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\def\at{adiabatic theorem}
\def\qm{Quantum Mechanics}
\def\e{\varepsilon}
\def\real{{\rm I\kern-.2em R}}
\def\complex{\kern.1em{\raise.47ex\hbox{ $\scriptscriptstyle
|$}}\kern-.40em{\rm C}}
\newtheorem{adiabatic}{Theorem}[section]
\newtheorem{pdotp}{Corollary}[section]
\newtheorem{friedrich}{Theorem}[section]
\newtheorem{qed}{Theorem}[section]
\title{An Adiabatic Theorem with Soft Photons}
\author{ J.~E.~Avron and A. Elgart
\\ Department of Physics, Technion, 32000 Haifa, Israel}
\begin{document}
\flushbottom
\maketitle {\begin{abstract} The \at\ of Quantum Mechanics disregards the soft
photons of QED. Why, then, is it so successful? To understand the role of soft
photons we prove an adiabatic theorem for the ground state of a simple model
Hamiltonians that allows soft photons. We show that for weak electron-photon
coupling, the adiabatic time scale is close to the
time scale in a theory without the quantized radiation field
up to a Lamb shift. The inclusion of
photons also leads to a logarithmic correction in three dimensions coming from
an infrared singularity characteristic of QED.
\end{abstract} }
%\newpage
\section{Introduction} The \at\ of Quantum Mechanics disregards the soft
photons of QED. Its success suggests that doing so must be, at least, a very
good approximation. In fact, there is evidence that this must be more than
just a very good approximation: In the theory of the integral quantum Hall
effect \cite{nt,as} the \qm\ \at\ is used to explain the precise quantization
of the Hall conductance, a quantization that, experimentally, is accurate to
a high order of
$\alpha$, the fine structure constant.
>From a mathematical and structural point of view QED does not seem all that
close to \qm . In fact, there is good a-priori reason to even doubt the
existence of a useful \at\ in QED. This is because, unlike \qm\, QED does
not have a large gap in its spectrum (because of far away red photons) to
protect the ground state. In the absence of such a gap, it is not clear that
there is a candidate for an
\at . Even if there was a gap, it is most probably much smaller than the
\qm\ gap. Since the size of the gap enters in adiabatic estimates the success
of the \qm\
\at\ is a mystery.
Let us elaborate on the significance of gaps in the common formulation of \at
s.
Adiabatic theorems need a time scale to fix how slow is really slow. In the
simplest of the \at s of classical mechanics, say for the Harmonic oscillator
\cite{arnold}, this time scale is set by the natural frequency of the (time
independent) oscillator. A perturbation is adiabatic if the oscillator
undergoes many oscillations during the time that the perturbation varies
appreciably. In quantum mechanics the analog time scale is set by the gaps in
the (unperturbed) spectrum. In the case of the Harmonic oscillator classical
and quantum mechanics actually have the same time scale:
$1/\omega$, classically, and
$\hbar/\hbar\omega$ quantum mechanically\footnote{\samepage {This relation
between
classical frequency and the distance between eigenvalues holds, in fact,
for any integrable system as one can see from the following argument
that we learned from S. Graffi: Semi-classically, the distance between
eigenvalues
$E_{j+\ell}-E_j\approx \hbar\omega\ell$ where
the frequency
is the classical frequency
$\omega=\frac{\partial H}{\partial I}$, where $H$ is the classical
Hamiltonian and
$I$ is the action
\cite{born}}}. The role of gaps
in the
\at\ has a natural and appealing physical interpretation: An adiabatic
perturbation acts on the quantum system like {\em external} soft photons. A
gap in the spectrum provides a protection against low energy excitations. This
intuition is indeed supported by the general
\at s of quantum mechanics
\cite{bf,kato,asy}.
In the absence of a gap, several things can happen to the adiabatic theorem.
The first is that there is no reasonable formulation of a useful \at. This is
the case for example, for a free particle on the line perturbed by, say, a slow
gauge field. Another thing that can happen is that the formulation of an
adiabatic theorem can be elusive. An example for that is the \at\ for a
quantum system with dense pure point (and singular continuous) spectrum
studied in
\cite{ahs}. But, in practice, the most interesting thing that can happen, is
that there is an adiabatic theorem, but the error is different than the error
when there is a gap. An example to this is the
\at\ for crossing eigenvalues studied first by Born and Fock. In the absence
of crossing they showed that the error in the \at\ was
$O\left(\frac{1}{\tau}\right)$ where
$\tau$ is the adiabatic time scale set by the gap. In the case of linear
crossing, they showed that the error is $O\left(\sqrt{\frac{1}{\tau}}\right)$.
There is an analog to this phenomenon in the classical
\at , where a rather radical change in a scenario manifests itself in error
estimate: Integrable and chaotic Hamiltonians lead to different errors in \at
\cite{br,ott}.
Here we study a class of examples of \at s for gapless Hamiltonians,
inspired and motivated by low energy QED. In these models the ground state is
at the threshold of the (absolutely) continuous spectrum. One such model is
the Friedrichs model
\cite{friedrichs}. The second, and more interesting model is a model
patterned after the Dicke model
\cite{d}. More precisely, it is the spin-boson model in the rotating wave
approximation
\cite{hs}. The model retains some of the infrared difficulties of low energy
QED but is sufficiently simple so that various things can be explicitly
calculated. Our aim is to show that the
\at\ for the model essentially reproduce the \qm\footnote{without
intrinsic photons} results up to a small corrections that renormalizes the
effective gap by a power of $\alpha$--the order of
(the analog of) the Lamb shift in the model. This explains why the \qm\ \at\
is right after-all.
