First commit

.gitignore

+*.log
+*.toc
+*.blg
+*.bbl
+*.aux
+*.out
+*.synctex.gz
+discussions

TODO

+=== 15 Juni 2022 ===
+
+- 2nd order diagrams
+- Luttinger results for self-energy
++ Compare with Romans limit: Wigner crystal works for small occupation
+- 4th cumulant - Ivanov paper J Phys. A: Math. Theor. 46 (2013) 085003
+- Luttinger theorem
+
+
+=== 8 June 2022 ===
+
++ Physical density
++ Calculate Eq. 78 for p_F L ~ 1 though p_F << 1
+- Compare with Romans limit
+
+
+=== 17 May 2022 ===
+
+- Physical density
++ Expansion over lambda
++ L dependence 
++ Summarize rezults
++ V_1 to K correspondence
++ Prove all cumulants are real
+- Check charge density wave in Giamarchi
\ No newline at end of file

determinant.tex

+% !TEX TS-program = pdflatex
+% !TEX encoding = UTF-8 Unicode
+% !TEX spellcheck = English UK
+
+\documentclass[10pt,fullpage,amsmath,amsthm]{amsart}
+\usepackage{fullpage,graphicx}
+\usepackage[colorlinks=false,urlcolor=blue,pdfborderstyle={/S/U/W 1}]{hyperref}
+\usepackage[plain]{fancyref}
+\usepackage{array,multirow}
+%\usepackage[dvipsnames]{xcolor}
+
+\title{Electrons counting}
+\author{Oleks\'iy Kash\'uba}
+\date{\today }
+
+
+
+
+%\def\fileversion{\InputIfFileExists{.git/current_version}{\it version~}{\it unversioned}%
+\def\fileversion{\it version~0.4\,%
+%(\pdffilesize{\jobname.tex} bytes)}
+\if\pdfstrcmp{\pdffilesize{\jobname.tex}}{18119}0{(original)}\else{(changed)}\fi}
+
+\DeclareMathOperator{\Tr}{Tr}
+\DeclareGraphicsExtensions{.pdf,.ai}
+\DeclareGraphicsRule{.ai}{pdf}{.ai}{}
+
+\makeatletter
+\let\@paragraph\paragraph
+\def\paragraph#1{\@paragraph{•#1}}
+\makeatother
+
+\newtheorem{theorem}{Theorem}
+\newcommand*{\fancyrefthmlabelprefix}{thm}
+\fancyrefaddcaptions{english}{\newcommand*{\Frefthmname}{Theorem}\newcommand*{\frefthmname}{\MakeLowercase{\Frefthmname}}}
+\frefformat{plain}{\fancyrefthmlabelprefix}{\MakeUppercase{\frefthmname}\fancyrefdefaultspacing#1}
+\Frefformat{plain}{\fancyrefthmlabelprefix}{\MakeUppercase{\Freffigname}\fancyrefdefaultspacing#1}
+
+
+\begin{document}
+
+
+\maketitle
+{\parskip-\baselineskip\par\noindent {\hfill\raisebox{-\baselineskip}[0pt][0pt]{\fileversion}\hfill}}
+
+
+
+
+%\tableofcontents
+
+
+\section{Intro}
+
+Let us calculate a trace for a non-interacting system
+\begin{equation}
+\Tr e^\mathcal{C}, \qquad \mathcal{C} = \sum_{mn}c_{m}^{\dagger} C_{mn}c_{n}
+\end{equation}
+Using the Schur decomposition~\cite{schdec} on a square complex matrix $C=U(\mathrm{diag}(\mu_{p}) + K)U^{+}$ where $K$ is upper triangular ($K_{pq}=0$ for $p\leq q$), and introducing new fermionic (since $U$ is unitary) operators $c_{p} = \sum_{n}U_{np}c_{p}$ we see that 
