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\topmatter
\title A Trace Formula for Multidimensional Schr\"odinger Operators
\endtitle
\rightheadtext{A Trace Formula for Multidimensional Schr\"odinger 
Operators}
\author F.~Gesztesy$^{1}$, H.~Holden$^{2}$, B.~Simon$^{3}$, and 
Z.~Zhao$^{1}$
\endauthor
\leftheadtext{F.~Gesztesy, H.~Holden, B.~Simon, and Z.~Zhao}
\thanks $^{1}$ Department of Mathematics, University of Missouri, 
Columbia, MO 65211.  E-mail for F.G.: mathfg\linebreak\@mizzou1.missouri.edu
\,E-mail for Z.Z.: mathzz\@mizzou1.missouri.edu
\endthanks
\thanks $^{2}$ Department of Mathematical Sciences, The Norwegian 
Institute of Technology, University of Trondheim, N-7034 Trondheim, 
Norway. E-mail: holden\@imf.unit.no
\endthanks
\thanks $^{3}$ Division of Physics, Mathematics, and Astronomy, 
California Institute of Technology, 253-37, \linebreak Pasadena, CA 
91125.  This material is based upon work supported by the National 
Science Foundation under Grant No.~DMS-9101715.  The Government has 
certain rights in this material.
\endthanks
\thanks To be submitted to {\it J.~Amer.~Math.~Soc.}
\endthanks
\endtopmatter
\bigpagebreak
\document
\block
\tenrm{\smc{Abstract.}} We prove multidimensional analogs of the trace formula 
obtained previously for one-dimensional Schr\"odinger operators.  For 
example, let $V$ be a continuous function on $[0, 1]^{\nu}\subset\Bbb 
R^{\nu}$.  For $A\subset\{1,\dots ,\nu\}$, let $-\Delta_{A}$ be the 
Laplace operator on $[0, 1]^{\nu}$ with mixed Dirichlet-Neumann 
boundary conditions
$$\alignat2
\varphi(x) &=0, &&\qquad x_{j}=0 \text{ or } x_{j}=1 \quad\text{for } 
j\in A, \\
\frac{\partial\varphi}{\partial x_{j}}(x) &= 0, &&\qquad x_{j}=0 
\text{ or } x_{j}=1 \quad\text{for } j\notin A.
\endalignat
$$
Let $|A|=$ number of points in $A$.  Then we'll prove that
$$
\text{Tr}\biggl(\sum_{A\subset\{1,\dots ,\nu\}} (-1)^{|A|} e^{-t(-
\Delta_{A}+V)}\biggr)=1-t\langle V\rangle +o(t) \quad\text{as }t\downarrow 0
$$
with $\langle V\rangle$ the average of $V$ at the $2^{\nu}$ corners of 
$[0, 1]^{\nu}$.
\endblock



\vskip 0.4in
\flushpar {\bf \S 1. Introduction}

This paper is devoted to extensions of the trace formula for the ODE 
$-\frac{d^2}{dx^2}+V(x)$ to the corresponding PDE, $-\Delta+V(x)$.  
The simplest of all the one-dimensional results is the trace formula 
[1,6] for the periodic case.  Suppose $V$ is a $C^1$ function on $\Bbb 
R$ obeying $V(x+1)=V(x)$.  Let $E_{1}<E_{4}\leq E_{5}<E_{8}\leq 
E_{9}<\cdots$ be the eigenvalues of the operator on $L^{2}([0, 1])$
$$
-\frac{d^2}{dx^2}+V \tag 1
$$
with periodic boundary conditions (call the operator $H^P$) and $E_{2}\leq 
E_{3}<E_{6}\leq E_{7}<\cdots$ the eigenvalues of (1) with antiperiodic 
boundary conditions (call the operator $H^A$).  Let 
$\mu_{1}(x)<\mu_{2}(x)<\cdots$ be the eigenvalues of the operator (1) 
on $L^{2}([x, x+1])$ with $u(x)=u(x+1)=0$ Dirichlet boundary 
conditions (call the operator $H^{D}_{x}$).  Then:
$$
V(x)=E_{0}+\sum^{\infty}_{n=1}\bigl[E_{2n}+E_{2n-1}-
2\mu_{n}(x)\bigr], \quad x\in [0, 1].
\tag 2
$$

One way (see, e.g., [8]) of proving (2) is to derive a heat kernel 
asymptotic relation
$$
\text{Tr}(e^{-tH^{P}}+e^{-tH^{A}}-2e^{-tH^{D}_{x}})=
1-tV(x)+o(t) \tag 3
$$
from which (2) follows from the known convergence of
$$
\sum^{\infty}_{n=1} |E_{2n}-E_{2n-1}|<\infty \tag 4
$$
and the relation
$$
E_{2n-1}\leq \mu_{n}(x)\leq E_{2n}.
$$

Equation (3) can be viewed as an Abelian summation method applied to 
(2) and holds even in cases where (4) diverges (e.g., if $V(x)=$ the 
characteristic function of\linebreak $\bigcup\limits^{\infty}_{n=-\infty}
[n-\frac14, n+\frac14]$).

Recently, we have described versions of (3) for arbitrary, 
not necessarily periodic, $V$'s by using other boundary conditions, see [3,2].  
For example, if $H$ is (1) on all of $(-\infty, \infty)$ and 
$H^{D}_{x}$ is (1) on $L^{2}(-\infty, x)\oplus L^{2}(x, \infty)$ with 
$u(x)=0$ Dirichlet boundary conditions, then we proved that
$$
2\text{Tr}\bigl(e^{-tH}-e^{-tH^{D}_{x}}\bigr)=1-tV(x)+o(t), \quad 
x\in\Bbb R \tag 5
$$
so as long as $x$ is a point of Lebesgue continuity of $V$ and $V$ is 
bounded from below and locally $L^1$.

One of our main accomplishments here is to extend (5) to higher 
dimensions. Let $H^{N}_{x}$ denote the operator with Neumann boundary 
conditions and suppose $V$ is even about $x$, that is, 
$V(x-y)=V(x+y)$, $y\in\Bbb R$.  Then
$$
e^{-tH^{Q}_{x}}(x+y, x+z)=e^{-tH}(x+y, x+z)\pm e^{-tH}(x+y, x-z)
$$
if $yz>0$ and the $-$ is used for $Q=D$ and $+$ for $Q=N$.  (Here and in
the remainder of this paper $e^{-tH}(x, y)$, $t>0$ denotes the integral
kernel of the semigroup $e^{-tH}$.)  From this it follows that 
$\text{Tr}(e^{-tH}-e^{-tH^{D}_{x}})=\text{Tr}(e^{-tH^{N}_{x}}-e^{-tH})$ 
and so (5) becomes
$$
\text{Tr}\bigl(e^{-tH^{N}_{x}}-e^{-tH^{D}_{x}}\bigr)=1-tV(x)+o(t),
\quad x\in\Bbb R. \tag 6
$$
But it is easy to see that (6) for $V$ even about $x$ implies it for 
arbitrary $V$ since the operators break up into direct sums (see Lemma 
2.1 below).