We conclude the introduction with a classical argument, which we owe to Amos
Ori, that suggests why radiation should not harm the \qm\ \at . The errors in
the \at\ are of two sources: As the Hamiltonian evolves in time, the ground
state may tunnel to the excited electronic state and it may radiate soft
photons. Tunneling is essentially \qm\ without photons. This process
contributes an error of order
$1/\tau$, as we know from Born and Fock. To estimate
photon emission, consider a slowly rotating dipole with frequency of rotation
$1/\tau$. Since dipole radiation is proportional to the square of
acceleration, the power radiated by the dipole is, classically,
$O\left(\frac{e^2}{c\tau^4}\right)$, the total energy radiated in time $\tau$
is
$O\left(\frac{e^2}{c\tau^3}\right)$ and the number of radiated photons
$O\left(\frac{\alpha}{\tau^2}\right)$. Classically,
radiation is always sub-dominant to quantum tunneling in the adiabatic limit.
This argument correctly predicts that there should be an \at . However, it
does not predict the correct error in the quantum model. As we shall see an
infrared divergence of QED lurks in the background and makes the error
$\frac{\log
\tau}{\tau}$ and not
$\frac{1}{\tau}$ as this the classical argument suggests. Second, and perhaps
more important, this argument misses an important role played by $\alpha$. In
this classical argument the number of emitted photons is sub-dominant to
quantum tunneling in the adiabatic limit for any value of $\alpha$. The
quantum result, in contrast, turns out to say that only when
$\alpha$ is small, when the Lamb shift is small on the scale of the
distance between eigenvalues, is there agreement between the
\at\ with and without photons.
\section{The Adiabatic Theorem and A Commutator Equation}
In this section we explain what we mean by ``\at'', and give a
condition for an
\at\ to hold. This condition is that the commutator
equation, Eq.~(\ref{commutators}) below, has solutions $X$,
$Y$ which are bounded operators\footnote{for $X$ we also need that its
derivative is bounded}. We also introduce notation, terminology, and collect
known facts that we need. To simplify the presentation, we shall stay away
from making optimal assertions.
We consider Hamiltonians that are bounded from below, and choose the origin
of the energy axis so that the spectrum begins at zero. Let
$H(s)\ge 0$ be a family of such self-adjoint Hamiltonians. The unitary
evolution generated by the Hamiltonian,
$U_\tau(s)$, is the solution of the initial value problem:
\begin{equation} i\,\dot U_\tau (s) = \tau H(s) U_\tau(s),\quad U_\tau(0)=1,
\quad s\in [0,1].\label{schrodinger}
\end{equation}
$\tau$ is the adiabatic time scale, and we are concerned with the limit of
large $\tau$. The physical time is
$t=\tau s\in[0,\tau]$. Since
$\tau$ is large $H(s)=H(t/\tau)$ varies adiabatically. We assume that all
operators are defined on some fixed dense
domain in the Hilbert space.
The (instantaneous)
ground state is in the range of the kernel of
$H(s)$ and we assume that the kernel is smooth and one-dimensional.
Let
$P(s)\neq 0$ be the projection on the kernel of
$H(s)$, i.e. $H(s)\, P(s)=0$, $dim\,P= Tr\,P=1$. By smoothness we mean that
$\dot P(s)$ a bounded operator.
The \at s we consider are are concerned with the large time
behavior of the evolution of the ground state where $t=O(\tau)$ or,
equivalently,
$s=O(1)$. The smoothness of the kernel implies that there is a natural
candidate for an
\at\ for the ground state, which is independent of whether $H(s)$ does or
does not have a gap in it spectrum. Namely, that if
$\psi(0)\in Range \, P(0)$ at time $s=0$, then it evolves in time so that,
$\psi_\tau(s)=U_\tau (s)\
\psi(0)$ lies in $Range\, P(s)$ at time $s$ in the adiabatic limit,
$\tau\to\infty$.
To formulate the \at\ with error estimates we need to get hold of {\em
adiabatic phases}
\cite{berry}. To do that we introduce the adiabatic evolution of Kato
\cite{kato}: Let $U_A(s)$ be the solution of the evolution equation
\begin{equation} \dot U_A(s) = [\dot P(s),P(s)]\, U_A(s),\quad U_A(0)=1,
\quad s\in [0,1].\label{kato}
\end{equation} It is known that
\begin{equation} U_A(s)\, P(0) =P(s)\, U_A(s).
\end{equation} That is $U_A(s)$ maps $Range\ P(0)$ onto $Range\ P(s)$. We can
now formulate the basic \at\ :
\begin{adiabatic} Let $H(s)P(s)=0$ for all $0\le s\le 1$, with $P$
differentiable projection on the ground state, with
$\Vert
\dot P(s)\Vert\le D$. Suppose that the commutator equation
\begin{equation} [\dot P(s), P(s)] = [H(s),X(s)] +Y(s),\label{commutators}
\end{equation} has operator valued solutions, $X(s)$ and $Y(s)$ so that for
$\varepsilon\searrow 0$
\begin{equation}
\Vert X(s)\Vert\ + \Vert \dot X(s)\Vert\ \le\ C\left\{ \begin{array}{c}
\,\varepsilon^{-\nu}\\ \vert\log \varepsilon\vert
\end{array} \right. ,\quad \Vert Y(s)\Vert
\le \hat C\,\varepsilon^{\mu},
\end{equation} with $\mu,\nu\ge 0$. Then
\begin{equation}
\Vert (U_\tau (s)-U_A(s))P(0)\Vert \le \tilde C \left\{
\begin{array}{c}\tau^{-\frac{\mu}{\nu+\mu}}
\\ \frac{\log \tau}{\tau}\end{array}\right. ,\quad s\in[0,1].