+\begin{equation}
+\Tr e^\mathcal{C} = \Tr e^{\sum_{p}\mu_{p}c_{p}^{\dagger}c_{p} + \sum_{p>q}K_{pq}c_{p}^{\dagger} c_{q}} = \Tr e^{\sum_{p}\mu_{p}c_{p}^{\dagger}c_{p}}
+=
+\Tr \prod_{p}e^{\mu_{p}c_{p}^{\dagger}c_{p}}
+=
+\begin{cases}
+\prod_{p} (1+ e^{\mu_{p}}) & \text{for fermions} \\
+\prod_{p} \left(\sum_{n=0}^{\infty}e^{\mu_{p}n}\right) & \text{for bosons} \\
+\end{cases}
+\end{equation}
+This gives us
+\begin{align*}
+&\Tr e^\mathcal{C} = \prod_{p} (1+ e^{\mu_{p}}) = \det (1+e^{\mathrm{diag}(\mu_{p}) + K}) = \det (1+e^C) 
+&& \text{for fermions}
+\\
+&\Tr e^\mathcal{C} = \prod_{p} \frac{1}{1- e^{\mu_{p}}} = \det (1-e^{\mathrm{diag}(\mu_{p}) + K})^{-1} = \det (1-e^C)^{-1} 
+&& \text{for bosons}
+\end{align*}
+or combining together,
+\begin{equation}
+\Tr \left(e^\mathcal{C}\right) = \det(1-\xi\,e^{C})^{-\xi},
+\end{equation}
+where $\xi=-1$ for fermions and $\xi=+1$ for bosons.
+
+
+\begin{theorem}
+\label{thm:expproddet}
+The product of two exponents satisfies the following relation:
+\begin{equation}
+\Tr \left(e^\mathcal{A}e^\mathcal{B}\right) = \det(1-\xi\,e^{A}e^{B})^{-\xi}.
+\end{equation}
+\end{theorem}
+
+\begin{proof}
+Further we consider fermions.
+We define two operators bilinear in $c^{\dagger}$ and $c$ and look at their commutator:
+\begin{equation}
+\mathcal{A} = \sum_{mn}c_{m}^{\dagger} A_{mn}c_{n}
+\quad
+\mathcal{B} = \sum_{kl}c_{k}^{\dagger} B_{kl}c_{l}, 
+\qquad [ \mathcal{A}, \mathcal{B} ]_{-} = \sum_{mnkl} A_{mn} B_{kl}[c_{m}^{\dagger}c_{n},c_{k}^{\dagger}c_{l}]_{-}
+=\sum_{mn} \left([A,B]_{-}\right)_{mn}c_{m}^{\dagger}c_{n}.
+\label{eq:commut2out1}
+\end{equation}
+For compactness let us introduce the notation 
+\begin{equation}
+A \oplus B = A+B+\frac{1}{2}[A,B]+\frac{1}{12}[A,[A,B]] - \frac{1}{12}[B,[A,B]]+\cdots
+\end{equation}
+and note also that due to the commutation of the bilinear in $c^\dagger$ and $c$ operators we get a relation
+\begin{equation}
+\mathcal{A} \oplus \mathcal{B} = \sum_{mn} \left( A \oplus B\right)_{mn}c_{m}^{\dagger}c_{n}.
+\label{eq:oplus}
+\end{equation}
+Then we use Baker–Campbell–Hausdorff formula~\cite{bakcamhau}, which tells that exists such a matrix $C$, so that $C = A \oplus B$, such that  $e^{A}e^{B}=e^{C}$.
+So we get
+\begin{equation}
+\Tr \left(e^\mathcal{A}e^\mathcal{B}\right) = \Tr \left(e^\mathcal{C}\right) =\det(1+e^{C}) =\det(1+e^{A}e^{B}).
+\end{equation}
+Since \fref{eq:commut2out1} works also for bosons, the same logic can be applied.
