Ludwig-Maximilians-Universität München
Fakultät für Physik
Fakultät für Mathematik, Informatik und Statistik
Technische Universität München
Fakultät für Physik
Fakultät für Mathematik

Master Thesis

submitted for the degree of

Master of Science
The Nelson Model on Static Spacetimes

by

Janik Kruse

Submission Date: June 25, 2021 Supervisors: Dr. Jonas Lampart (Université de
Bourgogne) Prof. Dr. Phan Thành Nam

Revised version of .

## Preface

The Nelson model describes the interaction between nonrelativistic quantum
particles and a relativistic quantum field of scalar bosons (e.g. interaction
between nonrelativistic nucleons and relativistic mesons in an atomic
nucleus). In the present work, we discuss the regularisation of this semi-
relativistic model. The formal Nelson Hamiltonian $H$ acting on the Hilbert
space
$\mathfrak{H}=L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}X)\otimes\Gamma_{\mathrm{s}}(L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x))$
is given by

$\displaystyle
H=-\Delta_{X}\otimes\mathbb{1}+\mathbb{1}\otimes\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega^{-\frac{1}{2}}\delta_{X}),$

where $\omega=\sqrt{-\Delta+m^{2}}$ is the relativistic dispersion relation
with bosonic field mass $m$, $\mathop{}\\!\mathrm{d}\Gamma(\omega)$ is its
second quantisation on the symmetric Fock space
$\Gamma_{\mathrm{s}}(L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x))$,
$\Phi(\omega^{-\frac{1}{2}}\delta_{X})$ is a field operator with form factor
$v_{X}=\omega^{-\frac{1}{2}}\delta_{X}/\sqrt{2}$, and $\delta_{X}$ is the
Dirac delta centred at the particle position $X$. The field operator
$\Phi(f)=(a^{*}(f)+a(f))/\sqrt{2}$ is a sum of creation and annihilation
operators. However, the operator $H$ is not well-defined because the form
factor $v_{X}\notin L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x)$ is too
singular.

We present two different techniques of regularising the Nelson Hamiltonian in
this work: renormalisation and regularisation by interior boundary conditions.
To renormalise the Hamiltonian $H$, we demonstrate that putting an ultraviolet
cut-off $\Lambda$ on the interaction (technically this amounts to replacing
$\delta_{X}$ by some smoother function $\rho_{\Lambda,X}$) leads to a well-
defined self-adjoint cut-off Hamiltonian $H_{\Lambda}$. The cut-off
Hamiltonian has a finite vacuum energy $-E_{\Lambda}$. We establish that once
the vacuum energy is removed from the cut-off Hamiltonian, a well-defined
limit of $H_{\Lambda}+E_{\Lambda}$ as $\Lambda\to\infty$ exists. This limit
operator is the renormalised Nelson Hamiltonian.

A weakness of this method is that it provides little information about the
renormalised Nelson Hamiltonian. The second regularisation method we present
is the method of interior boundary conditions (IBC).

The main idea behind the IBC method is to interpret the quantity
$a^{*}(v_{X})\Psi$ for suitable $\Psi\in\mathfrak{H}$ as a distribution. In
return, we also consider $H_{0}\Psi$ as a distribution and choose
$\Psi\in\mathfrak{H}$ in such a way that the singular parts of
$a^{*}(v_{X})\Psi$ and $H_{0}\Psi$ exactly cancel each other. Then,
$L=H_{0}+a^{*}(v_{X})$ is a well-defined operator. The main difficulty of this
approach is to find a physically reasonable extension $A$ of the annihilation
operator $a(v_{X})$ to the domain $D(L)$ of $L$. The IBC Hamiltonian $L+A$
constructed in this way is well defined, self-adjoint, and equivalent to the
renormalised Nelson Hamiltonian. However, this time, we obtain an explicit
description of the renormalised Nelson Hamiltonian and its domain, which was
not available before.

In the present work, we generalise both regularisation techniques to the
Nelson model on static spacetimes. In the Nelson model on static spacetimes
the Laplace operator is replaced by the Laplace–Beltrami operator, which is
the correct generalisation of the Laplace operator on pseudo-Riemannian
manifolds. We revisit a paper by Gérard et al. [5], who translated Nelson’s
renormalisation technique to the Nelson model on static spacetimes. The
original part of this work is to apply the IBC method to the Nelson model on
static spacetimes.

Usually, the Nelson model (on Euclidean spacetime) is discussed in Fourier
space because the Laplace operator is a simple multiplication operator in
Fourier space. The main difficulty of the present work lies in the fact that
no simple representation of the Nelson Hamiltonian on static spacetimes in
Fourier space exists because the Laplace–Beltrami operator is a partial
differential operator with variable coefficients. Instead, we rely on the
pseudo-differential calculus.

##### Acknowledgements

This work is the result of a thesis project in the master’s degree program
’Theoretical and Mathematical Physics’ at the Ludwig–Maximilians–Universität
München and Technische Universität München and was conducted in cooperation
with the Laboratoire Interdisciplinaire Carnot de Bourgogne (ICB) at the
Université de Bourgogne in Dijon. I very much appreciate the hospitality of
the ICB during my stay from October 2020 to April 2021.

I thank my supervisor Jonas Lampart for his helpful hints and discussions, and
for his help to find my way in Dijon. I also thank my second referee Phan Thàn
Nam who introduced me to mathematical quantum mechanics through his excellent
lectures. A special word of thanks for helpful comments on the first drafts of
this work goes to my fellow-students and friends Jakob Oldenburg, Jonas
Peteranderl, and Markus Frankenbach.

Paderborn, June 2021

Janik Kruse

### Statement of Authorship

I herewith assure that I wrote the present thesis independently and that I
have used no other means than the ones indicated. I have indicated all parts
of the work in which sources are used according to their wording or to their
meaning.

Paderborn,

Janik Kruse

###### Contents

  1. 1 Nelson Model
    1. 1.1 Fock Space Formalism
      1. 1.1.1 Fock Space
      2. 1.1.2 Second Quantisation
      3. 1.1.3 Field and Weyl Operators
    2. 1.2 Nelson Model with Constant Coefficients
    3. 1.3 Nelson Model with Variable Coefficients
  2. 2 Pseudo-differential Calculus
    1. 2.1 Symbols
    2. 2.2 Pseudo-differential Operators
    3. 2.3 Change in Quantisation
    4. 2.4 Symbolic Calculus
      1. 2.4.1 Adjoint Symbol
      2. 2.4.2 Moyal Product
      3. 2.4.3 Construction of a Parametrix
      4. 2.4.4 Functional Calculus
    5. 2.5 $L^{2}$-continuity
    6. 2.6 Operator-valued Symbols
  3. 3 Removal of the Cut-off
    1. 3.1 Pseudo-differential Operators in Nelson Model
    2. 3.2 Gross Transformation
    3. 3.3 Renormalisation
    4. 3.4 Proof of Nelson’s Theorem
  4. 4 IBC Method
    1. 4.1 IBC Hamiltonian
    2. 4.2 Mapping Properties of $G$
    3. 4.3 Extension of the Annihilation Operator
      1. 4.3.1 Construction of $T$
      2. 4.3.2 Mapping Properties of $T$
    4. 4.4 Regularity of Domain Vectors
    5. 4.5 Comparison with the Renormalised Hamiltonian
  5. A Some Functional Analysis

##  Chapter 1 Nelson Model

> First quantisation is a mystery, but second quantisation is a functor. —
> Edward Nelson

The Nelson model describes a physical system of nonrelativistic quantum
particles coupled to a relativistic quantum field of scalar bosons. Nelson
introduced this model in 1964 [13] and rigorously demonstrated the existence
of a well-defined Hamiltonian by renormalisation. In the present work, we are
interested in a generalised model on static spacetimes.

This introductory chapter is organised as follows. First, we explain the Fock
space formalism and second quantisation, which is essential for the present
work. Next, we give a mathematical description of the Nelson model on
Euclidean spacetime and collect some results from the literature. In the last
section, we define the Nelson model on static spacetimes.

###  1.1 Fock Space Formalism

In this section, we present the basic concepts of the Fock space formalism and
second quantisation. More elaborate and detailed expositions are available in
the literature [2, 11]. Furthermore, we include some results from the
appendices of [6, 7].

####  1.1.1 Fock Space

Let $\mathfrak{h}$ denote a complex separable Hilbert space. The state of a
quantum system with $n$ particles is usually described by an element of the
tensor product space $\mathfrak{h}^{\otimes n}$. If the number of particles is
not fixed, the Fock space $\Gamma(\mathfrak{h})$ is introduced as the direct
sum of all $n$-particle subspaces:

$\displaystyle\Gamma(\mathfrak{h})=\bigoplus_{n\in\mathbb{N}}\mathfrak{h}^{\otimes
n}.$ (1.1)

It is understood that $\mathfrak{h}^{\otimes 0}=\mathbb{C}$ and that the
direct (Hilbert space) sum is to be completed.

An element $\Psi\in\Gamma(\mathfrak{h})$ is a sequence
$\Psi=(\Psi^{(0)},\Psi^{(1)},\dots)$ with $\Psi^{(n)}\in\mathfrak{h}^{\otimes
n}$. One particular element is the vacuum vector $\Omega=(1,0,0,\dots)$.

The Fock space $\Gamma(\mathfrak{h})$ is a Hilbert space with the scalar
product

$\displaystyle\langle\Psi_{1},\Psi_{2}\rangle_{\Gamma(\mathfrak{h})}=\sum_{n=0}^{\infty}\langle\Psi_{1}^{(n)},\Psi_{2}^{(n)}\rangle_{\mathfrak{h}^{\otimes
n}}$ (1.2)

and the induced norm

$\displaystyle\lVert\Psi\rVert^{2}_{\Gamma(\mathfrak{h})}=\sum_{n=0}^{\infty}\lVert\Psi^{(n)}\rVert^{2}_{\mathfrak{h}^{\otimes
n}}.$ (1.3)

With respect to this norm, a dense subspace of $\Gamma(\mathfrak{h})$ is given
by the finite-particle vectors:

$\displaystyle\Gamma_{\mathrm{fin}}(\mathfrak{h})=\\{\Psi\in\Gamma(\mathfrak{h})\mid\exists
M\geq 0,\Psi^{(n)}=0\ \forall n\geq M\\}.$ (1.4)

For the rest of this section, let $f,f_{1},\dots,f_{n},g\in\mathfrak{h}$.

In physical applications, further restrictions must usually be imposed on the
Fock space due to quantum statistics. To extract a subspace relevant for the
description of bosonic systems, we introduce the symmetrisation operator
$P_{+}$. On pure tensor states, it is defined as

$\displaystyle P_{+}(f_{1}\otimes\dots\otimes
f_{n})=\frac{1}{n!}\sum_{\sigma\in S_{n}}f_{\sigma(1)}\otimes\dots\otimes
f_{\sigma(n)},$ (1.5)

where the sum is taken over all permutations $\sigma\in S_{n}$. The definition
is extended by linearity and continuity to $\Gamma(\mathfrak{h})$. It is easy
to verify that $P_{+}$ is an orthogonal projection operator. The symmetric
Fock space $\Gamma_{\mathrm{s}}(\mathfrak{h})$, which is the relevant Fock
space for bosonic systems, is the image of $P_{+}$, that is,

$\displaystyle\Gamma_{\mathrm{s}}(\mathfrak{h})=P_{+}\Gamma(\mathfrak{h}),$
(1.6)

and its symmetric $n$-particle subspaces are
$\mathfrak{h}^{\otimes_{\mathrm{s}}n}=P_{+}\mathfrak{h}^{\otimes n}$. An
antisymmetric Fock space can be defined by similar means. However, in the
following, we exclusively use the symmetric Fock space.

####  1.1.2 Second Quantisation

An operator $h$ defined on the Hilbert space $\mathfrak{h}$ can be lifted to
an operator $\mathop{}\\!\mathrm{d}\Gamma(h)$ on the (symmetric) Fock space:

$\displaystyle\mathop{}\\!\mathrm{d}\Gamma(h)\big{|}_{\mathfrak{h}^{\otimes_{\mathrm{s}}n}}=\sum_{j=1}^{n}\mathop{}\\!\mathbb{1}^{\otimes(j-1)}\otimes
h\otimes\mathop{}\\!\mathbb{1}^{\otimes(n-j)}.$ (1.7)

If $h$ is self-adjoint, $\mathop{}\\!\mathrm{d}\Gamma(h)$ is symmetric and
hence closable. Furthermore, a dense subset of analytic vectors of
$\mathop{}\\!\mathrm{d}\Gamma(h)$ is given by any finite sum of symmetrised
products of analytic vectors of $h$. Thus, the operator
$\mathop{}\\!\mathrm{d}\Gamma(h)$ is essentially self-adjoint according to
Nelson’s analytic vector theorem (see [14, Thm. X.39, Cor. 2]). Its self-
adjoint closure, for which we continue to write
$\mathop{}\\!\mathrm{d}\Gamma(h)$, is called the second quantisation of $h$.

######  Example 1.1.1.

The second quantisation of the identity operator is the number operator
$N=\mathop{}\\!\mathrm{d}\Gamma(\mathbb{1})$ with $N\Psi^{(n)}=n\Psi^{(n)}$.

######  Example 1.1.2.

Let $\mathfrak{h}=L^{2}(\mathbb{R}^{d})$ and let $\omega=\sqrt{-\Delta+m^{2}}$
be the relativistic dispersion relation of a boson with mass $m\geq 0$. Then
the action of the second quantisation $\mathop{}\\!\mathrm{d}\Gamma(\omega)$
on $\Psi\in D(\mathop{}\\!\mathrm{d}\Gamma(\omega))$ in Fourier space is given
by

$\displaystyle\widehat{(\mathop{}\\!\mathrm{d}\Gamma(\omega)\Psi)}^{(n)}(\xi_{1},\dots,\xi_{n})=\sum_{j=1}^{n}\omega(\xi_{j})\widehat{\Psi}^{(n)}(\xi_{1},\dots,\xi_{n})$
(1.8)

with $\omega(\xi)=\sqrt{|\xi|^{2}+m^{2}}$. The second quantisation
$\mathop{}\\!\mathrm{d}\Gamma(\omega)$ describes the total energy of a bosonic
system. In the Nelson model, the bosonic quantum field is such a system;
therefore, we call $\mathop{}\\!\mathrm{d}\Gamma(\omega)$ the free field
Hamiltonian.

Another important set of linear operators
$a(f):\mathfrak{h}^{\otimes_{\mathrm{s}}n}\to\mathfrak{h}^{\otimes_{\mathrm{s}}(n-1)}$
(set $\mathfrak{h}^{\otimes_{\mathrm{s}}-1}=0$) and
$a^{*}(f):\mathfrak{h}^{\otimes_{\mathrm{s}}n}\to\mathfrak{h}^{\otimes_{\mathrm{s}}(n+1)}$
is defined as follows. If $n=0$, we initially define $a(f)\Omega=0$ and
$a^{*}(f)\Omega=f$. For $n\geq 1$, we set

$\displaystyle a(f)f_{1}\otimes\dots\otimes f_{n}$
$\displaystyle=\sqrt{n}\langle
f,f_{1}\rangle_{\mathfrak{h}}f_{2}\otimes\dots\otimes f_{n},$ (1.9)
$\displaystyle a^{*}(f)\Psi^{(n)}$
$\displaystyle=\sqrt{n+1}P_{+}\left(f\otimes\Psi^{(n)}\right).$ (1.10)

These operators can be extended to operators on
$\Gamma_{\mathrm{s}}(\mathfrak{h})$. Because $a(f)$ decreases the number of
particles (i.e. $Na(f)\Psi^{(n)}=(n-1)a(f)\Psi^{(n)}$), the operator $a(f)$ is
called annihilation operator. The operator $a^{*}(f)$ increases the number of
particles, and therefore is called creation operator.

The commutation relations are well known and easy to verify:

$\displaystyle[a(f),a(g)]$ $\displaystyle=0,$ (1.11)
$\displaystyle[a^{*}(f),a^{*}(g)]$ $\displaystyle=0,$ (1.12)
$\displaystyle[a(f),a^{*}(g)]$ $\displaystyle=\langle f,g\rangle.$ (1.13)

Moreover, if $h$ is a self-adjoint operator on $\mathfrak{h}$ and $f\in D(h)$,
the commutators of the second quantisation $\mathop{}\\!\mathrm{d}\Gamma(h)$
with annihilation and creation operators are

$\displaystyle[\mathop{}\\!\mathrm{d}\Gamma(h),a^{*}(f)]$
$\displaystyle=a^{*}(hf),$ (1.14)
$\displaystyle[\mathop{}\\!\mathrm{d}\Gamma(h),a(f)]$ $\displaystyle=-a(hf).$
(1.15)

It is useful to have an explicit representation of annihilation and creation
operators for $\mathfrak{h}=L^{2}(\mathbb{R}^{d})$. Let
$x=(x_{1},\dots,x_{n})\in\mathbb{R}^{dn}$ and let
$\hat{x}_{j}\in\mathbb{R}^{d(n-1)}$ denote the vector $x$ with the entry $j$
removed. Then, for any $\Psi^{(n)}\in L^{2}(\mathbb{R}^{dn})$,
$\Psi^{(n-1)}\in L^{2}(\mathbb{R}^{d(n-1)})$, annihilation and creation
operators may be written as

$\displaystyle a(f)\Psi^{(n)}(\hat{x}_{n-1})$
$\displaystyle=\sqrt{n}\int_{\mathbb{R}^{d}}\overline{f(x_{n})}\Psi^{(n)}(x)\mathop{}\\!\mathrm{d}x_{n},$
(1.16) $\displaystyle a^{*}(f)\Psi^{(n-1)}(x)$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{j=1}^{n}f(x_{j})\Psi^{(n-1)}(\hat{x}_{j}).$
(1.17)

The following lemma provides useful norm estimates for annihilation and
creation operators.

######  Lemma 1.1.3.

Let $h\geq 1$ be a self-adjoint operator on $\mathfrak{h}$, and $f\in
D(h^{-\alpha})$ for an $\alpha\geq 1/2$. Then, for every $\Psi\in
D(\mathop{}\\!\mathrm{d}\Gamma(h)^{\alpha})$,

$\displaystyle\lVert a(f)\Psi\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\leq\lVert
h^{-\alpha}f\rVert_{\mathfrak{h}}\lVert\mathop{}\\!\mathrm{d}\Gamma(h)^{\alpha}\Psi\rVert_{\Gamma(\mathfrak{h})},$
(1.18) $\displaystyle\lVert a^{*}(f)\Psi\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\leq\lVert
h^{-\alpha}f\rVert_{\mathfrak{h}}\lVert\mathop{}\\!\mathrm{d}\Gamma(h)^{\alpha}\Psi\rVert_{\Gamma(\mathfrak{h})}+\lVert
f\rVert_{\mathfrak{h}}\lVert\Psi\rVert_{\Gamma(\mathfrak{h})}.$ (1.19)

Moreover, if $f,g\in D(h^{-\alpha/2})$, then, for every $\Psi\in
D(\mathop{}\\!\mathrm{d}\Gamma(h)^{\alpha})$,

$\displaystyle\lVert(N+1)^{-\frac{1}{2}}a(f)a(g)\Psi\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\leq\lVert h^{-\frac{\alpha}{2}}f\rVert_{\mathfrak{h}}\lVert
h^{-\frac{\alpha}{2}}g\rVert_{\mathfrak{h}}\lVert\mathop{}\\!\mathrm{d}\Gamma(h)^{\alpha}\Psi\rVert_{\Gamma(\mathfrak{h})}.$
(1.20)

###### Proof.

We can assume that $h\geq 1$ is a multiplication operator on
$L^{2}(\mathbb{R}^{d})$ due to the spectral theorem. Because $\Psi^{(n)}$ is
symmetric in $x=(x_{1},\dots,x_{n})\in\mathbb{R}^{nd}$, we obtain the
following equality:

$\displaystyle\langle\Psi^{(n)},\mathop{}\\!\mathrm{d}\Gamma(h^{2\alpha})\Psi^{(n)}\rangle_{\mathfrak{h}^{\otimes
n}}$
$\displaystyle=\sum_{i=1}^{n}\int_{\mathbb{R}^{nd}}h(x_{i})^{2\alpha}|\Psi^{(n)}(x)|^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle=n\int_{\mathbb{R}^{nd}}h(x_{n})^{2\alpha}|\Psi^{(n)}(x)|^{2}\mathop{}\\!\mathrm{d}x.$
(1.21)

From the above, it follows that

$\displaystyle\lVert a(f)\Psi^{(n)}\rVert^{2}_{\mathfrak{h}^{\otimes n}}$
$\displaystyle=n\int_{\mathbb{R}^{(n-1)d}}\left|\int_{\mathbb{R}^{d}}\overline{f(x_{n})}\Psi^{(n)}(x)\mathop{}\\!\mathrm{d}x_{n}\right|^{2}\mathop{}\\!\mathrm{d}\hat{x}_{n}$
$\displaystyle\leq n\lVert
h^{-\alpha}f\rVert^{2}_{\mathfrak{h}}\int_{\mathbb{R}^{nd}}h(x_{n})^{2\alpha}|\Psi^{(n)}(x)|^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle=\lVert
h^{-\alpha}f\rVert^{2}_{\mathfrak{h}}\lVert\mathop{}\\!\mathrm{d}\Gamma(h^{2\alpha})^{\frac{1}{2}}\Psi^{(n)}\rVert^{2}_{\mathfrak{h}^{\otimes
n}}.$ (1.22)

The condition $2\alpha\geq 1$ ensures that
$\mathop{}\\!\mathrm{d}\Gamma(h^{2\alpha})\leq\mathop{}\\!\mathrm{d}\Gamma(h)^{2\alpha}$.
This proves (1.18) if we sum (1.22) over $n$.

