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I don't believe that there are aliens. I believe there are really different people.
Take me to the height where success would seek my help to succeed!
I ARE I don't believe in fairies floating around, and I don't believe in telepathy, but there are ...
ISOBELLE CARMODY I don't believe in accidents. There are only encounters in history. There are no accidents.
PABLO PICASSO I don't believe there are heroes or saviours in politics.
ALEXIS TSIPRAS I don't believe that there are no spiritual beings around us. I don't know what to call them...
MARTHA BECK I really believe there are things nobody would see if I didn't photograph them.
DIANE ARBUS I'm a capitalist. I believe that people take risk, and there are rewards if they do well; they s...
HAROLD FORD, JR. I don't believe that competitions are important.
EDDIE IZZARD Oh yeah, I believe in God. I think there's much more evidence that there is a God than that ther...
DAVID ZUCKER I personally don't believe in aliens. But, I do believe that there is something out there that i...
J. T. WALSH One moment you are here, othere there, then there. Then you can't move!
DEYTH BANGER And I don't believe that I have to stay on one side of the fence or the other. I don't belie...
WHOOPI GOLDBERG Certainly there are people who like me, but then there are those who don't know me who gossip ab...
CARRIE FISHER I don't believe in trickle-down economics. I don't think that people who have the most are i...
DAVE MATTHEWS Ontarians don't want to believe that they are small people. They want to believe that they'r...
KATHLEEN WYNNE I don't like psychiatry. I don't believe it works. I believe psychiatrists are neurotic or p...
KIRSTIE ALLEY There are people who believe in the existence of angels, but do not believe in alien existence. Ther...
TOBA BETA I have a very open mind. I believe that there's a spirit world. I believe that there's a lot...
ANTHONY HEAD I don't believe in gutter politics. I don't believe in gutter journalism.
MARIA SHRIVER I don't really believe in regrets.
WAYNE NEWTON People don't believe that any politician is any different from any other one.
JUSTIN TRUDEAU I don't believe in an afterlife. I don't believe in a single or multiple godhead. I respect ...
TONY JUDT I am placed in situations with opportunities in those situations that others are not. I don't be...
RONALD PERELMAN I don't believe there is any finer mission on Earth than just to make people laugh.
FATTY ARBUCKLE I believe that the truest parts of people can be buried, and for many different reasons.
MARY GAITSKILL Do people really believe there's something different about the eyes of murderers?
BILL JAMES I don't believe that human beings are necessarily monogamous.
K. D. LANG Wherever you are, be all there.
CRAIG GROESCHEL I don't believe in stereotypes. Most of the time, stereotypes are just that.
JAVIER BARDEM I already know what things look like - I don't want description. People believe in appearances, ...
DUANE MICHALS I believe that a politician needs to be with people.
RAHUL GANDHI Fiscally, I'm very conservative. I don't believe in welfare states. I believe in giving peop...
GENE SIMMONS I believe that when things are easy, people stay dormant.
OLGA KURYLENKO I believe that government is the servant of the people and not their master.
DAVID ROCKEFELLER I don't believe in strong-arming people.
ZACHARY LEVI I don't push buttons to push buttons. Throwing the rebel card out there is really cheap.
KACEY MUSGRAVES I guess I don't really believe in retirement. I believe in shorter days and maybe in weekends!
ALICE WATERS I don't believe there should be any restrictions when it comes to firearms. None.
GARY JOHNSON If you are recording, you are recording. I don't believe there is such a thing as a demo or a te...
TOM WAITS I absolutely believe in assimilation. I don't believe I'm any different from straight people...
CHAZ BONO I don't believe, the president doesn't believe, that the high income tax cuts work, period. ...
AUSTAN GOOLSBEE Of course [I believe in aliens]. Are you so arrogant as to believe we are alone in this universe?
TOM CRUISE There are lots of people who believe that caricature of me the tabloids created, so they think they ...
JO BRAND Of course [I believe in aliens]. Are you so arrogant as to believe we are alone in this universe?
TOM CRUISE I don't know what it is about me. I don't know if there's something strange... but I con...
LAURA VANDERVOORT You have to believe in God before you can say there are things that man was not meant to know. I don...
DAVID CRONENBERG I think that it would be unwise of us not to believe that there is life outside of us, intelligent l...
SIA I can't believe there are people still out there like that.
FRANK PANO I believe in a lot of astrology. I believe in aliens... I look up into the stars and I imagine: '...
KATY PERRY I'm actually a really good cook, and people don't believe this about me!
MARIAH CAREY I don't really care how the Patriots are perceived, truthfully. I really don't. I really don...
TOM BRADY I don?t believe in titles, they make people think they are superior
HENRY FORD I believe women are the glue of everything.
WEILI DAI We are all worms. But I believe that I am a glow-worm.
WINSTON CHURCHILL You're always going to have people that are naysayers, that don't believe in your talent, th...
VANESSA WILLIAMS Of course I believe in aliens. I think it's very egocentric to think that there's nothing el...
LIZZIE BROCHERE I mean, there are times when you aren't working, but still believe that work will come.
ESTELLE PARSONS I don't believe in killing people. I believe in locking you up for the rest of your life.
YAHYA JAMMEH With the Hubble telescope and all the other things that are out there, I believe something would hav...
MARK GODDARD People don't understand computers. Computers are magical boxes that do things. People believe wh...
BRUCE SCHNEIER I don't really believe in cliques - I think everyone can be friends with everyone.
LUCY HALE I don't have any opinions about Russians. There are celebrity persons in each country. Different...
ALEXEI MORDASHOV I don't believe in angels and I have trouble with the whole God thing. I don't want to say I...
BILLY CONNOLLY I have a controversial opinion about evil, because I don't believe evil exists. I believe that a...
VICTOR SALVA There are only so many pitches in this old arm, and I don't believe in wasting them throwing to ...
EDDIE PLANK I don't believe love is fickle. I believe when you love someone, you are allowed to love from af...
KHLOE KARDASHIAN I told you I'm not going to criticize my successor. I'll just tell you that there are people...
GEORGE W. BUSH I don't believe in outing people. It's up to the individual, but there's nothing wrong p...
RICK MERCER There are those who believe that power is up above, and they don't notice that it's actually...
JOSE MUJICA I don't believe in guilty pleasures. If you enjoy something, there's nothing guilty about it...
BUSY PHILIPPS I don't believe in voodoo.
JENNIFER HUDSON I don't believe in icons.
THE WEEKND I don't believe in ghosts.
RACHEL HURD-WOOD I'm not really conservative. I'm conservative on certain things. I believe in less governmen...
CLINT EASTWOOD I don't believe in welfare.
CHARLES EVERS When I believe, I am crazy. When I don't believe, I suffer psychotic depression.
PHILIP K. DICK I don't believe in coincidence.
JULIETTE BINOCHE I don't believe in miracles.
SAM SIMON I don't believe in God.
GAIL PORTER I think they should have movies in restaurants. I can't believe that so many people get together...
ANDY WARHOL I don't believe in murder.
KABIR BEDI I don't believe in funerals. I believe in celebrating life, and showing people, while they'r...
CARMEN DELL'OREFICE I don't believe in funerals. I believe in celebrating life, and showing people, while they'r...
CARMEN DELL'OREFICE There are people who believe that a Dolce & Gabbana T-shirt is more stylish than a supermarket o...
DOMENICO DOLCE I don't believe in politics; I don't understand any of it.
MAIRA KALMAN I don't like gurus. I don't like people who ask you to follow or believe. I like people who ...
A. S. BYATT I definitely believe that you are drawn to certain things for inexplicable reasons, but in a very po...
RACHEL MCADAMS If something's not right, I don't believe in maintaining something for the sake of what'...
KATE HUDSON I actually don't believe in governance by referendum. I believe that we have a democracy, that w...
BETSY HODGES I'm
really worried. There are people in my country who believe I humiliated
Zimbabweans ...
MAKOSI MUSAMBASI I'm really worried. There are people in my country who believe I humiliated Zimbabweans by showing...
MAKOSI MUSAMBASI I don't believe there's two sides to every story. It's black and white. There's righ...
JOE WURZELBACHER I don't believe it happened that way. ... I just really don't believe that Chavis was there.
DAVID RIMMER I'm a great admirer of secularism. At its best, I think it's one of the best things that we ...
MARILYNNE ROBINSON I don't intimidate anyone. Instead, I try to get people to believe in themselves.
BOB ROSS I don't believe in God. I do believe in Judaism. I believe in ethics, morals.
EDGAR BRONFMAN, SR. I don't believe it. Prove it to me and I still won't believe it.
DOUGLAS ADAMS I don't believe anybody can really grasp everything that's even in one textbook.
JOSHUA LEDERBERG Prizes are like butterflies, colorful butterflies that fly away. I don't believe in prizes much.
LINA WERTMULLER I really believe there are things nobody would see if I didn't photograph them.
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ORSON SCOTT CARD Do not shout at me, Mr. Quill," said John [Adams]. "Justice may be blind, but she is not deaf.
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ORSON SCOTT CARD Laws against homosexual behavior should remain on the books.
ORSON SCOTT CARD ...it seemed a part of her life, to step from the ancient to the modern, back and forth. She felt ra...
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ORSON SCOTT CARD Don't launch it," said Bean into his microphone, head down. "Set it off inside your ship. God be wit...
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ORSON SCOTT CARD Life is full of grief, to exactly the degree we allow ourselves to love other people.
ORSON SCOTT CARD What could go wrong... Doing some stupid impulsive thing that caused the death of drowthers was prac...
ORSON SCOTT CARD Too many people in the American media have lost any concept of loyalty to their country -- if they e...
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ORSON SCOTT CARD There is more good writing and good acting in any ten minutes of Twister than in, say, all of Citize...