+\end{proof}
+
+
+
+
+\begin{theorem}
+\begin{equation}
+\Tr \left(\prod_{i=1}^{N} e^{\mathcal{A}^{(i)}}\right) = \det\left(1-\xi\,\prod_{i=1}^{N} e^{A^{(i)}}\right)^{-\xi}
+\end{equation}
+\end{theorem}
+\begin{proof}
+Let us use the mathematical induction procedure.
+A
+We proved in \fref{thm:expproddet} that for $N=2$ theorem works.
+Let us assume that it works for arbitrary $N\geq 2$, namely exists a matrix $C^{(N)}$ and corresponding bilinear in $c^{\dagger}$ and $c$ operator $\mathcal{C}^{(N)}$, such that
+\begin{equation}
+\Tr \left(\prod_{i=1}^{N} e^{\mathcal{A}^{(i)}}\right) = \Tr \left(e^{\mathcal{C}^{(N)}}\right) =\det(1-\xi \,e^{C^{(N)}})^{-\xi} =\det\left(1-\xi\,\prod_{i=1}^{N} e^{A^{(i)}}\right)^{-\xi}
+\end{equation}
+In order to prove the theorem, we need to show that it works for $N+1$ as well:
+\begin{equation}
+\Tr \left(\prod_{i=1}^{N+1} e^{\mathcal{A}^{(i)}}\right) = \Tr \left(e^{\mathcal{C}^{(N)}}e^{\mathcal{A}^{(N+1)}}\right)
+=\ldots
+\end{equation}
+According to \fref{thm:expproddet} exists a matrix $C^{(N+1)} = C^{(N)}\oplus A^{(N+1)}$ such that
+\begin{multline}
+\ldots=
+\Tr \left(e^{\mathcal{C}^{(N)}}e^{\mathcal{A}^{(N+1)}}\right) =
+\det(1-\xi\,e^{C^{(N+1)}})^{-\xi} =\\=
+\det\left(1-\xi\,e^{C^{(N)}} e^{A^{(N)}}\right)^{-\xi}
+=
+\det\left(1-\xi\,\prod_{i=1}^{N+1} e^{A^{(i)}}\right)^{-\xi}.
+\end{multline}
+Starting at $N=2$ we repeat the logic till the arbitrary value of $N$.
+\end{proof}
+
+
+%\begin{equation}
+%
+%\end{equation}•
+
+
+
+\section{Levitov's determinant}
+
+
+
+Following~\cite{Klich2002} we calculate the trace/determinant
+\begin{equation}
+\Tr\left(e^{-\beta \mathcal{H}} e^{i\lambda \mathcal{Q}}\right)= \det\left(1+ e^{-\beta H} e^{i\lambda Q}\right)
+\end{equation}
+where $H$ is a single-particle Hamiltonian ($N\times N$), while $Q$ is a sites projection matrix
+\begin{equation}
+H_{ij} = 
+\times
+\begin{cases}
+1/2 & |i-j|=1 \\
+0 & \text{otherwise}
+\end{cases}
+\qquad
+Q_{ij} = \delta_{ij}
+\times
+\begin{cases}
+1 & n \leq i < n+l \\
+0 & \text{otherwise}
+\end{cases}
+\end{equation}
+
+\paragraph{Case $l=1$} It can be solved easily using matrix determinant lemma\footnote{%
+Matrix determinant lemma\cite{matdetlem} tells us that 
+$$
+\det\left(\mathbf{A} + \mathbf{uv}^\textsf{T}\right)
+=
+\det\left(\mathbf{A}\right) \left(1 + \mathbf{v}^\textsf{T} \left(\mathbf{A}^{-1}\mathbf{u}\right)\right).
+$$}.
