text:
\begin{document}
\def\today{}
\title{A note on the Landauer principle\\ in quantum statistical mechanics}

\author{Vojkan  Jak\v{s}i\'c$^{1}$ and Claude-Alain  Pillet$^{2}$
\\ \\
$^1$Department of Mathematics and Statistics,
McGill University, \\
805 Sherbrooke Street West,
Montreal,  QC,  H3A 2K6, Canada
\\ \\
$^2$Universit\'e de Toulon, CNRS, CPT, UMR 7332, 83957 La Garde, France\\
Aix-Marseille Universit\'e, CNRS, CPT, UMR 7332, 13288 Marseille, France\\
FRUMAM
}
\maketitle
{\small
{\bf Abstract.} The Landauer principle asserts that the energy cost of erasure
of one bit of information by the action of a thermal reservoir in equilibrium at
temperature $T$ is never less than $k_{\rm B}T\log 2$. We discuss Landauer's
principle for quantum statistical models describing a finite level quantum
system $\cS$ coupled to an infinitely extended thermal reservoir $\cR$.
Using Araki's perturbation theory of KMS states and the Avron-Elgart
adiabatic theorem we prove, under a natural ergodicity assumption
on the joint system $\cS+\cR$, that Landauer's bound saturates for adiabatically
switched interactions. The recent work~\cite{ReWo} on the subject is discussed and compared.}
\thispagestyle{empty}

\section{Introduction}
Consider a quantum system $\cS$ described by a finite dimensional Hilbert
space $\cH_\cS$. Initially, $\cS$ is in a state described by a density
matrix $\rho_\i$. Let $\rho_\f$ be another density matrix on $\cH_\cS$.
The Landauer principle~\cite{La,Ma} sets a lower bound on the energetic cost
of the state transformation $\rho_\i\to\rho_\f$ induced by the action of a
reservoir $\cR$ in thermal equilibrium at temperature $T$. The principle can
be derived from the second law of thermodynamics, provided one accepts that
the (Clausius) entropy of the system $\cS$ in the state $\rho_{\i/\f}$
coincides with its von~Neumann entropy
\[
 S(\rho_{\i/\f})=-k_{\rm B}\tr(\rho_{\i/\f}\log\rho_{\i/\f}).
\]
Since this is only correct if both $\rho_{\i/\f}$ are equilibrium states,
such a derivation puts severe limits on the domain of validity of the
Landauer principle, in contrast to its supposed universality
and experimental verifications~\cite{Ber}.

The derivation goes as follows. The {\sl decrease} in the entropy of the system $\cS$ in the transition
$\rho_\i\to\rho_\f$ is
\[
\Delta S=S(\rho_\i)-S(\rho_\f).
\]
Let $\Delta\cQ$ denote the {\sl increase} in the energy of the reservoir $\cR$ in
the same process. Assuming that the joint system $\cS+\cR$ is isolated and that
the reservoir $\cR$ is large enough to remain in equilibrium at temperature $T$
during the whole process, the entropy of $\cR$ increases by
$\Delta S_\cR=\Delta\cQ/T$ and the entropy balance equation of the process
(see~\cite{dGM}) reads
$$
\Delta S+\sigma=\frac{\Delta\cQ}{T},
$$
where $\sigma$ is the entropy produced by the process. The second law of
thermodynamics stipulates that $\sigma\geq0$, with equality iff the
transition is the result of a reversible quasi-static process. Hence, the inequality
$$
\Delta\cQ\geq T\Delta S
$$
holds for arbitrary processes, with equality being achieved by reversible
quasi-static processes in which the change of the total entropy vanishes
$$
\Delta S_{\rm tot}=-\Delta S+\Delta S_\cR=0.
$$

With $d=\dim\cH_\cS$, $S(\rho_\i)$ is
maximal and equal to $k_{\rm B}\log d$ if $\rho_\i=\one/d$ is the chaotic state
and $S(\rho_\f)$ is minimal and equal to $0$ if
$\rho_\f=|\psi\rangle\langle \psi|$ is a pure state. It follows that
\beq\label{lan-pri}
\Delta\cQ\geq k_{\rm B}T\log d.
\eeq
If in addition  $d=2$, then $\Delta\cQ$ is the energy
cost of the erasure of the qubit of information stored in $\rho_\i$
and~\eqref{lan-pri} reduces to the Landauer bound.

The defects of the above ``derivation'' of the Landauer principle are manifest.
In spite of its importance, there are very few mathematically rigorous results
concerning the derivation of the Landauer bound from the first principles of
statistical mechanics.

In an interesting recent work, Reeb and Wolf~\cite{ReWo} point out that the lack
of mathematically precise formulation and proof of the Landauer principle in
the context of quantum statistical mechanics has led to a number of
controversies in the literature regarding its nature and validity. To remedy this
fact, in the same work they provide a derivation of the Landauer principle which
we will discuss in the next section.

One  of the values of the paper~\cite{ReWo} is that it has brought the Landauer
principle to the attention of researchers in quantum statistical mechanics.

In this note we shall examine the Landauer principle in the context of recent
developments in the mathematical theory of open quantum systems\footnote{
We shall discuss the Landauer principle only for microscopic Hamiltonian models
describing coupled system $\cS+\cR$. Repeated interaction systems
(see~\cite{BJM}) are an instructive and physically important class of models in
quantum statistical mechanics that also allow for  mathematically rigorous
analysis of the Landauer principle. This analysis is presented in~\cite{Raq}.}
(\cite{AS},\cite{ASF1}--\cite{ASF4}, \cite{AJPP1,AJPP2,BFS,DJ,DJP,dR,dRK,FM,FMU,FMSU,JOP1,JOP2,JOPP}, \cite{JP1}--\cite{JP7},
\cite{MMS1,MMS2,Pi1,Pi2,Ru1,Ru2,TM})\footnote{This is by no means a comprehensive
list of references. Some of the earlier works that motivated these developments
are~\cite{AM,BM,LeSp,McL,Rob,PW1,Sp1,Zu1,Zu2}.}, and compare the outcome with the
results of~\cite{ReWo}.

The paper is organized as follows. In Section~\ref{sec-RW} we will review the
work~\cite{ReWo}. The entropy balance equation in quantum statistical mechanics
and its implication regarding Landauer's principle are presented in
Section~\ref{sec-ep-bal}. We discuss the Landauer principle for instantaneously
switched interactions in Section~\ref{sec-isi} and for adiabatically
switched interactions in Section~\ref{sec-asi}. Section~\ref{sec-disc} is devoted
to the  discussion of the results presented in this note. The proofs are given in
Section~\ref{sec-proof}.

This note is similar in spirit to the recent work~\cite{JOPS}. It is an attempt
to bring together two directions of research which seem largely unaware of each
other, in the hope that they both may benefit from this connection.

\bigskip
{\bf Acknowledgments.} The research of V.J. was partly supported by NSERC.
C.A.P. is grateful to the Department of Mathematics and Statistics at McGill
University for its warm hospitality.

\section{The Reeb-Wolf derivation}
\label{sec-RW}

Suppose that $\cR$ is described by a finite dimensional Hilbert space $\cH_\cR$
(we shall call such reservoirs confined) and Hamiltonian $H_\cR$. Initially,
$\cR$ is in thermal equilibrium at temperature $T$, and its state is described by the
density matrix
\beq\label{Gibbs}
\nu_\i=\e^{-\beta H_\cR}/Z,
\eeq
where $Z=\tr(\e^{-\beta H_\cR})$ and $\beta=1/T$ (in the following, we shall
set Boltzmann's constant $k_{\rm B}$ to $1$). The Hilbert space of the coupled
system $\cS+\cR$  is
\[
\cH=\cH_\cS\otimes\cH_\cR,
\]
and its initial state has the product structure
\[
\omega_\i=\rho_\i\otimes\nu_\i.
\]
In what follows $\tr_{\cS/\cR}$ denotes the partial trace over $\cH_{\cS/\cR}$
and, whenever the meaning is clear within the context, we will denote operators
of the form $A\otimes\one$, $\one\otimes A$ by $A$. The relative entropy of two
positive linear maps $\zeta_1,\zeta_2$ is defined by
\beq\label{convention}
S(\zeta_1|\zeta_2)=\tr(\zeta_1(\log\zeta_1-\log \zeta_2)).
\eeq
If $\tr\,\zeta_1=\tr\,\zeta_2$, then $S(\zeta_1|\zeta_2)\geq 0$ with equality
iff $\zeta_1=\zeta_2$.

Let $U:\cH\rightarrow\cH$ be a unitary operator inducing the state
transformation
$$
\omega_U=U\omega_\i U^\ast.
$$
The transformed states of the subsystems $\cS$ and $\cR$ are given by
$$
\rho_{U}=\tr_\cR(\omega_U),\qquad\nu_{U}=\tr_\cS(\omega_U).
$$
In the literature, the relative entropy
\[
S(\omega_U|\rho_U \otimes \nu_U)=S(\rho_U)+S(\nu_U)-S(\omega_U)
\]
is sometimes called mutual information and the fact that it is non-negative
yields the subadditivity of entropy. The decrease in entropy of $\cS$ and the
increase in energy of $\cR$ in the transition process $\omega_\i\to\omega_U$
are respectively
$$
\Delta S=S(\rho_\i)-S(\rho_U),\qquad
\Delta\cQ=\tr(\nu_U H_\cR)-\tr(\nu_\i H_\cR).
$$
The unitarity of $U$ and the product structure of $\omega_\i$ imply
\[
S(\omega_{U})=S(\omega_\i)=S(\rho_\i)+S(\nu_\i),
\]
and Eq.\;\eqref{Gibbs} yields
$$
S(\nu_\i)=\beta\tr(\nu_\i H_\cR)+\log Z.
$$
It follows that
\begin{align*}
S(\omega_U|\rho_U\otimes\nu_\i)
&=-S(\omega_U)-\tr(\omega_U(\log\rho_U+\log\nu_\i))\\[2mm]
&=-S(\rho_\i)-S(\nu_\i)-\tr(\rho_U\log\rho_U)-\tr(\nu_U\log\nu_\i)\\[2mm]
&=-S(\rho_\i)-\beta\tr(\nu_\i H_\cR)-\log Z+S(\rho_U)+\beta\tr(\nu_U H_\cR)
+\log Z,
\end{align*}
and one arrives at the entropy balance equation
\beq\label{akin}
\Delta S+\sigma=\beta \Delta\cQ,
\eeq
where the entropy production term is given by
\begin{equation}\label{sigmaForm}
\sigma=S(\omega_U|\rho_U\otimes\nu_\i)\geq 0.
\end{equation}
This leads to the Landauer bound
\beq\label{bound-again}
\beta\Delta\cQ\geq \Delta S.
\eeq
Note that \eqref{sigmaForm} implies that $\sigma=0$ iff $\omega_U=\rho_U\otimes \nu_\i$.
The last relation yelds $\nu_U=\nu_\i$ and hence $\Delta\cQ=0$. Thus
equality holds in \eqref{bound-again} iff $\Delta\cQ=\Delta S=0$. In this case, it
further follows from the identities
$$
\tr(\rho_\i^{\alpha})\,\tr(\nu_\i^{\alpha})
=\tr(\omega_\i^{\alpha})=\tr(\omega_U^{\alpha})
=\tr(\rho_U^{\alpha})\,\tr(\nu_\i^{\alpha}),
$$
that $\tr(\rho_\i^{\alpha})=\tr(\rho_U^{\alpha})$ holds for all $\alpha\in\cc$.
One easily concludes from this fact that $\rho_\i$ and $\rho_U$ are unitarily
equivalent.

The following additional points are discussed in \cite{ReWo}.

\bigskip
{\bf Remark 1.} Given $\rho_\i$, $\beta$, $\cH_\cR$ and $H_\cR$, there are
many target states $\rho_\f$ for which there is no unitary $U$ such that
$\rho_U=\rho_\f$. Let
\[
\ell=e_{\rm max}-e_{\rm min},
\]
where $e_{{\rm max}/{\rm min}}$ is the maximal/minimal eigenvalue of $H_\cR$.
Then, for any unitary $U$,
\[
\e^{-\ell\beta}\rho_\i\le\rho_U\le\e^{\ell\beta}\rho_\i.
\]
This constrains the set of possible target states $\rho_\f$. To reach a given
$\rho_\f$, either exactly or up to a prescribed small error, one may need  to
adjust $\cH_\cR$, $H_\cR$, and $U$. The following example illustrates one trivial
way in which $\rho_\f$ can always be reached.

\medskip
{\bf Example 1.} Let $\rho_\f>0$ be the target state. Set $\cH_\cR=\cH_\cS$,
$\nu_\i=\rho_\f$, $H_\cR=-\log \rho_\f$. In this example,
$\beta=1$. Let $U$ be the flip map, $U(\phi\otimes\psi)=\psi\otimes\phi$.
Then $\rho_U=\rho_\f$, $\nu_U=\rho_\i$, the entropy production is
\[
\sigma=S(\rho_\i|\rho_\f),
\]
and $\Delta\cQ=\Delta S$ iff $\rho_\f=\rho_\i$.

\bigskip
{\bf Remark 2.} It turns out that Inequality~\eqref{bound-again}, as a lower
bound of $\Delta\cQ$ in terms of $\Delta S$, is not optimal.
This can be seen as follows, starting with the standard bound (see for example
Theorem~1.15 in~\cite{OP})
\[
S(\omega_U|\rho_U\otimes \nu_\i)
\geq\frac12\|\omega_U-\rho_U\otimes\nu_\i\|^2_1,
\]
where $\|X\|_1=\tr\,|X|=\sup_{A\not=0}|\tr(AX)|/\|A\|$ is the  trace norm.
With $e=(e_{\rm max}+e_{\rm min})/2$, we can estimate
\[
\|\omega_U-\rho_U\otimes\nu_\i\|_1
\geq\frac{|\tr[(H_\cR-e)(\omega_U-\rho_U\otimes\nu_\i)]|}{\|H_\cR-e\|}
=\frac{|\Delta\cQ|}{\ell/2},
\]
and so  the entropy production \eqref{sigmaForm} satisfies
\beq
\sigma\ge2\left(\frac{\Delta\cQ}{\ell}\right)^2.
\label{well}
\eeq
Combining  (\ref{akin}) and (\ref{well}) and solving the resulting quadratic inequality shows that the possible
entropy changes are restricted by the constraint
$\Delta S\le S_0=\beta^2\ell^2/8$
and that the corresponding energy cost satisfies the improved bound
$$
\beta\Delta\cQ\ge\left(1+\frac{1-\sqrt{1-\Delta S/S_0}}
{1+\sqrt{1-\Delta S/S_0}}\right)\Delta S.
$$

A part of the discussion in~\cite{ReWo} is devoted to the refinement and
optimization of the estimate~\eqref{bound-again} in the spirit of the above
argument.