ORSON SCOTT CARD I say this as a Democrat, for whom the Republican domination of government threatens many values tha...
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ORSON SCOTT CARD …I had seen the princess and let her lie there unawakened, because the happily ever after was so d...
ORSON SCOTT CARD We're like the wicked witch. We promise gingerbread, then eat the little brats alive.
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"A h...
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Thanks. Does this mean I get a raise?
No, just a medal. The budget isn't...
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ORSON SCOTT CARD In order to learn, one must change one's mind
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ORSON SCOTT CARD Whatever your gravity is when you get to the door, remember―the enemy's gate is down.
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ORSON SCOTT CARD I will never hurt you.
I will always help you.
If you are hungry
Ill give you my food...
ORSON SCOTT CARD From you I can learn things that nobody knows.
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ORSON SCOTT CARD Nero, you are an example to all the children on this shuttle. Because most of them are so foolish, t...
ORSON SCOTT CARD I'm not one to despise other people for their sins. I haven't found one yet, that I didn't say...
ORSON SCOTT CARD You get all the pleasures of the puppeteer.
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ORSON SCOTT CARD I don't care how much you eat, Ender, self-cannibalism won't get you out of this school.
ORSON SCOTT CARD How much time? Not as much as I had yesterday.
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"Am I the on...
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Danny was touched. "But t...
ORSON SCOTT CARD In my view, suicide is not really a wish for life to end.'
What is it then?'
It is the onl...
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ORSON SCOTT CARD Personal humiliation was painful. Humiliation of one's family was much worse. Humiliation of one's s...
ORSON SCOTT CARD because if you can't kill then you are always subject to those who can, and nothing and no one wil...
ORSON SCOTT CARD If only we could have talked to you, the hive-queen said in Ender's words. But since it could not be...
ORSON SCOTT CARD I'm not a liar, sir,' she said.
"'No, I'm sure you sincerely become whatever it is you're prete...
ORSON SCOTT CARD If words can be lethal weapons, I must provide them with an arsenal.
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ORSON SCOTT CARD If pigs could vote, the man with the slop bucket would be elected swineherd every time, no matter ho...
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ORSON SCOTT CARD Perhaps it's impossible to wear an identity without becoming what you pretend to be.
ORSON SCOTT CARD With false names, on the right nets, they could be anybody. Old men, middle-aged women, anybody, as ...
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ORSON SCOTT CARD Perhaps a physicist would know at once why this whole idea was absurd. But then, perhaps a physicist...
ORSON SCOTT CARD Human beings are free except when humanity needs them. Maybe humanity needs you. To do something. Ma...
ORSON SCOTT CARD Peter, you're twelve years old. I'm ten. They have a word for people our age. They call us children ...
ORSON SCOTT CARD Sometimes lies were more dependable than the truth.
ORSON SCOTT CARD Remember, the enemy's gate is down.
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ORSON SCOTT CARD I've watched through his eyes, I've listened through his ears, and I tell you he's the one.
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ORSON SCOTT CARD Changing the world is good for those who want their names in books. But being happy, that is for tho...
ORSON SCOTT CARD Well, I'm your man. I'm the bloody bastard you wanted when you had me spawned. I'm your tool, and wh...
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ORSON SCOTT CARD She worked her toes into the sand, feeling the tiny delicious pain of the friction of tiny chips of ...
ORSON SCOTT CARD Let's fly away and live forever
ORSON SCOTT CARD Besides," said Suriyawong. "This was not a rescue operation."
"What was it, target practice? Ch...
ORSON SCOTT CARD In a way she actually preferred Peter to other people because of this. He always acted out of intell...
ORSON SCOTT CARD Individual human beings are all tools, that the others use to help us all survive." "That's a ...
ORSON SCOTT CARD All is going well, very well, I couldn't ask for anything better—
So why do I hate my life?
ORSON SCOTT CARD Ender nodded. It was a lie, of course, that it wouldn't hurt a bit. But since adults always said i...
ORSON SCOTT CARD You don't know how to love people. You only know how to own them. And because people will never act ...
ORSON SCOTT CARD The only teacher that's worth anything to you is your enemy.
ORSON SCOTT CARD With life. Rooter says that life is how God gives purpose to the universe.
ORSON SCOTT CARD Ningún ser humano es indigno cuando se comprenden sus motivos. Ninguna vida deja de merecer la pena...
ORSON SCOTT CARD Thrower started toward them at a trot, then remembered his dignity and walked the rest of the way. T...
ORSON SCOTT CARD I came because I've spent my whole life in the company of the brother that I hated. Now I want a cha...
ORSON SCOTT CARD This is how humans are: We question all our beliefs, except for the ones that we really believe in, ...
ORSON SCOTT CARD The computer is also not famous for having mercy.
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ORSON SCOTT CARD There's only one thing that will make them stop hating you. And that's being so good at what you do ...
ORSON SCOTT CARD No human being, when you understand his desires, is worthless. No one's life is nothing. Even the mo...
ORSON SCOTT CARD Bean sighed inwardly. It never failed. Whenever he had any conversation with Ender, it turned into a...
ORSON SCOTT CARD What I want," he said softly, "is to stand in this meadow and walk in the light of the sun.
ORSON SCOTT CARD The Buggers have finally, finally learned that we humans value each and every individual human life....
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ORSON SCOTT CARD Poke gave him life. Ender gave it meaning.
ORSON SCOTT CARD The criminal misuse of time was pointing out the mistakes. Catching them―noticing them―that was ...
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ORSON SCOTT CARD Nights were so real that days began to seem dreamlike to him.
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ORSON SCOTT CARD It was possible, wasn't it, that he loved her, and that in this time of terrifying opportunity he wa...
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ORSON SCOTT CARD Humanity does not ask us to be happy. It merely asks us to be brilliant on its behalf.
ORSON SCOTT CARD Ender Wiggin isn't a killer. He just wins—thoroughly.
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ORSON SCOTT CARD Regardless of law, marriage has only one definition, and any government that attempts to change it i...
ORSON SCOTT CARD I listen to music constantly while writing.
ORSON SCOTT CARD But in the meantime I became accustomed to the writing life and it would be hard to change now - par...
ORSON SCOTT CARD God, our genes, our environment, or some stupid programmer keying in code at an ancient terminal - t...
ORSON SCOTT CARD Working is hard and distracts from having fun.
ORSON SCOTT CARD Metaphors have a way of holding the most truth in the least space.
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ORSON SCOTT CARD Human beings may be miserable specimens, in the main, but we can learn, and, through learning, becom...
ORSON SCOTT CARD Fiction, because it is not about somebody who actually lived in the real world, always has the possi...
ORSON SCOTT CARD I also remembered that you were beautiful."
"Memory does play tricks on us."
"No. Your fac...
ORSON SCOTT CARD I don't care if I pass your test, I don't care if I follow your rules. If you can cheat, so can I. I...
ORSON SCOTT CARD The story is one that you and I will construct together in your memory. If the story means anything ...
ORSON SCOTT CARD The story itself, the true story, is the one that the audience members create in their minds, guided...
ORSON SCOTT CARD We have to go. I'm almost happy here.
ORSON SCOTT CARD Ethan Wyeth: I hope you're thirsty."
Gideon Wyeth:"Why?"
Ethan: "Cause your dumb and ugly,...
ORSON SCOTT CARD You know how writers are... they create themselves as they create their work. Or perhaps they create...
ORSON SCOTT CARD Soldiers can sometimes make decisions that are smarter than the orders they've been given.
ORSON SCOTT CARD Our great civilizations are nothing more than social machines to create the ideal female setting, wh...
ORSON SCOTT CARD Happiness can depend as easily on useless things as on useful ones.
ORSON SCOTT CARD Hive Queen: They never know anything. They don't have enough years in their little lives to come to ...
ORSON SCOTT CARD It was not right, thought Han Fei-tzu, for his wife to die before him: her ancestor-of-the-heart had...
ORSON SCOTT CARD Others find humanity by looking in their own hearts. Only lost souls need to search for it outside t...
ORSON SCOTT CARD Quing-Jao: I am a slave to the gods, and I rejoice in it.
Jane: A slave who rejoices is a slave...
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ORSON SCOTT CARD I realized that everything important in sci-fi showed up in the magazines first. It's the proving gr...
ORSON SCOTT CARD Eko brushed a tear from her eye, and Immo jeered at her, but father held up a hand. "Never mock a te...
ORSON SCOTT CARD I wonder sometimes if the motivation for writers ought to be contempt, not admiration.
ORSON SCOTT CARD For he loved her, as you can only love someone who is an echo of yourself at your time of deepest so...
ORSON SCOTT CARD "Dread is the first and strongest of the...kinds of fear. It is that tension, that waiting that come...
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ORSON SCOTT CARD Anything can become a children's book if you give it to a child...Children are actually the best (an...
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ORSON SCOTT CARD Parents always make their worst mistakes with their oldest children. That's when parents know the le...
ORSON SCOTT CARD I want to be the kind of boy you are, thought Bean. But I don't want to go through what you've b...
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text:
\section{Introduction\label{section:introduction}}