+In this case
+\begin{equation}
+Q = v v^{T}, \qquad v_{i} = \delta_{in},
+\end{equation}
+so (note that $\det e^{-\beta H} = e^{-\beta\mathrm{Tr} H}$)
+\begin{multline}
+\det\left(1+ e^{-\beta H} e^{i\lambda v v^{T}}\right) =
+e^{-\beta\mathrm{Tr} H} \det\left(e^{\beta H} + 1 + (e^{i\lambda}-1)v v^{T} \right) = \\ =
+e^{-\beta\mathrm{Tr} H} \det\left(e^{\beta H} + 1 \right)\Bigl(1 + (e^{i\lambda}-1)v^{T}\left(e^{\beta H} + 1 \right)^{-1}v \Bigr)
+\label{eq:singlelem}
+\end{multline}
+Thus, with normalized density matrix we get
+\begin{equation}
+\frac{\det\left(1+ e^{-\beta H} e^{i\lambda v v^{T}}\right)}{\det\left(1+ e^{-\beta H}\right)} =
+1 + (e^{i\lambda}-1)\left[\left(e^{\beta H} + 1 \right)^{-1}\right]_{nn}
+\end{equation}
+
+%\paragraph{Induction} Let us define $P_{k}=v_{k} v_{k}^{T}$, $\gamma=e^{i\lambda}-1$ and $A_{n} = e^{-\beta H} \prod_{k=1}^{n}e^{i\lambda P_{k}}$.
+%Then we get
+%\begin{equation}
+%1+ A_{n} = 1+ A_{n-1} + \gamma A_{n-1} P_{n}
+%\end{equation}
+%
+%\begin{multline}
+%\det\left(1+ A_{n}\right) =
+%\det\left(1+ A_{n-1} e^{i\lambda P_{n}}\right) =
+%\det A_{n-1} \det\left(A_{n-1}^{-1} + 1 + \gamma P_{n} \right) = \\ =
+%\frac{1 + \gamma v_{n}^{T}\left( 1 - (1+A_{n-1})^{-1}\right)^{-1}v_{n}}{\det \left((1+A_{n-1})^{-1}\right)}
+%\end{multline}
+
+\paragraph{General $l$} Let us define rectangular matrix $N\times l$ such that $P_{ij}=Q_{ij}$ and $\gamma=e^{i\lambda}-1$.
+Then we get
+\begin{multline}
+\det\left(1+ e^{-\beta H} e^{i\lambda P P^{T}}\right) =
+e^{-\beta\mathrm{Tr} H}
+\det\left(e^{\beta H} + 1 + \gamma P P^{T}\right) = \\= 
+\det\left(1+ e^{-\beta H}\right)\det\left(1+ \gamma\left(1 + e^{\beta H}\right)^{-1} P P^{T} \right)
+\end{multline}
+So instead of the determinant $N\times N$ we calculate determinant $l\times l$ (see also~\cite{matdetlem}):
+\begin{equation}
+\frac{\det\left(1+ e^{-\beta H} e^{i\lambda v v^{T}}\right)}{\det\left(1+ e^{-\beta H}\right)} =
+\det\left(1+ \gamma P^{T}\left(1+ e^{\beta H}\right)^{-1} P  \right)
+\label{eq:average}
+\end{equation}
+
+\paragraph{Diagonalisation} 
+To diagonalise the $N$-sites chain with periodic boundary conditions we need unitary matrix
+\begin{equation}
+U = \frac{1}{\sqrt N} e^{2\pi i k m/N}, \qquad k,m = 1..N
+\end{equation}
+Thus the \fref{eq:average} is equal to
+\begin{equation}
+\det\left(1+ \gamma A \right),
+\qquad
+A_{mn} = \frac{1}{N} \sum_{k=1}^{N}\frac{e^{2\pi i k (n-m)/N}}{1+e^{\beta \epsilon_{k}}},
+\qquad
+n,m=1..l,
+\qquad
+\epsilon_{k} = \cos\frac{2\pi k}{N}-\epsilon_{F}
+\end{equation}
+
+\paragraph{Limits} In our case we take the limit $N\to \infty$ first, and only then $l\to \infty$.
