0\ll R}$. Determine effective horizontal force ${F_x}$ applied to the sphere from the surrounding fluid
\end{subpr}

\begin{subpr}{B11. \hfill 0.2pts.}
Now consider a similar case with a rigid sphere of radius ${R}$ placed inside fluid, where pressure ${P_f}$ changes with horizontal coordinate ${x}$ and time ${t}$ as

$$P_f = P_0 + c_2\sin kx\cos\omega t$$
where ${P_0}$, ${c_2},$ ${k}$ and ${\omega}$ are known constants\\
Evaluate average effective horizontal force ${\left<F_x\right>}$ exerted at the sphere, if known that radius of the sphere is much smaller than parameter ${1/k}$
\end{subpr}

\begin{subpr}{B12. \hfill 0.3pts.} Now consider an oscillating bubble in a standing wave, with fluid pressure described as a function of horizontal coordinate ${x}$ and time ${t}$ as
$$P = P_0 + c_2\sin kx\cos\omega t$$
where ${P_0}$, ${c_2}$, ${k}$, ${\omega}$ and ${\phi}$ are some positive constants\\
It can be shown that in such variable pressure, radius of the bubble ${R}$ would change as a function of time as
$$R = R_0 - \delta \sin kx\cos(\omega t +\phi)$$
where ${R_0}$ and ${\delta}$ are some positive constants\\
Let's call bubbles with size larger than critical radius  ${R_c}$ as "Large" and bubbles with radii smaller than ${R_c}$ as "Small". It can be shown that parameter ${\phi}$ is equal ${0}$ or ${\pi}$\\
Let ${\phi_L}$ and ${\phi_S}$ are coefficients ${\phi}$ for "Large" and "Small" bubbles respectively. Briefly justify in 1-2 sentences or in a few short equations, which values ${0}$ or ${\pi}$ should we use for parameters ${\phi_L}$ and ${\phi_S}$
\end{subpr}
\begin{subpr}{B13. \hfill 0.7pts.}
Consider bubble described in the question ${B12}$ with small oscillations such that ${\delta \ll R_0}$. Determine average horizontal force ${\left<F_x\right>}$ exerted at this oscillating bubble
\end{subpr}
\begin{subpr}{B14. \hfill 0.5pts.}
Consider that fluid pressure changes with time ${t}$ and coordinate ${x}$ as
$$	P = P_0 + c_3\sin kx\cos\omega t$$
This is an equation of a standing wave, which can be characterized with "nodes" and "antinodes"
\begin{center}
\includegraphics[height=3cm]{Images/22_bbl_prob_nodes_antinodes_grouping1.PNG}
\end{center}
Describe briefly in ${1-2}$ sentences where would gather "Large" and "Small" bubble in that standing wave (either close to nodes or in the vicinity of antinodes)
\end{subpr}

\clearpage


\chapter{Computers}

We recommend that you skip this chapter
until you use a computer to do physics calculations.

\section{Working with limited precision}

Computers can only store finitely many digits.
Computers can only represent a \emph{finite} subset of \(\Real\).
Computers can't represent most real numbers.

The take-home message is:
\emph{Mixing numbers with varying exponents increases errors}.

To understand these phenomena, we have to understand how computers represent numbers.

\section{Understanding how computers represent numbers}

There are many ways:
\UnorderedList{
\item floating-point integers
\item fixed-point integers
\item arbitrary-precision integers
\item symbolic representations
\item one's-complement signed integers
\item two's-complement signed integers
\item unsigned integers
\item sign-magnitude
}

The representation depends on the program.

Each representation has its benefits and drawbacks.