The inequality (1.19) for the creation operator follows from (1.18) and

$\displaystyle\lVert a^{*}(f)\Psi\rVert^{2}_{\Gamma(\mathfrak{h})}=\lVert
a(f)\Psi\rVert^{2}_{\Gamma(\mathfrak{h})}+\lVert
f\rVert^{2}_{\mathfrak{h}}\lVert\Psi\rVert^{2}_{\Gamma(\mathfrak{h})},$ (1.23)

which is a consequence of the commutation relation (1.13). Inequality (1.20)
can be proven using similar techniques. ∎

######  Example 1.1.4.

If $h=\mathbb{1}$ in Lemma 1.1.3, then

$\displaystyle\lVert a(f)\Psi\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\leq\lVert f\rVert_{\mathfrak{h}}\lVert
N^{\frac{1}{2}}\Psi\rVert_{\Gamma(\mathfrak{h})},$ (1.24) $\displaystyle\lVert
a^{*}(f)\Psi\rVert_{\Gamma(\mathfrak{h})}$ $\displaystyle\leq\lVert
f\rVert_{\mathfrak{h}}\lVert(N+1)^{\frac{1}{2}}\Psi\rVert_{\Gamma(\mathfrak{h})}.$
(1.25)

This shows that $D(a(f))\supset D(N^{1/2})$ and $D(a^{*}(f))\supset
D(N^{1/2})$ for the domains of the annihilation and creation operators.

####  1.1.3 Field and Weyl Operators

Annihilation and creation operators are adjoint to each other. Symmetric
operators like the (Segal) field operator

$\displaystyle\Phi(f)=\frac{1}{\sqrt{2}}\left(a^{*}(f)+a(f)\right)$ (1.26)

and the conjugate momentum operator

$\displaystyle\Pi(f)=\frac{\mathop{}\\!\mathrm{i}}{\sqrt{2}}\left(a^{*}(f)-a(f)\right)$
(1.27)

are constructed by combining annihilation and creation operators. Observe that
$\Phi(f)$ and $\Pi(f)$ are $\mathbb{R}$-linear but not $\mathbb{C}$-linear in
$f$. Moreover, $\Pi(f)=\Phi(\mathop{}\\!\mathrm{i}f)$. Considering the
following proposition, we denote the self-adjoint closures of $\Phi(f)$ and
$\Pi(f)$ by the very same symbol.

######  Proposition 1.1.5.

For every $f\in\mathfrak{h}$, the field operator $\Phi(f)$ and the conjugate
momentum operator $\Pi(f)$ are essentially self-adjoint on
$\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$.

###### Proof.

We prove the proposition for $\Phi(f)$. The proof for $\Pi(f)$ is identical.
For every $\Psi\in\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$ and $t\geq 0$, the
series

$\displaystyle\sum_{n=0}^{\infty}\frac{t^{n}}{n!}\lVert\Phi(f)^{n}\Psi\rVert_{\Gamma(\mathfrak{h})}<\infty$
(1.28)

is finite due to (1.24) and (1.25). Thus,
$\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$ is a (dense) set of analytic vectors
of $\Phi(f)$. The proposition follows from Nelson’s analytic vector theorem
(see [14, Thm. X.39, Cor. 2]). ∎

It is immediate from the commutation relations (1.11)–(1.13) for annihilation
and creation operators that

$\displaystyle[\Phi(f),\Phi(g)]$
$\displaystyle=\mathop{}\\!\mathrm{i}\Im\langle f,g\rangle_{\mathfrak{h}},$
(1.29) $\displaystyle[\Pi(f),\Pi(g)]$
$\displaystyle=\mathop{}\\!\mathrm{i}\Im\langle f,g\rangle_{\mathfrak{h}},$
(1.30) $\displaystyle[\Phi(f),\Pi(g)]$
$\displaystyle=\mathop{}\\!\mathrm{i}\Re\langle f,g\rangle_{\mathfrak{h}},$
(1.31)

where $\Im(z)$ denotes the imaginary part of $z\in\mathbb{C}$, and $\Re(z)$
the real part.

The Weyl operators

$\displaystyle V(f)=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(f)},\ \
W(f)=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Phi(f)}$ (1.32)

are unitary because the field operators $\Phi(f)$ are self-adjoint. It is not
difficult to verify that the Weyl operators satisfy the following algebraic
relations:

$\displaystyle V(f)V(g)$
$\displaystyle=\mathop{}\\!\mathrm{e}^{-\frac{\mathop{}\\!\mathrm{i}}{2}\langle
f,g\rangle_{\mathfrak{h}}}V(f+g),$ (1.33) $\displaystyle W(f)W(g)$
$\displaystyle=\mathop{}\\!\mathrm{e}^{-\frac{\mathop{}\\!\mathrm{i}}{2}\langle
f,g\rangle_{\mathfrak{h}}}W(f+g).$ (1.34)

Sometimes these relations are used as the defining property of the Weyl
algebra $\\{V(f),f\in\mathfrak{h}\\}=\\{W(f),f\in\mathfrak{h}\\}$. The
following proposition collects more interesting properties.

######  Proposition 1.1.6.

The following statements are true:

  1. 1.

The Weyl operator $V(g)$ leaves the domain $D(\Phi(f))$ invariant and

$\displaystyle V(g)\Phi(f)V(g)^{*}=\Phi(f)+\Re\langle
f,g\rangle_{\mathfrak{h}}.$ (1.35)

  2. 2.

If $h$ is a self-adjoint operator on $\mathfrak{h}$ and $g\in D(h)$, then
$V(g)$ leaves the domain $D(\mathop{}\\!\mathrm{d}\Gamma(h))$ invariant and

$\displaystyle
V(g)\mathop{}\\!\mathrm{d}\Gamma(h)V(g)^{*}=\mathop{}\\!\mathrm{d}\Gamma(h)+\Phi(hg)+\frac{1}{2}\langle
hg,g\rangle_{\mathfrak{h}}.$ (1.36)

  3. 3.

If $\Psi\in D(N^{1/2})$, then

$\displaystyle\lVert(V(f)-V(g))\Psi\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\leq\lVert\Pi(f-g)\Psi\rVert_{\Gamma(\mathfrak{h})}+\frac{1}{2}|\Im\langle
f,g\rangle_{\mathfrak{h}}|\lVert\Psi\rVert_{\Gamma(\mathfrak{h})}.$ (1.37)

###### Proof.

  1. 1.

According to Proposition 1.1.5 and its proof, every
$\Psi\in\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$ is an analytic vector for the
conjugate momentum operator $\Pi(g)$. Hence, we write $\Phi(f)V(g)^{*}\Psi$ as
a power series, that is,

$\displaystyle\Phi(f)V(g)^{*}\Psi=\sum_{k=0}^{\infty}\frac{(-\mathop{}\\!\mathrm{i})^{k}}{k!}\Phi(f)\Pi(g)^{k}\Psi.$
(1.38)

From the commutation relation (1.31) and induction, it follows that, for all
$k\geq 1$,

$\displaystyle[\Phi(f),\Pi(g)^{k}]=\mathop{}\\!\mathrm{i}k\Re\langle
f,g\rangle_{\mathfrak{h}}\Pi(g)^{k-1}.$ (1.39)

Using this commutator in the power series expansion above, we obtain

$\displaystyle\Phi(f)V(g)^{*}\Psi=V(g)^{*}(\Phi(f)+\Re\langle
f,g\rangle_{\mathfrak{h}})\Psi.$ (1.40)

This proves (1.35) on $\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$. Furthermore,
from (1.40), we derive the following inequality for
$\Psi\in\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$:

$\displaystyle\lVert\Phi(f)V(g)^{*}\Psi\rVert_{\Gamma(\mathfrak{h})}\leq\lVert\Phi(f)\Psi\rVert_{\Gamma(\mathfrak{h})}+|\Re\langle
f,g\rangle_{\mathfrak{h}}|\lVert\Psi\rVert_{\Gamma(\mathfrak{h})}.$ (1.41)

This inequality extends to $\Psi\in D(\Phi(f))$ because
$\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$ is dense in $D(\Phi(f))$. Because
$\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$ is a core for $\Phi(f)$, there is, for
every $\Psi\in D(\Phi(f))$, a sequence
$(\Psi_{k})_{k\in\mathbb{N}}\subset\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$ such
that
$\lVert\Phi(f)(\Psi_{k}-\Psi)\rVert_{\Gamma(\mathfrak{h})}+\lVert\Psi_{k}-\Psi\rVert_{\Gamma(\mathfrak{h})}\to
0$ as $k\to\infty$. Hence, $\Phi(f)V(g)^{*}\Psi_{k}\to\Phi(f)V(g)^{*}\Psi$
converges due to (1.41), and (1.40) extends to $D(\Phi(f))$.

Moreover, we deduce that $V(g)^{*}\Psi\in D(\Phi(f))$ because
$\Phi(f)V(g)^{*}\Psi$ is well defined. If we repeat the same argument but
replace $V(g)^{*}$ with $V(g)$, we obtain $V(g)\Psi\in D(\Phi(f))$ and
conclude that $V(g)D(\Phi(f))=D(\Phi(f))$.

  2. 2.

The proof is similar to that of the first item; therefore, we only sketch it.
For $\Psi\in
D(\mathop{}\\!\mathrm{d}\Gamma(h))\cap\Gamma_{\mathrm{s,fin}}(\mathfrak{h})$,
we expand $\mathop{}\\!\mathrm{d}\Gamma(h)V(g)^{*}\Psi$ in a power series:

$\displaystyle\mathop{}\\!\mathrm{d}\Gamma(h)V(g)^{*}\Psi=\sum_{k=0}^{\infty}\frac{(-\mathop{}\\!\mathrm{i})^{k}}{k!}\mathop{}\\!\mathrm{d}\Gamma(h)\Pi(g)^{k}\Psi.$
(1.42)

From (1.14) and (1.15), it follows that

$\displaystyle[\mathop{}\\!\mathrm{d}\Gamma(h),\mathop{}\\!\mathrm{i}\Pi(g)]=-\Phi(hg),$
(1.43)

and, by induction, for every $k\geq 2$,

$\displaystyle[\mathop{}\\!\mathrm{d}\Gamma(h),\mathop{}\\!\mathrm{i}^{k}\Pi(g)^{k}]=-k\mathop{}\\!\mathrm{i}^{k-1}\Pi(g)^{k-1}\Phi(hg)+\frac{k(k-1)}{2}\mathop{}\\!\mathrm{i}^{k-2}\Pi(g)^{k-2}\langle
hg,g\rangle_{\mathfrak{h}}.$ (1.44)

Inserting this commutator into (1.42), we obtain

$\displaystyle\sum_{k=0}^{\infty}\frac{(-\mathop{}\\!\mathrm{i})^{k}}{k!}\mathop{}\\!\mathrm{d}\Gamma(h)\Pi(g)^{k}\Psi=V(g)^{*}(\mathop{}\\!\mathrm{d}\Gamma(h)-\Phi(hg)+\frac{1}{2}\langle
hg,g\rangle_{\mathfrak{h}})\Psi.$ (1.45)

  3. 3.

For $\Psi\in D(N^{1/2})$, the following equality holds true:

$\displaystyle\lVert(\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(f)}-\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(g)})\Psi\rVert_{\Gamma(\mathfrak{h})}=\lVert(\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}\Pi(g)}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(f)}-1)\Psi\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\ \ \
=\lVert\int_{0}^{1}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\left(\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}t\Pi(g)}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\Pi(f)}\Psi\right)\mathop{}\\!\mathrm{d}t\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\ \ \
=\lVert\int_{0}^{1}\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}t\Pi(g)}\Pi(f-g)\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\Pi(f)}\Psi\mathop{}\\!\mathrm{d}t\rVert_{\Gamma(\mathfrak{h})}.$
(1.46)

From (1.30), we derive the commutator

$\displaystyle[\Pi(f-g),\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\Pi(f)}]=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\Pi(f)}t\Im\langle
f-g,f\rangle.$ (1.47)

Inserting this commutator into the above equality, we obtain the claim:

$\displaystyle\lVert(\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(f)}-\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(g)})\Psi\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle=\lVert\int_{0}^{1}\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}t\Pi(g)}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\Pi(f)}(\Pi(f-g)+t\Im\langle
f-g,f\rangle)\Psi\mathop{}\\!\mathrm{d}t\rVert_{\Gamma(\mathfrak{h})}$
$\displaystyle\ \ \ \leq\int_{0}^{1}\lVert\Pi(f-g)\Psi+t\Im\langle
f-g,f\rangle_{\mathfrak{h}}\Psi\rVert_{\Gamma(\mathfrak{h})}\mathop{}\\!\mathrm{d}t$
$\displaystyle\ \ \
\leq\lVert\Pi(f-g)\Psi\rVert_{\Gamma(\mathfrak{h})}+\frac{1}{2}|\Im\langle
f,g\rangle_{\mathfrak{h}}|\lVert\Psi\rVert_{\Gamma(\mathfrak{h})}.$ (1.48)

∎

###  1.2 Nelson Model with Constant Coefficients

After introducing the Fock space formalism, we can now describe the Nelson
model mathematically. As explained earlier, the Nelson model describes the
coupling of nonrelativistic quantum particles to a relativistic field of
bosons. We only discuss the case of a single nonrelativistic particle
(referred to simply as ’particle’ in the following text) because the general
case offers no more difficulties rather than complicating the notation.
Furthermore, we assume that the bosonic field mass $m>0$ is strictly positive
to avoid infrared divergences.

The underlying Hilbert space $\mathfrak{H}$ of the Nelson model is a tensor
product space. It consists of the Hilbert space
$\mathfrak{K}=L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}X)$ for the particle
and the symmetric Fock space $\Gamma_{\mathrm{s}}(\mathfrak{h})$ with
$\mathfrak{h}=L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x)$ for the field:

$\displaystyle\mathfrak{H}=\mathfrak{K}\otimes\Gamma_{\mathrm{s}}(\mathfrak{h})=L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}X)\otimes\Gamma_{\mathrm{s}}(L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x)).$
(1.49)

The $n$-boson subspaces are denoted by
$\mathfrak{H}^{(n)}=\mathfrak{K}\otimes\mathfrak{h}^{\otimes_{\mathrm{s}}n}$.

The free particle Hamiltonian $-\Delta_{X}$ and the free field Hamiltonian
$\mathop{}\\!\mathrm{d}\Gamma(\omega)$ with dispersion relation
$\omega=\sqrt{-\Delta+m^{2}}$ (see Example 1.1.2) combine into the free Nelson
Hamiltonian:

$\displaystyle
H_{0}=-\Delta_{X}\otimes\mathbb{1}+\mathbb{1}\otimes\mathop{}\\!\mathrm{d}\Gamma(\omega).$
(1.50)

Because it is the sum of two commuting self-adjoint operators, $H_{0}$ is
self-adjoint on $D(H_{0})=D(-\Delta_{X}\otimes\mathbb{1})\cap
D(\mathbb{1}\otimes\mathop{}\\!\mathrm{d}\Gamma(\omega))$.

If we formally perturb the free Nelson Hamiltonian by the field operator
$\Phi(\omega^{-\frac{1}{2}}\delta_{X})$ defined in (1.26), where $\delta_{X}$
is the Dirac delta centred at the particle position $X$, we obtain the formal
Nelson Hamiltonian:

$\displaystyle
H=-\Delta_{X}\otimes\mathbb{1}+\mathbb{1}\otimes\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega^{-\frac{1}{2}}\delta_{X}).$
(1.51)

However, the formal Nelson Hamiltonian is not a well-defined operator. The
problem is that the form factor $\omega^{-\frac{1}{2}}\delta_{X}$ is not an
element of $\mathfrak{h}=L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x)$. If
the creation operator $a^{*}(\omega^{-\frac{1}{2}}\delta_{X})$ is defined as
in (1.10) even though $\omega^{-\frac{1}{2}}\delta_{X}\notin\mathfrak{h}$,
then its natural domain is
$D(a^{*}(\omega^{-\frac{1}{2}}\delta_{X}))=\\{0\\}$. Hence, the field operator
$\Phi(\omega^{-\frac{1}{2}}\delta_{X})$ is not densely defined, and $H$ cannot
be self-adjoint as a sum of self-adjoint operators.

######  Remark 1.2.1.

The annihilation operator is typically easier to handle because the scalar
product $\langle\omega^{-\frac{1}{2}}\delta_{X},f\rangle_{\mathfrak{h}}$ in
the definition of the annihilation operator may be well defined for a dense
set of $f\in\mathfrak{h}$, even if
$\omega^{-\frac{1}{2}}\delta_{X}\notin\mathfrak{h}$. If
$a(\omega^{-\frac{1}{2}}\delta_{X})$ were defined on $D(H_{0}^{1/2})$, a
common method of interpreting (1.51) would be in the sense of quadratic forms.
However, this is not the case in the Nelson model.

The Nelson model has yet to be regularised. One possibility is to replace the
singular delta function with a smoother function $\rho_{\Lambda}$ which
depends on the scale $\Lambda$ in such a way that $\rho_{\Lambda}$ approaches
the delta function for large $\Lambda$. We realise this as follows. Assume
that $0\leq\rho\in\mathcal{S}(\mathbb{R}^{3})$ is a nonnegative symmetric
Schwartz function with $\int_{\mathbb{R}^{3}}\rho=1$ and set

$\displaystyle\rho_{\Lambda,X}(x)=\Lambda^{3}\rho(\Lambda(x-X)),\
\rho_{\Lambda}(x)=\rho_{\Lambda,0}(x).$ (1.52)

Then, for every $s<-3/2$, $\rho_{\Lambda,X}$ converges in
$H^{s}(\mathbb{R}^{3})$ to $\delta_{X}$ uniformly in $X$:

$\displaystyle\lVert\rho_{\Lambda,X}-\delta_{X}\rVert_{H^{s}}^{2}$
$\displaystyle=\lVert(1+|\cdot|^{2})^{\frac{s}{2}}(\widehat{\rho}_{\Lambda}-1)\rVert_{L^{2}}^{2}$
$\displaystyle=\int_{\mathbb{R}^{3}}\frac{\left|\widehat{\rho}\left(\xi/\Lambda\right)-1\right|^{2}}{(1+|\xi|^{2})^{-s}}\mathop{}\\!\mathrm{d}\xi\stackrel{{\scriptstyle\Lambda\to\infty}}{{\longrightarrow}}0.$
(1.53)

The convergence in the last step is justified because
$\widehat{\rho}(\xi/\Lambda)\to 1$ pointwise and because $s<-3/2$ is
sufficiently small to apply the dominated convergence theorem.

The cut-off Hamiltonian $H_{\Lambda}$ is obtained by replacing $\delta_{X}$
with $\rho_{\Lambda,X}$:

$\displaystyle
H_{\Lambda}=-\Delta_{X}\otimes\mathbb{1}+\mathbb{1}\otimes\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X}).$
(1.54)

Physically, this replacement suppresses energies above a certain scale
$\Lambda$, which are beyond the validity of the model. Therefore,
$\rho_{\Lambda}$ is called an ultraviolet cut-off. Equivalently, the cut-off
can be interpreted as assigning a nonzero radius to particles that are usually
point-like objects.

The following proposition proves that $H_{\Lambda}$ is a well-defined self-
adjoint operator.