\medskip
{\bf Example 2.} On physical grounds, one expects saturation of the Landauer bound
for quasi-static reversible processes. The following toy example
of~\cite{ReWo} illustrates this point. Let $\rho_\f>0$ be a given target state.
Let $\rr\ni t\mapsto\rho(t)$ be a twice continuously differentiable map with
values in density matrices on $\cH_\cS$ such that $\rho(0)=\rho_\i$,
$\rho(1)=\rho_\f$, and $\rho(t)>0$ for $t\in]0,1]$. Given a positive
integer $N$, set $\rho_n=\rho(n/N)$, $\cH_{\cR}=\bigotimes_{n=1}^N\cH_\cS$,
$ \nu_\i=\bigotimes_{n=1}^N\rho_n$. With $\beta=1$, it follows that $H_\cR=-\sum_{n=1}^N\log\rho_n$, $\omega_\i=\bigotimes_{n=0}^N\rho_n$.
Let $U:\cH\rightarrow\cH$ be defined by
\[
U(\psi_0\otimes\psi_1\otimes\cdots\otimes\psi_N)
=\psi_N\otimes\psi_0\otimes\cdots\otimes\psi_{N-1}.
\]
Then $\rho_U=\rho_\f$ and
$$
\Delta\cQ_N=\tr(\nu_U H_\cR)-\tr(\nu_\i H_\cR)
=\sum_{n=1}^N\tr [(\rho_{n}-\rho_{n-1})\log \rho_{n}].
$$
The differentiability assumption allows us to rewrite the
r.h.s.\;of the previous identity as a Riemann sum, leading to
\[
\lim_{N\rightarrow \infty}\Delta\cQ_N
=\int_0^1\tr(\dot\rho(t)\log\rho(t))\d t
=S(\rho_\i)-S(\rho_\f).
\]
In this example the number of steps $N$ plays the  role of an adiabatic parameter
and the limit $N\rightarrow\infty$ leads to a quasi-static process with optimal
Landauer bound.

\bigskip
{\bf Remark 3.} In the Landauer erasing principle $\rho_\i=\one/d$ and
$\rho_\f=|\psi\rangle\langle\psi|$. Pure target states are thermodynamically
singular and cannot be reached by the action of a thermal reservoir at strictly
positive temperature. It follows from Example~2 that for any $\epsilon >0$ one
can find $\rho_\f^\prime$, $\cH_\cR$, $\nu_\i$, and $U$ such that
$\|\rho_\f-\rho_\f^\prime\|_1<\epsilon$, $\rho_U=\rho_\f^\prime$,
and that the energy cost of the transformation
$\rho_\i\to\rho_\f^\prime$ satisfies $\beta\Delta\cQ\geq\log d-\epsilon$.
The last result can be refined by considering infinite dimensional
$\cH_\cR$'s and allowing for Hamiltonians $H_\cR$ with formally infinite energy
levels. An additional toy example discussed in Section~6 of \cite{ReWo}
illustrates this point.

\bigskip
With the exception of the toy example mentioned in Remark~3, the work
\cite{ReWo} is exclusively  concerned with finite dimensional thermal reservoirs.
The authors discuss several additional topics including  possible extensions of
the notion of Landauer processes. The paper contains  valuable discussions and
clarifications concerning the physics literature on the Landauer principle.
In the final Section~7 of the paper, the authors list a number of open
problems/conjectures, including the following, on which we will comment
later:
\begin{quote} {\bf Conjecture} \cite{ReWo}. {\em Landauer's Principle
can probably be formulated within the general statistical mechanical framework of
$C^\ast$ and $W^\ast$
dynamical systems~\cite{BR2,PW1,Th} and an equality version akin to~\eqref{akin}
can possibly be proven. Note that in this framework the mutual information can be
written as a relative entropy and the heat flow as a derivation w.r.t.\;the
dynamical semigroup.}
\end{quote}

We now turn to the discussion of the Landauer principle in the context of the existing mathematical theory of open quantum systems.

\section{The entropy balance equation}
\label{sec-ep-bal}

We start with the following remark regarding the derivation  of the previous
section. Let $\eta=\one\otimes\nu_\i$. Then
\[
S(\rho_\i)-S(\rho_U)+\sigma=S(\omega_U|\eta)-S(\omega_\i|\eta),
\]
and~\eqref{akin} can be written as
\beq\label{akin-2}
S(\omega_U|\eta)-S(\omega_\i|\eta)=\beta \Delta\cQ.
\eeq
The relation~\eqref{akin-2} is a special case of the general entropy
balance equation in quantum statistical mechanics. In the form~\eqref{akin-2} it
goes back at least to Pusz and Woronowicz (see the Remark at the end of
Section~2 in~\cite{PW1}) and was rediscovered in~\cite{JP3,JP7,Pi1}, see
also~\cite{LeSp,McL,O1,O2,OHI,Sp2,Ru2,TM,Zu1,Zu2} for related works on the
subject. To describe~\eqref{akin-2} in  full generality we assume that the
reader is familiar with basic definitions and results of algebraic quantum
statistical mechanics, and in particular with Araki's perturbation theory
of KMS structure. This material is standard and
can be found in the monographs~\cite{BR1,BR2}. A modern exposition of the
algebraic background  can be found in~\cite{BF,DJP,Pi2}. The interested reader
should also consult the fundamental paper~\cite{HHW}.
For definiteness we will work with $C^\ast$-dynamical systems. With only
notational changes all our results and proofs easily extend to $W^\ast$-dynamical systems
and we leave such generalizations to the reader.

In the algebraic framework, a quantum system is described by a $C^\ast$-dynamical
system $(\cO,\tau)$. $\cO$ is a $C^\ast$-algebra, with a unit $\one$, and
$\tau$ is a strongly continuous one-parameter group of $\ast$-automorphisms of
$\cO$. Elements of $\cO$ are observables, and their time evolution, in the
Heisenberg picture, is given by $\tau$. We denote by $\cO^\sa$ the set
of self-adjoint elements of $\cO$. A state of the system is a positive
linear functional $\omega$ on $\cO$ such that $\omega(\one)=1$. It is
$\tau$-invariant if $\omega\circ\tau^t=\omega$ for all $t\in\rr$.
A thermal equilibrium state at inverse temperature $\beta$ is a
$(\tau,\beta)$-KMS state. Such states are $\tau$-invariant.

Given a state $\omega$, the GNS construction provides a Hilbert space
$\cH_\omega$, a $\ast$-morphism $\pi_\omega:\cO\to\cB(\cH_\omega)$\footnote{Throughout
the paper $\cB(\cH)$ denotes the usual $C^\ast$-algebra of all bounded operators on a
Hilbert space $\cH$.} and a unit
vector $\Omega_\omega\in\cH_\omega$ such that $\pi_\omega(\cO)\Omega_\omega$ is
dense in $\cH_\omega$ and $\omega(A)=\langle \Omega_\omega,\pi_\omega(A)\Omega_\omega\rangle$
for all $A\in\cO$. A state given by $\zeta(A)=\tr(\rho\pi_\omega(A))$, where
$\rho$ is a density matrix on $\cH_\omega$, is said to be $\omega$-normal.
We denote by $\cN_\omega$ the set of all $\omega$-normal states on $\cO$. The state
$\omega$ is called ergodic for $(\cO,\tau)$ if, for all states
$\zeta\in\cN_\omega$ and all $A\in\cO$,
\[
\lim_{t\rightarrow\infty}\frac1t\int_0^t\zeta(\tau^s(A))\d s =\omega(A),
\]
and mixing if
\[
\lim_{t\rightarrow \infty}\zeta(\tau^t(A))=\omega(A).
\]

If $\omega$ is a $\tau$-invariant state, then there is a unique
self-adjoint operator $L_\omega$ on $\cH_\omega$ such that $L_\omega\Omega_\omega=0$
and $\pi_\omega(\tau^t(A))=\e^{\i tL_\omega}\pi_\omega(A)\e^{-\i tL_\omega}$ for all
$t\in\rr$ and $A\in\cO$. $L_\omega$ is called the $\omega$-Liouvillean of the dynamical
system $(\cO,\tau)$. If $\omega$ is a $(\tau,\beta)$-KMS state, then $L_\omega$ is
also the standard Liouvillean (see Section \ref{sec-proof}) of $(\cO,\tau)$.

The reservoir $\cR$ is described by a $C^\ast$-dynamical system
$(\cO_\cR,\tau_\cR,\nu_\i)$ in thermal equilibrium
at inverse temperature $\beta>0$. We denote by $\delta_\cR$ the
generator of $\tau_\cR$, $\tau_\cR^t=\e^{t\delta_\cR}$ and by $L_\cR$ its
standard Liouvillean. If the reservoir is
confined, then $\cO_\cR=\cB(\cH_\cR)$, $\delta_\cR(\,\cdot\,)=\i[H_\cR,\,\cdot\,]$,
and $\nu_\i=\e^{-\beta H_\cR}/\tr(\e^{-\beta H_\cR})$\footnote{If $\dim\cH<\infty$, we
shall not distinguish between positive linear functionals on $\cB(\cH)$ and positive
elements of $\cB(\cH)$. They are identified by $\zeta(A)=\tr(\zeta A)$.}.
The GNS Hilbert space $\cH_{\nu_\i}$ is $\cO_\cR$ equipped with the inner
product $\langle X,Y\rangle=\tr(X^\ast Y)$, the morphism $\pi_{\nu_\i}$ is defined
by $\pi_{\nu_\i}(A)X=AX$, $\Omega_{\nu_\i}=\nu_\i^{1/2}$ and $L_\cR X=[H_\cR,X]$.
However, in the remainder of this note we shall be concerned with
infinitely extended reservoirs.

The $C^\ast$-algebra of the system $\cS$, described by the finite dimensional Hilbert
space $\cH_\cS$, is  $\cO_\cS=\cB(\cH_\cS)$. The $C^\ast$-algebra of the joint system
$\cS+\cR$ is
$$
\cO=\cO_\cS\otimes\cO_\cR,
$$
and its initial state is
$$
\omega_\i=\rho_\i\otimes \nu_\i,
$$
where $\rho_\i$ denotes the initial state of $\cS$.
We continue with our notational convention of omitting tensored identity, hence $\delta_\cR=\mathrm{Id}\otimes\delta_\cR$, etc.

Let $S(\zeta_1|\zeta_2)$ be the relative entropy
of two positive linear functionals $\zeta_1, \zeta_2$ on $\cO$
\cite{Ar2, Ar3}, with the ordering convention of~\cite{BR2,Don,JP3,JP6,DJP} and
the sign convention of~\cite{Ar2,Ar3,OP} (with these conventions the relative
entropy of two density matrices is given by~\eqref{convention}). The basic
properties of the relative entropy are most easily deduced from  the
Pusz-Woronowicz-Kosaki variational formula~\cite{Ko,PW2}
$$
S(\zeta_1|\zeta_2)= \sup\int_0^\infty\left[\frac{\zeta_2(\one)}{1+t}- \zeta_2(y^\ast(t)y(t))-\frac{1}{t}\zeta_1(x(t)x^\ast(t))\right]\frac{\d t}{t},
$$
where the supremum is taken over all countably valued step functions
$[0,\infty[\ni t\mapsto x(t)\in{\cal O}$ vanishing in a neighborhood
of zero and satisfying $x(t)+y(t)=\one$. In particular, if $\zeta_1(\one)=\zeta_2(\one)$,
then $S(\zeta_1|\zeta_2)\geq 0$ with equality iff $\zeta_1=\zeta_2$.

Any unitary element $U\in\cO$ induces a $\ast$-automorphism
$$
A\mapsto\alpha_U(A)=U^\ast A U,
$$
and hence a state transformation $\omega\mapsto\omega\circ\alpha_U$.
Set $\eta=\one\otimes\nu_\i$. With this setup the entropy balance
equation of~\cite{JP3,JP7,Pi1,PW1} reads as:

\bet
Suppose that $U\in\Dom(\delta_\cR)$. Then
\beq
S(\omega\circ\alpha_U|\eta)=S(\omega_\i|\eta)-\i\beta \omega_\i(U^\ast\delta_\cR(U)).
\label{ep-form}
\eeq
\eet

Denote by $\rho_U$ and $\nu_U$ the restriction of the transformed state
$\omega\circ\alpha_U$ to $\cO_\cS$ and $\cO_\cR$ (i.e.,
$\rho_{U}(A)=\omega\circ\alpha_U(A\otimes\one)$ and
$\nu_U(B)=\omega\circ\alpha_U(\one\otimes B)$
for $A\in\cO_\cS$ and $B\in\cO_\cR$). If $\cR$ is confined, then
$$
-\i\omega_\i(U^\ast\delta_\cR(U))=-\i\omega_\i(U^\ast\i[H_\cR,U])
=\omega_\i(\alpha_U(H_\cR)-H_\cR)=\tr(\nu_UH_\cR)-\tr(\nu_\i H_\cR),
$$
and~\eqref{ep-form} reduces to~\eqref{akin-2}.

For any states $\rho$ on $\cO_\cS$ and $\omega$ on $\cO$, Araki's perturbation
formula for the relative entropy \cite{Ar1} (see also Proposition~6.2.32
in~\cite{BR2} and Appendix~A of~\cite{Don}) gives
\beq\label{ArakiForm}
S(\omega|\rho\otimes\nu_\i)=S(\omega|\eta)-\omega(\log\rho).
\eeq
Setting $\omega=\omega_\i=\rho_\i\otimes\nu_\i$, this implies in particular that
\beq\label{Srhoform}
S(\rho_\i)=-S(\omega_\i|\eta).
\eeq

The entropy balance equation~\eqref{ep-form} allows for an analysis of
Landauer's principle in the general setup of quantum statistical mechanics. The
decrease in entropy of $\cS$ and the increase in energy of $\cR$ in the
transition process $\omega_\i\rightarrow \omega\circ\alpha_U$ are
$$
\Delta S= S(\rho_\i)-S(\rho_U),\qquad
\Delta\cQ=-\i\omega_\i(U^\ast\delta_\cR(U)).
$$
Writing~\eqref{ep-form} as
\beq\label{barbados}
\Delta S +\sigma =\beta \Delta\cQ,
\eeq
and taking~\eqref{Srhoform} into account yields
\[
\sigma= S(\omega\circ\alpha_U|\eta)+S(\rho_U).
\]
Since $S(\rho_U)=-\omega\circ\alpha_U(\log\rho_U)$, Eq.\;\eqref{ArakiForm} further gives
\beq\label{sigmaOneForm}
\sigma=S(\omega\circ\alpha_U|\rho_U\otimes \nu_\i),
\eeq
and hence
\[
\sigma\geq0
\]
with equality iff $\omega\circ\alpha_U=\rho_U\otimes\nu_\i$. This implies the
Landauer bound
\[
\beta\Delta\cQ\geq\Delta S.
\]
for the state transformation induced by the inner $\ast$-automorphism $\alpha_U$.
This also settles the  conjecture of~\cite{ReWo} which has in fact been been known
for many years.

The analysis of the saturation of the Landauer bound is more delicate than in
the case of a confined reservoir. It relies on the spectral analysis of modular
operators. We shall give one result in this direction.

\bep\label{Saturation}
Assume that the point spectrum of the standard Liouvillean $L_\cR$ is finite.
Then $\Delta S=\beta\Delta\cQ$ if and only if $\Delta S=\Delta\cQ=0$ in which
case $\rho_U$ is unitarily equivalent to $\rho$ and $\nu_U=\nu_\i$
\eep

{\bf Remark.} If $\cR$ is confined, then the spectrum of $L_\cR$ is discrete and
finite so that the above proposition applies. It also applies to the physically
important class of ergodic extended reservoirs. Indeed, it follows
from Theorem~1.2 in \cite{JP4} that $0$ is the only eigenvalue of $L_\cR$ if
$\nu_\i$ is an ergodic state for $(\cO_\cR,\tau_\cR)$.
It is an interesting  structural question  to characterize all reservoir systems for
which the conclusions of Proposition~\ref{Saturation} holds.