It is (6) that we will generalize to $\nu$ dimensions.  Explicitly, 
given $A\subset\{1,\dots ,\nu\}$, let $H_{A; x}$ be defined as 
follows: Let $B_{\alpha}$, $\alpha\subset\{1,\dots,\nu\}$ be the $2^\nu$ 
blocks obtained by removing the hyperplanes $P^{(x)}_{j}:=\{y\in
\Bbb R^{\nu}\mid y_{j}=x_{j}\}$ from $\Bbb R^{\nu}$, that is, 
$B_{\alpha}=\{x\in\Bbb R^{\nu}\mid x_{i}>0\text{ if } i\in\alpha,\, 
x_{i}<0\text{ if } i\notin\alpha\}$. $H_{A; x}$ is then 
defined to be the operator on $\oplus L^{2}(B_{\alpha})$ with Dirichlet 
boundary conditions on $\{P^{(x)}_{j}\}_{j\in A}$ and Neumann boundary 
conditions on $\{P^{(x)}_{j}\}_{j\notin A}$. 

Explicitly, for each $\alpha=1,\dots,2^{\nu}$, let $\Cal 
D^{(A)}_{\alpha; x}$ be the set of functions, $\varphi$, on $B_{\alpha}$ 
which are $C^{\infty}$ on $B^{\text{int}}_{\alpha}$, with 
derivatives continuous up to $\partial B_{\alpha}$ with bounded 
support and which obey the boundary conditions:
$$\alignat2
\varphi(y) &=0, &&\qquad y\in P^{(x)}_{j} \quad\text{for } j\in A, \\
\frac{\partial\varphi}{\partial y_{j}}(y) &= 0, &&\qquad y\in P^{(x)}_{j}
\quad\text{for } j\notin A.
\endalignat
$$
Obviously, $\Cal D^{(A)}_{\alpha; x}$ is dense in $L^{2}(B_{\alpha})$.  
Then $\varphi\mapsto -\triangle\varphi$ is essentially self-adjoint on 
$\Cal D^{(A)}_{\alpha, x}$, $-\triangle_{A}$ is the direct sum 
of these operators on $\oplus L^{2}(B_{\alpha})$, and $H_{A; x}=-
\triangle_{A}\dot{+}V$ as a form sum.

We will prove (see Theorem 4.1) that
$$
\text{Tr}\biggl(\sum_{A\subset\{1,\dots ,\nu\}}(-1)^{|A|}\exp (-tH_{A; 
x})\biggr)=1-tV(x)+o(t), \quad x\in\Bbb R^{\nu}. \tag 7
$$
(Note that for $A=\{1,\dots,\nu\}$ resp.~$A=\emptyset$, $H_{A; x}$ has 
exclusively Dirichlet resp.~Neumann boundary conditions on the 
hyperplanes $P^{(x)}_{j}$, $1\leq j\leq\nu$.)

The paper is laid out as follows.  In \S 2, we'll use the method of 
images to reduce (7) to the study of integrals of the form
$$
2^{\nu}\int\limits_{\Bbb R^{\nu}} \exp (-tH)(y, -y)\, d^{\nu}y, \tag 8
$$
where $H=-\triangle+V$ is a Schr\"odinger operator in $L^{2}(\Bbb R^{\nu})$ 
without any boundary conditions.  In \S 3, we introduce a Gaussian process 
that provides a concise Feynman-Kac type formula for integrals of the form 
(8), and we'll prove (7) in \S 4.  We'll discuss the periodic version of (7) 
in \S 5, then prove an abelianized version of a recent conjecture of Lax [5] 
that motivated our work. Finally, in \S 6, we'll make a few remarks 
on the issue of going beyond Abelian sums in the periodic case.

\bigpagebreak
\flushpar {\bf \S 2. The Method of Images}

We begin with some small arguments to simplify notation and some later 
details.  First, without loss, we'll suppose $x=0$, that is, all 
boundary conditions are on planes through $0$ and the final formula is 
for $V(0)$.  Let then $P_j$ be the coordinate plane $\{x\in\Bbb R^{\nu}\mid 
x_{j}=0\}$ and let $H_{A}\equiv H_{A; 0}$.  Let $\pi_{j}$ be the 
reflection in $P_j$, that is,
$$\alignat2
(\pi_{j}x)_{i} &=x_{i}, &&\qquad i\neq j \\
&=-x_{j}, &&\qquad i=j.
\endalignat
$$
Call $V$ symmetric if and only if $V\circ\pi_{j}=V$ for all $j$.  For 
$\alpha\subset\{1, \dots ,\nu\}$, let $B_{\alpha}$ be the blocks introduced 
previously.  Let $V_\alpha$ be that symmetric function with $V_{\alpha}=V$ on 
$B_\alpha$.  Because of symmetry of $V_{\alpha}$ for each $A$, $\exp(-
t(-\Delta_{A}+V_{\alpha}))$ is a direct sum of $2^{\nu}$ pieces (acting on 
the different $L^{2}(B_{\beta})$) and each of the pieces is unitarily 
equivalent to a single operator, $P_{A; \alpha, t}$.  On the other 
hand, $\exp(-tH_{A})$ is also a direct sum, clearly unitarily 
equivalent to $\operatornamewithlimits{\oplus}
\limits_{\alpha\subset\{1,\dots , \nu\}} P_{A; \alpha, t}$.  It 
follows that:

\proclaim{Lemma 2.1}  To prove \rom{(7)}, it suffices to suppose $x=0$ 
and that $V$ is symmetric.
\endproclaim

So, henceforth, we can suppose that $V$ is symmetric, which we do in 
Lemma 2.3 and Theorems 2.4/2.5 below.