######  Proposition 1.2.2.

The cut-off Hamiltonian $H_{\Lambda}$ is self-adjoint on
$D(H_{\Lambda})=D(H_{0})$.

###### Proof.

Observe that the number operator $N$ is dominated by $H_{0}/m$ due to
$\mathop{}\\!\mathrm{d}\Gamma(\mathbb{1})\leq\mathop{}\\!\mathrm{d}\Gamma(\omega)/m$.
Furthermore, (1.24) and (1.25) imply that $D(N^{1/2})\subset
D(\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X}))$. Thus, $D(H_{0})\subset
D(H_{0}^{1/2})\subset D(N^{1/2})\subset
D(\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X}))$.

According to the Kato–Rellich theorem, it suffices to prove that the
perturbation $\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X})$ is relatively
bounded with respect to $H_{0}$ and that the relative bound is strictly
smaller than 1. Again, from (1.24) and (1.25), it follows that, for every
$\epsilon>0$,

$\displaystyle\lVert\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X})\Psi\rVert_{\mathfrak{H}}$
$\displaystyle\leq\sup_{X\in\mathbb{R}^{3}}\lVert\omega^{-\frac{1}{2}}\rho_{\Lambda,X}\rVert_{\mathfrak{h}}\lVert(N+1)^{\frac{1}{2}}\Psi\rVert_{\mathfrak{H}}$
$\displaystyle\leq\sup_{X\in\mathbb{R}^{3}}\lVert\omega^{-\frac{1}{2}}\rho_{\Lambda,X}\rVert_{\mathfrak{h}}(\epsilon\lVert
H_{0}\Psi\rVert_{\mathfrak{H}}+C_{\epsilon}\lVert\Psi\rVert_{\mathfrak{H}}).$
(1.55)

The factor
$\sup_{X}\lVert\omega^{-\frac{1}{2}}\rho_{\Lambda,X}\rVert_{\mathfrak{h}}=\lVert\omega^{-\frac{1}{2}}\rho_{\Lambda}\rVert_{\mathfrak{h}}$
is finite due to the cut-off $\rho_{\Lambda}$. In the second inequality, we
used $N\leq H_{0}/m$ and Young’s inequality. Choosing an $\epsilon$ that is
small enough yields the desired bound. ∎

The cut-off $\rho_{\Lambda}$ ensures that the cut-off Hamiltonian is well
defined and self-adjoint. However, the choice of a particular cut-off is
arbitrary, and it is expected that relevant physical quantities (e.g. energy
levels) heavily depend on how the cut-off is chosen.

The actual problem we face in the Nelson model without the cut-off is that the
Nelson Hamiltonian has infinite vacuum energy, whereas the cut-off Hamiltonian
has finite vacuum energy $-E_{\Lambda}$. It makes sense to remove the infinite
vacuum energy as we are usually interested in energy differences rather than
absolute values. Nelson [13] has proven the following theorem, which provides
a decent candidate for a well-defined Nelson Hamiltonian.

######  Theorem 1.2.3 (Nelson).

There are constants $E_{\Lambda}$ such that $H_{\Lambda}+E_{\Lambda}$
converges in the strong resolvent sense to a self-adjoint bounded from below
operator $H$ as $\Lambda\to\infty$.

A sequence $(A_{n})_{n\in\mathbb{N}}$ of self-adjoint operators on
$\mathfrak{H}$ converges in the strong resolvent sense to a self-adjoint
operator $A$ if $(A_{n}-\lambda)^{-1}\Psi\to(A-\lambda)^{-1}\Psi$ for every
(or equivalently for one) $\lambda\in\mathbb{C}\backslash\mathbb{R}$ and every
$\Psi\in\mathfrak{H}$.

The operator $H$ from Nelson’s theorem is the renormalised Nelson Hamiltonian.
We prove Nelson’s theorem in Chapter 3 for the Nelson model on static
spacetimes, which we introduce in the next section.

######  Remark 1.2.4.

A sequence of self-adjoint operators converges in the strong resolvent sense
if and only if the associated unitary groups converge strongly (see [15, Thm.
VIII.21]). That is, the convergence in Theorem 1.2.3 is equivalent to

$\displaystyle\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t(H_{\Lambda}+E_{\Lambda})}\Psi\to\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}tH}\Psi$
(1.56)

as $\Lambda\to\infty$ for all $t\in\mathbb{R}$ and $\Psi\in\mathfrak{H}$.
Although $\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}tH_{\Lambda}}\Psi$
diverges as $\Lambda\to\infty$, the state determined by it converges in the
sense that

$\displaystyle\langle\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}tH_{\Lambda}}\Psi,A\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}tH_{\Lambda}}\Psi\rangle\to\langle\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}tH}\Psi,A\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}tH}\Psi\rangle$
(1.57)

for any bounded self-adjoint operator $A$.

######  Remark 1.2.5.

Recently, Griesemer and Wünsch [7] proved that the convergence of
$H_{\Lambda}+E_{\Lambda}$ in Theorem 1.2.3 also holds in the norm resolvent
sense.

In summary, we defined the formal Nelson Hamiltonian in (1.51), realised that
the Hamiltonian is not well defined and that putting a cut-off on the
interaction solves the problem of self-adjointness. Then, we argued that the
cut-off can be removed by extracting the divergent vacuum energy
$-E_{\Lambda}$ of the system.

It remains an open question whether we have any chance of describing the
renormalised Hamiltonian $H$ because many interesting properties of the cut-
off Hamiltonians $H_{\Lambda}$ are not expected to survive the limit
$\Lambda\rightarrow\infty$. One example is that $D(H_{\Lambda})=D(H_{0})$ for
all $\Lambda<\infty$, but $D(H_{0})\cap D(H)=\\{0\\}$. This property was
conjectured by Nelson and proven by Griesemer and Wünsch [7] without
describing $D(H)$ explicitly.

In Chapter 4, we obtain an explicit description of the renormalised Nelson
Hamiltonian and its domain due to the novel IBC method introduced by Teufel
and Tumulka [20, 21]. Lampart and Schmidt [10] already demonstrated that the
IBC method applies to the Nelson-type Hamiltonians. The present work aims to
generalise these recently obtained results to the Nelson model on static
spacetimes, also known as the Nelson model with variable coefficients.

###  1.3 Nelson Model with Variable Coefficients

The Nelson model with variable coefficients is an extension of the Nelson
model, where the Laplace operator is replaced with a second-order partial
differential operator with variable coefficients. This extension describes the
Nelson model on static spacetimes, which is motivated by the fact that the
correct generalisation of the Laplace operator to pseudo-Riemannian manifolds
is the Laplace–Beltrami operator. The Nelson Hamiltonian with variable
coefficients was introduced in [4, 5], and we adopt some of the notation used
in these papers.

We begin with redefining the free particle Hamiltonian $K_{0}$ by

$\displaystyle K_{0}=-\sum_{1\leq j,k\leq
3}\partial_{X_{j}}g^{jk}(X)\partial_{X_{k}}=-\partial_{X}\cdot
g(X)\partial_{X},$ (1.58)

where the matrix $g(X):=[g^{jk}(X)]$ is assumed to be symmetric for every
$X\in\mathbb{R}^{3}$, which ensures that $K_{0}$ is symmetric. However, to
obtain self-adjointness, we must impose further regularity conditions on
$g(X)$. Later, we assume that the matrix coefficients $g^{jk}$ are smooth, but
for now it suffices to assume that they are Lipschitz continuous.
Additionally, we impose the following uniform ellipticity condition on the
matrix $g(X)$:

$\displaystyle C_{0}\leq g(X)\leq C_{1},\ C_{0}>0.$ (1.59)

The assumption of ellipticity is useful for applying well-known results from
the theory of elliptic operators. Furthermore, it guarantees that $K_{0}$ is
’similar’ to the Laplacian, which is easier to handle than $K_{0}$.

######  Proposition 1.3.1.

Assume that, for every $X\in\mathbb{R}^{3}$, the matrix $g(X)$ is symmetric.
Furthermore, assume that $g$ is uniform elliptic and that its matrix
coefficients are Lipschitz continuous. Then $K_{0}$ is self-adjoint on
$D(K_{0})=D(-\Delta)$.

###### Proof.

We define a quadratic form $q:H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})\to\mathbb{C}$ by

$\displaystyle q(u,v)=\sum_{1\leq j,k\leq
3}\langle\partial_{X_{j}}v,g^{jk}(X)\partial_{X_{k}}u\rangle=\langle\partial_{X}u,g(X)\partial_{X}v\rangle.$
(1.60)

The strategy of this proof is to associate a self-adjoint operator $B$ with
the quadratic form $q$ and prove that $B=K_{0}$. Clearly, $q$ is bounded from
below by 0:

$\displaystyle q(u,u)=\langle\partial_{X}u,g(X)\partial_{X}u\rangle\geq
C_{0}\lVert\partial_{X}u\rVert^{2}\geq 0.$ (1.61)

Furthermore, $q$ is closed, which follows from the fact that the quadratic
form norm

$\displaystyle\lVert u\rVert_{q}^{2}=q(u,u)+\lVert u\rVert^{2}$ (1.62)

is equivalent to the Sobolev $H^{1}$-norm by condition (1.59). Thus, by a
well-known theorem in the theory of quadratic forms (see [18, Thm. 2.14]), a
unique self-adjoint operator $B\geq 0$ exists with form domain
$Q(B)=H^{1}(\mathbb{R}^{3})$ such that $q(u,u)=\lVert
B^{\frac{1}{2}}u\rVert^{2}$. More explicitly, $B$ is given by

$\displaystyle D(B)$ $\displaystyle=\\{u\in H^{1}(\mathbb{R}^{3})\mid\exists
f\in L^{2}(\mathbb{R}^{3}),q(u,v)=\langle f,v\rangle\ \forall v\in
H^{1}(\mathbb{R}^{3})\\}$ (1.63) $\displaystyle Bu$ $\displaystyle=f.$ (1.64)

The condition $q(u,v)=\langle f,v\rangle$ for all $v\in H^{1}(\mathbb{R}^{3})$
is equivalent to calling $u\in H^{1}(\mathbb{R}^{3})$ a weak solution of the
partial differential equation $K_{0}u=f$. From the elliptic regularity theorem
(see [8, Thm. 17.2.7] and its proof) we deduce that $u\in
H^{2}(\mathbb{R}^{3})$. To apply [8, Thm. 17.2.7], $g(X)$ must be symmetric
and its matrix coefficients Lipschitz continuous.

Hence, $D(B)$ is contained in $H^{2}(\mathbb{R}^{3})$. As the other inclusion
is trivial (for $u\in H^{2}(\mathbb{R}^{3})$, choose $f=K_{0}u$),
$D(B)=H^{2}(\mathbb{R}^{3})$. On $H^{2}(\mathbb{R}^{3})$, the action of $B$
and $K_{0}$ coincide; thus, $B=K_{0}$. ∎

We add an exterior particle potential $W(X)$ to the free particle Hamiltonian
$K_{0}$. This potential should be chosen such that $K_{0}+W(X)$ is essentially
self-adjoint. The self-adjoint closure of this operator is denoted by $K$.

The free field Hamiltonian on static spacetimes is defined similarly to the
free particle Hamiltonian. We set

$\displaystyle h_{0}$ $\displaystyle=-\sum_{1\leq j,k\leq
3}\partial_{x_{j}}g^{jk}(x)\partial_{x_{k}}=-\partial_{x}\cdot
g(x)\partial_{x},$ (1.65) $\displaystyle h$ $\displaystyle=h_{0}+\mu(x)^{2}.$
(1.66)

We allow for a position-dependent field mass $0\leq\mu\in
L^{\infty}(\mathbb{R}^{3})$. Clearly, $h$ is a self-adjoint operator on
$D(h)=H^{2}(\mathbb{R}^{3})$ by the previous proposition and the fact that
$\mu$ is bounded.

The dispersion relation is $\omega=h^{1/2}$, and the free field Hamiltonian is
the second quantisation $\mathop{}\\!\mathrm{d}\Gamma(\omega)$. We denote by
$m$ the infimum of the spectrum of $\omega$. If $m>0$, we regard the model as
massive, and massless otherwise. As in the Nelson model with constant
coefficients, we assume that the Nelson model with variable coefficients is
massive to avoid infrared divergences.

Altogether, the Nelson Hamiltonian with variable coefficients and cut-off is
given by

$\displaystyle
H_{\Lambda}=K\otimes\mathbb{1}+\mathbb{1}\otimes\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X}).$
(1.67)

The cut-off function $\rho\in\mathcal{S}(\mathbb{R}^{3})$ is chosen as in the
previous section.

What makes the Nelson model with variable coefficients more challenging to
analyse than the Nelson model with constant coefficients is that the free
particle Hamiltonian $K_{0}$ or the operator $h_{0}$ cannot be replaced by a
multiplication operator in Fourier space. Instead, we rely on the pseudo-
differential calculus, which is introduced in the next chapter.

##  Chapter 2 Pseudo-differential Calculus

The pseudo-differential calculus defines an algebraic correspondence between
classical and quantum observables known as quantisation. The classical
observables in this calculus are represented by symbols that are smooth
functions defined on the physical phase space, a subset of
$\mathbb{R}^{d}\times\mathbb{R}^{d}$. Each symbol defines a pseudo-
differential operator, which is the corresponding quantum observable. However,
because of the ordering ambiguity in quantum mechanics (classical observables
commute, but their quantum mechanical counterparts generally do not), the
correspondence between classical and quantum observables is ambiguous.
Instead, different quantisation schemes exist.

From a mathematical viewpoint, pseudo-differential operators provide an
extended class of differential operators. To understand this, let $p(D)$ be a
linear differential operator of order $m$, that is,

$\displaystyle p(D)u(x)=\sum_{|\alpha|\leq m}g_{\alpha}D_{x}^{\alpha}u(x),$
(2.1)

where $g_{\alpha}\in\mathbb{C}$, $D_{x}=-\mathop{}\\!\mathrm{i}\partial_{x}$,
and the sum is taken over all multi-indices $\alpha\in\mathbb{N}^{d}$ with
$|\alpha|\leq m$. The operator $p(D)$ can alternatively be written as a
composition of a Fourier transformation, a multiplication by the symbol

$\displaystyle p(\xi)=\sum_{|\alpha|\leq m}g_{\alpha}\xi^{\alpha}$ (2.2)

and an inverse Fourier transformation:

$\displaystyle p(D)u(x)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}p(\xi)u(y)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\xi\rangle}p(\xi)\widehat{u}(\xi)\mathop{}\\!\mathrm{d}\xi.$ (2.3)

In Fourier space, the initially complicated differential operator $p(D)$ is a
simple multiplication by the symbol $p(\xi)$. Manipulations of the
differential operator (e.g. taking powers, inverses, etc.) are equivalent to
manipulations of its symbol.

A pseudo-differential operator $\mathrm{Op}(p)$ is an operator for which the
symbol $p=p(x,\xi)$ can depend on two phase space variables, $x$ and $\xi$,
representing position and momentum:

$\displaystyle\mathrm{Op}(p)u(x)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}p(x,\xi)u(y)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi.$
(2.4)

The main difficulty in the pseudo-differential calculus is that manipulations
of the symbol no longer correspond to the same manipulations of the pseudo-
differential operator (e.g. the inverse symbol (if existent) is not
necessarily the symbol of the inverse operator). Nevertheless, we demonstrate
that the correspondence holds at least in an approximate sense, and we obtain
estimates in how far the correspondence fails.

The analysis of pseudo-differential operators is a vast area of mathematics
with many different applications. It is impossible to provide a complete
overview of this topic, but we explain essential ideas and techniques relevant
to the present work. Several excellent references in the literature provide a
more detailed exposition. We follow [3, 8, 12, 16, 22] in this chapter.

###  2.1 Symbols

Generally, a symbol is a smooth function satisfying additional estimates. In
the literature, different classes of symbols are more or less useful in
different applications. To define appropriate symbol classes for the present
work, we first need the notion of an order function.

######  Definition 2.1.1.

A smooth function $M:\mathbb{R}^{k}\to(0,\infty)$ is an order function if
constants $C,m$ exist such that $M(z)\leq C\langle z-w\rangle^{m}M(w)$ for all
$z,w\in\mathbb{R}^{k}$, where $\langle z\rangle=\sqrt{1+|z|^{2}}$ is the
Japanese bracket.

######  Notation.

In this chapter, we write $z$ or $w$ for an element in $\mathbb{R}^{k}$
(unless stated otherwise). If $k=2d$, we replace $z$ with
$(x,\xi)\in\mathbb{R}^{d}\times\mathbb{R}^{d}$ and call $x$ the position
variable and $\xi$ the momentum variable. Sometimes we also deal with $k=3d$.
In this case, we write $z=(x,y,\xi)$, and $y$ takes the role of an integrated
variable.

######  Example 2.1.2.

Clearly, $M\equiv 1$ and $M(z)=\langle z\rangle$ are order functions, with the
latter being a consequence of Peetre’s inequality (see Theorem A.1.4).
Moreover, if $k=2d$, then $M(x,\xi)=\langle
x\rangle^{\rho}\langle\xi\rangle^{\sigma}$ is an order function for any
$\rho,\sigma\in\mathbb{R}$.

For a given order function $M$ and a Riemannian metric $g$, we define the
symbol class $S_{k}(M,g)$. A Riemannian metric on $\mathbb{R}^{k}$ is a smooth
map $g:\mathbb{R}^{k}\ni z\mapsto g_{z}$ with values in positive definite
quadratic forms on $\mathbb{R}^{k}$. Further regularity conditions must be
imposed on the Riemannian metric $g$ to obtain a well-behaved theory of
pseudo-differential operators (cf. [1] for a detailed discussion). In the
following text, we call these Riemannian metrics admissible.

######  Definition 2.1.3.

Let $M$ be an order function and $g$ be an admissible metric on
$\mathbb{R}^{k}$. The symbol class $S_{k}(M,g)$ is the set of all $a\in
C^{\infty}(\mathbb{R}^{k})$ that satisfy the following bound for all
$p\in\mathbb{N}$ and $v_{1},\dots,v_{p}\in\mathbb{R}^{k}$:

$\displaystyle\left|\prod_{i=1}^{p}(v_{i}\cdot\partial_{z})a(z)\right|\leq
C_{p}M(z)\prod_{i=1}^{p}|g_{z}(v_{i})|^{\frac{1}{2}}.$ (2.5)

We write $S_{k}(M,g)=S(M,g)$ if the dimension $k$ is irrelevant or clear from
the context.

The space $S(M,g)$ is a Fréchet space with semi-norms given by the best
constants $C_{p}$ satisfying the above bound.

It is easy to verify that the product of two symbols is, again, a symbol:

######  Proposition 2.1.4.

If $a_{1}\in S(M_{1},g)$ and $a_{2}\in S(M_{2},g)$, then $a_{1}a_{2}\in
S(M_{1}M_{2},g)$.

We rarely work with the symbol class $S(M,g)$ in its full generality but
restrict our attention to the following two examples.

######  Definition 2.1.5.

Let $M$ be an order function. Then
$S_{k}(M)=S_{k}(M,\mathop{}\\!\mathrm{d}z^{2})$ is the set of all $a\in
C^{\infty}(\mathbb{R}^{k})$ satisfying the bound
$|\partial_{z}^{\alpha}a(z)|\leq C_{\alpha}M(z)$ for every multi-index
$\alpha\in\mathbb{N}^{k}$.

In (quantum) mechanical applications, derivatives in the momentum variable
usually improve the order of a symbol $a\in S_{2d}(M)$. Therefore, we also
define the following subclass of $S(M)$.

######  Definition 2.1.6.

Let $m\in\mathbb{R}$. Then
$S^{m}=S_{2d}(\langle\xi\rangle^{m},\mathop{}\\!\mathrm{d}x^{2}+\langle\xi\rangle^{-2}\mathop{}\\!\mathrm{d}\xi^{2})\subset
S_{2d}(\langle\xi\rangle^{m})$ is the set of all $a\in
C^{\infty}(\mathbb{R}^{d}\times\mathbb{R}^{d})$ that satisfy the following
bound for all multi-indices $\alpha,\beta\in\mathbb{N}^{d}$:

$\displaystyle|\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|\leq
C_{\alpha\beta}\langle\xi\rangle^{m-|\beta|}$ (2.6)

We call $m$ the order of the symbol class, and write
$S^{-\infty}=\bigcap_{m}S^{m}$ and $S^{\infty}=\bigcup_{m}S^{m}$.

