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\begin{document}
\title{Exact Ground State Energy of the Strong-Coupling Polaron}
\author{Elliott~H.~Lieb$^1$ and Lawrence~E.~Thomas$^2$\\
\footnotesize \it $^1$Departments of Physics and Mathematics,
Jadwin Hall, Princeton University,\\
\footnotesize \it P.~O.~Box 708, Princeton, New Jersey 08544\\
\footnotesize \it
$^2$Department of Mathematics, University of Virginia,
Charlottesville, Virginia 22903}
\date{December 13, 1995}
\maketitle
\begin{abstract}
The polaron has been of
interest in condensed matter theory and field theory for about half a
century, especially its strong coupling limit. It was not until 1983,
however, that a proof of the asymptotic formula for the ground state
energy was finally given by using difficult arguments involving the large
deviation theory of path integrals. Here we derive the same
asymptotic result, $E_0\sim -C\alpha^2$, and with explicit error
bounds, by simple, rigorous methods applied directly to the
Hamiltonian. Our method is easily generalizable to other
settings, e.g., the excitonic and magnetic polarons.
\end{abstract}
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%\pacs{PACS numbers: 03.65.-w, 11.10.-z, 71.38.+i}
%\narrowtext
The polaron Hamiltonian of Fr\"ohlich \cite{HF} is a model for the
Coulomb interaction of one or more electrons with the phonons of an
ionic crystal. In the course of time it was also seen to be an
interesting field theory model of non-relativistic particles
interacting with a scalar boson field, and it was widely studied
\cite{P,Bog,Tyab,EG,JF,HS} in both contexts.
The model has one dimensionless coupling constant, $\alpha$, and it was
noticed very early that there seems to be a qualitative difference in
the ground state between the weak coupling regime, well described by
perturbation theory \cite{LLP}, in which the electron is spread out,
and the strong coupling regime in which the electron appears to be
trapped in a phonon hole of its own making. By now it seems doubtful
that such a trapping actually occurs \cite{HS,GL,PD}, but it remains
true that the calculation of the ground state energy, $E_0(\alpha)$,
is very different in the two regimes. The strong coupling theory was
studied by Pekar \cite{P} (see also \cite{Bog,Tyab,EG})
who hypothesized that in this limit the total
ground state wave function $\Psi$ could be taken to be a product of an
electronic function $\phi(x)$ and a phonon function $\vert {\bf \xi}
\rangle$.
Fr\"ohlich's Hamiltonian in appropriate units is
\begin{equation}\label{eq:1}
H= {\bf p}^2 + \sum_{\bf k}{a}_{\bf k}^*{a}_{\bf k}^{\phantom*} +
(\frac{4\pi\alpha}{V})^{1/2}\sum_{\bf k}
\left[\frac{a_{\bf k}^*}{|{\bf k}|}
e^{i{\bf k\cdot\bf x}} + h.c.\right]
\end{equation}
where ${\bf p}=-i{\bf \nabla}$ is the electron momentum operator, ${\bf
x}$ is its coordinate, and $V$ is the volume of the crystal, which
tends to infinity. The ${\bf k}$'s are the usual normal modes and, as
usual, $V^{-1}\sum_{\bf k} \rightarrow (2\pi)^{-3}\int d^3k$. When
$\langle \Psi \vert H \vert \Psi \rangle$ is computed with Pekar's
ansatz one easily finds that the phonon part $\vert \xi \rangle$ can be
easily evaluated by \lq completing the square\rq\ and the resulting
$\phi $ minimizes the energy
\begin{equation}
{\cal E}(\phi) \equiv \int
\vert {\bf \nabla} \phi \vert^2 d^3x -\alpha \int \int \frac {\phi({\bf
x})^2\phi({\bf y})^2} {\vert {\bf x-y}\vert}d^3xd^3y\label{Pekar}
\end{equation}
subject to $\int \phi({\bf x})^2 d^3x=1$.
It has been proved \cite{EL} that there is exactly one $\phi$, up to
translations, that minimizes ${\cal E(\phi)}$. By scaling, the minimum
energy (call it $E_P(\alpha)$) is proportional to $-\alpha^2$. The
question we address is this: {\it Is this energy, $E_P(\alpha)$,
asymptotically exact as $\alpha$ tends to infinity? } Since it is
given by a variational calculation, $E_P(\alpha)$ is an upper bound to
$E_0(\alpha)$; the problem is to find a lower bound that agrees with
$E_P(\alpha)$ to leading order. For a long time more or less the only
thing that could be done to validate $E_P(\alpha)$ was a lower bound
\cite{LY} of the right order, $-\alpha^2$, but which was about a
factor of 3 too large. It was not until 1983, with a precursor in
\cite{AGL},
that a complete proof was given \cite{DV}. The proof starts with an
expression \cite{F} for $E_0$ in terms of a self-interacting Brownian
path, obtained from the Feynman-Kac formula for $e^{-\beta H}$ by
integrating out the phonon variables. This proof is very far from
simple and requires an enormous knowledge of large deviation theory of
Brownian motion, a subject unknown to most physicists, and it does not
generalize easily to other models.
