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\vglue 1cm
\centerline{\twelvebf Ground state of a quantum particle}
\centerline{\twelvebf coupled to a scalar Bose field}
\bigskip
\medskip
\centerline{\bf Herbert Spohn}
\bigskip
\bigskip
\centerline{Theoretische Physik, Ludwig-Maximilians-Universit\"at
M\"unchen,}
\centerline{Theresienstr. 37, D-80333 M\"unchen, Germany}
\smallskip
\centerline{e-mail : spohn@stat.physik.uni-muenchen.de}
\vglue 2cm
\centerline{\bf Abstract}
\bigskip
\noindent {\parindent=1cm\narrower
We use methods from functional integration to prove existence and
uniqueness of the ground state of a confined quantum particle
coupled to a scalar massless Bose field. For an external potential
growing at infinity the ground state exists under fairly general
conditions while for a decaying potential an unphysical condition
on the coupling strength is still needed.
\par}
\vglue 2cm
\noindent {\bf 1. Introduction}
\medskip
We consider a quantum particle confined by an external potential
such that it has a unique state of minimal energy. The problem is
whether this ground state persists as the particle is coupled to a
massless scalar Bose field. Physically, the particle could be an
impurity and the Bose field the phonons in a solid. More
fundamental, perhaps, is the coupling of a hydrogen atom to the
radiation field \`a la Pauli-Fierz. In our note we have nothing to
say on the latter topic. To set the scene let us spell out the
Hamiltonian.
\smallskip
The particle is governed by
$$
H_e = -{1\over 2} \, \D + V \, . \eqno (1.1)
$$
The potential $V : \Rb^n \ra \Rb$ is such that $H_e$ is
defined as a self-adjoint operator on the domain $\Dc (H_e) = \Dc
(-\D) \sbs L^2 (\Rb^n)$. We have two standard examples in mind.
(i) The potential decays at infinity. One choice is $V \in L^p
(\Rb^n) + L^{\ify} (\Rb^n)$ with $p=2$ for $n \leq 3$, $p=n/2$ for
$n\geq 4$. Then the essential spectrum of $H_e$ is $[\Si ,\ify)$
with continuum edge $\Si \in \Rb$. We assume that $H_e$ has a
(unique) ground state $\psi_0$ with energy $E_0 < \Si$, $H_e \,
\psi_0 = E_0 \, \psi_0$, $\psi_0 \in L^2 (\Rb^n)$, and $\psi_0 >
0$. Of course, $H_e$ could have further bound states. We refer to
[1] for details. (ii) $V(x) \ra \ify$ as $\vert x \vert \ra \ify$,
e.g. by adding to a potential from (i) a continuous potential
bounded from below. Then $H_e$ has a purely discrete spectrum and
our ground state assumption is satisfied. Formally we subsume (ii)
under (i) by setting $\Si = \ify$.
\smallskip
The scalar Bose field $\{ a(k) , a^{\dagger} (k) , \, k \in \Rb^n
\}$ satisfies the commutation relations $[a(k) ,a^{\dagger} (k')] =
\d (k-k')$. Its energy is
$$
H_f = \int d k \, \om (k) \ a^{\dagger} (k) \, a(k) \eqno (1.2)
$$
defined on the bosonic Fock space, $\Fc$. For a massless field the
dispersion relation $\om (k) = \vert k\vert$. In fact, our method is
more general. It suffices to assume that $\om : \Rb^n \ra \Rb$ is
continuous with $\om (k) > 0$ for $k \not= 0$ and $\om (k) \geq
\om_0 > 0$ outside the unit ball.
\smallskip
We couple the particle and the field in a translation invariant
manner. Then
$$
H = H_e \ot 1 + 1 \ot H_f + \int d k \, (\lb (k) \, e^{ikx} \ot
a^{\dagger} (k) + \lb (k)^* \, e^{-ikx} \ot a(k)) \eqno (1.3)
$$
on $\Hc = L^2 (\Rb^n) \ot \Fc$. We assume $\lb (k) = \lb (-k)^*$.
