# Self-Adjointness of Toeplitz Operators
on the Segal-Bargmann Space

Wolfram Bauer  Wolfram Bauer, Leibniz Universität Hannover, Institut für
Analysis, Welfengarten 1, 30167 Hannover, Germany<EMAIL_ADDRESS>,  Lauritz van Luijk  Lauritz van Luijk, Institut für Theoretische Physik,
Leibniz Universität Hannover, Appelstraße 2, 30167 Hannover, Germany
<EMAIL_ADDRESS>,  Alexander Stottmeister  Alexander
Stottmeister, Institut für Theoretische Physik, Leibniz Universität Hannover,
Appelstraße 2, 30167 Hannover, Germany<EMAIL_ADDRESS>hannover.de and  Reinhard F. Werner  Reinhard F. Werner, Institut für
Theoretische Physik, Leibniz Universität Hannover, Appelstraße 2, 30167
Hannover, Germany<EMAIL_ADDRESS>
###### Abstract.

We prove a new criterion that guarantees self-adjointness of Toeplitz
operators with unbounded operator-valued symbols. Our criterion applies, in
particular, to symbols with Lipschitz continuous derivatives, which is the
natural class of Hamiltonian functions for classical mechanics. For this we
extend the Berger-Coburn estimate to the case of vector-valued Segal-Bargmann
spaces. Finally, we apply our result to prove self-adjointness for a class of
(operator-valued) quadratic forms on the space of Schwartz functions in the
Schrödinger representation.

##  1\. Introduction

Proving the self-adjointness of unbounded Hilbert space operators is a key
problem for many applications of functional analysis in quantum theory. For
von Neumann, who founded the theory [25], the principal distinction of self-
adjoint operators over merely hermitian operators was the existence of a
spectral resolution. This property was significant to him as the creator of
the mathematical formalism of quantum theory, not least because a partition of
unity into projections was needed for a quantum-probabilistic interpretation
as a quantum observable. Equally important in modern applications is the role
of self-adjoint operators as generators of dynamical evolution. According to
the rules of quantum mechanics, this must be given by a strongly continuous
unitary one-parameter group and, hence, be generated by a self-adjoint
operator. A hermitian operator built as a sum of terms motivated by a physical
model may very well fail this crucial property. This is then a sign that the
problem has not been sufficiently well defined, and a self-adjoint extension
must be singled out, typically by specifying suitable boundary conditions.
Once self-adjointness is achieved, and so the generator can be exponentiated,
the group property extends this to all times. The counterpart of the self-
adjointness problem for classical dynamical systems is, therefore, the
_global_ existence of the flow (possibly only for almost all initial values,
as the Coulomb potential $V(x)=\lvert x\rvert^{-2}$ shows). This connection is
not just superficial and can be studied case by case. Indeed, a folk “theorem”
states that a Schrödinger operator with given potential is self-adjoint on its
natural domain precisely if the Hamiltonian flow for the corresponding problem
of classical mechanics exists for all times. There are well-known
counterexamples to this statement in either direction [22, Ch. X.1].
Nevertheless, there is some truth in it, so the question may be which
additional regularity assumptions make it true. This paper establishes a
criterion for self-adjointness directly motivated by a classical counterpart.

The basic connection between the classical and the quantum world is provided
by a quantization $f\mapsto T_{f}$ relating classical phase space functions
$f$ to quantum operators $T_{f}$, e.g., the classical Hamilton function to the
quantum Hamiltonian. Even though the natural stage for quantum mechanical
operators $T_{f}$ of systems with $n$ canonical degrees of freedom is without
a doubt the Schrödinger representation, i.e., the Hilbert space
$L^{2}(\mathbb{R}^{n})$ together with the standard representation of the
canonical commutation relations in terms of position and momentum operators.
We will, however, work in the unitarily equivalent Segal-Bargmann
representation. The main reason for this is that we will use the coherent-
state quantization which takes on a more natural form in this representation,
known as the _Berezin-Toeplitz quantization_. It depends on a quantization
parameter $t>0$, which plays the role of Planck’s constant. Smaller values of
$t$ correspond to larger scales of _action_ and the limit $t\to 0$ (formally)
is the classical limit. A natural criterion to ensure global existence of the
solutions to Hamilton’s equations is provided by the Picard-Lindelöf theorem,
which requires the derivatives of the classical Hamiltonian function $f$ to be
globally Lipschitz continuous. We will show that precisely the same assumption
guarantees self-adjointness of the Berezin-Toeplitz operator $T_{f}$. As it
turns out, we can even weaken the assumption of Lipschitz continuity in that
we only need the derivatives of a heat transformed symbol to have bounded
oscillation. This comes in handy, as the heat transformed symbol is generally
better behaved.

In the following we denote by $T_{f}$ the Toeplitz operator with symbol $f$
acting on the Segal-Bargmann space $\mathcal{H}$ of Gaussian square integrable
analytic functions (s. Section 2). Applying the well-known reproducing kernel
function of $\mathcal{H}$ we express $T_{f}$ in an integral form and consider
it on its natural domain of functions for which this integral again defines an
element in the Segal-Bargmann space (see (12) for details).

######  Theorem 1.

Fix a quantization parameter $t>0$. Let
$f:\mathbb{R}^{2n}\\!\cong\mathbb{C}^{n}\to\mathbb{R}$ be a measurable
polynomially bounded symbol such that there is a time $s\in[0,\frac{t}{2})$
for which the first-order derivatives of the heat-transformed symbol
$\widetilde{f}^{(s)}$ at time $s$ have bounded oscillation, i.e., satisfy

$\lvert\partial_{j}\widetilde{f}^{(s)}(z)-\partial_{j}\widetilde{f}^{(s)}(w)\rvert\lesssim
1+\lvert z-w\rvert,\quad j=1,\ldots,2n,$

for all $z,w\in\mathbb{C}^{n}$. Then the Toeplitz operator $T_{f}$ is self-
adjoint on its natural domain.

The symbol “$\lesssim$” means that the inequality “$\leq$” holds provided that
the right-hand side is multliplied by some constant $c>0$, and
$\widetilde{f}^{(s)}$ is the heat transform of $f$ after time $s$. We will
extend this theorem to allow for operator-valued symbols and use it to prove a
self-adjointness criterion for (operator-valued) continuous quadratic forms on
the space of Schwartz functions in the Schrödinger representation (see
Theorems 14 and 20). One ingredient of the proof are the Berger-Coburn
estimates [6, 7] which we derive in the case of operator-valued symbols in
Section 4.

Previous work on the self-adjointness of unbounded Toeplitz operators was done
by J. Janas in [14, 15, 16]. Apart from considering the scalar-valued case
only (i.e., no operator-valued symbols), his work differs from ours in several
aspects. In particular, Janas investigates many aspects of unbounded
operators, e.g., closedness or computation of the adjoint, not just self-
adjointness. While Janas’ work often considers families of symbols with
specific properties such as (anti-)analyticity, we are interested in classes
of symbols that also serve as natural Hamiltonian functions for a classical
mechanical system. In [15, Thm. 3.3] it is proved (for $t=\frac{1}{2}$) that
the Toeplitz operators with measurable real-valued symbols $f$ are self-
adjoint if

$\lvert f(z)-f(w)\rvert\lesssim e^{\frac{1}{8t}\lvert z-w\rvert^{2}}$

for all $z,w\in\mathbb{C}^{n}$. By noting that this inequality still holds if
the right-hand side is replaced by $1+\lvert z-w\rvert$, we will show that 1
is a generalization of this result (see 17).

Finally, we collect some examples and applications in Section 6. Among others,
these include results on the self-adjointness of relativistic and non-
relativistic Schrödinger operators with operator-valued potentials, which are
new to the authors best knowledge.

##  2\. Notation

We denote by $\mathcal{H}$ the Segal-Bargmann space with quantization
parameter $t>0$, which is defined as the closed subspace consisting of entire
functions in $L^{2}(\mathbb{C}^{n},\mu)$, where

$\,\textup{d}\mu(z)=(2\pi t)^{-n}e^{-\frac{\lvert
z\rvert^{2}}{2t}}\,\textup{d}z.$

The norm and inner product on $\mathcal{H}$ are induced by
$L^{2}(\mathbb{C}^{n},\mu)$ and will be denoted by $\lVert\,\cdot\,\rVert$ and
$\langle\,\cdot\,,\,\cdot\,\rangle$, respectively. We will use these symbols
for the scalar products and norms of all Hilbert spaces, since it will be
clear from the context to which space the vectors belong. Nevertheless, we
will sometimes append the space as an index for emphasis when that seems
helpful. $\mathcal{H}$ is a reproducing kernel Hilbert space with kernel
$K(z,w)=e^{\frac{\overline{w}\cdot z}{2t}}$ where $z,w\in\mathbb{C}^{n}$,
i.e., $f(z)=\langle f,K(\,\cdot\,,z)\rangle$ for all $f\in\mathcal{H}$. The
normalized kernels are denoted by $k_{w}=\lVert
K(\,\cdot\,,w)\rVert^{-1}K(\,\cdot\,,w)$. The standard orthonormal basis of
$\mathcal{H}$ is given by the monomials

$e_{\nu}(z)=\frac{z^{\nu}}{\sqrt{\nu!(2t)^{\lvert\nu\rvert}}},\quad\nu\in\mathbb{N}_{0}^{n},$

where $z^{\nu}=(z_{1})^{\nu_{1}}\cdots(z_{n})^{\nu_{n}}$,
$\nu!=\nu_{1}!\cdots\nu_{n}!$ and $\lvert\nu\rvert=\nu_{1}+\dots+\nu_{n}$.
Denote by $P:L^{2}(\mathbb{C}^{n},\mu)\to\mathcal{H}$ the orthogonal
projection onto $\mathcal{H}$. The projection $P$ can be expressed in terms of
the reproducing kernel by

$Pf(z)=\int K(z,w)f(w)\,\textup{d}\mu(w).$

For a measurable function $f:\mathbb{C}^{n}\to\mathbb{C}$, the _Toeplitz
operator_ acts on a suitable domain in $\mathcal{H}$ by

$T_{f}g=P(f\cdot g).$ (1)

Using the reproducing kernel this leads to

$T_{f}g(z)=\int K(z,w)f(w)g(w)\,\textup{d}\mu(w).$ (2)

When denoting Toeplitz operators we will use “$z$” as a dummy variable, i.e.,
$T_{f(z)}=T_{f}$.