We are interested in writing the heat kernel for $H_A$ using the method of 
images.  Some group theoretic notation will be useful.  $\Cal P_{\nu}$ 
is the set of subsets of $\{1, \dots, \nu\}$ which forms a group under 
$(A, B)\mapsto A\bigtriangleup B$, the symmetric difference.  The identity 
is $\emptyset$, the empty set.  As a finite abelian group, $\Cal 
P_{\nu}$ is its own dual. The character $\chi_{A}$ associated to 
$A\in\Cal P_{\nu}$ acts by
$$
\chi_{A}(B)=(-1)^{|A\cap B|}. \tag 9
$$
In particular, orthogonality of characters implies

\proclaim{Lemma 2.2} $$\sum_{B\in\Cal P_{\nu}} 
\chi_{A}(B)\chi_{C}(B)=2^{\nu}\delta_{AC}.$$
\endproclaim

Given $B\in\Cal P_{\nu}$, define the reflection $R_{B}$ acting on 
$\Bbb R^{\nu}$ by
$$
R_{B}=\prod\limits_{j\in B} \pi_{j}.
$$
The method of images formula then says

\proclaim{Lemma 2.3}  If $x, y$ are in the same orthant, then
$$
\exp(-tH_{A})(x, y)=\sum_{B\in\Cal P_{\nu}}\chi_{B}(A)\exp(-tH)(x, 
R_{B}y)
$$
and the integral kernel is zero if $x, y$ are in different orthants.
\endproclaim

This immediately yields:

\proclaim{Theorem 2.4}  Let $C_{t}=\sum\limits_{A\in\Cal P_{\nu}}
(-1)^{|A|} \exp(-tH_{A})$.  Then the integral kernel of $C_{t}$ is
$$
C_{t}(x, y)=\cases 2^{\nu}\exp(-tH)(x, -y), & \,x, y \text{\rom{ in the same 
orthant}} \\
0, & \,x, y \text{\rom{ in different orthants}}.
\endcases
$$
\endproclaim

\demo{Proof}  Let $B_{0}=\{1, \dots, \nu\}$ so $R_{B_0}y=-y$ and 
$\chi_{B_0}(A)=(-1)^{|A|}$ so if $x, y$ lie in the same orthant:
$$\alignat2
C_{t}(x, y) &=\sum_{A\in\Cal P_{\nu}}\chi_{B_0}(A)\exp(-tH_{A})(x, y)\\
&=\sum\Sb A\in\Cal P_{\nu} \\ B\in\Cal P_{\nu} \endSb \chi_{B_0}(A)
\chi_{B}(A)\exp(-tH_{A})(x, R_{B}y)&&\qquad \text{(by Lemma 2.3)} \\
&=2^{\nu}\sum_{B\in\Cal P_{\nu}} \delta_{B_{0}B} \exp(-tH)(x, R_{B}y) &&\qquad
\text{(by Lemma 2.2)} \\
&= 2^{\nu} \exp(-tH)(x, -y). \qed
\endalignat
$$
\enddemo

\proclaim{Theorem 2.5}  Let $V$ be bounded below and $t>0$. Then the 
operator $C_t$ of Theorem \rom{2.4} is a trace class opertor in 
$L^{2}(\Bbb R^{\nu})$ and
$$
\text{\rom{Tr}}(C_{t})=2^{\nu}\int\limits_{\Bbb R^{\nu}} \exp(-tH)(x, -x)\, 
d^{\nu}x.
$$
\endproclaim

\demo{Proof} Let $S_{t}=\exp(-tH)$.  For $\alpha\in\tilde{\Bbb 
Z}^{\nu}\equiv\Bbb Z^{\nu}+(\frac12, \dots, \frac12)$ let 
$\chi_{\alpha}$ be the characteristic function of $\{x\mid |x_{i}-
\alpha_{i}| <\frac12,\text{ all }i\}$, and let $P_\alpha$ be the projection 
which is multiplication by $\chi_\alpha$.  Let $\tilde{\Bbb 
Z}^{\nu}_{+}=\{\alpha\in\tilde{\Bbb Z}^{\nu}\mid \alpha_{i}>0\}$.

$C_t$ is a direct sum of $2^\nu$ operators, each unitarily equivalent 
to $\tilde C_{t}:= C_{t} \restriction\{x\mid x_{i}>0\}$.  Let 
$R_{B_0}$ be the reflection $x\to -x$.  Then by Theorem 2.4:
$$
\tilde C_{t}=2^{\nu}\sum_{\alpha, \beta\in\tilde{\Bbb Z}^{\nu}_{+}}
P_{\alpha} R_{B_0} S_{t} P_{\beta}.
$$
By the lemma below, $P_{\alpha}R_{B_0}S_{t}P_{\beta}$ is trace class 
with trace norm bound by \linebreak$C_{1}\exp(-C_{2}|\alpha+\beta|)$ 
and trace given by the integral of the diagonal integral kernel.  Since 
$\sum\limits_{\alpha, \beta\in\tilde{\Bbb Z}^{\nu}_{+}} \exp(-C_{2}
|\alpha +\beta|)<\infty$, the result follows. \qed
\enddemo

\proclaim{Lemma 2.6} In the notation of the last proof, 
$P_{\alpha} R_{B_0} S_{t} P_{\beta}$ is trace class with trace norm 
bounded by $C_{1}\exp(-C_{2}|\alpha+\beta|)$ and trace given by the 
integral of the diagonal of the \rom(continuous\rom) integral kernel.
\endproclaim

\demo{Proof}  $P_{\alpha} R_{B_0} S_{t} P_{\beta}=\sum
\limits_{\gamma\in\tilde{\Bbb Z}^{\nu}} R_{B_0} (P_{-\alpha} S_{t/2} 
P_{\gamma})(P_{\gamma} S_{t/2} P_{\beta})$.  By a standard estimate 
(see, e.g., [9]):
$$
|S_{u}(x, y)|\leq C_{1, u}\exp(-C_{2, u}|x-y|)
$$
(of course one can even have $|x-y|^{2}$ but we don't need that), so 
by integrating the square of the integral kernel:
$$
\|P_{\alpha} S_{t/2} P_{\gamma}\|_{2} \leq C_{3}\exp(-C_{2}|\alpha-
\gamma|),
$$
where $\|\,\cdot\,\|_{2}$ is the Hilbert-Schmidt norm.  Summing over $\gamma$, 
we obtain the trace class result and bound since
$$
\sum_{\gamma\in\tilde{\Bbb Z}^{\nu}} \exp(-C_{2}|\alpha+\gamma|) 
\exp(-C_{2}|\beta-\gamma|) \leq C_{5} \exp(-C_{4}|\alpha+\beta|).
$$
Since the trace of a product of Hilbert-Schmidt operators is given by 
the integral of the diagonal integral kernel, we obtain the trace 
result. \qed
\enddemo

\bigpagebreak
\flushpar {\bf \S 3. A Gaussian Process}

In this section, we present a Feynman-Kac type formula for 
$\text{Tr}(C_{t})$ where $C_t$ is the operator of Theorem 2.4.  If $V$ 
is bounded, (7) is an immediate consequence of this formula.  For $V$ 
unbounded (from above) at infinity, we will need an additional 
estimate on the Gaussian process, $\bold L(t)$, used in this formula and 
that estimate appears at the end of this section.  (We shall employ 
the notation used in [8], i.e., $E(f)=\int\limits_{\Omega}f\,d\mu$, 
$E(A)=\int\limits_{A}d\mu=\mu(A)$, $E(f; A)=\int\limits_{A}f\,d\mu$, 
etc., where $(\Omega, \Cal F, \mu)$ denotes a probability space, 
$A\in\Cal F$, $f:\Omega\to\Bbb R$ is $\Cal F$-measurable.)

