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Commit d9e03fad authored by Alex Kashuba's avatar Alex Kashuba
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The logarithmic term in Szego theorem

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......@@ -10,3 +10,4 @@ data*.dat
determinant.pdf
fig?.jpg
perturbative.pdf
determinant-v1.1.pdf
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\def\fileversion{\it version~0.4\,%
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%(\pdffilesize{\jobname.tex} bytes)}
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\DeclareMathOperator{\Tr}{Tr}
\DeclareGraphicsExtensions{.pdf,.ai}
......@@ -41,7 +41,7 @@
\maketitle
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{\parskip-\baselineskip\par\noindent {\hfill\raisebox{-\baselineskip}[0pt][10pt]{\fileversion}\hfill}}
......@@ -457,11 +457,78 @@ In this case the solution is easy and equal
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}
%\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}
%\section{The finite temperature case}
%
%For finite temperature the function $f(e^{i\theta})$ is analytic:
%\begin{multline}
%-\log f(e^{i\theta}) = -\log \frac{1}{e^{2\beta\cos\theta - \alpha}} =
%\sum_{n=1}^{\infty}\frac{1}{n}e^{-\alpha n} e^{2n\beta\cos\theta} =
%\sum_{n=1}^{\infty}\frac{e^{-\alpha n}}{n}\sum_{m=0}^{\infty} \frac{n^{m}\beta^{m}}{m!} (2\cos\theta)^{m}
%=\\=
%\sum_{n=1}^{\infty}\frac{e^{-\alpha n}}{n}\sum_{m=0}^{\infty} \frac{n^{m}\beta^{m}}{m!} \sum_{k=0}^{m} C^{k}_{m} e^{i(2k-m)\theta}
%=
%\sum_{n=1}^{\infty}\frac{e^{-\alpha n}}{n}\sum_{m=0}^{\infty} \sum_{k=0}^{m} \frac{n^{m}\beta^{m}}{k!(m-k)!} e^{i(2k-m)\theta}
%\end{multline}
%so that for $\log f(z)=\sum_{l=-\infty}^{\infty}V_{l} z^{l} $ we get
%\begin{equation}
%V_{l} = \sum_{n=1}^{\infty}\sum_{s=0}^{\infty} \frac{n^{|l|+2s-1}\beta^{|l|+2s}e^{-\alpha n}}{s!(|l|+s)!}
%=\sum_{n=1}\frac{e^{-\alpha n}}{n}I_{l}(2\beta n)
%\end{equation}
%where $I_{l}(z)$ is modified Bessel function of the first kind.
\section{The logarithmic term in Szeg\"o theorem}
Let us consider more accurately the Fisher-Hartwig formula.
We follow the designations given in \cite{Deift2011} and \cite{Abanov2011}
In our case
\begin{equation}
f(e^{i\theta}) = 1 + (e^{i\lambda}-1)\frac{1}{e^{2\beta\cos\theta - \alpha}+1}
\end{equation}
It is crucial to be able to expand into Laurent series:
\begin{equation}
\log f(z) = \sum_{l=-\infty}^{\infty}V_{l}z^{l}
\end{equation}
For finite $\beta$ the $V_{l}$ gets a cutoff $l\sim \beta$ and Szeg\"o theorem gives us a formula
for the determinant of $L\times L$ matrix
\begin{equation}
\exp\left( V_{0} L + \sum_{l=1}^{\infty} l V_{l}V_{-l}\right)
\end{equation}
Note the infinity in the sum.
Aristov \cite{Aristov1998} proposes formally cut off this series by $l=L$ (referring to work \cite{Luttinger1963}) and indeed, it gives a result in the limit of $T=0$
\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}
-\frac{\lambda^{2}}{2\pi^{2}}\left(\log (2L\sin p_{F})+\gamma_\text{Euler}\right)
\end{equation}
Nevertheless, the Fisher-Hartwig formula given in \cite{Deift2011}, Eq.~(1.12) does not imply the dependence on $L$ (notation in paper $n$) for non-singular $f$.
The cutoff is made automatically for $l\sim\beta$, or in other words, the Eq.~(1.12) does not imply the finite $L$ if $T>0$.
For zero temperature the discontinuity forces us to this more general formula in Eq.~(1.12).
The particular case that interests us is described in \cite{Abanov2011} in Eq.~(B.9) and afterwards.
The substitution into Eq.~(1.12) of \cite{Deift2011} gives us Eq.~(B.10) in \cite{Abanov2011}.
Expansion of Eq.~(B.10) in \cite{Abanov2011} for small $\lambda$ gives us very similar result
\begin{equation}
-\frac{\lambda^{2}}{2\pi^{2}}\left(\log (2L\sin p_{F})+\gamma_\text{Euler}+1\right)
\end{equation}
This result is also mentioned in \cite{LeHur2012}, right after Eq.~(2.49).
Taking into account the interaction-dependence obtained from the perturbative approach, we get
\begin{equation}
-\frac{\lambda^{2}}{2\pi^{2}}\left(K\left[\log (2L\sin p_{F})+\gamma_\text{Euler}\right]+1\right)
\end{equation}
The numerical result given by Andreas speaks in favour of this formula:
\begin{figure}[h]
\centering
\includegraphics[scale=.5]{figKG}
\caption{Analytics and numerics comparison for $p_{F}=\pi/2$ (half-filling)}
\label{fig:figKG}
\end{figure}
\bibliographystyle{amsplain}
\bibliography{../../literature/quantinfo, localcites}
......
figKG.pdf 0 → 100644
File added
%% This BibTeX bibliography file was created using BibDesk.
%% http://bibdesk.sourceforge.net/
%% Created for Alex Kashuba at 2022-06-03 16:14:16 +0200
%% Created for Alex Kashuba at 2022-10-17 21:18:58 +0200
%% Saved with string encoding Unicode (UTF-8)
......@@ -61,10 +61,10 @@
url = {https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula}}
@inbook{McCoyWu1973,
author = {B. M. McCoy and T. T. Wu,},
author = {B. M. McCoy and T. T. Wu},
chapter = {X},
date-added = {2022-02-02 15:14:29 +0100},
date-modified = {2022-06-03 16:12:10 +0200},
date-modified = {2022-10-17 21:18:57 +0200},
publisher = {Harvard University Press, Cambridge},
title = {The two-dimensional {I}sing model},
year = {1973}}
......
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