The symbol classes $S^{m}$ have the practical advantage that we can perform
computations in $S^{n}$ modulo $S^{l}$ because $S^{l}\subset S^{n}$ for $l\leq
n$. It is also possible to expand elements of $S^{m}$ in asymptotic sums. To
assign a precise meaning to asymptotic sums of symbols, two symbols $a,b\in
S^{\infty}$ are called asymptotically equivalent, written as $a\sim b$, if
$a-b\in S^{-\infty}$. A surprising feature is that although a formal series of
the type $\sum_{j}a_{j}$, where the order of $a_{j}$ is decreasing to
$-\infty$ for larger $j$, has no reason to be convergent, a symbol that is
asymptotically equivalent to this series can always be found.

######  Theorem 2.1.7.

Let $a_{j}\in S^{m-j}$ for any $j\in\mathbb{N}$. Then a symbol $a\in S^{m}$
uniquely determined up to asymptotic equivalence (i.e. up to a symbol in
$S^{-\infty}$) exists such that

$\displaystyle a-\sum_{j<l}a_{j}\in S^{m-l}$ (2.7)

for every $l\in\mathbb{N}$.

###### Sketch of the Proof.

If $a$ exists, it is uniquely determined by (2.7) up to a symbol in
$S^{-\infty}$. To prove the existence of $a$, we define a resummation of the
formal series $\sum_{j}a_{j}$.

We take a $\chi\in C_{0}^{\infty}(\mathbb{R}^{d})$, which is equal to $1$ in a
neighbourhood of $0$, and set
$A_{j}(x,\xi)=(1-\chi(\epsilon_{j}\xi))a_{j}(x,\xi)$, where $\epsilon_{j}\to
0$ rapidly. Then $A_{j}\sim a_{j}$ because $A_{j}(x,\cdot)=a_{j}(x,\cdot)$
outside a compact set. Clearly, $a(x,\xi)=\sum_{j}A_{j}(x,\xi)$ defines an
element in $C^{\infty}(\mathbb{R}^{2d})$ because the sum is finite near any
fixed $\xi_{0}\in\mathbb{R}^{d}$. One can verify that the symbol $a$ defined
in this way indeed satisfies (2.7). We refer to [8, Thm. 18.1.3] for the full
proof. ∎

######  Example 2.1.8.

The symbol $\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
D_{x},D_{\xi}\rangle}a$ with $a\in S^{\infty}$ has the following asymptotic
expansion:

$\displaystyle\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
D_{x},D_{\xi}\rangle}a(x,\xi)$
$\displaystyle\sim\sum_{j}\frac{1}{j!}(\mathop{}\\!\mathrm{i}\langle
D_{x},D_{\xi}\rangle)^{j}a(x,\xi)=\sum_{\alpha}\frac{1}{\alpha!}\partial_{\xi}^{\alpha}D_{x}^{\alpha}a(x,\xi).$
(2.8)

A resummation of the sum on the right-hand side is found in [8, Thm. 18.1.7].

######  Remark 2.1.9.

The nonzero symbol of the highest order in the asymptotic sum $\sum_{j}a_{j}$
is called the principal symbol. Typically, it is easy to determine the
principal symbol (e.g., in (2.8), the principal symbol of
$\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle D_{x},D_{\xi}\rangle}a$
is simply $a$), whereas the lower order terms are usually harder to compute.

It is also possible to define asymptotic expansions for general symbols in
$S(M)$ at the cost of introducing an additional order parameter $h$. In this
case, we call $a\in S(M)$ asymptotically equivalent to the formal series
$\sum_{j}h^{j}a_{j}$ with $a_{j}\in S(M)$ if

$\displaystyle\left|\partial^{\alpha}\left(a-\sum_{j<l}h^{j}a_{j}\right)\right|\leq
C_{l\alpha}h^{l}M$ (2.9)

for all $l\in\mathbb{N}$, $\alpha\in\mathbb{N}^{k}$ and all $h>0$ that are
sufficiently small. With a similar resummation argument as in the proof of
Theorem 2.1.7, we associate each formal series $\sum_{j}h^{j}a_{j}$ with a
symbol $a\in S(M)$ that is asymptotically equivalent to this series.

It is useful to have both concepts of asymptotic sums. For simplicity, we
restrict our attention to symbols in $S^{\infty}$ when we construct asymptotic
sums in the following sections.

###  2.2 Pseudo-differential Operators

Having defined appropriate classes of symbols, we introduce pseudo-
differential operators. We assume that $M$ is an order function on
$\mathbb{R}^{3d}$ that depends only on the momentum variable (i.e.
$M(x,y,\xi)=M(\xi)$).111This assumption is not strictly necessary for defining
pseudo-differential operators, but it simplifies the discussion. The general
case is of no use for the present work. In this case, an $m\in\mathbb{R}$
exists such that $M(x,\xi)\leq CM(0)\langle\xi\rangle^{m}$.

######  Definition 2.2.1.

Let $M$ be an order function specified as above and $a\in S_{3d}(M)$. Then we
call the operator $\mathop{}\\!\mathrm{Op}(a)$ defined on
$C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})$ by

$\displaystyle\mathrm{Op}(a)u(x)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a(x,y,\xi)u(y)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi$
(2.10)

a pseudo-differential operator of order $m$.

We defined $\mathop{}\\!\mathrm{Op}(a)$ on
$C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})$ to make sense of (2.10) as an
oscillatory integral. To verify that the integral (2.10) is well defined, we
assume first that $m<-d$. Then,

$\displaystyle|a(x,y,\xi)|\leq CM(x,y,\xi)\leq C\langle\xi\rangle^{m},$ (2.11)

and the integral on the right-hand side of (2.10) is absolutely convergent. If
$m\geq-d$, we exploit the oscillatory behaviour of
$\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle x-y,\xi\rangle}$. We
observe the following:

$\displaystyle L\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle},\ L=\frac{1-\xi D_{y}}{1+|\xi|^{2}}.$ (2.12)

Thus, if $L^{T}$ is the transpose of $L$, we obtain via integration by parts,
for any $k\in\mathbb{N}$,

$\displaystyle\mathrm{Op}(a)u(x)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}(L^{T})^{k}(a(x,y,\xi)u(y))\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi\equiv
I_{k}(a)u(x).$ (2.13)

If $k$ is chosen sufficiently large, the integral $I_{k}(a)u(x)$ is absolutely
convergent; hence, $\mathop{}\\!\mathrm{Op}(a)$ is well defined on
$C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})$.

######  Theorem 2.2.2.

If $a\in S(M)$, then $\mathop{}\\!\mathrm{Op}(a)$ defines a continuous map
from $C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})$ to
$C^{\infty}(\mathbb{R}^{d})$.

###### Proof.

We take $l\in\mathbb{N}$ and $k=[m]+d+l+1$, where $[m]$ is the smallest
integer such that $m\leq[m]$. Then $\mathop{}\\!\mathrm{Op}(a)u=I_{k}(a)u$ has
$l$ continuous derivatives by the dominated convergence theorem. As $l$ is
arbitrary, $\mathop{}\\!\mathrm{Op}(a)u\in C^{\infty}(\mathbb{R}^{d})$.

Furthermore, it is not difficult to bound
$\partial^{\alpha}\mathop{}\\!\mathrm{Op}(a)u$ by a finite sum of semi-norms
of $u$ (see [12, Thm. 2.4.3]). This proves that
$\mathop{}\\!\mathrm{Op}(a):C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})\to
C^{\infty}(\mathbb{R}^{d})$ is continuous. ∎

The continuity of $\mathop{}\\!\mathrm{Op}(a)$ on
$C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})$ and the Schwartz kernel theorem (see
Theorem A.1.1) imply the existence of a unique distributional kernel $K$ such
that $\langle\mathop{}\\!\mathrm{Op}(a)u,v\rangle=K(u\otimes v)$ for all
$u,v\in C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})$. It is suggestive to denote
the kernel $K$ by

$\displaystyle
K(x,y)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a(x,y,\xi)\mathop{}\\!\mathrm{d}\xi$ (2.14)

and write
$\mathop{}\\!\mathrm{Op}(a)u(x)=\int_{\mathbb{R}^{d}}K(x,y)u(y)\mathop{}\\!\mathrm{d}y$
as an integral operator.

The following examples of pseudo-differential operators are easy to verify.

######  Example 2.2.3.

  1. 1.

If $a(x,y,\xi)=\xi^{\alpha}$, then $\mathrm{Op}(a)u(x)=D^{\alpha}u(x)$.

  2. 2.

If $a(x,y,\xi)=a(x)$ or $a(x,y,\xi)=a(y)$, then $\mathrm{Op}(a)u(x)=a(x)u(x)$.

  3. 3.

If $a(x,y,\xi)=\sum_{|\alpha|\leq m}g_{\alpha}(x)\xi^{\alpha}$, then
$\mathop{}\\!\mathrm{Op}(a)u(x)=\sum_{|\alpha|\leq
m}D^{\alpha}(g_{\alpha}u)(x)$.

  4. 4.

If $a(x,y,\xi)=\sum_{|\alpha|\leq m}g_{\alpha}(y)\xi^{\alpha}$, then
$\mathrm{Op}(a)u(x)=\sum_{|\alpha|\leq m}g_{\alpha}(x)D^{\alpha}u(x)$.

######  Remark 2.2.4.

The last two examples establish that replacing $g_{\alpha}(x)$ by
$g_{\alpha}(y)$ commutes the order in which the differential $D^{\alpha}$ and
multiplication by $g_{\alpha}$ act on $u$. This is connected to the ordering
ambiguity of quantum mechanics (cf. Section 2.3 below).

Thus far, we defined pseudo-differential operators only on
$C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d})$, which is insufficient for practical
purposes. In the rest of this section, we continuously extend pseudo-
differential operators to Schwartz functions and, subsequently, to tempered
distributions.

######  Theorem 2.2.5.

If $a\in S(M)$, then $\mathop{}\\!\mathrm{Op}(a)$ defines a continuous map
from $\mathcal{S}(\mathbb{R}^{d})$ to $\mathcal{S}(\mathbb{R}^{d})$.

###### Proof.

For simplicity, we assume that the symbol does not depend on the integration
variable $y$. This assumption is justified in the next section. The reason for
this choice is that $\mathop{}\\!\mathrm{Op}(a)u$ simplifies to

$\displaystyle\mathop{}\\!\mathrm{Op}(a)u(x)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\xi\rangle}a(x,\xi)\widehat{u}(\xi)\mathop{}\\!\mathrm{d}\xi.$ (2.15)

In this simplified form, it is easy to verify that
$x^{\beta}D_{x}^{\alpha}\mathop{}\\!\mathrm{Op}(a)u(x)$ is a linear
combination of

$\displaystyle\mathop{}\\!\mathrm{Op}(x^{\beta-\delta}D_{x}^{\alpha-\gamma}a)(x^{\delta}D_{x}^{\gamma}u)(x).$
(2.16)

It follows that $x^{\beta}D_{x}^{\alpha}\mathop{}\\!\mathrm{Op}(a)u(x)$ is
bounded by a finite sum of semi-norms of $u$. This proves that
$\mathop{}\\!\mathrm{Op}(a)u\in\mathcal{S}(\mathbb{R}^{d})$ if
$u\in\mathcal{S}(\mathbb{R}^{d})$ and that
$\mathop{}\\!\mathrm{Op}(a):\mathcal{S}(\mathbb{R}^{d})\to\mathcal{S}(\mathbb{R}^{d})$
is a continuous map. ∎

To establish the same result for the dual space
$\mathcal{S}^{\prime}(\mathbb{R}^{d})$, we define the distributional pairing

$\displaystyle\langle\mathrm{Op}(a)u,v\rangle=\langle
u,\mathrm{Op}(b)v\rangle$ (2.17)

for $u\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$,
$v\in\mathcal{S}(\mathbb{R}^{d})$, where
$\mathop{}\\!\mathrm{Op}(b)=\mathop{}\\!\mathrm{Op}(a)^{*}$ is the formal
adjoint of $\mathop{}\\!\mathrm{Op}(a)$. The symbol $b$ of the formal adjoint
is defined by $b(x,y,\xi)=\overline{a(y,x,\xi)}$.

######  Theorem 2.2.6.

If $a\in S(M)$, then $\mathop{}\\!\mathrm{Op}(a)$ defines a continuous map
from $\mathcal{S}^{\prime}(\mathbb{R}^{d})$ to
$\mathcal{S}^{\prime}(\mathbb{R}^{d})$.

###### Proof.

If $u\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$, then $\mathrm{Op}(a)u$ defined
in (2.17) is an element of $\mathcal{S}^{\prime}(\mathbb{R}^{d})$ because
$\mathrm{Op}(b)v\in\mathcal{S}(\mathbb{R}^{d})$ for all
$v\in\mathcal{S}(\mathbb{R}^{d})$ by the previous theorem. Moreover,
$\mathrm{Op}(a):\mathcal{S}^{\prime}(\mathbb{R}^{d})\to\mathcal{S}^{\prime}(\mathbb{R}^{d})$
is continuous because $\langle\mathop{}\\!\mathrm{Op}(a)u_{n},v\rangle=\langle
u_{n},\mathop{}\\!\mathrm{Op}(b)v\rangle\to 0$ for all
$v\in\mathcal{S}(\mathbb{R}^{d})$ if $u_{n}\to 0$ in
$\mathcal{S}^{\prime}(\mathbb{R}^{d})$. ∎

###  2.3 Change in Quantisation

In Definition 2.2.1, we assumed that the underlying symbol of a pseudo-
differential operator depends on the position variable $x$, integration
variable $y$, and momentum variable $\xi$ (almost) arbitrarily. In this
section, we demonstrate that it is not a restriction to replace the arbitrary
dependence on $(x,y)$ by the convex combination $tx+(1-t)y$ for any
$t\in[0,1]$, where the parameter $t$ distinguishes different quantisations.

For a symbol $a\in S_{2d}(M)$, where the order function $M$ satisfies the same
assumptions as in Definition 2.2.1, we set

$\displaystyle\mathop{}\\!\mathrm{Op}_{t}(a)u(x)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a(tx+(1-t)y,\xi)u(y)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi,$
(2.18)

and call $\mathop{}\\!\mathrm{Op}_{t}(a)$ a pseudo-differential operator in
quantisation $t$.

######  Theorem 2.3.1.

Let $a\in S_{3d}(M)$. Then, for every $t\in[0,1]$, a unique $a_{t}\in
S_{2d}(M)$ exists such that
$\mathop{}\\!\mathrm{Op}(a)=\mathop{}\\!\mathrm{Op}_{t}(a_{t})$. The symbol
$a_{t}$ is given by

$\displaystyle a_{t}(x,\xi)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle\theta,\eta-\xi\rangle}a(x+(1-t)\theta,x-t\theta,\eta)\mathop{}\\!\mathrm{d}\eta\mathop{}\\!\mathrm{d}\theta$
$\displaystyle=\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}\langle
D_{\theta},D_{\xi}\rangle}a(x+(1-t)\theta,x-t\theta,\xi)\big{|}_{\theta=0}.$
(2.19)

###### Proof.

To determine the symbol $a_{t}$, we compare the integral kernels of
$\mathop{}\\!\mathrm{Op}_{t}(a_{t})$ and $\mathop{}\\!\mathrm{Op}(a)$:

$\displaystyle\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a_{t}(tx+(1-t)y,\xi)\mathop{}\\!\mathrm{d}\xi=\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a(x,y,\xi)\mathop{}\\!\mathrm{d}\xi.$ (2.20)

If $\theta=x-y$ and $z=tx+(1-t)y=x-(1-t)\theta$, then

$\displaystyle\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle\theta,\xi\rangle}a_{t}(z,\xi)\mathop{}\\!\mathrm{d}\xi=\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle\theta,\xi\rangle}a(z+(1-t)\theta,z-t\theta,\xi)\mathop{}\\!\mathrm{d}\xi.$
(2.21)

Hence, an inverse Fourier transformation with respect to $\theta$ on both
sides yields

$\displaystyle
a_{t}(z,\eta)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle\theta,\xi-\eta\rangle}a(z+(1-t)\theta,z-t\theta,\xi)\mathop{}\\!\mathrm{d}\xi\mathop{}\\!\mathrm{d}\theta,$
(2.22)

which is the desired result if we rename $x\leftrightarrow z$ and
$\eta\leftrightarrow\xi$. ∎

Due to their appearance in applications, three values of $t$ are particularly
important:

  * •

Standard quantisation ($t=1$): $a(x,D)=\mathrm{Op}(a)=\mathrm{Op}_{1}(a)$;

  * •

Weyl quantisation ($t=\frac{1}{2}$):
$a^{\mathrm{w}}(x,D)=\mathrm{Op}_{\frac{1}{2}}(a)$;

  * •

Right quantisation ($t=0$): $\mathrm{Op}_{0}(a)$.

The Weyl quantisation has the pleasant property that $a^{\mathrm{w}}(x,D)$ is
a symmetric operator if $a$ is real-valued. However, the standard quantisation
and, sometimes, the right quantisation are easier to handle. For example, a
pseudo-differential operator in standard quantisation has the following form:

$\displaystyle
a(x,D)u(x)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\xi\rangle}a(x,\xi)\widehat{u}(\xi)\mathop{}\\!\mathrm{d}\xi.$ (2.23)

To understand the different quantisations better, we consider the following
example relevant for the Nelson model with variable coefficients:

$\displaystyle a(x,\xi)=\sum_{1\leq j,k\leq d}\xi_{j}g^{jk}(x)\xi_{k}\in
S^{2}.$ (2.24)

Then $\mathop{}\\!\mathrm{Op}_{t}(a)$ in standard, Weyl, and right
quantisation is given by

$\displaystyle a(x,D)u(x)$ $\displaystyle=\sum_{1\leq j,k\leq
d}g^{jk}(x)D_{x_{j}}D_{x_{k}}u(x),$ (2.25) $\displaystyle
a^{\mathrm{w}}(x,D)u(x)$ $\displaystyle=\sum_{1\leq j,k\leq
d}D_{x_{j}}g^{jk}(x)D_{x_{k}}u(x),$ (2.26)
$\displaystyle\mathop{}\\!\mathrm{Op}_{0}(a)u(x)$ $\displaystyle=\sum_{1\leq
j,k\leq d}D_{x_{j}}D_{x_{k}}g^{jk}(x)u(x).$ (2.27)

We observe that the choice of a particular quantisation (i.e. the choice of a
particular $t\in[0,1]$) resolves the ordering ambiguity of the classical
position observable $x$ and momentum observable $\xi$. Furthermore, we deduce
from the commutator $[D_{x},g(x)]=(D_{x}g)(x)$ that the difference between the
standard, Weyl, and right quantisation is an operator of order $1$, whereas
$a$ is a symbol of order $2$. This observation can be generalised.

######  Theorem 2.3.2.

Let $a\in S_{2d}(M)$. Then
$\mathop{}\\!\mathrm{Op}_{t}(a_{t})=\mathop{}\\!\mathrm{Op}_{0}(a)$ for
$a_{t}=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\langle
D_{x},D_{\xi}\rangle}a(x,\xi)$.

###### Proof.

From Theorem 2.3.1 it follows that

$\displaystyle a_{t}(x,\xi)$
$\displaystyle=\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}\langle
D_{\theta},D_{\xi}\rangle}a(x-t\theta,\xi)\big{|}_{\theta=0}$
$\displaystyle=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\langle
D_{x},D_{\xi}\rangle}a(x-t\theta,\xi)\big{|}_{\theta=0}$
$\displaystyle=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}t\langle
D_{x},D_{\xi}\rangle}a(x,\xi).$ (2.28)

This proves the theorem. ∎

######  Corollary 2.3.3.

If $a\in S^{m}$ and $s,t\in[0,1]$, then the pseudo-differential operators
$\mathrm{Op}_{t}(a)$ and $\mathrm{Op}_{s}(a)$ are identical up to a pseudo-
differential operator of order $m-1$.

###### Proof.

Theorem 2.3.2 indicates that $\mathrm{Op}_{t}(a)=\mathrm{Op}_{s}(b)$ for

$\displaystyle b(x,\xi)$
$\displaystyle=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}(s-t)\langle
D_{x},D_{\xi}\rangle}a(x,\xi)$
$\displaystyle\sim\sum_{\alpha}\frac{(s-t)^{\alpha}}{\alpha!}D_{x}^{\alpha}\partial_{\xi}^{\alpha}a(x,\xi)$
$\displaystyle=a(x,\xi)+S^{m-1}.$ (2.29)

Thus,
$\mathrm{Op}_{t}(a)=\mathrm{Op}_{s}(b)=\mathrm{Op}_{s}(a)+\mathrm{Op}(S^{m-1})$.
∎

###  2.4 Symbolic Calculus

Thus far, we formulated the basic notions of symbols and pseudo-differential
operators. In this section, we explore how manipulations of a pseudo-
differential operator affect its symbol.