The unitary _Weyl-operators_ act by

$W_{w}f(z)=k_{w}(z)\,f(z-w),$

where $w,z\in\mathbb{C}^{n}$. The reproducing kernel is connected to the Weyl-
operators through $k_{z}=W_{z}1$. The Weyl-operators are generated by
unbounded Toeplitz operators with linear symbols

$W_{w}=e^{\frac{1}{2t}(T_{w\cdot\overline{z}}-T_{\overline{w}\cdot
z})}=e^{\frac{i}{t}T_{\omega(w,z)}}$ (3)

where $\omega(z,w)=\operatorname{Im}(\overline{z}\cdot w)$. The mapping
$z\mapsto W_{z}$ is a strongly continuous, projective unitary representation
of the group $(\mathbb{C}^{n},+)$ of phase space translations. They satisfy
the so-called _Weyl-relations_ ,
$W_{w}W_{z}=e^{\frac{i}{2t}\omega(z,w)}W_{z+w}$ and, hence, define a proper
unitary representation of the Heisenberg group $\mathbb{H}_{n}$. The Weyl-
operators implement phase space translations on $\mathcal{H}$ [9]. To see
this, we define $\alpha_{z}(A)=W_{z}AW_{-z}$ and $\alpha_{z}f(w)=f(w+z)$ for
operators $A$ on $\mathcal{H}$ and functions $f$ on $\mathbb{C}^{n}$. Then it
follows that

$\alpha_{z}T_{f}=T_{\alpha_{z}f}.$

The _Berezin transform_ of an operator $A:D(A)\to\mathcal{H}$ is
$\widetilde{A}(z)=\langle Ak_{z},k_{z}\rangle$, provided $k_{z}\in D(A)$ for
all $z\in\mathbb{C}^{n}$. While the Toeplitz operator is a form of covariant
quantization, the Berezin transform provides a covariant dequantization, i.e.,
$\alpha_{z}\widetilde{A}=(\alpha_{z}A)^{\sim}$. The Berezin transform of a
Toeplitz operator $T_{f}$ is the heat transform of the symbol $f$ after time
$t$ [7].

By the _Schrödinger representation_ we mean the Hilbert space
$L^{2}(\mathbb{R}^{d})$, which is isometrically isomorphic to $\mathcal{H}$
via the _Bargmann isometry_ (cf. [11, Chap. 1.6])

$B:\mathcal{H}\to L^{2}(\mathbb{R}^{d}).$

The Bargmann isometry maps the standard basis $\\{e_{\nu}\\}$ to the basis of
(scaled) Hermite functions: $Be_{\nu}(x)=t^{-n/4}\psi_{\nu}(x/\sqrt{t})$,
where $\psi_{\nu}$ is the $\nu$-th Hermite function.

Let $\mathcal{K}$ be a separable complex Hilbert space. The _vector-valued
Segal-Bargmann space_ $\mathcal{H}(\mathcal{K})$ is the closed subspace of
analytic functions in $L^{2}(\mathbb{C}^{n},\mu;\mathcal{K})$, the space of
square integrable $\mathcal{K}$-valued functions with inner product

$\langle f,g\rangle_{L^{2}(\mathbb{C}^{n},\mu;\mathcal{K})}=\int\langle
f(z),g(z)\rangle\,\textup{d}\mu(z)$

for $f,g\in L^{2}(\mathbb{C}^{n},\mu;\mathcal{K})$. The norm and inner product
on $\mathcal{H}(\mathcal{K})$ are, as in the scalar-valued case, induced by
$L^{2}(\mathbb{C}^{n},\mu;\mathcal{K})$ and are again denoted
$\lVert\,\cdot\,\rVert$ and $\langle\,\cdot\,,\,\cdot\,\rangle$, respectively.
The space $\mathcal{H}(\mathcal{K})$ can also be viewed as the Hilbert space
tensor product $\mathcal{H}\otimes\mathcal{K}$ by setting $(f\otimes
x)(z)\coloneqq f(z)x$. We denote by $\mathcal{B}(\mathcal{V})$ the algebra of
bounded operators on a Hilbert space $\mathcal{V}$. For a weakly measurable
(see 5) symbol $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$, the Toeplitz
operator $T_{f}$ is defined on a suitable domain as in the scalar-valued case:
$T_{f}g\coloneqq P(f\cdot g)$, $g\in\mathcal{H}(\mathcal{K})$, where $(f\cdot
g)(z)=f(z)g(z)$ and
$P:L^{2}(\mathbb{C}^{n},\mu;\mathcal{K})\to\mathcal{H}(\mathcal{K})$ is the
orthogonal projection onto $\mathcal{H}(\mathcal{K})$. The integral in (2)
exists as a Bochner-integral in $\mathcal{K}$. In the case that
$f=\lambda\otimes A$, i.e., $f(z)x=\lambda(z)Ax$, $x\in\mathcal{K}$, with
$\lambda\in L^{\infty}(\mathbb{C}^{n})$ and $A\in\mathcal{B}(\mathcal{K})$,
one has $T_{f}=T_{\lambda}\otimes A$.

##  3\. Schwartz Elements

In [2] the set of analytic functions $f\in\mathcal{H}$ that are mapped to
Schwartz-functions by the Bargmann isometry, i.e.,
$Bf\in\mathscr{S}(\mathbb{R}^{n})$, was characterized by a growth condition
(cf. 2.3). We refer to such functions as _Schwartz elements_ of the Segal-
Bargmann space.

The space of Schwartz elements in $\mathcal{H}$, which we denote by
$\mathcal{S}$, will serve us as an invariant domain for a class of unbounded
Toeplitz operators and as a form domain for quadratic forms. Especially for
the latter it is convenient (for our purposes) that $\mathcal{S}$ is a Fréchet
space with respect to a natural topology, and that $\mathcal{S}$ is directly
connected to the action of the Weyl-operators (cf. 2.6).

The following proposition is a collection of several equivalent definitions of
the space $\mathcal{S}$.

######  Proposition 2.

Let $f\in\mathcal{H}$. Then the following are equivalent

  1. (1)

$f\in\mathcal{S}$, i.e., $Bf\in\mathscr{S}(\mathbb{R}^{d})$,

  2. (2)

$(\langle f,e_{\nu}\rangle)_{\nu\in\mathbb{N}_{0}^{n}}$ is rapidly decreasing,

  3. (3)

$\lvert f(z)\rvert\lvert z\rvert^{N}\lesssim e^{\frac{1}{4t}\lvert
z\rvert^{2}}$ for all $N\in\mathbb{N}_{0}$,

  4. (4)

$f(z)\,e^{-\frac{1}{4t}\lvert z\rvert^{2}}\in\mathscr{S}(\mathbb{C}^{n})$,

  5. (5)

$z^{\alpha}\partial^{\beta}f(z)\in\mathcal{H}$ for all
$\alpha,\beta\in\mathbb{N}_{0}^{n}$,

  6. (6)

$\mathbb{C}^{n}\ni z\mapsto W_{z}f\in\mathcal{H}$ is smooth.

For the proof, we need the following lemma which follows readily from the fact
that $(W_{sw})_{s\in\mathbb{R}}$ is a strongly continuous unitary one-
parameter group. We denote by $\nabla f(z)$ and $\overline{\nabla}f(z)$ the
vectors $(\partial_{1}f(z),\dots,\partial_{n}f(z))$ and
$(\overline{\partial}_{1}f(z),\dots,\overline{\partial}_{n}f(z))$,
respectively. Furthermore, we use the notation
$\overline{z}=(\overline{z}_{1},\dots,\overline{z}_{n})$.

######  Lemma 3.

Let $w\in\mathbb{C}^{n}$. Then $s\mapsto W_{sw}f$ is in
$C^{1}(\mathbb{R},\mathcal{H})$ if and only if
$T_{\omega(w,\,\cdot\,)}f(z)=w\cdot\nabla f(z)-\overline{w}\cdot
z\,f(z)\in\mathcal{H}$. For such $f$, the derivative of $W_{z}f$ at $z=0$ in
direction $w$ is given by $\frac{i}{t}T_{\omega(w,\,\cdot\,)}f$.

###### Proof of 2.

We only prove $n=1$ for the sake of better readability.

2 $\Leftrightarrow$ 1: Put $\psi_{m}=Be_{m}$, for $t=1$ this is the $m$-th
Hermite function. Applying [21, Thm. V.13] finishes this step.

5 $\Rightarrow$ 2: Define $N=\frac{1}{2}T_{z}T_{\overline{z}}$ which acts by
$Nf(z)=tzf^{\prime}(z)$. One has that $Ne_{m}=me_{m}$. Thus, 2 is equivalent
to $f\in D(N^{\infty})\coloneqq\bigcap_{k\geq 1}D(N^{k})$. The assumption of 5
implies $(z\partial)^{m}f=t^{-m}N^{m}f\in\mathcal{H}$ for all $m$, i.e., $f\in
D(N^{\infty})$.

1 $\Leftrightarrow$ 3 was proved by Bargmann in [2].

2 $\Rightarrow$ 5: We use the expansion in terms of the monomial basis

$\displaystyle\lVert z^{a}\partial^{b}f\rVert^{2}=\sum\lvert\langle
f,e_{m}\rangle\rvert^{2}\lVert z^{a}\partial^{b}e_{m}\rVert^{2}$
$\displaystyle=c\sum\lvert\langle
f,e_{m}\rangle\rvert^{2}\prod_{j=0}^{b-1}(m-j)\prod_{k=1}^{a}(m-b+k)$
$\displaystyle\leq c\sum\lvert\langle
f,e_{m}\rangle\rvert^{2}(m+a)^{a+b}<\infty$

for a constant $c=c(a,b,t)$. Summability follows from rapid decay of $\langle
f,e_{m}\rangle$.

4 $\Rightarrow$ 5: Consider the estimate

$(2\pi t)^{n}\lVert z^{a}\partial^{b}f\rVert\leq\lVert
z^{a}\partial^{b}\big{[}f(z)e^{-\frac{\lvert
z\rvert^{2}}{4t}}\big{]}\rVert_{L^{2}(\mathbb{C}^{n})}+(4t)^{-b}\lVert
z^{a}\overline{z}^{b}f(z)e^{-\frac{\lvert
z\rvert^{2}}{4t}}\rVert_{L^{2}(\mathbb{C}^{n})}<\infty.$

Both summands are $L^{2}$-norms of Schwartz functions on $\mathbb{C}^{n}$ and
thus finite.

5 $\Rightarrow$ 4: It is clear that $g(z)\coloneqq(2\pi
t)^{-n/2}f(z)e^{-\frac{\lvert z\rvert^{2}}{4t}}$ is smooth. It suffices to
show uniform boundedness $\lVert
z^{a}\partial^{b}\overline{\partial}^{c}g(z)\rVert_{L^{2}(\mathbb{C}^{n})}<\infty$
for all $a,b,c\in\mathbb{N}_{0}$. We can drop the $\overline{\partial}^{c}$
because it effectively multiplies by $z^{c}$. Similar to the above, the result
follows from

$(2\pi t)^{n}\lVert z^{a}\partial^{b}g\rVert_{L^{2}(\mathbb{C}^{n})}\leq\lVert
z^{a}\partial^{b}f\rVert+(4t)^{-b}\lVert z^{a+b}f\rVert<\infty.$

6 $\Rightarrow$ 5: The Wirtinger derivatives of $W_{w}f$ exist up to any order
by assumption. By equation (3) they are given by

$\partial_{w}W_{w}f|_{w=0}=f^{\prime}=\frac{1}{2t}T_{\overline{z}}f\quad\text{and}\quad\overline{\partial}_{w}W_{w}f|_{w=0}=-\frac{1}{2t}zf=-\frac{1}{2t}T_{z}f.$

5 $\Rightarrow$ 6: This follows from 3. We know that, by assumption, $W_{z}f$
is differentiable and that the derivative at $0$ in direction $w$ is given by
$\frac{i}{t}T_{\omega(w,\,\cdot\,)}f\in\mathcal{H}$. As this function again
satisfies condition 5, the result follows by induction. ∎

In the Schrödinger representation the equivalences 1 $\Leftrightarrow$ 2 and 1
$\Leftrightarrow$ 6 are well-known ([21, Thm. V.13] and [13, p. 827],
respectively). The equivalence of 4 and 1 can be deduced from the formalism
developed in [17] (the proof given here is different). The characterization 3
was shown in [2].