+The first limit allow us to define $2\pi k /N \to p \in [0..2\pi]$, so that
+\begin{equation}
+A_{mn} = A_{n-m},
+\qquad
+A_{s}=\int_{-\pi}^{\pi}\frac{dp}{2\pi} e^{i p s} n_{p},
+\qquad
+n_{p}=\frac{1}{1+e^{\beta \varepsilon_{p}}},
+\qquad
+s=-l..l,
+\qquad
+\varepsilon_{p} = \varepsilon_{F} - \cos p,
+\quad
+\varepsilon_{F} = \cos p_{F}
+\end{equation}
+The matrix element $c_{n}=\delta_{n0}+\gamma A_{n}$ used in book \cite{McCoyWu1973}, the function $C(e^{ip})$ is
+\begin{equation}
+C(e^{ip}) = 1+\gamma n_{p}
+\end{equation}
+and the determinant is
+\begin{gather}
+\det(1+\gamma A) \approx \exp\left( g_{0} l + \sum_{l'=1}^{l} l' g_{l'}g_{-l'}\right),
+\qquad
+g_{l} = \int_{-\pi}^{\pi}\frac{dp}{2\pi} e^{i p l} \log(1+\gamma n_{p}),
+\end{gather}
+For $\beta\to\infty$ we get $n_{F} = \langle c_{n}^{\dagger}c_{n}\rangle = p_{F}/\pi$, so
+\begin{equation}
+g_{l} = i\lambda \int_{-p_{F}}^{p_{F}}\frac{dp}{2\pi} e^{i p l} = 
+i\lambda
+\begin{cases}
+\displaystyle \frac{p_{F}}{\pi} & l=0
+\\[10pt]
+\displaystyle \frac{\sin(p_{F} l)}{\pi l} & l\ne 0
+\end{cases}
+\end{equation}
+The latter sum in exponent gives us
+\begin{equation}
+\sum_{l'=1}^{l} l' g_{l'}g_{-l'} = -\frac{\lambda^{2}}{2\pi^{2}} \sum_{l'=1}^{l} \frac{1-\cos(2p_{F} l')}{l'}
+\approx \gamma_\text{Euler} + \log(2 l \sin p_{F})
+\end{equation}
+So
+\begin{equation}
+\det(1+\gamma A) \approx e^{i\lambda c l - \frac{\lambda^{2}}{2\pi^{2}}\log(l/l_{0})},
+\qquad
+l_{0} = \frac{e^{-\gamma_\text{Euler}}}{2 \sin p_{F}}
+\end{equation}
+
+\begin{figure}
+\centering
+\includegraphics[scale=0.5]{l0cplot}
+\caption{Length scale $l_{0}$ as a function of occupation $n_{F}\equiv \langle c^{+}_{n}c_{n}\rangle=p_{F}/\pi$}
+\label{fig:l0cplot}
+\end{figure}
+
+\section{Corellator}
+
+Let us calculate the trace
+\begin{equation}
+\Tr\left[ e^{-\beta \mathcal{H}}c_{n}^{\dagger}c_{m} e^{i\lambda\mathcal{Q}}\right]
+\end{equation}
+The pair of creation/destruction operators can be written as
+\begin{equation}
+e^{\alpha c_{n}^{\dagger}c_{n}} = 1+(e^{\alpha}-1) c_{n}^{\dagger}c_{n}
