# The massive modular Hamiltonian

Roberto Longo , Gerardo Morsella
Dipartimento di Matematica, Università di Roma Tor Vergata,
Via della Ricerca Scientifica, 1, I-00133 Roma, Italy Supported by the ERC
Advanced Grant 669240 QUEST “Quantum Algebraic Structures and Models”, MIUR
FARE R16X5RB55W QUEST-NET and GNAMPA-INdAM. E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We compute the vacuum local modular Hamiltonian associated with a space ball
region in the free scalar massive Quantum Field Theory. We give an explicit
expression on the one particle Hilbert space in terms of the massive Legendre
differential operator and the Green integral for the Helmholtz operator. The
quadratic form of the massive modular Hamiltonian is expressed in terms of an
integral of the energy density with parabolic distribution and of a Yukawa
potential, that here appears intrinsically. We then get the formula for the
local entropy of a Klein-Gordon wave packet. This gives the vacuum relative
entropy of a coherent state on the double cone von Neumann algebras associated
with the free scalar QFT. Among other points, we provide the passivity
characterisation of the modular Hamiltonian within the standard subspace set
up.

##  1 Introduction

In this paper we provide the formula for the vacuum modular Hamiltonian
associated with the free massive scalar quantum field in a bounded spacetime
region (double cone), an open problem in Quantum Field Theory that is standing
up since the Tomita-Takesaki modular theory uncovered in the 70’s the
intrinsic modular evolution associated with a normal state of a von Neumann
algebra, see [35].

The Bisognano-Wichmann theorem [6] showed the geometric nature of the modular
flow for the wedge region case, in a general framework. The bounded region
problem has then been subject of work by several researchers over the years,
see e.g. [3, 13, 15, 33, 34, 37]. The underlying picture in our approach is to
consider the massive set up as a deformation of the massless one. Our analysis
is valid in arbitrary spacetime dimension $d+1$, with $d>1$.

Background. Let ${\mathcal{M}}$ be a von Neumann algebra and $\varphi$ a
faithful normal state on ${\mathcal{M}}$. As is well known, the Tomita-
Takesaki modular theory provides us with a canonical one-parameter group of
${\mathcal{M}}$ associated with $\varphi$, the modular group
$\sigma^{\varphi}$. Thus the quantum system ${\mathcal{M}}$ is equipped with
an intrinsic evolution $\sigma^{\varphi}$, that is characterised by the KMS
thermal equilibrium condition [35]. In the GNS representation, the modular
group is implemented by a unitary one-parameter group $\Delta_{\varphi}^{is}$,
whose generator $\log\Delta_{\varphi}$ is the modular Hamiltonian associated
with $\varphi$ (see [26] for a discussion on the matter).

In Quantum Field Theory, for each spacetime region $O$, we have the von
Neumann algebra ${\mathcal{A}}(O)$ of observables localised in $O$. By the
Reeh-Schlieder theorem, the vacuum vector is cyclic and separating for
${\mathcal{A}}(O)$ if both $O$ and its causal complement have non empty
interiors, see [18]. Thus the restriction of vacuum state $\varphi$ to
${\mathcal{A}}(O)$ is faithful and the modular Hamiltonian associated with
${\mathcal{A}}(O)$ and $\varphi$ gives us a local Hamiltonian
$\log\Delta_{O}$.

The problem of computing $\log\Delta_{O}$ is then natural. Among other
motivations, the modular Hamiltonian is related to Araki’s relative entropy
[2], that is recently playing an important role in Quantum Field Theory, also
in relation with quantum entropy/energy inequalities; an account of this wide
research topic goes beyond the purpose of this introduction (see [3, 10, 24,
29, 28, 40] and refs. therein).

As said, if $W$ is a wedge region, there is an important model independent
result [6]: $\Delta_{W}^{-is}$ is identified with the $2\pi$-rescaled boost
unitary transformations leaving $W$ globally invariant, a result rich of
consequences, see [18]. For a different spacetime region $O$, the
understanding of the local modular structure is definitely more problematic
because, in general, $\Delta_{O}^{is}$ has no geometric description due to the
lack of enough spacetime symmetries. A basic issue concerns the case where $O$
is a double cone, the causal envelop of a time-zero ball. In the free massless
QFT, the geometric description of the double cone vacuum modular group was
derived in [19]. The general conformal case was later analysed solely in terms
of the local von Neumann algebras [17, 7]. Among other results, we mention
here the forward light cone case in the free massless QFT [8] and the
approximate local estimates in [16].

The modular theory has a version for standard subspaces, see [21, 32, 22]. Let
${\mathcal{H}}$ be a complex Hilbert space and $H\subset{\mathcal{H}}$ a
standard subspace, i.e. $H$ is a real linear, closed subspace of
${\mathcal{H}}$ such that $H\cap iH^{\prime}=\\{0\\}$ and
${\overline{H+iH}}={\mathcal{H}}$, with $H^{\prime}$ the symplectic complement
of $H$. The modular operator $\Delta_{H}$ associated with $H$ is a canonical
positive, non-singular selfadjoint operator on ${\mathcal{H}}$ associated with
$H$ that satisfies

$\Delta^{is}_{H}H=H\,,\quad s\in\mathbb{R}\,.$

In a free scalar QFT, the vacuum modular (unitary) group associated with the
von Neumann algebra of a region $O$ is the second quantisation of a unitary
one-parameter group in the one-particle Hilbert space, indeed of the modular
group of the local standard subspace $H(O)$. We shall henceforth denote by
$\Delta_{O}$ the modular operator $\Delta_{H(O)}$ on the one particle Hilbert
space. Also, we set $H(B)=H(O)$, $\Delta_{B}=\Delta_{O}$ if $O$ is the double
cone with basis the unit space ball centred at origin $B$. Hence, in this
paper

$\text{Modular Hamiltonian of $B$}=\log\Delta_{B}\,.$

Massless Hamiltonian. In the massless case, $\Delta_{B}^{is}$ is associated
with a one parameter group of conformal transformations that globally preserve
$O$ [19]. Nonetheless, the description of $\log\Delta_{B}$ has not been worked
out so far. This will be our starting point.

In terms of the wave Cauchy data, we shall see that the local massless modular
Hamiltonian is given by

$\log\Delta_{B}=-2\pi i\left[\begin{matrix}0&\frac{1}{2}(1-r^{2})\\\
\frac{1}{2}(1-r^{2})\nabla^{2}-r\partial_{r}-D&0\end{matrix}\right]\,,$ (1)

with $B$ the unit space ball and $D=(d-1)/2$ the scaling dimension of the free
scalar field. Namely

$\log\Delta_{B}=-2\pi i\left[\begin{matrix}0&M\\\ L&0\end{matrix}\right]\,,$
(2)

with

$\displaystyle M=\ \text{Multiplication operator by}\
\,\frac{1}{2}(1-r^{2})\,,$ (3) $\displaystyle L=\ \text{Legendre operator}\
\,\frac{1}{2}(1-r^{2})\nabla^{2}-r\partial_{r}-D\,.$ (4)

As we shall see, the right hand side of (1) gives indeed an essentially
selfadjoint operator on the one particle Hilbert space on the smooth,
compactly supported function domain.

We shall derive the following formula for the massless modular Hamiltonian of
$B$ in terms of the classical stress-energy tensor $T$:

$-(\Phi,\log\Delta_{B}\Phi)=2\pi\int_{x_{0}=0}\frac{1-r^{2}}{2}\langle
T_{00}\rangle_{\Phi}(x)dx+\pi D\int_{x_{0}=0}\Phi^{2}dx\,,$ (5)

with $\Phi$ a real wave with smooth, compactly supported Cauchy data.

The right hand side of (5) is similar to a formula for the modular Hamiltonian
sketched by Casini, Huerta and Myers [9, (2.23)] in terms of the QFT stress-
energy tensor, when the Cauchy data are supported in $B$.

Massive Hamiltonian. The above formula (5) gives a first hint for the
structure of the massive modular Hamiltonian: one would be tempted to replace
the massless stress energy tensor $T$ with the massive stress energy tensor
$T^{(m)}$. This would amount to replace in (1) the Legendre operator with its
massive perturbation $\frac{1}{2}(1-r^{2})(\nabla^{2}-m^{2})-r\partial_{r}-D$.

This guess leads only to a part of the formula. Here we are at a delicate
point. The quadratic form of $K_{m}$ restricts to the quadratic form of
$\frac{i}{2\pi}\log\Delta_{B}$ on $C_{0}^{\infty}(B)$. As we shall see, in the
massive case, on $C^{\infty}_{0}(B)$, we have
$\frac{i}{2\pi}\log\Delta_{B}=K_{m}^{B}$

$K^{B}_{m}=\left[\begin{matrix}0&M\\\ L_{m}&0\end{matrix}\right]\,,$ (6)

with

$L_{m}=\frac{1}{2}(1-r^{2})(\nabla^{2}-m^{2})-r\partial_{r}-D-\frac{1}{2}m^{2}G^{B}_{m}\,,$
(7)

here $G^{B}_{m}$ is the operator associated with the quadratic form of the
restriction of the inverse Helmholtz operator ${\rm H}_{m}$

${\rm H}_{m}=-\nabla^{2}+m^{2}\,.$

$G^{B}_{m}$ is given by a Green function integral; in particular, in $d=3$
space dimensions,

$G^{B}_{m}f({\bf x})=\frac{1}{4\pi}\int_{B}\frac{e^{-m|{\bf x}-{\bf
y}|}}{|{\bf x}-{\bf y}|}f({\bf y})d{\bf y}\,.$

Then $\log\Delta_{B}$ is determined by complex linearity on a dense set of the
Hilbert space which is a domain of essential selfadjointness. One important
point is that $K^{B}_{m}$ is a compact perturbation of $K^{B}_{0}$.

Due to the presence of the term proportional to $G^{B}_{m}$, $i\log\Delta_{B}$
does not act locally on $B$. Yet it acts locally on a large subspace of waves,
those with Cauchy data $f,g$ supported in $B$ with $f$ is the propagator
$f=(\nabla^{2}-m^{2})h$ of a compactly supported function $h$.

One remarkable point here is the intrinsic appearance of the Yukawa potential
$e^{-mr}/4\pi r$, a fact that deserves further physical insight [41].

The quadratic form $-(\Phi,\log\Delta_{B}\Phi)$ on $H(B)$ is the entropy
$S_{\Phi}$ and is expressed in terms of the energy density of $\Phi$ as we
shall now explain.

The measure of information. One main consequence of our analysis is a formula
for the entropy density carried by a wave packet. Let’s explain the framework.

Suppose $\Phi$ is a scalar wave packet, a solution of the wave or Klein-Gordon
equation $(\square+m^{2})\Phi=0$. At given time, we can measure the signal
contained say in space ball $B$. The quantity $S_{\Phi}$ that represents the
mean information stored by $\Phi$ in $B$ at that time is called the entropy of
$S_{\Phi}$ with respect of $B$ and has been introduced in [24, 25, 11],
although different space regions (wedges) were there considered.