$\mathcal{S}$ is a Fréchet space with the family of semi-norms
$q_{\mu,\nu}(f)=\lVert z^{\mu}\partial^{\nu}f(z)\rVert$ indexed by
$\mu,\nu\in\mathbb{N}_{0}^{n}$. The Bargmann isometry is a topological
isomorphism $B:\mathcal{S}\to\mathscr{S}(\mathbb{R}^{n})$. Convenient
subspaces of $\mathcal{S}$ are

$\mathcal{P}=\operatorname{lh}\\{e_{\nu}:\nu\in\mathbb{N}_{0}^{n}\\}\equiv\mathbb{C}[z_{1},\dots,\overline{z}_{n}]\quad\text{and}\quad\mathcal{E}=\operatorname{lh}\\{K(\,\cdot\,,w):w\in\mathbb{C}^{n}\\}.$
(4)

In fact, $\mathcal{S}$ is invariant under the whole metaplectic
representation. This follows from the fact that the generators of the
metaplectic representation are $\frac{i}{t}$ times Toeplitz operators with
real homogenous polynomial symbols of degree two [11, Ch. 4]. $\mathcal{S}$
is, however, not closed under pointwise multiplication, as
$f(z)=e^{\frac{1}{8t}z^{2}}$ satisfies 4 but $f(z)^{2}$ does not. Note that
the set $\mathcal{S}$ depends on the value of $t$.

We can also make sense of Schwartz elements in the vector-valued case.
$\mathcal{S}(\mathcal{K})$ is the space of $f\in\mathcal{H}(\mathcal{K})$,
such that $\langle f(\cdot),x\rangle\in\mathcal{S}$ for all $x\in\mathcal{K}$.
The vector-valued Schwartz elements are topologized by the “same” family of
semi-norms: $q_{\mu,\nu}(f)=\lVert z^{\mu}\partial^{\nu}f(z)\rVert$. Since
$\mathcal{S}$ is a nuclear Fréchet space, the injective and projective tensor
products of $\mathcal{S}$ and $\mathcal{K}$ coincide and we denote them by
$\mathcal{S}\otimes\mathcal{K}$ [24, Ch. 50]. The tensor product
$\mathcal{S}\otimes\mathcal{K}$ is naturally isomorphic to
$\mathcal{S}(\mathcal{K})$ with the correspondence determined by $(f\otimes
x)(z)=f(z)x$. 2 holds mutatis mutandis in the vector-valued case.

The (vector-valued) Schwartz elements can be used as a domain for unbounded
Toeplitz operators (with operator-valued symbols). We have the following

######  Proposition 4.

Let $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$ be (weakly) measurable and
polynomially bounded, i.e., $\lVert f(z)\rVert\lesssim 1+\lvert z\rvert^{N}$
for some $N>0$. Then $f\cdot g\in L^{2}(\mathbb{C}^{n},\mu;\mathcal{K})$ for
any $g\in\mathcal{S}(\mathcal{K})$. Thus, the Toeplitz operator
$T_{f}g=P(f\cdot g)$ is well-defined on $\mathcal{S}(\mathcal{K})$. In fact,
$T_{f}$ leaves $\mathcal{S}(\mathcal{K})$ invariant and is a continuous
operator $T_{f}:\mathcal{S}(\mathcal{K})\to\mathcal{S}(\mathcal{K})$.

######  Remark 5.

As we assume that the Hilbert space $\mathcal{K}$ is separable, weak
measurability of a function
$g:\mathbb{C}^{n}\rightarrow\mathcal{B}(\mathcal{K})$ is equivalent to its
(strong) measurability according to Pettis’ Theorem [20]. To ensure weak
measurability of an operator-valued function
$f:\mathbb{C}^{n}\rightarrow\mathcal{B}(\mathcal{K})$ is sufficient to prove
the weak and, therefore, strong measurability of the point-wise product
$f\cdot g:\mathbb{C}^{n}\rightarrow\mathcal{K}$ because

$\displaystyle\langle(f\cdot g)(z),x\rangle$ $\displaystyle=\sum_{i\in
I}\langle g(z),v_{i}\rangle\langle f(z)v_{i},x\rangle,$

is measurable for any $x\in\mathcal{K}$ and some (countable) basis
$\\{v_{i}\\}_{i\in I}$ of $\mathcal{K}$ as the point-wise limit of sums of
products of measurable functions.

###### Proof of 4.

The assumption of polynomial boundedness together with 2, 3 immediately
implies that $f\cdot g$ is in $L^{2}(\mathbb{C}^{n},\mu;\mathcal{K})$.

We work in the Schrödinger representation to show that $T_{f}$ is a continuous
operator on $\mathcal{S}(\mathcal{K})$. Here $T_{f}$ acts as the Weyl-
pseudodifferential operator with symbol $\widetilde{f}^{(t/2)}$, which is the
heat transform of $f$ after time $\frac{t}{2}$ (cf. [7]). It is clear that
$\widetilde{f}^{(t/2)}$ is smooth and polynomially bounded. We pick the order
function $m:\mathbb{C}^{n}\to[1,\infty),\,z\mapsto(1+\lvert
z\rvert^{2})^{1/2}$. Let $N>0$ be such that $\lVert f(z)\rVert\lesssim
m(z)^{N}$. Using Peetre’s inequality, we get for
$\alpha,\beta\in\mathbb{N}_{0}^{n}$

$\displaystyle\lVert\partial^{\alpha}\overline{\partial}^{\beta}\widetilde{f}^{(t/2)}(z)\rVert$
$\displaystyle\leq(\pi t)^{n}\int\lVert
f(z-w)\rVert\Big{|}\partial^{\alpha}_{w}\overline{\partial}^{\beta}_{w}e^{-\frac{\lvert
w\rvert^{2}}{4t}}\Big{|}\,\textup{d}w$ $\displaystyle\lesssim\int
m(z-w)^{N}\lvert
w\rvert^{\lvert\alpha\rvert+\lvert\beta\rvert}e^{-\frac{\lvert
w\rvert^{2}}{4t}}\,\textup{d}w$ $\displaystyle\lesssim m(z)^{N}\\!\int
m(w)^{N+\lvert\alpha\rvert+\lvert\beta\rvert}e^{-\frac{\lvert
w\rvert^{2}}{4t}}\,\textup{d}w\lesssim m(z)^{N}.$

Proposition A.4 in [23] finishes the proof. ∎

######  Remark 6.

If one is interested in finding a dense subspace $\mathcal{D}$ such that for a
larger class of symbols, the Toeplitz operators
$T_{f}:\mathcal{D}\to\mathcal{D}$ are well-defined, then one can modify the
construction. For example, the approach presented in [4] allows for symbols
with certain exponential growth and the domain $\mathcal{D}$ used therein is
included in $\mathcal{S}$. The invariance of the common domain is necessary
for a discussion of products (or commutators) of Toeplitz operators. In the
extension of the Berger-Coburn estimates in Section 4, we will admit symbols
with a significantly weaker growth condition (cf. 10).

It is clear that the Weyl-operators $W_{z}$ leave $\mathcal{S}$ invariant. In
fact, we can show that the action of the Weyl-operators is strongly smooth on
$\mathcal{S}$.

######  Lemma 7.

For any $f\in\mathcal{S}$ the map $z\mapsto W_{z}f$ is in
$C^{\infty}(\mathbb{C}^{n},\mathcal{S})$.

###### Proof.

The product formula for Toeplitz operators with polynomial symbols (cf. [8,
Thm. 2]) implies that the topology of $\mathcal{S}$ is induced by the
equivalent family of semi-norms $q_{p}(f)\coloneqq\lVert T_{p}f\rVert$ where
$p$ is a polynomial in
$z_{1},\dots,z_{n},\overline{z}_{1},\dots,\overline{z}_{n}$. Note that

$W_{w}^{*}T_{p}W_{w}=T_{p(z+w)}=\sum_{\alpha,\beta\in\mathbb{N}_{0}^{n}}\frac{1}{\alpha!\beta!}\,{w^{\alpha}\overline{w}^{\beta}}\,T_{\partial^{\alpha}\overline{\partial}^{\beta}p}.$

Furthermore, we have that
$[T_{\omega(w,\,\cdot\,)},T_{p}]=\frac{t}{2i}T_{\overline{w}\cdot\overline{\nabla}p-w\cdot\nabla
p}$. Therefore,

$\displaystyle
T_{p}\Big{(}\frac{1}{s}(W_{sw}-\mathds{1})-\frac{i}{t}T_{\omega(w,z)}\Big{)}$
$\displaystyle=$
$\displaystyle\frac{1}{s}\Big{(}W_{sw}W_{sw}^{*}T_{p}W_{sw}-T_{p}\Big{)}-\frac{i}{t}T_{\omega(w,z)}T_{p}+\frac{i}{t}[T_{\omega(w,z)},T_{p}]$
$\displaystyle=$
$\displaystyle\Big{(}\frac{1}{s}(W_{w}-\mathds{1})-\frac{i}{t}T_{\omega(w,z)}\Big{)}T_{p}+\text{terms
with $s,s^{2},\cdots,s^{\text{deg}(p)-1}$}.$

Applying this to an element $f\in\mathcal{S}$ and using 2 proves that

$q_{p}\Big{(}\frac{1}{s}(W_{sw}f-f)-\frac{i}{t}T_{\omega(w,z)}f\Big{)}=\Big{\|}\frac{1}{s}(W_{w}g-g)-\frac{i}{t}T_{\omega(w,z)}g\Big{\|}+\mathbf{o}(s)\overset{w\to
0}{\longrightarrow}0$

with $g=T_{p}f\in\mathcal{S}$. As the derivatives are again in $\mathcal{S}$,
smoothness follows inductively. ∎

As anounced earlier, we use $\mathcal{S}$ as a domain for quadratic forms. Let
$E$ be a topological vector space. A continuous $E$-valued quadratic form on
$\mathcal{S}$ is a separately continuous and sesquilinear map
$A:\mathcal{S}\times\mathcal{S}\to E$. The principle of uniform boundedness
[21, Thm. V.7] and the fact that $\mathcal{S}$ is a nuclear Fréchet space
imply that separate continuity is equivalent to joint continuity. We will not
always mention the space $E$ and the domain $\mathcal{S}$ and simply refer to
these as “continuous forms”.