Recent experiments on thin films of doped polymers such as PEDOT-PSS, PEDOT-Tos, and PBTTT have found both high conductivity and a large thermopower \cite{Yuen2009,Pipe2013,Bubnova2013,Chabinyc2015}.  There are some possible explanations for the source of conductive behavior in polymers \cite{Heeger1988,Sai2007}, but they are not yet definitive; in this paper we will bypass this question and instead analyze a different facet of the problem.  Namely, since we take for granted the conductive nature of individual polymers, we can describe each polymer as a Luttinger liquid, and we look for possible signatures of the Luttinger liquid behavior that survive even when the one-dimensional systems are coupled to form a quasi-two-dimensional material.

The Luttinger liquid model \cite{Haldane1981Luttinger,GiamarchiBook} represents a one-dimensional electron gas modified by interaction between the electrons and can therefore be viewed as the one-dimensional analogue of the more well-known Fermi liquid model, though the generic behavior of the system is quite different.  In the Fermi liquid theory for two- and three-dimensional systems, the interacting system actually behaves very much like the corresponding non-interacting electron gas---the excitations are fermionic quasiparticles which behave qualitatively like electrons even if specific properties like mass are renormalized to new values.

By contrast, in one dimension interactions between electrons have a strong qualitative effect on the behavior of the system.  Schematically, one can picture electrons in higher-dimensional systems having space to ``go around'' each other and thus they still remain roughly independent (noninteracting), while in one dimension this is impossible, and so electrons will move together, forming collective (bosonic) excitations.  The Luttinger liquid theory and the technique of bosonization make this intuitive idea concrete.

There are numerous convincing experimental results on one-dimensional systems that confirm various predictions of the Luttinger liquid theory.  For instance, Luttinger liquids are expected to exhibit spin-charge separation, where charge and spin degrees of freedom act independently \cite{GiamarchiBook}; spin-charge separation has been convincingly observed via photoemission experiments in artificially created one-dimensional structures \cite{Segovia1999}.  Likewise, the density of states around the Fermi level is predicted to show a distinctive power law behavior \cite{Braunecker}; this was also observed in an artificially created 1D chain \cite{Blumenstein2011}.  Other observations of Luttinger liquid-like behavior, however, have been made not on actual one-dimensional chains but rather on two-dimensional collections of one-dimensional systems such as in the polymer films that motivated this work\cite{Yuen2009} or on highly anisotropic three-dimensional crystals\cite{Moser1998,Wakeham2011,Bangura2013}; it is not immediately clear that the results of these experiments should be directly compared to theories of single Luttinger liquids.  Rather, the coupling of 1D chains to form a quasi-2D material may modify or destroy altogether the distinctive signatures of Luttinger liquid behavior.  A theory of coupled Luttinger liquids would thus be very helpful.

While the theory of weakly coupled Luttinger liquids has been considered in the past by many different authors, there are very few results for thermal transport in a system of infinitely many coupled chains.  Some results deal with ``ladders'' consisting of just two coupled chains \cite{ClarkeStrong1996,GiamarchiBook}, while some of the most well-known treat coupling two half-infinite chains at their ends as a way of modeling an impurity in the Luttinger liquid\cite{KaneFisher1992,KaneFisher1992b,KaneFisher1996}.  Both electrical and thermal transport have also been computed for many impurities on a single chain\cite{Li2002}.  Papers that do consider an infinite array of weakly coupled Luttinger liquids have mostly focused only on the electrical conductivity \cite{Moser1998,Anderson1995,Giamarchi2000_2013} and not on any kind of thermal transport.  There is one recent paper on the off-diagonal terms of the thermopower tensor for infinitely many coupled chains\cite{Schattner2016Arxiv}, but we are not aware of any existing results for the thermal conductivity or Lorenz number in the type of model we consider.  This is the gap we intend to fill.

In this work we consider a model of coupled one-dimensional systems in which each 1D chain is treated as a (spinless) Luttinger liquid, and the individual chains are coupled by a perturbatively weak interchain hopping.  We refer to this situation of strong interactions within 1D chains and weak, incoherent coupling between them as the ``quasi-atomic limit.''  The approximations and assumptions inherent in this model, as well as some justifications of their validity, are discussed in section \ref{section:assumptions}.

We consider two somewhat different versions of the model, which incorporate Luttinger liquid behavior at different stages of the calculation.  In both cases, we calculate transport coefficients using the Kubo formalism.  In the first model, discussed in section \ref{section:nonint_model}, the electronic system is initially assumed to be noninteracting so that the state of the system can be described by occupation of single-particle orbitals; we introduce Luttinger behavior via the electronic density of states.  In the second model (section \ref{section:LL}), we use the full Luttinger liquid correlation functions.  In section \ref{section:discussion}, we summarize our key results and their applicability to experimental systems, and we further discuss the comparison between the two models.

We find that both models predict the same power law dependence on temperature for the transport coefficients, $\sigma \propto T^{2\gamma-3}$ and $\kappa \propto T^{2\gamma-2}$, where $\gamma$ is a measure of interaction strength as defined in equation \eqref{eq:def_gamma}, but that the precise values of the transport coefficients (as measured by the Lorenz number) vary with electron-electron interaction strength more strongly in the second, more complete, calculation.  In the generalized noninteracting model (section \ref{section:nonint_model}) we find that the Lorenz number is larger than the value predicted by the Wiedemann-Franz law by a factor between $\gamma^2$ and $\gamma^{2.4}$.  In the full Luttinger liquid model (section \ref{section:LL}), we find an even larger violation, with the Lorenz number augmented by as much as $\gamma^{3.6}$.

\section{Assumptions and approximations: the quasi-atomic limit\label{section:assumptions}}

In the Hubbard model, the ``atomic limit'' is the limit as the hopping between lattice sites vanishes while electron-electron interaction is held constant \cite{Beni1974,Mukerjee2005}.  We study the problem of weakly coupled chains with a similar approach, in which we do a perturbative calculation to lowest order in the interchain hopping while treating each one-dimensional chain as a single coherent quantum system.  This limit of full coherence in one direction (along chains) and weak incoherent hopping in the other direction (between chains) we call the ``quasi-atomic limit.''\footnote{Note that the term ``quasi-atomic limit'' has been used in the past to describe situations between full coherence and the atomic limit\cite{Siringo1991,Rojas2012}; we use it instead to indicate a system that is fully in the atomic limit in one direction and not at all in the other.}

To be more precise, we make the following assumptions:
\begin{enumerate}[(1)]
\item \label{assumption:independent_polymers}There is no electron-electron interaction between the 1D chains.
\item \label{assumption:atomic_limit}The different chains are perturbatively coupled through a weak hopping of electrons between adjacent chains.
\item \label{assumption:polymer_array}The 1D chains are located at evenly spaced points along a one-dimensional line, meaning that electrons may hop from one polymer to adjacent ones on either side of it and that the hopping strength between any pair of adjacent polymers is the same.
\label{assumption:final}
\end{enumerate}

\noindent We will briefly justify the applicability of these assumptions to real physical systems, beginning with assumption \eqref{assumption:atomic_limit}.  To measure transport properties for a macroscopic object (like a polymer film) we really want to use not the microscopic model of the system but rather an effective theory that results from a renormalization group flow.  At zero temperature, any coupling between chains is a relevant perturbation in the renormalization group sense, but fortunately this is not the case at finite temperature\cite{Anderson1994,ClarkeStrong1996,Biermann2002}.  This means that, so long as the temperature is much higher than the energy scale of the interchain coupling, the atomic limit will be valid.  For any particular material, this sets a lower bound on the temperature regime in which our results are applicable.

In this temperature regime of validity, the thermalization time within each chain (proportional to $1/T$) will be much less than the interchain hopping time (proportional to the inverse hopping strength), so that each individual one-dimensional chain will thermalize between hopping events.  We can therefore intuitively think of the interchain hopping as incoherent, though we do not explicitly use that fact anywhere in our calculations.

Assumption \eqref{assumption:polymer_array} is an accurate description for the case of anisotropic crystals.  The application to polymer films is less direct, as they are known to have regions where the polymers are relatively aligned in some organized array (as in assumption \ref{assumption:polymer_array}), as well as amorphous regions \cite{Stanford_experiment_paper,Bubnova2013,Steyrleuthner2014}. In the latter regions, which may account for a significant fraction of the overall film, as long as the polymers form a single two-dimensional layer and do not cross, at a sufficiently small scale the polymers should still form a neat array and our assumption will apply.  We can therefore approximately treat the film as consisting of a collection of randomly oriented domains, each of which individually satisfies the assumption.  We discuss this further in section \ref{section:discussion}.

\section{Generalized noninteracting model\label{section:nonint_model}}

The first version of our model is intended to capture the key Luttinger liquid behavior while still being simple enough to provide helpful physical intuition about the system we study.  We thus use a noninteracting  model for most of the calculation, finally substituting the Luttinger liquid density of states at the end.