All Gaussian processes considered in this paper have mean zero and we 
will suppose that without explicitly saying it each time.  Recall [8] 
that the Brownian bridge $\{\alpha(s)\}_{0\leq s\leq 1}$ is the 
Gaussian process with covariance:
$$\align
E_{\alpha}(\alpha (s)\alpha (t)) &=\min(s, t)(1-\max(s, t)) \\
&=\frac12 (s+t-|s-t|)-st. \tag 10
\endalign
$$
If $b(s)$ is Brownian motion, then $\alpha (s)=b(s)-sb(1)$ is an 
explicit realization of the Brownian bridge.  The $\nu$-dimensional 
Brownian bridge is $\nu$ independent copies of $\alpha (s)$ thought of 
as a vector valued process and one still has for the $\nu$-dimensional 
objects
$$
\boldsymbol\alpha (s)=\bold b (s)-s \bold b (1). \tag 11
$$
Let $\bold g_{\bold x, \bold y} (s)=s \bold x +(1-s) \bold y$ be the straight 
line from $\bold x$ to $\bold y$.  Then, [8] shows that for any $V$ 
bounded from below (and locally bounded from above, say), if $H=-
\Delta+V$ in $L^{2}(\Bbb R^{\nu})$,
$$
\exp(-tH)(\bold x, \bold y)=\exp(t\Delta)(\bold x, \bold y)E_{\alpha}\biggl(
\exp\biggl(-\int\limits^{t}_{0} V\biggl(\bold g_{\bold x, \bold y} 
\biggl(\frac{s}{t}\biggr)+\sqrt{2t}\,\boldsymbol\alpha 
\biggl(\frac{s}{t}\biggr) \biggr)\, ds \biggr)\biggr). \tag 12
$$

We have $\sqrt{2t}$ rather than the $\sqrt{t}$ on pg.~54 of [8] 
because we use $-\Delta$ rather than $-\frac12\Delta$.  Plugging (12) 
into Theorem 2.5 we find that
$$
\text{Tr}(C_{t})=\int\limits_{\Bbb R^{\nu}} d^{\nu}x N_{t}(\bold x) 
E_{\alpha}\biggl(\exp \biggl(-\int\limits^{t}_{0} V
\biggl(\bold g_{\bold x, - \bold x} \biggl(\frac{s}{t}\biggr) +\sqrt{2t}\, 
\boldsymbol\alpha\biggl(\frac{s}{t}\biggr)\biggr)\, ds \biggr)\biggr),
$$
where $N_{t}(\bold x)=2^{\nu}\exp(t\Delta)(\bold x, -\bold x)=2^{\nu}
\exp(t\Delta)(2\bold x, 0)\equiv\prod\limits^{\nu}_{i=1} \tilde N_{t}(x_{i})$.  
Notice that $\tilde{N_{t}}(x_{i})$ is a Gaussian probability distribution 
with variance $\langle x^{2}_{t,i}\rangle=\frac{2t}{(2)^{2}}=
\frac{t}{2}=\frac14 (\sqrt{2t})^{2}$, so if we let $x_{0,i}$ be a Gaussian 
variable of variance $\langle x^{2}_{0,i}\rangle =\frac14$, then 
$x_{t,i}=\sqrt{2t}\,x_{0,i}$.  Note that $g_{\bold x, -\bold x}(\frac{s}{t})=
\sqrt{2t}\,\bold x_{0}(2(\frac{s}{t})-1)$.

This suggests we define a new process
$$
\bold L(s)=\bold x_{0} (2s-1)+\boldsymbol\alpha(s), \qquad 0\leq s\leq 1, 
\tag 13
$$
where the components of $\bold x_0$ are independent Gaussian variables with
$\langle x^{2}_{0, i}\rangle=\frac14$ and independent of $\boldsymbol\alpha$ 
so
$$\align
E(\bold L_{i}(s) \bold L_{j}(w)) 
&=E_{\alpha}(\boldsymbol\alpha_{i} (s)\boldsymbol\alpha_{j} (w))
+\frac{\delta_{ij}}{4} (2s-1)(2w-1)  \\
&=\delta_{ij}\biggl[\frac14 - \frac12 |s-w|\biggr]. \tag 14
\endalign
$$

We have thus proven that:

\proclaim{Theorem 3.1}  Let $\bold L$ be the Gaussian process with 
covariance \rom{(14)}.  Then \rom(with $C_t$ given by Theorem 
\rom{2.4)}:
$$
\text{\rom{Tr}}(C_{t})=E\biggl(\exp\biggl(-\int\limits^{t}_{0}
V\biggl(\sqrt{2t}\, \bold L\biggl(\frac{s}{t}\biggr)\biggr)\, ds \biggr)
\biggr).
$$
\endproclaim

We need the following estimate on $\bold L$:

\proclaim{Theorem 3.2}  $E\bigl(\bigl\{\bigl[\sup\limits_{0\leq s\leq 1}
|\bold L(s)|\bigr]\geq a\bigr\}\bigr)\leq C_{1}\exp(-C_{2}a^{2})$ 
for some $C_{1}, C_{2} >0$.
\endproclaim

\demo{Proof}  By the realizations (13) and (11), $\bold L(s)=\bold b(s)-
s\bold b(1)+\bold x_{0}(2s-1)$ so
$$
\sup\limits_{0\leq s\leq 1} |\bold L(s)| \leq |\bold x_{0}| + |\bold b(1)| +
\sup\limits_{0\leq s\leq 1} |\bold b(s)| \tag 15
$$
and for the $\sup\limits_{0\leq s\leq 1} |\bold L(s)|$ to be larger than $a$, 
one or more of the three terms on the right side of (15) must be larger than 
$a/3$.  Each has a Gaussian bound since $\bold x_0$ and $\bold b(1)$ are 
Gaussian and $\sup\limits_{0\leq s \leq 1}|\bold b(s)|$ has a Levy inequality 
estimate (see [8], pp.~64 ff). \qed
\enddemo

\remark{Remarks}  1.  Each component of $\bold L(t)$ is an independent 
copy of the one-dimensional $L(t)$.  $L(t)$ is intimately related to 
the xi process, $\omega$, we introduced in [2]; namely
$$\alignat2
L(t) &= \omega(t), &&\qquad t\leq T_{\omega} \\
&= -\omega(t), &&\qquad t\geq T_{\omega},
\endalignat
$$
where $T_\omega$ is the first time that $\omega(t)=0$, that is, $\omega$ 
and $L$ are related by reflection at a first hitting time.  Theorem 3.2 is 
thus another proof of the estimate we proved on the xi process in [2].