\end{equation}
\end{adiabatic} Remarks: 1. In the case that there is a gap in the spectrum,
one can always find $X(s)$ bounded so
$\nu=0$, and $Y=0$, see \cite{asy}. $X$, and therefor also $\tilde C$, is of
the order of (gap)$^{-1}$. This gives error of
$1/\tau$, and generalizes the \at\ of Born and Fock and Kato for discrete
spectra, to more complicated spectra provided there is a gap.
2. The theorem says that the physical evolution clings to the instantaneous
spectral subspace. In particular, if $P$ is one dimensional, it says that the
physical evolution of the ground state remains close to the instantaneous
ground state.
3. Here, and throughout, we are concerned only with the \at\ to lowest order.
If $s$ is chosen outside the support of $\dot P$ then much stronger results
can be obtained. See e.g. \cite{ks}.
4. The adiabatic time scale $\tau_0$ set by this theorem is $\tau_0= O(
(2+D)C)$.
Proof: Let $W(s) = U_A^\dagger(s) U_\tau(s))$, with $W(0)=1$.
From the equation of motion, and the commutator equation,
Eq.~(\ref{commutators}),
\begin{eqnarray}
P(0)\,\dot W(s)\,&=& -P(0) U^\dagger_A(s)\Big(i\,\tau\,H(s)+[\dot
P(s),P(s)]\Big) U_\tau(s)\nonumber
\\ &=&-U^\dagger_A(s)\,P(s)\Big(i\,\tau\,H(s)+[\dot P(s),P(s)]\Big)U_\tau(s)
\nonumber
\\ &=&\,-U^\dagger_A(s)\,P(s)\,[\dot P(s),P(s)]\, U_\tau(s)
\nonumber
\\ &=&\,-U^\dagger_A(s)\,P(s)\,\Big([H(s),X(s)]+Y(s)\Big)\,U_\tau(s)
\nonumber
\\ &=&\,-U^\dagger_A(s)\,P(s)\,\Big(-X(s)\,H(s)+Y(s)\Big)\,U_\tau(s)
\nonumber
\\ &=&\frac{i}{\tau}\,P(0)\, U^\dagger_A(s)\,X(s)\,
\dot U_\tau(s)-P(0)\,U^\dagger_A(s)\,Y(s)\, U_\tau(s).
\end{eqnarray} To get rid of derivatives of $U_\tau$, which are large by the
equation of motion, we rewrite the first term on the rhs (up to the $P(0)$ on
the right) as :
\begin{eqnarray}
U^\dagger_A(s)\,X(s)\,\dot U_\tau(s) &=&\dot{\left( U^\dagger_A(s)\,X(s)
U_\tau(s)\right)}- U^\dagger_A(s)\,\dot X(s)\, U_\tau(s)-
\dot U^\dagger_A(s)\,X(s)\,U_\tau(s) \\&=&
\dot{\left( U^\dagger_A(s)\,X(s)\, U_\tau(s)\right)} - U^\dagger_A(s)\,\dot
X(s)\, U_\tau(s)+ U^\dagger_A(s)\,[\dot P(s),P(s)]\,X(s)\, U_\tau(s).\nonumber
\end{eqnarray} From this it follows, by integrating, that for $s\in[0,1]$
\begin{eqnarray}
\Vert (U_\tau(s)-U_A(s))\,P(0)\Vert &=& \Vert
P(0)(U^\dagger_\tau(s)-U^\dagger_A(s))\,\Vert\nonumber \\
\Vert P(0)(1-W(s))\,\Vert&\le& \hat
C\e^\mu +\frac{(2+D)\, C}{\tau}\left\{
\begin{array}{c}{\e^{-\nu}}
\\ \vert\log\e\vert.\end{array}\right.
\end{eqnarray} Choosing
$\e= \tau^{-\frac{1}{\mu+\nu}}$ gives
\begin{equation}
\Vert (U_\tau(s)-U_A(s))\,P(0)\Vert \le \tilde C \left\{
\begin{array}{c}\tau^{-\frac{\mu}{\nu+\mu}}
\\ \frac{\log \tau}{\tau}\end{array}\right. .
\end{equation}
This concludes the proof of the theorem.\hfill$\Box$
It is convenient to rewrite this solvability condition in a way that one needs
to solve for a fixed $X$ and $Y$ rather than functions $X(s)$ and $Y(s)$. This
is accomplished by
\begin{pdotp} Let $P(s)$ be the family
\begin{equation} P(s)=V(s)\, P\,V^\dagger(s),\quad V(s)=\exp (i\,s\,\sigma).
\end{equation} It is enough to solve for the commutator equation
\begin{equation}
i\, K=[H, X] +Y, \quad K= \{\sigma,P\} -2 P\sigma P,\label{commute}
\end{equation} for fixed $X$ and $Y$
so that for $\varepsilon\searrow 0$
\begin{equation}
\Vert X\Vert\ \le\ C\left\{ \begin{array}{c}
\,\varepsilon^{-\nu}\\ \vert\log \varepsilon\vert
\end{array} \right. ,\quad \Vert Y\Vert
\le \hat C\,\varepsilon^{\mu},
\end{equation} with $\mu,\nu\ge 0$, and $\Vert \dot P(s)\Vert\le D$.
$X(s)$ and $Y(s)$ are then determined by the obvious unitary conjugation.
\end{pdotp} Proof:
Since
$P(s)=V(s)\, P\,V^\dagger(s)$, we have
\begin{equation}
\dot P(s)=i\,V(s)\, [\sigma,P]\,V^\dagger(s),
\end{equation} and
\begin{eqnarray} [\dot P(s),P(s)]&=&i\,V(s)\,
\Big[[\sigma,P],P\Big]\,V^\dagger(s)\nonumber \\ &=&i\,V(s)\,
\Big(\{\sigma,P\}-2P\,\sigma\,P\Big)\,V^\dagger(s).