For a continuous form $A$ and a continuous operator
$B:\mathcal{S}\to\mathcal{S}$ the commutator $[A,B](f,g)\coloneqq
A(Bf,g)-A(f,Bg)$ is well-defined as a continuous form (vector-valued whenever
$A$ is). We introduce phase space derivatives of continuous forms to be the
forms $\partial_{j}A$ and $\overline{\partial}_{j}A$ with

$\partial_{j}A\coloneqq-\frac{1}{2t}[A,T_{\overline{z}_{j}}]\quad\text{and}\quad\overline{\partial}_{j}A\coloneqq\frac{1}{2t}[A,T_{z_{j}}].$
(5)

These are again continuous forms because of 2. Higher-order derivatives are
defined iteratively
$\partial^{\beta}\overline{\partial}^{\gamma}A=(\partial_{1})^{\beta_{1}}\dots(\overline{\partial}_{n})^{\gamma_{n}}A$,
where $\beta,\gamma\in\mathbb{N}_{0}^{n}$. In fact, these derivatives commute
with each other because $T_{z_{j}}$ and $T_{\overline{z}_{j}}$ commute with
their commutator.

For a continuous form on $\mathcal{S}$, we define the continuous form
$\alpha_{z}A$ by $\alpha_{z}A(f,g)\coloneqq A(W_{-z}f,W_{-z}g)$ and the
Berezin transform $\widetilde{A}(z):\mathbb{C}^{n}\to E$ by
$\widetilde{A}(z)\coloneqq A(k_{z},k_{z})=\alpha_{-z}A(1,1)$. The suggestive
definition of phase space derivatives of continuous forms is justified by the
next result and the fact that the derivatives of a sufficiently regular
function $f$ on phase space coincide with the derivatives of
$z\mapsto\alpha_{z}f$ at $z=0$.

######  Corollary 8.

Let $E$ be a Fréchet space and let $A$ be an $E$-valued continuous quadratic
form on $\mathcal{S}$. Then $z\mapsto\alpha_{z}A(f,g)$ is a smooth map
$\mathbb{C}^{n}\to E$ for all $f,g\in\mathcal{S}$. The derivatives are given
by

$\frac{\partial^{\lvert\beta\rvert+\lvert\gamma\rvert}}{\partial
z^{\beta}\partial\overline{z}^{\gamma}}\alpha_{z}A(f,g)\big{|}_{z=0}=\partial^{\beta}\overline{\partial}^{\gamma}A(f,g),\quad\beta,\gamma\in\mathbb{N}_{0}^{n}.$

In particular, the Berezin transform $\widetilde{A}:\mathbb{C}^{n}\to E$ is
smooth.

###### Proof.

For any continuous semi-norm $q$ on $E$, pick a continuous semi-norm $p$ on
$\mathcal{S}$ such that $q(A(f,g))\leq p(f)p(g)$ for all $f,g\in\mathcal{S}$.
By 2 the commutator $\frac{i}{t}[A,T_{\omega(w,z)}]$ is again a continuous
$E$-valued form on $\mathcal{S}$. Differentiability of $\alpha_{z}A(f,g)$ now
follows from applying the triangle inequality

$\displaystyle\
q\Big{(}\frac{1}{s}\big{(}A(W_{sw}f,W_{sw}g)-A(f,g)\big{)}-\frac{i}{t}[A,T_{\omega(w,z)}](f,g)\Big{)}$
$\displaystyle\leq$ $\displaystyle\
q\Big{(}A\Big{(}\frac{1}{s}(W_{sw}f-f)-\frac{i}{t}T_{\omega(w,z)}f,W_{sw}g\Big{)}\Big{)}+q\Big{(}A\big{(}f,\frac{1}{s}(W_{sw}g-g)-\frac{i}{t}T_{\omega(w,z)}g\big{)}\Big{)}$
$\displaystyle+\frac{1}{t}q\Big{(}A(T_{\omega(w,z)}f,W_{sw}g-g)\Big{)}.$

The first term is bounded by
$p\big{[}(W_{sw}f-f)/s-(i/t)T_{\omega(w,z)}f\big{]}p(W_{tw}g)$ which tends to
zero as $s\to 0$. The second and third term behave similarly. ∎

If $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$ is (weakly) measurable and
and polynomially bounded, then the Toeplitz operator
$T_{f}:\mathcal{S}(\mathcal{K})\to\mathcal{S}(\mathcal{K})$ can be viewed as a
continuous $\mathcal{B}(\mathcal{K})$-valued quadratic form on $\mathcal{S}$
by setting

$\langle T_{f}(g,h)x,y\rangle\coloneqq\langle T_{f}(g\otimes x),h\otimes
y\rangle$

for $x,y\in\mathcal{K}$ and $g,h\in\mathcal{H}$. We can now look at
derivatives of Toeplitz operators. We expect that
$\partial^{\beta}\overline{\partial}^{\gamma}T_{f}=T_{\partial^{\beta}\overline{\partial}^{\gamma}f}$
holds for sufficiently regular $f$ because of the covariance
$\alpha_{z}T_{f}=T_{\alpha_{z}f}$. In fact, the following holds:

######  Lemma 9.

Let $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$ be $k$-times differentiable
w.r.t. the weak operator topology on $\mathcal{B}(\mathcal{K})$ and assume
that $f$ and its derivatives up to $k$-th order are polynomially bounded. Then
the phase space derivatives
$\partial^{\beta}\overline{\partial}^{\gamma}T_{f}$ up to $k$-th order are
again Toeplitz operators on $\mathcal{S}(\mathcal{K})$. In fact,

$\partial^{\beta}\overline{\partial}^{\gamma}T_{f}=T_{\partial^{\beta}\overline{\partial}^{\gamma}f}\quad\text{if
$\lvert\beta\rvert+\lvert\gamma\rvert\leq k$}.$

###### Proof.

It is sufficient to consider the case $k=1$. We use 7:

$\displaystyle\partial_{j}T_{f}(g,h)=\frac{\partial}{\partial z_{j}}\langle
T_{f}W_{-z}g,W_{-z}h\rangle\big{|}_{z=0}$
$\displaystyle=\frac{\partial}{\partial z_{j}}\langle
T_{\alpha_{z}f}g,h\rangle\big{|}_{z=0}$
$\displaystyle=\frac{\partial}{\partial z_{j}}\int\langle
f(z+w)g(w),h(w)\rangle\,\textup{d}\mu(w)\big{|}_{z=0}$
$\displaystyle=\int\langle(\partial_{j}f\cdot
g)(w),h(w)\rangle\,\textup{d}\mu(w)=\langle T_{\partial_{j}f}g,h\rangle,$

where we used the polynomial boundedness of the derivative and dominated
convergence. The same argument works for the $\overline{\partial}_{j}$
derivatives. ∎

Continuous quandratic forms on $\mathcal{S}$ have been considered in a very
similar manner in [17] in the context of quantum harmonic analysis [26] which
has recently been used in the framework of the Segal-Bargmann space in [12]
(see also [18, 19]).

##  4\. Berger-Coburn Estimates: Vector-Valued Case

This section is about extending the Berger-Coburn estimates [6, 7] to Toeplitz
operators with operator-valued symbols. This is achieved by the following
theorem. For an operator-valued symbol
$f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$ we use the notation $\lVert
f\rVert_{\infty}\coloneqq\sup_{u\in\mathbb{C}^{n}}\lVert f(u)\rVert$.

######  Theorem 10.

Let $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$ be weakly measurable such
that $K(\,\cdot\,,w)\lVert f(\,\cdot\,)\rVert\in L^{2}(\mathbb{C}^{n},\mu)$
for all $w\in\mathbb{C}^{n}$. Let $s\in(0,\frac{t}{2})$. Then there is a
$C(s,t)>0$ (depending on $s,t$ only) such that

$\lVert T_{f}\rVert\leq C(s,t)\lVert\widetilde{f}^{(s)}\rVert_{\infty}.$ (6)

An off-diagonal extension of the heat transform of a weakly measurable
function $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$ is

$\widetilde{f}^{(t)}(z,w)\coloneqq\int
f(u)k_{z}(u)\overline{k_{w}(u)}\,\textup{d}\mu(u),$ (7)

where $z,w\in\mathbb{C}^{n}$. Note that (7) defines an
$\mathcal{B}(\mathcal{K})$-valued function on
$\mathbb{C}^{n}\times\mathbb{C}^{n}$ and that the heat transform is given by
$\widetilde{f}^{(t)}(z)=\widetilde{f}^{(t)}(z,z)$ for all
$z\in\mathbb{C}^{n}$. A direct calculation shows the norm estimate

$\lVert\widetilde{f}^{(t)}(w,z)\rVert\leq
e^{\frac{-|z|^{2}}{4t}}\bigg{(}\int\|f(u)\|^{2}|K(u,z)|^{2}\,\textup{d}\mu(u)\bigg{)}^{\frac{1}{2}}.$

The next lemma shows a useful integral representation of $T_{f}$.

######  Lemma 11.

Let $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$ be as in 10 and assume that
$g\in\mathcal{H}(\mathcal{K})$ fulfills

$\|g(\,\cdot\,)\|e^{\frac{|\,\cdot\,|^{2}}{4t}}\in L^{1}(\mathbb{C}^{n},\mu).$

With $z\in\mathbb{C}^{n}$ we have

$T_{f}g(z)=\int
e^{\frac{1}{4t}|z-w|^{2}+\frac{1}{2t}\operatorname{Re}(z\cdot\overline{w})}\widetilde{f}^{(t)}(w,z)g(w)\,\textup{d}\mu(w).$

###### Proof.

Note that the integral on the right-hand side exists under the assumptions on
$f$ and $g$ and the above estimate. Let $x\in\mathcal{K}$ and
$z\in\mathbb{C}^{n}$ be arbitrary. Then (2) implies

$\big{\langle}T_{f}g(z),x\big{\rangle}=\int\big{\langle}f\cdot
g(w),x\big{\rangle}K(z,w)\,\textup{d}\mu(w).$ (+)

On the other hand

$\displaystyle\big{\langle}\widetilde{f}^{(t)}(w,z)g(w),x\big{\rangle}$
$\displaystyle=\int\big{\langle}f(u)g(w),x\big{\rangle}k_{w}(u)\overline{k_{z}(u)}\,\textup{d}\mu(u)$
$\displaystyle=e^{-\frac{|z|^{2}+|w|^{2}}{4t}}\int\big{\langle}g(w),f(u)^{*}x\big{\rangle}K(u,w)K(z,u)\,\textup{d}\mu(u).$