2.  The covariance (14) associated with $L(t)$ is just the zero energy 
Green's function for $-\frac{d^2}{dx^2}$ on $L^{2}([0, 1])$ with 
antiperiodic boundary conditions, just as (10) is the Dirichlet Green's 
function.  Notice that $L(1)=-L(0)$ is related to the antiperiodicity.
\endremark

\bigpagebreak
\flushpar {\bf \S 4. The Main Result in Unbounded Space}

Given Theorems 2.5, 3.1, and 3.2, the proof of (7) is easy following 
the methods of [2,8]:

\proclaim{Theorem 4.1} Let $V$ be bounded from below and continuous on 
$\Bbb R^{\nu}$.  Let $C_{t}=\sum\limits_{A\in\Cal P_{\nu}} 
(-1)^{|A|}\mathbreak\exp(-tH_{A})$, $t>0$.  Then $C_t$ is a trace 
class operator in $L^{2}(\Bbb R^{\nu})$ and
$$
\text{\rom{Tr}}(C_{t})=1-tV(0)+o(t)\qquad\text{as } t\downarrow 0.
$$
\endproclaim

\demo{Proof}  As noted in Lemma 2.1, we can suppose that $V$ is symmetric.  
By Theorems 2.5 and 3.1:
$$\align
\text{Tr}(C_{t}) &= E\biggl(\exp\biggl(-\int\limits^{t}_{0}V\biggl(
\sqrt{2t}\, L \biggl(\frac{s}{t}\biggr)\biggr) \, ds \biggr)\biggr) \\
&=T_{1}(V)+T_{2}(V),
\endalign
$$
where
$$
T_{i}(V)=E\biggl(\chi_{i,t}(L)\exp\biggl(-\int\limits^{t}_{0} V
\biggl(\sqrt{2t}\, L\biggl(\frac{s}{t}\biggr)\biggr)\, ds \biggr)\biggr)
$$
and $\chi_{1,t}$ is the characteristic function of 
$\{L\mid\sup\limits_{0\leq s\leq 1} |L(s)|<t^{-1/3}\}$ and $\chi_{2,t}=
1-\chi_{1,t}$.

By Theorem 3.2, $|T_{2}(V)|\leq C_{1}\exp(-t\,\inf|V(x)|)\exp(-
C_{2}/t^{2/3})=o(t)$ and $E(\chi_{1,t}(L))\mathbreak=1+o(t)$, thus
$$\align
\lim\limits_{t\downarrow 0} (T_{1}(V)-1)/t &=\lim\limits_{t\downarrow 
0} E\biggl(\chi_{1,t}(L)t^{-1} \biggl\{\exp\biggl(-\int\limits^{t}_{0} V
\biggl(\sqrt{2t}\, L\biggl(\frac{s}{t}\biggr)\biggr)\, ds -1\biggr\}
\biggr) \\
&= V(0)
\endalign
$$
by continuity and dominated convergence (since $V$ is bounded near 0). 
\qed
\enddemo

\remark{Remark}  Because we use path integral estimates, continuity of 
$V$ is not needed.  One only needs conditions that can be stated in 
terms of the Kato class $K_{\nu}$ defined as
$$
K_{\nu}=\biggl\{V\mid\lim\limits_{\beta\downarrow 0}\,\biggl[\sup\limits_{x
\in\Bbb R^{\nu}}\int\limits_{|x-y|\leq\beta} |x-y|^{-(\nu -2)} |V(y)|\, 
d^{\nu}y=0\biggr]\biggr\}
$$
(if $\nu=2$, $|x-y|^{-(\nu -2)}$ is replaced by $\ln(|x-y|^{-1})$) and 
in terms of the Stummel class $M_{2-\alpha, 1}$ (see, e.g., [9]):
$$
M_{2-\alpha, 1}=\biggl\{V\mid \sup\limits_{x\in\Bbb R^{\nu}} 
\int\limits_{|x-y|\leq 1} |x-y|^{-(\nu-2)-\alpha}
|V(y)|\, d^{\nu}y <\infty\biggr\}.
$$
By using the methods of [2], one can prove Theorem 4.1 if
\roster
\item"{(a)}" $\min(V, 0)$ is in the Kato class $K_{\nu}$. 
\item"{(b)}" $V$ is in a local Stummel class $M_{2-\alpha,1}$ for 
some $\alpha >0$ (i.e., $V\chi_{R}\in M_{2-\alpha, 1}$ for all $R>0$, 
where $\chi_{R}$ denotes the characteristic function of the ball 
$\{x\in\Bbb R^{\nu}\mid |x|\leq R\}$).
\item"{(c)}" 0 is a point of Lebesgue continuity for $V$ for averaging 
over balls shrinking to zero.
\endroster
\endremark

\medpagebreak
Just as we could write the one-dimensional trace formula as either (5) 
or (6), in low dimension one can use the method of images to write $C_t$ in 
alternate ways that avoid mixed boundary conditions.  For example, in 
two dimensions, let $H^D$ be what we called $H_{A=\{1, 2\}}$ resp. 
$H^{N}=H_{A=\emptyset}$ correspond to Dirichlet resp.~Neumann 
boundary conditions on both axes.  Then by the method of images (i.e., 
Lemma 2.3):
$$
e^{-tH^{D}}(x, x)+e^{-tH^{N}}(x, x)=2e^{-tH}(x, x)+2e^{-tH}(x, -x),
\qquad x\in\Bbb R^{2}
$$
so we have

\proclaim{Proposition 4.2}  In two dimensions
$$
\text{\rom{Tr}}(C_{t})=2\int\limits_{\Bbb R^2}\bigl[e^{-tH^{D}}+
e^{-tH^{N}}-2e^{-tH}\bigr] (x, x)\, d^{2}x.
$$
\endproclaim

\remark{Remark}  Note we have not stated in the proposition that $[\dots]$ 
in the last integral is trace class because it is not in general.  For 
example, if $V=0$, it is not even Hilbert-Schmidt because of the 
contribution of the integral kernel $-2e^{-tH}(x, y)$ with 
$x=(u_{1}$, $v_{1})$, $y=(-u_{2}$, $v_{2})$, $0<u_{i}<1$, 
$0<v_{i}<\infty$, $|v_{1}-v_{2}|<1$.
\endremark

\remark{Remark}  There are also results for Dirichlet conditions only.  
Explicitly, for $A\subset \{1,\dots,\nu\}$, let $\tilde H_{A; x}$ be 
the operator with Dirichlet boundary conditions on the planes 
$P^{(x)}_{j}$ with $j\in A$ but no conditions on the planes with 
$j\notin A$ (i.e., free boundary conditions, so, e.g., if 
$A=\emptyset$, the $\tilde H_{A; x}$ is just $-\triangle +V$).  Then 
we can show that if
$$
B_{t}=\sum_{A\subset\{1,\dots,\nu\}} (-1)^{|A|} \exp (-t\tilde{H}_{A; 
x})
$$
has a continuous integral kernel, then
$$
\int\limits_{\Bbb R^{\nu}} B_{t}(y, y)\, d^{\nu}y=2^{-\nu} [1-
tV(x)+o(t)].
$$
This generalizes (5). We believe that $B_{t}$ is always trace class 
but have only proven that if $V$is symmetric under reflection in each 
of the planes $P^{(x)}_{j}$.
\endremark