\end{eqnarray}
\hfill$\Box$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ An Adiabatic Theorem for Ground State at Threshold} As a warmup, let
us prove an \at\ in a situation where the ground state lies at the threshold
of the ac spectrum\footnote{ac=absolutely continuous} of a one particle
Schr\"odinger operator. The model we consider is a simple version of the
Friedrichs model, and we do not make any pretense that it describes an
interesting physical situation. Rather, the simplicity of its mathematical
structure and the fact that it displays the basic elements relevant to \at\
without gaps is its motivation.
There is an inherent difficulty in the situation of a bound state at threshold
in general, and in the Friedrichs model \cite{friedrichs,friedrichs1,howland}
in particular,
namely, that a bound state at threshold is not a stable situation. Under a
small deformation of the Hamiltonian, the ground state will, generically,
split away from the ac spectrum and a gap develops. Since our aim is to study
\at\ in the absence of a gap we force the ground state to stay at threshold
of $H(s)$ for all $s$. A rather trivial way of implementing this is to
consider
$H(s)=V(s)\,H\,V^\dagger(s)$ where $H$ has a bound state at threshold and
$V(s)$ is a smooth family of unitaries.
\subsection{The Friedrichs Model}
We shall consider a family of Hamiltonians, closely related to the
standard Friedrichs model
\cite{friedrichs}, parameterized by the
scaled time $s$, a real number $d>0$ that plays the role of dimension, and a
function $f$ that describes the deformation of the family. Since we are only
interested in the low energy behavior of the family we shall introduce an
``ultraviolet cutoff'' to avoid inessential.
The Hilbert space of the Friedrichs model (with an ultraviolet cutoff) is
${\cal H}=
\complex\oplus L^2([0,1],k^{d-1}\,dk)$. A vector $\psi\in{\cal H}$ is
normalized by
\begin{equation}
\psi=\left(\begin{array}{c}
\beta\\ f(k)
\end{array}\right)\, \quad \Vert \psi\Vert^2= |\beta\vert^2+\int_0^1 \vert
f(k)\vert^2 k^{d-1}\,dk,\quad
\beta\in\complex.
\end{equation} We choose a special, and trivial, case of a diagonal Hamiltonian
whose action on a vector $\psi$ is as follows:
\begin{equation} H\,\psi=\left(\begin{array}{ll} 0&0\\ 0&k
\end{array}\right)\,\left(\begin{array}{c}
\beta\\ f(k)
\end{array}\right)=\left(\begin{array}{c} 0\\ k\,f(k)
\end{array}\right).
\end{equation}
$H$ has a ground state at zero energy with projection
\begin{equation} P=\left(\begin{array}{ll} 1&0\\ 0&0
\end{array}\right).
\end{equation} The rest of the spectrum is the unit interval $[0,1]$, and is
absolutely continuous with formal (delta normalized) eigenvectors
\begin{equation}
\psi_E=k^{\frac{d-1}{2}}\,\,\left(\begin{array}{c} 0\\
\delta (k-E)
\end{array}\right),\quad 0< E< 1.
\end{equation}
The density of states in this model is proportional to $E^{d-1}$.
We construct the family $H(s)$ by conjugating $H$ with a family of unitaries:
\begin{equation} V_f(s)=\exp is\sigma(f), \quad
\sigma(f)=\left(\begin{array}{ll} 0&\langle f\vert \\
\vert f \rangle&0
\end{array}\right),
\end{equation} where $f$ is a vector in $L^2([0,1],k^{d-1}\, dk)$.
\begin{friedrich} Let $H(s;d,f)$ be the family of Friedrichs models with a
ground state at threshold for all s
\begin{equation} H(s;d,f)= V_f(s)\, H\,V^\dagger_f(s).
\end{equation}
Suppose that
\begin{equation} g(k)=i\, k^{-1}\, f(k)\in L^2([0,1],k^{d-1} dk), \quad
V_f(s)=\exp\, i\,s\,\sigma(f)
\end{equation} then the quantum evolution of the ground state of $H(s;d,f)$ is
adiabatic and its deviation from the instantaneous ground state is, at most,
$O(1/\tau)$.
\end{friedrich} Remarks: 1. Note that if the conditions in the theorem hold in
dimension $d_0$, then they hold in all dimensions $d\ge d_0$. The physical
interpretation of that is that the density of states at low energies decreases
with $d$. So, even though there is spectrum near zero, there is only very
little of it.
2. If $g$ is not in $L^2$ there may still be an \at\ with slower falloff in
$\tau$ by accommodating $Y\neq 0$. An example will be discussed in the next
section.
3. The Friedrichs model is vanilla: $H$ has no interesting energy scale to
fix the adiabatic time scale. The scale is set by the perturbation alone:
$\tau_0= O((1+\Vert f\Vert^2)\,\Vert g\Vert )$. This is
quite unlike the case in the usual \at .
Proof: In this case $K$ of Corollary 2.1 is $K=\sigma(f)$. With $g\in L^2$,
$\sigma(g)$ is a bounded (in fact, finite rank) operator and an easy
calculation gives
\begin{equation} [H, \sigma(g)]=\left(\begin{array}{lr}0&\langle -kg|\\
|kg\rangle&0\end{array}\right)=i
\sigma(f).
\end{equation} Hence \begin{equation} X= \sigma(g) ,\quad Y=0
\end{equation} solve the commutator equation, Eq.~(\ref{commute}), with a
bounded $X(s)$ and $Y(s)=0$.\hfill$\Box$
If $g$ is not in $L^2$, $\sigma(s)$ and hence $X(s)$ are not bounded
operators. Even then one can often get an \at\ but with a slower decay rate.
%%%%%%%%%%%%
\section{A Two Level System in a Photon Field}
In this section we describe an \at\ for a model patterned after the Dicke
model \cite{d}. The Schr\"odinger operator is replaced by a two level
system and the photons are considered as a field theory in Fock space. In
the standard Dicke model one considers a single mode for the photon field.