In any event, there is the following problem: The Pekar ansatz is based
on the physically appealing notion that at large coupling the phonons
cannot follow the rapidly moving electron (as they do at weak coupling)
and so resign themselves to interacting only with the \lq
mean\rq\ electronic density, $\phi({\bf x})^2$. What, exactly, do these
somewhat anthropomorphic words mean? If they are so very physical,
they should be quantifiable and it should not take four decades to
build a proof of their correctness. Moreover, one should be able to
find a proof that is relatively simple (since the idea is a simple
one), and that yields some kind of quantitative error
estimate. It should also be robust enough to allow an easy extension to
some variations of $H$, such as having several electrons instead of one (the
polaronic exciton), inclusion of magnetic fields, an electron kinetic
energy other than ${\bf p}^2$, accommodation of ${\bf k}$-dependence and
nonlinearity in the phonon self
energies, electron-phonon interaction energies other than $1/\vert {\bf
k} \vert$, etc.
The method we present here satisfies, we believe, the criteria of
simplicity and robustness. For this
reason it might be of general interest
in condensed matter theory or field theory. Despite its generality we
shall, for clarity, restrict our discussion to the original
Hamiltonian, $H$. Our bound on the error is $O(\alpha^{9/5})$,
to be compared with the main term $O(\alpha^2)$.
Before going into the details, however, it is helpful to give an
overview of our method. The first thing to notice is that Pekar's
ansatz amounts, mathematically, to saying that we can replace the
operators $a_{\bf k}$ by c-numbers $\xi_{\bf k}$. One might think, at
first, that such a replacement always leads to a lower bound for the
energy, but this is false. (If it were true, the small $\alpha$ energy
would have to be proportional to $-\alpha^2$ instead of the correct
value $-\alpha$.) To pursue this idea, nevertheless, the natural tool
to use is coherent states for each phonon mode ${\bf k}$. As is well
known, these states are indexed by complex numbers $\xi =p+iq$ and, if
$ \vert \xi \rangle \langle \xi \vert$ is the projector onto the
coherent state $\vert \xi \rangle$, we have that $\int \vert \xi
\rangle \langle \xi \vert \ d\xi d\xi^* = I$ and $\int \xi \ \vert \xi
\rangle \langle \xi \vert \ d\xi d\xi^* = a $ and $\int (\vert
\xi\vert^2 -1)\vert \xi \rangle \langle \xi \vert \ d\xi d\xi^* =
a^*a$. {\it It is the extra term $-1$ that kills the lower bound and
gives an unwanted energy $-1$ for each phonon mode} (and hence a
negative contribution to the energy that is infinite both because
there is no infrared and no ultraviolet cutoff). In other words,
coherent states would give us what we want (effectively replacing the
operator $a$ by the number $\xi$), were it not for the unfortunate
fact that the positive operator $a^*a$ is represented by the
nonpositive function $\vert \xi\vert^2 -1$.
Our remedy will be to reduce the effective number of phonon modes to a
{\it finite} number of $O(\alpha^{9/5})$, independent of $V$. These
modes will be quite different from the original $a_{\bf k}$ modes;
indeed they will be the $a_{\bf k}$ modes summed over boxes in ${\bf
k}$-space of size $\alpha^{3/5}$. Thus, our physical description of
strong coupling will be a little different from the conventional one.
{\it Instead of saying that the phonons cannot follow the electron, our
point of view will be that the electron significantly excites only
finitely many field modes.}
This mode reduction is accomplished in several steps which can be
outlined as follows.
$\bullet$ {\bf I.} Using a simple commutator estimate (as in the elementary
proof of Heisenberg's uncertainty principle) we can show that ${\bf \vert
k\vert}$ values larger than $K \approx \alpha^{6/5}$ can be ignored
with an energy cost of only $\alpha^{9/5}$. This eliminates the
ultraviolet problem, i.e., the fact that ${\bf k}^{-2}$ is not summable.
$\bullet$ {\bf II.} With the same energy error we can localize the electron
to a cube of side length $\alpha^{-9/10}$.