Physically $\lb (k) = \a \, \hat{\rho} (k) /$ $\sqrt{2\om (k)}$,
where $\a$ is a coupling constant and $\rho$, the Fourier
transform of $\hat{\rho}$, is the normalized real-valued
``charge'' distribution. In particular $\build \lim_{k \ra 0}^{}
\, \hat{\rho} (k) =1$. To avoid trivial degeneracies we assume
$\lb \not= 0$ a.s.. We also require that $\int \vert \lb \vert^2 \,
(1+\om^{-1}) \, dk < \ify$. Then $H$ is self-adjoint on $D(H_e)
\ot D(H_f)$ and bounded from below [2].
\smallskip
For $\lb = 0$ $H$ has the unique ground state $\psi_0 \, \ot \,
\Om$ with $\Om$ the Fock vacuum. We would like to have conditions
which guarantee that for $\lb \not= 0$ $H$ still has a unique
ground state. In fact, for a suitable realization of Fock space,
the kernel of $e^{-tH}$ is positivity preserving and ergodic [2].
Thus the uniqueness comes from a Perron-Frobenius argument [1,
Theorems XIII.44, XIII.46] and the real issue is {\it existence}.
Note that if $\om (0) = 0$, then for $\lb = 0$ the continuous
spectrum starts at $E_0$ and a perturbative argument will not work.
\smallskip
Physically, one argues that the coupling to the field makes the
particle more heavy. This can be seen by taking $V=0$ and
considering the ground state energy $E(P)$ as a function of the
total momentum, $P$. The effective mass is usually defined by
$m=[\part_P^2 \, E(0)]^{-1}$. One can prove that $m\geq 1$
in our units [3]. Approximately the particle is then governed by the
effective Hamiltonian $H_{\rm eff} = - {1\over 2m} \, \D +V$. If
$H_e$ has a ground state, so does $H_{\rm eff}$. Thus no matter how
strong the interaction the coupled system should always have a
ground state.
\smallskip
There is one proviso. The coupling could be so strong that $m=\ify$.
Then tunneling is suppressed and the ground state sits at one of
the absolute minima of the potential. Such a phenomenon has been
proved for simplified versions of (1.3) [4]. The strong reduction
in fluctuations is always associated with an infinite number of
bosons. Thus in our mathematical framework it would mean that
there is {\it no} ground state for $H$ in $\Hc$, i.e. existence is
lost.
\smallskip
Recently Bach, Fr\"ohlich, and Sigal [2] obtained very sharp
estimates on the resonances for the Hamiltonian (1.3). As an
aside, they also investigate the ground state problem by using
operator methods originally developed in the context of quantum
field theory. One first replaces $\om (k) = \vert k \vert$ by
$\om_m (k) = \sqrt{k^2 +m^2}$ and proves that $H(\om_m)$ has a
ground state provided $\Si - E_0 -\int \vert \lb \vert^2 \,
\om^{-1} \, dk \geq 4m > 0$. The true ground state at $m=0$ is
constructed through the weak limit $m \ra 0$ of the ground states
of $H(\om_m)$. One important ingredient is an estimate on the
average number of bosons in the ground state uniformly in $m$.
Their final result reads: $H$ has a ground state provided (i) $\Si
- E_0 \geq \int \vert \lb \vert^2 \, \om^{-1} \, dk$ and (ii)
$\left( \int \vert \lb \vert^2 \, \om^{-2} \, dk \right)$
$\left[1+4 \, (\Si - E_0)^{-2} \, \int \vert \lb \vert^2 \, dk
\right] < 1$. (In [2] a more general class of Hamiltonians is
studied and I specialized here to the translation invariant model
(1.3).) Such an approximation technique was developed
independently by Arai and Hirokawa [5] and applied to a variety of
related models.