+\qquad
+e^{\alpha c_{n}^{\dagger}c_{m\ne n}} = 1 + \alpha c_{n}^{\dagger}c_{m}
+\end{equation}
+or, vice verse,
+\begin{equation}
+c_{n}^{\dagger}c_{n} = \frac{e^{\alpha c_{n}^{\dagger}c_{n}}-1}{e^{\alpha}-1}
+\qquad
+c_{n}^{\dagger}c_{m\ne n} = \frac{e^{\alpha c_{n}^{\dagger}c_{m}}-1}{\alpha}
+\end{equation}
+For the $n=m$ we get using \fref{eq:singlelem}, and the commutation of the exponents in determinant\footnote{%
+$$
+\det(1+AB) = \det A \, \det (A^{-1}+B) = \frac{1}{\det A^{-1}} \, \det (A^{-1}+B) = \det (A^{-1}A+BA) = \det (1+BA)
+$$}
+\begin{multline}
+\Tr\left[ e^{-\beta \mathcal{H}}c_{n}^{\dagger}c_{n} e^{i\lambda\mathcal{Q}}\right]
+=
+\frac{1}{e^{\alpha}-1}\left( \Tr\left[ e^{-\beta \mathcal{H}} e^{\alpha \mathcal{Q}'} e^{i\lambda\mathcal{Q}}\right] -\Tr\left[ e^{-\beta \mathcal{H}} e^{i\lambda\mathcal{Q}}\right] \right)
+=\\=
+\frac{1}{e^{\alpha}-1}\left( \Tr\left[ e^{i\lambda\mathcal{Q}} e^{-\beta \mathcal{H}} e^{\alpha \mathcal{Q}'} \right] -\Tr\left[ e^{i\lambda\mathcal{Q}} e^{-\beta \mathcal{H}} \right] \right)
+=\\=
+\frac{1}{e^{\alpha}-1}\left( \det\left[1+ e^{i\lambda Q}e^{-\beta H} e^{\alpha vv^{T}}\right] -\det\left[1+ e^{i\lambda Q} e^{-\beta H}\right] \right)
+=\\=
+\frac{1}{e^{\alpha}-1}\left( 
+\det\left[1 + e^{i\lambda Q}e^{-\beta H} + (e^{\alpha}-1) e^{i\lambda Q}e^{-\beta H} vv^{T} \right] - \det\left[1+e^{i\lambda Q} e^{-\beta H} \right]\right)
+=\\=
+\frac{1}{e^{\alpha}-1}\left( \det\left[e^{i\lambda Q}e^{-\beta H}\right]
+\det\left[e^{\beta H}e^{-i\lambda Q} + 1 + (e^{\alpha}-1)  vv^{T} \right] - \det\left[1+e^{i\lambda Q} e^{-\beta H} \right]\right)
+=\\=
+\frac{1}{e^{\alpha}-1}\left( 
+\det\left[1 + e^{i\lambda Q}e^{-\beta H} \right]\Bigl(1 + (e^{\alpha}-1)v^{T}\left(e^{\beta H}e^{-i\lambda Q} + 1 \right)^{-1}v \Bigr) - \det\left[1+e^{i\lambda Q} e^{-\beta H} \right]\right)
+=\\=
+\det\left[1 + e^{i\lambda Q}e^{-\beta H} \right]
+\Bigl[\left(e^{\beta H}e^{-i\lambda Q} + 1 \right)^{-1} \Bigr]_{nn}
+\end{multline}
+where $\mathcal{Q}' = c^\dagger_{n}c_{n}$.