\bigpagebreak
\flushpar {\bf \S 5.  The Main Result in a Box}

Our main goal here is to prove:

\proclaim{Theorem 5.1}  Let $V$ be continuous on $[0, 1]^{\nu}$.  For 
$A\subset\{1, \dots, \nu\}$, let $H_A$ be $-\Delta+V$ on $L^{2}([0, 
1]^{\nu})$ with Dirichlet boundary condition on the hyperplanes with 
$x_{j}=0$ or $x_{j}=1$ and $j\in A$ and Neumann boundary condition on 
the hyperplanes with $x_{j}=0$ or $x_{j}=1$ and $j\notin A$.  Let 
$\langle V\rangle$ be the average of $V$ at the $2^\nu$ corners of 
$[0, 1]^{\nu}$.  Then
$$
\sum_{A\subset\{1, \dots, \nu\}} (-1)^{|A|}\text{\rom{Tr}}(e^{-tH_{A}})=
1-t\langle V\rangle +o(t)\qquad\text{as } t\downarrow 0.
$$
\endproclaim

\remark{Remarks} 1.  We take a unit cube for notational simplicity 
only.  The result holds without change on 
$\operatornamewithlimits{\times}\limits^{\nu}_{i=1}[a_{i}, b_{i}]$.  But 
rectangular symmetry is critical.  It may be possible to extend the 
result to the unit cells for some other space groups with enough 
reflections.

2.  If $V$ is continuous and periodic with period one in each 
direction, then of course $\langle V\rangle =V(0)$.

3.  All one needs is that $V$ lies in the Kato class and suitable 
Lebesgue continuity of $V$ at each corner.
\endremark

\medpagebreak

In the group theoretical part of the proof we gave of (7), the key was 
that $\Cal P_{\nu}$ was the group generated by the $\pi_{j}$'s.  Let 
$\rho_{j}$ be the reflection in $x_{j}=1$, that is,
$$\alignat2
[\rho_{j}(x)]_{i} &= x_{i}, &&\qquad i=j \\
&=2-x_{j}, &&\qquad i\neq j.
\endalignat
$$
Let $\Cal G_{\nu}$ be the group of actions on $\Bbb R^{\nu}$ generated by 
the $\pi_{j}$'s and $\rho_{j}$'s.  Then it is easy to see that $V$ has a 
unique extension to $\Bbb R^{\nu}$ which is $\Cal G_{\nu}$ invariant and 
this extension, which we'll also call $V$, is continuous on 
$\Bbb R^{\nu}$.  Let $H=-\Delta +V$ on $L^{2}(\Bbb R^{\nu})$.

$\Cal G_{\nu}$ is a semidirect product $2\Bbb Z^{\nu}\circledS\Cal 
P_{\nu}$, so every $G\in\Cal G_{\nu}$ can be written as $G=(\bold a, B)$ 
where $\frac{1}{2}\bold a\in\Bbb Z^{\nu}$ and $G$ acts by
$$
Gx=R_{B}x+a.
$$
Define for $A\in\Cal P^{\nu}$,
$$
\chi_{A}(G)=(-1)^{|A\cap B|}.
$$

Then the method of images formula for $e^{-tH_{A}}$ is:

\proclaim{Proposition 5.2}  If $x, y\in [0, 1]^{\nu}$, $A\in\Cal P_{\nu}$, then
$$
e^{-tH_{A}}(x, y)=\sum_{G\in\Cal G_{\nu}} \chi_{A}(G) e^{-tH}(x, Gy).
$$
\endproclaim

As in \S 2, let $R_{B_0}$ be the inversion in 0, that is, $R_{B_0}x=-x$.  
Then Lemma 2.2 implies
$$
\sum_{A\in\Cal P_{\nu}}(-1)^{|A|}\chi_{A}(G)=\cases 2^{\nu} &\text{if 
$G=(\bold a, B_{0})$} \\
0 &\text{if $G=(\bold a, B)$ with $B\neq B_{0}$}
\endcases
$$
and we conclude that

\proclaim{Proposition 5.3}  $$\sum_{A\in\Cal P_{\nu}} (-1)^{|A|}
\text{\rom{Tr}}(e^{-tH_{A}})=\sum_{a\in\Bbb Z^{\nu}} M(a, t),$$
where
$$
M(a, t)\equiv 2^{\nu}\int\limits_{[0, 1]^{\nu}} e^{-tH}(x, 2a-
x)\, d^{\nu}x.
$$
\endproclaim

With these preliminaries we are ready for the

\demo{Proof of Theorem \rom{5.1}}  Let $Q$ be the set of $2^{\nu}$ corners 
of $[0, 1]^{\nu}$ as points in $\Bbb Z^{\nu}$.  If $a\in\Bbb 
Z^{\nu}\backslash Q$, then
$$
\min\limits_{x\in [0, 1]^{\nu}} \, \text{dist}(x, 2a-x) =2\min\limits_
{x\in [0, 1]^{\nu}} \|x-a\|\geq 2. \tag 16
$$
Since (see, e.g., [9])
$$
|e^{-tH}(x, y)|\leq C_{1} \exp(-C_{2}(x-y)^{2}/t), \tag 17
$$
(16) implies
$$
\sum_{a\in\Bbb Z^{\nu}\backslash Q} M(a, t)=O(e^{-c/t})
$$
so by Proposition 5.3, it suffices to prove that for $a\in Q$:
$$
M(a, t)=2^{-\nu}[1-V(a)t+o(t)]. \tag 18
$$

By translation and reflection symmetry, we need only prove (18) for 
$a=0$.  But by (17):
$$\align
M(0, t) &=2^{\nu}\int\limits_{x_{i}\geq 0} e^{-tH}(x, -x)\, d^{\nu}x 
+O(e^{-c/t}) \\
&=\int\limits_{\Bbb R^{\nu}} e^{-tH}(x, -x)\, d^{\nu}x +O(e^{-c/t})
\endalign
$$
so (18) is precisely (7) given Theorem 2.5. \qed
\enddemo

\remark{Remark}  Because $V$ is bounded, we do not need the estimate 
in Theorem 3.2 to prove Theorem 5.1.
\endremark

The proof of Theorem 5.1 shows:

\proclaim{Theorem 5.4}  Let $V$ be continuous on $[0, 1]^{\nu}$.  For 
$A\subset\{1, \dots, \nu\}$, let $\tilde H_{A}$ be $-\Delta +V$ on 
$L^{2}([0, 1]^{\nu})$ with Dirichlet boundary conditions on the 
hyperplanes with $x_{j}=0$ for $j\in A$ and Neumann boundary 
conditions on the hyperplanes with $x_{j}=0$ for $j\notin A$ or $x_{k}=1$ 
for all $k\in\{1, \dots, \nu\}$.  Then
$$
\sum_{A\subset\{1, \dots, \nu\}} (-1)^{|A|}\text{\rom{Tr}}(e^{-t\tilde 
H_{A}})=2^{-\nu}[1-tV(0)+o(t)]\qquad\text{as } t\downarrow 0.
$$
\endproclaim