\smallskip
In this note I use methods from functional integration. If
applicable, as it happens to be the case here, functional
integrals typically yield stronger results and I wanted to
understand how far they reach. The use of functional integration
in our context has been described elsewhere [6] and I take it for
granted that certain matrix elements can be written in terms of a
suitable expectation. E.g.
$$
\displaylinesno{
(\psi_0 \ot \Om , e^{-TH} \, \psi_0 \ot \Om) = \cr
\Eb_{[0,T]} \, (\psi_0 (x_0) \, \psi_0 (x_t) \, \exp[-\int_0^T dt
V (x_t) -S_{[0,T]} (x_{\cdot})]) \, . &(1.4) \cr
}
$$
Here $\Eb$ refers to the expectation in the standard Wiener
measure. $t \mpo x(t)$ is a continuous path on $[0,T]$. The first
piece is the ground state process for $-{1\over 2} \, \D + V$. The
action $S$ originates from integrating over the Bose field. In our
case it is given by
$$
S_{[0,T]} (x_{\cdot}) = \int_0^T dt \, \int_0^T ds \, W(x_t - x_s
, t-s) \, , \eqno (1.5)
$$
$$
W(x,t) = -{1\over 2} \, \int \vert \lb \vert^2 \, e^{-\om \vert
t\vert} \, \cos kx \, dk \, . \eqno (1.6)
$$
By assumption $\vert W(x,t)\vert \leq \int \vert \lb \vert^2 \, dk
< \ify$. Therefore the integral (1.4) is well defined.
\smallskip
In the form (1.4) the existence of the ground state is reduced to
a problem of statistical mechanics in one dimension (i.e. the
Euclidean time axis). Ground state means limit $T \ra \ify$. First
note that $W$ has its minimum at $x=0$ and that $S$ is invariant
under the global shift $x_t$ to $x_t + a$, which reflects the
translation invariance of the coupling. If $V=0$ and if $W(\cdot
,t)$ decays sufficiently fast in $t$ we may expect that on a large
scale the path measure looks effectively as Brownian motion with
diffusion constant $1/m$ [3]. Since $m>1$, the paths fluctuate
less compared to $S \eqv 0$, which should improve binding. If
$m=\ify$, the suppression could be so strong that in the limit $T
\ra \ify$ the path no longer fluctuates at all and the
distribution of $x_t$ concentrates at a single point, which means
there is no ground state in $\Hc$. To have a crude estimate, one
compares energy and entropy. Let me assume that $x_t =0$ and let
me study an excitation of the form $\tilde{x}_t = a$ for $\vert t
\vert \leq R$ and $\tilde{x}_t = 0$ for $\vert t \vert \geq R$. The
entropy of such an excitation grows as $\log R$. To be on the safe
side, if the energy of this excitation remains bounded as $R \ra
\ify$, then the entropy will outweigh the energy and there should be
sufficient fluctuations even as $T \ra \ify$. This leads to the
condition
$$
\int_{-\ify}^0 dt \, \int_0^{\ify} ds \, (W(\tilde{x}_t -
\tilde{x}_s , t-s)-W(0,t-s)) \qquad \hbox{essentially bounded.}
\eqno (1.7)
$$
If we just use $\vert 1 - \cos kx \vert \leq 2$, a sufficient
condition is
$$
\int \vert \lb \vert^2 \, \om^{-2} \, dk < \ify \, , \eqno (1.8)
$$
which will reappear later. For $\om (k) = \vert k \vert$ and the
physical coupling the integrand of (1.8) becomes $\a^2 \, \vert
\hat{\rho} \vert^2 / \om^3$ which diverges logarithmically at
$k=0$ in three dimensions $(n=3)$. To improve we observe that
because of the external potential $\vert x_t - x_s \vert$ should be
essentially bounded. Thus we may expand at small $k$ as $1-\cos kx
= {1\over 2} \, k^2 \, x^2$. Then (1.7) leads to the condition
$$
\int \vert \lb \vert^2 \, \om^{-2} \, k^2 (1+k^2)^{-1} \, dk <
\ify \, . \eqno (1.9)
$$
We conjecture that if (1.9) holds then $H$ has a unique ground
state.
\smallskip
Unfortunately our methods do not yield such a strong result.
\bigskip
\noindent {\it Theorem 1.} Let $\int \vert \lb \vert^2 \,
(1+\om^{-1}) \, dk < \ify$ and $\lb \not= 0$ a.s.. If $\int \vert
\lb \vert^2 \, \om^{-2} \, dk < \ify$ and if
$$
\Si - E_0 > \int \vert \lb \vert^2 \, {1\over 2} \, k^2 \, \left(
\om + {1\over 2} \, k^2 \right)^{-1} \, dk \, , \eqno (1.10)
$$
then $H$ has a unique ground state.