If you are doing physics with computers,
then there's a high chance that the numbers you see are \emph{IEEE 754 double-precision floating-point integers},
which are often shortened to \emph{doubles}.%
\footnote{\url{https://en.wikipedia.org/wiki/Double-precision_floating-point_format}}

\footnote{\url{https://en.wikipedia.org/wiki/Binary_number}}%
\footnote{\url{https://en.wikipedia.org/wiki/Signed_number_representations}}

\subsection{Understanding \emph{doubles}}

A \emph{double} represents a number as the base-2 scientific notation
\(
(-1)^s \times 1 . d_{51} d_{50} d_{49} \ldots d_2 d_1 d_0 \times 2^{p - 1023}
\)
Note that the first significant digit is always one and is not stored in memory.
Each \(d_k\) is a binary digit (zero or one).

Bit 63 is the sign bit \(s\):
0 means positive;
1 means negative.

Bits 62\textendash{}52 (11 bits) are the \emph{biased exponent} \(p\).

Bits 51\textendash{}0 (52 bits) are called the \emph{significand} or \emph{mantissa}.
Bit 51 is \(d_{51}\), and so on; bit 0 is \(d_0\).

\subsection{Understanding the problem with doubles}

If a fraction has a denominator that is not a power of two,
then a double can't represent the fraction exactly.
For example, a double can't represent even a simple fraction such as \(1/3\) exactly
because the base-2 expansion of \(1/3\) doesn't terminate,
in the same way that the decimal expansion of \(1/3\) (\(0.333\ldots\)) doesn't terminate.

A double has only 53 significant binary digits.
A multiplication of two 53-digit numbers may produce up to a 106-digit number.
The result is rounded to 53 digits; thus at most 53 digits are lost.

There is also a website\footnote{\url{http://floating-point-gui.de/}} that explains the issue.

\subsection{Understanding computer algebra systems}

Computer algebra systems can represent numbers like \(\sqrt{2}\) and \(1/3\) exactly because
it does not represent numbers as strings of digits.
It stores \(\sqrt{2}\) as something like \verb@(sqrt 2)@.
It does not evaluate \(\sqrt{2}\) to \(1.4142\ldots\) before storing it.

\subsection{Understanding ulp: units of least precision}

The standard is IEEE 754.

For example, in IEEE 754 double-precision floating-point integers,
\( 2^{53} + 1 = 2^{53} \).

In some browsers, you can verify this. Press Ctrl+Shift+J to open its Console,
and then enter \verb@Math.pow(2,53)@, and then enter \verb@Math.pow(2,53)+1@,
and see that they give the same number.

\section{Solving a system of linear equations}

You can use GNU Octave.

Use left division:
\begin{verbatim}
A \ C
\end{verbatim}

This also works for Matlab.

\section{Related fields of study}

Related fields of study are
\emph{scientific computing} (also known as \emph{computational science})
and \emph{numerical analysis}.%
\footnote{\url{https://en.wikipedia.org/wiki/Computational_science}}%
\footnote{\url{https://en.wikipedia.org/wiki/Numerical_analysis}}