One can continue with the abstract analysis of the Landauer principle in the
above framework. As in the finite dimensional case, the bound
\beq
\sigma\geq\frac12\|\omega\circ\alpha_U-\rho_U\otimes\nu_\i\|^2
\label{lungs}
\eeq
follows from Eq.\;\eqref{sigmaOneForm}, the norm on the right hand side
being dual to the $C^\ast$-norm of $\cO$. Since
\begin{align*}
\|\nu_U-\nu_\i\|&=\sup_{0\not=A\in\cO_\cR}
\frac{|\omega\circ\alpha_U(\one\otimes A)-\rho_U\otimes\nu_\i(\one\otimes A)|}
{\|\one\otimes A\|}\\[4pt]
&\le\sup_{0\not=A\in\cO}\frac{|\omega\circ\alpha_U(A)
-\rho_U\otimes\nu_i(A)|}{\|A\|}
=\|\omega\circ\alpha_U-\rho_U\otimes\nu_\i\|,
\end{align*}
(\ref{lungs}) gives
\beq\label{maiden}
\sigma\ge\frac12\|\nu_U-\nu_\i\|^2.
\eeq

Suppose that $\rho_\i>0$ and let $\rho_\f>0$ be a target state. Set
$\omega_\f=\rho_\f\otimes\nu_\i$. Another application of Araki's
perturbation formula gives
\beq\label{ever}
\sigma=S(\omega\circ\alpha_U|\rho_U\otimes\nu_\i)
=S(\omega\circ\alpha_U|\omega_\f)-S(\rho_{U}|\rho_\f).
\eeq
Assume that there exists a sequence $U_n$ of unitary elements of $\cO$ such that $U_n\in\Dom(\delta_\cR)$ and
\beq\label{tra-wed}
\lim_{n\to\infty}\omega\circ\alpha_{U_n}(A)=\omega_\f(A)
\eeq
for all $A\in\cO$.
Since this implies that $\rho_{U_n}\to\rho_\f$, it follows from the
entropy balance relation~\eqref{barbados} that
\[
\liminf_{n\to\infty}\beta\Delta\cQ_n
=\liminf_{n\to\infty}\sigma_n+S(\rho_\i)-S(\rho_\f)
\geq S(\rho_\i)-S(\rho_\f).
\]
Moreover,
\[
\lim_{n\to\infty}\beta\Delta\cQ_n=S(\rho_\i)-S(\rho_\f)
\]
if and only if
\[
\lim_{n\to\infty}\sigma_n=0,
\]
which, by~\eqref{ever}, is equivalent to
\beq\label{tra-quasistatic}
\lim_{n\to\infty}S(\omega\circ\alpha_{U_n}|\omega_\f)= 0.
\eeq
The relation~\eqref{tra-quasistatic} quantifies the notion of quasi-static
transition process. If~\eqref{tra-quasistatic} holds, then
Inequality~\eqref{maiden} implies
\beq\label{meet}
\lim_{n\to\infty}\|\omega\circ\alpha_{U_n}-\omega_\f\|=0.
\eeq
On the other hand, the  norm convergence~\eqref{meet} does not
imply~(\ref{tra-quasistatic}). Sufficient conditions for~\eqref{tra-quasistatic}
are discussed in the foundational papers~\cite{Ar1, Ar2}. For example, if in
addition to~\eqref{meet} there is $\lambda>0$ such that
\beq\label{improve}
\lambda\omega\circ\alpha_{U_n}\geq\omega_\f
\eeq
for all $n$, then~\eqref{tra-quasistatic} holds. A sufficient condition
for~\eqref{improve} is that
\[
\sup_{n}\|\e^{-\i\delta_\cR/2\beta}(U_n)\|<\infty.
\]

{\noindent\bf Remark.} If the quantum dynamical system $(\cO_\cR,\tau_\cR,\nu_\i)$ describes an infinitely extended reservoir in
thermal equilibrium at positive temperature and in a pure phase, then on physical grounds it is natural to  assume
that the enveloping von Neumann algebra
$\pi_{\nu_\i}(\cO_\cR)''$ is an injective factor of type ${\rm III}_1$
(see, e.g., \cite{Ar4,ArW,Hu}). In this case, it is a simple consequence of a
result of Connes and St\o rmer (Theorem 4 in~\cite{CSt}) and Kaplansky's density
theorem (Corollary 5.3.7 in \cite{KR}) that there is a sequence
of unitaries $U_n\in\Dom(\delta_\cR)$ such that~\eqref{meet} and
hence~\eqref{tra-wed} holds.

\bigskip
Although one can go quite far by continuing the above structural analysis of the
Landauer principle, we shall not pursue this direction  further. Instead,
we shall focus on physically relevant realizations of $\alpha_U$'s by
considering the dynamics of the coupled system $\cS+\cR$ and we shall
analyze the Landauer principle in this context. Non-trivial dynamics are
characterized by interactions that allow energy/entropy flow between $\cS$
and $\cR$. We shall distinguish between instantaneously and adiabatically
switched interactions.

\section{Instantaneously switched interactions}

\label{sec-isi}

\subsection{Setup}
For $K\in\cO^\sa$, the $\ast$-derivation
$$
\delta_K=\delta_\cR+\i[K,\,\cdot\,]
$$
generates a strongly continuous group $\tau_K^t=\e^{t\delta_K}$ of
$\ast$-automorphisms of $\cO$. Self-adjoint elements of $\cO$ are
called local perturbations and the group $\tau_K$ is the local perturbation of
$\tau_\cR$ induced by $K$. For example, if $H_\cS$ is the Hamiltonian of $\cS$
and $V$ describes the interaction of $\cS$ with $\cR$, then the dynamics of the
interacting system $\cS+\cR$ is given by $\tau_K$, with $K=H_\cS+V$.
In this section, we investigate the Landauer principle for the
dynamical system $(\cO,\tau_K)$.

The interacting dynamics can be expressed as
\[
\tau_K^t(A)=\tau_\cR^t(U^\ast_K(t)A U_K(t)),
\]
where the interaction picture propagator $U_K(t)$ is a family of unitary
elements of $\cO$ satisfying
\begin{equation}\label{CocycleODE}
\i\partial_t U_K(t)
=U_K(t)\tau_\cR^{-t}(K),\qquad U_K(0)=\one.
\end{equation}
Hence, we have
\[
\omega_\i\circ\tau_K^t=\omega_\i\circ\alpha_{U_K(t)},
\]
and we can apply the results of the previous section.
Assuming $K\in\Dom(\delta_\cR)$, it follows from the Dyson expansion
$$
U_K(t)=\one+\sum_{n=1}^\infty(-\i)^n
\int\limits_{0\le s_1\le\cdots\le s_n\le t}
\tau_\cR^{-s_1}(K)\cdots\tau_\cR^{-s_n}(K)\d s_1\cdots\d s_n
$$
that $U_K(t)\in\Dom(\delta_\cR)$ for all $t\in\rr$ and
Eq.\;\eqref{barbados} gives
\beq\label{barbados-no}
\Delta S(K,t)+\sigma(K,t)=\beta \Delta\cQ(K,t),
\eeq
where
\[
\Delta\cQ(K,t)
=-\i\omega_\i(U_K^\ast(t)\delta_\cR(U_K(t))),
\]
and
\[
\Delta S(K,t)=S(\rho_\i)-S(\rho_K(t)),\qquad
\sigma(K,t)=S(\omega_\i\circ\tau_K^t|\rho_K(t)\otimes\nu_\i),
\]
$\rho_K(t)$ denoting the restriction of $\omega_\i\circ\tau_K^t$
to $\cO_\cS$.

\bigskip
{\noindent\bf Remark.} One easily checks that
$T(t)=\i U^\ast_K(t)\delta_\cR(U_K(t))+\tau_\cR^{-t}(K)$
satisfies the Cauchy problem
$$
\partial_t T(t)=\i[\tau_\cR^{-t}(K),T(t)],\qquad T(0)=K.
$$
Comparing with Eq.\;\eqref{CocycleODE}, we infer
$T(t)=U_K^\ast(t)K U_K(t)$, so that
$$
-\i U^\ast_{K}(t)\delta_\cR(U_{K}(t))
=\tau_\cR^{-t}(K-\tau_{K}^t(K)),
$$
and therefore
$$
\Delta\cQ(K,t)=\omega_\i(K-\tau_{K}^t(K)).
$$
Conservation of the total energy leads to the conclusion that $\Delta\cQ(K,t)$ is
indeed the change of the reservoir energy. Since
$$
\partial_t(K-\tau_{K}^t(K))
=\tau_{K}^t(-\delta_\cR(K)),
$$
one  can  further write
\[
\Delta\cQ(K,t)=-\int_0^t\omega_\i(\tau_{K}^s(\Phi))\d s,
\]
where
\[
\Phi=\delta_\cR(K),
\]
is the observable describing the instantaneous energy flux out of $\cR$.

\subsection{The Landauer principle in the  large time limit}
\label{sec-inst}
We shall now consider realizations of the  state transition
$\rho_\i\to\rho_\f$ and the corresponding entropic balance as a limiting case
of the transition $\rho_\i\to\rho_K(t)$ as $t\to\infty$. To simplify the
discussion, we shall assume here and in the following that the equilibrium state
$\nu_\i$ describes a pure thermodynamic phase of $\cR$, i.e., that it is an
extremal
$(\tau_\cR,\beta)$-KMS state. This implies that for $K\in\cO^\sa$ there is
a unique $(\tau_K,\beta)$-KMS state in $\cN_{\omega_\i}$ which we denote
by $\mu_K$. Let $\varrho_K$ be its restriction to $\cO_\cS$. We observe
that $\cN_{\mu_K}=\cN_{\omega_\i}$.

The following proposition shows that by an appropriate choice of $K$
we can reach any faithful\footnote{The cases where the
target state is not faithful are handled by an additional limiting argument
that we will describe later.} target state of $\cO_\cS$ with the
$(\tau_K,\beta)$-KMS state $\mu_K$.

\bep\label{HTarget}
Let $\rho_\f>0$ be a state on $\cO_\cS$ and $V\in\cO^\sa$. Then there exists
$\delta>0$ and a real analytic function
$]-\delta,\delta[\ni\lambda\mapsto H_\lambda\in\cO_\cS^\sa$ such that $H_0=-\beta^{-1}\log\rho_\f$
and $\varrho_{K_\lambda}=\rho_\f$ for $K_\lambda=H_\lambda+\lambda V$ and any $\lambda\in ]-\delta, \delta[$.
\eep

Our main dynamical assumption is:

\begin{quote} {\bf Assumption A.} There exists  $\gamma\in ]-\delta,\delta[$ such that the KMS state $\mu_{K_\gamma}$ is mixing for the
dynamical system $(\cO,\tau_{K_\gamma})$.
\end{quote}

We now explore the  consequences of this assumption on the long time asymptotics
of entropy balance (note that obviously $\gamma\not=0$). The first is
$$
\lim_{t\to\infty}\rho_{K_\gamma}(t)=\rho_\f.
$$
Furthermore,
\beq\label{rainysat}
\begin{split}
\Delta S&=\lim_{t\to\infty}\Delta S(K_\gamma,t)
=S(\rho_\i)-S(\rho_\f),\\[2mm]
\Delta\cQ(\gamma)&=\lim_{t\to\infty}\Delta\cQ(K_\gamma,t)
=\omega_\i(K_\gamma)-\mu_{K_\gamma}(K_\gamma).
\end{split}
\eeq

It follows from~\eqref{barbados-no} that
\[
\sigma(\gamma)=\lim_{t\rightarrow \infty}\sigma(K_\gamma, t)
\]
also exists and that
\beq\label{dog}
\Delta S+\sigma(\gamma)=\beta\Delta\cQ(\gamma).
\eeq
Clearly, $\sigma(\gamma)\geq 0$, and the relation (\ref{dog}) gives the Landauer principle for the transition process
$\rho_\i\rightarrow\rho_\f$ realized  by the large time limit
$t\rightarrow \infty$.

One does not expect that  the Landauer bound can be saturated by an  instantaneously
switched   interaction and that is indeed the case.
\bep
\beq\label{strict-po}
\sigma(\gamma)>0.
\eeq
\label{roof}
\eep

This completes our analysis of the Landauer principle for instantaneously switched
interactions.

\bigskip
{\noindent\bf Remark 1.} The above analysis  extends with no changes to $W^\ast$-dynamical
systems. Unbounded interactions $V$ satisfying the general assumptions of~\cite{DJP}
are also allowed.

\medskip

{\noindent\bf Remark 2.} In the Landauer erasing principle, $\rho_\i=\one/d$
and $\rho_\f=|\psi\rangle\langle\psi|$. Pure target states are
thermodynamically singular and cannot be directly reached by the action of a
thermal reservoir unless the reservoir is at zero temperature. The proper way to
formulate the Landauer principle for pure states is to examine the stability of
the entropy balance equation of the processes with faithful target states
$\rho_\f^\prime$ in a vicinity of $\rho_\f$. For instantaneously
switched interactions there is no stability. As
$\rho_\f^\prime\to\rho_\f$, $S(\rho_\f^\prime)\to S(\rho_\f)=0$.
However, in this limit $\sigma(\gamma)\to\infty$ and
$\Delta\cQ(\gamma)\to\infty$. This singularity is due to an
instantaneous change of the Hamiltonian.
As we shall see in the next section, if the change of the Hamiltonian is
adiabatic, this singularity is
absent.

\medskip

\noindent{\bf Remark 3.} It follows from
Araki's perturbation theory of KMS states that the map
\[ ]-\delta, \delta[ \ni \lambda\mapsto\Delta\cQ(\lambda)= \omega_\i(K_\lambda)-\mu_{K_\lambda}(K_\lambda)\]
is  real analytic and that

$$
\Delta\cQ(0)=\rho_\i(H_0)-\rho_\f(H_0).
$$
The relation (\ref{dog}) defines $\sigma(\lambda)$ for
$\lambda \in ]-\delta, \delta[$ and
\[
\sigma(0)=S(\rho_\i|\rho_\f).
\]

\medskip

\medskip
{\noindent\bf Remark 4.} For many models, Assumption {\bf A} is satisfied in a stronger form:

\begin{quote} {\bf Assumption A'.} There exists $\lambda_0>0$ such that for $0 <|\lambda|<\lambda_0$  the KMS state $\mu_{K_\lambda}$ is mixing for the
dynamical system $(\cO,\tau_{K_\lambda})$.
\end{quote}

In this case the entropy balance equation
\beq \Delta S + \sigma(0)=\beta \Delta {\cal Q}(0)
\label{clouds}
\eeq
gives the Landauer principle for the transition process $\rho_\i \rightarrow \rho_\f$ realized by the double limit $t\rightarrow \infty$,
$\lambda\rightarrow 0$. The relation (\ref{clouds}) is certainly
expected in view  of the Lebowitz-Spohn weak coupling
limit thermodynamics of open quantum systems~\cite{LeSp,JPW}. Under suitable
assumptions, in the van Hove scaling limit $\lambda\rightarrow0$,
$t\rightarrow\infty$ with $\bar t=\lambda^2 t$ fixed, the reduced dynamics of
$\cS$ is described by a quantum dynamical semigroup on $\cO_\cS$,
\[T_{\bar t}(A)=\e^{\bar t {\cal K}}(A),\]

where $\cK$ is the so-called {\em Davies generator} in the Heisenberg picture.
Under the usual effective coupling assumptions one has\footnote{The adjoint is
taken with respect to the inner product $\langle A,B\rangle=\tr(A^\ast B)$ on $\cO_\cS$}
\[
\lim_{\bar t\to\infty}\e^{\bar t\cK^\dagger}(\rho)=\rho_\f
\]
for any state $\rho$ on $\cO_\cS$.
This relation defines the transition process $\rho_\i\to\rho_\f$ in the van Hove
scaling limit. The Lebowitz-Spohn entropy balance equation is
\[
S(\rho_\i)-S(\e^{\bar t\cK^\dagger}(\rho_\i))
+S(\rho_\i|\e^{\bar t\cK^\dagger}(\rho_\i))=\beta\Delta\bar\cQ(\bar t),
\]
where
\[
\Delta\bar\cQ(\bar t)=\rho_\i(\e^{\bar t\cK}(H_0)- H_0).
\]
It follows that
\[
\Delta\bar\cQ=\lim_{{\bar t}\to\infty}\Delta\bar\cQ(\bar t)
=\rho_\f(H_0)-\rho_\i(H_0),
\]
and one derives
\beq
\beta\Delta\bar\cQ=S(\rho_\i|\rho_\f)+S(\rho_\i)-S(\rho_\f).
\label{corrections1}
\eeq
Since the van Hove weak coupling limit is expected to yield the first
non-trivial contribution (in the coupling constant $\lambda$) to the microscopic
thermodynamics, the identity~\eqref{clouds}=\eqref{corrections1} is
certainly not surprising. A somewhat surprising fact is that Assumption {\bf A'}
is only vaguely related to the assumptions of the weak coupling limit theory
\cite{Dav,DF,LeSp}.