The model we consider allow all modes. The model has a conservation law
that makes it amenable to explicit analysis. We also makes
other sacrifices which are less important, such as the fact that real photons,
in three dimensions, come with two polarization. The helicity of the photon
does not play an interesting role in the questions we want to study, but
clutters the notation. We disregard such realistic aspects. For the Dicke
model, we prove an adiabatic theorem that says that the evolution of the
ground state adheres to the instantaneous ground state and the time scale, at
least in three dimensions, is essentially the time scale fixed by \qm\ without
photons, but with a logarithmic correction.
\subsection{The Dicke Model}
The Dicke model is the Spin-Boson Hamiltonian in the
rotating wave approximation. The rotating
wave approximation, can indeed be motivated in the single-mode Dicke
model. In the multi-mode case we consider the rotating wave approximation
is not really motivated by an argument. It is a name that describes which
terms in the Hamiltonian are kept and which are not. For a mathematical
discussion of the spin-boson Hamiltonian, and the rotating wave model see
\cite{hs,dg,ms}.
The model describes a two level system coupled to a massless boson field in $d$
dimensions. For the kind of applications we are interested in, i.e. low
energy QED, it is natural to use atomic units where
$e=\hbar=1$ and the energy scale is Rydberg. In these units
$\alpha=1/ c$, where
$c$ is the velocity of light, is the small parameter of the theory. The
Hamiltonian is:
\begin{equation} H(m,d,f,\alpha)=m\,(1 -P)\otimes {\mathbf 1} + \alpha^{-1}\,
{\mathbf 1}\otimes E +
\sqrt{\alpha}\,\sigma_+\otimes a^{\dagger}(f) +
\sqrt{\alpha}\,\sigma_-\otimes a(f),\label{eq1}
\end{equation}acting on the Hilbert space $\complex^2\otimes {\cal F}$ with
${\cal F}$ being the symmetric
Fock space over $L^2(R^d,d^dk)$. Here
\begin{equation} P=\left(\begin{array}{ll} 1&0\\ 0&0\end{array}\right),\quad
\sigma_+=\left(\begin{array}{ll} 0&1\\ 0&0\end{array}\right),
\quad
\sigma_-=\left(\begin{array}{ll} 0&0\\ 1&0\end{array}\right)\quad E=\int{|k|\,
a^{\dagger}(k)a(k)d^dk}.
\end{equation}
$m>0$ is the gap in the quantum Hamiltonian (without photons). $a(f)$ and
$a^\dagger(f)$ are the usual creation and annihilation operators on ${\cal F}$
obeying the canonical commutation relations
\begin{equation} [a(f),a^{\dagger}(g)]=\langle f\vert g\rangle.
\end{equation} We denote by $|0\rangle$ the field vacuum and by $\Omega$
the projection on the vacuum.
It may be worthwhile to explain where the various powers of $\alpha$ in
$H$ come from. For the radiation field the $\alpha^{-1}$ comes from
$\hbar\omega =\hbar c |k|$ which explains why the field energy comes with a
large coupling constant. The $\sqrt{\alpha}$ has one
inverse power of $c$ from minimal coupling,
$\frac{e}{2mc} (p\cdot A+A\cdot p)$. Half a power comes from the standard
formula for the vector potential
\begin{equation} {A}({x}):=
\int d^3k\,
\sqrt{\frac{2\pi c}{|k|}}\,\Big(e^{-i{k}\cdot {x}}\,a^{\dagger}({k})
+e^{i{k}\cdot{x}} a({k})\Big).
\end{equation} Compare e.g. \cite{dicke}.
A useful formula we shall need is
\begin{equation} E\, a^\dagger(g) \Omega =a^\dagger(|k|g)\Omega.
\end{equation}
The model has the pleasant feature, not shared by the general spin-boson
model, that photon emission is associated with the electron cascading down,
and photon absorption with the electron excitation.
It is instructive to trace the origin of $f$ in the model so that one can get
a feeling for what are reasonable assumptions to impose on it. The
electron-photon interaction is $\alpha\,(p\cdot A + A\cdot
p)$ with
$p=-i\,\nabla$. If we denote by $\psi_1(x)$ and $\psi_2(x)$ the atomic wave
functions of the electronic two level system, then the matrix element of the
vector potential between the two states gives
$f$ as
\begin{equation} f(k)=-i\,\sqrt{\frac{2\pi}{|k|}}
\int\,\left(\psi_1^*(x)\,e^{-i{k}\cdot {x}}(\nabla\psi_2)(x)\,
-(\nabla\psi_1)^*(x)\,e^{-i{k}\cdot {x}}\,\psi_2(x)\right)\,d^dx.
\end{equation} Clearly, with reasonable atomic eigenfunctions,
$f(k)$ has fast decay at infinity and the model is ultraviolet regular. In the
infrared limit this behaves like
\begin{equation}
f(k)\to-i\,\sqrt{\frac{2\pi}{|k|}} \int\,\Big(\psi_1^*(x)\,(\nabla\psi_2)(x)\,
-(\nabla\psi_1)^*(x)\,\psi_2(x)\Big)\,d^dx.
\end{equation} In particular we see that for small $k$
\begin{equation} f(k)=K\,\sqrt{\frac{1}{ |k|}}.\label{K}
\end{equation} The square root singularity is a characteristic infrared
divergence of QED, and it has consequences for the
\at\ as we shall see. Note that with $f$ having a square root singularity the
model makes sense (as an operator) provide
$d>1$, for otherwise $a^\dagger(f)$ is ill defined since $f$ is not in $L^2$.
An important parameter in the model is
\begin{equation} {\cal E}= \left\langle f \left\vert \frac{1}{|k|}\right\vert
f\right\rangle.