To be precise, we add two more simplifying assumptions to those given in section \ref{section:assumptions} above:
\begin{enumerate}[(1)]
\setcounter{enumi}{3
\item \label{assumption:noninteracting}Each individual 1D chain can be described by a set of non-interacting single-particle orbitals, given by the Fourier modes of the localized on-site orbitals; the orbitals' energies are distributed according to the tunneling density of states of a Luttinger liquid, and each chain's orbitals are the same.
\item \label{assumption:hopping}Electrons hop from a well-defined single-particle eigenstate on one chain to an eigenstate with approximately the same energy and momentum on an adjacent chain.  The hopping strength is sharply peaked in $|k-k'|$ where $k$ and $k'$ are the wavenumbers on the two chains, and the value at $k=k'$ is independent of $k$.  (In practice, we assume the hopping is Gaussian in $k-k'$, but this assumption is only needed when we compare the two versions of our model, see Appendix \ref{appendix:correspondence}.)
\label{assumption:nonint_final}
\end{enumerate}

\noindent The \numberstringnum{\getrefnumber{assumption:nonint_final}} assumptions above lead to a specific interpretation of the standard tight-binding Hamiltonian
\begin{equation}
H = \sum_{j,k} E_k c^\dagger_{jk} c_{jk} - \sum_{jkk'} \left(t_{kk'}c^\dagger_{j,k} c_{j+1,k'} + \text{h.c.}\right)\label{eq:nonint_H}
\end{equation}

\noindent The index $j$ labels 1D chains, while $k$ and $k'$ label extended (Fourier state) orbitals on each chain.  $c^\dagger$ and $c$ are the usual fermion creation and annihilation operators, while $E_k$ is the single-particle energy corresponding to the orbital $k$.

In the noninteracting limit, the $E_k$ are just the energies of a one-dimensional tight-binding model $H_0 = -t_{\!/\!/\!}\sum_i c^\dagger_i c_{i+1} +\text{h.c.}$; if the lattice spacing is $a$, the energy levels are $E_k = -2t_{\!/\!/\!}\cos(ka)$, which are then linearized around the Fermi points $k=\pm k_F$.  When interactions are introduced, there are no longer well-defined single-particle orbitals, so we cannot give an explicit formula for the energies $E_k$.  Instead, we will derive an expressions for the transport coefficients in which the energy spectrum only appears via the density of states, for which we can use the well-defined single-particle tunneling density of states of a Luttinger liquid.

\subsection{Calculation of transport coefficients}

We calculate the transport coefficients in this model using the Kubo formalism.  For consistency with standard references, we use the conventions of reference \onlinecite{Mahan3},  in which case the electrical conductivity, thermal conductivity, and thermopower are given by
\begin{subequations}
\begin{align}
\sigma & = \frac{e^2}{T}L^{(11)}\label{eq:sigma_formula}\\
\kappa & = \frac{1}{T^2}\left[L^{(22)} - \frac{(L^{(12)})^2}{L^{(11)}}\right]\label{eq:kappa_formula}\\
S & = - \frac{1}{eT}\frac{L^{(12)}}{L^{(11)}}\label{eq:S_formula}
\end{align}
\label{eq:coeffs_from_L}
\end{subequations}

\noindent In a two dimensional material, each of these coefficients is actually a 2x2 matrix; the diagonal entries give the response in the direction of an applied field, while the off-diagonal entries give the response in a perpendicular direction (e.g., the Hall conductivity).  We will specifically focus on the longitudinal response in the interchain direction.

The $L^{(il)}$ coefficients in the transport coefficient formulas are defined by\cite[eqs. 3.487, 3.488]{Mahan3}
\begin{subequations}
\begin{align}
J & = -\frac{1}{T}L^{(11)}\nabla(eV) + L^{(12)}\nabla\left(\frac{1}{T}\right)\\
J_E & = -\frac{1}{T}L^{(21)}\nabla(eV) + L^{(22)}\nabla\left(\frac{1}{T}\right)
\end{align}
\end{subequations}
where $J$ is the particle current, or electrical current divided by the charge per particle, and $J_E$ is the energy current.  Note that $L^{(12)}=L^{(21)}$.  In practice, we find the $L^{(il)}$ coefficients in terms of current-current correlation functions as\footnote{Equation \eqref{eq:L_Kubo_formula} is a corrected version of (3.518) from reference \onlinecite{Mahan3}; see the Supplemental Material\cite{SuppMat} for details.}
\begin{align}
L^{(il)} & = \lim_{\omega\rightarrow 0} \lim_{\d\rightarrow 0} \frac{1}{\omega}\left[\frac{-i}{\Omega\b}\int_0^\b d\tau e^{i\omega_n \tau} \langle T_\tau j_l(\tau)j_i(0)\rangle\right]
\label{eq:L_Kubo_formula}\\
&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ^{i\omega_n\rightarrow \omega+i\d}\nonumber
\end{align}
where $j_1$ is the particle current operator $J$ and $j_2$ is the energy current operator $J_E$.  Both are the current operators for the interchain direction.  $\Omega$ is the volume of the system.  Because we calculate the transport coefficients at finite temperature, we perform the calculation using the Matsubara formalism.  $\tau$ is the imaginary time, $\omega_n=2\pi n/\b$ for $n=0,1,2,\cdots$ are the discrete (bosonic) Matsubara frequencies, and $i\omega_n\rightarrow \omega+i\d$ indicates analytic continuation from the positive imaginary axis to just above the positive real axis.  In practice we will take only the real part of $L^{(il)}$, since we are interested specifically in transport.

The current operators we find using \cite{Mukerjee2005}
\begin{subequations}
\begin{align}
J & = \lim_{k\rightarrow 0} \frac{1}{k} \sum_j [N_j,H]e^{ika_c j}\\
J_E & = \lim_{k\rightarrow 0} \frac{1}{k} \sum_j [H_j,H]e^{ika_c j}\label{eq:def_JE_operator}
\end{align}\label{eq:def_J_operators}
\end{subequations}

\noindent in units where $\hbar=1$.  Here $a_c$ is the distance between 1D chains and $N_j$ is the total number operator on chain $j$, $N_j = \sum_k c^\dagger_{jk}c_{jk}$.  $H_j$ is the part of the Hamiltonian associated with chain $j$, which includes both the on-chain portion
\begin{subequations}
\begin{equation}
h_j = \sum_{k} E_k c^\dagger_{jk} c_{jk}
\end{equation}
and the hopping portion
\begin{equation}
h'_j = - \frac{1}{2}\sum_{kk'} t_{kk'}\left(c^\dagger_{j,k} c_{j+1,k'} + c^\dagger_{j-1,k} c_{j,k'} \right) +  \text{h.c.}
\end{equation}
\end{subequations}
This leads, after some algebra, to the expressions
\begin{subequations}
\begin{align}
J & = ia_c\sum_{jkk'}t_{kk'}c^\dagger_{j-1,k}c_{j,k'} - t_{kk'}^\ast c^\dagger_{j,k'}c_{j-1,k}\label{eq:nonint_J_e}\\
J_E & = ia_c\sum_{jkk'} \left[\left(\frac{E_k+E_{k'}}{2}\right) \left(t_{kk'}c^\dagger_{j-1,k} c_{jk'}-\text{h.c.}\right)\right]\label{eq:nonint_J_Q}
\end{align}
\end{subequations}
From these current operators and equation \eqref{eq:L_Kubo_formula}, we derive (see Appendix \ref{appendix:nonint_Kubo}) the expression
\begin{equation}
\text{Re}\left[L^{(il)}\right] = \frac{A a_c t^2 v \b^{-n_{il}}}{2\pi^4}\int \frac{g^2(y/\b)y^{n_{il}}}{\left(1+e^y\right)\left(1+e^{-y}\right)}\,\,dy\label{eq:nonint_L}
\end{equation}
where $n_{il}=i+l-2$ (e.g., 0 for $L^{(11)}$), $v$ is the (possibly renormalized by interactions) Fermi velocity, $A$ is a dimensionless number, $t$ is the peak value of the interchain hopping $t=t_{kk}$, $\b$ as usual is $1/T$ (we use units of $k_B=1$), and $g(E)$ is the electronic density of states.  The integral over the dimensionless variable $y=\b E$ runs from $-\infty$ to $\infty$.

The form of the integrand can be intuitively understood from a semiclassical perspective.  If a particle is hopping from an orbital at energy $E$ on one chain to an orbital at energy $E$ on another, then the number of ways that can happen is the number of orbitals at that energy on the first chain, $g(E)$, multiplied by the fraction that are occupied, $(1+e^{\b E})^{-1}$, times the number of orbitals at that energy on the second chain, $g(E)$, multiplied by the fraction that are unoccupied, $(1+e^{-\b E})^{-1}$.  Multiplying all of these factors and integrating over the energy gives
\begin{equation}
\int \frac{g^2(E)}{(1+e^{\b E})(1+e^{-\b E})} dE.\label{eq:semiclassical_intuition}
\end{equation}
This should be proportional to the hopping rate, and therefore to the electrical conductivity.  Indeed, equation \eqref{eq:semiclassical_intuition} looks just like the integrand in equation \eqref{eq:nonint_L} for $L^{(11)}$, which is proportional to the electrical conductivity.  The fact that a semiclassical picture is helpful in understanding equation \eqref{eq:nonint_L} is not too surprising given that our weak hopping assumption is only valid when the temperature is high enough for the interchain hopping to be incoherent.