\medskip

{\noindent\bf Remark 5.}  Specific physically relevant models (spin-boson model,
spin-fermion model, electronic black box model, locally interacting fermionic
systems) for which Assumption {\bf A'} holds   are discussed in
\cite{AM,AJPP1,AJPP2,BFS,BM,dRK,DJ,FMU,FMSU,JOP1,JOP2,JP1,JP6,MMS1,MMS2}.

\section{Adiabatically switched interactions}
\label{sec-asi}

Our next topic is the optimality of the Landauer bound in the context of time
dependent Hamiltonian dynamics of $\cS+\cR$. We shall assume that the reader is
familiar with basic results concerning non-autonomous perturbations of
$C^\ast$-dynamical systems (see Section~5.4.4 in~\cite{BR2} and the Appendix
to Section~IV.5 in~\cite{Si}).

\subsection{Setup}

Let $K:[0,1]\to\cO^\sa\cap\Dom(\delta_\cR)$ be a continuous function which we
assume to be twice continuously differentiable on $]0,1[$ with uniformly bounded
first and second derivatives. For $T>0$, we define the rescaled function $K_T$ by
$$
K_T(t)=K(t/T).
$$
Let $[0,T]\ni t\mapsto\alpha_{K_T}^t$ be the non-autonomous $C^\ast$-dynamics
defined by the Cauchy problem
\beq\label{AlphaDef}
\partial_t \alpha_{K_T}^t(A)
=\alpha_{K_T}^t(\delta_\cR(A)+\i[K_T(t),A]),\qquad\alpha_{K_T}^0(A)=A.
\eeq
We recall that $\{\alpha_{K_T}^t\}_{t\in[0,T]}$ is a strongly continuous family
of $\ast$-automorphisms of $\cO$ given by
$$
\alpha_{K_T}^t(A)=\tau_\cR^t(A) +\sum_{n=1}^\infty\,
\int\limits_{0\le s_1\le\cdots\le s_n\le t}
\i[\tau_\cR^{s_1}(K_T(s_1)),\i[
\cdots,\i[\tau_\cR^{s_n}(K_T(s_n)),\tau_\cR^t(A)]]]
\d s_1\cdots\d s_n.
$$
Moreover, the interaction representation
\beq\label{InterGamma}
\tau_\cR^{-t}\circ\alpha_{K_T}^t(A)
=\tau_\cR^{-t}(\Gamma_{K_T}(t))^\ast A\tau_\cR^{-t}(\Gamma_{K_T}(t)),
\eeq
holds with a family of unitaries $\Gamma_{K_T}(t)\in\cO$ satisfying the
Cauchy problem
\beq\label{GammaK}
\i\partial_t\Gamma_{K_T}(t)=\tau_\cR^t(K_T(t))\Gamma_{K_T}(t),\qquad
\Gamma_{K_T}(0)=\one.
\eeq

We denote by $\rho_T$ the restriction of $\omega_\i\circ\alpha_{K_T}^T$ to
$\cO_\cS$. Our assumptions ensure that $\Gamma_{K_T}(T)\in\Dom(\delta_\cR)$ and
Eq.\;\eqref{barbados} gives
\beq\label{ep-ad}
\Delta S_T+\sigma_T=\beta\Delta\cQ_T,
\eeq
where
$$
\Delta S_T=S(\rho_\i)-S(\rho_T),\qquad
\Delta\cQ_T
=-\i\omega_\i\left(\Gamma_{K_T}^\ast(T)\delta_\cR(\Gamma_{K_T}(T))\right),
$$
and
$$
\sigma_T=S(\omega_\i\circ\alpha_{K_T}^T|\rho_T\otimes\nu_\i).
$$

To interpret the right hand side of Eq.\;\eqref{ep-ad}, we write
$$
\Delta\cQ_T=Q(T)-Q(0)=\int_0^T\partial_t Q(t)\d t,
$$ with
$Q(t)=-\i\omega_\i(\Gamma_{K_T}^\ast(t)\delta_\cR(\Gamma_{K_T}(t)))$. It follows
from the differential equation~\eqref{GammaK} that
$$
\partial_t Q(t)
=-\omega_\i\left(
\Gamma_{K_T}^\ast(t)\tau_\cR^t(\delta_\cR(K_T(t)))\Gamma_{K_T}(t)\right),
$$
and Eq.\;\eqref{AlphaDef}-\eqref{InterGamma} give
\[
\begin{split}
\partial_t Q(t)&=-\omega_\i\circ\alpha_{K_T}^t\left(\delta_\cR(K_T(t))\right)\\[3mm]

&=-\frac{\d\ }{\d t}\,\omega_\i\circ\alpha_{K_T}^t(K_T(t))
+\omega_\i\circ\alpha_{K_T}^t\left(\frac{\d\ }{\d t}\,K_T(t)\right).
\end{split}
\]
This leads to
\beq\label{FirstLaw}
\Delta\cQ_T+\omega_\i\circ\alpha_{K_T}^T(K_T(T))-\omega_\i(K_T(0))
=\int_0^T\omega_\i\circ\alpha_{K_T}^t\left(P_T(t)\right)\d t,
\eeq
where $P_T(t)=\partial_t K_T(t)$ is the instantaneous power injected into the
system $\cS+\cR$. Energy conservation yields that $\Delta\cQ_T$ is the
total change in the energy of the subsystem $\cR$ from time $t=0$ to time $t=T$.

\subsection{The Landauer principle in the adiabatic limit}
\label{AL}

We shall now consider the adiabatic limit $T\to\infty$. Our main assumption in
this section concerns the instantaneous $C^\ast$-dynamics
$\tau_{K(\gamma)}$.

\begin{quote}{\bf  Assumption B.} For $0<\gamma<1$, the
$(\tau_{K(\gamma)},\beta)$-KMS state
$\mu_{K(\gamma)}$ is ergodic for the dynamical system $(\cO,\tau_{K(\gamma)})$.
\end{quote}

The Avron-Elgart adiabatic theorem~\cite{AE,Teu} and Araki's perturbation theory
of KMS states give (\cite{ASF1}-\cite{ASF3}, \cite{JP8}):

\bet\label{adiabatic}
Suppose that Assumption {\bf B} holds. Then one has
\[
\lim_{T\to\infty}\|\mu_{K(0)}\circ\alpha_{K_T}^{\gamma T}-\mu_{K(\gamma)}\|=0
\]
for all $\gamma\in[0,1]$.
\eet
For completeness and the reader's convenience the proof of Theorem \ref{adiabatic}
is given in Section \ref{sec-proof}.

Let $\rho_\f$ be a given faithful target state of $\cS$ and set
\[
\omega_\f=\rho_\f\otimes\nu_\i.
\]

According to Theorem~\ref{adiabatic}, to achieve the transition
$\rho_\i\to\rho_\f$ in the limit $T\to\infty$,  it suffices to assume that Assumption
{\bf B} holds for $K(\gamma)$ satisfying the boundary conditions
\[
K(0)=-\beta^{-1}\log \rho_\i,\qquad
K(1)=-\beta^{-1}\log \rho_\f.
\]
Indeed, these conditions ensure that $\mu_{K(0)}=\omega_\i$ and $\mu_{K(1)}=\omega_\f$ so that
$$
\lim_{T\to\infty}\omega_\i\circ\alpha_{K_T}^T
=\lim_{T\to\infty}\mu_{K(0)}\circ\alpha_{K_T}^T
=\mu_{K(1)}=\omega_\f.
$$
Theorem~\ref{adiabatic} further implies that
\[
\Delta S=\lim_{T\to\infty}\Delta S_T=S(\rho_\i)-S(\rho_\f).
\]
Moreover, rewriting Eq.\;\eqref{FirstLaw} as
$$
\Delta\cQ_T=\int_0^1\omega_\i\circ\alpha_{K_T}^{\gamma T}\left(
\partial_\gamma K(\gamma)\right)\d\gamma
-\beta^{-1}\omega_\i\circ\alpha_{K_T}^T(\log\rho_\f)+\beta^{-1}\omega_\i(\log\rho_\i),
$$
we get
$$
\Delta\cQ=\lim_{T\rightarrow \infty}\Delta\cQ_T
=\int_0^1\mu_{K(\gamma)}(\partial_\gamma K(\gamma))\d\gamma
+\beta^{-1}\Delta S.
$$
The balance equation~\eqref{ep-ad} yields that
\[
\Delta S +\sigma=\beta \Delta\cQ
\]
with
\[
\sigma=\lim_{T\to\infty}\sigma_T
=\beta\int_0^1\mu_{K(\gamma)}(\partial_\gamma K(\gamma))\d\gamma.
\]
Clearly, $\sigma \geq 0$. The adiabatic limit is a quasi-static process and one may expect the optimality of the Landauer bound. This is indeed the case.
\bep\label{thm-td-1}
\[
\sigma=0.
\]
\eep
The proof of the last result  requires modular theory. Note however that for finite reservoirs  the relation
\[
\int_{0}^1\mu_{K(\gamma)}(\partial_\gamma K(\gamma))\d\gamma=0
\]
is easily derived\footnote{On the other hand,  Theorem \ref{adiabatic} and relation $\lim_{T\rightarrow \infty}\sigma_T=\sigma$ {\em cannot} hold for finite reservoirs.}:
\begin{align*}
\int_{0}^1\mu_{K(\gamma)}(\partial_\gamma K(\gamma))\d\gamma
&=\int_{0}^1
\frac{\tr\left(\e^{-\beta(H_\cR+K(\gamma))}\partial_\gamma K(\gamma)\right)}
{\tr\left(\e^{-\beta(H_\cR+K(\gamma))}\right)}\d\gamma\\[2mm]
&=-\frac1\beta\int_0^1\partial_\gamma\log\tr\left(
\e^{-\beta(H_\cR+K(\gamma))}\right)\d\gamma\\[2mm]
&=-\frac{1}{\beta}\left(\log\tr (\omega_\f)-\log\tr(\omega_\i)\right)=0.
\end{align*}
This completes our mathematical analysis of the Landauer principle for
adiabatically switched interactions.

\bigskip
\noindent{\bf Remark 1.} Regarding the remarks at the end of Section~\ref{sec-inst},
Remark 1 applies to the results of this section as well. In the adiabatic case
the entropy production term vanishes and the instability discussed in
Remark 2 is absent. Remark 4 also extends to the adiabatic setting (see~\cite{DS}
for the discussion of the adiabatic theorem and~\cite{AHHH} for a discussion of
the Landauer principle in the van Hove weak coupling limit).
Since mixing implies ergodicity, the physically relevant models
for which Assumption {\bf B} has been verified are listed in Remark 5.

\medskip
\noindent{\bf Remark 2.} The Narnhoffer-Thirring adiabatic theorem of quantum
statistical mechanics~\cite{NT} is based on $C^\ast$-scattering and requires
$L^1$-asymptototic Abelianess which is stronger than our ergodicity assumption
{\bf B}. The physically relevant models satisfying $L^1$-asymptotic Abelianess
are discussed in~\cite{AM,AJPP1,AJPP2,BM,FMU,FMSU,JOP2}. If $L^1$-asymptotic
Abelianess holds, then the Landauer principle for adiabatically switched
interactions can be further refined. The result of this analysis is given
in~\cite{Han}.

\section{Discussion}
\label{sec-disc}
In this section we comment on the key ingredients involved in the analysis of the
Landauer principle presented in Sections~\ref{sec-ep-bal}--\ref{sec-asi}, and on
their relation with the work~\cite{ReWo}.

\bigskip
\noindent{\em The entropy balance equation.}
Relation~\eqref{ep-form} is  a model-independent structural identity linked to
the KMS condition and modular theory. It is tautological in the finite dimensional
case. The general case follows from Araki's perturbation theory of the
modular structure. The mathematical analysis of the second law of thermodynamics
starts with the entropy balance equation but certainly does not end
there\footnote{In the literature, the entropy balance equation~\eqref{ep-form} is
sometimes called "finite time second law of thermodynamics" reflecting the fact
that in typical application $U$ is a unitary cocycle describing time evolution
over a finite time period.}\cite{ASF1,ASF2,ASF3}. The thermodynamic behavior of the
coupled system $\cS+\cR$ emerges only in the thermodynamic limit in which the reservoir
$\cR$ becomes infinitely extended. In the large time limit the coupled system settles
into a steady state, substantiating the zeroth law of thermodynamics
\cite{BFS,DJ,FM,JP1}.

These two limiting processes, large reservoir size and large time, have been pillars
of the mathematical theory of open quantum systems since its
foundations~\cite{Rob,BR1,BR2}. In a sense, the same applies to the Landauer
principle and this is the main message of this note: the control of the entropy
balance equation for open quantum systems with infinitely extended reservoirs
in the large time (or adiabatic) limit is one of the central issues in the analysis
of the Landauer principle within quantum statistical mechanics. This brings us to our
second point.

\bigskip
\noindent{\em Confined reservoirs.}
A typical physical example of a confined reservoir is a Fermi gas or a Bose gas
in thermal equilibrium confined to a finite box. Confined reservoirs are not ergodic
and lead to quasi-periodic dynamics when coupled to a finite system $\cS$.
The analysis of the large time asymptotics of such systems requires some
time averaging which is not compatible with the formulation of Landauer's
principle. In this context,
one may say that the main contribution of~\cite{ReWo} concerns estimates
regarding the accuracy of the Landauer principle for confined reservoirs.

\bigskip
\noindent{\em Ergodicity.}
The large time asymptotics of the microscopic system $\cS$ coupled to the thermal
reservoir $\cR$ is critically linked to the ergodic properties  (Assumptions {\bf A}
and {\bf B}) of the dynamical system which describes the joint system $\cS+\cR$ in the
framework of statistical mechanics.
As we have shown, ergodicity allows for arbitrary transition $\rho_\i\to\rho_\f$
of the system $\cS$ in the adiabatic limit with the minimal energy dissipation
predicted by Landauer.
Needless to say, Assumptions {\bf A} and {\bf B}, which are part of the zeroth law of
thermodynamics,  are notoriously difficult to prove for physically relevant
models. In particular, they cannot hold in the framework of~\cite{ReWo}, where the
reservoirs are confined.