+For the $n\ne m$ we get (here $u_{i} = \delta_{im}$)
+\begin{multline}
+\Tr\left[ e^{-\beta \mathcal{H}}c_{n}^{\dagger}c_{m} e^{i\lambda\mathcal{Q}}\right]
+=
+\frac{1}{\alpha}\left( \det\left[ e^{-\beta H} e^{\alpha vu^{T}} e^{i\lambda Q}\right] -\det\left[ e^{-\beta H} e^{i\lambda Q}\right] \right)
+=\\=
+\frac{1}{\alpha}\left( 
+\det\left[1 + e^{i\lambda Q}e^{-\beta H} + \alpha e^{i\lambda Q}e^{-\beta H} vu^{T} \right] - \det\left[1+e^{i\lambda Q} e^{-\beta H} \right]\right)
+=\\=
+\frac{1}{\alpha}\left( 
+\det\left[1 + e^{i\lambda Q}e^{-\beta H} \right]\Bigl(1 + \alpha u^{T}\left(e^{\beta H}e^{-i\lambda Q} + 1 \right)^{-1}v \Bigr) - \det\left[1+e^{i\lambda Q} e^{-\beta H} \right]\right)
+=\\=
+\det\left[1 + e^{i\lambda Q}e^{-\beta H} \right]
+\Bigl[\left(e^{\beta H}e^{-i\lambda Q} + 1 \right)^{-1} \Bigr]_{mn}
+\end{multline}
+Thus, for arbitrary $m$ and $n$ we get (note the oder of indices)\footnote{We used
+$\langle c_{m}c_{n}^{\dagger}e^{i\lambda\mathcal{Q}}\rangle = \langle e^{i\lambda\mathcal{Q}}\rangle \delta_{mn} - \langle c_{n}^{\dagger}c_{m}e^{i\lambda\mathcal{Q}}\rangle$}
+\begin{align}
+\left< c_{n}^{\dagger}c_{m} e^{i\lambda\mathcal{Q}}\right> &=
+\left< e^{i\lambda\mathcal{Q}}\right>\Bigl[\left(e^{\beta H}e^{-i\lambda Q}+1\right)^{-1} \Bigr]_{mn},
+\\
+\left< c_{m}c_{n}^{\dagger} e^{i\lambda\mathcal{Q}}\right> &=
+\left< e^{i\lambda\mathcal{Q}}\right>\Bigl[1-\left(e^{\beta H}e^{-i\lambda Q}+1\right)^{-1} \Bigr]_{mn},
+\end{align}
+
+
+\newpage
+\appendix
+
+\section{First Szeg\"o theorem}
+
+Let say we have a Toeplitz matrix $N\times N$ thus $T^{(N)}_{nm} = c_{n-m}$.
+Imagine we have a matrix equation 
+\begin{equation}
+\sum_{m=0}^{N-1}c_{n-m}x^{(N)}_{m} = \delta_{n0},\qquad n=0..N-1  
+\label{eq:auxmateq}
+\end{equation}
+Then
+\begin{equation}
+x_{0}^{(N)} = \frac{\det T^{(N-1)}}{\det T^{(N)}}
+\end{equation}
+If summation in Eq.~\eqref{eq:auxmateq} were from $m = -\infty .. +\infty$, the Fourier transform would be very useful.
+Let us try it anyway:
+\begin{equation}
+x_{m} = \frac{1}{ N}\sum_{q=0}^{N-1} e ^{-2\pi i q m/N} y_{q}
+\end{equation}
+Substitutin and taking reverse transform we get
+\begin{equation}
+\sum_{q=0}^{N-1} \left[ \frac{1}{ N} \sum_{n,m=0}^{N-1}e ^{2\pi i (p n - q m)/N} c_{n-m} \right]   y_{q} = 1
+\end{equation}
+
+The sum can be rewritten as
+\begin{equation}
+\sum_{n,m=0}^{N-1} X_{nm}= \sum_{n=0}^{N-1}X_{nn} + \sum_{s=1}^{N-1} \sum_{n=0}^{N-1-s} X_{n,n+s} + \sum_{s=1}^{N-1} \sum_{n=s}^{N-1} X_{n,n-s}
+\end{equation}
+Since $c_{n-m}$ is symmetric,
+\begin{multline}
+\frac{1}{N} \sum_{n,m=0}^{N-1}e ^{2\pi i (p n - q m)/N} c_{n-m} = \\ =