Applying the reproducing kernel formula shows

$\displaystyle\int
e^{\frac{1}{4t}|z-w|^{2}+\frac{1}{2t}\operatorname{Re}(z\cdot\overline{w})}$
$\displaystyle\big{\langle}\widetilde{f}^{(t)}(w,z)g(w),x\big{\rangle}\,\textup{d}\mu(w)$
$\displaystyle=$
$\displaystyle\iint\big{\langle}g(w),f(u)^{*}x\big{\rangle}K(u,w)K(z,u)\,\textup{d}\mu(u)\,\textup{d}\mu(w)$
$\displaystyle=$ $\displaystyle\int K(z,u)\big{\langle}(f\cdot
g)(u),x\big{\rangle}\,\textup{d}\mu(u).$

According to our assumptions on $f$ and $g$ the above double integral
converges absolutely which justifies the change of order in integration. Since
$x\in\mathcal{K}$ was chosen arbitrarily we obtain the result by comparing
with (+ ‣ 4). ∎

Now, we can prove the operator-valued version of [6, Lem. 3.2], which in the
case $z=w$ reduces to the semi-group property of the heat transform:

######  Lemma 12.

Let $z,w\in\mathbb{C}^{n}$ and $0<s<t$. With
$f:\mathbb{C}^{n}\rightarrow\mathcal{B}(\mathcal{K})$ as above we have

$\widetilde{f}^{(t)}(w,z)=e^{\frac{s}{4t(t-s)}{|w-z|^{2}}-\frac{is}{2t(t-s)}\operatorname{Im}(z\cdot\overline{w})}\bigl{(}\widetilde{f}^{(s)}\widetilde{\bigr{)}\,}^{(t-s)}(w,z).$

###### Proof.

We reduce the statement to the scalar-valued case (see [6]). Let
$x,y\in\mathcal{K}$ be arbitrary and define

$f_{x,y}:\mathbb{C}^{n}\rightarrow\mathbb{C}:f_{x,y}(z)\coloneqq\langle
f(z)x,y\rangle.$

We need the (scalar-valued) Segal-Bargmann space $\mathcal{H}_{t-s}$ with
quantization parameter $(t-s)$. We denote the inner product of this space by
$\langle\,\cdot\,,\,\cdot\,\rangle_{t-s}$ and the normalized reproducing
kernels by $k_{z}^{t-s}\in\mathcal{H}_{t-s}$ with $z\in\mathbb{C}^{n}$.
According to [6, Lemma 3.2] we obtain

$\displaystyle\big{\langle}\widetilde{f}^{(t)}(w,z)x,y\big{\rangle}=\widetilde{f_{x,y}}^{(t)}(w,z)=e^{\frac{s}{4t(t-s)}{|w-z|^{2}}-\frac{is}{2t(t-s)}\operatorname{Im}(z\cdot\overline{w})}\big{\langle}\widetilde{f_{x,y}}^{(s)}\cdot
k_{w}^{t-s},k_{z}^{t-s}\big{\rangle}_{t-s}.$

A straightforward calculation shows that

$\big{\langle}\widetilde{f_{x,y}}^{(s)}\cdot
k_{w}^{t-s},k_{z}^{t-s}\big{\rangle}_{t-s}=\Big{\langle}\bigl{(}\widetilde{f}^{(s)}\widetilde{\bigr{)}\,}^{(t-s)}(w,z)\,x,y\Big{\rangle}.$

This finishes the proof, since $x,y\in\mathcal{K}$ were arbitrary. ∎

We need the following estimate:

######  Lemma 13.

Let $f:\mathbb{C}^{n}\rightarrow\mathcal{B}(\mathcal{K})$ be uniformly
bounded, then

$\big{\|}\widetilde{f}^{(t)}(w,z)\big{\|}\leq\lVert
f\rVert_{\infty}\>e^{-\frac{1}{8t}|w-z|^{2}}.$ (8)

###### Proof.

We estimate the norm of the operator-valued integral as follows

$\displaystyle\Big{\|}\int
f(u)k_{w}(u)\overline{k_{z}(u)}\,\textup{d}\mu(u)\Big{\|}\leq
e^{-\frac{|z|^{2}+|w|^{2}}{4t}}\lVert
f\rVert_{\infty}\int\Big{|}e^{\frac{u\cdot\overline{w}}{2t}+\frac{\overline{u}\cdot
z}{2t}}\Big{|}\,\textup{d}\mu(u).$

The integral on the right hand side can be calculated explicitly and one
obtains (8). ∎

Using the above lemmas we can prove 10:

###### Proof of 10.

The space of $g\in\mathcal{H}(\mathcal{K})$ that satisfy the assumption in 13
is dense, hence it suffices to show $\lVert T_{f}g\rVert\leq
C(s,t)\lVert\tilde{f}^{(s)}\rVert_{\infty}\lVert g\rVert$ for such $g$. From
the integral representation in 11 and standard estimates, one has for
$z\in\mathbb{C}^{n}$

$\displaystyle\big{\|}T_{f}g(z)\big{\|}\leq\int
e^{\frac{1}{4t}|z-w|^{2}+\frac{1}{2t}\operatorname{Re}(z\cdot\overline{w})}\Big{\|}\widetilde{f}^{(t)}(w,z)g(w)\Big{\|}\,\textup{d}\mu(w).$

12 and 13 with $f:\mathbb{C}^{n}\rightarrow\mathcal{B}(\mathcal{K})$ replaced
by $\widetilde{f}^{(s)}$ show for any $z,w\in\mathbb{C}^{n}$

$\displaystyle\big{\|}\widetilde{f}^{(t)}(w,z)g(w)\big{\|}$
$\displaystyle=e^{\frac{s}{4t(t-s)}|z-w|^{2}}\Big{\|}\big{(}\widetilde{f}^{(s)}\widetilde{\big{)}}^{(t-s)}(w,z)g(w)\Big{\|}$
$\displaystyle\leq
e^{\big{(}\frac{s}{4t(t-s)}-\frac{1}{8(t-s)}\big{)}|z-w|^{2}}\|\widetilde{f}^{(s)}\|_{\infty}\|g(w)\|.$

Inserting this estimate into the above integral shows

$\big{\|}T_{f}g(z)\big{\|}\leq\|\widetilde{f}^{(s)}\|_{\infty}\int
e^{\frac{1}{8(t-s)}|z-w|^{2}+\frac{1}{2t}\operatorname{Re}(z\cdot\overline{w})}\|g(w)\|\,\textup{d}\mu(w).$

We now estimate the integral on the right-hand side by applying the Cauchy-
Schwarz inequality. Define the symmetric kernel

$L(z,w)\coloneqq
e^{\frac{|z-w|^{2}}{8(t-s)}+\frac{1}{2t}\operatorname{Re}(z\cdot\overline{w})}>0.$

We obtain

$\int L(z,w)\|g(w)\|\,\textup{d}\mu(w)\\\ \leq\left(\int
L(z,w)e^{\frac{|w|^{2}}{4t}}\,\textup{d}\mu(w)\right)^{\frac{1}{2}}\left(\int\|g(w)\|^{2}L(z,w)e^{-\frac{|w|^{2}}{4t}}\,\textup{d}\mu(w)\right)^{\frac{1}{2}}.$
(9)

From $t-s>\frac{t}{2}$ we conclude that there is $\varepsilon>0$ depending on
$s$ such that

$\frac{1}{8(t-s)}\leq\frac{1}{4t}-\varepsilon.$

Therefore, we have for all $z,w\in\mathbb{C}^{n}$

$\frac{|z-w|^{2}}{8(t-s)}+\frac{\operatorname{Re}(z\cdot\overline{w})}{2t}\leq\frac{|z|^{2}}{4t}+\frac{|w|^{2}}{4t}-\varepsilon|z-w|^{2},$

implying the pointwise estimate

$0<L(z,w)\leq e^{\frac{|z|^{2}+|w|^{2}}{4t}-\varepsilon|z-w|^{2}}.$ ($**$)

We can now bound the first integral on the right hand side in (9)

$\int L(z,w)e^{\frac{|w|^{2}}{4t}}\,\textup{d}\mu(w)\leq
C_{\varepsilon,t}e^{\frac{|z|^{2}}{4t}}\hskip 12.91663pt\text{with}\hskip
12.91663ptC_{\varepsilon,t}\coloneqq\frac{1}{(2\pi t)^{n}}\int
e^{-\varepsilon|w|^{2}}\,\textup{d}w.$

Therefore

$\|T_{f}g(z)\|^{2}\leq
C_{\varepsilon,t}\|\widetilde{f}^{(s)}\|^{2}_{\infty}e^{\frac{|z|^{2}}{4t}}\int\|g(w)\|^{2}L(z,w)e^{-\frac{|w|^{2}}{4t}}\,\textup{d}\mu(w).$

Integration of this inequality, Fubini’s theorem and a second application of
the estimate ($**$ ‣ 4) show

$\displaystyle\|T_{f}g\|^{2}$
$\displaystyle=\int\|T_{f}g(z)\|^{2}\,\textup{d}\mu(z)$
$\displaystyle\leq\frac{C_{\varepsilon,t}}{(2\pi
t)^{n}}\|\widetilde{f}^{(s)}\|_{\infty}^{2}\int\|g(w)\|^{2}\int
L(z,w)e^{-\frac{|z|^{2}+|w|^{2}}{4t}}\,\textup{d}z\,\textup{d}\mu(w)$
$\displaystyle\leq
C_{\varepsilon,t}^{2}\|\widetilde{f}^{(s)}\|^{2}_{\infty}\lVert g\rVert^{2}.$

Hence, we obtain $\|T_{f}\|\leq
C_{\varepsilon,t}\|\widetilde{f}^{(s)}\|_{\infty}$ as required. Note that
$C_{\varepsilon,t}\rightarrow\infty$ as $s\uparrow\frac{t}{2}$. ∎

##  5\. Self-Adjointness

Recall that we denote the heat transform at time $s$ of a polynomially bounded
function $f$ on $\mathbb{C}^{n}$ by $\widetilde{f}^{(s)}$, i.e.,

$\widetilde{f}^{(s)}(z)=(2\pi s)^{-n/2}\\!\int e^{-\frac{\lvert
w\rvert^{2}}{2s}}f(z+w)\,\textup{d}w.$ (10)

This makes sense as a Bochner-integral if $f$ is
$\mathcal{B}(\mathcal{K})$-valued, and we set $\widetilde{f}^{(0)}=f$. We
denote by $\mathcal{B}(\mathcal{K})_{h}$ the (real) Banach space of hermitian
bounded operators.

Let $(E,\lVert\,\cdot\,\rVert)$ be some normed space. A function
$f:\mathbb{C}^{n}\to E$ has _bounded oscillation_ [5], if it is continuous and
satisfies

$\lVert f(z)-f(w)\rVert\lesssim 1+\lvert z-w\rvert,\quad\text{for all
$z,w\in\mathbb{C}^{n}$}.$ (11)

This can be seen to be equivalent to $\sup\\{\lVert
f(z)-f(w)\rVert:z,w\in\mathbb{C}^{n},\lvert w\rvert<1\\}<\infty$ (cf. [3]).