This is the point in the calculation where the fact that each 1D chain is a Luttinger liquid becomes important.  The density of states for a Luttinger liquid is given by Eq. (61) of reference \onlinecite{Braunecker} as
\begin{equation}
g_{LL}(E)=2\frac{|E/W|^{\gamma-1}}{2\pi v \Gamma(\gamma)},\label{eq:LL_dos}
\end{equation}
valid for $E\ll W$, where $W=v/a$ is proportional to the Fermi energy ($E_F\propto k_F^2/m = (k_F/m)/k_F^{-1}\propto v/a$) or bandwidth of the underlying 1D model and $\gamma$ is a measure of interaction strength in the Luttinger liquid defined by
\begin{equation}
\gamma=\frac{K+K^{-1}}{2}.\label{eq:def_gamma}
\end{equation}
$K$ is the usual Luttinger liquid interaction parameter, as defined for the Luttinger liquid Hamiltonian below (equation \ref{eq:LL_H}).  (Note that using $K$ for this parameter is a relatively standard convention, used for instance in the book by Giamarchi\cite{GiamarchiBook}, though some authors refer to it as $g$ or $K^2$.\cite{KaneFisher1992,KaneFisher1992b,KaneFisher1996,finiteLLpaper})  $K=1$ corresponds to noninteracting electrons, while $K<1$ corresponds to repulsive interactions and $K>1$ corresponds to attractive interactions.  We have introduced the new parameter $\gamma$, which is symmetric in $K$ and $K^{-1}$ and thus is independent of whether the interactions happen to be attractive or repulsive.  It always satisfies $\gamma \geq 1$, and $\gamma=1$ if and only if the system is noninteracting.

Substituting equation \eqref{eq:LL_dos} into equation \eqref{eq:nonint_L} and using that result in equations \eqref{eq:coeffs_from_L}, we find the following results for the transport coefficients:
\begin{subequations}
\begin{align}
\sigma & = \frac{a_c e^2 t^2}{v T}\left(\frac{T}{W}\right)^{2(\gamma-1)}\!\!\!\!\!\!\!\!\!\!\times\frac{A}{2\pi^6\Gamma(\gamma)^2}\int_{y}\frac{|y|^{2(\gamma-1)}}{\left(1+e^y\right)\left(1+e^{-y}\right)}dy\label{eq:nonint_sigma_LLg}\\
\kappa & = \frac{a_c t^2}{v}\left(\frac{T}{W}\right)^{2(\gamma-1)}\!\!\!\!\!\!\!\!\times\frac{A}{2\pi^6\Gamma(\gamma)^2}\int_{y}\frac{y^2|y|^{2(\gamma-1)}}{\left(1+e^y\right)\left(1+e^{-y}\right)}dy\label{eq:nonint_kappa_LLg}\\
S & = 0
\end{align}
\label{eq:nonint_LLg_results}
\end{subequations}

\noindent Both the thermopower and the second term of equation \eqref{eq:kappa_formula} for the thermal conductivity vanish because $L^{(12)}$ is 0 when the density of states is particle-hole symmetric.  Mathematically this follows because the integrand in equation \eqref{eq:nonint_L} is odd when $g(E)$ is an even function.

\subsection{Correction for nonzero thermopower}

To model a real material and get nonzero thermopower, we can introduce an asymmetry in the band structure.  In particular, the Tomonaga-Luttinger model begins by linearizing a typical 1D band structure around the Fermi points, so we adopt the picture that the Luttinger liquid arises from adding interactions to a 1D electron gas with a typical dispersion $E=\frac{\hbar^2k^2}{2m}\propto k^2$.  In that case, the density of states is $dk/dE\propto E^{-1/2}$.  In our calculations above we have set the Fermi level to $E=0$, in which case the noninteracting density of states becomes
\begin{equation}
g_\text{1D}(E)\propto (E_F+E)^{-1/2}.
\end{equation}
The Fermi energy is proportional to $v/a$, so for consistency with equation \eqref{eq:LL_dos} we can write it as $E_F = b \, W$ for a dimensionless constant $b$.  Using this 1D density of states as a correction to the Luttinger liquid one gives
\begin{equation}
g(E) = \frac{g_\text{LL}(E)}{\sqrt{1+E/(b \, W)}}\approx g_\text{LL}(E)\left(1-\frac{1}{2}\frac{E}{b\,W}\right).\label{eq:modified_DOS}
\end{equation}
This density of states is a phenomenological way of capturing the real physical behavior of the system which should be accurate enough to find how the thermopower depends on temperature.  The most important features are the violation of particle-hole symmetry by the introduction of a bandwidth and the preservation of the density of states to lowest order in $E/W$ when $E$ is small (near the Fermi energy).

If we calculate $L^{(12)}$ with equation \eqref{eq:modified_DOS} replacing equation \eqref{eq:LL_dos} as the density of states, we find for the thermopower
\begin{equation}
S = \frac{k_B^2 T}{W e} \times\frac{\int\frac{y^2|y|^{2(\gamma-1)}}{\left(1+e^{y}\right)\left(1+e^{-y}\right)}dy}{b\int\frac{|y|^{2(\gamma-1)}}{\left(1+e^{y}\right)\left(1+e^{-y}\right)}dy}
\end{equation}
where Boltzmann's constant has been restored to get the correct final units.

Note that in principle we could also use the same correction for the conductivities, equations \eqref{eq:nonint_LLg_results}, but any additional terms would be higher order in $k_B T/W$ than those given above.  $k_B T/W$ must be small, otherwise the Tomonaga-Luttinger model, which is based on a linearized band structure (i.e., $W\rightarrow \infty$), would not be applicable at that temperature.

\subsection{Lorenz number}

The expressions for the conductivities, equations \eqref{eq:nonint_LLg_results}, are clear and understandable, but they do contain material-dependent parameters like $a_c$, $v$, and $W$.  To find a robust result that can be tested experimentally, we would like a quantity in which these material-dependent quantities do not appear.  One such parameter is the Lorenz number,
\begin{equation}
L = \frac{\kappa}{\sigma T}.
\end{equation}
This is a particularly useful quantity to consider, since the Wiedemann-Franz law states that for a noninteracting system or for a Fermi liquid, the Lorenz number should take a specific value, namely
\begin{equation}
L_0 = \frac{\pi^2}{3}\left(\frac{k_B}{e}\right)^2.
\end{equation}
The Lorenz number for our model can be found by dividing the results from equations \eqref{eq:nonint_LLg_results} to get
\begin{equation}
L = \frac{k_B^2}{e^2} \frac{\int\frac{y^2|y|^{2(\gamma-1)}}{\left(1+e^{y}\right)\left(1+e^{-y}\right)}dy}{\int\frac{|y|^{2(\gamma-1)}}{\left(1+e^{y}\right)\left(1+e^{-y}\right)}dy}.\label{eq:nonint_L_LLg}
\end{equation}
As expected from the Wiedemann-Franz law, in the noninteracting limit of $\gamma=1$ we get precisely $L_0$.  At $\gamma>1$, this expression for $L$ can be evaluated via numerical integration.  With interactions, $\gamma>1$, we find that $L>L_0$, violating the Wiedemann-Franz law.  The Lorenz number is plotted as a function of the interaction strength $\gamma$ in the lower curve in figure \ref{fig:combined_Lorenz_number}.

\begin{figure}
\centering
\includegraphics[width=\columnwidth]{Lorenz_number_combined_detail_inset_eps.eps}
\caption{(Color online) Lorenz number, $L$, as calculated in the generalized noninteracting (GN) and Luttinger liquid (LL) models.  The Lorenz number is plotted as a function of the interaction strength $\gamma$ in units of $L_0$, the value expected from the Wiedemann-Franz law.  For both models, we find that $L=L_0$ in the noninteracting case $\gamma=1$.  Electron-electron interactions ($\gamma>1$) lead to a violation of the Wiedemann-Franz law; the violation is stronger in the LL model than in the GN model.  The Lorenz number is evaluated at discrete points in the LL model; error bars indicate the precision of numerical results as described in the text.  Lines connecting the data points for the LL model show linear interpolation between adjacent points, and the dashed line below $\gamma=1.005$ in the inset shows extrapolation to $\gamma=1$.}
\label{fig:combined_Lorenz_number}
\end{figure}

The Lorenz number should scale approximately as $\gamma^2$ in this model, since the extra two powers of $y$ in equation \eqref{eq:nonint_L} that appear for $L^{(22)}$ (and therefore $\kappa$) but not for $L^{(11)}$ (and therefore $\sigma$) become derivatives with respect to $x$ if the expression is rewritten via Fourier transform; these derivatives act on the Green's function that looks roughly like $f(x)^{-\gamma}$ and thus pull down two factors of $\gamma$.  To test that it is indeed the case that $L\approx L_0 \gamma^2$, we define $a(\gamma)$ by $L=L_0\gamma^{a(\gamma)}$ in which case\begin{equation}
a(\gamma)=\frac{\log(L/L_0)}{\log(\gamma)}.\label{eq:def_a_gamma}
\end{equation}
This quantity is plotted in figure \ref{fig:nonint_L_exponent}.  From the plot we see that the exponent $a$ is between $2.35$ and $2$ for all interaction strengths $\gamma$.  For large $\gamma$, the scaling of the Lorenz number is close to $\gamma^2$; for small $\gamma$, expanding around $\gamma=1$ gives
\begin{align}
a(\gamma\approx 1) & = 1-2\log(\pi)+\frac{6}{\pi^2}\!\left(\gamma_1{\kern 0.001em}'\left(\frac{1}{2}\right)-\gamma_1{\kern 0.001em}'(1)\!\right)\nonumber\\
&\approx 2.3432
\end{align}
where $\gamma_1(\nu)$ is a generalized Stieltjes constant\cite{Blagouchine2014}\textsuperscript{,}\footnote{For purposes of calculation, the generalized Stieltjes constant is implemented in the commercial software Wolfram Mathematica as $\gamma_n(\nu)=\text{StieltjesGamma}[n,\nu]$\cite{WolframSGDef}}.