\end{equation} Bearing
in mind the square root singularity of
$f$
we see that
\begin{equation} {\cal E}\sim \int \frac{d^dk}{|k|^2},
\end{equation} is finite for all $d>2$.
%%%%%%%%%%%%%%%%
\subsection {Spectral Properties} What makes the Dicke model simple is that it
has a constant of motion
\cite{hs}. If we let
$N=\int a^\dagger(k) \, a(k) d^dk$ be the photon number operator, then one
checks that ${\cal N}$ commutes with $H$ where
\begin{equation} {\cal N}=\left(\begin{array}{ll} N&0 \\
0&N+1\end{array}\right)={\mathbf 1}\otimes N + P\otimes {\mathbf 1}.
\end{equation} The spectrum of ${\cal N}$ is the non-negative integers. The
spectral properties of $H(m,d,f,\alpha)$ can be studied by restricting to
subspaces
of ${\cal N}$.
\paragraph{${\cal N}=0$}: The kernel of
${\cal N}$ is one dimensional and is associate with the projection
\begin{equation} P=\left(\begin{array}{ll}
\Omega&0 \\ 0&0\end{array}\right).
\end{equation}
$\Omega$ is the projection on the field vacuum. It is easy to see that
$P\,H(\nu,d,f)P=0$, so the model always has a state at zero energy. This state
may or may not be the ground state. It is the ground state if
$\alpha^2 {\cal E} >\alpha^2 $ in atomic units, (about
$10^{-3}$ eV), the inequality holds. In two dimensions the left hand side is
log divergent, and the spectrum in the
${\cal N}=1$ sector has a bound state at negative energy. This state lies
below the bound state of the
${\cal N}=0$ sector. We do not consider this situation and henceforth stick
to
$d\ge 3$.
\paragraph{${\cal N}\ge 2$}: It is known \cite{hs} that the bottom of the
spectrum in all these sectors is at zero if (\ref{standard}) holds.
\subsection{Resonance and Lamb Shift}
The ${\cal N}=1$ sector has a resonance that serves to define
the Lamb shift in the model and plays a role in the \at . The resonance is a
solutions of the analytically extended eigenvalue equation, given below, which
for
$\alpha\to 0$ converge to the eigenvalue $m$ and lies in the lower half plane,
see
\cite{friedrichs,howland,ms}. The real part of the shift is, by definition, the
Lamb shift of the model, while the imaginary shift is the life time. It
turns out
that, for
$d\ge 3$, the Lamb shift is dominant and the life time is a higher order in
$\alpha$. For the application to the adiabatic theorem we need only the dominant
contribution, i.e. only the Lamb shift. Computing the Lamb shift is easy.
Computing the life time is harder. For the sake of
completeness we compute both, even though we only need one.
The eigenvalue equation is
\begin{equation}
E-m=\alpha^2\, G(\alpha E)\label{res}
\end{equation}
where $G(e)$ is defined as the analytic continuation from the
upper half plane of
\begin{equation}
\ G(e)=\int_{\real^d}\,\frac
{|f|^2}{e-{|k|}}
\, d^d k,\quad \Im\, e\ge 0.
\end{equation}
By taking the imaginary part, it is easy to see that Eq~(\ref{res}) has no
solution in the upper half plane as it should (since the Hamiltonian is
self-adjoint). To solve the equation in the lower half plane one needs an
explicit expression, at least for small
$\alpha$, and
$e$ near $\alpha m$, of this analytic continuation.
Now, suppose we found this analytic continuation, then, we can solve
Eq.~(\ref{res}) by iteration, and to lowest order we have
\begin{equation}
E_r\approx m+\alpha^2\, G(\alpha m).\label{iterate}
\end{equation}
Clearly $G(\alpha m)\to - {\cal E}$, in the limit $\alpha\to 0$, so to leading
order
\begin{equation}
E_r\approx m-\alpha^2\, {\cal E}.
\end{equation}
To this order, one does not see the imaginary part of the resonance energy.
The real part of the shift in the energy, $\alpha^2\,{\cal E}$ is, by
convention,\cite{dicke}, the Lamb shift of the model.
As we shall see $m-\alpha^2 {\cal E}$ will play the role of the gap ($m$ in
the theory without photons) in the adiabatic theorem of the model. It may be
worthwhile to point out that the Lamb shift for the Hydrogen atom,
\cite{bethe}, is
actually of {\em higher} order, namely,
$\alpha^3
\log(\alpha^{-1})$. Since the Lamb shift of Hydrogen also involves an
ultraviolet
regularization, while the present model is ultraviolet regular, it is not
surprising that the order of the two is different. What is surprising is that
the order of Hydrogen is higher rather than lower.
Estimating the life time is, as we noted, considerably harder, and, because of
this, irrelevant to the adiabatic theorem. So a reader will loose little by
skipping the rest of this section. However, for the benefit of the reader who
is interested in how the computation of the life time goes, it is given below.
We shall show below that for $d\ge 3$, and $|e-\alpha m|<\alpha m$, the
analytic continuation of
$G(e)$ to the lower half plane, and to the next relevant order in $\alpha$, is
given by
\begin{equation}
G(e)=-{\cal E}-i\pi \,K\Omega^d e^{d-2}\,,\quad \Im\, e\le 0,\label{analytic}
\end{equation}
where $K$ is as in Eq.~(\ref{K}), and $\Omega^d$ is the surface area of the
unit ball in d dimensions.
>From Eq.~(\ref{iterate}), and taking into account Eq.~(\ref{K}),
we get for the Lamb shift and the life-time:
\begin{eqnarray} E_r&\approx& m
-\alpha^2 {\cal E} -i
\alpha^2 \pi \,K\,\Omega^d (m\alpha)^{d-2} \nonumber \\ &=& m -\alpha^2 {\cal
E} -i
\alpha^2 \pi \Omega^d (m\alpha)^{d-1} \left\vert f(\alpha m)\right\vert^2.