######  Theorem 14.

Let $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})_{h}$ be a weakly measurable
polynomially bounded symbol such that there is a time $s\in[0,\frac{t}{2})$
for which the first-order derivatives of $\widetilde{f}^{(s)}$ have bounded
oscillation. Then $T_{f}$ is self-adjoint on its natural domain

$D(T_{f})=\bigl{\\{}g\in\mathcal{H}(\mathcal{K}):\text{the integral
\eqref{eq:toeplitz_kernel} exists $\forall z\in\mathbb{C}^{n}$ and
$T_{f}g\in\mathcal{H}(\mathcal{K})$}\bigr{\\}}.$ (12)

The algebraic tensor products $\mathcal{P}\otimes\mathcal{K}$,
$\mathcal{E}\otimes\mathcal{K}$, and thus $\mathcal{S}(\mathcal{K})$, are
cores for $T_{f}$.

If $s=0$ then the symbol $f$ has to be continuously differentiable w.r.t. the
weak operator topology on $\mathcal{B}(\mathcal{K})$. The spaces
$\mathcal{P}$, and $\mathcal{E}$ are defined in (4). We divide the proof in
two steps: In the first step we treat the case $s=0$ by use of the commutator
theorem, and in the second step we use the operator-valued Berger-Coburn
estimate in 10 to reduce the statement to the case $s=0$.

The main tool in the proof of 14 is the _commutator theorem_ (cf. [22, Thm.
X.37]).

######  Lemma 15 (Commutator Theorem).

Let $N\geq c\,\mathds{1}$, $c>0$, be a self-adjoint operator on a separable
Hilbert space $\mathcal{V}$ with core $\mathcal{D}$. Let
$A:\mathcal{D}\to\mathcal{V}$ be a symmetric operator, such that

  1. (i)

$\lVert Ax\rVert\lesssim\lVert Nx\rVert$ for all $x\in\mathcal{D}$.

  2. (ii)

$\lvert\langle Nx,Ax\rangle-\langle Ax,Nx\rangle\rvert\lesssim\langle
Nx,x\rangle$ for all $x\in\mathcal{D}$.

Then $A$ is essentially self-adjoint, $D(N)\subset D(\overline{A})$ and any
core for $N$ is also a core for the self-adjoint extension $\overline{A}$.

###### Proof of 14. Step 1.

We prove the case $s=0$ under the assumption that $f$ is differentiable w.r.t.
the weak operator topology on $\mathcal{B}(\mathcal{K})$.

We will apply 15 with $N=N_{1}\otimes\mathds{1}_{\mathcal{K}}$ where
$N_{1}=\frac{1}{2}T_{\lvert z\rvert^{2}}$,
$\mathcal{D}\in\\{\mathcal{P}\otimes\mathcal{K},\mathcal{E}\otimes\mathcal{K},\mathcal{S}(\mathcal{K})\\}$
and $A=T_{f}:\mathcal{D}\to\mathcal{H}$. With the composition formula for
Toeplitz operators with polynomial symbols (cf. [8, Thm. 2]) it can be checked
that

$N_{1}=\frac{1}{2}T_{z}T_{\overline{z}}+nt\mathds{1}_{\mathcal{H}}\quad\text{
and }\quad N_{1}e_{\gamma}=t\Big{(}\lvert\gamma\rvert+n\Big{)}e_{\gamma}.$

Thus, $N_{1}\geq{nt}\mathds{1}$ is self-adjoint on
$D(N_{1})=\\{f\in\mathcal{H}:\sum_{\gamma}\lvert\langle
f,e_{\gamma}\rangle\rvert^{2}\lvert\gamma\rvert<\infty\\}$. The operator
$N_{1}$ is known as the quantum harmonic oscillator. For the proof we need the
following easy observations:

  1. (a)

$\langle T_{h}g,g\rangle\geq 0$ for all $g\in\mathcal{S}(\mathcal{K})$ and all
polynomially bounded positive operator-valued
$h:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$.

  2. (b)

$\langle T_{h}^{2}g,g\rangle\leq\langle T_{h^{2}}g,g\rangle$ for all
$g\in\mathcal{S}(\mathcal{K})$ and all polynomially bounded symbols
$h:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})_{h}$.

  3. (c)

For any $a_{0},\dots,a_{m}\in\mathbb{R}$, $m\in\mathbb{N}$, there is a $c>0$
such that $\sum_{k=0}^{m}a_{k}N^{k}\leq cN^{m}$.

As is well known, the harmonic oscillator is one of the infinitesimal
generators of the metaplectic representation (cf. [11, Chap. 4]). Namely, it
corresponds to the classical Hamiltonian function
$N_{\text{cl}}(x,\xi)=\frac{1}{2}(\lvert x\rvert^{2}+\lvert\xi\rvert^{2})$. In
the complex phase space setting, the symplectic flow that is generated by
$N_{\text{cl}}$ is $z\mapsto e^{i\theta}z$, where $\theta\in\mathbb{R}$ is the
time parameter. The unitary one-parameter group generated by $N$ is

$e^{i\theta N/t}g(z)=e^{in\theta}g(e^{i\theta}z),\quad
g\in\mathcal{H}(\mathcal{K}).$

Obviously, the dense domain $\mathcal{D}$ is invariant under $e^{i\theta
N/t}$. From the core theorem it follows that $\mathcal{D}$ is a core for $N$
(cf. [10, Prop. II.1.7]).

We start by checking condition (i) of the commutator theorem. The assumptions
guarantee that the first-order derivatives of $f$ are linearly bounded, which
in turn implies that $f$ is quadratically bounded, i.e., $\lVert
f(z)\rVert\lesssim(1+\lvert z\rvert^{2})$ for all $z$. Applying b, we estimate

$\lVert T_{f}g\rVert^{2}=\langle T_{f}^{2}g,g\rangle\leq\langle
T_{f^{2}}g,g\rangle\leq\langle(\mathds{1}_{\mathcal{H}}+2T_{\lvert
z\rvert^{2}}+T_{\lvert
z\rvert^{4}})\otimes\mathds{1}_{\mathcal{K}}g,g\rangle.$

The Toeplitz composition formula [4] shows that $T_{\lvert
z\rvert^{4}}=(T_{\lvert z\rvert^{2}})^{2}+2tT_{\lvert
z\rvert^{2}}-4nt^{2}\mathds{1}$ which is a polynomial in $N$. Inserting this
into the above equation and using c shows that $\lVert
T_{f}g\rVert\lesssim\langle N^{2}g,g\rangle^{1/2}=\lVert Ng\rVert$.

To check the second condition of the commutator theorem, we use (5) to get
$e^{i\theta N/t}T_{f}e^{-i\theta N/t}=T_{f(e^{i\theta}\,\cdot\,)}$. This
implies

$\frac{i}{t}\langle[N,T_{f}]g,g\rangle=\frac{\mathrm{d}}{\mathrm{d}\theta}\langle
e^{i\theta N/t}T_{f}e^{-i\theta
N/t}g,g\rangle\Big{|}_{\theta=0}=\frac{\mathrm{d}}{\,\textup{d}\theta}\int\langle
f(e^{i\theta}z)g(z),g(z)\rangle\,\textup{d}\mu(z)\Big{|}_{\theta=0},\\\ $

where
$\partial_{\theta}f(z)=\frac{\mathrm{d}}{\mathrm{d}\theta}f(e^{i\theta}z)\big{|}_{\theta=0}=i(z\cdot\overline{\nabla}f(z)-\overline{z}\cdot\nabla
f(z))\in\mathbb{R}$. For the last equality we used dominated convergence to
exchange differentiation with the integral sign. By assumption the derivatives
of $f$ have bounded oscillation. Thus,
$\lVert\partial_{\theta}f(z)\rVert\lesssim(1+\lvert z\rvert^{2})$. Applying a
and c yields

${\langle[N,T_{f}]g,g\rangle}\lesssim\langle(\mathds{1}+T_{\lvert
z\rvert^{2}})\otimes\mathds{1}_{\mathcal{K}}g,g\rangle\lesssim\langle
Ng,g\rangle.$

Applying the same argument to $-f$ proves (ii).

We now denote by $T_{f}$ the unique self-adjoint extension of
$T_{f}:\mathcal{D}\to\mathcal{H}$ which is independent of the choice of
$\mathcal{D}\in\\{\mathcal{P}\otimes\mathcal{K},\mathcal{E}\otimes\mathcal{K},\mathcal{S}(\mathcal{K})\\}$.
The domain $D(T_{f})$ is just the domain of the adjoint operator
$(T_{f}|_{\mathcal{E}\otimes\mathcal{K}})^{*}$ which is easily checked to be
characterized as claimed (cf. [14, Prop. 1.4]). ∎

For the second step we need the following

######  Lemma 16.

Let $(E,\lVert\,\cdot\,\rVert)$ be a Banach space and $F:\mathbb{R}^{n}\to E$
be a differentiable function such that its derivatives have bounded
oscillation. Then the function

$R_{x}(y)=F(y+x)-F(y)-x\cdot\nabla F(y),\quad x,y\in\mathbb{R}^{n}$

is uniformly bounded with $\lVert R_{x}\rVert_{\infty}\leq c(\lvert
x\rvert+\lvert x\rvert^{2})$ for any $c>0$ such that
$\lVert\partial_{j}F(x)-\partial_{j}F(y)\rVert\leq c(1+\lvert x-y\rvert)$ for
all $j=1,\dots,n$.

The case $n=1$ follows directly from the fundamental theorem of calculus. The
case $n>1$ can be reduced to the one-dimensional situation by considering all
one-dimensional linear submanifolds of $\mathbb{R}^{n}$ and by noting that the
constants discussed in the assertion work for all of these simultaneously,
which proves the claim.

###### Proof of 14. Step 2.

This part of the proof is based on the Berger-Coburn estimate [7], which
yields

$\lVert T_{f}-T_{\widetilde{f}^{(s)}}\rVert\leq
c(s)\lVert\widetilde{f}^{(s)}-\widetilde{f}^{(2s)}\rVert_{\infty},$

where $\lVert f\rVert_{\infty}=\sup\\{\lVert f(z)\rVert:z\in\mathbb{C}^{n}\\}$
for $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})$. In Section 4 it was shown
that this estimate also makes sense in the vector-valued case.