With ${\mathcal{H}}$ a complex Hilbert space and $H\subset{\mathcal{H}}$ a
factorial standard subspace, one defines the entropy of a vector
$k\in{\mathcal{H}}$ with respect to $H$ as

$S_{k}=\Im(k,P_{H}i\log\Delta_{H}\,k)\ .$ (8)

Here, $P_{H}:H+H^{\prime}\to H$, $P_{H}:h+h^{\prime}\mapsto h$, is the cutting
projection associated with $H$, $\Delta_{H}$ is the modular operator
associated with $H$ and $H^{\prime}$ is the symplectic complement of $H$.

Motivated by Quantum Field Theory, one equips the waves’ real linear space
with a complex Hilbert space structure, where the imaginary part of the scalar
product is given by the time independent symplectic form

$\Im(\Phi,\Psi)=\frac{1}{2}\int_{x_{0}=t}\big{(}\Psi\partial_{0}\Phi-\Phi\partial_{0}\Psi\big{)}d{\bf
x}\,.$

Then one considers the local net of standard subspaces associated of the
resulting ${\mathcal{H}}$: $H(B)$ is the closure of the real linear space of
waves with Cauchy data supported in $B$. So one defines $S_{\Phi}$ as the
entropy of the vector $\Phi$ with respect to $H(B)$. We have

$S_{\Phi}=\pi\int_{B}\big{(}\Psi\Phi^{\prime}-\Phi\Psi^{\prime}\big{)}d{\bf
x}\ ,$ (9)

(time zero integral) with $\Psi=i\log\Delta_{B}\Phi$ and the prime denotes the
time derivative and $\Delta_{H(B)}=\Delta_{B}$.

One then have to compute (9). In [25, 11], this computation has been worked
out in the case of a Klein-Gordon wave for a half-space region (whose causal
envelop is a wedge), a case motivated by the a study of the Quantum Null
Energy Condition inequality.

From the point of view of information theory, it is however natural to
consider the case $B$ is a bounded region. Note that the computation in [25,
11] relies on the explicit knowledge of the modular Hamiltonian
$\log\Delta_{W}$ [6].

Now, $P_{H(B)}$ acts by cutting the Cauchy data [11], so we have at our
disposal all the ingredients to compute the local entropy $S_{\Phi}(R)$ of the
wave packet $\Phi$ in the (causal envelop of) the radius $R$ space ball
$B_{R}(\bar{\bf x})$ around the space point $\bar{\bf x}$, at time $t$. We
shall see that $S_{\Phi}(R)$ is the sum of three terms:

$\displaystyle S_{\Phi}(R)$ $\displaystyle=\pi\int_{B_{R}(\bar{\bf
x})}\frac{R^{2}-r^{2}}{R}\langle T^{(m)}_{00}(t,{\bf x})\rangle_{\Phi}d{\bf
x}\quad$ stress-energy tensor term
$\displaystyle+\pi\frac{d-1}{2R}\int_{B_{R}(\bar{\bf x})}\Phi^{2}(t,{\bf
x})d{\bf x}\quad$ mass independent normalisation
$\displaystyle+\pi\frac{m^{2}}{R}\int_{B_{R}(\bar{\bf x})\times B_{R}(\bar{\bf
x})}{\rm G}_{m}({\bf x}-{\bf y})\Phi(t,{\bf x})\Phi(t,{\bf y})d{\bf x}d{\bf
y}\quad$ Yukawa potential term

with $r=|{\bf x}-\bar{\bf x}|$ and $T_{00}^{(m)}$ the energy density of
$\Phi$:

$T^{(m)}_{00}=\frac{1}{2}\big{(}(\partial_{0}\Phi)^{2}+|\nabla_{{\bf
x}}\Phi|^{2}+m^{2}\Phi^{2}\big{)}\ .$

The first term is the main one. The last term is a sort of boundary effect, it
does not survive when the region $B_{R}(\bar{\bf x})$ approaches a half-space.
We however expect further physical insight to be spelled out in this regard.

We note here the appearance of the parabolic distribution
$\frac{1}{2}(1-r^{2})$ in stress-energy term of the formula for the modular
Hamiltonian. We wonder about possible deep roots for this, somehow similarly
with the appearance of Wigner semicircular distribution [39] in the Free
Probability framework [38]. The parabolic distribution in three dimensional
space is a higher-dimensional generalisation of the Wigner semicircular
distribution and is related to the marginal distribution of a spherical
distribution.

Content of this paper. Our paper is organised as follows. First, we collect
functional analytic results. Among these, we have a standard subspace version
of the Pusz and Woronowicz [31] complete passivity characterisation, up to a
proportionality constant, of the modular Hamiltonian. This reflects the second
principle of thermodynamics. By our results, our local Hamiltonian is
completely passive in our standard subspace sense.

Then we describe the massless modular group in the wave set up, so we get an
explicit description of the massless Hamiltonian. Next, we procede to the
massive case by a series of deformation arguments. We then provide our local
entropy formula for a Klein-Gordon wave packet and discuss some of the
implications in Quantum Field Theory.

Our general references for Operator Algebras and Quantum Field Theory are [14,
18, 35].

##  2 Abstract set up

We begin with abstract, functional analytical results.

###  2.1 Stability of skew-selfadjointness

Let $H$ be a real Hilbert space and $T:D(T)\subset H\to H$ a (real) linear
operator with $D(T)$ dense in $H$.

We shall say that $T$ is skew-Hermitian if $T\subset-T^{*}$ and skew-
selfadjoint if $T=-T^{*}$.

Let $H_{\mathbb{C}}$ be the usual complexification of $H$, so
$H_{\mathbb{C}}=H\oplus iH$ with the usual complex structure. Then $T$ extends
to a complex linear operator $T_{\mathbb{C}}$ on $H_{\mathbb{C}}$ and one can
apply most results for complex linear operators.

In particular, $T$ is the infinitesimal generator of a one-parameter group of
orthogonal operators on $H$ iff $T$ is skew-Hermitian and ran$(K\pm 1)=H$.

Recall that $T$ is a Fredholm operator if $T$ is closed and
${\hbox{dim}}\ker(T),\operatorname{codim}\operatorname{ran}(T)<\infty$; then
the index is defined by

$\operatorname{ind}(T)={\hbox{dim}}\ker(T)-\operatorname{codim}\operatorname{ran}(T)\,.$

We shall use the following stability result [20, Chapter IV]: if $P:H\to H$ is
a linear operator:

$T\ \text{Fredholm},\ P\ \text{compact}\implies T+P\ \text{Fredholm}\ \&\
\operatorname{ind}(T+P)=\operatorname{ind}(T)\,.$

Let $H^{\prime}$ be a Hilbert space equivalent to $H$, namely $H$ and
$H^{\prime}$ are the same linear space with equivalent norms. Clearly, we may
regard $T$ also as an operator on $H^{\prime}$.

######  Proposition 2.1.

Let $H,H^{\prime}$ be as above, $K$ a skew-selfadjoint operator on $H$ and
$P:H\to H$ a compact linear operator.

If $K+P$ is skew-Hermitean on $H^{\prime}$, then $K+P$ is skew-selfadjoint on
$H^{\prime}$.

  * Proof.

By considering the complexification $K_{\mathbb{C}}$, we may assume that our
operators act on a complex Hilbert space and are complex linear.

As $K$ is skew-selfadjoint on $H$, $K\pm 1$ is invertible. As $P$ is compact,

$\operatorname{ind}(K+P\pm 1)=\operatorname{ind}(K\pm 1)=0\,.$

Clearly the index as operator on $H$ and on $H^{\prime}$ is the same. Now,
$K+P$ is Hermitian as operator on $H^{\prime}$, so ker$(K+P\pm 1)=\\{0\\}$.
Therefore $\operatorname{ran}(K+P\pm 1)=H$. We conclude that $K+P\pm 1$ is
invertible, thus $K+P$ is skew-selfadjoint on $H^{\prime}$.  $\square$

###  2.2 Real linear invariant subspaces

We now characterise the closed real linear subspaces of a complex Hilbert
space ${\mathcal{H}}$ that are invariant for a one parameter unitary group.

######  Proposition 2.2.

Let ${\mathcal{H}}$ be a Hilbert space, $H\subset{\mathcal{H}}$ a closed, real
linear subspace and $A:D(A)\subset{\mathcal{H}}\to{\mathcal{H}}$ a selfadjoint
operator. With $V(s)=e^{isA}$, $s\in{\mathbb{R}}$, $K=iA$ and ${\mathcal{C}}$
(resp. ${\mathcal{B}}$) the real algebra of complex, continuous functions $g$
on $\mathbb{R}$ vanishing at $\infty$ (resp. complex, bounded Borel functions
on ${\mathbb{R}}$) such that $g(-t)=\bar{g}(t)$, the following are equivalent:

  * $(i)$

$V(s)H=H\,,\quad s\in\mathbb{R}$,

  * $(ii)$

$g(A)H\subset H$, $g\in{\mathcal{C}}$,

  * $(iii)$

$g(A)H\subset H$, $g\in{\mathcal{B}}$,

  * $(iv)$

$(K^{2}-1)^{-1}H\subset H$ and $K(K^{2}-1)^{-1}H\subset H$,

  * $(v)$

$(K\pm 1)^{-1}H\subset H$,

  * $(vi)$

$K|_{H}$ is skew-selfadjoint on $H$, namely $D(K)\cap H$ is dense in $H$,
$K(D(K)\cap H)\subset H$ and $K:(D(K)\cap H)\subset H\to H$ is skew-
selfadjoint.

  * Proof.

$(i)\Rightarrow(v)$: Let $h$ be a real $L^{1}$-function on ${\mathbb{R}}$ and
$g=\hat{h}$ its Fourier transform. Then

$g(A)=\int_{\mathbb{R}}h(s)V(s)ds\,.$

If $(i)$ holds true, then $g(A)H\subset H$ for such $g$. In particular $(v)$
holds, similarly as in (10).

$(v)\Rightarrow(iv)$ follows by the identities

$2K(K^{2}-1)^{-1}=(K+1)^{-1}+(K-1)^{-1}\,,\quad
2(K^{2}-1)^{-1}=(K+1)^{-1}-(K-1)^{-1}\,.$

$(iv)\Rightarrow(ii)$: The complexification ${\mathcal{C}}_{\mathbb{C}}$ of
${\mathcal{C}}$ is the $C^{*}$-algebra $C$ of continuous functions on
$\mathbb{R}$ vanishing at infinity, namely every $F\in C$ is written uniquely
as

$F=f+ig\,,\quad f,g\in{\mathcal{C}}\ ,\qquad
f(t)=\frac{1}{2}[F(t)+\bar{F}(-t)],\ g(t)=\frac{1}{2i}[F(t)-\bar{F}(-t)].$

Note that ${\mathcal{C}}$ contains the real algebra ${\mathcal{C}}_{0}$ of
functions of the form $f(t)=p((t^{2}+1)^{-1},it(t^{2}+1)^{-1})$ with $p$ a
two-variables polynomial with real coefficients and zero constant coefficient.
Note that ${\mathcal{C}}$ and ${\mathcal{C}}_{0}$ are closed with respect to
complex conjugation.