\bigskip

\noindent{\em Conclusion.} The claim of the authors in \cite{ReWo} that they
have proven the Landauer principle in quantum statistical mechanics may lead to a
confusion regarding some foundational aspects of mathematical theory of open
quantum systems and we have attempted to clarify this point. The complementary
analysis of the Landauer principle presented in
Sections~\ref{sec-ep-bal}--\ref{sec-asi} relies on the entropy balance equation,
Araki's perturbation theory of KMS states, and the Avron-Elgart adiabatic
theorem. It is a simple consequence of well-known and deep structural results.
The workers in quantum information theory appear unaware of this fact.
From the
point of view of state-of-the-art quantum statistical mechanics, the
interesting aspect of the Landauer principle concerns the verifications of
Assumptions {\bf A} and {\bf B}. The models for which this has been achieved are discussed in \cite{AM,AJPP1,AJPP2,BFS,BM,dRK,DJ,FMU,FMSU,JOP1,JOP2,JP1,JP6,MMS1,MMS2}. One may say that one  of the main challenges of quantum statistical mechanics at
the moment is to extend  the class of physically relevant models for which
Assumptions {\bf A} and {\bf B} can be proved. The progress in this direction
requires novel ideas and techniques in the study of the large time dynamics of
infinitely extended Hamiltonian quantum statistical models.

\section{Proofs}
\label{sec-proof}

\paragraph{Preliminaries.}
We start with some general properties of the GNS representation
$(\fH,\pi,\Omega)$ of $\cO$ associated to the state
$\omega_\i=\rho_\i\otimes\nu_\i$. This material is standard and we refer the
reader to \cite{BR1, BR2,DJP} for a detailed exposition and proofs. We denote by
$\fM=\pi(\cO)^{\prime\prime}$ the enveloping von Neumann algebra and by
$\cP\subset\fH$ and
$J$ the natural cone and modular conjugation of the pair $(\fM,\Omega)$.
Any state $\omega\in\cN_{\omega_\i}$ has a unique standard representative,
a unit vector $\Psi\in\cP$ such that $\omega(A)=\langle \Psi,\pi(A)\Psi\rangle$ for all
$A\in\cO$. The standard Liouvillean of a strongly continuous group $\varsigma$ of
$\ast$-automorphisms of $\cO$ is the unique self-adjoint operator $L$ on $\fH$
such that
$$
\pi(\varsigma^t(A))=\e^{\i tL}\pi(A)\e^{-\i tL},\qquad\e^{\i tL}\cP\subset\cP,
$$
for all $t\in\rr$ and all $A\in\cO$.

Let $L_0$ be the standard Liouvillean of a group $\varsigma_0^t=\e^{t\delta_0}$
of $\ast$-automorphisms of $\cO$ and $\Phi_0$ the standard representative of a
$(\varsigma_0,\beta)$-KMS state $\omega_0\in\cN_{\omega_\i}$.
If $Q\in\cO^\sa$ and $\delta_Q=\delta_0+\i[Q,\,\cdot\,]$, then the standard
Liouvillean of the locally perturbed group $\varsigma_Q^t=\e^{t\delta_Q}$ is
$$
L_Q=L_0+\pi(Q)-J\pi(Q)J.
$$
Moreover, $\Phi_0\in\Dom(\e^{-\beta(L_0+\pi(Q))/2})$ and the vector
$$
\Psi_Q=\frac{\Phi_Q}{\|\Phi_Q\|},\qquad
\Phi_Q=\e^{-\beta(L_0+\pi(Q))/2}\Phi_0,
$$
is the standard representative of a $(\varsigma_Q,\beta)$-KMS state. In particular,
one has $\Psi_Q\in\Ker(L_Q)$.

We shall need the following perturbative expansion of the unnormalized
KMS vector $\Phi_Q$.
For any $Q_1,\ldots,$ $Q_n\in\fM$ and $(\beta_1,\ldots,\beta_n)\in T_{\beta,n}
=\left\{(\beta_1,\ldots,\beta_n)\in\rr_+^n,|\,
\beta_1+\cdots+\beta_n\leq\beta/2\right\}$ one has
\[
\Phi_0\in\Dom\left(\e^{-\beta_1L_0} Q_1\cdots\e^{-\beta_n L_0}Q_n\right).
\]
Moreover, the map
\[
T_{\beta,n}\ni(\beta_1,\ldots,\beta_n)
\mapsto\e^{-\beta_1L_0} Q_1\cdots\e^{-\beta_n L_0}Q_n\Phi_0\in{\mathfrak H},
\]
is continuous and satisfies
\beq
\sup_{(\beta_1,\ldots,\beta_n)\in T_{\beta,n}}
\|\e^{-\beta_1L_0}Q_1\cdots\e^{-\beta_n L_0}Q_n\Phi_0\|
\leq \|Q_1\|\cdots\|Q_n\|.
\label{araki-est}
\eeq
The vector $\Phi_Q$ has the norm convergent expansion
\beq
\Phi_Q=\sum_{n=0}^\infty (-1)^n \int_{T_{\beta, n}}
\e^{-\beta_1L_0}\pi(Q)\cdots\e^{-\beta_n L_0}
\pi(Q)\Phi_0\,\d\beta_1\cdots\d\beta_n.
\label{araki-exp}
\eeq

For $\delta Q\in\cO^\sa$, the following chain rule applies
$$
\Phi_{Q+\delta Q}=\e^{-\beta(L_Q+\pi(\delta Q))/2}\Phi_Q,
$$
(see Theorem 5.1 (6) in \cite{DJP}).
It follows from the expansion~\eqref{araki-exp} and the
estimate~\eqref{araki-est} that the map $\cO^\sa\ni Q\mapsto\Phi_Q\in\fH$ is
differentiable. Its derivative at $Q$ is the map
\beq\label{PhiPrime}
\Phi'_Q:\delta Q\mapsto-\int_0^{\beta/2}\e^{-sL_Q}\pi(\delta Q)\Phi_Q\d s.
\eeq
The same argument shows that if $\alpha\mapsto Q(\alpha)$ is a real analytic
function from some open subset of $\rr^n$ to $\cO^\sa$, then the function
$\alpha\mapsto\Phi_{Q(\alpha)}$ is also real analytic.

\bigskip
\paragraph{Proof of Proposition~\ref{Saturation}} To simplify the notation, we write
$\rho=\rho_\i$, $\nu=\nu_\i$, $\omega=\omega_\i$. We denote by $\Omega_U\in\cP$ the
standard representative of the state $\omega_U$.
The modular operator $\Delta_\omega$ and the relative modular operator
$\Delta_{\omega_U|\omega}$ are positive operators on $\fH$ satisfying
$$
J\Delta_\omega^{1/2}\pi(A)\Omega=\pi(A)^\ast\Omega,\qquad
J\Delta_{\omega_U|\omega}^{1/2}\pi(A)\Omega=\pi(A)^\ast\Omega_U,
$$
for all $A\in\cO$. It follows from $\omega_U=\omega\circ\alpha_U$ that
 $\Omega_U=\pi(U)J\pi(U)J\Omega$. Since $J\pi(U)J\in\fM'$, one has
\begin{align*}
J\Delta_{\omega_U|\omega}^{1/2}\pi(A)\Omega&=\pi(A)^\ast\pi(U)J\pi(U)J\Omega\\
&=J\pi(U)J\pi(A)^\ast\pi(U)\Omega\\
&=J\pi(U)JJ\Delta_\omega^{1/2}\pi(U)^\ast\pi(A)\Omega,
\end{align*}
and the cyclic property of $\Omega$ allows us to conclude that
\beq
\Delta_{\omega_U|\omega}=\pi(U)\Delta_\omega\pi(U)^\ast.
\label{DeltaU}
\eeq

The product structure of the state $\omega$
induces the factorization $\fH=\fH_\cS\otimes\fH_\cR$ where $\fH_\cS=\cO_\cS$ equipped
with the inner product $\langle X,Y\rangle=\tr(X^\ast Y)$ and the Hilbert
space $\fH_\cR$ carries a GNS representation of $\cO_\cR$ induced by the state $\nu$.
Moreover, one has
\beq
\Delta_{\omega}=\Delta_\rho\otimes\Delta_\nu,\qquad
\Delta_{\omega_U|\omega}=\Delta_{\rho_U|\rho}\otimes\Delta_{\nu},
\label{DeltaFact}
\eeq
where $\Delta_\rho$, $\Delta_{\nu}$ and $\Delta_{\rho_U|\rho}$ are respectively
the modular operator of the state $\rho$, the modular operator
of the state $\nu$, and the relative modular operator of the state $\rho_U$
w.r.t.\;$\rho$. The operators $\Delta_\rho$ and $\Delta_{\rho_U|\rho}$ act on $\fH_\cS$
according to
\beq
\Delta_\rho X=\rho X\rho^{-1},\qquad\Delta_{\rho_U|\rho}X=\rho_U X\rho^{-1},
\label{DeltaForm}
\eeq
(see, e.g., Section~2.12 in \cite{JOPP}). In particular, they have discrete spectra.

Denote by $\Delta_{\omega,\rm p}$ and $\Delta_{\omega_U|\omega,\rm p}$ the pure point
parts of $\Delta_\omega$ and $\Delta_{\omega_U|\omega}$. Eq.\;\eqref{DeltaFact} implies that
\beq
\Delta_{\omega,\rm p}=\Delta_\rho\otimes\Delta_{\nu,{\rm p}},\qquad
\Delta_{\omega_U|\omega,\rm p}=\Delta_{\rho_U|\rho}\otimes\Delta_{\nu,{\rm p}},
\label{DeltaP}
\eeq
where $\Delta_{\nu,{\rm p}}$ is the pure point part of $\Delta_\nu$.
Since $\Delta_\nu=\e^{-\beta L_\cR}$, the operators $\Delta_{\omega,\rm p}^{\i\alpha}$
and $\Delta_{\omega_U|\omega,\rm p}^{\i\alpha}$ are trace class by assumption and
it follows from Eq.\;\eqref{DeltaU} that these two operators are
unitarily equivalent so that
$$
\tr(\Delta_{\omega,\rm p}^{\i\alpha})=\tr(\Delta_{\omega_U|\omega,\rm p}^{\i\alpha}),
$$
for all $\alpha\in\cc$. Using Eq.\;\eqref{DeltaForm} and \eqref{DeltaP}, an
explicit calculation yields
$$
\tr(\Delta_{\omega,\rm p}^{\i\alpha})=
\tr(\Delta_\rho^{\i\alpha})\,\tr(\Delta_{\nu,{\rm p}}^{\i\alpha})
=\tr(\rho^{\i\alpha})\,\tr(\rho^{-\i\alpha})
\,\tr(\Delta_{\nu,{\rm p}}^{\i\alpha}),
$$
$$
\tr(\Delta_{\omega_U|\omega,\rm p}^{\i\alpha})
=\tr(\Delta_{\rho_U|\rho}^{\i\alpha})\,\tr(\Delta_{\nu,{\rm p}}^{\i\alpha})
=\tr(\rho_U^{\i\alpha})\tr(\rho^{-\i\alpha})\,\tr(\Delta_{\nu,{\rm p}}^{\i\alpha}).
$$
Thus, we conclude that
$$
\tr(\rho^{\i\alpha})=\tr(\rho_U^{\i\alpha}),
$$
for all $\alpha\in\cc$,
which implies that $\rho$ and $\rho_U$ are unitarily equivalent.
\hfill\qed

\bigskip

\paragraph{Proof of Proposition \ref{HTarget}} The proof is based on an application of
the real analytic implicit function theorem.
Denote by $\mathfrak X$ the real vector space $\{X\in\cO_\cS^\sa\,|\,\tr(X)=0\}$
equipped with the inner product $( X,Y)=\tr(XY)$. Let
$$
\rr\times\mathfrak{X}\ni(\lambda,X)\mapsto F(\lambda,X)
=\rho_{X+\lambda V}-\rho_\f\in\mathfrak{X}.
$$
First, note  that $F$ is real analytic.
Moreover, for any $X\in\mathfrak{X}$, one has
$$
F(0,X)=\frac{\e^{-\beta X}}{\tr(\e^{-\beta X})}-\rho_\f,
$$
and  $F(0,X)=0$ iff $X=H_0$.
Let $L$ be the standard Liouvillean of the group $\tau_{H_0+\lambda V}$
and $\Phi\in\cP$ the standard representative of the KMS state $\mu_{H+\lambda V}$.
For $X,Y\in\mathfrak{X}$ one has
$$
( \rho_{H+X+\lambda V},Y)=\mu_{H+X+\lambda V}(Y\otimes\one)=
\frac{\langle \e^{-\beta(L+\pi(X))/2}\Phi,\pi(Y\otimes\one)\e^{-\beta(L+\pi(X))/2}\Phi\rangle}
{\|\e^{-\beta(L+\pi(X))/2}\Phi\|^2}.
$$
Using Eq.\;\eqref{PhiPrime}, an explicit calculation yields that the derivative
$F'(\lambda, H_0)$ of the function $F$ with respect to its second argument  is the
symmetric linear map on $\mathfrak{X}$ given by
$$
( F'(\lambda,H_0)X,Y)
=-2\int_0^{\beta/2}\Re\langle \e^{-sL/2}\pi(\hat X\otimes\one)\Psi,
\e^{-sL/2}\pi(\hat Y\otimes\one)\Psi\rangle \d s,
$$
where $\hat X=X-\rho_{H_0+\lambda V}(X)\one$. Since $\hat X=0$ iff $X=0$, it
follows that
$$
(F'(\lambda,H_0)X,X)
=-2\int_0^{\beta/2}\|\e^{-sL/2}\pi(\hat X\otimes\one)\Psi\|^2\d s<0,
$$
for all $0\not=X\in\mathfrak{X}$, and the implicit function theorem yields the conclusions of Proposition \ref{HTarget}.
\hfill\qed

\paragraph{Proof of Proposition \ref{roof}}  Suppose that $\sigma(\gamma)=0$.
The weak-$\ast$ lower semicontinuity of relative entropy yields
\[
0=\sigma(\gamma)
=\lim_{t\to\infty}S(\omega_\i\circ\tau_{K_\gamma}^t|\rho_{K_\gamma}(t)\otimes\nu_\i)
\geq S(\mu_{K_\gamma}|\rho_\f\otimes\nu_\i),
\]
which implies $\mu_{K_\gamma}=\rho_\f\otimes\nu_\i$ and hence $\mu_{K_\gamma}\circ\tau_\cR^t=\mu_{K_\gamma}=\mu_{K_\gamma}\circ\tau_{K_\gamma}^t$
for all $t\in\rr$. It follows that
$$
\mu_{K\gamma}(\delta_\cR(A))=0=\mu_{K_\gamma}(\delta_\cR(A)+\i[K_\gamma,A])
$$
for all $A\in\Dom(\delta_\cR)$, from which we conclude that $K_\gamma$ belongs
to the centralizer of $\mu_{K_\gamma}$. It follows from the KMS property of $\mu_{K_\gamma}$ that $\tau_{K_\gamma}^t(K_\gamma)=K_\gamma$ for all $t\in\rr$
(see, e.g., Proposition~5.3.28 in~\cite{BR2}).