\section{Using an equation to program a computer}


\section{Manipulate Conductivity in Quantum Hall System}

Now using the relations derived in Eq. \eqref{5.34} and Eq. \eqref{7.17} we can derive that
\begin{equation} \label{8.1}
  \Gamma_n = \qty(\frac{\hbar }{2\tau(\varepsilon_n)}) =
  \frac{1}{2}\Gamma^{00}_{N}(\tilde{I}),
\end{equation}
and using \eqref{5.32}, this can be re-write as follows
\begin{equation} \label{8.2}
   \Gamma_n = \frac{1}{2} \Lambda_N (\tilde{I}) \Gamma^{00}_{N=0}\big|_{E=0}.
\end{equation}
Let $\Gamma^{00}_{N=0}\big|_{E=0} = \Gamma_0$ and this leads to
\begin{equation} \label{8.3}
   \gamma_n = \frac{\Gamma_n}{\hbar \omega_0} =
   \frac{\Gamma_0}{2\hbar \omega_0} \Lambda_N (\tilde{I})
\end{equation}
Let $\alpha \equiv \flatfrac{\Gamma_0}{2\hbar \omega_0}$ and we can obtain
\begin{equation} \label{8.4}
   \gamma_n = \alpha \Lambda_N (\tilde{I})
\end{equation}
Thus we can write the transverse conductivity as a function of fermi energy and intensity
\begin{equation} \label{8.5}
  \begin{aligned}
    \sigma^{xx}(X_F,\tilde{I}) & =
    \frac{e^2}{\pi\hbar A}
    \sum_{n}
    \frac{\qty(n+1)}{\alpha^2 \Lambda_n \Lambda_{n+1}}
    \qty[
      \frac{1}
      {
        1 + \qty(\frac{X_F - n -1}{\alpha\Lambda_n})^2
      }
    ]
    \qty[
      \frac{1}
      {
        1 + \qty(\frac{X_F - n}{\alpha\Lambda_{n+1}})^2
      }
    ].
  \end{aligned}
\end{equation}
Then we can define a normalized transverse conductivity as follows
\begin{equation} \label{8.6}
  \begin{aligned}
    \widetilde{\sigma}^{xx}(X_F,\tilde{I}) =
    \flatfrac{\sigma^{xx}}{\sigma_0} =
    \sum_{n}
    \frac{\qty(n+1)}{\alpha^2 \Lambda_n \Lambda_{n+1}}
    \qty[
      \frac{1}
      {
        1 + \qty(\frac{X_F - n -1}{\alpha\Lambda_n})^2
      }
    ]
    \qty[
      \frac{1}
      {
        1 + \qty(\frac{X_F - n}{\alpha\Lambda_{n+1}})^2
      }
    ],
  \end{aligned}
\end{equation}
where $\sigma_0 = \flatfrac{e^2}{(\pi\hbar A)}$.

Above parameters can be evaluated for our experiment variables($\Gamma_0 = 0.24 \text{meV}$) as follows
\begin{equation} \label{8.2}
  \Gamma_n(\tilde{I}) = 0.12\times
  \qty[
  \frac
  {\int_{-\infty}^{\infty} d {k}_1 \;
  J_0^2\qty(2.090\sqrt{\tilde{I}}\times{k}_1)
  \qty|
  \int_{-\infty}^{\infty} d{k}_2 \;
  \tilde{\chi}_{N}\qty(2.342 \times k_2)
  \tilde{\chi}_{N}\qty(2.342 \times \qty[{k}_1 - {k}_2])|^2}
  {\int_{-\infty}^{\infty} d {k}_1 \;
  \qty|
  \int_{-\infty}^{\infty} d{k}_2 \;
  \tilde{\chi}_{0}\qty(2.342 \times k_2)
  \tilde{\chi}_{0}\qty(2.342 \times \qty[{k}_1 - {k}_2])|^2}
  ]^{1/2} \text{meV}
\end{equation}

\noindent
In addition we can calculate the cyclotron energy as
\begin{equation} \label{8.3}
  \hbar\omega_0 = 1.95663 \;\text{meV}
\end{equation}
and
\begin{equation} \label{8.4}
  \gamma_n = \frac{\Gamma_n}{\hbar \omega_0} = 0.06133 \times \Lambda_n(\tilde{I}) \approx 0.061 \Lambda_n(\tilde{I})
\end{equation}