Since ${\mathcal{C}}_{0}$ separates the points of ${\mathbb{R}}$, given any
$f\in{\mathcal{C}}$, there exists a sequence $F_{n}=f_{n}+ig_{n}$ with
$f_{n},g_{n}\in{\mathcal{C}}_{0}$ such that $F_{n}\to f$ uniformly on
${\mathbb{R}}$, thus $f_{n}\to f$ uniformly on ${\mathbb{R}}$. We conclude
that ${\mathcal{C}}_{0}$ is norm dense in ${\mathcal{C}}$. Therefore we have
$(ii)$.

$(ii)\Rightarrow(iii)$: Let $g\in{\mathcal{B}}$. Given $h\in H,\,k\in
H^{\perp}$ (real orthogonal), by Lusin’s theorem there exists a bounded
sequence of continuous functions $g_{n}$ such that $g_{n}\to g$ almost
everywhere w.r.t. the spectral measure of $A$ associated with $h,k$. By
replacing $g_{n}(t)$ with $g_{n}(t)+\bar{g}_{n}(-t)$, we may assume that
$g_{n}\in{\mathcal{C}}$. By Lebesgue’s dominated convergence theorem, we have
$(k,g_{n}(A)h)\to(k,g(A)h)$. As $\Re(k,g_{n}(A)h)=0$, we have
$\Re(k,g(A)h)=0$, that implies $g(A)H\subset H$ because $h,k$ can be
arbitrarily chosen.

$(vi)\Rightarrow(v)$: $K|_{H}$ generates a one-parameter group $V_{H}$ of
orthogonal operators on $H$, therefore

$(1\pm K)^{-1}H=-\int_{0}^{\infty}e^{-s}V_{H}(\mp s)ds\,H\subset H\,.$ (10)

Clearly $(i)\Rightarrow(vi)$, $(iii)\Rightarrow(ii)$ and
$(iii)\Rightarrow(i)$.  $\square$

If $A$ and $H$ satisfy the condition is the above proposition, we shall say
that $H$ is $iA$-invariant.

Let ${\mathcal{H}}$ be a complex Hilbert space. A standard subspace $H$ of
${\mathcal{H}}$ is a closed, real linear subspace of ${\mathcal{H}}$ with

$\overline{H+iH}={\mathcal{H}}\,,\quad H\cap iH=\\{0\\}\,.$

With $H$ a standard subspace of ${\mathcal{H}}$, let $K_{H}:D(K_{H})\subset
H\to H$ be a real linear, densely defined operator. Then we can extend $K_{H}$
to a complex linear operator $K:D(K)\subset{\mathcal{H}}\to{\mathcal{H}}$,
with dense domain $D(K)=D(K_{H})+iD(K_{H})$, defined by

$K(h+ik)=K_{H}h+iK_{H}k\,,\quad h,k\in D(K_{H})\,.$

We shall say that $K_{H}$ is $\partial$-symplectic if the real quadratic form
$\beta(h,K_{H}k)$, $h,k\in D(K_{H})$, is symmetric, with
$\beta(h,k)=\Im(h,k)$. Clearly, the infinitesimal generator of a one-parameter
continuous group of symplectic maps is $\partial$-symplectic.

######  Corollary 2.3.

With $K_{H}$ and $K$ as above, $K$ is skew-hermitian on ${\mathcal{H}}$ iff
$K_{H}$ is skew-Hermitean on $H$ w.r.t. the real part of the scalar product
and $\partial$-symplectic on $D(K_{H})$.

In this case, if $K_{H}$ is essentially skew-selfadjoint on $H$, $K$ is
essentially skew-selfadjoint on ${\mathcal{H}}$ and the skew-selfadjoint
closure $\bar{K}$ of $K$ leaves $H$ invariant as in Prop. 2.2.

  * Proof.

Note that $(h,K_{H}k)=-(K_{H}h,k)$ for all $h,k\in D(K_{H})$ iff $K_{H}$ is
skew-Hermitean and $\partial$-symplectic on $D(K_{H})$. In this case, we now
check that $K$ is skew-hermitian on $D(K)$. Indeed,

$(h+ik,K(h+ik))=(h+ik,K_{H}h+iK_{H}k)=(h,K_{H}h)+(k,K_{H}k)-i(k,K_{H}h)+i(h,K_{H}k)\\\
=i(k,K_{H}h)-i(h,K_{H}k)=-i(K_{H}k,h)+i(K_{H}h,k)=-(K(h+ik),h+ik)\,.$

Concerning the last part of the statement, assume that $K_{H}$ is essentially
skew-selfadjoint on $H$ and $\partial$-symplectic on $D(K_{H})$. Then, the
range of $K_{H}\pm 1$ is dense in $H$, which implies that the range of $K\pm
1$ is dense in ${\mathcal{H}}$, so $K$ is essentially skew-selfadjoint and
$\bar{K}$ is skew-selfadjoint. Since $K$ extends $K_{H}$, we have that $(K\pm
1)^{-1}$ extends $(K_{H}\pm 1)^{-1}$, therefore

$(K\pm 1)^{-1}{\rm ran}(K_{H})=(K_{H}\pm 1)^{-1}{\rm
ran}(K_{H})=D(K_{H})\subset H\,.$

As ran$(K_{H})$ is dense in $H$, we then have $(\bar{K}\pm 1)^{-1}H\subset H$,
thus the corollary follows by Prop. 2.2.  $\square$

Note now that, if $T_{1},T_{2}$ are bounded linear operator on a Hilbert space
$H$ with $\ker(T_{1})=\ker(T_{1})=\\{0\\}$ and $R(T_{1})\cap R(T_{2})$ dense,
then

$T_{1}-T_{2}=T_{1}(T^{-1}_{2}-T^{-1}_{1})T_{2}$

on a dense domain, so

$T^{-1}_{2}-T^{-1}_{1}\ {\rm compact}\implies T_{1}-T_{2}\ {\rm compact}\,.$
(11)

Suppose then that $H$ is a real Hilbert space with two embeddings
$H\subset{\mathcal{H}}_{1}$, $H\subset{\mathcal{H}}_{2}$ in the complex
Hilbert spaces ${\mathcal{H}}_{1},{\mathcal{H}}_{2}$, isometric with respect
to the real part of the scalar product, and
$A_{k}:D(A_{k})\subset{\mathcal{H}}_{k}\to{\mathcal{H}}_{k}$ a selfadjoint
operator such that $H$ left invariant by $K_{k}=-\imath_{k}A_{k}$ as in Prop.
2.2, with $\imath_{k}$ the complex structure of ${\mathcal{H}}_{k}$. We have:

######  Proposition 2.4.

Let $A_{1},A_{2}$ be as above, with $D(A_{1})\cap D(A_{2})$ dense. Then

$\imath_{1}A_{1}|_{H}-\imath_{2}A_{2}|_{H}\quad{\rm compact}\implies
g(A_{1})|_{H}-g(A_{2})|_{H}\quad{\rm compact}$

for every continuous complex function $g\in{\cal C}$ (see Prop. 2.2).

  * Proof.

By the identity (11), we have

$\imath_{1}A_{1}|_{H}-\imath_{2}A_{2}|_{H}\quad{\rm
compact}\implies\imath_{1}(A_{1}\pm\imath_{1})^{-1}|_{H}-\imath_{2}(A_{2}\pm\imath_{2})^{-1}|_{H}\quad{\rm
compact}\,;$

therefore, if $\imath_{1}A_{1}|_{H}-\imath_{2}A_{2}|_{H}$ is compact, then

$p\big{(}\imath_{1}(A_{1}+\imath_{1})^{-1},\imath_{1}(A_{1}-\imath_{1})^{-1}\big{)}\big{|}_{H}-p\big{(}\imath_{2}(A_{2}+\imath_{2})^{-1},\imath_{2}(A_{2}-\imath_{2})^{-1}\big{)}\big{|}_{H}$

is compact, for every two-variable polynomial $p$ such that
$p(-x,-y)=\overline{p(y,x)}$.

The proposition then follows by uniformly approximating $g(\lambda)$ by
$p\big{(}i(\lambda+i)^{-1},i(\lambda-i)^{-1}\big{)}$ on ${\mathbb{R}}$, with
$p$ a polynomial as above, similarly as in the proof of Prop. 2.2.  $\square$

###  2.3 Orthogonal and cutting projections

Let $H\subset{\mathcal{H}}$ be a standard subspace of the complex Hilbert
space ${\mathcal{H}}$ and $\Delta_{H}$, $J_{H}$ be the modular operator and
conjugation of $H$, namely

$S_{H}=J_{H}\Delta_{H}$

is the polar decomposition of the antilinear involution $S_{H}:h+ik\mapsto
h-ik$, $h,k\in H$, (Tomita’s operator). Then

$\Delta_{H}^{is}H=H\,,\quad J_{H}H=H^{\prime}\,,$

$s\in{\mathbb{R}}$, where $H^{\prime}=iH^{\perp}$ the symplectic complement of
$H$, with $\perp$ denoting the orthogonal w.r.t. the real part of the scalar
product. $\Delta_{H}^{is}$ and $\log\Delta_{H}$ are called the modular unitary
group and the modular Hamiltonian of $H$.

In this section, and in the following one, we assume $H$ to be factorial, i.e.
$H\cap H^{\prime}=\\{0\\}$, that is $H+H^{\prime}$ is dense in
${\mathcal{H}}$; equivalently $1$ is not an eigenvalue of $\Delta_{H}$.

Associated with $H$, there are three other standard subspaces: the real
orthogonal $H^{\perp}$, the symplectic complement $H^{\prime}$, and $iH$. We
shall consider the three associated real linear projections

$\displaystyle E_{H}:H+H^{\perp}\to H\ ,\quad h+h^{\perp}\mapsto h\ ,$
$\displaystyle P_{H}:H+H^{\prime}\to H\ ,\quad h+h^{\prime}\mapsto h\ ,$
$\displaystyle Q_{H}:H+iH\to H,\quad h+ik\mapsto h\ .$

Note that $E_{H},P_{H},Q_{H}$ are closed, densely defined real linear
operators, $E_{H}$ is bounded and $P_{H}$ is the cutting projection [11]. We
have

$\displaystyle
E_{H}=(1+\Delta_{H})^{-1}+J_{H}\Delta_{H}^{1/2}(1+\Delta_{H})^{-1}\ ,$ (12)
$\displaystyle
P_{H}=(1-\Delta_{H})^{-1}+J_{H}\Delta_{H}^{1/2}(1-\Delta_{H})^{-1}\ ,$ (13)
$\displaystyle Q_{H}=\frac{1}{2}(1+S_{H})\ ;$ (14)

more precisely, (13) means that $P_{H}$ is the closure of the operator
$a(\Delta_{H})+b(\Delta_{H})J_{H}$ on $D(a(\Delta_{H}))\cap D(b(\Delta_{H}))$,
with

$a(\lambda)=(1-\lambda)^{-1}\ ,\quad b(\lambda)=\lambda^{1/2}(1-\lambda)^{-1}\
,\ \lambda\in(0,+\infty)\smallsetminus\\{1\\}\,.$

Formulas (12) and (13) were obtained respectively in [15] (see also [30]) and
in [11]; formula (14) is straightforward. We then have

$E_{H}=2Q_{H}(1+\Delta_{H})^{-1}\ ,\quad P_{H}=2Q_{H}(1-\Delta_{H})^{-1}\ ,$

hence

$P_{H}=E_{H}(1+\Delta_{H})(1-\Delta_{H})^{-1}=E_{H}\coth\Big{(}\frac{1}{2}\log\Delta_{H}\Big{)}$
(15)

(denoting by the same symbol an operator and its closure).