\begin{figure}
\centering
\includegraphics[width = \columnwidth]{Noninteracting_Lorenz_number_exponent_eps.eps}
\caption{The Lorenz number scales as $L/L_0=\gamma^{a(\gamma)}$.  For the generalized noninteracting model we find $2<a(\gamma)<2.35$ for all $\gamma$, with $L\approx L_0\gamma^2$ for large $\gamma$.}
\label{fig:nonint_L_exponent}
\end{figure}

\subsection{Summary of generalized noninteracting model}

Our most robust predictions are those that do not depend on any material-dependent parameter but the interaction strength.  These are (a) the power law dependencies of $\sigma$, $\kappa$, and $S$ on temperature and (b) the Lorenz number.  We find that
\begin{subequations}
\begin{align}
\sigma & \propto T^{2\gamma-3}\label{eq:nonint_sigma_T_dep}\\
\kappa & \propto T^{2\gamma-2}\label{eq:nonint_kappa_T_dep}\\
S & \propto T
\end{align}\label{eq:nonint_T_dependence}
\end{subequations}
and
\begin{align}
L & = \frac{k_B^2}{e^2} \frac{\int\frac{y^2|y|^{2(\gamma-1)}}{\left(1+e^{y}\right)\left(1+e^{-y}\right)}dy}{\int\frac{|y|^{2(\gamma-1)}}{\left(1+e^{y}\right)\left(1+e^{-y}\right)}dy}\nonumber\\
& \approx L_0 \gamma^2
\end{align}
In the noninteracting case, $\gamma=1$, the Lorenz number agrees with the usual Wiedemann-Franz Law.  With either attractive or repulsive interactions, the Wiedemann-Franz law is violated as shown in figure \ref{fig:combined_Lorenz_number}.

\section{Luttinger liquid model\label{section:LL}}

In the second version of our model, we introduce Luttinger liquid physics much earlier in the analysis.  To do so, we replace assumptions \eqref{assumption:noninteracting} and \eqref{assumption:hopping} with two new, corresponding assumptions:
\begin{enumerate}[(1')]
\setcounter{enumi}{3
\item \label{assumption:LL_interacting}Each individual 1D chain is described by the Luttinger liquid Hamiltonian\cite{GiamarchiBook},
\begin{equation}
H = \frac{1}{2\pi}\int dx \left[vK(\nabla\t)^2 + \frac{v}{K}(\nabla\phi)^2\right]\label{eq:LL_H}
\end{equation}
where again we have set $\hbar=1$.  As above, $K$ is a parameter that measures interaction strength and $v$ is the renormalized Fermi velocity. $\phi$ and $\theta$ are bosonic field operators related to the fermion operators by\cite{GiamarchiBook}
\begin{subequations}
\begin{align}
\psi_\a(x) & = U_\a \lim_{a\rightarrow 0} \frac{1}{\sqrt{2\pi a}}e^{i\a k_F x}e^{-i(\a\phi(x)-\t(x))}\\
\psi^\dagger_\a(x) & = U_\a^\dagger \lim_{a\rightarrow 0} \frac{1}{\sqrt{2\pi a}}e^{-i\a k_F x}e^{i(\a\phi(x)-\t(x))}
\end{align}\label{eq:LL_psi_from_pt}
\end{subequations}

\noindent where $\a$ can be $R$ or $L$ (labeling right-movers versus left-movers) when used as an index and 1 or $-1$, respectively, when used as a multiplicative factor.  The $U_r$ operators are called Klein factors, and are included to make sure that the fermion operators anticommute and that they do not conserve particle number.
\item \label{assumption:LL_hopping}Electrons hop between real-space localized orbitals.  The hopping strength is sharply peaked in $|x-x'|$, where $x$ and $x'$ are the locations along the two chains, measured from the same ``center'' point (so that all the ``$x=0$'' points lie on a line perpendicular to the chains).  In the thermodynamic limit, a delta-function hopping in real space is consistent with the sharply peaked hopping in Fourier space from assumption \eqref{assumption:hopping} from the first version of our model (see Appendix \ref{appendix:correspondence}).  We also assume that right-movers on one chain can only hop to right-movers on the adjacent chain and the same for left-movers; this is needed for consistency with the approximate momentum conserving hopping in the generalized noninteracting model.
\label{assumption:LL_final}
\end{enumerate}

\noindent Including both on-chain and hopping terms, the Hamiltonian for this second version of our model is:
\begin{widetext}
\begin{align}
H & = \sum_j \, H_j = \sum_j \, h_j + h'_j\nonumber\\
h_j & = \frac{1}{2\pi}\int \! dx \left[vK\left(\nabla \t_j\right)^2 + \frac{v}{K}\left(\nabla \phi_j\right)^2\right]\label{eq:LL_model_H}\\
h'_j & =\! -\frac{1}{2}\!\sum_{\a\b}\!\int\! dx\,dx'\!\left[t_{\a\b}(x-x')\left(\psi_{j\a}^\dagger(x)\psi_{j+1,\b}(x')+\, \psi_{j-1,\a}^\dagger(x)\psi_{j\b}(x')\right) + \,\, \text{h.c.}\right]\nonumber
\end{align}

\subsection{Calculation of transport coefficients}

\noindent As in the generalized noninteracting model, to find the transport coefficients we first find operators for the electrical and energy currents.  This can be done using equations \eqref{eq:def_J_operators} just as before, but with the new definitions for $h_j$ and $h'_j$.  The results (for some details of the calculation, see Appendix \ref{appendix:LL}) are
\begin{subequations}
\begin{align}
J & = -ia_c\sum_j\sum_{\a\b = R,L} \int t_{\a\b}(x-x')\left[\psi_{j\a}^\dagger(x)\psi_{j-1,\b}(x') - \psi_{j-1,\a}^\dagger(x)\psi_{j\b}(x')\right]\,dx\,dx'\label{eq:LL_J_e}\\
J_E & = -\frac{ia_c v}{2}\sum_{j\a\b}\int t_{\a\b}(x-x')\left[\left([\nabla_j]^\a_{x} +[\nabla_{j-1}]^\b_{x'}\right)\psi^\dagger_{j\a}(x)\psi_{j-1,\b}(x') - \left([\nabla_{j-1}]^\a_{x} +[\nabla_j]^\b_{x'}\right)\psi^\dagger_{j-1,\a}(x)\psi_{j\b}(x')\right]\,dx\,dx'
\label{eq:LL_J_E}
\end{align}
\end{subequations}
where
\begin{equation}
[\nabla_j]^\a_y = \a K\nabla\t_j(y) - K^{-1}\nabla\phi_j(y).\label{eq:def_nabla}
\end{equation}

\noindent Unlike in the generalized noninteracting model, we do not find a single simple formula like equation \eqref{eq:nonint_L} that gives all the transport coefficients.  Instead, the particularly nice expressions that we find are for the current-current correlators in terms of the Green's function for a single Luttinger liquid:
\begin{subequations}
\begin{align}
\langle J(\tau) J\rangle & = -2N_c L\left(\frac{a_c t}{2\pi}\right)^2\sum_\a\int dx\, G_\a(x,\tau)G_\a(-x,-\tau)\label{eq:LL_JeJeCorrelator_unspecified_Greens}\\
\langle J_E(\tau) J_E\rangle & = -2N_cL\gamma^2\left(\frac{a_c v t}{2\pi}\right)^2\sum_\a\int dx\,\big[(k_F+i\a\partial_x)G_\a(x,\tau)\big]\times\big[(k_F-i\a\partial_x)G_\a(-x,-\tau)\big]\label{eq:LL_JEJECorrelator_unspecified_Greens}\\
\langle J_E(\tau) J\rangle & = 2v\gamma N_c L\left(\frac{a_c t}{2\pi}\right)^2 \sum_{\a}\int dx\,
G_\a(x,\tau)(k_F-i\a\partial_x) G_\a(-x,-\tau)\label{eq:LL_JEJeCorrelator_unspecified_Greensv2}
\end{align}\label{eq:JJCorrelators_unspecified_Greens}
\end{subequations}