For $\zeta\in\rr$, set
$S_\zeta=\e^{\zeta K_\gamma}/\mu_{K_\gamma}(\e^{2\zeta K_\gamma})^{1/2}$
and note that $\xi_\zeta(A)=\mu_{K_\gamma}(S_\zeta A S_\zeta)$ defines a
state in $\mathcal N_{\omega_\i}$. The mixing property and the fact that
$\tau_{K_\gamma}^t(S_\zeta)=S_\zeta$ yield
$$
\mu_{K_\gamma}(A)=\lim_{t\to\infty}\xi_\zeta\circ\tau_{K_\gamma}^t(A)=
\lim_{t\to\infty}\mu_{K_\gamma}(S_\zeta\tau_{K_\gamma}^t(A)S_\zeta)=
\xi_\zeta(A),
$$
from which we conclude that $\mu_{K_\gamma}(S_\zeta A S_\zeta-A)=0$ for
all $A\in\cO$. Setting $A=S_\zeta^2-\one$ further yields
$\mu_{K_\gamma}((S_\zeta^2-\one)^2)=0$. Since $\mu_{K_\gamma}$ is faithful
we conclude that $S_\zeta^2=\one$ and hence that $K_\gamma$ is a multiple
of $\one$. This implies that $\delta_{K_\gamma}=\delta_\cR$ and contradicts
Assumption~{\bf A}. \qed

\paragraph{Proof of Theorem \ref{adiabatic}}
Denote by $L$ the standard Liouvillean of the group $\tau_{K(0)}$. Let $\Psi(0)$ be
the standard vector
representative of the KMS state $\mu_{K(0)}$.
For $t\in[0,T]$, set
\[
L_T(t)=L+\pi(\widehat K_T(t))-J\pi(\widehat K_T(t))J,
\]
with $\widehat K_T(t)=K_T(t)-K_T(0)$. The family $\{W_T(t)\}_{t\in[0,T]}$ of
unitary operators on $\fH$ satisfying
\[
\i\partial_t W_T(t)=L_T(t)W_T(t), \qquad W_T(0)=I,
\]
implements the dynamics $\alpha_{K_T}$ and preserves the natural cone, i.e.,
\beq\label{Conalpha}
\pi(\alpha_{K_T}^t(A))=W_T^\ast(t)\pi(A)W_T(t),\qquad
W_T(t)\cP\subset\cP,
\eeq
for all $t\in[0,T]$ and all $A\in\cO$.
With $\widehat{K}(\gamma)=K(\gamma)-K(0)$, the standard Liouvillean of the
instantaneous dynamics $\tau_{K(\gamma)}$ is
$$
L(\gamma)=L+\pi(\widehat K(\gamma))- J\pi(\widehat K(\gamma))J,
$$
and the standard representative of the KMS state $\mu_{K(\gamma)}$ is
\[
\Psi({\gamma})=\frac{\e^{-\beta(L+\pi(\widehat K(\gamma)))/2}\Psi(0)}
{\|\e^{-\beta(L+\pi(\widehat K(\gamma)))/2}\Psi(0)\|}.
\]
By construction, the orthogonal projection
$$
P(\gamma)=|\Psi(\gamma)\rangle\langle\Psi(\gamma)|
$$
is such that $\Ran(P(\gamma))\subset\Ker(L(\gamma))$ for $\gamma\in[0,1]$.
Moreover, Assumption {\bf B} implies $\Ran(P(\gamma))=\Ker(L(\gamma))$ for
$\gamma\in]0,1[$. Since the function $]0,1[\ni \gamma\mapsto\widehat{K}(\gamma)$ is
$C^2$ in norm  with uniformly bounded first and second derivative, the expansion~\eqref{araki-exp}, the estimate~\eqref{araki-est}, and an
obvious telescoping argument show that the map
\[
]0,1[\ni\gamma\mapsto P(\gamma)\in\cB(\fH)
\]
is also $C^2$ in norm with uniformly bounded first and second derivative.

One easily checks that the adiabatic evolution $\cW_T(t)$ defined by
\[
\i\partial_t\cW_{T}(t)
=(L_T(t)+T^{-1}\i[\dot P(t/T),P(t/T)])\cW_{T}(t),
\qquad \cW_{T}(0)=\one.
\]
intertwines $P(0)$ and $P(t/T)$, i.e., that
\beq\label{Intertw}
\cW_{T}(t)P(0)=P(t/T)\cW_{T}(t),
\eeq
holds for $t\in[0,T]$.

With these preliminaries, the Avron-Elgart adiabatic  theorem \cite{AE,Teu,ASF1} gives:

\bet\label{adiabatic-teufel}
Suppose that Assumption {\bf B} holds. Then
\[
\lim_{T\rightarrow \infty}\sup_{t\in[0,T]}\|W_T(t)-\cW_{T}(t)\|=0.
\]
\eet
For $\gamma\in[0,1]$, it follows from Eq.\;\eqref{Conalpha} that
\[
\mu_{K(0)}\circ\alpha_{K_T}^{\gamma T}(A)
=\langle W_T(\gamma T)\Omega,\pi(A)W_T(\gamma T)\Omega\rangle,
\]
while the intertwining relation~\eqref{Intertw} yields
\[
\mu_{K(\gamma)}(A)
=\langle \cW_{T}(\gamma T)\Omega,\pi(A)\cW_{T}(\gamma T)\Omega\rangle.
\]
Thus, we have the estimate
\[
|\mu_{K(0)}\circ\alpha_{K_T}^{\gamma T}(A)-\mu_{K(\gamma)}(A)|
\leq 2\|\A\|\sup_{t\in[0,T]}\|W_T(t)-\cW_{T}(t)\|,
\]
which, together with Theorem~\ref{adiabatic-teufel}, yields
Theorem \ref{adiabatic}.\hfill\qed
\bigskip

\paragraph{Proof of Proposition \ref{thm-td-1}}
\label{proof-thm-td-1}

We use the same notation as in the proof of Theorem \ref{adiabatic}.  Set $\Phi(\gamma)=\e^{-\beta(L+\pi(\widehat{K}(\gamma)))/2}\Psi(0)$. Araki's
perturbation formula yields
$$
S(\omega|\omega_{K(\gamma)})
=S(\omega|\omega_\i)+\beta\omega(\widehat{K}(\gamma))+\log\|\Phi(\gamma)\|^2,
$$
for any $\omega\in\cN_{\omega_\i}$. Setting $\omega=\omega_{\i/\f}$ and
$\gamma=0/1$ we derive $\|\Phi(0)\|=\|\Phi(1)\|=1$. Next, we claim that
\beq
\mu_{K(\gamma)}(\partial_\gamma K(\gamma))
=-\frac{1}{\beta}
\partial_\gamma\log\|\Phi(\gamma)\|^2,
\label{really}
\eeq
which clearly implies Proposition~\ref{thm-td-1}.

The identity
\[
\partial_\gamma\log\|\Phi(\gamma)\|^2
=\frac{\langle \partial_\gamma\Phi(\gamma),\Phi(\gamma)\rangle
+\langle \Phi(\gamma),\partial_\gamma\Phi(\gamma)\rangle}{\|\Phi(\gamma)\|^2},
\]
implies that (\ref{really}) follows from
$$
\langle\Phi(\gamma),\partial_\gamma\Phi(\gamma)\rangle
=-\frac{\beta}{2}
\langle \Phi(\gamma),\pi(\partial_\gamma K(\gamma))\Phi(\gamma)\rangle.
$$
The last identity is a  direct consequence of Eq.\;\eqref{PhiPrime} and the fact
that $L(\gamma)\Phi(\gamma)=0$.\hfill\qed


TITLE: Learning Differential Topology
QUESTION [2 upvotes]: So, I'm currently at the point of learning about derivatives and integrals in Elementary Real Analysis, I'm also learning Point set topology and I've already gone over the main definition of topology, basis, subspace topology, quotient topology, product topology up to continuous functions thus far. I also understand elementary linear algebra and know enough Abstract Algebra.
My question is whether or not I can start reading Differential Topology by Guillemin and Pollack. It states a full year of analysis and semester of linear algebra but I'm not exactly sure if my current level understanding is sufficient to start studying differential topology.
If not should I start going over Hatcher's book on Algebraic Topology? I've read that knowing algebraic topology before differential topology is a good idea.

REPLY [1 votes]: I would say no. In my experience, in order to really study differential topology you need to have a firm ground in multivariable calculus. In particular, things like understanding the derivative is a linear map (best linear approximation), implicit function theorem, inverse function theorem, etc.
Ted Shifrin has a really awesome book that will give you the relevant linear algebra and multi- calculus to begin studying differential topology here. I would also recommend Tu's introduction to manifolds, or Lee who develop a bit more geometric understanding of tangent space (than G & P) which is the first topic to digest. A lot of users may not like this but I think the order of things follows best if you go,
$$\textbf{linear algebra + multivariable calculus} \to \textbf{differential geometry} \to \textbf{differential topology}$$


TITLE: Generating functions for election results
QUESTION [1 upvotes]: This is from my textbook which does not give solutions to problems (i.e. not homework).

Find the generating series for the number of election results involving 4 candidates with respect to the number of voters.

This is easy and is obviously $(1+x+x^2+\cdots)^4$. However, the next question is hard:

Do the same, except assume that all 4 candidates vote for themselves.

How would I solve this problem then?

REPLY [3 votes]: In the first question, each of the four candidates can have $0$ or more people who vote for them. In the second question, each candidate is guaranteed at least $1$ vote (themselves). Hence, we obtain:
$$
(x + x^2 + x^3 + \cdots)^4
$$


TITLE: Use Euler's Formula to prove the following equation $ y = Ae^{\sqrt{\lambda}ix} + Be^{-\sqrt{\lambda}ix} $
QUESTION [0 upvotes]: I'm having a difficult time understanding how one would take $ y = Ae^{\sqrt{\lambda}ix} + Be^{-\sqrt{\lambda}ix} $ and end up with $ y = A \cos{\sqrt{\lambda}x} + B\sin{\sqrt{\lambda}x} $ knowing that $ \lambda > 0 $.
Would it be easiest to start out by just plugging in Euler's Formulas $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ and $e^{-i\theta} = \cos(\theta) - i\sin(\theta)$? The only problem I have is I end up with this very long equation of $ y = A \cos{\sqrt{\lambda}x} + Ai\sin{\sqrt{\lambda}x} + B\cos{\sqrt{\lambda}x} - Bi\sin{\sqrt{\lambda}x}$. I'm really confused as to how this long expression can be reduced to the $ y = A \cos{\sqrt{\lambda}x} + B\sin{\sqrt{\lambda}x} $ it is asking for.
Any help is appreciated!

REPLY [1 votes]: You can just add the complex coeffiecients together so that  $$y = A \cos(t) + Ai\sin(t) + B\cos(t) - Bi \sin(t)= (A + B)\cos(t) + (Ai - Bi)\sin(t).$$


TITLE: Present Value of loan payment
QUESTION [1 upvotes]: You value a car to be $30,000. If you plan to make continuous payments over 5 years
and at an interest rate of r = 10%.
1) How much should you pay per year so that the present value of your total payments
in 30; 000?
The formula that i think i should use is
$$PV = PMT\frac{1-(1+i)^{-n}}{i}$$
So solving for $PMT$ I got 7913.48. Did i do that correctly?
2)What if instead you decided to let your payments increase with time and pay at a
rate of $6000 + t*1000$ per year, where t is measured in years. How long would it
take you to pay off the car ? (Note the equation you get might be difficult to solve,
so you can use a graphing calculator to estimate.)
I'm not too sure how to approach this. Do i use the same equation and solve for t?
I assume the $t$ is the same as the $n$ I used in my $pv$ formula, is this correct?
Thanks

REPLY [0 votes]: I do not know what $r$ reprents, so will assume that it is the effective annual rate.
If that is the case, then a dollar grows in a year to $e^k$ dollars, where $e^k=1.10$. Thus $k=\ln(1.10)$. But we will keep on using $k$.
Let $c$ be the amount paid per year. So we are paying at a yearly rate of $c$. The present value of these payments is
$$\int_0^5 c e^{-kt}\,dt.$$
Integrate. The result should be $30000$. Thus we obtain the equation
$$c\frac{1}{k}\left(1-e^{-5k}\right)=30000.$$
In the above equation, we know everything except $c$, so can solve for $c$.
For the second problem, the setup is roughly similar, except that we do not know the number $y$ of years, but we do know the rate of payment. We end up with the equation
$$\int_0^y (6000+1000t)e^{-kt}\,dt=30000.$$
We can do the integration using integration by parts. However, we end up with an equation in $y$ that we cannot solve analytically. However, the equation can be solved by using a numerical method. It can also be solved to adequate accuracy with a graphing calculator.
Remark: Conceivably the $r$ could refer to the force of interest. In that case we would have $k=0.10$. I chose to interpret it as the effective annual rate, since that seemed more plausible.


TITLE: When does zero divergence imply a vector potential exists?
QUESTION [1 upvotes]: From electrodynamics we know that $\boldsymbol{\nabla}\mathbf{B}=\mathbf 0$ hence we can introduce a vector potential such that $\mathbf{B}=[\boldsymbol \nabla\times \mathbf{A}]$.
What is the general mathematical proof that a vector field with zero divergence can be expressed as a curl of another vector field (that this field would necessarily exist)? What are the restrictions on the fields?
I suspect this question was asked before, I just didn't manage to find a clean answer.

REPLY [0 votes]: The framework of vector-analysis provides certain concepts and identities regarding how 'vectors' can be manipulated.
One of them being: a divergence-less $[\nabla.\vec{X}=0]$ vector field should wind upon itself, or simply be solenoidal $ [\vec{X} \text{ is curl of some other field}\implies \vec{X}=\nabla\times \vec{Y}]$ since $\forall \vec{Y}\,\nabla . (\nabla \times \vec{Y})=0$. (Unlike many other identities in vector analysis, this one's quite easy to visualize.)
And since EM deals with vector fields representing E and B in space, the divergence-lessness of $\vec{B}$ gives its solenoidal character, being $\vec{B}=\nabla \times \vec{A}$
But as @symplectomorphic mentioned in the comment, the restrictions are quite intricate indeed, in general.


TITLE: Does this game have infinite expected payout?
QUESTION [1 upvotes]: Consider the following game: Suppose the initial value of the pot is $ S $. Our player Josephine then rolls a fair $n$-sided die. If the roll is not $1$, then the pot is multiplied by that roll, and she rolls again; if it's $1$, then she receives whatever was in the pot, and the game is over. The game continues until Josephine sees a $1$. Am I right that the "expected value" of this game should be infinite?
My reasoning is about this: Payment only occurs when the sequence of rolls hits $1$ for the first time. So we can divide things up into the families $W_{k}$ sequences $w_{1} \ldots w_{k - 1} 1$, where $w_{j} \in \{2, 3, , \ldots, n \}$. Since the die is fair, these are all equiprobable. So
\begin{align*}
E & = \sum_{k = 1}^{\infty} \frac{\sum_{w \in W_{k}} w_{1} \cdots w_{k - 1}}{n^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{\sum_{w_{1}, \ldots, w_{k - 1} = 2}^{n} w_{1} \cdots w_{k - 1}}{n^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{(2 + \cdots + n)^{k}}{n^{k}} \\
& = \frac{1}{2} \sum_{k = 1}^{\infty} \frac{(n(n + 1) - 1)^{k}}{n^{k}} ,
\end{align*}
which should diverge.
But at the same time, I have a sneaking suspicion I'm doing something wrong. Can anyone make sure?
Moreover, if I'm right, and the payout is infinite, then could an unfair die get it under control?