####  2.4.1 Adjoint Symbol

We mentioned at the end of Section 2.2 that the adjoint operator
$\mathop{}\\!\mathrm{Op}(a)^{*}=\mathop{}\\!\mathrm{Op}(b)$ is a pseudo-
differential operator with symbol $b(x,y,\xi)=\overline{a(y,x,\xi)}$. In terms
of the pseudo-differential operator $\mathop{}\\!\mathrm{Op}_{t}(a)$ in
quantisation $t$, this means that

$\displaystyle\mathop{}\\!\mathrm{Op}_{t}(a)^{*}=\mathop{}\\!\mathrm{Op}_{1-t}(\overline{a}).$
(2.30)

A slight obstacle is that $\mathop{}\\!\mathrm{Op}_{1-t}(\overline{a})$ is not
in the same quantisation as $\mathop{}\\!\mathrm{Op}_{t}(a)$ except if
$t=1/2$. However, based on our results from Section 2.3, it is not difficult
to construct the adjoint symbol $a^{*}$ satisfying
$\mathop{}\\!\mathrm{Op}_{t}(a)^{*}=\mathop{}\\!\mathrm{Op}_{t}(a^{*})$. The
following theorem immediately follows from Theorem 2.3.2.

######  Theorem 2.4.1.

Let $a\in S(M)$ and set

$\displaystyle a^{*}(x,\xi)$
$\displaystyle=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}(2t-1)\langle
D_{x},D_{\xi}\rangle}\overline{a(x,\xi)}.$ (2.31)

Then $a^{*}$ defines a symbol in $S(M)$, and
$\mathop{}\\!\mathrm{Op}_{t}(a)^{*}=\mathop{}\\!\mathrm{Op}_{t}(a^{*})$.

####  2.4.2 Moyal Product

Determining a symbol for the composition
$\mathop{}\\!\mathrm{Op}_{t}(a_{1})\mathop{}\\!\mathrm{Op}_{t}(a_{2})$ with
$a_{1}\in S(M_{1})$ and $a_{2}\in S(M_{2})$ is more difficult. For simplicity,
we discuss this construction only for the standard quantisation. Expanding the
expression for
$\mathop{}\\!\mathrm{Op}(a_{1})\mathop{}\\!\mathrm{Op}(a_{2})u$, we obtain

$\displaystyle\mathrm{Op}(a_{1})\mathrm{Op}(a_{2})u(x)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\eta\rangle}a_{1}(x,\eta)\mathcal{F}[\mathop{}\\!\mathrm{Op}(a_{2})u](\eta)\mathop{}\\!\mathrm{d}\eta$
$\displaystyle=\frac{1}{(2\pi)^{2d}}\int_{\mathbb{R}^{3d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\eta\rangle}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
y,\xi-\eta\rangle}a_{1}(x,\eta)a_{2}(y,\xi)\widehat{u}(\xi)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi\mathop{}\\!\mathrm{d}\eta$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\xi\rangle}a_{1}\\#a_{2}(x,\xi)\widehat{u}(\xi)\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\mathop{}\\!\mathrm{Op}(a_{1}\\#a_{2})u(x),$ (2.32)

where we defined the Moyal product

$\displaystyle a_{1}\\#a_{2}(x,\xi)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}\langle
x-y,\xi-\eta\rangle}a_{1}(x,\eta)a_{2}(y,\xi)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\eta$
$\displaystyle=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
D_{y},D_{\eta}\rangle}a_{1}(x,\eta)a_{2}(y,\xi)\big{|}_{\eta=\xi,y=x}.$ (2.33)

A similar construction is possible for the other quantisations. For
completeness, we state the following theorem (see [12, Thm. 2.7.4]).

######  Theorem 2.4.2.

Let $a_{1}\in S(M_{1})$, $a_{2}\in S(M_{2})$, and set

$\displaystyle a_{1}\\#_{t}a_{2}$
$\displaystyle=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}(\langle
D_{\eta},D_{v}\rangle-\langle
D_{u},D_{\xi}\rangle)}a_{1}(tx+(1-t)u,\eta)a_{2}((1-t)x+tv,\xi)\big{|}_{u=v=x,\eta=\xi}.$
(2.34)

Then $a_{1}\\#_{t}a_{2}\in S(M_{1}M_{2})$ and
$\mathop{}\\!\mathrm{Op}_{t}(a_{1})\mathop{}\\!\mathrm{Op}_{t}(a_{2})=\mathop{}\\!\mathrm{Op}_{t}(a_{1}\\#_{t}a_{2})$.

######  Example 2.4.3.

We construct a symbol for the commutator $[\cdot,\cdot]$ of two pseudo-
differential operators in standard quantisation. If $a_{1}\in S^{m_{1}}$ and
$a_{2}\in S^{m_{2}}$, then

$\displaystyle[\mathop{}\\!\mathrm{Op}(a_{1}),\mathop{}\\!\mathrm{Op}(a_{2})]=\mathop{}\\!\mathrm{Op}(a_{1}\\#a_{2}-a_{2}\\#a_{1}).$
(2.35)

In leading order, the asymptotic expansion of $a_{1}\\#a_{2}-a_{2}\\#a_{1}$ is
given by

$\displaystyle a_{1}\\#a_{2}-a_{2}\\#a_{1}$
$\displaystyle=-\mathop{}\\!\mathrm{i}(\partial_{\xi}a_{1}\partial_{x}a_{2}-\partial_{\xi}a_{2}\partial_{x}a_{1})+S^{m_{1}+m_{2}-2}$
$\displaystyle=\mathop{}\\!\mathrm{i}\\{a_{1},a_{2}\\}+S^{m_{1}+m_{2}-2},$
(2.36)

where $\\{\cdot,\cdot\\}$ is the Poisson bracket. This is the well-known
correspondence between the Poisson bracket in classical theory and the
commutator in quantum theory.

####  2.4.3 Construction of a Parametrix

We use the Moyal product to construct inverses of elliptic pseudo-differential
operators.

######  Definition 2.4.4.

A symbol $a\in S_{k}(M)$ is elliptic if a constant $C>0$ exists such that
$|a(z)|\geq CM(z)$ for all $z\in\mathbb{R}^{k}$. A pseudo-differential
operator is elliptic if its underlying symbol is elliptic.

The inverse function $M^{-1}$ of an order function $M$ is an order function.
Hence, if $a\in S(M)$ is elliptic, then $a^{-1}\in S(M^{-1})$. The following
theorem reveals that every elliptic symbol has an inverse with respect to the
Moyal product.

######  Theorem 2.4.5.

Let $a\in S(M)$ be elliptic. Then a symbol $b\in S(M^{-1})$ uniquely
determined up to asymptotic equivalence exists such that $a\\#b\sim 1$ and
$b\\#a\sim 1$.

###### Proof.

We prove the theorem for $\\#=\\#_{1}$ and $a\in S^{m}$, $m\in\mathbb{R}$. The
proof for arbitrary quantisations and general $a\in S(M)$ uses similar ideas.

The first candidate for the inverse symbol of $a$ with respect to the Moyal
product is $a^{-1}$. Modulo a symbol $r\in S^{-1}$, the symbol $a^{-1}\in
S^{-m}$ is the inverse of $a$ because

$\displaystyle a\\#a^{-1}\sim 1+\sum_{|\alpha|\geq
1}\frac{1}{\alpha!}\partial_{\xi}^{\alpha}aD_{x}^{\alpha}a^{-1}\equiv 1-r.$
(2.37)

We define the symbol $b_{1}$ to be asymptotically equivalent to

$\displaystyle b_{1}\sim a^{-1}\\#r+a^{-1}\\#r\\#r+\cdots.$ (2.38)

This symbol exists by Theorem 2.1.7, and $b_{1}$ is a right inverse of $a$ up
to asymptotic equivalence:

$\displaystyle a\\#b_{1}\sim(1-r)\\#(r+r\\#r\cdots)=1+S^{-\infty}.$ (2.39)

In similar fashion, we construct a left inverse $b_{2}$ satisfying
$b_{2}\\#a=1+S^{-\infty}$. From $b_{1}\sim b_{2}\\#a\\#b_{1}\sim b_{2}$, it
follows that $b_{1}$ is asymptotically equivalent to $b_{2}$. ∎

An immediate consequence of the above theorem is that, for every pseudo-
differential operator
$A=\mathop{}\\!\mathrm{Op}(a)\in\mathop{}\\!\mathrm{Op}(S^{m})$, a pseudo-
differential operator $B$ exists such that
$AB=BA=1+\mathop{}\\!\mathrm{Op}(S^{-\infty})$. The operator $B$ is called an
approximate inverse of $A$. Because pseudo-differential operators in
$\mathop{}\\!\mathrm{Op}(S^{-\infty})$ are smoothing (see Theorem 2.5.3
below), the approximate inverse $B$ is also called a parametrix of $A$.

####  2.4.4 Functional Calculus

The symbolic calculus developed so far culminates in the functional calculus
of self-adjoint pseudo-differential operators. Although we do not prove the
functional calculus, it satisfies all properties we expect intuitively. If $f$
is a sufficiently smooth function, then $f(\mathop{}\\!\mathrm{Op}(a))$ is a
pseudo-differential operator and its principal symbol is $f(a)$. ’Sufficiently
smooth’ in this case means that ’$f$ is a symbol of order $p$’, that is, an
element of

$\displaystyle S^{p}(\mathbb{R})=\\{f\in
C^{\infty}(\mathbb{R})\mid|f^{(k)}(t)|\leq C_{k}\langle t\rangle^{p-k}\
\forall k\in\mathbb{N}\\}.$ (2.40)

For a full discussion of the functional calculus we refer to [1]. The
functional calculus is usually formulated for pseudo-differential operators in
Weyl quantisation because in this quantisation a pseudo-differential operator
is symmetric if the underlying symbol is real-valued.

######  Theorem 2.4.6 (Functional calculus).

Let $a\in S(M,g)$ be real-valued and elliptic such that its Weyl quantisation
$a^{\mathrm{w}}$ is essentially self-adjoint on $\mathcal{S}(\mathbb{R}^{d})$.
If $f\in S^{p}(\mathbb{R})$, the operator $f(a^{\mathrm{w}})$ is a pseudo-
differential operator, and the symbol of $f(a^{\mathrm{w}})$ is an element of
$S(M^{p},g)$. If $a\in S^{m}$, then
$f(a^{\mathrm{w}})-f(a)^{\mathrm{w}}\in\mathrm{Op}(S^{mp-1})$.

###  2.5 $L^{2}$-continuity

For applications in quantum mechanics, it is vital to know how pseudo-
differential operators act on
$L^{2}(\mathbb{R}^{d})\subset\mathcal{S}^{\prime}(\mathbb{R}^{d})$. It is well
known from Fourier analysis that the Fourier transformation on
$\mathcal{S}^{\prime}(\mathbb{R}^{d})$ restricts to a bounded operator on
$L^{2}(\mathbb{R}^{d})$. The Calderon–Vaillancourt theorem generalises this
statement for pseudo-differential operators with bounded symbols. In this
section, we present a simplified but instructive proof of the
Calderon–Vaillancourt theorem for symbols in $S^{0}$. In the next section, we
present a more general proof for symbols in $S(1)$.

######  Theorem 2.5.1 (Calderon–Vaillancourt).

For every $a\in S^{0}$, the pseudo-differential operator
$\mathop{}\\!\mathrm{Op}(a)$ defines a bounded operator on
$L^{2}(\mathbb{R}^{d})$.

###### Proof.

First, we assume that $a\in S^{-d-1}$. In standard quantisation, the integral
kernel of $\mathrm{Op}(a)$ is

$\displaystyle
K(x,y)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a(x,\xi)\mathop{}\\!\mathrm{d}\xi,$ (2.41)

and, due to $a\in S^{-d-1}$,

$\displaystyle|(x-y)^{\alpha}K(x,y)|$
$\displaystyle=\frac{1}{(2\pi)^{d}}\left|\int_{\mathbb{R}^{d}}(\partial_{\xi}^{\alpha}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle})a(x,\xi)\mathop{}\\!\mathrm{d}\xi\right|$
$\displaystyle\leq\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}|\partial_{\xi}^{\alpha}a(x,\xi)|\mathop{}\\!\mathrm{d}\xi$
(2.42)

is bounded for every multi-index $\alpha\in\mathbb{N}^{d}$. Hence,
$(1+|x-y|^{2})^{d}|K(x,y)|$ is bounded; therefore,

$\displaystyle\sup_{y}\int_{\mathbb{R}^{d}}|K(x,y)|\mathop{}\\!\mathrm{d}x\leq\sup_{y}\int_{\mathbb{R}^{d}}\frac{C}{(1+|x-y|^{2})^{d}}\mathop{}\\!\mathrm{d}x\leq
C$ (2.43)

and

$\displaystyle\sup_{x}\int_{\mathbb{R}^{d}}|K(x,y)|\mathop{}\\!\mathrm{d}y\leq\sup_{x}\int_{\mathbb{R}^{d}}\frac{C}{(1+|x-y|^{2})^{d}}\mathop{}\\!\mathrm{d}y\leq
C.$ (2.44)

It follows from the Schur test (see Theorem A.1.2 with $B=\mathbb{C}$) that
the integral operator $\mathop{}\\!\mathrm{Op}(a)$ is bounded on
$L^{2}(\mathbb{R}^{d})$.

Next, we prove that $\mathop{}\\!\mathrm{Op}(a)$ is bounded for symbols of
negative order. It suffices to prove that $\mathop{}\\!\mathrm{Op}(a)$ is
bounded if $a\in S^{-1/2^{k}}$ for all $k\in\mathbb{\mathbb{Z}}$ because
$S^{m}\subset S^{n}$ for $m\leq n$. This proof is done by induction. According
to the first step, a symbol $a\in S^{-1/2^{k}}$ defines a bounded operator if
$k$ is sufficiently small (e.g. $k\leq-d$). Assume we proved the claim for all
$k\leq l$. If $a\in S^{-1/2^{l+1}}$, then $a^{*}\\#a\in S^{-1/2^{l}}$, and

$\displaystyle\lVert\mathop{}\\!\mathrm{Op}(a)u\rVert^{2}=\langle\mathop{}\\!\mathrm{Op}(a^{*}\\#a)u,u\rangle\leq\lVert\mathop{}\\!\mathrm{Op}(a^{*}\\#a)\rVert\lVert
u\rVert^{2}.$ (2.45)

By the induction hypothesis,
$\lVert\mathop{}\\!\mathrm{Op}(a^{*}\\#a)\rVert<\infty$; hence,
$\mathop{}\\!\mathrm{Op}(a)$ is bounded.

It remains to prove the theorem for $a\in S^{0}$. Define

$\displaystyle M^{2}=\sup_{(x,\xi)}|a(x,\xi)|^{2}<\infty.$ (2.46)

Then $M^{2}-|a(x,\xi)|^{2}\in S^{0}$ is a nonnegative symbol. Furthermore, if
$F\in C^{\infty}(\mathbb{C})$ is any smooth function such that
$F(z)=\sqrt{1+z}$ for $z\in\mathbb{R}_{+}$, then

$\displaystyle b(x,\xi)=F(M^{2}-|a(x,\xi)|^{2})$ (2.47)

defines a symbol $b\in S^{0}$. The symbol $b$ is chosen such that
$a^{*}\\#a+b^{*}\\#b=1+M^{2}+c$ for a $c\in S^{-1}$. Thus, it follows that

$\displaystyle\lVert\mathop{}\\!\mathrm{Op}(a)u\rVert^{2}$
$\displaystyle\leq\lVert\mathop{}\\!\mathrm{Op}(a)u\rVert^{2}+\lVert\mathop{}\\!\mathrm{Op}(b)u\rVert^{2}$
$\displaystyle=\langle\mathop{}\\!\mathrm{Op}(a^{*}\\#a+b^{*}\\#b)u,u\rangle$
$\displaystyle=\langle(1+M^{2})u,u\rangle+\langle\mathop{}\\!\mathrm{Op}(c)u,u\rangle$
$\displaystyle\leq(1+M^{2}+\lVert\mathop{}\\!\mathrm{Op}(c)\rVert)\lVert
u\rVert^{2}.$ (2.48)

Hence, $\mathop{}\\!\mathrm{Op}(a)$ is bounded because
$\lVert\mathop{}\\!\mathrm{Op}(c)\rVert<\infty$ by the previous step. ∎

######  Corollary 2.5.2.

Let $a\in S^{m}$. Then, for every $s\in\mathbb{R}$,
$\mathop{}\\!\mathrm{Op}(a):H^{s}\to H^{s-m}$ is a bounded operator from the
Sobolev space $H^{s}$ to $H^{s-m}$.

###### Proof.

Set $b=\langle\cdot\rangle^{s-m}\\#a\\#\langle\cdot\rangle^{-s}\in S^{0}$.
Then

$\displaystyle\lVert\mathrm{Op}(a)u\rVert_{H^{s-m}}$
$\displaystyle=\lVert\langle
D\rangle^{s-m}\mathop{}\\!\mathrm{Op}(a)u\rVert_{L^{2}}=\lVert\mathrm{Op}(b)\langle
D\rangle^{s}u\rVert_{L^{2}}$ $\displaystyle\leq C\lVert\langle
D\rangle^{s}u\rVert_{L^{2}}=C\lVert u\rVert_{H^{s}}$ (2.49)

proves the claim. ∎

The Calderon–Vaillancourt theorem has further consequences for mapping
properties of pseudo-differential operators. We remind the reader of the
following notions. If $u\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$ is a tempered
distribution, the singular support of $u$ denoted by
$\mathop{}\\!\mathrm{sing\ supp}(u)$ is the complement of all
$x\in\mathbb{R}^{d}$ that have an open neighbourhood $U$ such that $\varphi
u\in C_{\mathrm{c}}^{\infty}(U)$ for all $\varphi\in
C_{\mathrm{c}}^{\infty}(U)$. An operator $A$ on
$\mathcal{S}^{\prime}(\mathbb{R}^{d})$ is called pseudo-local if
$\mathop{}\\!\mathrm{sing\ supp}(Au)\subset\mathop{}\\!\mathrm{sing\ supp}(u)$
for all $u\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$ and hypoelliptic if
$\mathop{}\\!\mathrm{sing\ supp}(Au)=\mathop{}\\!\mathrm{sing\ supp}(u)$ for
all $u\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$. Furthermore, $A$ is called
smoothing if it maps tempered distributions to smooth functions.

######  Theorem 2.5.3.

If $a\in S^{-\infty}$, then $\mathop{}\\!\mathrm{Op}(a)$ is smoothing.

###### Proof.

Let $u\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$ be a tempered distribution. To
prove that $\mathop{}\\!\mathrm{Op}(a)u$ is smooth, we demonstrate that
$D^{\alpha}\mathop{}\\!\mathrm{Op}(a)u=\mathop{}\\!\mathrm{Op}(\xi^{\alpha}\\#a)u\equiv\mathop{}\\!\mathrm{Op}(b)u$
is a continuous function for every multi-index $\alpha\in\mathbb{N}^{d}$. As
$u$ is a tempered distribution, an $N\in\mathbb{N}$ exists such that
$v=(1+|\cdot|^{2})^{-N}u\in H^{-N}$ (see [16, Lem. 1.16]). A simple
computation indicates that

$\displaystyle\mathop{}\\!\mathrm{Op}(b)u(x)$
$\displaystyle=\mathop{}\\!\mathrm{Op}(b)(1+|x|^{2})^{N}v(x)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}b(x,\xi)\sum_{\beta\in\mathbb{N}^{d}}\frac{1}{\beta!}\partial_{x}^{\beta}(1+|x|^{2})^{N}(y-x)^{\beta}v(y)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\sum_{\beta\in\mathbb{N}^{d}}\frac{1}{\beta!}D_{x}^{\beta}(1+|x|^{2})^{N}\mathop{}\\!\mathrm{Op}(\partial_{\xi}^{\beta}b)v(x).$
(2.50)

Clearly, $\partial_{\xi}^{\beta}b\in S^{-\infty}$ for all
$\beta\in\mathbb{N}^{d}$ because $b\in S^{-\infty}$. It follows that
$\mathop{}\\!\mathrm{Op}(\partial_{\xi}^{\beta}b)v\in H^{d}$ by Corollary
2.5.2. Because $H^{d}$ is a subspace of the space of continuous functions,
$\mathop{}\\!\mathrm{Op}(\partial_{\xi}^{\beta}b)v$ is continuous, and so is
$\mathop{}\\!\mathrm{Op}(b)u$. ∎

######  Theorem 2.5.4.

If $a\in S^{m}$, then $\mathop{}\\!\mathrm{Op}(a)$ is pseudo-local. If $a\in
S^{m}$ is elliptic, then $\mathop{}\\!\mathrm{Op}(a)$ is hypoelliptic.