\smallskip
In the remainder of the paper we give a proof, but before I add
some comments.
\smallskip
\noindent (i) If $V(x) \ra \ify$ as $\vert x \vert \ra \ify$, then
$\Si = \ify$ and there is no condition on the coupling strength.
But we still cannot allow for the physical case $\om (k) = \vert k
\vert$, $\lb = \a \, \hat{\rho} / \sqrt{2\om}$, $n=3$.
\smallskip
\noindent (ii) An instructive simplification is the spin-boson
Hamiltonian
$$
H_{SB} = -\ve \, \s_x \ot 1 + 1 \ot H_f + \s_z \ot \int (\lb (k)
\, a^{\dagger} (k) + \lb (k)^* \, a(k)) \, dk \eqno (1.11)
$$
with $\om (k) = \vert k \vert$ and $\s_x$, $\s_z$ the Pauli spin
matrices. The corresponding functional integral is a
ferromagnetic Ising model over $\Rb$ [4]. $H_{SB}$ has a ground
state in $\Cb^2 \, \ot \, \Fc$ if and only if the spontaneous
magnetization, $m^*$, of the Ising model vanishes [7]. One has
fairly sharp estimates on $m^*$. E.g. one sufficient condition is
$\build \lim_{t\ra \ify}^{} \, t^2 \int \vert \lb \vert^2 \,
e^{-\om \vert t \vert} \, dk =0$. On the other hand if we replace
$-\s_x$ by, say, $-\s_x + a \s_y$, $a\in \Rb$, then in the $\s_z$
representation the kernel $\exp \, [-t \, (-\s_x + a\s_y )]_{\s
\s'}$ is no longer positive for $t > 0$, which prevents a
functional integral representation at least in the obvious way. The
only methods available are then [2,5].
\vglue 1cm
\noindent {\bf 2. Existence}
\medskip
We prove Theorem 1. The strategy is to show that the sequence
$$
\psi_T = e^{-TH} \, \psi_0 \ot \Om / \Vert e^{-TH} \, \psi_0 \ot
\Om \Vert \eqno (2.1)
$$
has a non-zero weak limit as $T \ra \ify$. For this purpose we
consider
$$
\displaylinesno{
(\psi_T , e^{-\tau H_e} \ot P_{\Om} \, \psi_T) = [(\psi_0 \ot \Om ,
e^{-TH} \, e^{-\tau H_e} \ot P_{\Om} \, e^{-TH} \, \psi_0 \ot \Om)
/ \cr
(\psi_0 \ot \Om , e^{-(2T+\tau)H} \, \psi_0 \ot \Om)] \, [(\psi_0
\ot \Om , e^{-(2T+\tau)H} \, \psi_0 \ot \Om )/ \cr
(\psi_0 \ot \Om , e^{-2TH} \, \psi_0 \ot \Om )] &(2.2) \cr
}
$$
with $\tau > 0$ and $P_{\Om}$ the projection onto the Fock vacuum.
Let $E_m$ be the infimum of the spectrum of $H$.
\bigskip
\noindent {\it Lemma 2.} We have
$$
\build \lim_{T\ra \ify}^{} (\psi_0 \ot \Om , e^{-(2T+\tau)H} \,
\psi_0 \ot \Om)/(\psi_0 \ot \Om , e^{-2TH} \, \psi_0 \ot \Om) =
e^{-\tau E_m} \, . \eqno (2.3)
$$
\bigskip
\noindent {\it Proof.} Let $\mu$ be the spectral measure for
$\psi_0 \ot \Om$. We claim that $\hbox{inf supp} \, \mu = E_m$. If
so, (2.3) follows from the spectral theorem.