As in 16, we define
$R_{w}(z)=\widetilde{f}^{(s)}(w+z)-\widetilde{f}^{(s)}(z)-w\cdot\nabla\widetilde{f}^{(s)}(z)-\overline{w}\cdot\overline{\nabla}\widetilde{f}^{(s)}(z)$.
We estimate the right-hand side of (5):

$\displaystyle\lVert\widetilde{f}^{(s)}(z)-\widetilde{f}^{(2s)}(z)\rVert_{\infty}$
$\displaystyle=(2\pi s)^{-n}\Big{\|}\int\\!e^{-\frac{\lvert
w\rvert^{2}}{2s}}(\widetilde{f}^{(s)}(z)-\widetilde{f}^{(s)}(z+w))\,\textup{d}w\Big{\|}_{\infty}$
$\displaystyle=(2\pi s)^{-n}\Big{\|}\int e^{-\frac{\lvert
w\rvert^{2}}{2s}}\Big{(}w\cdot\nabla\widetilde{f}^{(s)}(z)+\overline{w}\cdot\overline{\nabla}\widetilde{f}^{(s)}(z)+R_{w}(z)\Big{)}\,\textup{d}w\Big{\|}_{\infty}$
$\displaystyle\lesssim\int(\lvert w\rvert+\lvert
w\rvert^{2})\,e^{-\frac{\lvert w\rvert^{2}}{2s}}\,\textup{d}w<\infty.$

We used that odd moments of a normal distribution vanish together with 16.
This proves that $T_{f}$ is a bounded (and hermitian) perturbation of
$T_{\widetilde{f}^{(s)}}$. The latter is essentially self-adjoint since we
assume the first derivatives to have bounded oscillation and can apply the
case $s=0$ of the theorem, which has been proved in step one. ∎

######  Remark 17.

We want to discuss the assumption of bounded oscillation. Let $E$ be a normed
space and let $b:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be any continuous increasing
function. Let $f:\mathbb{C}^{n}\to E$ be a function such that

$\lVert f(z)-f(w)\rVert\lesssim b(\lvert z-w\rvert)$

for all $z,w$. Then the same is true for $b(r)=1+r$, i.e., $f$ has bounded
oscillation provided that $f$ is continuous. To see this, put $a=z-w$ and let
$N=\min(\mathbb{N}\cap[\lvert a\rvert,\infty))$. Since
$f(z)-f(w)=\sum_{k=0}^{N-1}f(w+(k/N)a)-f(z+((k+1)/N)a)$ we get that $\lVert
f(z)-f(w)\rVert\leq Nb(\lvert a\rvert/N)\leq b(1)(1+\lvert z-w\rvert)$.

This applies, for example, to [15, Thm. 3.3] which states that $T_{f}$ is
self-adjoint if $\lVert f(z)-f(w)\rVert\lesssim\exp(\lvert z-w\rvert^{2}/8t)$,
and we learn that this is the case precisely if $f$ satisfies the oscillation
bound (11) (which is independent of $t$). Using the Berger-Coburn estimate as
in step two of the proof of 14, it follows that the oscillation bound needs
only hold for the heat transformed symbol $\widetilde{f}^{(s)}$ for time
$s\in[0,\frac{t}{2})$. Standard properties of the heat transform now show that
14 is a generalization of [15, Thm. 3.3].

######  Corollary 18.

Let $f:\mathbb{C}^{n}\to\mathcal{B}(\mathcal{K})_{h}$ be continuously
differentiable w.r.t. the weak operator topology. We denote by
$\partial_{\theta}f$ the function

$\partial_{\theta}f(z)\coloneqq
z\cdot\overline{\nabla}f(z)-\overline{z}\cdot\nabla
f(z)=\frac{\mathrm{d}}{\mathrm{d}\theta}f(e^{i\theta}z)\big{|}_{\theta=0}.$

If $\lVert f(z)\rVert$ and $\lVert\partial_{\theta}f(z)\rVert$ are
quadratically bounded functions, then the Toeplitz operator $T_{f}$ is self-
adjoint on the domain (12) and the algebraic tensor products
$\mathcal{P}\otimes\mathcal{K}$ and $\mathcal{E}\otimes\mathcal{K}$ as well as
$\mathcal{S}(\mathcal{K})$ are cores for $T_{f}$.

###### Proof.

Step one of the proof of 14 actually only uses these weaker assumptions and
thus shows the result. ∎

This criterion applies, for example, to hermitian product functions
$f(z)=f_{1}(\operatorname{Re}z)f_{2}(\operatorname{Im}z)$ of symbols $f_{i}\in
C^{1}(\mathbb{R}^{n},\mathcal{B}(\mathcal{K}))$ with bounded derivatives (see
Example 3).

We can generalize 14 to a class of continuous hermitian quadratic forms on
$\mathcal{S}$. The idea is to show that for sufficiently regular forms $A$ the
Berezin transform $\widetilde{A}$ is differentiable and that $A$ is a bounded
perturbation of the operator $T_{\widetilde{A}}$ to which we may apply 14.

Let $A$ be a continuous $\mathcal{B}(\mathcal{K})$-valued quadratic form on
$\mathcal{S}$, i.e.,
$A:\mathcal{S}\times\mathcal{S}\to\mathcal{B}(\mathcal{K})$ is sesquilinear
and jointly continuous in the topology of $\mathcal{S}$. We call $A$ hermitian
if for all $f\in\mathcal{S}$ the operator $A(f,f)$ is self-adjoint. As usual
$\lVert A\rVert$ denotes the infimum of all $C\in(0,\infty]$ with $\lVert
A(f,g)\rVert\leq C\lVert f\rVert\lVert g\rVert$ for all $f,g\in\mathcal{S}$,
and the form $A$ is said to be bounded if $\lVert A\rVert<\infty$. If $A$ is
bounded there is a unique $\hat{A}\in\mathcal{B}(\mathcal{H}(\mathcal{K}))$,
such that $\langle A(f,g)x,y\rangle=\langle\hat{A}f\otimes x,g\otimes
y\rangle$ for all $f,g\in\mathcal{S}$ and $x,y\in\mathcal{K}$.

In terms of the phase space translations $\alpha_{z}f=f(\,\cdot\,+z)$, we can
express bounded oscillation of a function $f$ on $\mathbb{C}^{n}$ as
$\lVert\alpha_{z}f-f\rVert_{\infty}\lesssim 1+\lvert z\rvert$. Similarly, we
say that a quadratic form $A$ on $\mathcal{S}$ has bounded oscillation if
$\alpha_{z}A-A$ is bounded with $\lVert\alpha_{z}A-A\rVert\lesssim 1+\lvert
z\rvert$.

######  Theorem 19.

Let $A$ be a hermitian $\mathcal{B}(\mathcal{K})$-valued continuous quadratic
form on $\mathcal{S}$ with first derivatives that satisfy the oscillation
bound

$\lVert\alpha_{z}\partial_{j}A-\partial_{j}A\rVert\lesssim 1+\lvert
z\rvert\quad\text{and}\quad\lVert\alpha_{z}\overline{\partial}_{j}A-\overline{\partial}_{j}A\rVert\lesssim
1+\lvert z\rvert$ (12)

for all $z\in\mathbb{C}^{n}$. In particular, the forms
$\alpha_{z}\partial_{j}A-\partial_{j}A$ and
$\alpha_{z}\overline{\partial}_{j}A-\overline{\partial}_{j}A$ are bounded for
all $j=1,\dots,n$. Then there is an essentially self-adjoint operator
$\hat{A}:\mathcal{S}(\mathcal{K})\to\mathcal{H}(\mathcal{K})$ with $\langle
A(f,g)x,y\rangle=\langle\hat{A}f\otimes x,g\otimes y\rangle$ for all
$f,g\in\mathcal{S}$ and $x,y\in\mathcal{K}$.

###### Proof.

We can express the quadratic form $A-T_{\widetilde{A}}$ as an integral

$A(f,g)-\langle
T_{\widetilde{A}}f,g\rangle=\int\big{[}A-\alpha_{w}A\big{]}(f,g)\,\textup{d}\mu(w)\quad\text{
for $f,g\in\mathcal{S}$.}$ (++)

We will show that (++ ‣ 5) is a bounded quadratic form on $\mathcal{S}$ and
that the first-order derivatives derivatives of the Berezin transform
$\widetilde{A}$ have bounded oscillation. This finishes the proof because it
follows that $A$ is a bounded perturbation of $T_{\widetilde{A}}$ which is
self-adjoint by 14 and 16.

For fixed $f,g\in\mathcal{S}$, put $F(z)=\alpha_{z}A(f,g)\in
C^{\infty}(\mathbb{C}^{n},\mathcal{B}(\mathcal{K}))$. It follows that

$\frac{\partial^{\lvert\beta\rvert+\lvert\gamma\rvert}}{\partial
z^{\beta}\partial\overline{z}^{\gamma}}F(z)\big{|}_{z=0}=\partial^{\beta}\overline{\partial}^{\gamma}A(f,g).$

Thus, the assumption implies that the functions
$\partial^{\beta}\overline{\partial}^{\gamma}F$ with
$\lvert\beta\rvert+\lvert\gamma\rvert=1$, and hence also
$\partial_{j}\widetilde{A}$ as well as $\overline{\partial}_{j}\widetilde{A}$,
have bounded oscillation. We define $R_{z}$ as in 16 with
$F(z)=\alpha_{z}A(f,g)$, the lemma implies that the remainder is bounded by
$\lVert R_{z}\rVert_{\infty}\lesssim\lVert f\rVert\lVert g\rVert(\lvert
z\rvert+\lvert z\rvert^{2})$. Consequently, (++ ‣ 5) yields

$\displaystyle\lVert A(f,g)-\langle T_{\widetilde{A}}f,g\rangle\rVert$
$\displaystyle=\Big{\|}\int\bigg{(}w\cdot\nabla
F(0)+\overline{w}\cdot\overline{\nabla}F(0)+R_{w}(0)\bigg{)}\,\textup{d}\mu(w)\Big{\|}$
$\displaystyle=\Big{\|}0+\int R_{w}(0)\,\textup{d}\mu(w)\Big{\|}\lesssim\lVert
f\rVert\lVert g\rVert\int(\lvert w\rvert+\lvert
w\rvert^{2})\,\textup{d}\mu(w)$

for a suitable constant $c>0$. We used the triangle inequality, the estimate
on the norm of the remainder and the fact that all odd moments of a normal
distribution vanish. ∎

The theorem applies, in particular, if the second derivatives of $A$, i.e.,
the forms $\partial^{\beta}\overline{\partial}^{\gamma}A$ for
$\beta,\gamma\in\mathbb{N}_{0}^{n}$ with
$\lvert\beta\rvert+\lvert\gamma\rvert=2$, are already bounded.