\noindent For the Green's function we use the expression\cite{finiteLLpaper,Braunecker}
\begin{equation}
G_\a(x,\tau) = -\frac{e^{i\a k_F x}}{2\pi a}\left[\frac{-ia}{\frac{v\b}{\pi}\sinh\left(\frac{x-iv\tau}{v\b/\pi}\right)}\right]^{\frac{\gamma-\a}{2}}\left[\frac{ia}{\frac{v\b}{\pi}\sinh\left(\frac{x+iv\tau}{v\b/\pi}\right)}\right]^{\frac{\gamma+\a}{2}}\label{eq:LL_Greens}
\end{equation}
and we are then able to perform the integration over $x$ exactly, getting results in terms of the Appell hypergeometric function $F_1$ as defined in \S 16.13 of reference \onlinecite{NIST:DLMF}.\footnote{For purposes of calculation, the function $F_1$ is implemented in the commercial software Wolfram Mathematica as $F_1(a;b_1,b_2;c;x,y)=\text{AppellF1}[a,b1,b2,c,x,y]$\cite{WolframAppellDef}}  As an example, the result for $\langle J(\tau)J\rangle$ is
\begin{equation}
\langle J(\tau') J\rangle= 4N_c L\left(\frac{a_c t}{2\pi}\right)^2\frac{2a}{(2\pi a)^2}\left(\frac{2\pi a}{v\b}\right)^{2\gamma-1}\bigg(2 f(\gamma,\tau',1,1)-\cos(2\tau')\big(f(\gamma,\tau',0,1) + f(\gamma,\tau',2,1)\big)\bigg)
\label{eq:LL_JeJeCorrelatorFull}
\end{equation}
where $\tau'$ is a scaled version of the imaginary time, $\tau' = \tau\pi/\b$, and
\begin{equation}
f(\gamma,\tau,n,m) = \frac{F_1(\gamma+n;\gamma+m,\gamma+m;\gamma+n+1;e^{2i\tau},e^{-2i\tau})}{\gamma+n}.\label{eq:f_from_F1}
\end{equation}
\end{widetext}
The analogous expressions for the other two current-current correlators are longer and more complex, and thus proportionally less enlightening.  We present them in Appendix \ref{appendix:LL} for the edification of the interested reader.

The next step is to evaluate each of the $L^{(il)}$ coefficients using equation \eqref{eq:L_Kubo_formula}.  In the previous model it was possible to perform the Fourier transform and analytic continuation analytically, but here we must perform the $\tau$ integrals of the current-current correlators numerically for each Matsubara frequency and then numerically perform the analytic continuation and limits.  The procedure we follow is discussed further in Appendix \ref{appendix:LL}, and in great detail in the Supplemental Material\cite{SuppMat}.

For each transport coefficient, we get a numerical part from the procedure mentioned above and a prefactor that contains all the dimensionful quantities, notably the dependence on temperature.  Including for now only the dimensionful quantities, we find
\begin{subequations}
\begin{align}
\sigma & \propto \frac{a_c a \,q^2 t^2}{v^2}\left(\frac{a}{v\b}\right)^{2\gamma-3}\\
\kappa & \propto \frac{a_c t^2}{v}\left(\frac{a}{v\b}\right)^{2\gamma-2}\\
S & = 0
\end{align}\label{eq:LL_transport_dimensions}
\end{subequations}

\noindent Recalling that the energy scale $W$ introduced in the Luttinger liquid density of states, equation \eqref{eq:LL_dos}, was $W=v/a$, the dependence of the transport coefficients on the material-dependent parameters $a_c$, $a$, and $v$ in this model (equations \ref{eq:LL_transport_dimensions}) precisely matches what we found in the generalized noninteracting model (equations \ref{eq:nonint_LLg_results}).

In the generalized noninteracting model, we introduced a correction to the density of states to find a nonzero thermopower.  Due to the complexity of the full Luttinger liquid model, we consider the equivalent correction here to be beyond the scope of this paper.

\subsection{Lorenz number}
The numerical analytic continuation has not yet been needed for the results presented above.  We would like, however, to evaluate the Lorenz number numerically as a function of the interaction strength, $\gamma$, just as in the generalized noninteracting model.  For that calculation, the full numerics are needed.

To compute the precise transport coefficients, for each interaction strength $\gamma$ we must separately evaluate the Fourier transform of the current-current correlation functions at a number of Matsubara frequencies, fit an analytic function to these results, analytically continue the function, and then take the limits as the frequency $\omega$ and the infinitesimal parameter $\d$ go to 0.  (For details, see Appendix \ref{appendix:LL}.)

Due to the complexity of the correlation functions (for instance equation \ref{eq:LL_JeJeCorrelatorFull}), the calculation of each Fourier transform, and thus the calculation of transport coefficients for each interaction strength $\gamma$, is very computationally expensive.  We therefore evaluate the Lorenz number for a limited number of values of the interaction strength, with a higher density around $\gamma=1$ to make sure that the results in the noninteracting limit are reliable.  The results are shown for $\gamma$ in the range 1 to 3 by the discrete data points in figure \ref{fig:combined_Lorenz_number} (connected by linear interpolation for visual clarity).  An inset shows a detail of $\gamma\in[1,1.05]$; from the inset it is clear that in the noninteracting limit the Lorenz number approaches the expected value from the Wiedemann-Franz law.

The error bars on the Luttinger liquid model data in figure \ref{fig:combined_Lorenz_number} indicate the numerical precision of the Lorenz number for each $\gamma$.  We compute the numerical integral for each Fourier transform with a relative precision of $10^{-10}$, and allowing the values of the Fourier transforms to vary within this range and recomputing the Lorenz number gives a sharply peaked distribution of possible values of $L$.  The error bars in the figure show one standard deviation of this distribution for each interaction strength $\gamma$.

Comparing the results of the full Luttinger liquid model with the corresponding results for the generalized noninteracting model, as shown in the upper and lower curves respectively in figure \ref{fig:combined_Lorenz_number}, we see that the full Luttinger liquid model exhibits a stronger violation of the Wiedemann-Franz law with increasing interaction strength.  We argued that in the generalized noninteracting model the Lorenz number should scale as $\gamma^2$ because the two extra factors of energy for $L^{(22)}$ relative to $L^{(11)}$ in equation \eqref{eq:nonint_L} act, in a real-space representation, as derivatives of the Green's function.  For the full Luttinger liquid model, we can make a similar argument that $L/L_0\approx \gamma^4$.  There are indeed two derivatives acting on the Green's function in the expression for $\langle J_E(\tau)J_E\rangle$, equation \eqref{eq:LL_JEJECorrelator_unspecified_Greens}, that are not present in the expression for $\langle J(\tau)J\rangle$, equation \eqref{eq:LL_JeJeCorrelator_unspecified_Greens}, giving rise to the same two factors of $\gamma$ as in the generalized noninteracting model.

There are additionally two factors of $\gamma$ in the prefactor in the expression for $\langle J_E(\tau)J_E\rangle$, which come from the $[\nabla_j]^\a_x$ operators in the expression for the energy current operator, equation \eqref{eq:LL_J_E}, and are thus missing from the generalized noninteracting model because there the energy current operator was derived in the noninteracting limit where $\gamma=1$.  With these two additional factors of $\gamma$ included, we find that the Lorenz number should scale approximately as $L/L_0\approx \gamma^4$.

This argument neglects the full complexity of the correlation functions, so to find more precisely how the Lorenz number scales with $\gamma$ we again introduce the function $a(\gamma)$ defined by equation \eqref{eq:def_a_gamma}, $L/L_0 = \gamma^{a(\gamma)}$.  This is plotted in figure \ref{fig:LL_L_exponent}.  We find that $L/L_0$ satisfies $\gamma^{3.2}<L/L_0<\gamma^{3.7}$ for $\gamma\leq 3$.  This is a slightly weaker dependence than the predicted $\gamma^4$, but it is still a much stronger violation of the Wiedemann-Franz law than $L/L_0\approx \gamma^2$ from the generalized noninteracting model.

\begin{figure}
\centering
\includegraphics[width=\columnwidth]{LL_Lorenz_number_exponent_connected_eps.eps}
\caption{\label{fig:LL_L_exponent}Exponent $a$ in $L/L_0=\gamma^{a(\gamma)}$ for the Luttinger liquid model.  The dependence on $\gamma$ is stronger than in the generalized noninteracting model. Lines are given by linear interpolation between adjacent data points, and error bars are omitted for clarity.}
\end{figure}

\subsection{Summary of Luttinger liquid model}

As in the generalized noninteracting model, it is useful to summarize those results that do not depend on any material-dependent parameter apart from the interaction strength.  For the dependence of the conductivities on temperature, we find the same power laws as in the generalized noninteracting model, namely $\sigma \propto T^{2\gamma-3}$ and $\kappa \propto T^{2\gamma-2}$.  For the Lorenz number we find a stronger violation of the Wiedemann-Franz law than in the generalized noninteracting model.  We analytically estimate that
\begin{equation}
L \approx L_0 \gamma^4
\end{equation}
and numerically observe that
\begin{equation}
L_0\gamma^{3.2}<L<L_0\gamma^{3.7}
\end{equation}
The precise dependence of the Lorenz number on the interaction strength is shown in figure \ref{fig:combined_Lorenz_number}.  In the noninteracting case, $\gamma=1$, the Lorenz number is $L_0$, the expected value from the Wiedemann-Franz Law.

\section{Discussion and analysis}\label{section:discussion}

We have analyzed two different models for weakly coupled Luttinger liquids, finding in both cases the electrical and thermal conductivity.  In terms of the interaction parameter $\gamma$, the conductivities scale in both models as $\sigma \propto T^{2\gamma-3}$ and $\kappa \propto T^{2\gamma-2}$.  In both cases we find a violation of the Wiedemann-Franz law with increasing interaction strength; for the generalized noninteracting model $L\approx L_0\gamma^2$ as shown in figures \ref{fig:combined_Lorenz_number} and \ref{fig:nonint_L_exponent}, while for the Luttinger liquid model $L\approx L_0\gamma^{3.2}$ as shown in figures \ref{fig:combined_Lorenz_number} and \ref{fig:LL_L_exponent}.  This type of violation of the Wiedemann-Franz law as a power of the interaction strength is similar to the result of Kane and Fisher\cite{KaneFisher1996}, although the precise dependence is of course different since our models describe a different physical system.  In the generalized noninteracting model we also find a nonzero expression for the thermopower if we correct the density of states to account for particle-hole symmetry breaking, in which case $S\propto T$.  This linear dependence of thermopower on temperature, which matches the expected behavior in a Fermi liquid, was also found by Kane and Fisher in their coupled chain model\cite{KaneFisher1996}.