######  Proposition 2.5.

$P_{H}i$ restricts to a skew-selfadjoint operator on $H$ and

$P_{H}i|_{H}=i\coth\Big{(}\frac{1}{2}\log\Delta_{H}\Big{)}\Big{|}_{H}\,.$ (16)

  * Proof.

$H$ is $i\tanh(\frac{1}{2}\log\Delta_{H})$-invariant by Prop. 2.2, because the
hyperbolic tangent is a bounded, odd function; so
$i\tanh(\frac{1}{2}\log\Delta_{H})|_{H}$ is bounded, skew-selfadjoint on $H$.
Therefore, its inverse $-i\coth(\frac{1}{2}\log\Delta_{H})|_{H}$ is a skew-
selfadjoint operator on $H$. Namely, $H$ is
$i\coth(\frac{1}{2}\log\Delta_{H})$-invariant. Equation (15) then gives (16)
by restriction.  $\square$

We end this subsection by noting the following relations

$\displaystyle P_{H}^{*}=-iP_{H}i\,;$ (17) $\displaystyle
P_{H}i\log\Delta_{H}=i\log\Delta_{H}\,P_{H}\,.$ (18)

The first relation is proved in [11], the second one is valid on
$D(P_{H}i\log\Delta_{H})\cap D(i\log\Delta_{H}\,P_{H})$ because $H$ is
$i\log\Delta_{H}$-invariant.

###  2.4 Entropy and quadratic forms

Recall the entropy of a vector $k$ in a Hilbert space ${\mathcal{H}}$ with
respect to a factorial standard subspace $H$ is defined by

$S_{k}=\Im(k,P_{H}i\log\Delta_{H}\,k)=\Re(k,P^{*}_{H}\log\Delta_{H}\,k)\,,$
(19)

where $P_{H}$ is the cutting projection $P_{H}:H+H^{\prime}\to H$. Formula
(19) is to be understood in the sense of quadratic forms, namely
$S_{k}=S(k,k)$ as we now define.

Let ${\mathcal{D}}=D(\sqrt{|\log\Delta_{H}|}\,F)$, with $F$ the spectral
projection of $\Delta_{H}$ relative to the interval $(0,1)$. Then
${\mathcal{D}}$ is a dense linear subspace of ${\mathcal{H}}$. With
$h,k\in{\mathcal{D}}$, we set

$S(h,k)=\Re(h,P^{*}_{H}\log\Delta_{H}\,k)=\Im(h,P_{H}i\log\Delta_{H}k)\,.$
(20)

More precisely, following the discussion il [11], let $E(\lambda)$ be the
spectral family of $\Delta_{H}$, the right hand side of (20) is defined by on
${\mathcal{D}}$ by

$\Im(h,P_{H}i\log\Delta_{H}k)=\int_{0}^{+\infty}a(\lambda)\log\lambda\,d(h,E(\lambda)k)-\int_{0}^{+\infty}b(\lambda)\log\lambda\,d(h,J_{H}E(\lambda)k)\
.$ (21)

Note that $b(\lambda)\log\lambda$ is a bounded function, so the right integral
is always finite. Moreover, $a(\lambda)\log\lambda$ is bounded on
$(1,+\infty)$ and positive on $(0,1)$. So the above formula is well defined by
the spectral theorem, provided $h,k\in{\mathcal{D}}$. In particular,

$S_{k}<\infty\ \Longleftrightarrow\ k\in{\mathcal{D}}\,,$

[11, Prop. 2.4].

Set ${\mathcal{D}}_{0}=D(a(\Delta_{H})\log\Delta_{H})\cap
D(b(\Delta_{H})\log\Delta_{H})\subset
D(\log\Delta_{H}\,P_{H})\cap{\mathcal{D}}$.

######  Lemma 2.6.

$P^{*}_{H}\log\Delta_{H}$ is essentially selfadjoint on ${\mathcal{D}}_{0}$
and
$P^{*}_{H}\log\Delta_{H}|_{{\mathcal{D}}_{0}}=\log\Delta_{H}\,P_{H}|_{{\mathcal{D}}_{0}}$.

  * Proof.

By equations (17), we have

$(P^{*}_{H}\log\Delta_{H})^{*}h=\log\Delta_{H}\,P_{H}h=-ii\log\Delta_{H}\,P_{H}h=-iP_{H}i\log\Delta_{H}h=P^{*}_{H}\log\Delta_{H}h$

for all $h\in{\mathcal{D}}_{0}$, namely
$(P^{*}_{H}\log\Delta_{H})^{*}|_{{\mathcal{D}}_{0}}=P^{*}_{H}\log\Delta_{H}|_{{\mathcal{D}}_{0}}=\log\Delta_{H}\,P_{H}|_{{\mathcal{D}}_{0}}$,
so it suffices to show that the closure $\overline{\log\Delta_{H}\,P_{H}}$ of
$\log\Delta_{H}\,P_{H}|_{{\mathcal{D}}_{0}}$ is selfadjoint.

With $0<\varepsilon<1$, let $E_{\varepsilon}$ be the spectral projection of
$\log\Delta_{H}$ relative to the set
$(-\varepsilon^{-1},-\varepsilon)\cup(\varepsilon,\varepsilon^{-1})$. Then
$E_{\varepsilon}(H+H^{\prime})\subset H+H^{\prime}$ by using Prop. 2.2;
moreover $E_{\varepsilon}(H+H^{\prime})\subset{\mathcal{D}}_{0}$. So
$E_{\varepsilon}{\mathcal{H}}\subset D(\overline{\log\Delta_{H}\,P_{H}})$
because $\log\Delta_{H}\,P_{H}\,E_{\varepsilon}$ is bounded by formula (13)
and $E_{\varepsilon}$ commutes with $\log\Delta_{H}\,P_{H}$, thus also with
$(\log\Delta_{H}\,P_{H})^{*}$.

We conclude that
$(\log\Delta_{H}\,P_{H})^{*}E_{\varepsilon}=\overline{\log\Delta_{H}\,P_{H}}E_{\varepsilon}$
and the lemma follows since $E_{\varepsilon}\nearrow 1$ as
$\varepsilon\searrow 0$.  $\square$

######  Proposition 2.7.

$S$ is a real linear, closed, symmetric, positive quadratic form on
${\mathcal{D}}$.

  * Proof.

$S$ is real linear and positive. By Lemma 2.6, $S$ is also symmetric on
${\mathcal{D}}$. Thus $S$ closable on ${\mathcal{D}}_{0}$, being associated
with the closable, real linear operator $P^{*}_{H}\log\Delta_{H}$ by eq. (20),
cf. the proof of [20, Thm. 1.27]. Indeed, $S$ is closed on ${\mathcal{D}}$
because ${\mathcal{D}}_{0}$ is a core for $\overline{\log\Delta_{H}\,P_{H}}$,
so a form core for $S$.  $\square$

###  2.5 Standard subspaces and passivity

We provide here a standard subspace version of Pusz-Woronowicz’s complete
passivity characterisation of the modular Hamiltonian in the $C^{*}$-algebraic
setting [31].

Let ${\mathcal{H}}$ be a complex Hilbert space and $H$ standard subspace. In
this section, $A$ is a selfadjoint linear operator on ${\mathcal{H}}$ such
that $e^{isA}H=H$, $s\in\mathbb{R}$, namely $H$ is $iA$-invariant as in Prop.
2.2.

We shall say that $A$ is active/passive with respect to $H$ if

$\pm(\xi,A\xi)\geq 0\,,\quad\xi\in D(A)\cap H\,.$

$A$ is $n$-active/passive w.r.t. $H$ if the generator of $e^{itA}\otimes
e^{itA}\cdots\otimes e^{itA}$ is active/passive with respect to the $n$-fold
tensor product $H\otimes H\otimes\cdots\otimes H$, (closed real linear span of
monomials $h_{1}\otimes h_{2}\otimes\cdots\otimes h_{n}$, $h_{i}\in H$, cf.
[27]). $A$ is completely active/passive if $A$ is $n$-active/passive for all
$n\in\mathbb{N}$.

Note that $A$ is active/passive iff

$\pm(\xi,A\xi)\geq 0\,,\quad\xi\in{\cal D}\,,$

with ${\cal D}\subset D(A)\cap H$ a real linear space such that the closure of
${\cal D}$ in the graph topology of $A$ is equal to $D(A)\cap H$.

Since $e^{itA}$ leaves $H$ globally invariant, $e^{itA}$ commutes with
$\Delta_{H}^{is}$ and $J_{H}$. We have

$J_{H}\Delta_{H}J_{H}=\Delta_{H}^{-1}\,\quad J_{H}AJ_{H}=-A\,.$ (22)

######  Proposition 2.8.

$\log\Delta_{H}$ is completely passive w.r.t. $H$.

  * Proof.

$\log\Delta_{H}$ is passive w.r.t. $H$, see [25]. Hence it is completely
passive because the modular unitary group of $H\otimes H\cdots\otimes H$ is
$\Delta_{H}^{is}\otimes\Delta_{H}^{is}\cdots\otimes\Delta_{H}^{is}$, see [27].
$\square$

Let ${\mathcal{D}}_{\rm an}\subset{\mathcal{H}}$ be the subspace of vectors
with bounded spectrum with respect to both $A$ and $\log\Delta$.

######  Lemma 2.9.

If $A$ is passive w.r.t. $H$, then $A\log\Delta_{H}$ is a positive selfadjoint
operator on ${\mathcal{H}}$.

  * Proof.

Denote by $S_{H}$ the Tomita operator of ${\mathcal{H}}$. As $S_{H}$ commutes
with $e^{itA}$, $\Delta_{H}^{is}$ and $J_{H}$, we have
$S_{H}{\mathcal{D}}_{\rm an}={\mathcal{D}}_{\rm an}$. Let
$\xi\in{\mathcal{D}}_{\rm an}$, so $(1+S_{H})\xi\in H$. Then, by passivity,

$0\geq((1+S_{H})\xi,A(1+S_{H})\xi)\ ,$ (23)

thus

$0\geq((1+S_{H})i\xi,A(1+S_{H})i\xi)=((1-S_{H})\xi,A(1-S_{H})\xi)\ .$ (24)

Summing up (23) and (24) we get

$0\geq(\xi,A\xi)+(S_{H}\xi,AS_{H}\xi)=(\xi,A\xi)+(J_{H}\Delta_{H}^{1/2}\xi,AJ_{H}\Delta_{H}^{1/2}\xi)=(\xi,A\xi)-(J_{H}\Delta_{H}^{1/2}\xi,J_{H}A\Delta_{H}^{1/2}\xi)\\\
=(\xi,A\xi)-(A\Delta_{H}^{1/2}\xi,\Delta_{H}^{1/2}\xi)=(\xi,A(1-\Delta_{H})\xi)\,,$

thus $A(1-\Delta_{H})\leq 0$ because ${\mathcal{D}}_{\rm an}$ is a core for
$A(1-\Delta_{H})$. But $A(1-\Delta_{H})\leq 0$ is equivalent to
$A\log\Delta_{H}\geq 0$.  $\square$

We shall say that a standard subspace is abelian if $\Delta_{H}=1$, see [22].