Of course, the game of “phase space derivatives of quadratic forms” can also
be played in the Schrödinger representation. Here, one thinks of the phase
space as the symplectic vector space
$\mathbb{R}^{n}\times\mathbb{R}^{n}\equiv\mathbb{C}^{n}$. For simplicity we
set $t\equiv\hbar=1$. Denote by $Q_{j}$ and $P_{j}$ the $j$-th position and
momentum operator, respectively, i.e., $Q_{j}\psi(x)=x_{j}\psi(x)$ and
$P_{j}\psi(x)=-i\partial_{j}\psi(x)$. The Weyl-operators are given by
$W_{x,\xi}=\exp\\{i\sum_{j=1}^{n}(\xi_{j}Q_{j}-x_{j}P_{j})\\}$, where
$(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}$, and define a strongly
continuous representation of the Heisenberg group $\mathbb{H}_{n}$. As before
we put $\alpha_{x,\xi}A(\psi,\varphi)=A(W_{x,\xi}^{*}f,W_{x,\xi}^{*}f)$. The
phase space derivatives of continuous forms on $\mathscr{S}(\mathbb{R}^{n})$
are given by $\partial_{j}A=-i[P_{j},A]$ and $\partial_{j+n}A=i[Q_{j},A]$ for
$j=1,\dots,n$. We get the following version of 19 in the Schrödinger
representation:

######  Theorem 20.

Let $A$ be a hermitian $\mathcal{B}(\mathcal{K})$-valued continuous quadratic
form on $\mathscr{S}(\mathbb{R}^{n})$ with derivatives that satisfy the
oscillation bound

$\lVert\alpha_{x,\xi}\partial_{j}A-\partial_{j}A\rVert\lesssim 1+\lvert
x\rvert+\lvert\xi\rvert$

for all $x,\xi\in\mathbb{R}^{n}$ and $j=1,\dots,2n$. In particular, the forms
$\alpha_{x,\xi}\partial_{j}A-\partial_{j}A$ are bounded. Then there is an
essentially self-adjoint operator
$\hat{A}:\mathscr{S}(\mathbb{R}^{n};\mathcal{K})\to
L^{2}(\mathbb{R}^{n};\mathcal{K})$ such that $\langle
A(\psi,\varphi)x,y\rangle=\langle\hat{A}\psi\otimes x,\varphi\otimes y\rangle$
for all $\psi,\varphi\in L^{2}(\mathbb{R}^{n})$, $x,y\in\mathcal{K}$.

##  6\. Examples and Applications

We collect some examples and applications of our results. In Examples 2 and 3
we prove self-adjointness of physically relevant operators, such as
(relatvistic and non-relativistic) Schrödinger operators. To the authors best
knowledge these results have not previously appeared in the literature.

We start by giving a counterexample which shows that a naive generalization of
our main result cannot be true. This example was already discussed in [14].

######  Example 1.

14 does not hold if one only asks for the second-order derivatives of $f$ or
$\widetilde{f}^{(s)}$ to have bounded oscillation. In [14, Ex. 3.6] it is
proved that the Toeplitz operator $T_{\operatorname{Re}z^{3}}$ is not
essentially self-adjoint on the domain
$\\{f\in\mathcal{H}:f(z)\cdot\operatorname{Re}z^{3}\in
L^{2}(\mathbb{C},\mu)\\}$. Since this domain contains $\mathcal{S}$ (cf. 2.3)
it follows that $T_{\operatorname{Re}z^{3}}$ is not essentially self-adjoint
on any of the domains $\mathcal{S}$, $\mathcal{P}$ or $\mathcal{E}$.

20 has immediate applications to Schrödinger operators because
$\partial_{j}(-\Delta+V(x))=\partial_{x_{j}}V(x)$ and
$\partial_{j+n}(-\Delta+V(x))=2i\partial_{x_{j}}$ for sufficiently regular
potentials $V:\mathbb{R}^{n}\to\mathbb{R}$. We may even include operator-
valued potentials, which describe an interaction between the canonical degrees
of freedom (i.e., position and momentum) and inner degrees of freedom such as
the particle’s spin.

######  Example 2.

Let $V:\mathbb{R}^{n}\to\mathcal{B}(\mathcal{K})_{h}$ be differentiable with
derivatives of bounded oscillation. Then the non-relativistic and relativistic
Schrödinger operators

$-\Delta+V(x)\quad\text{ and }\quad\sqrt{-\Delta+m^{2}}+V(x)\mathds{1}_{2}$

with $m>0$ are essentially self-adjoint operators
$\mathscr{S}(\mathbb{R}^{n};\mathcal{K})\to
L^{2}(\mathbb{R}^{n};\mathcal{K})$.

Another type of interaction with inner degrees of freedom is through magnetic
fields. Our theorems do, however, not directly apply to Schrödinger operators
with magnetic fields or, more generally, to Pauli operators. For a potential
$V(x)$ and a magnetic field $\vec{B}$ with vector potential $\vec{A}$ (i.e.,
$\nabla\times\vec{A}=\vec{B}$), the Pauli operator is of the form

$H_{\text{Pauli}}=\big{(}it\nabla-\vec{A}\big{)}^{2}\,\mathds{1}_{2}-\vec{B}\cdot\vec{\sigma}+V(x):\mathscr{S}(\mathbb{R}^{3};\mathbb{C}^{2})\to
L^{2}(\mathbb{R}^{3};\mathbb{C}^{2}),$

where $\vec{\sigma}=(\sigma_{x},\sigma_{y},\sigma_{z})$ is the vector of Pauli
matrices and $\vec{B}\cdot\vec{\sigma}\psi(x)=\sum B_{j}(x)\sigma_{j}\psi(x)$.
We can however use the relaxed assumptions in 18 to show essential self-
adjointness of the Toeplitz version of the Pauli operator:

######  Example 3.

Let $\vec{A}:\mathbb{R}^{3}\to\mathbb{R}^{3}$ be a $C^{2}$-function with
bounded derivatives and $V(x)$ a potential as in Example 2. Then the Toeplitz
operator $T_{f}:\mathcal{S}(\mathbb{C}^{2})\to\mathcal{S}(\mathbb{C}^{2})$
with

$f(x+i\xi)=|\xi-\vec{A}(x)|^{2}\mathds{1}_{2}-\vec{B}(x)\cdot\vec{\sigma}+V(x)\mathds{1}_{2},$

where $x,y\in\mathbb{R}^{n}$, is essentially self-adjoint in
$\mathcal{H}(\mathbb{C}^{2})$.

###### Proof.

Expanding yields
$f(x+i\xi)=\lvert\xi\rvert^{2}\mathds{1}_{2}-2\xi\cdot\vec{A}(x)+\lvert\vec{A}(x)\rvert^{2}\mathds{1}_{2}-t\vec{B}(x)\cdot\vec{\sigma}$.
We check the assumptions discussed in 18: The symbol is obviously
quadratically bounded. Furthermore,
$\partial_{\theta}f=x\cdot\nabla_{\xi}f-\xi\cdot\nabla_{x}f$ is clearly
linearly bounded. ∎

######  Example 4.

Put $\mathcal{K}=\ell^{2}(\mathbb{N})$. Let $M\in\mathcal{B}(\mathcal{K})$ be
the multiplication operator $M(x_{n})_{n}=(-n^{-2}x_{n})_{n}$. Denote by $S$
the right shift, i.e., $(Sx)_{n}=x_{n-1}$ with $x_{0}\coloneqq 0$. Now
consider the function $f:\mathbb{C}\to\mathcal{B}(\ell^{2}(\mathbb{N}))$ given
by

$f(z)=g(z)\mathds{1}+\alpha(zS+\overline{z}S^{*})+M$

for some $\alpha>0$ and some $g:\mathbb{C}\to\mathbb{R}$ that satisfies the
assumption of 14. Then $T_{f}$ is essentially self-adjoint. For
$g=\frac{1}{2}\lvert z\rvert^{2}$ this can be used to model the interaction of
a quantum harmonic oscillator with the bounded states of an atom.

######  Example 5.

Put $\mathcal{K}=L^{2}(\mathbb{C})$ . Let $f:\mathbb{C}^{2}\to\mathbb{R}$ be
measurable and assume that $f(\,\cdot\,,w)$ is differentiable with derivatives
of bounded oscillation for almost all $w$. Furthermore, assume that
$f(z,\,\cdot\,)\in L^{\infty}$ for all $z$. Then the Toeplitz operator $T_{F}$
with symbol $F(z)=M_{f(z,\,\cdot\,)}$, i.e., $[F(z)g](w)=f(z,w)g(w)$, is
essentially self-adjoint.

##  7\. Outlook

We want to discuss the _problem of dynamical completeness_ for the Toeplitz
quantization.

Let $f:\mathbb{R}^{2n}\to\mathbb{R}$ be continuously differentiable on phase
space. Then $f$ is said to be _classically complete_ (up to a null-set of
initial values) if the Hamiltonian dynamics generated by $f$, i.e., the flow
on $\mathbb{R}^{2n}$ generated by the ODE

$(\dot{x},\dot{\xi})=-J\nabla f(x,\xi),$ (13)

where $J(x,\xi)=(\xi,-x)$, exists for all times and (almost) all initial
values. Existence of the flow for almost all initial values and all times is
sufficient to define an isometric one-parameter group on
$L^{1}(\mathbb{R}^{2n})$ where isometricity follows from Liouvilles theorem in
symplectic geometry [1, Prop. 3.3.4]. A symmetric operator $H$ on a Hilbert
space is said to be _quantum complete_ if it is essentially self-adjoint.

Let $\mathcal{Q}$ be some quantization scheme which maps a classical
Hamiltonian function $f$ on phase space to a symmetric operator
$\mathcal{Q}(f)$ on some Hilbert space $\mathcal{H}$. Examples are the
Berezin-Toeplitz quantization $f\mapsto T_{f}$ and the Weyl-quantization which
sends a symbol $f$ to its Weyl-pseudodifferential operator $op^{w}(f)$. The
problem of dynamical completeness asks whether quantum completeness of
$\mathcal{Q}(f)$ is equivalent to classical completeness of $f$.

For the Weyl-quantization well-known counterexamples in both directions exist
already on the level of Schrödinger operators. See, for instance, the examples
on $L^{2}(\mathbb{R}_{+})$ in [22, Chap. X.1] which can be easily extended to
the whole real line by considering the potentials $V(\lvert x\rvert)$.

What makes these counterexamples work is the fact that the chosen potentials
$V(x)$ become “more singular” as $x\to\infty$ but it is straightforward to see
that the counterexamples fail if one considers heat-transformed potentials
$\widetilde{V}^{(t)}(x)$ instead, even for arbitrarily short times $t>0$.
This, however, is precisely what happens if one uses the Toeplitz quantization
$T_{f}$ with $f(x,\xi)=\xi^{2}+V(x)$ instead of the Schrödinger operator
$-\Delta+V(x)$ because
$T_{f}\equiv-\Delta+\widetilde{V}^{(t/2)}(x)+\frac{t}{2}\mathds{1}$ for
measurable and polynomially bounded $V$. This leads us to the following
conjecture:

######  Conjecture.

For a reasonably regular Hamiltonian function
$f:\mathbb{R}^{2n}\equiv\mathbb{C}^{n}\to\mathbb{R}$, the following are
equivalent

  1. (1)

The Hamiltonian function $f$ is classically complete up to a null-set of
inital values.

  2. (2)

The Berezin-Toeplitz quantization $T_{f}$ is quantum complete for all
sufficiently small quantization parameters $t>0$.

14 implies that this is true for the class of differentiable symbols with
Lipschitz continuous derivatives but, of course, this class consists only of
classically complete functions.

It is not hard to generate counterexamples if one would ask for classical
completeness to hold everywhere: For $n>1$ one can pick $f\in L^{1}$ such that
classical completeness fails on a null-set. Still the corresponding
quantization $T_{f}$ is a hermitian trace-class operator and thus self-
adjoint.

### Acknowledgments

We thank Robert Fulsche for helpful discussions and suggestions. The second
named author acknowledges support by the Quantum Valley Lower Saxony.

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