\demo{Proof} We still use the same group $\Cal G_{\nu}$.  Define with 
$G=(\bold a, B)\in\Cal G_{\nu}$
$$
\tilde{\chi}_{A}(G)=(-1)^{|A\cap B|} (-1)^{(1/2)\operatornamewithlimits
{\Sigma}\limits_{j\in A}|a_{j}|}.
$$   
$\Cal G_{\nu}$ is generated by $\{\pi_{j}\}\cup\{\rho_{j}\}$ and
$$\alignat2
\tilde{\chi}_{A}(\pi_{j}) &= -1\quad (\text{resp.~$1$}) &&\qquad 
\text{for } j\in A \quad (\text{resp.~$j\notin A$}) \\
\tilde{\chi}_{A}(\rho_{j})&= 1 &&\qquad \text{for all } j=1,\dots,\nu
\endalignat
$$
and $\tilde{\chi}$ is a character.  Hence we have our method of images 
formula
$$
e^{-t\tilde{H}_{A}}(x, y)=\sum\limits_{G\in\Cal G_{\nu}} 
\tilde{\chi}_{A} (G)\,e^{-tH}(x, Gy)
$$
with $H$ associated to the $G$ invariant extension of $V$ as in 
Proposition 5.2.

Note that for $G\in\Cal G_{\nu}$ fixed, $A\mapsto\tilde{\chi}_{A}(G)$ 
is a character on $\Cal P_{\nu}$.  By the orthogonality of characters
$$
\sum\limits_{A\in\Cal P_{\nu}}(-1)^{|A|}\tilde{\chi}_{A}(G) = \cases
2^{\nu} \quad &\text{if } \tilde{\chi}_{A}(G)\equiv (-1)^{A} \\
0 \quad &\text{if }\tilde{\chi}_{A}(G)\not\equiv (-1)^{|A|}. 
\endcases
$$

But $\tilde{\chi}_{A}(G)\equiv (-1)^{|A|}$ if and only if $G=(\bold a, 
B)$ with $B=B_{0}$ and each $\frac{1}{2}a_{i}$ is even. Thus, as in the 
proof of Proposition 5.3,
$$
\sum\limits_{A\in\Cal P_{\nu}}(-1)^{|A|}\text{Tr}(e^{-t\tilde H_{A}})=
\sum_{a\in 2\Bbb Z^{\nu}}M(a, t)
$$
with the same $M$'s. This completes the proof.  \qed
\enddemo

\medpagebreak

Our next result, an analog of Proposition 4.2, is an abelianized 
version of a formula that Lax [5] derived formally in two dimensions:

\proclaim{Theorem 5.5}  Let $V$ be a continuous periodic function on 
$\Bbb R^{2}$ with $V(x_{1}+n_{1}, x_{2}+n_{2})=V(x_{1}, x_{2})$ for
$\bold n =(n_{1}, n_{2})\in\Bbb Z^{2}$.  Let $H_{P}, H_{A}, H_{AP}, 
H_{PA}, H_{N}, H_{D}$ be the operators on $L^{2}([0, 1]^{2})$ with periodic, 
antiperiodic, $AP$, $PA$, Neumann, and Dirichlet boundary conditions 
where $AP$ \rom(resp.~$PA$\rom) means antiperiodic in the $x_1$ 
\rom(resp.~$x_2$\rom) direction and periodic in the $x_{2}$ 
\rom(resp.~$x_1$\rom) direction.  Then
$$\multline
\text{\rom{Tr}}(e^{-tH_{P}}+e^{-tH_{A}}+e^{-tH_{PA}}+e^{-tH_{AP}}-
2e^{-tH_{N}}-2e^{-tH_{D}}) \\
=-1+tV(0)+o(t)\qquad\text{as } t\downarrow 0. \endmultline \tag 19
$$
\endproclaim

\demo{Proof}  Let $\tilde V$ be the extension of $V$ given by 
reflection, and $\tilde H=-\Delta+\tilde V$.  By the method of images as 
above,
$$
2\text{Tr}(e^{-tH_{N}}+e^{-tH_{D}})=\sum_{a\in\Bbb Z^{2}}\tilde M(a, 
t)+\sum_{a\in\Bbb Z^{2}}\tilde N(a, t),
$$
where $\tilde M$ is defined in Proposition 5.3 (with $H$ replaced by 
$\tilde H$) and
$$
\tilde N(a, t)=4\int\limits_{[0, 1]^{2}} e^{-t\tilde H} (x, x+2a)\, d^{2}x.
$$
By (17) and (18)
$$
2\text{Tr}(e^{-tH_{N}}+e^{-tH_{D}})=1-tV(0)+\tilde N(0, t)+O(e^{-c/t}).
$$
Now let $H=-\Delta+V$.  Then
$$
4\text{Tr}(e^{-tH_{P}})=\sum_{a\in\Bbb Z^{2}} N\biggl(\frac12 a, 
t\biggr)
$$
with
$$
N\biggl(\frac12 a, t\biggr)=4\int\limits_{[0, 1]^{2}} e^{-tH}(x, 
x+a)\, d^{2}x.
$$
By (17) again
$$
4\text{Tr}(e^{-tH_{P}})=N(0, t)+O(e^{-c/t}).
$$
Similar formulae show that
$$
4\text{Tr}(e^{-tH_{A}})=N(0, t)+O(e^{-c/t})
$$
and for $H_{AP}$ and $H_{PA}$.  Thus
$$
\text{LHS of (19)}=-1+tV(0)+o(t)+[N(0, t)-\tilde N(0, t)]
$$
and (19) follows from the following assertion (20):
$$
N(0, t)-\tilde N(0, t)=o(t). \tag 20
$$

The proof of (20) is a little subtle because the integral kernel $e^{-
tH}(x, x)$ is $O(t^{-1})$ as $t\downarrow 0$ in two dimensions.  In terms 
of a path integral expansion, only points $O(t^{1/2})$ from the border 
contribute.  Thus to get an $o(t)$ error, we must have complete 
cancellation of the $O(t)$ terms in the expansion of the exponentials 
in the path integral.

So let $\bold w_{t}(s)$ be $\bold x+\sqrt{2t}\, \boldsymbol\alpha
(\frac{s}{t})$ where $\boldsymbol\alpha$ is the Brownian bridge and $x$ is 
independent of $\boldsymbol\alpha$ and uniformly distributed on $[0, 
1]^{2}$.  Define
$$
g(\bold w_{t}, t):= \int\limits^{t}_{0} V(\bold w_{t}(s))\, ds
$$
and $\tilde g$ with $V$ replaced by $\tilde V$ and let
$$
f(y)=e^{-y}-1+y.
$$
Then
$$\align
N(0, t)-\tilde N(0, t) &=(4\pi t)^{-1} E(e^{-g}-e^{-\tilde g}) \\
&\equiv A+B,
\endalign
$$
where
$$\gather
{\align A&=-(4\pi t)^{-1} E(g-\tilde g),\\
B&=(4\pi t)^{-1} E(f(g)-f(\tilde g)). \endalign}
\endgather
$$
We'll prove that $A=0$ and $B=O(t^{3/2-\epsilon})$, $\epsilon>0$.