######  Theorem 2.10.

$A$ is completely active with respect to $H$ iff $\log\Delta_{H}=\lambda A$
for some $\lambda\leq 0$.

  * Proof.

Assume that $A$ is completely active with respect to $H$. Let
$\Lambda\subset\mathbb{R}^{2}$ be the joint spectrum of $A$ and
$\log\Delta_{H}$. By Lemma 2.9, $\Lambda$ is contained is the region
$Q=\\{(a,b)\in\mathbb{R}^{2}:\,ab\geq 0\\}$. By (22) we have
$-\Lambda=\Lambda$ and by complete passivity we have
$\Lambda+\Lambda+\cdots+\Lambda\subset Q$ (finite sum).

Let $C_{1},C_{2}$ be two different points in $\Lambda$. Then
$n_{1}C_{1}+n_{2}C_{2}\in Q$ for all integers $n_{1},n_{2}\in{\mathbb{Z}}$. So
$C_{1}$ and $C_{2}$ must belong to a same straight line $Z$ through the origin
$(0,0)$. Thus $\Lambda\subset Z$. If $Z$ is vertical, then $\log\Delta_{H}=0$,
namely $H$ is abelian, so $\log\Delta_{H}=\lambda A$ with $\lambda=0$. If $Z$
is not vertical then $A=\lambda\log\Delta_{H}$ with $\lambda>0$.

For the converse, it remains to show that $A$ is completely active if
$\Delta_{H}=1$, namely if $H$ is abelian. In this case, the scalar product of
${\mathcal{H}}$ is real on $H$, and $iH$ is the real orthogonal complement of
$H$. As $A$ maps $H$ into $iH$, we have $\Re(h,Ah)=0$ for $h\in D(A)\cap H$,
thus $(h,Ah)=0$ because $A$ is selfadjoint. So $A$ is active, thus completely
active by repeating this argument for $H\otimes H\cdots\otimes H$.  $\square$

##  3 Preliminaries on the waves’ space

Let ${\mathcal{S}}$ denote the real linear space of smooth, compactly
supported real functions on $\mathbb{R}^{d}$. We shall always assume $d\geq 2$
unless otherwise specified.

As is known, if $f,g\in{\mathcal{S}}$, there is a unique smooth real function
$\Phi$ on $\mathbb{R}^{d+1}$ which is a solution $\Phi$ of the Klein-Gordon
equation with mass $m\geq 0$

$(\square+m^{2})\Phi=0$

(a wave packet or, briefly, a wave) with Cauchy data $\Phi|_{x_{0}=0}=f$,
$\partial_{0}\Phi|_{x_{0}=0}=g$. We set $\Phi=w_{m}(f,g)$ and denote by
${\mathcal{T}}_{m}$ the real linear space of these $\Phi$’s; we will often use
the idenfication

${\mathcal{S}}^{2}\longleftrightarrow{\mathcal{T}}_{m}\,,\qquad\langle
f,g\rangle\longleftrightarrow w_{m}(f,g)\,.$ (25)

The one particle Hilbert space is
${\mathcal{H}}_{m}=L^{2}\big{(}{\mathfrak{H}}_{m},\delta(p^{2}-m^{2})\big{)}$
with $\mathfrak{H}_{m}$ the mass $m$ hyperboloid. We denote by
$(\cdot,\cdot)_{m}$ the scalar product of ${\cal H}_{m}$. The Fourier
transform of a $\Phi\in{\mathcal{T}}_{m}$ is a distribution of the form
$\hat{\Phi}(p)=\delta(p^{2}-m^{2})F(p)$ with a compactly supported smooth
function $F:{\mathfrak{H}}_{m}\cup(-{\mathfrak{H}}_{m})\to{\mathbb{C}}$, so
that ${\mathcal{T}}_{m}$ real linearly embeds into ${\mathcal{H}}_{m}$ by
$\Phi\mapsto(2\pi)^{d/2}F|_{\mathfrak{H}_{m}}$. We may thus consider
${\mathcal{T}}_{m}$ as a dense subset of ${\mathcal{H}}_{m}$.

Consider the symplectic form on ${\mathcal{T}}_{m}$

$\beta(\Phi,\Psi)=\frac{1}{2}\int_{x_{0}=0}\big{(}\Psi\partial_{0}\Phi-\Phi\partial_{0}\Psi)dx\,.$
(26)

This is the imaginary part of the restriction of the scalar product of ${\cal
H}_{m}$ to ${\mathcal{T}}_{m}$:

$\Im(\langle f_{1},g_{1}\rangle,\langle f_{2},g_{2}\rangle)_{m}=\beta(\langle
f_{1},g_{1}\rangle,\langle f_{2},g_{2}\rangle)\ .$

In particular, $\Im\langle\cdot,\cdot\rangle_{m}$ is therefore actually
independent of $m$. We denote by $H_{m}^{s}$ the real Hilbert space of real
valued tempered distributions $f\in S^{\prime}({\mathbb{R}}^{d})$ such that

$||f||_{s}^{2}=\|f\|^{2}_{s;m}=\int_{{\mathbb{R}}^{d}}{(|{\bf
p}|^{2}+m^{2})}^{s}|\hat{f}({\bf p})|^{2}d{\bf p}<+\infty\,,\quad
s\in{\mathbb{R}}\,.$ (27)

It is clear that ${\mathcal{S}}$ is dense in $H_{m}^{\pm 1/2}$ and that
$\mu_{m}:H_{m}^{1/2}\to H_{m}^{-1/2}$ with

$\widehat{\mu_{m}f}({\bm{p}})=\sqrt{{|\bm{p}|}^{2}+m^{2}}\hat{f}({\bm{p}})\,.$
(28)

is a unitary operator.

Then

$\imath_{m}=\left[\begin{matrix}0&\mu_{m}^{-1}\\\
-\mu_{m}&0\end{matrix}\right]\,,$ (29)

namely $\imath_{m}\langle f,g\rangle=\langle\mu_{m}^{-1}g,-\mu_{m}f\rangle$,
is a unitary operator $\imath_{m}$ on $H_{m}=H_{m}^{1/2}\oplus H_{m}^{-1/2}$.

As $\imath_{m}^{2}=-1$, the unitary $\imath_{m}$ defines a complex structure
(multiplication by the imaginary unit) on $H_{m}$ that becomes a complex
Hilbert space with scalar product

$(\Phi,\Psi)=(\Phi,\Psi)_{m}=\beta(\Phi,\imath_{m}\Psi)+i\beta(\Phi,\Psi)\ .$

In terms of the Cauchy data, the scalar product is given by

$(\langle f,g\rangle,\langle
h,k\rangle)_{m}=\frac{1}{2}\big{(}(f,\mu_{m}h)+(g,\mu_{m}^{-1}k)+i[(h,g)-(f,k)]\big{)},\qquad\langle
f,g\rangle,\langle h,k\rangle\in H_{m}^{1/2}\oplus H_{m}^{-1/2},$

(where $(\cdot,\cdot)$ is the standard $L^{2}$ scalar product). Furthermore
the map

$\langle f,g\rangle\in H_{m}\mapsto w_{m}(f,g)\in{\mathcal{H}}_{m}$

is isometric and complex linear, so it extends to a unitary operator. We will
then identify these spaces; namely the one particle Hilbert space is

${\mathcal{H}}_{m}=\text{real Hilbert space}\ H_{m}\ \text{with complex
structure given by}\ \imath_{m}\,.$

Note that, when $m=1$, $H_{m}^{\pm 1/2}$ is a Sobolev space with usual
definition. For $m>m^{\prime}>0$, the norms $\|\cdot\|_{\pm 1/2,m}$ are
equivalent on the same linear space:

$\displaystyle\|f\|_{-1/2,m}\leq\|f\|_{-1/2,m^{\prime}}$
$\displaystyle\leq\sqrt{\frac{m}{m^{\prime}}}\|f\|_{-1/2,m}$ (30)
$\displaystyle\|f\|_{1/2,m^{\prime}}\leq\|f\|_{1/2,m}$
$\displaystyle\leq\sqrt{\frac{m}{m^{\prime}}}\|f\|_{1/2,m^{\prime}}\,.$ (31)

$H_{m}^{1/2}$ and $H_{m}^{-1/2}$ are naturally a dual pair under the bilinear
form

$\langle f,g\rangle\in H_{m}^{1/2}\times
H_{m}^{-1/2}\mapsto\int_{{\mathbb{R}}^{d}}\bar{\hat{f}}\hat{g}\ .$ (32)

Let $Z$ be an open, non-empty subset of $\mathbb{R}^{d}$ and denote by
$Z^{\prime}$ the interior of its complement. We shall denote by $H_{m}(Z)$ the
closed, real linear subspace of ${\mathcal{H}}_{m}$

$H_{m}(Z)=\\{\Phi\in{\mathcal{T}}_{m}(Z)\\}^{-}\,,$

where

${\mathcal{T}}_{m}(Z)=\\{\Phi=w_{m}(f,g)\in{\mathcal{T}}_{m}:{\rm
supp}(f),\,{\rm supp}(g)\subset Z\\}\,.$

If $Z$ and $Z^{\prime}$ are non-empty, then $H_{m}(Z)$ is a standard subspace
of ${\mathcal{H}}_{m}$, the one associated with $Z$. We shall be mainly
interested in the case $Z$ or $Z^{\prime}$ is the unit ball $B$ of
$\mathbb{R}^{d}$. Clearly, setting

$H^{\pm 1/2}_{m}(Z)=\text{closure of $C_{0}^{\infty}(Z)$ in $H^{\pm
1/2}_{m}$}\,,$

we have

$H_{m}(Z)=H_{m}^{1/2}(Z)\oplus H_{m}^{-1/2}(Z)\,.$

Here and in the following, $C^{\infty}_{0}(Z)$ denotes the space of real
$C^{\infty}$ function on $\mathbb{R}^{d}$ with compact support in $Z$. By
duality [1], see also [13, 21], we have

$H_{m}(B)^{\prime}=H_{m}(B^{\prime})\,.$ (33)

######  Lemma 3.1.

([15, Prop. A.2]) $H_{m}^{\pm 1/2}(B)$ and $H_{0}^{\pm 1/2}(B)$ are the same
linear space with equivalent norms, $m\geq 0$.

  * Proof.

Note first that $H_{0}^{-1/2}(B)\subset H_{m}^{-1/2}(B)$, indeed
$H_{0}^{-1/2}(B)$ has a bounded embedding into $H_{m}^{-1/2}(B)$. Let’s show
the reverse inclusion, namely

$f\in H_{m}^{-1/2}(B)\implies f\in H_{0}^{-1/2}(B)\,.$

As $f$ is compactly supported, $\hat{f}$ is analytic; let $C>0$ be such that
$|\hat{f}({\bf p})|^{2}\leq C$ if $|{\bf p}|\leq 1$. Then we have

$\int_{|{\bf p}|<1}\frac{|\hat{f}({\bf p})|^{2}}{|{\bf p}|}d{\bf p}\leq
C\int_{|{\bf p}|<1}\frac{d{\bf p}}{|{\bf p}|}<\infty$

($d\geq 2$). On the other hand,

$\int_{|{\bf p}|\geq 1}\frac{|\hat{f}({\bf p})|^{2}}{|{\bf p}|}d{\bf
p}<\infty$

because $f\in H_{m}^{-1/2}(B)$; therefore $f\in H_{0}^{-1/2}(B)$. The norms on
$H_{0}^{-1/2}(B)$ and $H_{m}^{-1/2}(B)$ are then equivalent by the open
mapping theorem, as can be checked directly too.