\end{eqnarray}
The life time is higher order in $\alpha$ than the Lamb shift, and is of
order
$\alpha^d$. For
$d=3$ this is, indeed, the order of the life time of atomic levels that decay
by dipole transition. For small
$\alpha$ the Lamb shift dominates the life time, both in the Dicke
model and in Hydrogen.
It remains to show that the analytic continuation of $G(e)$ to the lower half
plane in a neighborhood of $m\alpha$, is indeed given by Eq.~(\ref{analytic}).
This can be done as follows: Let
$B_r$ be a ball of radius $r=2 m\alpha$. Then, in the upper half plane
\begin{equation}
\ G(e)=\left(\int_{B_r}+\int_{B_r^c}\right) \frac
{|f|^2}{e-{|k|}}
\, d^d k=G_r(e)+G^c_r(e).
\end{equation}
Clearly, $G^c_r(e)$ extends analytically to a half circle in the lower
half plane
$|e-\alpha m|<\alpha m$. In the
limit of $\alpha \to 0$, by continuity,
\begin{equation}
G^c_r(0)\to -{\cal E}.
\end{equation}
This is the dominant piece, and it is real.
Consider the analytic continuation
of
$G_r(e)$ for
$|e-\alpha m|\le \alpha m$. Since, for small argument $f(k)$ is given by
Eq.~(\ref{K}), one has (in the upper half plane)
\begin{equation}
G_r(e)=K\Omega^d\int_0^{2m\alpha} \frac{k^{d-2}}{e-k}\, dk=
K\Omega^d\int_\gamma \frac{k^{d-2}}{e-k}\, dk,
\end{equation}
where $\gamma$ is the obvious semi-circle in the complex $k$ plane and
$\Omega^d$ the surface area of the unit ball in $d$ dimensions. The right hand
side is analytic in
$e$ in the lower half plane provided
$|e-\alpha m|<\alpha m$, and so gives the requisite analytic continuation.
Since $e$ is small, and of order $\alpha$, to leading order, we have
\begin{eqnarray}
G_r(e)&=&K\Omega^d\int_\gamma
\frac{(k-e +e)^{d-2}}{e-k}\,dk\nonumber \\
&=&-K\Omega^d\, \sum_{j=0}^{d-2} \left(\begin{array}{c} d-2\\
j\end{array}\right)\,\,e^{d-j-2}\int_\gamma (k-e)^{j-1}\,dk
\nonumber \\
&\approx&-K\Omega^d\, \,e^{d-2}\int_\gamma \frac{dk}{k-e}\nonumber \\
&=&-\, K\Omega^d\, e^{d-2}
\left(i\,\pi\,+\log\left(\frac{2\alpha
m}{e}\right)\,+\,O(\alpha\log\alpha)\right)
\end{eqnarray}
and the error term in approximation that we did not compute is real and
being sub-dominant to
${\cal E}$ is irrelevant.
\paragraph{Adiabatic Deformations}
Suppose, that the two level system of the Dicke model describes two Zeeman
split energy levels of an atom in constant external magnetic field $B$. In
this case, $m$
is proportional to the $|B|$ (and so of order $\alpha$ in atomic units.
Note that in this case Eq.~(\ref{standard}) holds if $\alpha$ is small).
Following Berry
\cite{berry}, let us consider adiabatically rotating the magnetic field
keeping $|B|$ constant, then the family of Dicke models that we obtain is
related by unitary transformation describing the effect of rotation in the two
dimensional Hilbert space of the atom. This motivates looking at the family of
Hamiltonians obtained by conjugating $H(m,d,f,\alpha)$ with a unitary
$V(s)$. We choose
\begin{equation} V(s) =\exp\, (i\,s\, \sigma),\quad \sigma
=\left(\begin{array}{ll} 0&1\\ 1&0\end{array}\right)\otimes {\mathbf
1},\label{v}
\end{equation} which corresponds to a rotation of the atom, leaving the
(transverse) photons alone. This choice of $\sigma$ does not commute with
${\cal N}$,
\begin{equation}[{\cal N},\sigma]=\left(\begin{array}{rl}0&1\\
-1&0\end{array}\right)=J.
\end{equation} Had we chosen
$\sigma$ that commuted with ${\cal N}$ the \at\ for the ground state would
hold for trivial reasons: One can reduce the problem to the ${\cal N}=0$
sector where the ground state is an isolated eigenvalue.
To have an interesting \at\ we need to choose $\sigma$ that
fails to commute with ${\cal N}$. This is the choice we have made.
\subsection{The Adiabatic Theorem}
\begin{qed} Let $H(s;m,d,f,\alpha)=V(s) H(m,d,f,\alpha )V^\dagger(s), \
s\in[0,1]$ be the family of time dependent Dicke models with
$f$ square integrable, with square root singularity at $k=0$;
$m>\alpha^2\,\langle f|\frac{1}{|k|}|f\rangle$; $d\ge 3$ and
$V(s)=\exp \, (i\,s\,\sigma)$ as in Eq.~(\ref{v}). Then, $U_A$, the adiabatic
evolution associated with the kernel of $H(s;m,d,f,\alpha )$, and $U_\tau$,
the Schr\"odinger evolution. are close in the sense that
\begin{equation}
\Vert (U_A(s)- U_\tau(s))P(0)\vert \le
C\left\{\begin{array}{lr}\frac{1}{\tau}&\mbox{if $d>3$}\\
\frac{\log \tau}{\tau}& d=3\end{array}\right.