\smallskip
Let $W(f) = \exp \, \left[ \int (f(k) \, a^{\dagger} (k) - f(k)^* \,
a(k)) \, dk \right]$ be the Weyl operator with test function $f\in
L^2$. For bounded functions $g$ we have
$$
\displaylinesno{
(g \, \psi_0 \ot W (f) \, \Om , e^{-TH} \, g \, \psi_0 \ot W(f) \,
\Om) \cr
= \Eb_{[0,T]} \, \Biggl(\psi_0 (x_0) \, \psi_0 (x_t) \, \exp \,
\left[ -\int_0^T dt \, V(x_t) -S_{[0,T]} \right] \, g(x_0) \,
g(x_t) &(2.4) \cr
\exp \left[ -{1\over 2} \int \vert f \vert^2 dk + {i\over 2}
\int_0^T dt \int (f^* \lb e^{ikx_t} e^{-\om t} + f
\lb^* e^{-ikx_t} e^{-\om (T-t)}) dk \right] \Biggl) \, . \cr
}
$$
Let $\k = \hbox{inf supp} \, \mu$. By the spectral theorem
$$
\build \lim_{T\ra \ify}^{} \, -{1\over T} \, \log \, (\psi_0 \ot
\Om , e^{-TH} \, \psi_0 \ot \Om )=\k \, . \eqno (2.5)
$$
By (2.4) the ratio $(g \, \psi_0 \ot W(f) \, \Om , e^{-TH} \, g \,
\psi_0 \ot W(f) \, \Om) / (\psi_0 \ot \Om , e^{-TH} \, \psi_0 \ot
\Om)$ is bounded uniformly in $T$. Thus also
$$
\build \lim_{T\ra \ify}^{} -{1\over T} \, \log \, (g \, \psi_0 \ot
W(f) \, \Om , e^{-TH} \, g \, \psi_0 \ot W(f) \, \Om)=\k \, .
\eqno (2.6)
$$
The $\hbox{lsp} \, \{ g \, \psi_0 \ot W(f) \, \Om \mid g \
\hbox{bounded,} \ f \in L^2 \}$ is dense in $\Hc$ and each vector
has a spectral measure with infimum $\kappa$. This is possible only
if $\k = E_m$. \xx
\bigskip
We write the first factor in (2.4) as
$$
\displaylinesno{
\Eb_{[0,2T+\tau]} \Biggl( \exp \Bigl[ -\int_0^{2T+\tau} V(x_t) \, dt
- S_{[0,T]} - S_{[T+\tau ,2T+\tau]} \, + \cr
S_{[0, 2T+\tau]} - S_{[0,2T+\tau]}\Bigl] \psi_0 (x_0) \, \psi_0
(x_{2T+\tau}) \Biggl) / \cr
\Eb_{[0,2T+\tau]} \left( \exp \left[ - \int_0^{2T+\tau} V(x_t)
\, dt - S_{[0,2T+\tau]} \right] \psi_0 (x_0) \, \psi_0 (x_{2T+\tau})
\right) \, . &(2.7) \cr
}
$$
Using $\vert \cos kx \vert \leq 1$ we obtain the pathwise bound
$$
\displaylinesno{
S_{[0,T]} + S_{[T+\tau , 2T+\tau]} - S_{[0,2T+\tau]} \leq \cr
{1\over 2} \int \vert \lb \vert^2 \left(\left( 2\int_{-\ify}^0
\int_{\tau}^{\ify} + \, 4\int_0^{\tau} \int_{\tau}^{\ify} +
\int_0^{\tau} \int_0^{\tau} \right) e^{-\om \vert t-s\vert}
\, dt \, ds \right) \, dk \leq \cr
\int \vert \lb \vert^2 \, (\tau \om^{-1} + \om^{-2} \,
(3-2e^{-\tau \om}) \, dk = \ell (\tau) \, . &(2.8) \cr
}
$$
Let $E(d\lb)$ be the spectral resolution of $H_e$. By assumption
for sufficiently small $\d$ the spectral projection $E([E_0 ,\Si
-\d])$ is finite dimensional. (If $\Si = \ify$, we take $\Si -\d <
\ify$.) From the argument above and by Lemma 1 we conclude
$$
\displaylinesno{
\build{\lim \inf}_{T\ra \ify}^{} \Biggl( \int_{E_0}^{\Si -\d}
e^{-\lb \tau} \, (\psi_T , E(d\lb) \ot P_{\Om} \, \psi_T ) \, + \cr
\int_{\Si -\d}^{\ify} e^{-\lb \tau} \, (\psi_T , E(d\lb) \ot P_{\Om}