Similarly, we have $H_{m}^{1/2}(B)\subset H_{0}^{1/2}(B)$. For the converse
inclusion, let $f\in H_{0}^{1/2}(B)$ and $C>0$ be such that $|{\bf
p}||\hat{f}({\bf p})|^{2}\leq C$ if $|{\bf p}|\leq 1$.

$\frac{1}{\sqrt{1+m^{2}}}\int_{|{\bf p}|<1}|\hat{f}({\bf p})|^{2}\sqrt{|{\bf
p}|^{2}+m^{2}}\,d{\bf p}\leq\int_{|{\bf p}|<1}|{\bf p}|\frac{|\hat{f}({\bf
p})|^{2}}{|{\bf p}|}d{\bf p}\leq C\int_{|{\bf p}|<1}\frac{d{\bf p}}{|{\bf
p}|}<\infty\,.$

$\square$

##  4 Massless Hamiltonian

We start with the analysis of the modular Hamiltonian in the massless case for
two reasons. Firstly, the formula for the massless Hamiltonian has not been
worked out so far and it already shows new and interesting aspects. Secondly,
the massive Hamiltonian will be derived by a deformation of the massless one.

###  4.1 The modular group, $m=0$

In the following, $O$ will denote the double cone on the Minkowski spacetime
$\mathbb{R}^{d+1}$ with base the open unit ball $B$ centred at the origin in
the time zero hyperplane $\mathbb{R}^{d}$. As above, $H_{0}(O)=H_{0}(B)$ is
the standard subspace in the massless, scalar one-particle Hilbert space
${\cal H}_{0}$. We assume $d\geq 2$.

The modular group $\Delta_{B}^{is}$ associated with $H_{0}(B)$ has been
computed in [19] in terms of the action on the field or, equivalently, on the
spacetime test-functions. $\Delta_{B}^{is}$ is associated with a one parameter
group of conformal transformation that preserve $O$, namely

$(u,v)\mapsto\big{(}(Z(u,s),Z(v,s)\big{)}\,,$ (34)

where $Z$ is given by

$Z(z,s)=\frac{g(z,s)}{f(z,s)}$

with

$f(z,s)=\frac{(1+z)+e^{-s}(1-z)}{2}\ ,\quad
g(z,s)=\frac{(1+z)-e^{-s}(1-z)}{2}\ .$

Here we need to compute $\Delta_{B}^{is}$ it in terms of waves. Let $\Phi$ be
a wave and set

$(V(s)\Phi)(u,v)=\gamma(u,v;s)\Phi\big{(}Z(u,s),Z(v,s)\big{)}\,,$ (35)

with

$u=x_{0}+r,\quad v=x_{0}-r,\quad r=|{\bf
x}|\equiv\sqrt{x_{1}^{2}+\cdots+x_{d}^{2}}$

and we omit the remaining spherical coordinates as the action is trivial on
them.

The cocycle $\gamma$ given by

$\gamma(u,v;s)=F(u,s)F(-v,-s),\quad F(z,s)\equiv f^{-D}(z,s)\ ,$

with

$D=\frac{d-1}{2}\,;$ (36)

$D$ is the scaling dimension. We have:

######  Theorem 4.1.

The modular group of $H_{0}(B)$ is given by

$\Delta_{B}^{-is}=V(2\pi s)\,.$

  * Proof.

That $\Delta_{B}^{-is/2\pi}$ is associated with the flow $Z$ (34) and the
cocycle $\gamma$ follows rather directly from the test function formula [19],
but for the determination of the value of the constant $D$. Now, as in [19],
$H_{0}(B)$ is equivalent to $H_{0}(W)$, with $W$ a wedge region, by a unitary
operator that is obtained by composing unitaries associated with translations
and ray inversion map. The cocycle is associated with the Jacobian of the ray
inversion map. Thus, by computation similar to those in [19], the cocycle is
of the form $f^{-D}$ for some constant $D>0$.

However, it will follows from Lemma 4.9 that $D=(d-1)/2$ is the only value
compatible with the complex linearity of the generator of $V$. Of course, $D$
can be determined by direct calculations too.  $\square$

###  4.2 The generator of the modular group, $m=0$

We now compute the $K_{0}=\frac{d}{ds}V(s)\big{|}_{s=0}$, the generator of $V$
in Theorem 4.1, that is proportional to the modular Hamiltonian.

With the above notations, denoting by a prime the derivative with respect to
$s$-parameter, we have (Identity 7.2)

$Z(z,0)=z\,,\qquad Z^{\prime}(z,0)=(1-z^{2})/2$

and

$\gamma(u,v;0)=1\,,\qquad\gamma^{\prime}(u,v;0)=-\frac{D}{2}(u+v)=-D\,x_{0}\
.$

Therefore

$(V(s)\Phi)(u,v)^{\prime}=\gamma^{\prime}(u,v;s)\Phi\big{(}Z(u,s),Z(v,s)\big{)}\\\
+\gamma(u,v;s)\Big{(}\partial_{u}\Phi\big{(}Z(u,s),Z(v,s)\big{)}Z^{\prime}(u,s)+\partial_{v}\Phi\big{(}Z(u,s),Z(v,s)\big{)}Z^{\prime}(v,s)\Big{)}\,.$

We then have:

######  Proposition 4.2.

We have

$(K_{0}\Phi)(u,v)=-\frac{D}{2}(u+v)\Phi-\frac{1}{2}u^{2}\partial_{u}\Phi-\frac{1}{2}v^{2}\partial_{v}\Phi+\frac{1}{2}\partial_{u}\Phi+\frac{1}{2}\partial_{v}\Phi\
.$

In terms of the $x_{0},r$ coordinates

$(K_{0}\Phi)(x_{0},r)=\frac{1}{2}\big{(}1-(x^{2}_{0}+r^{2})\big{)}\partial_{0}\Phi-
x_{0}r\partial_{r}\Phi-D\,x_{0}\Phi\ .$ (37)

  * Proof.

We compute:

$\displaystyle(K_{0}\Phi)(u,v)$
$\displaystyle=(V(s)\Phi)(u,v)^{\prime}\big{|}_{s=0}$
$\displaystyle=\gamma^{\prime}(u,v;0)\Phi(u,v)+\partial_{u}\Phi(u,v)Z^{\prime}(u,0)+\partial_{v}\Phi(u,v)Z^{\prime}(v,0)$
$\displaystyle=-\frac{D}{2}(u+v)\Phi(u,v)+\frac{1}{2}\partial_{u}\Phi(u,v)(1-u^{2})+\frac{1}{2}\partial_{v}\Phi(u,v)(1-v^{2})$
$\displaystyle=-\frac{D}{2}(u+v)\Phi-\frac{1}{2}u^{2}\partial_{u}\Phi-\frac{1}{2}v^{2}\partial_{v}\Phi+\frac{1}{2}\partial_{u}\Phi+\frac{1}{2}\partial_{v}\Phi\
.$

Now

$\partial_{u}=\frac{1}{2}(\partial_{0}+\partial_{r}),\quad\partial_{v}=\frac{1}{2}(\partial_{0}-\partial_{r})\,,$

so

$(K_{0}\Phi)(x_{0},r)=-\frac{1}{4}\big{(}(x_{0}+r)^{2}(\partial_{0}+\partial_{r})\Phi+(x_{0}-r)^{2}(\partial_{0}-\partial_{r})\big{)}\Phi-{D}x_{0}\Phi+\frac{1}{2}\partial_{0}\Phi\\\
=-\frac{1}{2}(x^{2}_{0}+r^{2})\partial_{0}\Phi-
x_{0}r\partial_{r}\Phi-{D}x_{0}\Phi+\frac{1}{2}\partial_{0}\Phi\\\
=\frac{1}{2}\big{(}1-(x^{2}_{0}+r^{2})\big{)}\partial_{0}\Phi-
x_{0}r\partial_{r}\Phi-D\,x_{0}\Phi\ .$ (38)

$\square$

######  Corollary 4.3.

We have

$\displaystyle(K_{0}\Phi)|_{x_{0}=0}=\frac{1}{2}(1-r^{2})\partial_{0}\Phi|_{x_{0}=0}\,,$
(39)
$\displaystyle(\partial_{0}K_{0}\Phi)|_{x_{0}=0}=\frac{1}{2}(1-r^{2})\nabla^{2}\Phi-r\partial_{r}\Phi-D\Phi|_{x_{0}=0}\,.$
(40)

  * Proof.

The first equality follows immediately from (37). Again from (37) we have

$\partial_{0}K_{0}\Phi|_{x_{0}=0}=\frac{1}{2}(1-r^{2})\partial^{2}_{0}\Phi-r\partial_{r}\Phi-D\Phi|_{x_{0}=0}$

thus (40) holds.  $\square$

The above corollary translates into the following:

######  Proposition 4.4.

We have

$K_{0}:w_{0}(f,g)\mapsto
w_{0}\Big{(}\frac{1}{2}(1-r^{2})g,\frac{1}{2}(1-r^{2})\nabla^{2}f-r\partial_{r}f-D\,f\Big{)}\,.$

In other words, $K_{0}$ is the operator on ${\mathcal{S}}^{2}$ given by the
$2\times 2$ matrix

$K_{0}=\left[\begin{matrix}0&\frac{1}{2}(1-r^{2})\\\
\frac{1}{2}(1-r^{2})\nabla^{2}-r\partial_{r}-D&0\end{matrix}\right]\,.$ (41)

Now the symplectic form $\beta$ on ${\mathcal{T}}_{0}$ (26) is given, in terms
of Cauchy data, as

$\beta(\Phi,\Psi)=\frac{1}{2}\int_{\mathbb{R}^{d}}\big{(}f_{2}g_{1}-f_{1}g_{2}\big{)}d{\bf
x}\ .$ (42)

with $\Phi=w_{0}(f_{1},g_{1})$, $\Psi=w_{0}(f_{2},g_{2})$.

Thus

$2\beta(\Phi,K_{0}\Phi)=\int_{x_{0}=0}(K_{0}\Phi)\partial_{0}\Phi-\Phi\partial_{0}(K_{0}\Phi)dx\\\
=\int_{\mathbb{R}^{d}}\frac{1}{2}(1-r^{2})g^{2}d{\bf
x}-\int_{\mathbb{R}^{d}}\frac{1}{2}(1-r^{2})f\nabla^{2}fd{\bf
x}+\int_{\mathbb{R}^{d}}rf\partial_{r}fd{\bf
x}+D\int_{\mathbb{R}^{d}}f^{2}d{\bf x}\ .$ (43)

So, taking into account Identity (7.1), we have

$2\beta(\Phi,K_{0}\Phi)=\frac{1}{2}\int_{\mathbb{R}^{d}}(1-r^{2})\big{(}g^{2}+|\nabla
f|^{2}\big{)}d{\bf x}+D\int_{\mathbb{R}^{d}}f^{2}d{\bf x}\ ,$

namely

$\beta(\Phi,K_{0}\Phi)=\frac{1}{4}\int_{x_{0}=0}(1-r^{2})\big{(}(\partial_{0}\Phi)^{2}+|\nabla\Phi|^{2}\big{)}dx+\frac{D}{2}\int_{x_{0}=0}\Phi^{2}dx\,.$
(44)

So we have obtained the following proposition.