###### Proof.

Let $u\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$ be a tempered distribution and
$U=\mathbb{R}^{d}\backslash\mathop{}\\!\mathrm{sing\ supp}(u)$ be the
complement of the singular support of $u$. Then, $\psi u\in
C_{\mathrm{c}}^{\infty}(U)$ for all $\psi\in C_{\mathrm{c}}^{\infty}(U)$. To
prove $\mathop{}\\!\mathrm{sing\
supp}(\mathop{}\\!\mathrm{Op}(a)u)\subset\mathop{}\\!\mathrm{sing\ supp}(u)$,
we show that $\varphi\mathop{}\\!\mathrm{Op}(a)u\in
C_{\mathrm{c}}^{\infty}(U)$ for all $\varphi\in C_{\mathrm{c}}^{\infty}(U)$.
If $\varphi\in C_{\mathrm{c}}^{\infty}(U)$, a $\psi\in
C_{\mathrm{c}}^{\infty}(U)$ exists such that $\psi=1$ on
$\mathop{}\\!\mathrm{supp}(\varphi)$. We write
$\varphi\mathop{}\\!\mathrm{Op}(a)u$ as

$\displaystyle\varphi\mathop{}\\!\mathrm{Op}(a)u=\varphi\mathop{}\\!\mathrm{Op}(a)(\psi
u)+\varphi\mathop{}\\!\mathrm{Op}(a)((1-\psi)u).$ (2.51)

The function $\mathop{}\\!\mathrm{Op}(a)(\psi u)$ is a Schwartz function
because $\psi u\in
C_{\mathrm{c}}^{\infty}(U)\subset\mathcal{S}(\mathbb{R}^{d})$ and because
$\mathop{}\\!\mathrm{Op}(a)$ maps Schwartz functions to Schwartz functions
(see Theorem 2.2.5). Hence, $\varphi\mathop{}\\!\mathrm{Op}(a)(\psi u)\in
C_{\mathrm{c}}^{\infty}(U)$. The second summand may be written as
$\varphi\mathop{}\\!\mathrm{Op}(a)((1-\psi)u)=\mathop{}\\!\mathrm{Op}(b)u$
with $b=\varphi a\\#(1-\psi)\in S^{-\infty}$ because the supports of $\varphi
a$ and $1-\psi$ are disjoint. From the previous theorem it follows that
$\mathop{}\\!\mathrm{Op}(b)u$ is smooth; thus, $\mathop{}\\!\mathrm{Op}(b)u\in
C_{\mathrm{c}}^{\infty}(U)$.

If $a\in S^{m}$ is elliptic, then, according to Theorem 2.4.5, a $b\in S^{-m}$
and $r\in S^{-\infty}$ exist such that $b\\#a=1+r$. The function
$\mathop{}\\!\mathrm{Op}(r)u$ is smooth because $\mathop{}\\!\mathrm{Op}(r)$
is smoothing; hence, $\mathop{}\\!\mathrm{sing\
supp}(u)=\mathop{}\\!\mathrm{sing\
supp}(\mathop{}\\!\mathrm{Op}(b\\#a)u)\subset\mathop{}\\!\mathrm{sing\
supp}(\mathop{}\\!\mathrm{Op}(a)u)$. ∎

###  2.6 Operator-valued Symbols

In this section, we discuss pseudo-differential operators with operator-valued
symbols as presented in [19, Appendix A]. This calculus does not provide any
more difficulties than what we discussed so far, as most of the proofs can be
adapted without modification. In Chapter 4, where we apply the IBC method to
the Nelson model with variable coefficients, a calculus with operator-valued
symbols simplifies our analysis.

Let $\mathcal{H}_{1}$, $\mathcal{H}_{2}$ be two separable Hilbert spaces. We
write $\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})$ for the space of all
bounded operators from $\mathcal{H}_{1}$ to $\mathcal{H}_{2}$. The following
definition is a direct generalisation of the symbol class $S_{k}(M)$.

######  Definition 2.6.1.

Let $M$ be an order function. Then
$S_{k}(M,\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$ is the space of all
$a\in
C^{\infty}(\mathbb{R}^{k},\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$ that
satisfy, for all $\alpha\in\mathbb{N}^{k}$, the following bound:

$\displaystyle\lVert\partial^{\alpha}_{z}a(z)\rVert_{\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})}\leq
C_{\alpha}M(z).$ (2.52)

We recover the symbol class $S_{k}(M)$ if
$\mathcal{H}_{1}=\mathcal{H}_{2}=\mathbb{C}$. Again,
$S_{k}(M,\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$ is a Fréchet space
with semi-norms given by the best constants $C_{\alpha}$ satisfying the above
estimate. The operator-valued pseudo-differential operator
$\mathop{}\\!\mathrm{Op}(a)$ for a symbol $a\in
S_{3d}(M,\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$ is defined on
$C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d},\mathcal{H}_{1})$ in an analogy to
Definition 2.2.1 by

$\displaystyle\mathrm{Op}(a)u(x)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a(x,y,\xi)u(y)\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}\xi,$
(2.53)

where the integral is a Bochner integral. The results of Section 2.3
generalise without modification to operator-valued pseudo-differential
operators. Thus, it is not a restriction to assume that
$a(x,y,\xi)=a(tx+(1-t)y,\xi)$ for some $t\in[0,1]$.

The only theorem we re-prove for operator-valued pseudo-differential operators
is the Calderon–Vaillancourt theorem because of its central importance for the
present work and because the proof of Theorem 2.5.1 fails for symbols in
$S(1)$. The proof we present follows [12, Thm. 2.8.1], which we translate into
the setting of operator-valued symbols.

######  Theorem 2.6.2 (Calderon–Vaillancourt).

For every $a\in S(1,\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$, the
operator-valued pseudo-differential operator $\mathop{}\\!\mathrm{Op}(a)$
extends to a bounded operator from $L^{2}(\mathbb{R}^{d},\mathcal{H}_{1})$ to
$L^{2}(\mathbb{R}^{d},\mathcal{H}_{2})$. Moreover, there are constants $C,N>0$
independent of $a$ such that

$\displaystyle\lVert\mathop{}\\!\mathrm{Op}(a)u\rVert_{L^{2}(\mathbb{R}^{d},\mathcal{H}_{2})}\leq
C\sum_{|\alpha|\leq
N}\sup_{z\in\mathbb{R}^{2d}}\lVert\partial_{z}^{\alpha}a(z)\rVert_{\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})}\lVert
u\rVert_{L^{2}(\mathbb{R}^{d},\mathcal{H}_{1})}$ (2.54)

for all $u\in L^{2}(\mathbb{R}^{d},\mathcal{H}_{1})$.

###### Proof.

The key idea of the proof is to apply the Cotlar–Stein Lemma with
$H_{1}=L^{2}(\mathbb{R}^{d},\mathcal{H}_{1})$ and
$H_{2}=L^{2}(\mathbb{R}^{d},\mathcal{H}_{2})$ (see Theorem A.1.3).

We define a partition of the operator $A=\mathop{}\\!\mathrm{Op}(a)$ by
selecting a $\chi\in C_{\mathrm{c}}^{\infty}(\mathbb{R}^{2d})$ such that

$\displaystyle\sum_{\mu\in\mathbb{Z}^{2d}}\chi_{\mu}=1,$ (2.55)

where $\chi_{\mu}(z)=\chi(z-\mu)$ is the function $\chi$ shifted by the
lattice point $\mu\in\mathbb{Z}^{2d}$. We set $a_{\mu}=a\chi_{\mu}$ and
$A_{\mu}=\mathop{}\\!\mathrm{Op}(a_{\mu})$. The operator $A_{\mu}:H_{1}\to
H_{2}$ is bounded for every $\mu\in\mathbb{Z}^{2d}$:

$\displaystyle\lVert A_{\mu}u\rVert_{H_{2}}^{2}$
$\displaystyle=\int_{\mathbb{R}^{d}}\lVert
A_{\mu}u(x)\rVert^{2}_{\mathcal{H}_{2}}\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{1}{(2\pi)^{2d}}\int_{\mathbb{R}^{d}}\left(\int_{\mathbb{R}^{d}}\lVert
a_{\mu}(x,\xi)\widehat{u}(\xi)\rVert_{\mathcal{H}_{2}}\mathop{}\\!\mathrm{d}\xi\right)^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{1}{(2\pi)^{2d}}\int_{\mathbb{R}^{d}}\left(\int_{\mathbb{R}^{d}}\lVert
a_{\mu}(x,\xi)\rVert_{\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})}^{2}\mathop{}\\!\mathrm{d}\xi\right)\left(\int_{\mathbb{R}^{d}}\lVert\widehat{u}(\xi)\rVert_{\mathcal{H}_{1}}^{2}\mathop{}\\!\mathrm{d}\xi\right)\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{1}{(2\pi)^{2d}}\sup_{z\in\mathbb{R}^{2d}}\lVert
a(z)\rVert_{\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})}^{2}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|\chi_{\mu}(x,\xi)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}\xi\
\lVert\widehat{u}\rVert_{H_{1}}^{2}$ $\displaystyle\leq
C\sup_{z\in\mathbb{R}^{2d}}\lVert
a(z)\rVert_{\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})}^{2}\lVert
u\rVert_{H_{1}}^{2}.$ (2.56)

The constant $C$ does not depend on $\mu$ because the integral over $(x,\xi)$
in the fourth line is translation invariant.

Next, we estimate the operator norms of $A_{\mu}A_{\nu}^{*}$ and
$A_{\mu}^{*}A_{\nu}$ for
$\mu=(\mu_{1},\mu_{2}),\nu=(\nu_{1},\nu_{2})\in\mathbb{Z}^{2d}$ and illustrate
that

$\displaystyle\sup_{\mu\in\mathbb{Z}^{2d}}\sum_{\nu\in\mathbb{Z}^{2d}}\lVert
A_{\mu}A_{\nu}^{*}\rVert_{\mathfrak{B}(H_{2})}^{\frac{1}{2}}\leq
C\sum_{|\alpha|\leq
N}\sup_{z\in\mathbb{R}^{2d}}\lVert\partial_{z}^{\alpha}a(z)\rVert_{\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})}\equiv
C(a),$ (2.57)
$\displaystyle\sup_{\mu\in\mathbb{Z}^{2d}}\sum_{\nu\in\mathbb{Z}^{2d}}\lVert
A_{\mu}^{*}A_{\nu}\rVert_{\mathfrak{B}(H_{1})}^{\frac{1}{2}}\leq
C\sum_{|\alpha|\leq
N}\sup_{z\in\mathbb{R}^{2d}}\lVert\partial_{z}^{\alpha}a(z)\rVert_{\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2})}\equiv
C(a)$ (2.58)

for some constants $C,N>0$ which do not depend on $a$.

We already observed that $A_{\mu}$ is bounded by $C(a)$. Hence, for every
$\mu\in\mathbb{Z}^{2d}$, it suffices to bound the operator norms of
$A_{\mu}A_{\nu}^{*}$ and $A_{\mu}^{*}A_{\nu}$ for almost every $\nu$ and
demonstrate that the operator norms are summable. Therefore, we assume
$|\mu-\nu|$ is sufficiently large such that the supports of $\chi_{\mu}$ and
$\chi_{\nu}$ are disjoint. To proceed further, we note that

$\displaystyle
A_{\mu}A_{\nu}^{*}u(x)=\int_{\mathbb{R}^{d}}K_{\mu\nu}(x,y)u(y)\mathop{}\\!\mathrm{d}y$
(2.59)

with the integral kernel

$\displaystyle
K_{\mu\nu}(x,y)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}a_{\mu}(x,\xi)a_{\nu}(y,\xi)^{*}\mathop{}\\!\mathrm{d}\xi\in\mathfrak{B}(\mathcal{H}_{2}).$
(2.60)

Furthermore, we observe that

$\displaystyle L\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle},\ L=\frac{1+(x-y)D_{\xi}}{1+|x-y|^{2}}.$ (2.61)

Hence, via integration by parts, it follows that, for any $k\in\mathbb{N}$,

$\displaystyle
K_{\mu\nu}(x,y)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-y,\xi\rangle}(L^{\mathrm{T}})^{k}(a_{\mu}(x,\xi)a_{\nu}(y,\xi)^{*})\mathop{}\\!\mathrm{d}\xi.$
(2.62)

If
$(x,y)\in\mathop{}\\!\mathrm{supp}(a_{\mu}(\cdot,\xi)a_{\nu}(\cdot,\xi)^{*})$,
then $x\neq y$ because we assumed that the supports of $\chi_{\mu}$ and
$\chi_{\nu}$ are disjoint. Thus, a constant $c>0$ independent of $\mu,\nu$,
and $\xi$ exists such that, for every
$(x,y)\in\mathop{}\\!\mathrm{supp}(a_{\mu}(\cdot,\xi)\overline{a}_{\nu}(\cdot,\xi))$,

$\displaystyle c^{-1}|\mu-\nu|\leq|x-y|\leq c|\mu-\nu|.$ (2.63)

With a similar argument as in the first step of the proof of Theorem 2.5.1, we
obtain the following bound:

$\displaystyle\int_{\mathbb{R}^{d}}\lVert
K_{\mu\nu}(x,y)\rVert_{\mathfrak{B}(\mathcal{H}_{2})}\mathop{}\\!\mathrm{d}y\leq
C(a)\int_{B_{\mu\nu}(x)}\int_{\begin{subarray}{c}|\xi-\mu_{2}|\leq C_{\chi}\\\
|\xi-\nu_{2}|\leq
C_{\chi}\end{subarray}}\frac{1}{(1+|x-y|)^{k}}\mathop{}\\!\mathrm{d}\xi\mathop{}\\!\mathrm{d}y,$
(2.64)

where $B_{\mu\nu}(x)=\\{y\in\mathbb{R}^{d}\mid c^{-1}|\mu-\nu|\leq|x-y|\leq
c|\mu-\nu|\\}$. The constant $C_{\chi}$ depends only on the support of $\chi$.
The integral over $\xi$ in (2.64) is bounded and independent of $\mu$ and
$\nu$; hence,

$\displaystyle\int_{\mathbb{R}^{d}}\lVert
K_{\mu\nu}(x,y)\rVert_{\mathfrak{B}(\mathcal{H}_{2})}\mathop{}\\!\mathrm{d}y$
$\displaystyle\leq
C(a)\int_{B_{\mu\nu}(x)}\frac{1}{(1+|x-y|)^{k}}\mathop{}\\!\mathrm{d}y$
$\displaystyle\leq
C(a)\int_{\mathbb{R}^{d}}\frac{(1+|\mu-\nu|)^{d+1-k}}{(1+|x-y|)^{d+1}}\mathop{}\\!\mathrm{d}y$
$\displaystyle\leq C(a)(1+|\mu-\nu|)^{d+1-k}.$ (2.65)

Analogously, we confirm that integrating $\lVert
K_{\mu\nu}(x,y)\rVert_{\mathfrak{B}(\mathcal{H}_{2})}$ over $x$ yields the
same bound:

$\displaystyle\int_{\mathbb{R}^{d}}\lVert
K_{\mu\nu}(x,y)\rVert_{\mathfrak{B}(\mathcal{H}_{2})}\mathop{}\\!\mathrm{d}x\leq
C(a)(1+|\mu-\nu|)^{d+1-k}.$ (2.66)

This proves that the integral kernel $K_{\mu\nu}(x,y)$ passes the Schur test
(see Theorem A.1.2); thus,

$\displaystyle\lVert A_{\mu}A_{\nu}^{*}\rVert_{\mathfrak{B}(H_{2})}\leq
C(a)(1+|\mu-\nu|)^{d+1-k}.$ (2.67)

Similarly, the operator norm of $A_{\mu}^{*}A_{\nu}$ is bounded:

$\displaystyle\lVert A_{\mu}^{*}A_{\nu}\rVert_{\mathfrak{B}(H_{1})}\leq
C(a)(1+|\mu-\nu|)^{d+1-k}.$ (2.68)

If $k\in\mathbb{N}$ is chosen sufficiently large, $(1+|\mu-\nu|)^{d+1-k}$ is
summable over $\nu\in\mathbb{Z}^{2d}$ uniformly in $\mu\in\mathbb{Z}^{2d}$.
From the Cotlar–Stein Lemma it follows that $\sum_{\mu}A_{\mu}$ converges in
the strong operator topology to a bounded operator
$A:L^{2}(\mathbb{R}^{d},\mathcal{H}_{1})\to
L^{2}(\mathbb{R}^{d},\mathcal{H}_{2})$ with $\lVert A\rVert\leq C(a)$. The
action of $A$ on $C_{\mathrm{c}}^{\infty}(\mathbb{R}^{d},\mathcal{H}_{1})$
coincides with that of $\mathop{}\\!\mathrm{Op}(a)$. Thus,
$\mathop{}\\!\mathrm{Op}(a)$ extends to a bounded operator from
$L^{2}(\mathbb{R}^{d},\mathcal{H}_{1})$ to
$L^{2}(\mathbb{R}^{d},\mathcal{H}_{2})$. ∎

######  Corollary 2.6.3.

If $a\in S(M,\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$, then
$\mathop{}\\!\mathrm{Op}(a)$ is relatively bounded with respect to $M(D)$.
Furthermore, if $a$ is elliptic, then $\mathop{}\\!\mathrm{Op}(a^{-1})$ is
relatively bounded with respect to $M(D)^{-1}$.

###### Proof.

It is not a restriction to assume that $\mathop{}\\!\mathrm{Op}(a)$ is given
in standard quantisation. A simple computation demonstrates that

$\displaystyle\mathop{}\\!\mathrm{Op}(a)u(x)$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\xi\rangle}a(x,\xi)M(\xi)^{-1}M(\xi)\widehat{u}(\xi)\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x,\xi\rangle}a(x,\xi)M(\xi)^{-1}\widehat{M(D)u}(\xi)\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\mathop{}\\!\mathrm{Op}\left(aM^{-1}\right)M(D)u(x);$ (2.69)

hence, the first claim is a consequence of Theorem 2.6.2 because $aM^{-1}\in
S(1,\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$.

If $a$ is elliptic, then $a^{-1}\in
S(M^{-1},\mathfrak{B}(\mathcal{H}_{1},\mathcal{H}_{2}))$. Because
$a^{-1}(x,\xi)M(\xi)$ is bounded by 1, a similar computation as in (2.69)
verifies that $\mathop{}\\!\mathrm{Op}(a^{-1})$ is relatively bounded with
respect to $M(D)^{-1}$. ∎

##  Chapter 3 Removal of the Cut-off

In Section 1.2, we argued that the formal Nelson Hamiltonian must be
regularised to obtain a well-defined Hamiltonian. We proved in Proposition
1.2.2 that putting an ultraviolet cut-off $\Lambda$ on the interaction leads
to a self-adjoint Hamiltonian $H_{\Lambda}$. In this chapter, we remove the
cut-off using renormalisation to obtain the self-adjoint renormalised Nelson
Hamiltonian.

######  Notation.

We write $f\in C^{\infty}_{\mathrm{b}}(\mathbb{R}^{3})$ if $f$ is a smooth
function and if all derivatives of $f$ (including the function $f$ itself) are
bounded.

######  Theorem 3.0.1.

Let $H_{\Lambda}$ be the Nelson Hamiltonian with variable coefficients and
cut-off defined in Section 1.3. Assume that $g^{jk},\mu^{2},W\in
C^{\infty}_{\mathrm{b}}(\mathbb{R}^{3})$. Then particle potentials
$-E_{\Lambda}(X)$ exist such that $H_{\Lambda}+E_{\Lambda}(X)$ converges in
the norm resolvent sense to a self-adjoint bounded from below operator $H$ as
$\Lambda\to\infty$.

The theorem in the above formulation was first proven by Gérard et al. [4],
and the proof we present follows their paper. Furthermore, we include ideas
from Nelson’s original work [13] and from Griesemer and Wünsch [6, 7]. The
latter proved that the convergence holds in the norm resolvent sense in the
Nelson model with constant coefficients, and their technique applies to the
Nelson model with variable coefficients. The main steps in the proof of
Theorem 3.0.1 are the following.

  1. 1.