\, \psi_T ) \Biggl) \geq e^{-\tau E_m} \, e^{-\ell(\tau)} \, . &(
2.9) \cr
}
$$
The second term on the left hand side is bounded by $e^{-\tau (\Si
-\d)}$. Finally we optimize with respect to $\tau$ with the result
\bigskip
\noindent {\it Lemma 3.} If $\int \vert \lb \vert^2 \, \om^{-2} \,
dk < \ify$ and if $\int \vert \lb \vert^2 \, \om^{-1} \, dk < \Si
- E_m$, then for sufficiently small $\d > 0$
$$
\build{\lim \inf}_{T \ra \ify}^{} (\psi_T , E([E_0 ,\Si -\d ]) \ot
P_{\Om} \, \psi_T ) > 0 \, . \eqno (2.10)
$$
We still have to convert the condition in Lemma 3 to a condition
on $\Si - E_0$. This can be achieved through a variational bound
on $E_m$.
\bigskip
\noindent {\it Lemma 4.} We have the bound
$$
E_m \leq E_0 - \int \vert \lb \vert^2 \, \left( \om + {1\over 2}
\, k^2 \right)^{-1} \, dk \, . \eqno (2.11)
$$
\bigskip
\noindent {\it Proof.} Let $P = \int dk \, k \ a^{\dagger} (k) \,
a(k)$ be the total momentum of the field. We choose as trial
function $\psi_f = \psi_0 (x) \, e^{ixP} \, W(f) \, \Om$ with
$f$ yet to be determined. Then
$$
\displaylinesno{
E_m \leq (\psi_f , H \, \psi_f) = E_0 + {1\over 2} \left( \int k
\vert f \vert^2 \, dk \right)^2 + \int \left( \om + {1\over 2} \,
k^2 \right) \, \vert f \vert^2 \, dk \, + \cr
\int (\lb f^* + \lb^* f) \, dk \, . &( 2.12) \cr
}
$$
The minimizing $f=\lb \left( \om + {1\over 2} \, k^2 \right)^{-1}$
yields the bound (2.11). \xx
\bigskip
Lemma 3 and Lemma 4 imply the existence of a ground state provided
$$
\Si - E_0 \geq \int \vert \lb \vert^2 \, \om^{-1} \, dk - \int
\vert \lb \vert^2 \, \left( \om + {1\over 2} \, k^2 \right)^{-1}
\eqno (2.13)
$$
as claimed in Theorem 1.
\bigskip
\noindent {\it Acknowledgement.} I thank V. Bach for illuminating
discussions.
\vglue 2cm
\noindent {\bf References}
\medskip
\item{[1]} M. Reed and B. Simon, Methods of Modern Mathematical
Physics : Analysis of Operators, Vol. 4. Academic Press, 1978.
\smallskip
\item{[2]} V. Bach, J. Fr\"ohlich and I. Sigal, Quantum
electrodynamics of confined non-relativistic particles. Preprint,
1996.
\smallskip
\item{[3]} H. Spohn, The effective mass of the polaron -- a
functional integral approach, {\it Annals of Physics} {\bf 175},
278-318 (1987).
\smallskip
\item{[4]} H. Spohn, Ground state(s) of the spin-boson
Hamiltonian, {\it Comm. Math. Phys.} {\bf 123}, 277-304 (1988).
\smallskip
\item{[5]} A. Arai and M. Hirokawa, On the existence and
uniqueness of ground states of a generalized spin-boson model.
Preprint, 1996.
\smallskip
\item{[6]} E. Nelson, Schr\"odinger particles interacting with a
quantized scalar field. Theory and Applications of Analysis in
Function Space, ed. by W.T. Martin and I. Segal. MIT Press,
Cambridge, Mass., 1964.
\smallskip
\item{[7]} H. Spohn, unpublished note, 1991.
\bye