The violation of the Wiedemann-Franz law that we observe in both models is an indication that Luttinger liquid behavior survives when 1D chains are coupled to form a two-dimensional material.  Just how large is the violation in practice?  Experimental measurements\cite{Vescoli1998} and theoretical calculations\cite{Schulz1990,Ejima2005,Ejima2006,Sirker2012,Affleck2013} have found Luttinger parameters in a typical range of about $0.2$ through $1.5$, corresponding to values of $\gamma$ up to about 3.  In both our models, $\gamma=3$ would lead to a large violation of the Wiedemann-Franz law by an order of magnitude or more, an easily measurable effect that could be observed in experiments.

The results summarized here are all independent of any material-dependent parameters apart from the Luttinger liquid interaction parameter, which makes them good candidates for experimental testing and verification on any system with strong anisotropy that might lead to quasi-one-dimensional behavior.  One very direct application of our theory would be to highly anisotropic crystals, as they typically have electron hopping strength along one axis which is at least an order of magnitude stronger than the hopping along the other two axes\cite{Vescoli1998}.  For temperatures between the two hopping scales, it would be reasonable to treat the system as a collection of weakly coupled 1D chains as we have done here, and by the nature of the crystal they form an ordered array, again matching our model.  Such anisotropic crystals are known to show strong violations of the Wiedemann-Franz law, especially in the Hall direction in a magnetic field~\cite{Wakeham2011}.  By comparing the measured violations of the Wiedemann-Franz law in these systems with our predictions, it should be possible to estimate the effective Luttinger parameter $K$ for the constituent one-dimensional chains.  Conversely, if $K$ is independently known then such measurements would serve as a verification of our predictions.

Applying our theory to polymer films, the original motivation of the work, requires some additional work since the films are partially amorphous.  One approach would be to treat the polymer film as a polycrystal, consisting of randomly oriented grains; within each grain, the polymers form an ordered array to which our theory directly applies.  The overall transport properties of the polymer film could then be found by averaging using methods like those discussed in reference \onlinecite{Isichenko1992}.  The precise level of alignment of polymers can also vary significantly between films\cite{Stanford_experiment_paper,Bubnova2013,Steyrleuthner2014}, and more work is needed to properly take this into account.  One experimental result on polymer films which is clearly consistent with our calculations is the fact that some polymers show conductivity increasing with temperature, while others show the opposite behavior\cite{Bubnova2013}.  We find that $\sigma$ increases with temperature if the interaction strength is large enough, $\gamma>3/2$, but decreases with increasing temperature for $1\leq\gamma<3/2$.

Numerical studies of transport and other dynamical properties in quasi-one-dimensional systems have made great progress since the advent of matrix product state algorithms for time dependence~\cite{vidalrg,whitetdmrg,schollwoeck}.  In the case of a single chain, it is possible to see the characteristic power laws of Luttinger liquid behavior~\cite{huangkarrasch}, and while coupled chains are considerably more demanding, it has been possible to access at least some excited-state properties~\cite{jameskonik}.  Coupled-chain numerical studies could in principle provide a more precise and tunable ``numerical laboratory'' to test our predictions than current polymer experiments.

There are a number of ways that our models could be extended for future work.  We have dealt only with spinless Luttinger liquids, so a spin sector could be added.  Due to the spin-charge separation in Luttinger liquids, this would be a relatively simple change and would just result in extra additive contributions to some $L^{(il)}$ coefficients.  The models could also be made more complete via the addition of disorder and by going to higher order in the perturbation theory in the interchain hopping strength.  The latter two corrections would be potentially quite difficult, though disorder could be added at a relatively late stage in the calculation by modifying the density of states as used in equation \eqref{eq:nonint_L} or the Green's function in equations \eqref{eq:JJCorrelators_unspecified_Greens}.

To implement these or other extensions of our model, if the goal is only to find how transport properties depend on temperature then it will apparently be sufficient to use a noninteracting model for most of the calculation as in section \ref{section:nonint_model}; if the precise values of the transport coefficients are needed, such as for calculating the Lorenz number, then a more complete calculation, as in section \ref{section:LL}, will be required.

\begin{acknowledgments}
We would like to acknowledge generous support from the Air Force Office of Scientific Research Multidisciplinary Research Program of the University Research Initiative (AFOSR MURI FA9550-12-1-0002), as well as travel support from a Simons Investigatorship.  We also want to thank Christoph Karrasch, Takahiro Morimoto, Snir Gazit, Benjamin Ponedel, and Len Evans for helpful conversations.
\end{acknowledgments}

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Lumiconger arafura är en fiskart som beskrevs av Castle och Paxton, 1984. Lumiconger arafura ingår i släktet Lumiconger och familjen havsålar. Inga underarter finns listade i Catalogue of Life.

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Under the hood, the Galaxy S10 packs an octa-core Exynos 9820 processor along with the Mali-G76 MP12 GPU. The SoC is backed by 8GB of RAM and 128GB of internal storage. The memory of the phone can be further expanded via a micro-sd card up to 512GB.
In the camera side of things, the Galaxy S10 packs a triple camera set up. Among the three, the primary is a 12MP sensor with a variable aperture of f/1.5-2.4. Next, we have a 12MP telephoto lens with f/2.4 aperture and a 16MP ultra-wide angle sensor with an f/2.2 aperture. The phone also comes with an f/1.9 10MP front camera sitting snug in a punch hole cut out.
Moving on, at the software side of things, the phone runs the latest One UI based on Andoird 9.0 Pie. It also comes with a fingerprint scanner, using ultrasonic technology, embedded into the display of the phone. And finally powering the device all up is a 3400mAh non-user removable battery. Samsung this time around as thrown in Fast Battery Charging, Fast Wireless charging, and Reverse Wireless charging capabilities. Amazing inclusions!
Samsung Galaxy S10 Price in Nepal: Rs 1,10,900
Samsung Galaxy S10+ Overview
The Samsung Galaxy S10+ is the true flagship smartphone from Samsung. The phone packs a 6.4-inch Quad HD (1440 x 3040 pixels) Dynamic AMOLED display with a 19:9 aspect ratio. The panel this time around comes with Corning Gorilla Glass 5 and HDR10+ compatibility. The new Infinity-O design makes an appearance residing at the top right of the panel. The notch for the S10+ is more of a pill shape than an O shape. And encasing the other internals of the phone is a glass-on-glass design encased by a metallic frame. The phone also has a ceramic back option which will be available later in Nepal.
Under the hood, the Galaxy S10+ packs an octa-core Exynos 9820 processor along with the Mali-G76 MP12 GPU. The SoC is backed by 8GB of RAM and an option of 128GB or 512GB of internal storage. The memory of the phone is expandable via a micro-sd card up to 512GB.
In the camera department, the Galaxy S10+ packs a triple camera set up. Among the three, the primary is a 12MP sensor with a variable aperture of f/1.5-2.4. Next, we have a 12MP telephoto lens with f/2.4 aperture and a 16MP ultra-wide angle sensor with an f/2.2 aperture. The phone also comes with an f/1.9 10MP front camera sitting snug in a punch hole cut out.
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Moving on, at the software side of things, the phone runs One UI based on Andoird 9.0 Pie. It also comes with a fingerprint scanner, using ultrasonic technology, embedded into the display of the phone. And finally powering the device all up is a large 4100mAh non-user removable battery. Samsung this time around as thrown in Fast Battery Charging, Fast Wireless charging, and Reverse Wireless charging capabilities. It also comes with an IP68 Dust and Water Resistant body which we have come to expect from a flagship Samsung smartphone.
Samsung Galaxy S10+ Price in Nepal (128GB): Rs 1,23,500
Steps to Pre-order the Galaxy S10 series
Go to www.s10preorder.com/outlet and find a Samsung Nepal Outlet near you.
Visit the outlet and fill up a form with all your personal details.
Pay an advance of Rs 5,000 to confirm your Galaxy S10 series purchase.
Receive a confirmation e-mail and wait for the official launch date.
Verdict: Samsung Galaxy S10 Price in Nepal
It is amazing how quickly Samsung Nepal was able to bring the Galaxy S10 series here. Hats off to that! That said, we feel that the price is a bit expensive compared to the international rate. To counter this increase in price, we suggest you pre-book the phone if you are going to get the Galaxy S10 series anyway. At least then you will be able to get a free Galaxy Bud worth Rs 14,500 ($150). Other offers when buying the Galaxy S10 series are 1-year breakage insurance, a slew of galaxy gifts, and a chance to participate in a lucky draw.
Thank you for reading! For more the Samsung Galaxy S10 price in Nepal, like our Facebook page and subscribe to our Youtube channel. And also, if you have any questions or confusions, please feel free to ask them in the comments section down below. We will answer them as soon as possible.
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