Realise that the relevant operators in the Nelson model with variable
coefficients are pseudo-differential operators.

  2. 2.

Define a suitable unitary Gross transformation $U_{\Lambda}$ and compute the
transformed cut-off Hamiltonian $U_{\Lambda}H_{\Lambda}U_{\Lambda}^{*}$.

  3. 3.

Identify a particle potential $-E_{\Lambda}(X)$ that is divergent as
$\Lambda\to\infty$. Remove this divergent potential from the transformed cut-
off Hamiltonian.

  4. 4.

Show that
$H_{\Lambda}^{\prime}=U_{\Lambda}H_{\Lambda}U_{\Lambda}^{*}+E_{\Lambda}(X)$
defines a family $(q_{\Lambda})_{\Lambda\geq 0}$ of quadratic forms. Prove
that the limit $q_{\infty}$ exists and that the corresponding operator
$H_{\infty}^{\prime}$ defines the required self-adjoint operator
$H=U_{\infty}^{*}H_{\infty}^{\prime}U_{\infty}$.

###  3.1 Pseudo-differential Operators in Nelson Model

The particle Hamiltonian $K=K_{0}+W(X)$ and the square of the dispersion
relation $\omega^{2}=h$ can be treated as pseudo-differential operators:

$\displaystyle K=K_{0}+W(X)$
$\displaystyle=K_{0}^{\mathrm{w}}(X,D_{X})+W(X)=K^{\mathrm{w}}(X,D_{X}),$
(3.1) $\displaystyle h=h_{0}+\mu(x)^{2}$
$\displaystyle=h_{0}^{\mathrm{w}}(x,D_{x})+\mu(x)^{2}=h^{\mathrm{w}}(x,D_{x}).$
(3.2)

The corresponding symbols

$\displaystyle K(X,\Xi)$ $\displaystyle=K_{0}(X,\Xi)+W(X)=\Xi\cdot
g(X)\Xi+W(X),$ (3.3) $\displaystyle h(x,\xi)$
$\displaystyle=h_{0}(x,\xi)+\mu(x)^{2}=\xi\cdot g(x)\xi+\mu(x)^{2}$ (3.4)

are under the assumptions of Theorem 3.0.1 elements of $S^{2}$. Furthermore,
these symbols are elliptic because $g$ is uniformly elliptic.

##### Dispersion relation

The dispersion relation $\omega$ is the positive square root of $h$. The
square root of a pseudo-differential operator is not necessarily a pseudo-
differential operator because the square root is not smooth in $0$. However,
$h\geq m^{2}>0$ is bounded from below by the square of the (by assumption)
positive bosonic field mass $m^{2}$. If $f\in S^{1/2}(\mathbb{R})$ (see (2.40)
for the definition of $S^{p}(\mathbb{R})$) is such that $f(t)=\sqrt{t}$ for
$t\geq m^{2}/2$, then $f$ and $t\mapsto\sqrt{t}$ agree on the spectrum of $h$;
hence,

$\displaystyle\omega=(h^{\mathrm{w}})^{\frac{1}{2}}=f(h^{\mathrm{w}}).$ (3.5)

According to the functional calculus (see Theorem 2.4.6), the operator
$\omega=f(h^{\mathrm{w}})$ is a pseudo-differential operator of order 1. Its
principal symbol is $f(h(x,\xi))=h(x,\xi)^{1/2}$, that is,
$f(h)^{\mathrm{w}}-\omega=f(h)^{\mathrm{w}}-f(h^{\mathrm{w}})\in\mathrm{Op}(S^{0})$.
If we prefer to use the standard quantisation instead, we can also write
$\mathrm{Op}(f(h))-\omega\in\mathop{}\\!\mathrm{Op}(S^{0})$ due to Corollary
2.3.3. Similarly, we confirm that $\omega^{k}$ is a pseudo-differential
operator of order $k$ for every $k\in\mathbb{R}$.

##### Free Nelson Hamiltonian

The free Nelson Hamiltonian
$H_{0}=K\otimes\mathbb{1}+\mathbb{1}\otimes\mathop{}\\!\mathrm{d}\Gamma(\omega)\equiv
K+\mathop{}\\!\mathrm{d}\Gamma(\omega)$ restricted to the $n$-boson subspace
$\mathfrak{H}^{(n)}$ is a pseudo-differential operator.111Hereafter, if it is
clear from the context, we omit tensor products with the identity operator.
Its symbol is elliptic and an element of the symbol class $S(M,g)$ with the
order function

$\displaystyle
M(\Xi,\xi)=\langle\Xi\rangle^{2}+\sum_{j=1}^{n}\langle\xi_{j}\rangle\equiv\langle\Xi\rangle^{2}+\Omega(\xi),$
(3.6)

where $\Omega(\xi)=\sum_{j=1}^{n}\langle\xi_{j}\rangle$, and the metric

$\displaystyle
g_{(X,\Xi,\xi,x)}=\mathop{}\\!\mathrm{d}X^{2}+\langle\Xi\rangle^{-2}\mathop{}\\!\mathrm{d}\Xi^{2}+\sum_{j=1}^{n}(\mathop{}\\!\mathrm{d}x_{j}^{2}+\langle\xi_{j}\rangle^{-2}\mathop{}\\!\mathrm{d}\xi_{j}^{2}).$
(3.7)

A peculiarity of the Nelson model is that the symbol is quadratic in $\Xi$ but
linear in $\xi$. In a fully relativistic theory, the symbol is of linear order
in $\Xi$ and $\xi$. The additional order in $\Xi$ is vital for our analysis
because it makes some otherwise divergent integrals convergent.

##### Form factor

We denote the regularised form factor of the field operator
$\Phi(\omega^{-1/2}\rho_{\Lambda,X})=a^{*}(v_{\Lambda,X})+a(v_{\Lambda,X})$ by
$v_{\Lambda,X}=\omega^{-1/2}\rho_{\Lambda,X}/\sqrt{2}$, where the cut-off
function $\rho_{\Lambda,X}$ is defined in (1.52). Above, we observed that
$\omega^{-1/2}$ is a pseudo-differential operator of order $-1/2$. Let
$\sigma(\omega^{-1/2})\in S^{-1/2}$ be its symbol in standard quantisation.
Then the form factor $v_{\Lambda,X}$ can be written as follows:

$\displaystyle
v_{\Lambda,X}(x)=\frac{1}{\sqrt{2}}\int_{\mathbb{R}^{3}}\frac{\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-X,\xi\rangle}}{(2\pi)^{3}}\sigma(\omega^{-\frac{1}{2}})(x,\xi)\widehat{\rho}(\xi/\Lambda)\mathop{}\\!\mathrm{d}\xi.$
(3.8)

We extract the leading contribution of the above integral by expanding the
symbol $\sigma(\omega^{-\frac{1}{2}})(x,\xi)$ in $x=X$:

$\displaystyle\sigma(\omega^{-\frac{1}{2}})(x,\xi)=\sigma(\omega^{-\frac{1}{2}})(X,\xi)-\sqrt{2}(x-X)\cdot
r(x,\xi),$ (3.9)

where $r\in[S^{-1/2}]^{3}=S^{-1/2}\times S^{-1/2}\times S^{-1/2}$. The form
factor decomposes:

$\displaystyle v_{\Lambda,X}=u_{\Lambda,X}+\tilde{u}_{\Lambda,X}$ (3.10)

with

$\displaystyle
u_{\Lambda,X}(x)=\frac{1}{\sqrt{2}}\int_{\mathbb{R}^{3}}\frac{\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-X,\xi\rangle}}{(2\pi)^{3}}\sigma(\omega^{-\frac{1}{2}})(X,\xi)\widehat{\rho}(\xi/\Lambda)\mathop{}\\!\mathrm{d}\xi$
(3.11)

and

$\displaystyle\tilde{u}_{\Lambda,X}(x)$
$\displaystyle=-\int_{\mathbb{R}^{3}}\frac{\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-X,\xi\rangle}}{(2\pi)^{3}}(x-X)\cdot
r(x,\xi)\widehat{\rho}(\xi/\Lambda)\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\int_{\mathbb{R}^{3}}\frac{\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-X,\xi\rangle}}{(2\pi)^{3}}(D_{\xi}\cdot
r)(x,\xi)\widehat{\rho}(\xi/\Lambda)\mathop{}\\!\mathrm{d}\xi+\frac{1}{\Lambda}\int_{\mathbb{R}^{3}}\frac{\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-X,\xi\rangle}}{(2\pi)^{3}}r(x,\xi)\cdot(D_{\xi}\widehat{\rho})(\xi/\Lambda)\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\mathop{}\\!\mathrm{Op}(D_{\xi}\cdot
r)\rho_{\Lambda,X}(x)+\frac{1}{\Lambda}\mathop{}\\!\mathrm{Op}(r)\tau_{\Lambda,X}(x),$
(3.12)

where $\tau(x)=x\rho(x)$ is a vector of Schwartz functions.

###  3.2 Gross Transformation

We proceed with the second step of the proof and find a unitary transformation
of the cut-off Hamiltonian that reveals a divergent potential in the Nelson
Hamiltonian.

Assume for a moment that the nonrelativistic particle has infinite mass (i.e.
its position in space is fixed, and its kinetic energy is negligible). The
cut-off Hamiltonian, in this case, reduces to

$\displaystyle
H^{\circ}_{\Lambda}=\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X}).$
(3.13)

Recall the definition and properties of the unitary Weyl operators
$V(f)=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(f)}$ from Subsection
1.1.3. We compute the Gross transformed Hamiltonian
$V(f)H_{\Lambda}^{\circ}V(f)^{*}$. From Proposition 1.1.6 it follows that

$\displaystyle\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(f)}H^{\circ}_{\Lambda}\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}\Pi(f)}$
$\displaystyle=\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega
f+\omega^{-\frac{1}{2}}\rho_{\Lambda,X})+\frac{1}{2}\lVert\omega^{\frac{1}{2}}f\rVert^{2}_{\mathfrak{h}}+\Re\langle
f,\omega^{-\frac{1}{2}}\rho_{\Lambda,X}\rangle_{\mathfrak{h}}.$ (3.14)

If $f=-\omega^{-\frac{3}{2}}\rho_{\Lambda,X}$, then

$\displaystyle\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(\omega^{-\frac{3}{2}}\rho_{\Lambda,X})}H^{\circ}_{\Lambda}\mathop{}\\!\mathrm{e}^{-\mathop{}\\!\mathrm{i}\Pi(\omega^{-\frac{3}{2}}\rho_{\Lambda,X})}=\mathop{}\\!\mathrm{d}\Gamma(\omega)-\frac{1}{2}\lVert\omega^{-1}\rho_{\Lambda,X}\rVert^{2}_{\mathfrak{h}}.$
(3.15)

Thus, the Gross transformation reveals the term
$-\lVert\omega^{-1}\rho_{\Lambda,X}\rVert^{2}_{\mathfrak{h}}/2$, which
diverges as $\Lambda\to\infty$. This term must be removed to renormalise the
Nelson model with a particle of infinite mass.

We perform a similar Gross transformation for the full cut-off Hamiltonian
$H_{\Lambda}$. The correct generalisation of the function
$f=-\omega^{-\frac{3}{2}}\rho_{\Lambda,X}$ turns out to be

$\displaystyle
B_{\Lambda,X}^{\sigma}(x)=-(K+\omega)^{-1}\omega^{-\frac{1}{2}}\rho_{\Lambda,X}^{\sigma}(x).$
(3.16)

We added a superscript $\sigma$ in the cut-off function $\rho_{\Lambda}$. This
is an infrared cut-off and is defined via

$\displaystyle\widehat{\rho_{\Lambda}^{\sigma}}(\xi)=\chi(\xi)\widehat{\rho}(\xi/\Lambda),$
(3.17)

where $\chi\in C^{\infty}(\mathbb{R}^{3})$, $0\leq\chi\leq 1$, is any smooth
symmetric function equal to $0$ if $|\xi|\leq\sigma$ and equal to $1$ if
$|\xi|\geq 2\sigma$. It becomes clear why this cut-off is needed at the end of
the proof.

We collect analytical properties of the function $B_{\Lambda,X}^{\sigma}$.
First, for every $X\in\mathbb{R}^{3}$, $B_{\Lambda,X}^{\sigma}$ is a Schwartz
function because pseudo-differential operators map Schwartz functions to
Schwartz functions (see Theorem 2.2.5). Moreover, $B_{\Lambda,X}^{\sigma}$ is
real-valued because $\rho_{\Lambda,X}^{\sigma}$ is real-valued and because the
operators $K$ and $\omega$ are real operators.

######  Lemma 3.2.1.

For every $\alpha<1$, $\omega^{\alpha}B_{\Lambda,X}^{\sigma}$,
$\omega^{\alpha-1}\partial_{X}B_{\Lambda,X}^{\sigma}\in\mathfrak{h}=L^{2}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x)$,
and an $s<-3/2$ exists such that

$\displaystyle\lVert\omega^{\alpha}B_{\Lambda,X}^{\sigma}\rVert_{\mathfrak{h}}$
$\displaystyle\leq C\lVert\rho_{\Lambda}^{\sigma}\rVert_{H^{s}},$ (3.18)
$\displaystyle\lVert\omega^{\alpha-1}\partial_{X}B_{\Lambda,X}^{\sigma}\rVert_{\mathfrak{h}}$
$\displaystyle\leq C\lVert\rho_{\Lambda}^{\sigma}\rVert_{H^{s}}$ (3.19)

for a constant $C$ independent of $\Lambda$, $\sigma$, and $X$. Here $H^{s}$
is the Sobolev space $H^{s}(\mathbb{R}^{3},\mathop{}\\!\mathrm{d}x)$.

###### Proof.

Let $t^{-1}\in S(M^{-1},g)$ be the symbol of the pseudo-differential operator
$(K+\omega)^{-1}$ in standard quantisation with order function $M$ and metric
$g$ defined in (3.6) and (3.7) with $n=1$. Then,

$\displaystyle(K+\omega)^{-1}\rho_{\Lambda,X}^{\sigma}(x)$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
X,\Xi\rangle+\mathop{}\\!\mathrm{i}\langle
x,\xi\rangle}t^{-1}(X,\Xi,x,\xi)\delta(\Xi+\xi)\widehat{\rho^{\sigma}}(\xi/\Lambda)\mathop{}\\!\mathrm{d}\Xi\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int_{\mathbb{R}^{3}}\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\langle
x-X,\xi\rangle}t^{-1}(X,-\xi,x,\xi)\widehat{\rho^{\sigma}}(\xi/\Lambda)\mathop{}\\!\mathrm{d}\xi.$
(3.20)

For $X\in\mathbb{R}^{3}$ fixed, $t_{X}^{-1}(x,\xi)=t^{-1}(X,-\xi,x,\xi)$
defines a symbol $t_{X}^{-1}\in S^{-2}$. The form of the symbol $t_{X}^{-1}$
is irrelevant for this proof, but for later applications, we note that
$t_{X}^{-1}$ is given by

$\displaystyle t^{-1}_{X}(x,\xi)$
$\displaystyle=(K(X,\xi)+\sigma(\omega)(x,\xi))^{-1}+S^{-3}$
$\displaystyle=(K_{0}(X,\xi)+1)^{-1}+S^{-3}.$ (3.21)

Because $\omega^{-1/2}\in\mathop{}\\!\mathrm{Op}(S^{-1/2})$, a symbol
$b_{X}\in S^{-5/2}$ exists such that

$\displaystyle
B_{\Lambda,X}^{\sigma}=(K+\omega)^{-1}\omega^{-\frac{1}{2}}\rho_{\Lambda,X}^{\sigma}=\omega^{-\frac{1}{2}}\mathop{}\\!\mathrm{Op}(t_{X}^{-1})\rho_{\Lambda,X}^{\sigma}=\mathop{}\\!\mathrm{Op}(b_{X})\rho_{\Lambda,X}^{\sigma}.$
(3.22)

Due to Corollary 2.5.2, we obtain the following inequality for every
$s\in\mathbb{R}$:

$\displaystyle\lVert\omega^{\alpha}\mathop{}\\!\mathrm{Op}(b_{X})\rho_{\Lambda,X}\rVert_{H^{s-m}}\leq
C\lVert\rho_{\Lambda,X}^{\sigma}\rVert_{H^{s}}=C\lVert\rho_{\Lambda}^{\sigma}\rVert_{H^{s}},$
(3.23)

where $m=-5/2+\alpha<-3/2$. The inequality (3.18) is proven if we choose
$s=m$.

To prove (3.19), we observe that
$\partial_{X}\rho_{\Lambda,X}^{\sigma}(x)=-\partial_{x}\rho_{\Lambda,X}^{\sigma}(x)$
and

$\displaystyle\partial_{X}B_{\Lambda,X}^{\sigma}=\mathop{}\\!\mathrm{Op}(\partial_{X}b_{X})\rho_{\Lambda,X}^{\sigma}-\mathop{}\\!\mathrm{Op}(b_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma}$
(3.24)

with $\partial_{X}b_{X}\in S^{-5/2}$ and
$\lVert\partial_{x}\rho_{\Lambda,X}^{\sigma}\rVert_{H^{s}}\leq\lVert\rho_{\Lambda,X}^{\sigma}\rVert_{H^{s+1}}$.
The rest of the proof is similar to the proof of the first inequality. ∎

We return to the transformation of the cut-off Hamiltonian $H_{\Lambda}$. The
Gross transformation of $H_{\Lambda}$ is realised using the unitary operator

$\displaystyle
U_{\Lambda}=V(B_{\Lambda,X}^{\sigma})=\mathop{}\\!\mathrm{e}^{\mathop{}\\!\mathrm{i}\Pi(B_{\Lambda,X}^{\sigma})}.$
(3.25)

Computing the transformed operator $U_{\Lambda}H_{\Lambda}U_{\Lambda}^{*}$,
however, is more tedious than for $H^{\circ}_{\Lambda}$. At the end of a
careful computation, we arrive at

$\displaystyle
U_{\Lambda}H_{\Lambda}U_{\Lambda}^{*}=H_{0}+R_{\Lambda}(X)+V_{\Lambda}(X)$
(3.26)

with $R_{\Lambda}(X)$ and $V_{\Lambda}(X)$ to be specified below.

######  Notation.

For notational convenience, we drop the subscripts in $\rho=\rho_{\Lambda,X}$,
$\rho^{\sigma}=\rho_{\Lambda,X}^{\sigma}$, and
$B^{\sigma}=B_{\Lambda,X}^{\sigma}$ for the rest of this section.

The transformation of the free field Hamiltonian and the interacting field are
known from the beginning of this section:

$\displaystyle U_{\Lambda}\mathop{}\\!\mathrm{d}\Gamma(\omega)U_{\Lambda}^{*}$
$\displaystyle=\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega
B^{\sigma})+\frac{1}{2}\lVert\omega^{\frac{1}{2}}B^{\sigma}\rVert^{2}_{\mathfrak{h}},$
(3.27) $\displaystyle
U_{\Lambda}\Phi(\omega^{-\frac{1}{2}}\rho)U_{\Lambda}^{*}$
$\displaystyle=\Phi(\omega^{-\frac{1}{2}}\rho)+\langle
B^{\sigma},\omega^{-\frac{1}{2}}\rho\rangle_{\mathfrak{h}}.$ (3.28)

The particle Hamiltonian is defined as $K=K_{0}+W(X)$. Because the potential
$W(X)$ does not depend on the boson variables, it commutes with $U_{\Lambda}$
and remains invariant under the unitary transformation. To transform the free
particle Hamiltonian $K_{0}$, observe that

$\displaystyle
U_{\Lambda}\partial_{X}U_{\Lambda}^{*}=\partial_{X}-\mathop{}\\!\mathrm{i}\Pi(\partial_{X}B^{\sigma}).$
(3.29)

We used the fact that $\Pi(B^{\sigma})$ and $\Pi(\partial_{X}B^{\sigma})$
commute according to the commutation relation (1.30) because $B^{\sigma}$ and
$\partial_{X}B^{\sigma}$ are real-valued functions. Thus,

$\displaystyle U_{\Lambda}K_{0}U_{\Lambda}^{*}$
$\displaystyle=-[\partial_{X}-\mathop{}\\!\mathrm{i}\Pi(\partial_{X}B^{\sigma})]\cdot
g(X)[\partial_{X}-\mathop{}\\!\mathrm{i}\Pi(\partial_{X}B^{\sigma})]$
$\displaystyle=K_{0}-\mathop{}\\!\mathrm{i}\Pi(K_{0}B^{\sigma})+\Pi(\partial_{X}B^{\sigma})\cdot
g(X)\Pi(\partial_{X}B^{\sigma})$ $\displaystyle\ \ \
+2\mathop{}\\!\mathrm{i}\Pi(\partial_{X}B^{\sigma})\cdot g(X)\partial_{X}.$
(3.30)

We expand and simplify the right-hand side of the above equality. First, the
conjugate momentum operator $\Pi$ and field operator $\Phi$ are related by

$\displaystyle-\mathop{}\\!\mathrm{i}\Pi(K_{0}B^{\sigma})=\Phi(K_{0}B^{\sigma})-\sqrt{2}a(K_{0}B^{\sigma}).$
(3.31)

Moreover, expanding the third summand in (3.30) leads to

$\displaystyle\Pi(\partial_{X}B^{\sigma})\cdot
g(X)\Pi(\partial_{X}B^{\sigma})$ $\displaystyle\ \ \
=-\frac{1}{2}a^{*}(\partial_{X}B^{\sigma})\cdot
g(X)a^{*}(\partial_{X}B^{\sigma})-\frac{1}{2}a(\partial_{X}B^{\sigma})\cdot
g(X)a(\partial_{X}B^{\sigma})$ $\displaystyle\ \ \ \ \ \
+a^{*}(\partial_{X}B^{\sigma})\cdot
g(X)a(\partial_{X}B^{\sigma})+\frac{1}{2}\langle\partial_{X}B^{\sigma},g(X)\partial_{X}B^{\sigma}\rangle_{\mathfrak{h}}.$
(3.32)

######  Remark 3.2.2.

In (3.32), we use the commutation relation (1.13) to obtain normal ordering,
that is, creation operators are to the left of annihilation operators. This
ordering is practical because it avoids the appearance of creation operators
if (3.32) is interpreted in the sense of quadratic forms. The additional term
$\langle\partial_{X}B^{\sigma},g(X)\partial_{X}B^{\sigma}\rangle_{\mathfrak{h}}$,
however, is divergent as $\Lambda\to\infty$.