\noindent
Now we can use this into our conductivity expression derived in Eq. \eqref{7.23} and present the normalized transverse conductivity as a function of fermi energy and intensity of the dressing field
\begin{equation} \label{8.5}
  \begin{aligned}
    \sigma^{xx}(X_F,\tilde{I}) & =
    \sum_{n}
    \frac{\qty(n+1)}{0.0037\Lambda_n \Lambda_{n+1}}
    \qty[
      \frac{1}
      {
        1 + \qty(\frac{X_F - n -1}{0.06\Lambda_n})^2
      }
    ]
    \qty[
      \frac{1}
      {
        1 + \qty(\frac{X_F - n}{0.06\Lambda_{n+1}})^2
      }
    ]
  \end{aligned}
\end{equation}
where
\begin{equation} \label{8.6}
    \Lambda_n  =
    \qty[
    \frac
    {\int_{-\infty}^{\infty} d {k}_1 \;
    J_0^2\qty(2.090\sqrt{\tilde{I}}\times{k}_1)
    \qty|
    \int_{-\infty}^{\infty} d{k}_2 \;
    \tilde{\chi}_{n}\qty(2.342 \times k_2)
    \tilde{\chi}_{n}\qty(2.342 \times \qty[{k}_1 - {k}_2])|^2}
    {\int_{-\infty}^{\infty} d {k}_1 \;
    \qty|
    \int_{-\infty}^{\infty} d{k}_2 \;
    \tilde{\chi}_{0}\qty(2.342 \times k_2)
    \tilde{\chi}_{0}\qty(2.342 \times \qty[{k}_1 - {k}_2])|^2}
    ]^{1/2}.
\end{equation}
\hfill$\blacksquare$

\noindent
Now using Eq. \eqref{8.6} we can calculate the normalized broadening of each Landau levels as given in Table \ref{tab:8.1}.

\begin{table}[ht!]
\begin{center}
\begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c| }
 \hline
 ${\tilde{I}}$
 & $n=0$ & $n=1$ & $n=2$ & $n=3$ & $n=4$ & $n=5$ & $n=6$ & $n=7$ & $n=8$ & $n=9$
 & $n=10$
 \\ [0.5ex] \hline\hline
 0  & $1.0000$ & $0.8660$ & $0.8004$ & $0.7578$ & $0.7266$
 & $0.7021$ & $0.6821$ & $0.6653$ & $0.6507$ & $0.6380$& $0.6267$ \\ \hline
 1  & $0.8546$ & $0.7037$ & $0.6502$ & $0.6114$ & $0.5810$
 & $0.5584$ & $0.5416$ & $0.5280$ & $0.5160$ & $0.5050$ & $0.4948$ \\ \hline
 4  & $0.6875$ & $0.6345$ & $0.5902$ & $0.5528$ & $0.5307$
 & $0.5118$ & $0.4958$ & $0.4840$ & $0.4730$ & $0.4629$  & $0.4547$ \\ \hline
 9  & $0.5936$ & $0.5539$ & $0.5333$ & $0.5180$ & $0.5006$
 & $0.4812$ & $0.4685$ & $0.4564$ & $0.4469$ & $0.4377$  & $0.4305$ \\ \hline
\end{tabular}
\caption {\label{tab:8.1} $\Lambda_n$ values for each Landau level by changing  applied dressing field intensity ($\tilde{I}$). }
\end{center}
\end{table}

\noindent
Using above values we can analyse the changes can be done to the transverse conductivity using apllied dressing field. As given in Figure \ref{fig:8.1} and \ref{fig:8.2} we can manipulate the conductivity using external dressing field. When the applied field's intensity increase the broadening of energy bands of Landau levels get reduced the conductivity also get decrease all the regions except the peak point of the energy band. Using this manipulation we can filter the conductivity which is change with the Fermi level. Since Fermi level can be change with the applied gate voltage of the material this can be used as a 2D switch for optoelectonic applications. Using the applied dressing field we can fine tune the switching mechanism.

\begin{figure}[ht!]
  \centering
  \includegraphics[scale=1]{figures/fig81.pdf}
  \caption{Normalized transverse conductivity against Fermi level ($X_F$) with different intensities ($\tilde{I}$) of dressing field.}
  \label{fig:8.1}
\end{figure}
\begin{figure}[ht!]
  \centering
  \includegraphics[scale=1]{figures/fig82.pdf}
  \caption{$3$rd Landau level's normalized transverse conductivity against Fermi level ($X_F$) with different intensities ($\tilde{I}$) of dressing field.}
  \label{fig:8.2}
\end{figure}