######  Proposition 4.5.

With $A_{0}=-\imath_{0}K_{0}$, we have

$(\Phi,A_{0}\Phi)=\beta(\Phi,K_{0}\Phi)=\frac{1}{2}\int_{x_{0}=0}(1-r^{2})\langle
T_{00}\rangle_{\Phi}dx+\frac{D}{2}\int_{x_{0}=0}\Phi^{2}dx$ (45)

with $\langle
T_{00}\rangle_{\Phi}=\frac{1}{2}\big{(}(\partial_{0}\Phi)^{2}+|\nabla\Phi|^{2}\big{)}$
the energy density given by classical stress-energy tensor $T$.

  * Proof.

Since $A_{0}$ is selfadjoint, $(\Phi,A_{0}\Phi)$ is real, so

$(\Phi,A_{0}\Phi)=\Im
i(\Phi,A_{0}\Phi)=\Im(\Phi,\imath_{0}A_{0}\Phi)=\Im(\Phi,K_{0}\Phi)=\beta(\Phi,K_{0}\Phi)\,,$

so the proposition follows from (44).  $\square$

Since

$\beta(K_{0}\Phi,\Psi)+\beta(\Phi,K_{0}\Psi)=0\,,$

by the polarisation identity we then get

$\beta(\Phi,K_{0}\Psi)=\frac{1}{2}\Big{(}\beta\big{(}\Phi+\Psi,K_{0}(\Phi+\Psi)\big{)}-\beta\big{(}\Phi-\Psi,K_{0}(\Phi-\Psi)\big{)}\Big{)}\\\
=\frac{1}{2}\int_{x_{0}=0}(1-r^{2})\langle
T_{00}\rangle_{\Phi,\Psi}dx+\frac{D}{2}\int_{x_{0}=0}\Phi\Psi dx$ (46)

###  4.3 $K_{0}$ acting on ${\mathcal{H}}_{0}$

Set $H_{0}(B)=H_{0}^{1/2}(B)\oplus H_{0}^{-1/2}(B)$ for the standard subspace
associated with $B$ as before.

Let

$M_{0}:D(M_{0})\subset H_{0}^{-1/2}\to H_{0}^{1/2}\,,\quad
L_{0}:D(L_{0})\subset H_{0}^{1/2}\to H_{0}^{-1/2}$

be the closures of the operators

$\displaystyle M_{0}$ $\displaystyle=\frac{1}{2}(1-r^{2})$ $\displaystyle
L_{0}$ $\displaystyle=\frac{1}{2}(1-r^{2})\nabla^{2}-r\partial_{r}-D$

on ${\mathcal{S}}$. ($M_{0}$ is the multiplication operator by
$\frac{1}{2}(1-r^{2})$). We shall indeed see that the both operator $M_{0}$
and $L_{0}$ on ${\mathcal{S}}$ are closable as $M_{0}^{*}$ contains $L_{0}$ on
${\mathcal{S}}$.

Denote by $K_{0}=-\frac{\imath_{0}}{2\pi}\log\Delta_{B}$ the modular
Hamiltonian relative to $B$ ($m=0$).

######  Lemma 4.6.

${\mathcal{S}}^{2}$ is a core for $K_{0}$ (as real linear operator).

  * Proof.

By the geometrical action of the modular group $\Delta_{B}^{is}$, and duality,
one sees that $C_{0}^{\infty}(B)^{2}+C_{0}^{\infty}(B^{\prime})^{2}$ is a
dense, real linear subspace of $H_{0}$, globally $\Delta_{B}^{is}$-invariant,
contained in the domain of the generator $K_{0}$; thus it is a core for
$K_{0}$. So ${\mathcal{S}}^{2}$ is a core too being a larger subspace still
contained in $D(K_{0})$.  $\square$

So we have:

######  Theorem 4.7.

The massless Hamiltonian $\log\Delta_{B}$ is given by $-2\pi
A_{0}=\log\Delta_{B}$ with

$K_{0}=\left[\begin{matrix}0&M_{0}\\\
L_{0}&0\end{matrix}\right]=\left[\begin{matrix}0&M_{0}\\\
-M_{0}^{*}&0\end{matrix}\right]\,.$

and $A_{0}=-\imath_{0}K_{0}$.

  * Proof.

By Lemma 4.6, we have the first equality. As $A_{0}$ is selfadjoint, we have
$K_{0}^{*}=-K_{0}$, thus the second equality holds.  $\square$

Next corollary will be generalised in Section 5.5.

######  Corollary 4.8.

$A_{0}$ is completely active with respect to $H_{0}(B)$.

  * Proof.

Immediate by Theorem 2.10.  $\square$

######  Corollary 4.9.

We have $K_{0}=-\imath_{0}K_{0}\imath_{0}$. Thus

$\mu_{0}M_{0}\mu_{0}=M_{0}^{*}=-L_{0}\,.$

  * Proof.

Immediate by the complex linearity of $K_{0}$.  $\square$

Note, in particular, that the constant $D$ in the expression of the operator
$L_{0}$ is fixed by the above corollary.

######  Corollary 4.10.

$K_{0}^{B}=K_{0}|_{H_{0}(B)}$ is skew-selfadjoint on $H_{0}(B)$.

  * Proof.

Since $H_{0}(B)$ is globally invariant for $e^{sK_{0}}$, we may apply Prop.
2.2.  $\square$

######  Remark 4.11.

By Cor. 4.10, we have indirectly solved a Dirichlet problem for the degenerate
elliptic operators $L_{0}M_{0}-1$ and $M_{0}L_{0}-1$, see [4]. In the next
section we shall see a more general degenerate Dirichlet problem for an
integro-differential operator.

##  5 Massive Hamiltonian

Let $M_{m}$ be the real linear operator

$M_{m}:D(M_{m})\subset H_{m}^{-1/2}\to H_{m}^{1/2}\,,\quad
M_{m}f=\frac{1-r^{2}}{2}f\,,$ (47)

with domain $D(M_{m})=\\{f\in H_{m}^{-1/2}:M_{m}f\in H_{m}^{1/2}\\}$ and
$M_{m}^{*}$ the adjoint of $M$ with respect to the real Hilbert space scalar
products

$M_{m}^{*}:D(M_{m}^{*})\subset H_{m}^{1/2}\to H_{m}^{-1/2}\,.$

Note that ${\mathcal{S}}$ is a core for $M_{m}$; indeed the transpose of
$M_{m}|_{\mathcal{S}}$ under the duality (32) is $M_{m}$. While the formal
expression $M_{m}$ for $M$ does not depend on $m$, the operator $M_{m}^{*}$
will have a different expression depending on $m$.

Let also $L_{m}:D(L_{m})\subset H_{m}^{1/2}\to H_{m}^{-1/2}$ be the closure of
the real linear operator on ${\mathcal{S}}$

$L_{m}=\frac{1}{2}(1-r^{2})\big{(}\nabla^{2}-m^{2}\big{)}-r\partial_{r}-D+\frac{1}{2}m^{2}(\nabla^{2}-m^{2})^{-1}$
(48)

The actions of $L_{m}$ and $M_{m}$ on ${\mathcal{S}}$ are naturally defined in
Fourier transform.

###  5.1 The operator $\widetilde{K}_{m}$

As a first step towards the formula for the massive Hamiltonian, we are going
to single out a natural deformation operator $\widetilde{K}_{m}$ of the
massless skew-adjoint massless Hamiltonian $K_{0}$. As we shall see, the
quadratic form associated with $\widetilde{K}_{m}$ will restrict to the
quadratic form of the massive skew-adjoint Hamiltonian on the ball $B$.

######  Proposition 5.1.

$\mu_{m}M_{m}\mu_{m}=-L_{m}$ on the Schwartz space $S(\mathbb{R}^{d})$.

  * Proof.

For simplicity, denote by $\hat{\mu}$ the function $\hat{\mu}({\bf
p})=\sqrt{|{\bf p}|^{2}+m^{2}}$. We have $\nabla_{\bf
p}\hat{\mu}_{m}=\frac{{\bf p}}{\hat{\mu}_{m}}$, so

$\nabla_{\bf p}^{2}\hat{\mu}_{m}=\nabla_{\bf p}\cdot\frac{{\bf
p}}{\hat{\mu}_{m}}=\sum_{k}\frac{\hat{\mu}_{m}-p_{k}\frac{p_{k}}{\hat{\mu}_{m}}}{\hat{\mu}_{m}^{2}}=\frac{1}{\hat{\mu}_{m}}\sum_{k}\frac{\hat{\mu}_{m}^{2}-p^{2}_{k}}{\hat{\mu}_{m}^{2}}=\frac{1}{\hat{\mu}_{m}}\Big{(}d-\frac{|{\bf
p}|^{2}}{\hat{\mu}_{m}^{2}}\Big{)}\,,$ $\nabla_{\bf
p}^{2}\hat{\mu}_{m}^{2}=d-1\,.$

With $f\in S(\mathbb{R}^{d})$, we the have

$\displaystyle\frac{1}{2}\big{(}\mu_{m}(1-r^{2})\mu_{m}f\big{)}\hat{\,}$
$\displaystyle=\frac{1}{2}{\hat{\mu}_{m}}(1+\nabla_{\bf
p}^{2})({\hat{\mu}_{m}}\hat{f})$
$\displaystyle=\frac{1}{2}{\hat{\mu}_{m}}^{2}+\frac{1}{2}{\hat{\mu}_{m}}\nabla_{\bf
p}^{2}({\hat{\mu}_{m}}\hat{f})$
$\displaystyle=\frac{1}{2}{\hat{\mu}_{m}}^{2}(1+\nabla_{\bf
p}^{2})\hat{f}+{\bm{p}}\cdot{\nabla}_{\bf
p}\hat{f}+\frac{1}{2}\Big{(}d-\frac{|{\bf
p}|^{2}}{\hat{\mu}_{m}^{2}}\Big{)}\hat{f}$
$\displaystyle=\frac{1}{2}{\hat{\mu}_{m}}^{2}(1+\nabla_{\bf
p}^{2})\hat{f}+{\bm{p}}\cdot{\nabla}_{\bf
p}\hat{f}+D\hat{f}+\frac{1}{2}\frac{m^{2}}{\hat{\mu}_{m}^{2}}\hat{f}$
$\displaystyle=\frac{1}{2}{|{\bf p}|}^{2}(1+\nabla_{\bf
p}^{2})\hat{f}+{\bm{p}}\cdot{\nabla}_{\bf
p}\hat{f}+D\hat{f}+\frac{1}{2}m^{2}(1+\nabla_{\bf
p}^{2})\hat{f}+\frac{1}{2}\frac{m^{2}}{\hat{\mu}_{m}^{2}}\hat{f}$
$\displaystyle=\Big{(}\big{(}-L+\frac{1}{2}m^{2}(1-r^{2})+\frac{1}{2}m^{2}\mu_{m}^{-2}\big{)}f\Big{)}^{\widehat{}}=(-L_{m}f)\,\hat{}$

as desired.  $\square$

######  Lemma 5.2.

With $M_{m}$ and $M_{m}^{*}$ as above, we have
$\mu_{m}^{*}M_{m}^{*}=M_{m}\mu_{m}$, namely $M\mu_{m}$ is a selfadjoint
operator on $H_{m}^{1/2}$. Here $\mu_{m}:H_{m}^{1/2}\to H_{m}^{-1/2}$ is the
unitary operator given by (28).