\end{equation} The effective gap is renormalized to
${m}\left(1+ O(\alpha^2)\right)$, and coincides with the gap without photons
up to a correction of the order of the Lamb shift.
\end{qed} Proof: From Corollary 2.1 we find $K=\sigma\otimes\Omega$. We will
first show that a solution of the commutator equation, Eq.~(\ref{commute}),
for $d>0$, is
\begin{equation} X=\frac{i\,X_1-X_2(g)}{m-\alpha^2\,{\cal E}},\quad
Y=0,\label{x}
\end{equation} where
\begin{equation} {X}_1 =J\otimes \Omega,
\quad X_2(g) = P\otimes(
a^{\dagger}(g)\,\Omega + \mathrm{h.c}), \quad g= i\,\alpha^{\frac{3}{2}}\,
\frac{f}{|k|}.\label{g}
\end{equation} Note that the gap of the two level system $m$ is renormalized to
${m+i\,\alpha\,\langle f|g\rangle}$, which is just the Lamb shift. This is a
small correction, of order $\alpha^2$.
Let us compute the commutators of
${X}_1$, ${X}_2$ with $H$:
\begin{eqnarray} [H,{X_1}] &=&
\left[\left( \begin{array}{cc} \frac{E}{\alpha}&
\sqrt{\alpha}\,{a^\dagger(f)}
\\ \sqrt{\alpha}\, {a(f)}&m + \frac{E}{\alpha} \end{array} \right),
\left( \begin{array}{cc} {0}&-\Omega \\ \Omega & {0} \end{array}
\right)\right] \nonumber \\ &=&m\,\sigma\otimes \Omega+\sqrt{\alpha}\,
P\otimes
\Big(a^\dagger(f) \Omega+
\Omega a(f)\Big).
\end{eqnarray} For the second commutator
\begin{eqnarray} [H,{X_2}] &=& \left[\left( \begin{array}{cc}
\frac{E}{\alpha}& {\sqrt{\alpha}\,a^\dagger(f)}
\\ \sqrt{\alpha}\, {a(f)}& m + \frac{E}{\alpha} \end{array}
\right),\left ( \begin{array}{cc} { a^{\dagger}(g)\,\Omega +
\mathrm{h.c}}& {0} \\ {0}& 0 \end{array} \right)\right] \nonumber \\
&=&\frac{1}{\alpha}\, P\otimes (a^{\dagger}(|k| g)\Omega - \Omega a(|k|
g)+\sqrt{\alpha}\,\left(\begin{array}{lr}0&-\langle g|f
\rangle\\ \langle f|g \rangle&0\end{array}\right)
\otimes \Omega.\label{second}
\end{eqnarray} So, if we take $g$ of Eq.~(\ref{g}) then
\begin{equation} [H,i\,{X_1}-{X_2}]=i\,\left(m-\alpha^2\,{\cal E}\right)
\sigma\otimes
\Omega.
\end{equation} We see that we can formally solve for the commutator equation,
Eq.~(\ref{commute}) provided
${\cal E}$ is finite.
This is, however, not the only condition.
$X$ is a bounded operator in the Hilbert space provided $g\in L^2$, for
otherwise
$a^\dagger(g)$ is ill defined:
\begin{equation}
\int{\frac{|f|^2}{|k|^2}d^dk}\sim \int{\frac{1}{|k|^3}d^dk}<\infty.
\end{equation} The integral is finite if $d\ge 4$ but is logarithmically
divergent in $d=3$.
So the $d>3$ part of the theorem is done. For $d=3$ we need to squeeze $X$
back to the bounded operators. We do that by allowing for $Y\neq 0$.
Let $\chi_\e$ be the characteristic function of a ball of radius $\e$ and
$\chi_\e^c=1-\chi_\e$ and let $g_\e^c =\chi_\e^c g$ and $g_\e =\chi_\e g$.
Let us take $X_2(g_\e^c)$ is well defined and it norm is
$O(\alpha^{\frac{3}{2}}\log\e )$. For $X$ we take, as before,
\begin{equation} X=\frac{i\,X_1-X_2(g^c_\e)}{m+i\sqrt{\alpha}\langle
f|g^c_\e\rangle }.
\end{equation} From this
\begin{equation}
\Vert X\Vert = O\left(\frac{1 +\alpha^{\frac{3}{2}}|\log\e|}{|m-\alpha^2 {\cal
E} |}.
\right)
\end{equation}
For $Y$ we take
\begin{eqnarray}
(m+i\, \sqrt{\alpha} \langle f| g^c_\e
\rangle)\,Y&=&[H,X_2(g)-X_2(g_\e^c)]=[H,X_2(g_\e)]
\\ &=&\frac{1}{\alpha}\, P\otimes \Big(a^{\dagger}(|k| g_e)\Omega -
\Omega a(|k| g_e)\Big)+\sqrt{\alpha}\,\left(\begin{array}{lr}0&-\langle
g_e|f
\rangle\\ \langle f|g_e \rangle&0\end{array}\right)\otimes\Omega,\nonumber
\end{eqnarray} and we used the computation of the commutator
Eq.~(\ref{second}). With $f$ having a square root singularity,
\begin{equation}
\Vert Y\Vert = O\left(\frac{\sqrt{\alpha}
\,\e +\alpha^2\,\e}{|m-\alpha^2{\cal E} |}
\right) .
\end{equation}
This puts us in the frame of theorem 2.1 and establishes the main
result.\hfill $\Box$
\section*{Acknowledgments} We are grateful to I.M. Sigal for a conversation
that stimulated this work, and to S. Graffi and A. Ori for illuminating
observations and to C. Brif for spotting an embarrassing mistake. This work was
partially supported by a grant from the Israel Academy of Sciences, the
Deutsche Forschungsgemeinschaft, and by the Fund for Promotion of Research at
the Technion.
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\end{document}