That $B=O(t^{3/2-\epsilon})$ follows by noting first that 
$|f(y)|\leq Cy^{2}$ on $[-1, 1]$ and that $|g(\bold w_{t}, t)|\leq Ct$ 
uniformly in $\bold w$ so $|f(g)-f(\tilde g)|\leq Ct^{2}$ uniformly in 
$\bold w$.  On the other hand, if $\text{dist}(x, \partial [0, 1]^{2})
\geq t^{1/2-\epsilon}$, $\text{Prob}(x+\sqrt{2t}\, \alpha (\frac{s}{t})
\notin [0, 1]^{2}\text{ for some } s)\leq\exp(-C/t^{2\epsilon})$, so, 
since $f(g)=f(\tilde g)$ if $\bold w_{t}(s)\in [0, 1]^{2}$ for all $s$, 
we have
$$
E(|f(g)-f(\tilde g)|)\leq O(\exp(-t^{-2\epsilon}))+Ct^{2}t^{1/2-
\epsilon}
$$
and we conclude that $B=O(t^{3/2-\epsilon})$ as required.

$A=0$ because of a cancellation.  Let $\rho_{t}(x)$ be the probability 
density of $\bold w_{t}(s)$ where $s$ is uniformly distributed in $[0, 1]$.  
Then
$$
A=-(4\pi)^{-1} \int\limits_{\Bbb R^{2}}\rho_{t}(x) [V(x)-\tilde V(x)]\, 
d^{2}x. \tag 21
$$
For each $\alpha\in\Bbb Z^{2}$, let $\square_{\alpha} $ be the square 
centered at $(\frac12, \frac12)+\alpha$.  There is a symmetry 
$S_{\alpha}:\square_{\alpha}\to\square_{-\alpha}$ so that $V\circ 
S_{\alpha}=\tilde V$ and $\rho_{t}\circ S_{\alpha}=\rho_{t}$ so that the 
contribution of $V$ over $\square_{\alpha}$ in (21) cancels the 
contribution of $\tilde V$ over $\square_{-\alpha}$. \qed
\enddemo

A different kind of a two-dimensional trace formula for $V(x)$ by 
comparing heat kernels for $H=-\triangle +V$ and $H_{o}=-\triangle$ with 
Dirichlet boundary conditions on a rectangular box was recently studied in [7].

\bigpagebreak
\flushpar {\bf \S 6. Sums Without Abelian Summation}

An interesting issue on which we haven't much to report is the extent 
to which a formula like (2) holds.  We note:

\proclaim{Theorem 6.1}  In the context of Theorem \rom{5.1}, let 
$\{E_{n}\}^{\infty}_{n=0}$ be a listing of all the eigenvalues of 
$\{H_{A}\mid |A| \text{ is even}\}$ and $\{\tilde 
E_{n}\}^{\infty}_{n=1}$ for $\{H_{A}\mid |A| \text{ is odd}\}$ ordered so 
$E_{n}\leq E_{n+1};\tilde E_{n}\leq\tilde E_{n+1}$.  Suppose that
$$
\sum^{\infty}_{n=1} |E_{n}-\tilde E_{n}| < \infty. \tag 22
$$
Then
$$
\langle V\rangle =E_{0}+\sum^{\infty}_{n=1} (E_{n}-\tilde E_{n}).
$$
\endproclaim

\remark{Remark}  Note that the counting of $E_n$ starts at 0, but for 
$\tilde E_n$ at 1.
\endremark

\demo{Proof}  Theorem 5.1 says that
$$
\langle V\rangle =\lim\limits_{t\downarrow 0}\, \biggl[t^{-1}(1-e^{-tE_{0}})+
\sum^{\infty}_{n=1} t^{-1} (e^{-t\tilde E_{n}}-e^{-tE_{n}})\biggr].
$$
If $0\leq t\leq 1$, then
$$
|e^{-ta}-e^{-tb}| \leq [e^{-t\min(a, b)}+1] |a-b|
$$
and
$$
\lim\limits_{t\downarrow 0}\, t^{-1} (e^{-ta}-e^{-tb}) = b-a
$$
so the result follows by dominated convergence. \qed
\enddemo

When $V=0$, it is easy to see that $E_{n}=\tilde E_{n}$, $n\in\Bbb N$ so 
(22) holds.  It remains to be seen if one can prove it for sufficiently 
smooth $V$'s.

\vskip 0.5in

\example{Acknowledgments}  We would like to thank Peter Lax for 
telling us of his work prior to publication.  F.G.~is indebted to 
M.~Aschbacher and G.~Neugebauer for the hospitality at Caltech where 
some of this work was done.  F.G.~and H.H.~were also supported in part by 
the Norwegian Research Council for Science and the Humanities (NAVF).
\endexample

\vskip 0.5in

\Refs

\endRefs
\bigpagebreak

\item{[1]} H.~Flaschka, {\it On the inverse problem for Hill's 
operator}, Arch.~Rat.~Mech.~Anal.~ {\bf 59} (1975), 293--309.
\item{[2]} F.~Gesztesy, H.~Holden, B.~Simon, and Z.~Zhao, {\it Higher 
order trace relations for Schr\"o-dinger operators}, to be submitted 
to Commun.~Pure Appl.~Math.
\item{[3]} F.~Gesztesy and B.~Simon, {\it The xi function}, to be submitted 
to Ann.~Math.
\item{[4]} H.~Hochstadt, {\it On the determination of a Hill's 
equation from its spectrum}, Arch.~Rat. Mech.~Anal. {\bf 19} (1965),
353--362.
\item{[5]} P.~Lax, {\it The trace formula for the Schr\"odinger 
operator}, to appear in Commun.~Pure Appl.~Math.
\item{[6]} H.P.~McKean and P.~van Moerbeke, {\it The spectrum of Hill's 
equation}, Invent.~Math. {\bf 30} (1975), 217--274.
\item{[7]} V.G.~Papanicolaou, {\it Trace formulas and the behavior of 
large eigenvalues}, to appear in SIAM J.~Math.~Anal.
\item{[8]} B.~Simon, {\it Functional Integration and Quantum Physics}, 
Academic Press, New York, 1979.
\item{[9]} B.~Simon, {\it Schr\"odinger semigroups}, 
Bull.~Amer.~Math.~Soc. {\bf 7} (1982), 447--526.

\enddocument