Combining the fourth term in (3.30) with
$-\sqrt{2}a(K_{0}B^{\sigma})=\sqrt{2}a(\partial_{X}\cdot
g(X)\partial_{X}B^{\sigma})$ from the second term yields

$\displaystyle 2\mathop{}\\!\mathrm{i}\Pi(\partial_{X}B^{\sigma})\cdot
g(X)\partial_{X}-\sqrt{2}a(K_{0}B^{\sigma})$ $\displaystyle\ \ \
=-\sqrt{2}a^{*}(\partial_{X}B^{\sigma})\cdot
g(X)\partial_{X}+\sqrt{2}a(\partial_{X}B^{\sigma})\cdot
g(X)\partial_{X}-\sqrt{2}a(K_{0}B^{\sigma})$ $\displaystyle\ \ \
=-\sqrt{2}a^{*}(\partial_{X}B^{\sigma})\cdot
g(X)\partial_{X}+\sqrt{2}\partial_{X}\cdot g(X)a(\partial_{X}B^{\sigma}).$
(3.33)

Altogether, we obtain the following for the transformed cut-off Hamiltonian:

$\displaystyle U_{\Lambda}H_{\Lambda}U_{\Lambda}^{*}$
$\displaystyle=K+\mathop{}\\!\mathrm{d}\Gamma(\omega)+\Phi(\omega^{-\frac{1}{2}}\rho+(K_{0}+\omega)B^{\sigma})$
$\displaystyle\ \ \ -\sqrt{2}a^{*}(\partial_{X}B^{\sigma})\cdot
g(X)\partial_{X}+\sqrt{2}\partial_{X}\cdot g(X)a(\partial_{X}B^{\sigma})$
$\displaystyle\ \ \ -\frac{1}{2}a^{*}(\partial_{X}B^{\sigma})\cdot
g(X)a^{*}(\partial_{X}B^{\sigma})-\frac{1}{2}a(\partial_{X}B^{\sigma})\cdot
g(X)a(\partial_{X}B^{\sigma})$ $\displaystyle\ \ \
+a^{*}(\partial_{X}B^{\sigma})\cdot g(X)a(\partial_{X}B^{\sigma})$
$\displaystyle\ \ \
+\frac{1}{2}\lVert\omega^{\frac{1}{2}}B^{\sigma}\rVert^{2}_{\mathfrak{h}}+\langle
B^{\sigma},\omega^{-\frac{1}{2}}\rho\rangle_{\mathfrak{h}}+\frac{1}{2}\langle\partial_{X}B^{\sigma},g(X)\partial_{X}B^{\sigma}\rangle_{\mathfrak{h}}.$
(3.34)

The form factor of the field operator $\Phi$ is
$\omega^{-\frac{1}{2}}(\rho-\rho^{\sigma})$ according to the definition of
$B^{\sigma}$ in (3.16). We combine this field operator with the second to
fourth line to $R_{\Lambda}(X)$ (reinstating subscripts $\Lambda$ and $X$):

$\displaystyle R_{\Lambda}(X)=$
$\displaystyle-\sqrt{2}a^{*}(\partial_{X}B_{\Lambda,X}^{\sigma})\cdot
g(X)\partial_{X}+\sqrt{2}\partial_{X}\cdot
g(X)a(\partial_{X}B_{\Lambda,X}^{\sigma})$
$\displaystyle-\frac{1}{2}a^{*}(\partial_{X}B_{\Lambda,X}^{\sigma})\cdot
g(X)a^{*}(\partial_{X}B_{\Lambda,X}^{\sigma})$
$\displaystyle-\frac{1}{2}a(\partial_{X}B_{\Lambda,X}^{\sigma})\cdot
g(X)a(\partial_{X}B_{\Lambda,X}^{\sigma})$
$\displaystyle+a^{*}(\partial_{X}B_{\Lambda,X}^{\sigma})\cdot
g(X)a(\partial_{X}B_{\Lambda,X}^{\sigma})+\Phi(\omega^{-\frac{1}{2}}(\rho_{\Lambda,X}-\rho_{\Lambda,X}^{\sigma})).$
(3.35)

In Section 3.4, we show that $R_{\Lambda}(X)$ defines a quadratic form on
$D(H_{0}^{1/2})$. We abbreviate the fifth line in (3.34) by $V_{\Lambda}(X)$:

$\displaystyle V_{\Lambda}(X)$
$\displaystyle=\frac{1}{2}\lVert(K+\omega)^{-1}\rho^{\sigma}_{\Lambda,X}\rVert^{2}_{\mathfrak{h}}-\langle\rho^{\sigma}_{\Lambda,X},(K+\omega)^{-1}\omega^{-1}\rho_{\Lambda,X}\rangle_{\mathfrak{h}}$
$\displaystyle\ \ \
+\frac{1}{2}\langle\omega^{-1}\partial_{X}(K+\omega)^{-1}\rho^{\sigma}_{\Lambda,X},g(X)\partial_{X}(K+\omega)^{-1}\rho^{\sigma}_{\Lambda,X}\rangle_{\mathfrak{h}}.$
(3.36)

The potential $V_{\Lambda}(X)$ is chosen such that it contains all divergent
terms. In the next section, we identify these divergent terms.

###  3.3 Renormalisation

The discussions in this section are not strictly necessary for the proof of
Theorem 3.0.1 if we choose $-E_{\Lambda}(X)=V_{\Lambda}(X)$. Nevertheless,
this section provides a physical interpretation of the term $V_{\Lambda}(X)$
and some useful mathematical techniques.

In Section 1.2, we argued that it makes sense to remove the vacuum energy of
the formal Nelson Hamiltonian. The minimum of the spectrum of the free Nelson
Hamiltonian $H_{0}$ is 0, and $0\in\sigma_{\mathrm{ess}}(H_{0})$ lies in the
essential spectrum of $H_{0}$. If $(\psi_{k})_{k\in\mathbb{N}}\subset
L^{2}(\mathbb{R}^{3})$ is a singular Weyl sequence for $(-\Delta_{X},0)$, that
is, $(\psi_{k})_{k\in\mathbb{N}}$ is an orthonormal family and
$\lVert-\Delta_{X}\psi_{k}\rVert_{L^{2}}\to 0$ as $k\to\infty$, then
$(\psi_{k}\otimes\Omega)_{k\in\mathbb{N}}$ is a singular Weyl sequence for
$(H_{0},0)$, where $\Omega$ is the Fock space vacuum. We interpret
$\psi_{k}\otimes\Omega$ as an approximate ground state of the free Nelson
Hamiltonian for a sufficiently large $k$. Perturbation theory in leading order
suggests that the perturbation $\Phi(\omega^{-1/2}\rho_{\Lambda,X})$ of the
cut-off Hamiltonian $H_{\Lambda}=H_{0}+\Phi(\omega^{-1/2}\rho_{\Lambda,X})$
causes the following energy shift:

$\displaystyle\langle\psi_{k}\otimes\Omega,\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X})\psi_{k}\otimes\Omega\rangle_{\mathfrak{H}}-\langle\psi_{k}\otimes\Omega,\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X})H_{0}^{-1}\Phi(\omega^{-\frac{1}{2}}\rho_{\Lambda,X})\psi_{k}\otimes\Omega\rangle_{\mathfrak{H}}$
$\displaystyle\ \ \
=-\frac{1}{2}\int_{\mathbb{R}^{3}}\langle\Omega,a(\omega^{-\frac{1}{2}}\rho_{\Lambda,X})H_{0}^{-1}a^{*}(\omega^{-\frac{1}{2}}\rho_{\Lambda,X})\Omega\rangle_{\mathfrak{h}}|\psi_{k}(X)|^{2}\mathop{}\\!\mathrm{d}X$
$\displaystyle\ \ \
=-\frac{1}{2}\int_{\mathbb{R}^{3}}\langle\omega^{-\frac{1}{2}}\rho_{\Lambda,X},(K+\omega)^{-1}\omega^{-\frac{1}{2}}\rho_{\Lambda,X}\rangle_{\mathfrak{h}}|\psi_{k}(X)|^{2}\mathop{}\\!\mathrm{d}X.$
(3.37)

We call
$\langle\omega^{-\frac{1}{2}}\rho_{\Lambda,X},(K+\omega)^{-1}\omega^{-\frac{1}{2}}\rho_{\Lambda,X}\rangle_{\mathfrak{h}}/2$
the (boson) vacuum energy. Using only the leading order part $u_{\Lambda,X}$
in (3.11) of the form factor
$v_{\Lambda,X}=\omega^{-\frac{1}{2}}\rho_{\Lambda,X}/\sqrt{2}$ and the symbol
$t_{X}^{-1}$ in (3.21), the boson vacuum energy in leading order is equal to

$\displaystyle-E_{\Lambda}(X)$
$\displaystyle=-\frac{1}{2(2\pi)^{3}}\int_{\mathbb{R}^{3}}\frac{|\sigma(\omega^{-\frac{1}{2}})(X,\xi)|^{2}}{K_{0}(X,\xi)+1}\left|\widehat{\rho}\left(\frac{\xi}{\Lambda}\right)\right|^{2}\mathop{}\\!\mathrm{d}\xi$
$\displaystyle=-\frac{1}{2(2\pi)^{3}}\int_{\mathbb{R}^{3}}\frac{(h_{0}(X,\xi)+1)^{-\frac{1}{2}}}{K_{0}(X,\xi)+1}\left|\widehat{\rho}\left(\frac{\xi}{\Lambda}\right)\right|^{2}\mathop{}\\!\mathrm{d}\xi+F_{\Lambda}(X),$
(3.38)

where $(F_{\Lambda})_{\Lambda\geq 0}\subset
C_{\mathrm{b}}^{\infty}(\mathbb{R}^{3})$ is a family of functions converging
uniformly to some $F\in C_{\mathrm{b}}(\mathbb{R}^{3})$ as $\Lambda\to\infty$.
In the following, any family of functions (not necessarily the same on
different lines) that has this property is denoted by
$(F_{\Lambda})_{\Lambda\geq 0}$.

######  Theorem 3.3.1.

A bounded continuous potential $V_{\mathrm{ren}}\in
C_{\mathrm{b}}(\mathbb{R}^{3})$ exists such that
$V_{\Lambda}(X)+E_{\Lambda}(X)$ converges uniformly in $X$ to
$V_{\mathrm{ren}}(X)$ as $\Lambda\to\infty$.

###### Proof.

From the proof of Lemma 3.2.1 it follows that a symbol $t^{-1}_{X}\in S^{-2}$
exists such that

$\displaystyle(K+\omega)^{-1}\rho_{\Lambda,X}^{\sigma}$
$\displaystyle=\mathop{}\\!\mathrm{Op}(t_{X}^{-1})\rho_{\Lambda,X}^{\sigma},$
(3.39) $\displaystyle\partial_{X}(K+\omega)^{-1}\rho_{\Lambda,X}^{\sigma}$
$\displaystyle=\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\rho_{\Lambda,X}^{\sigma}-\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma}.$
(3.40)

We insert these identities in (3.36):

$\displaystyle V_{\Lambda}(X)$
$\displaystyle=\frac{1}{2}\lVert\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\rho_{\Lambda,X}^{\sigma}\rVert^{2}_{\mathfrak{h}}-\langle\rho_{\Lambda,X}^{\sigma},\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\omega^{-1}\rho_{\Lambda,X}\rangle_{\mathfrak{h}}$
$\displaystyle\ \ \
+\frac{1}{2}\langle\omega^{-1}\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\rho_{\Lambda,X}^{\sigma},g(X)\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\rho_{\Lambda,X}^{\sigma}\rangle_{\mathfrak{h}}$
$\displaystyle\ \ \
-\langle\omega^{-1}\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\rho_{\Lambda,X}^{\sigma},g(X)\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma}\rangle_{\mathfrak{h}}$
$\displaystyle\ \ \
+\frac{1}{2}\langle\omega^{-1}\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma},g(X)\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma}\rangle_{\mathfrak{h}}.$
(3.41)

Some of these terms converge uniformly to a bounded continuous function as
$\Lambda\to\infty$. For the first summand, this is a consequence of the
inverse triangle inequality and Lemma 3.2.1 with $\alpha=1/2$:

$\displaystyle|\lVert\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\rho_{\Lambda,X}^{\sigma}\rVert_{\mathfrak{h}}-\lVert\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\delta_{X}^{\sigma}\rVert_{\mathfrak{h}}|$
$\displaystyle\leq\lVert\mathop{}\\!\mathrm{Op}(t^{-1}_{X})(\rho_{\Lambda,X}^{\sigma}-\delta_{X}^{\sigma})\rVert_{\mathfrak{h}}$
$\displaystyle\leq
C\lVert\rho_{\Lambda}^{\sigma}-\delta^{\sigma}\rVert_{H^{s}}$ (3.42)

for an $s<-3/2$. Because
$\lVert\rho_{\Lambda}^{\sigma}-\delta^{\sigma}\rVert_{H^{s}}\to 0$ as
$\Lambda\to\infty$ for every $s<-3/2$,
$\lVert\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\rho_{\Lambda,X}^{\sigma}\rVert_{\mathfrak{h}}$
converges to
$\lVert\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\delta_{X}^{\sigma}\rVert_{\mathfrak{h}}$.
The expression
$\lVert\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\delta_{X}^{\sigma}\rVert_{\mathfrak{h}}$
depends continuously on $X$ and is bounded by Lemma 3.2.1; thus, it defines an
element of $C_{\mathrm{b}}(\mathbb{R}^{3})$.

In similar fashion, we demonstrate the convergence of the second line in
(3.41):

$\displaystyle|\langle\omega^{-1}\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\rho_{\Lambda,X}^{\sigma},g(X)\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\rho_{\Lambda,X}^{\sigma}\rangle_{\mathfrak{h}}-$
$\displaystyle\ \ \ \ \ \ \
\langle\omega^{-1}\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\delta_{X}^{\sigma},g(X)\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})\delta_{X}^{\sigma}\rangle_{\mathfrak{h}}|$
$\displaystyle\ \ \
=|\langle\omega^{-1}\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})(\rho_{\Lambda,X}^{\sigma}-\delta_{X}^{\sigma}),g(X)\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})(\rho_{\Lambda,X}^{\sigma}+\delta_{X}^{\sigma})\rangle_{\mathfrak{h}}|$
$\displaystyle\ \ \ \leq
C\lVert\omega^{-\frac{1}{2}}\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})(\rho_{\Lambda,X}^{\sigma}-\delta_{X}^{\sigma})\rVert_{\mathfrak{h}}\lVert\omega^{-\frac{1}{2}}\mathop{}\\!\mathrm{Op}(\partial_{X}t^{-1}_{X})(\rho_{\Lambda,X}^{\sigma}+\delta_{X}^{\sigma})\rVert_{\mathfrak{h}}$
$\displaystyle\ \ \ \leq
C\lVert\rho_{\Lambda}^{\sigma}-\delta^{\sigma}\rVert_{H^{s}}\sup_{\Lambda}\lVert\rho_{\Lambda}^{\sigma}+\delta^{\sigma}\rVert_{H^{s}}$
(3.43)

for an $s<-3/2$. Analogously, we confirm that the third line in (3.41)
converges. However, for the remaining two terms in (3.41),

$\displaystyle-\langle\rho_{\Lambda,X}^{\sigma},\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\omega^{-1}\rho_{\Lambda,X}\rangle_{\mathfrak{h}}$
$\displaystyle+\frac{1}{2}\langle\omega^{-1}\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma},g(X)\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma}\rangle_{\mathfrak{h}},$
(3.44)

the above line of reasoning fails because the order of the involved pseudo-
differential operators is not sufficiently negative. These are the divergent
terms in $V_{\Lambda}(X)$.

We use the pseudo-differential calculus to extract the vacuum energy
$-E_{\Lambda}(X)$ from (3.44). From the proof of Lemma 3.2.1, the symbol of
$t_{X}^{-1}$ is given by

$\displaystyle t^{-1}_{X}(x,\xi)=(K_{0}(X,\xi)+1)^{-1}+S^{-3}.$ (3.45)

The symbol $\sigma(\omega^{-1})$ of $\omega^{-1}$ in standard quantisation is

$\displaystyle\sigma(\omega^{-1})(x,\xi)$
$\displaystyle=(h_{0}(x,\xi)+\mu(x)^{2})^{-\frac{1}{2}}+S^{-2}$
$\displaystyle=(h_{0}(x,\xi)+1)^{-\frac{1}{2}}+S^{-2}.$ (3.46)

Relevant operators for the remaining terms in (3.44) are compositions of
$\mathop{}\\!\mathrm{Op}(t^{-1}_{X})$ and $\omega^{-1}$, which can be written
as

$\displaystyle\mathop{}\\!\mathrm{Op}(t^{-1}_{X})\omega^{-1}$
$\displaystyle=\mathop{}\\!\mathrm{Op}(c_{X})+\mathrm{Op}(\tilde{c}_{X})$
(3.47)
$\displaystyle\mathop{}\\!\mathrm{Op}(t^{-1}_{X})^{*}\omega^{-1}\mathop{}\\!\mathrm{Op}(t^{-1}_{X})$
$\displaystyle=\mathop{}\\!\mathrm{Op}(d_{X})+\mathrm{Op}(\tilde{d}_{X})$
(3.48)

with $\tilde{c}_{X}\in S^{-4}$, $\tilde{d}_{X}\in S^{-6}$ uniformly in $X$,
and

$\displaystyle c_{X}(x,\xi)$
$\displaystyle=(h_{0}(x,\xi)+1)^{-\frac{1}{2}}(K_{0}(X,\xi)+1)^{-1}\in
S^{-3},$ (3.49) $\displaystyle d_{X}(x,\xi)$
$\displaystyle=(h_{0}(x,\xi)+1)^{-\frac{1}{2}}(K_{0}(X,\xi)+1)^{-2}\in
S^{-5}.$ (3.50)

We can neglect the pseudo-differential operators of order $-4$ and $-6$ in
(3.47) and (3.48), respectively, because

$\displaystyle\langle\rho_{\Lambda,X}^{\sigma},\mathop{}\\!\mathrm{Op}(\tilde{c}_{X})\rho_{\Lambda,X}\rangle_{\mathfrak{h}}+\frac{1}{2}\langle\partial_{x}\rho_{\Lambda,X}^{\sigma},g(X)\mathop{}\\!\mathrm{Op}(\tilde{d}_{X})\partial_{x}\rho_{\Lambda,X}^{\sigma}\rangle_{\mathfrak{h}}$
(3.51)

converges uniformly to a bounded continuous function as $\Lambda\to\infty$ by
the same arguments as above. It remains to analyse the terms with