$\bullet$ {\bf III.} We decompose the ball $\vert {\bf k}\vert 0$, and with $K\equiv
8\alpha/\pi \varepsilon$,
\begin{eqnarray}
\lefteqn{ -(\frac{4\pi\alpha}{V})^{1/2}\sum_{|{\bf k}|\geq
K}\big[\langle\frac{{a}_{\bf k}^*}{|{\bf k}|}e^{i{\bf k}\cdot{\bf
x}}\rangle+c.c.\big]}\nonumber\\ &\leq&
4(\frac{4\pi\alpha}{V})^{1/2}\langle{\bf p}^{2}\rangle^{1/2}
\sum_{|{\bf k}|\geq K}|{\bf k}|^{-2}\langle{a}_{\bf k}^*{a}_{\bf
k}\rangle^{1/2}\nonumber\\ &\leq&
4(\frac{4\pi\alpha}{V})^{1/2}\langle{\bf p}^2\rangle^{1/2}(\sum_{|{\bf
k}|\geq K} |{\bf k}|^{-4})^{1/2}(\sum_{|{\bf k}|\geq K}\!\!\! \langle
\ {a}_{\bf k}^*{a}_{\bf k}^{\phantom *}\rangle)^{1/2}\nonumber\\ &=&
2\varepsilon^{1/2}\langle{\bf p}^2\rangle^{1/2}(\sum_{|{\bf k}|\geq K}
\langle \
{a}_{\bf k}^*{a}_{\bf k}{\phantom *}\rangle)^{1/2}\nonumber\\
&\leq& \varepsilon\langle{\bf p}^2\rangle +\sum_{|{\bf k}|\geq K} \
\langle \ {a}_{\bf k}^*{a}_{\bf k}{\phantom *}\rangle,\label{uvbound}
\end{eqnarray}
again by the Schwarz inequality. (We have taken the limit
$V\rightarrow \infty$ in the sum.)
The above inequality (\ref{uvbound}) is an ultraviolet bound. It shows
that the Hamiltonian $H$ is bounded below by a new one $H_{K}$, i.e.,
$H\geq H_{K}$, with
\begin{equation}
H_{K}\equiv(1-\frac{8\alpha}{\pi K}){\bf p}^2
+\sum_{|{\bf k}|0$,
there exists a function $\phi$ of the electron coordinate ${\bf
x}$ alone (but depending on $\Psi$), with support in a cube of side length
$L= \pi(3/\Delta E)^{1/2}$
such that
\begin{equation}
\langle\phi\Psi|H_{K}|\phi\Psi\rangle \!
\Bigm/ \! \langle\phi\Psi|\phi\Psi\rangle\leq E+\Delta E.
\end{equation}
To see this, let $\phi({\bf x})=\prod_{j=1}^3\cos((\Delta
E/3)^{1/2}x_j)$ inside the cube of side length $\pi (3/\Delta E)^{1/2}$
centered at the origin and $\phi= 0$ outside the cube. Let $\phi_{\bf
y}({\bf x})= \phi({\bf x}-{\bf y})$ and consider the integral
\begin{eqnarray}
\int\!\!\!&\Big(&\!\!\langle\phi_{\bf y}\Psi|H_{K}|\phi_{\bf
y}\Psi\rangle
-(E+
\Delta E)\langle\phi_{\bf y}\Psi|\phi_{\bf y}\Psi\rangle\Big)d^3
y\nonumber\\
& =&\int(|\nabla\phi|^2-\Delta E|\phi|^2) \ d^3 y = 0.\label{local}
\end{eqnarray}
Evidently, there must be a point
${\bf y}$ such that at this point, the integrand on the left side of
(\ref{local}) is non-positive (and where $\langle\phi_{\bf
y}\Psi|\phi_{\bf y}\Psi\rangle$ is non-zero). This $\phi_{\bf
y}$ is the $\phi$ that we need. The electron localization is now complete.
For our purposes, we take $\Delta E= c_2\alpha^{9/5}$, and the
assertion above together with the ultraviolet bound then implies that the
ground state energy $E_{0}$ of $H$ satisfies
\begin{equation}
E_{0}\geq {\inf_{\Psi}}\,
^\prime\langle\Psi|H_{K}|\Psi\rangle-c_2\alpha^{9/5}\label{Eo}
\end{equation}
where the infimum is taken over all normalized $\Psi$'s but having
their ${\bf x}$-support in a cube of side length no larger than $L=
(3/c_2)^{1/2}\pi\alpha^{-9/10}$ somewhere in the large volume $V$.
{\bf III.} The next step is to group the phonon modes together into blocks,
which we take to be cubes with sides of length $P= c_3\alpha^{3/5}$ or,
more precisely, the portion of those cubes lying in the
big ball $\{{\bf k}:|{\bf k}|0$, we have the inequality
\begin{eqnarray}
\lefteqn{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\sum_B\sum_{{\bf k}\in B}\left[\delta{a}_{\bf k}^*{a}_{\bf k}^{\phantom
*} +(\frac{4\pi\alpha}{V})^{1/2}\big[\frac{{a}_{\bf k}^*}{|{\bf k}|}(
e^{i{\bf k}\cdot{\bf x}}\!- e^{i{\bf k}_B\cdot{\bf x}}) +
h.c.\big]\right]}\nonumber\\
&&\geq -\frac{4\pi\alpha}{\delta
V}\big(\frac{3\pi c_3}{2}(\frac{3}{c_2})^{1/2}\alpha^{-3/10}\big)^2\sum_{|{\bf
k}|