+\frac{1}{N}\sum_{n=0}^{N-1}e ^{2\pi i (p - q)n/N} c_{0} + \frac{1}{N}\sum_{s=1}^{N-1} \sum_{n=0}^{N-1-s} e ^{2\pi i ((p - q)n - q s)/N} c_{s} + \frac{1}{N}\sum_{s=1}^{N-1} \sum_{n=s}^{N-1} e ^{2\pi i ((p - q)n + q s)/N} c_{s} = \\ =
+\frac{1}{N}\sum_{n=0}^{N-1}e ^{2\pi i (p - q)n/N} c_{0} + \frac{1}{N}\sum_{s=1}^{N-1} \sum_{n=0}^{N-1-s} e ^{2\pi i ((p - q)n - q s)/N} c_{s} + \frac{1}{N}\sum_{s=1}^{N-1} \sum_{n=s}^{N-1} e ^{2\pi i ((p - q)n + q s)/N} c_{s}
+\end{multline}
+For $p=q$ we get
+\begin{equation}
+\text{for $p=q$}\quad c_{0} + 2 \sum_{s=1}^{N-1} \frac{N-s}{N} \cos \left(\frac{2\pi q s}{N}\right) c_{s}
+\end{equation}
+For $p\ne q$ we get
+\begin{multline}
+\text{for $p\ne q$}\quad 
+\frac{1}{N}\sum_{s=1}^{N-1} \frac{e ^{2\pi i(p - q)(N-s)/N}-1}{e ^{2\pi i(p - q)/N}-1} e ^{-2\pi i q s/N} c_{s} + \frac{1}{N}\sum_{s=1}^{N-1} \frac{e^{2\pi i(p - q)N/N}-e ^{2\pi i(p - q)s/N}}{e^{2\pi i(p - q)/N}-1}  e ^{2\pi i q s/N} c_{s}
+= \\ =
+\frac{1}{N}
+\frac{1}{e ^{2\pi i(p - q)/N}-1}
+\sum_{s=1}^{N-1} \left[ e ^{-2\pi ips/N}-e ^{-2\pi i q s/N} + e ^{2\pi i q s/N} - e ^{2\pi ips/N} \right] c_{s}
+=\\=
+\frac{-2i}{N}
+\frac{1}{e ^{2\pi i(p - q)/N}-1}
+\sum_{s=1}^{N-1} \left[ \sin(2\pi ps/N) - \sin(2\pi qs/N) \right] c_{s}
+\end{multline}
+
+Together this can be written as 
+\begin{equation}
+\delta_{pq} \left(c_{0} + 2 \sum_{s=1}^{\infty} \cos \left(\frac{2\pi q s}{N}\right) c_{s} \right)
++
+(1-\delta_{pq})\frac{-2i}{N}
+\frac{1}{e ^{2\pi i(p - q)/N}-1}
+\sum_{s=1}^{\infty} \left[ \sin(2\pi ps/N) - \sin(2\pi qs/N) \right] c_{s}
+\end{equation}
+
+
+\subsection{Aristov case}
+
+In the Aristov's case we have 
+\begin{equation}
+c_{s} = \frac{\sin \pi c s}{\pi s},
+\qquad
+\sum_{s=1}^{\infty} e^{2\pi i p s /N} c_{s} = \frac{i}{2\pi}\log\frac{1-e^{i\pi c + 2\pi i p /N}}{1-e^{-i\pi c + 2\pi i p /N}}
+\end{equation}
+
+get
+\begin{equation}
+\delta_{pq} \left(1+\gamma c + \frac{i}{\pi} \gamma \log
+\frac{1-e^{i\pi c + 2\pi i p /N}}{1-e^{-i\pi c + 2\pi i p /N}}
+\frac{1-e^{i\pi c - 2\pi i p /N}}{1-e^{-i\pi c - 2\pi i p /N}} \right)
+\end{equation}
+In this case the solution is easy and equal
+\begin{equation}
+x_{0} = \left(1+\gamma c + \frac{i}{\pi} \gamma \log\frac{1-e^{i\pi c}}{1-e^{-i\pi c}} \right)^{-1}
+\end{equation}
+
+\section{Hubbart-Stratanovich with auxiliary fermionic field}
+
+\begin{equation}
+\mathrm{Tr}_{d} \left[e^{i \lambda d^{+} d} e^{i \frac{\pi}{2} (d^{+} a + a^{+} d)}\right] = e^{i\lambda a^{+}a}
+\end{equation}
+
+\bibliographystyle{amsplain}
+\bibliography{../../literature/quantinfo, localcites}
+
+
+\end{document}
+