REPLY [2 votes]: It's even easier than that.  When $n = 2$ the expected payout of the game is $2^{-1}\sum_{k=0}^{\infty} 2^k / 2^k = \infty$, and this is a lower bound on the expected payout for general $n$.


TITLE: Is the set of sinusoidal functions with period $2\pi$ closed under addition?
QUESTION [2 upvotes]: Let $w: \mathbb{R} \to \mathbb{R}$ be an element of B iff there exist $A \in \mathbb{R}_{ \geq 0}$ and $\phi \in [0, 2\pi)$ such that $w(t) = A \sin(t+\phi)$. Is B a linear subspace of $\mathbb{R}^{\mathbb{R}}$ ? By a theorem, B is a subspace of $\mathbb{R}^{\mathbb{R}}$ iff:

The zero function is in B ;
If $w_{1} (t)$ and $w_{2} (t)$ are elements of B, then $w_{1} (t) + w_{2} (t)$ is an element of B, too ;
If $w (t)$ is an element of B and $c(t)$ is a scalar function of $\mathbb{R}^{\mathbb{R}}$, then $(c \cdot w) (t) $ is an element of B as well.

I'd like to verify the second point. So I define $w_{1} (t) := A_{1} \sin (t + \phi_{1})$ and $w_{2} (t) := A_{2} \sin( t + \phi_{2})$ and see if we can find an $A_{3}$ and $\phi_{3}$ such that $w_{1} (t) + w_{2} (t) = A_{3} \sin (t + \phi_{3} ) \qquad (3)$.
We can rewrite the left side of $(3)$ by writing it as $$ A_{1} (\sin(t)\cos(\phi_{1}
) + \cos(t)\sin(\phi_{1})) + A_{2} (\sin(t)\cos(\phi_{2}) + \cos(t)\sin(\phi_{2})) \quad .$$
Furthermore, we can rewrite the right side in a way that it is almost similar: (provided we set $A_{3} = A_{1} + A_{2}$)  $$ A_{1} (\sin(t)\cos(\phi_{3}) + \cos(t)\sin(\phi_{3})) + A_{2} (\sin(t)\cos(\phi_{3}) + \cos(t)\sin(\phi_{3})) \quad.  $$
However, I'm not sure I've shown closure under addition now. Because the way I did it, the two expressions are only equal when $\phi_{1} = \phi_{3}$ and $\phi_{2} = \phi_{3}$, so $\phi_{1} = \phi_{2}$. But I thought equality ought to be able to hold for arbitrarily chosen $\phi_{1}$ and $\phi_{2}$, so not necessarily when they're equal.
Have I shown additivity now? If not, can you help me to do it?

REPLY [3 votes]: I'm not totally sure how to make your approach work. But you have, by expanding $A_1\sin(t+\phi_1)+A_2\sin(t+\phi_2)$,
\begin{align}
&A_1\cos(\phi_1)\sin(t)+A_1\sin(\phi_1)\cos(t)+A_2\cos(\phi_2)\sin(t)+A_2\sin(\phi_2)\cos(t)\\ \ \\
=&(A_1\cos(\phi_1)+A_2\cos(\phi_2))\,\sin(t)+(A_1\sin(\phi_1)+A_2\sin(\phi_2))\,\cos(t).
\end{align}
Now, to simplify notation, we need to show that $\alpha\,\sin t+\beta\cos t$ is of the form you want. The key observation is now that, for any $\alpha,\beta\in\mathbb R$, we can always find $\phi$ with
$$
\cos \phi=\frac\alpha{(|\alpha|^2+|\beta|^2)^{1/2}},\ \ \ \
\sin \phi=\frac\alpha{(|\alpha|^2+|\beta|^2)^{1/2}}.
$$
Then
\begin{align}
\alpha\sin t+\beta\cos t
&=(|\alpha|^2+|\beta|^2)^{1/2}\,(\cos\phi\,\sin t+\sin\phi\,\cos t)\\ \ \\
&=(|\alpha|^2+|\beta|^2)^{1/2}\,\sin(t+\phi).
\end{align}

Note that the same argument can be used to show that linear combinations of sines of any period are still multiples of $\sin(t+\phi)$.


TITLE: a simple inequality for a monotone function
QUESTION [0 upvotes]: To prove that the sequence $a_n = \sqrt{n^2+1} - n$ is monotone I used $a_{n+1}-a_n$ and got $\sqrt{(n+1)^2+1}-\sqrt{n^2+1}-1$, which is always negative, but that's as far as I'm able to process it, how do I show formally that it's less than $0$?
($n$ is in $\mathbb N_+$)

REPLY [2 votes]: You have $a_n=1/b_n$ with $b_n=\sqrt{n^2+1}+n$, and $b_n$ is clearly $>0$ and increasing, so $a_n$ is decreasing.


TITLE: If sequence converges, then terms converge to zero?
QUESTION [1 upvotes]: I've only seen this theorem for series:

If series converges, then the terms converge to 0.

However, I've seen it applied in sequences (e.g. [1]), so does this same apply to sequences? I cannot seem to find a proof or the theorem for that matter.

REPLY [0 votes]: A sequence is an indexed and ordered list of terms. When it is said that an infinite sequence converges, this means the sequence reaches a finite limit (which can be zero or nonzero), i.e. $\displaystyle |\lim_{k \to \infty} a_k| < \infty$.
A series is the sum of the individual terms of a sequence. When it is said that an infinite series converges, this means the limit of the partial sums is finite (again it can be zero or nonzero) , i.e. $\displaystyle |\lim_{n \to \infty} \sum_{k=1}^{\infty} a_k| < \infty$.
In the latter case, it is necessary but not sufficient that the sequence being summed has a limit of zero - an example of this occurs with the harmonic series. Apart from the the sequence of terms having a limit of zero, other criteria need to be satisfied - this is the basis of the tests of series convergence.
An example of a sequence that converges to a nonzero limit is $\displaystyle a_1 = 1, a_{k+1}= 1+\frac{1}{a_k}$, which converges to the golden ratio, $\phi$.


TITLE: How to check if some ring is an integral domain?
QUESTION [4 upvotes]: Let $k$ be a field, $a\in k$, and for all $b\in k$, $a\neq b^2$. I want to check if the ring $k[x,y,z]/(x^2-ay^2)$ is an integral domain.

I need to show that $(x^2-ay^2)$ is a prime ideal. If $fg\in (x^2-ay^2)$, I don't know how to show $(x^2-ay^2)$ contains $f$ or $g$ ?

REPLY [4 votes]: $k[x, y, z]$ is a unique factorization domain which helps: Now one can just show that $x^2-ay^2$ is an irreducible element.
You may approach this in various ways. One way that is quite convenient is to note that in $L[x,y,z]$ where $L=k(\sqrt{a}),$ one has $x^2-ay^2=(x-\sqrt{a}y)(x+\sqrt{a}y),$
and the two factors are irreducible. This implies that if $x^2-ay^2$ were to decompose in $k[x, y, z]$ as well, its irreducible factors would have to be $x-\sqrt{a}y, x+\sqrt{a}y$ at least up to $L$-scalar multiple. Now one concludes, since no nonzero $L$-scalar multiple of the polynomials have coefficients in $k$.


TITLE: Compute $\oint\frac{e^{2z}+\sin z}{(z^2+1)^3}dz$, over the curve C
QUESTION [0 upvotes]: Compute $I = \oint\frac{e^{2z}+\sin z}{(z^2+1)^3}dz$, over the curve $C:|z+i|=r, r\neq2$

So what I understood from my classes I have to find what it's poles are and then apply the residual theorem but I have some problems when trying to determine in what interval $r$ is and use that..
So what I've done so far is:
$z^2+1=0\implies z_{1,2}=\pm i$ and then I have to check for what $r$ are these roots in the $IntC$, right?
Then apply the theorem that $I=2\pi i \sum_{z_k\space singular \space point}Rez(f(z), z_k)$
Say for $z_1\in IntC\implies |i+i|< r\leftrightarrow 2 < r$, but for $z_2\in IntC\implies |-i+i|< r\leftrightarrow 0 < r$, how do I deal with this?
And we have $\pm i$ poles of order $3$ for that polynomial.. how do I solve it from here? And how do I prove that the integral is $0$ if the singular points are not in the interior of the curve?

REPLY [0 votes]: You are on the right track...
If $r < 2$ the only pole in the interior of $C$ is $z=i$ and the integral is given by
$$
\int_C f(z) dz = 2\pi i Res(f,-i)
$$
If $r > 2$ there are two poles inside the curve and the integral is given by
$$
\int_C f(z) dz = 2 \pi i (Res(f,-i) + Res (f,i))
$$


text:
\begin{document}
\keywords{Topological complexity, configuration spaces.}
\subjclass[2000]{Primary 55M99, 55R80; Secondary 68T40.}
\thanks{The research was supported by grants from the EPSRC and from the Royal Society.}

\title[Topological complexity]{Topological complexity of configuration
spaces}

\author[M. ~Farber and M. ~Grant]{Michael Farber and Mark Grant}
\address{Department of Mathematical Sciences, Durham University\\
South Road, Durham, DH1 3LE, UK}

\email{michael.farber@durham.ac.uk}
\email{mark.grant@durham.ac.uk}

\maketitle
\begin{abstract}
The topological complexity $\tc(X)$ is a homotopy invariant which reflects the complexity of the problem of constructing a motion planning algorithm in the space $X$, viewed as configuration space of a mechanical system. In this paper we complete the computation of the topological complexity of the configuration space of
$n$ distinct points in Euclidean $m$-space for all $m\ge 2$ and $n\ge 2$; the answer was previously known in the cases $m=2$ and $m$ odd. We
also give several useful general results concerning sharpness of upper bounds for the
topological complexity.
\end{abstract}

\section{Introduction}
The motion planning problem is a central theme of robotics \cite{La}. Given a mechanical
system $S$, a motion planning algorithm for $S$ is a function associating with any
pair of states $(A,B)$ of $S$ a continuous motion of the system starting at $A$ and
ending at $B$. If $X$ denotes the configuration space of the system, one considers the
path fibration
\begin{equation}\label{fibration}
\pi: PX \to X\times X, \quad \pi(\gamma) = (\gamma(0), \gamma(1)),
\end{equation}
where $PX=X^I$ is the space of all continuous paths $\gamma: I=[0,1]\to X$.
In these terms, a motion planning algorithm for $S$ is a section (not necessarily
continuous) of $\pi$.

The topological complexity of a topological space $X$, denoted $\tc(X)$, is defined to
be the genus, in the sense of Schwarz \cite{Sch}, of fibration (\ref{fibration}).
More explicitly, $\TC(X)$ is the minimal integer $k$ such that $X\times X$ admits a cover
by $k$ open subsets, on each of which there exists a continuous
local section of fibration (\ref{fibration}). One of the basic properties of $\tc(X)$ is its homotopy invariance \cite{Fa03}.
If $X$ is a Euclidean Neighbourhood Retract then the number $\tc(X)$ can be equivalently characterized (see \cite{Finv}, Proposition 4.2)
as the minimal integer $k$ such that
there exists a section $s: X \times X\to PX$ of (\ref{fibration}) and a decomposition
$$X\times X= G_1\cup \dots\cup G_k, \quad G_i\cap G_j =\emptyset, \quad i\not=j$$
where each $G_i$ is locally compact and such that the restriction $s|G_i: G_i \to PX$ is continuous for $i=1, \dots,k$.
A section $s$ as above can be viewed as a motion planning algorithm: given a pair of states $(A, B)\in X\times X$ the path $s(A, B)(t)$ represents a continuous motion of the system starting from $A$ and ending at $B$. The number $\tc(X)$ is a measure of the complexity of motion planning algorithms for a system whose configuration space is $X$.

The concept $\tc(X)$ was
introduced and studied in \cite{Fa03}, \cite{Fa04}.  We refer the reader to surveys \cite{Fa06}, \cite{Finv} for detailed treatment of the invariant $\tc(X)$.
Computation of $\tc(X)$ in various practically interesting examples has received much
recent interest, see for instance papers \cite{CP1}, \cite{CP2}, \cite{FG07},
\cite{Gon}, \cite{G}.

In this paper we study the topological complexity $\tc(F(\R^m, n))$ of the space of
configurations of $n$ distinct points in Euclidean $m$-space. Here $m, n \ge 2$, and
$$F(\R^m,n)=\left\{ (\x_1,\ldots ,\x_n)\in (\R^m)^{\times n};
 \x_i\neq \x_j \, \mbox{for} \, i\neq j \right\},$$
topologised as a subspace of the Cartesian power $(\R^m)^{\times n}$.
This space appears in robotics when one controls multiple objects simultaneously trying to avoid collisions between them.
 Our main result in this paper is the following.
\begin{theorem}\label{thm1}
One has
\begin{equation}
\TC(F(\R^m,n))=\left\{\begin{array}{ll}
2n-1 & \textrm{for all }m\textrm{ odd}, \\
2n-2 & \textrm{for all }m\textrm{ even.}
\end{array}\right.
\end{equation}
\end{theorem}
The cases $m=2$ and $m\geq 3$ odd of Theorem \ref{thm1} were proven by Farber and
Yuzvinsky in \cite{FY04}, where it was conjectured that $\TC(F(\R^m,n))=2n-2$ for all even
$m$. Here we settle this conjecture in the affirmative. Note that the methods employed in \cite{FY04} are not applicable in the case when $m > 2$ is even. We therefore suggest an alternative approach based on sharp upper bounds for the topological complexity.

The plan of the paper is as follows. In the next section we state Theorems \ref{thm2} \and \ref{thm3} about sharp upper bounds; their proofs appear in section \S \ref{prfs}. The concluding section \S \ref{prf} contains the proof of Theorem \ref{thm1}.

\section{Sharp upper bounds for the topological complexity}\label{sharp}

Let $X$ be a CW-complex of finite dimension $\dim(X)=n\geq 1$.
We denote by $\Delta_X\subset X\times X$ the diagonal $\Delta_X=\left\{(\x, \x); \x \in X\right\}$.
Let $A$ be a
local system of coefficients on $X\times X$.
A cohomology class $$u\in H^*(X\times X;A)$$ is called {\it a zero-divisor} if
its restriction to the diagonal is trivial, i.e. $u|{\Delta_X}=0\in H^\ast(X; A|X)$. The importance of zero-divisors stems from the following fact (see \cite{Finv}, Corollary 4.40):

{\it If the cup-product of $k$ zero-divisors $u_i\in H^\ast(X\times X; A_i)$, where $i=1, \dots, k$, is nonzero then $\tc(X) >k$.}

Theorem \ref{thm2} below supplements the general dimensional upper bound of \cite{Fa03} by giving necessary and sufficient conditions for its sharpness.

\begin{theorem}\label{thm2}
For any $n$-dimensional cell complex $X$ one has
 \bi
\item[(a)] $\TC(X)\leq 2n+1$;
\item[(b)] $\TC(X)=2n+1$ if and only if there exists a local coefficient
 system $A$ on $X\times X$ and a zero-divisor $\xi\in
  H^1(X\times X;A)$ such that the $2n$-fold cup product
  $$\xi^{2n}=\xi\cup \dots\cup \xi \neq   0\in H^{2n}(X\times X; A^{2n})$$
  is nonzero. Here $A^{2n}$ denotes the tensor product of $2n$ copies
  $A\otimes \dots \otimes A$
  of $A$ (over $\Z$).
\ei
\end{theorem}
Next we state a similar sharp upper bound result for $(s-1)$-connected spaces $X$ where $s>1$.
We use the following notation.
 If $B$ is an abelian group and $v\in
H^r(X;B)$ is a cohomology class then the class
\begin{eqnarray}\label{bar}
\bar{v} =v\times 1 - 1\times v\in
H^r(X\times X;B)\end{eqnarray}
is a zero-divisor, where $1\in H^0(X;\Z)$ is the unit and $\times$ denotes the cohomological cross-product.

We say that
a finitely generated abelian group is {\it
  square-free} if it has no subgroups isomorphic to $\Z_{p^2}$, where
$p$ is a prime.

\begin{theorem}\label{thm3} Let $X$ be a $(s-1)$-connected $n$-dimensional finite cell complex where $s\ge 2$. Assume additionally that $2n=rs$ where $r$ is an integer.\footnote{This last assumption is automatically satisfied (with $r=n$) for $s=2$, i.e. when $X$ is simply connected.}
Then
\bi
\item[(a)] $\TC(X)\leq r+1$;
\item[(b)] $\TC(X)= r+1$ if and only if there exists a finitely generated abelian group $B$ and a cohomology class $v\in H^s(X;B)$
such that the $n$-fold cup-product of the corresponding zero-divisors (\ref{bar}) is nonzero
$${\bar v}^r=\bar v\cup \dots\cup \bar v \ne 0 \, \in H^{2n}(X\times X; B^r).$$
Here $B^r$ denotes the $r$-fold tensor power
    $B\otimes \dots \otimes B$;
\item[(c)] If $H_\ast(X;\Z)$ is square-free, then $\TC(X)=r+1$ if and
  only if there exists a field $\kk$
 and cohomology classes $v_1,\ldots,v_r\in H^s(X;\kk)$ such that
 $$\bar v_1\cup \dots \cup \bar v_r \ne 0 \in H^{2n}(X\times X; \kk);$$
\item[(d)] If $H_s(X;\Z)$ is free abelian, then $\TC(X)=r+1$ if and
  only if there exist classes $v_1,\ldots ,v_r\in H^s(X;\Z)$ such that
   $$\bar v_1\cup \dots \cup \bar v_r \ne 0 \in H^{2n}(X\times X; \Z).$$
\ei
\end{theorem}

\section{Proofs of Theorems \ref{thm2} and \ref{thm3}}\label{prfs}

\begin{proof}[Proof of Theorem \ref{thm2}] The first statement follows from \cite{Fa03}, Theorem 4.
If there exists a local system coefficient system $A$ and a zero-divisor $\xi\in H^1(X\times X; A)$ such that $\xi^{2n}\not=0$ then $\tc(X) \ge 2n+1$, by
Corollary 4.40 of \cite{Finv}. The remaining part of Theorem \ref{thm2} was proven in \cite{CF}, Theorem 7. More precisely, let $G=\pi_1(X, x_0)$ denote the fundamental group of $X$ and let $I\subset \Z[G]$ denote the augmentation ideal. $I$ can be viewed as a left $\Z[G\times G]$-module via the action
$$(g, h)\cdot \sum n_ig_i= \sum gg_ih^{-1},$$ where $g,h, \in G$ and $\sum n_ig_i\in I$; this defines a local system with stem $I$ on $X\times X$, see \cite{Wh}, chapter 6. A crossed homomorphism $f: G\times G\to I$ given by the formula $$f(g, h)=gh^{-1} -1, \quad g, h \in G$$ determines a cohomology class $\vv\in H^1(X\times X;I)$. This class is a zero-divisor and has the property that $\vv^{2n}\not=0$ assuming that $\tc(X)=2n+1$ according to Theorem 7 from \cite{CF}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3}]
Statement (a) follows directly from Theorem 5.2 of \cite{Fa04} which states that
\begin{equation}
\TC(X)<\frac{2n+1}{s}+1.
\end{equation}
for any $(s-1)$-connected CW-complex  $X$ of dimension $n$.

(b) One part of statement (b) follows from Corollary 4.40 of \cite{Finv}; indeed if $\bar v^{r}\not=0$ then $\tc(X)\ge r+1$ since each $\bar v$ is a zero-divisor.

The proof of the remaining part of statement (b) is derived from obstruction theory and results of
A.\ S.\ Schwarz \cite{Sch} centered around the notion of genus of a fibration.
We assume that $X$ is $(s-1)$-connected, $s\ge 2$, and $n$-dimensional and $2n=rs$ where $r$ is an integer. The case $n=1$ is trivial, therefore we will assume that $n\ge 2$. We want to show that $\tc(X)=r+1$ implies that $\bar v^r\not=0 \in H^{2n}(X\times X; B^r)$ for some class $v\in H^s(X;B)$.

Recall that $\tc(X)$ is defined as the genus of the path fibration (\ref{fibration}) and according to Theorem 3 from \cite{Sch} one has $\tc(X)\le r$ if and only if the $r$-fold fiberwise join
\begin{eqnarray}\label{fibrationr}
\pi_r: P_rX\to X\times X
\end{eqnarray}
of the original fibration $\pi: PX\to X\times X$ admits a continuous section. Hence our assumption $\tc(X)=r+1$ implies that
$\pi_r$ has no continuous sections.
The fibre $F_r$ of (\ref{fibrationr}) is the $r$-fold join
\begin{eqnarray}\label{join}F_r \, = \, \Omega X\ast \Omega X\ast \dots\ast \Omega X\end{eqnarray}
where $\Omega X$ denotes the space of loops in $X$ starting and ending at the base point $x_0\in X$.
Note that $\Omega X$ is $(s-2)$-connected and therefore the fibre $F_r$ is $(2n-2)$-connected since\footnote{One knows that the join a $p$-connected complex and a $q$-connected complex is $(p+q+2)$-connected.}
$r(s-2)+2(r-1)=  2n-2.$

The primary obstruction to the existence of a section of (\ref{fibrationr}) is an element $\theta_r \in H^{2n}(X\times X; \pi_{2n-1}(F_r))$.
It is in fact the only obstruction since the higher obstructions land in zero groups. Thus we obtain that $\theta_r\not=0$.
By the Hurewicz theorem $$\pi_{2n-1}(F_r) = H_{2n-1}(F_r)= B\otimes B\otimes \dots \otimes B = B^r$$
where $B$ denotes the abelian group $H_{s-1}(\Omega X) = H_s(X)$.
Here we have used the K\"unneth theorem for joins, see for instance \cite{Sch}, chapter 1, \S 5.
By Theorem 1 from \cite{Sch} the obstruction $\theta_r$ equals the $r$-fold cup-product

$$\theta_r = \theta\cup \dots \cup \theta = \theta^r$$
where $\theta\in H^s(X\times X; B)$ is the primary obstruction to the existence of a section of $\pi: PX \to X\times X$.
Writing $\theta = v\times 1 + 1\times w$ and observing that $\theta|\Delta_X=0$ (since there is a continuous section of (\ref{fibration}) over the diagonal
$\Delta_X\subset X\times X$) shows that $v+w=0$ and therefore $\theta = v\times 1 - 1\times v=\bar v$. Hence we have found a cohomology class $v\in H^s(X;B)$ with
 $\bar v^r\not=0$.

(c) In one direction the statement of (c) follows from the upper bound (a) and \cite{Fa03}, Thm. 7, i.e. the existence of classes $v_1, \dots, v_r \in H^s(X;\kk)$ with
$\bar v_1\cup \dots\cup \bar v_r \not=0$ combined with (a) gives $\tc(X)=r+1$.
Suppose now that $H_\ast(X)$ is square free. Write $B=H_s(X)$ as a direct sum $$B=\oplus_{i\in I} B_i$$ where each $B_i$ is either $\Z$ or a cyclic group of prime order $\Z_p$ and $I$ is an index set. The $r$-fold tensor power $B^r=B\otimes \dots \otimes B$ is a direct sum
$$B^r= \bigoplus_{(i_1, \dots, i_r)\in I^r}B_{i_1}\otimes B_{i_2} \otimes \dots \otimes B_{i_r}$$
and each tensor product $B_{i_1}\otimes B_{i_2} \otimes \dots \otimes B_{i_r}$ is either $\Z$, $\Z_p$ or trivial. As we know from the proof of (b) there is a class $v\in H^s(X;B)$ such that $\bar v^r\not=0\in H^{2n}(X\times X; B^r)$. For any index $i\in I$ denote by $v_i$ the image of $v$ under the coefficient projection $B\to B_i$.
Since $\bar v^r\not=0$ there exists a sequence $(i_1, \dots, i_r)\in I^r$ such that the product
$$\bar v_{i_1} \cup \dots \cup \bar v_{i_r}\in H^{2n}(X\times X; B_{i_1}\otimes B_{i_2} \otimes \dots \otimes B_{i_r}).$$
is nonzero. If the product $B_{i_1}\otimes B_{i_2} \otimes \dots \otimes B_{i_r}$ is $\Z_p$ then each $B_{i_j}$ is either $\Z$ or $\Z_p$ and
taking $\kk=\Z_p$ and reducing all these classes $v_{i_k}$ mod $p$ we obtain that (c) is satisfied. In the case
when the product $B_{i_1}\otimes B_{i_2} \otimes \dots \otimes B_{i_r}$ is infinite cyclic each of the groups $B_{i_k}$ is $\Z$ and the class
\begin{eqnarray}\label{prod}
\bar v_{i_1}\cup \dots \cup \bar v_{i_r}\not= 0 \in H^{2n}(X\times X; \Z)
\end{eqnarray}
is integral and nonzero.

Since the group $H^{2n}(X\times X; \Z)$ is square-free the cup-product (\ref{prod}) is indivisible by some prime $p$.
Indeed, the group $H^{2n}(X\times X; \Z)$ is direct sum of cyclic groups of prime order and infinite cyclic groups and the product (\ref{prod}) has a nontrivial component in at least one of these groups. A nonzero element of $\Z$ is divisible by finitely many primes and a nonzero element of $\Z_p$ is divisible by all primes except $p$.

Therefore, as follows from the exact sequence
 $$\dots \to H^{2n}(X\times X;\Z) \stackrel{p}\to H^{2n}(X\times X;\Z) \to H^{2n}(X\times X;\Z_p)\to \dots, $$
 the mod $p$ reduction of the product (\ref{prod}) is nonzero.
Now, taking $\kk=\Z_p$ and reducing the classes $v_{i_k}$ mod $p$ gives a sequence of classes $w_{j_k}\in H^s(X;\kk)$ such that $\prod \bar w_{j_k}\not =0$ where
$k=1, \dots, r$.

(d) The proof of statement (d) of Theorem \ref{thm3} is similar to that of (c), with the simplification that all the groups $B_i$ are in this case infinite cyclic.
\end{proof}

\section{Proof of Theorem 1}\label{prf}

The cases $m=2$ and $m\geq 3$ odd of Theorem 1 were dealt with by Farber
and Yuzvinsky in \cite{FY04}. Their arguments also show that if $m\geq
4$ is
even, then $\TC(F(\R^m,n))$ equals
either $2n-1$ or $2n-2$. Hence to prove Theorem 1 it suffices to show
that $\TC(F(\R^m,n))\neq 2n-1$ when $m\geq 4$ is even.

Fix $n\geq 2$. For any $m\geq 2$ the space $F(\R^m,n)$ is $(m-2)$-connected, since it is the complement
of an arrangement of codimension $m$ subspaces of $\R^{mn}$. Its
integral cohomology ring is shown in \cite{FH} to be graded-commutative
algebra over $\Z$ on generators $$e_{ij}\in H^{m-1}(F(\R^m,n)), \quad 1\leq i<j\leq
n,$$ subject to the relations $$e_{ij}^2=0, \quad e_{ij}e_{ik}=(e_{ij}-e_{ik})e_{jk}$$ for
any triple $1\leq i<j<k\leq n$. In particular, $H^*(F(\R^m,n))$ is nonzero only in dimensions $i(m-1)$ where $i=0, 1, \dots, (n-1)$.
Applying the result of Eilenberg and Ganea \cite{EG} we obtain that for $m\ge 3$ the space $F(\R^m, n)$ is homotopy equivalent to a finite complex of dimension $\le (m-1)(n-1)$. Now we may apply statement (d) of Theorem \ref{thm3}, which gives, firstly, that $\tc(F(\R^m, n))\le 2n-1$ and, secondly,
$\tc(F(\R^m, n))= 2n-1$ if and only if there exist cohomology classes $$v_1, \dots,v_{2(n-1)}\in H^{m-1}(F(\R^m, n))$$ such that the product of the corresponding zero-divisors $$\bar v_1\cup \bar v_2\cup \dots \cup \bar v_{2(n-1)}$$  is nonzero; recall that the notation $\bar v$ is introduced in (\ref{bar}).
We show below that such classes $v_1, \dots, v_{2(n-1)}$ do not exists if $m\ge 4$ is even.

We recall the result of \cite{FY04} stating that $\tc(F(\C, n))=2n-2$. It is shown in the proof of Theorem 6 in \cite{FY04},
that
$F(\C, n)$ is homotopy equivalent to the product
$X\times S^1$ where $X$ is a finite polyhedron of dimension $\le n-2$. This argument uses the algebraic structure of
$\C=\R^2$ and does not generalize to $F(\R^m,n)$ with $m>2$.
Using the product inequality (Theorem 11 in \cite{Fa03}) one obtains
\begin{eqnarray*}\tc(F(\C, n)) &\le& \tc(X) +\tc(S^1)-1\\
&\le& (2(n-2)+1)+2-1=2n-2.\end{eqnarray*}
Hence there exist no $2(n-1)$ cohomology classes $v_1, \dots, v_{2(n-1)}\in H^1(F(\C, n))$ such that the product of the zero-divisors
$\bar v_1 \cup \dots \cup \bar v_{2(n-1)}$ is nonzero,
as this would contradict Theorem 7 from \cite{Fa03}.

Now we observe that for any even $m\geq 2$ there is an algebra isomorphism
\begin{eqnarray}
\phi: H^*(F(\C;n)) \to H^{*(m-1)}(F(\R^m,n))
\end{eqnarray}
mapping classes of degree $i$ to classes of degree $(m-1)i$ where $i=0, 1, \dots, n-1$, see \cite{FH}. Thus we conclude that there exist no cohomology classes
$w_1, \dots, w_{2(n-1)}\in H^{m-1}(F(\R^m, n))$ such that the product of the corresponding zero-divisors
$\bar w_1 \cup \dots \cup \bar w_{2(n-1)}$ is nonzero. Theorem \ref{thm3} (statement (d)) gives now that $\tc(F(\R^m,n))\le 2(n-1)$.

On the other hand, it is proven in \cite{FY04} that one may find $2n-3$ cohomology classes $v_1, \dots, v_{2n-3}\in H^1(F(\C,n))$ such that the cup-product
$\bar v_1 \cup \dots\cup \bar v_{2n-3}$ is nonzero. Hence, repeating the above argument we see that for $m$ even there exists classes
$w_1, \dots, w_{2n-3}\in H^{m-1}(F(\R^m,n))$ (where $w_i=\phi(v_i)$) with nonzero product $\bar w_1 \cup \dots \cup \bar w_{2n-3}$; this gives the opposite inequality
$\tc(F(\R^m, n)) \ge 2n-2$.

Hence, $\tc(F(\R^m, n))=2n-2$ as stated.
\qed

config: arxiv