  * Proof.

$g\in H_{m}^{1/2}$ belongs to $D((M_{m}\mu_{m})^{*})$ iff there exists
$g^{\prime}\in H_{m}^{1/2}$ such that

$(g,M_{m}\mu_{m}f)_{1/2}=(g^{\prime},f)_{1/2}\quad\forall f\in
D(M_{m}\mu_{m})\,.$

Now denote by $h$ the function $h({\bf x})=\frac{1}{2}(1-r^{2})$; we have

$(g,M_{m}\mu_{m}f)_{1/2}=\int(\mu_{m}g)M_{m}\mu_{m}fd{\bf
x}=\int(\mu_{m}g)h(\mu_{m}f)d{\bf x}=\int h(\mu_{m}g)(\mu_{m}f)d{\bf x}$

thus $g\in D((M_{m}\mu_{m})^{*})$ iff $h\mu_{m}g\in H_{m}^{1/2}$, namely iff
$g\in D(M_{m}\mu_{m})$, and then we have $(M_{m}\mu_{m})^{*}g=M_{m}\mu_{m}g$.
$\square$

As $\mu_{m}$ is unitary, we have the formula

$M_{m}^{*}=\mu_{m}M_{m}\mu_{m}\,.$

Let
$\widetilde{K}_{m}:D(\widetilde{K}_{m})\subset{\mathcal{H}}_{m}\to{\mathcal{H}}_{m}$
be the real linear operator

$\widetilde{K}_{m}=\left[\begin{matrix}0&M_{m}\\\
-M_{m}^{*}&0\end{matrix}\right]\,.$

The domain of $\widetilde{K}_{m}$ is given by
$D(\widetilde{K}_{m})=D(M_{m}^{*})\oplus D(M_{m})\subset H_{m}^{1/2}\oplus
H_{m}^{-1/2}$.

We then have:

######  Proposition 5.3.

$\widetilde{K}_{m}=\left[\begin{matrix}0&M_{m}\\\
-M_{m}^{*}&0\end{matrix}\right]=\left[\begin{matrix}0&M_{m}\\\
-\mu_{m}M_{m}\mu_{m}&0\end{matrix}\right]=\left[\begin{matrix}0&M_{m}\\\
L_{m}&0\end{matrix}\right]\,.$

  * Proof.

By the above discussion, we have the first equality and the second equality
holds on $S(\mathbb{R}^{d})^{2}$. So it remains to show that
$S(\mathbb{R}^{d})$ is a core for $\mu_{m}M_{m}\mu_{m}$, namely that
$\mu_{m}S(\mathbb{R}^{d})$ is a core for $M_{m}$, which is true because
$\mu_{m}S(\mathbb{R}^{d})=S(\mathbb{R}^{d})$ as is manifest in Fourier
transform.  $\square$

######  Remark 5.4.

It is useful to compare with the corresponding structure in the case of the
half-space $x_{1}>0$, whose casual envelop is the wedge region $W_{1}$. In
this case, the modular group is given by the rescaled pure Lorentz
transformations in the $x_{1}$-direction. A computation analogous to the one
given in Sect. 4 then gives

$2\pi\log\Delta_{W_{1}}=\imath_{m}\left[\begin{matrix}0&x_{1}\\\
x_{1}(\nabla^{2}-m^{2})+\partial_{1}&0\end{matrix}\right]\,.$

Note that the upper-right entry in the above matrix is independent of the mass
$m\geq 0$.

###  5.2 The operator $K^{B}_{m}$

Recall that $H_{m}(B)=H_{m}^{1/2}(B)\oplus H_{m}^{-1/2}(B)$ . Let $E_{m}$ be
the orthogonal projection ${\cal H}_{m}\to H_{m}(B)$. Thus

$E_{m}=\left[\begin{matrix}E_{m}^{+}&0\\\ 0&E_{m}^{-}\end{matrix}\right]\,,$

with $E_{m}^{\pm}$ the orthogonal projection $H_{m}^{\pm 1/2}\to H_{m}^{\pm
1/2}(B)$.

We define $K_{m}^{B}$ as the operator on $H_{m}(B)$ given by the compression
of $\widetilde{K}_{m}$ on $H_{m}(B)$, namely by restricting the quadratic form
of $\widetilde{K}_{m}$ to $H_{m}(B)$:

$K_{m}^{B}=E_{m}\widetilde{K}_{m}|_{H_{m}(B)}\,;$

thus $K_{m}^{B}:D(K_{m}^{B})\subset H_{m}(B)\to H_{m}(B)$ with
$D(K_{m}^{B})=D(\widetilde{K}_{m})\cap H_{m}(B)$.

As the only non local term in $\widetilde{K}_{m}$ is
$\frac{1}{2}m^{2}\mu_{m}^{-2}$, we have on $H_{m}(B)$

$K_{m}^{B}=\left[\begin{matrix}0&\frac{1}{2}(1-r^{2})\\\
\frac{1}{2}(1-r^{2})(\nabla^{2}-m^{2})-r\partial_{r}-D&0\end{matrix}\right]-\frac{1}{2}m^{2}\left[\begin{matrix}0&0\\\
G^{B}_{m}&0\end{matrix}\right]$ (49)

with

$G^{B}_{m}=E^{-}_{m}\mu_{m}^{-2}\,.$

That is to say

$K_{m}^{B}=\left[\begin{matrix}0&M_{m}\\\
L^{B}_{m}&0\end{matrix}\right]\quad{\rm on}\ H_{m}(B)\,$ (50)

with $L_{m}^{B}$ the sum of the local, massive Legendre operator
$\frac{1}{2}(1-r^{2})(\nabla^{2}-m^{2})-r\partial_{r}-D$ and the operator
$-\frac{1}{2}m^{2}G^{B}_{m}$, that we further discuss in Sect. 5.3.

Note that

$(\Phi,K_{m}^{B}\Psi)=(\Phi,K_{m}\Psi)\,,\quad\Phi\in H_{m}\,,\Psi\in
D(K_{m}^{B})\,,$

thus

$(\Phi,K^{B}_{m}\Psi)=-(K^{B}_{m}\Phi,\Psi)\,,\quad\Phi,\Psi\in
D(K_{m}^{B})\,,$

namely $K^{B}_{m}$ is skew-Hermitian on its domain.

######  Proposition 5.5.

$K_{m}^{B}$ is skew-selfadjoint on $H_{m}(B)$, $m\geq 0$.

  * Proof.

$K_{0}^{B}$ is skew-selfadjoint on $H_{0}(B)$ by Corollary 4.10. Moreover

$K^{B}_{m}=K_{0}^{B}-P_{m}-V^{B}_{m}$ (51)

with

$P_{m}=m^{2}\left[\begin{matrix}0&0\\\
\frac{1}{2}(1-r^{2})&0\end{matrix}\right]\,,\quad
V^{B}_{m}=\frac{1}{2}m^{2}\left[\begin{matrix}0&0\\\
G^{B}_{m}&0\end{matrix}\right]\,,$ (52)

and $P_{m}$, $V^{B}_{m}$ are compact operators $H_{m}(B)\to H_{m}(B)$ by
Proposition 5.6.

We now apply Proposition 2.1. Since $K_{0}^{B}$ is skew-selfadjoint in
$H_{0}(B)$ and $K^{B}_{m}$ is skew-Hermitian in $H_{m}(B)$, it follows that
$K^{B}_{m}$ is skew-selfadjoint in $H_{m}(B)$.  $\square$

###  5.3 More on the operator $G^{B}_{m}$

With $m>0$, we have considered the operator

$G^{B}_{m}=E_{m}^{-}\cdot(-\nabla^{2}+m^{2})^{-1}|_{H_{m}^{1/2}(B)}:H_{m}^{1/2}(B)\to
H_{m}^{-1/2}(B)\,.$ (53)

The operator $V^{B}_{m}$ in (52) acts on $H_{m}(B)$ and is a term in the
expression for $K^{B}_{m}$.

Let $f,g\in C_{0}^{\infty}(B)\subset H_{m}^{-1/2}(B)$ and $\Lambda$ be the
embedding $H_{m}^{-1/2}(B)\hookrightarrow H_{m}^{-3/2}(B)$. We have

$(g,G_{m}^{B}f)_{-1/2}=(g,\mu_{m}^{-2}f)_{-1/2}=(\Lambda g,\Lambda
f)_{-3/2}=(g,\Lambda^{*}\Lambda f)_{-1/2}$

thus $G_{m}^{B}=\Lambda^{*}\Lambda$ as operator $H_{m}^{-1/2}(B)\to
H_{m}^{-1/2}(B)$, in particular it is positive selfadjoint.

In the following proposition, the multiplication operator by
$\frac{1}{2}(1-r^{2})$ acting on different spaces is denoted by the same
symbol $M$.

######  Proposition 5.6.

$M_{m}:H_{m}^{1/2}(B)\to H_{m}^{-1/2}(B)$ and $G_{m}^{B}:H_{m}^{1/2}(B)\to
H_{m}^{-1/2}(B)$ are compact operators. Therefore, both the operators $P_{m}$
and $V^{B}_{m}$ given by (52) are compact on $H_{m}(B)$.

  * Proof.

First we show that the multiplication operator
$\frac{1}{2}(1-r^{2}):H_{m}^{-1/2}(B)\to H_{m}^{-1/2}(B)$ is bounded. Let $h$
be a real, smooth, compactly supported function on $\mathbb{R}^{d}$ that
agrees with $\frac{1}{2}(1-r^{2})$ on $B$. Then $\frac{1}{2}(1-r^{2})f=hf$ if
$h\in H_{m}^{-1/2}(B)$. As the multiplication by $h$ is a bounded operator
$H_{m}^{s}\to H_{m}^{s}$ [36], we see that
$\frac{1}{2}(1-r^{2}):H_{m}^{s}(B)\to H_{m}^{s}(B)$ is bounded for every
$s\in{\mathbb{R}}$, in particular with $s=-1/2$.

Now, we have a factorization

${H_{m}^{1/2}(B)}$${H_{m}^{-1/2}(B)}$${H_{m}^{-1/2}(B)}$$\scriptstyle{M_{m}}$$\scriptstyle{\\!\\!\frac{1}{2}(1-r^{2})}$

where the upright arrow operator is bounded and the vertical embedding is
compact by Rellich’s theorem.

Similarly, $G_{m}^{B}$ factors as:

${H_{m}^{1/2}(B)}$${H_{m}^{-1/2}(B)}$${H_{m}^{-1/2}(B)}$$\scriptstyle{G_{m}^{B}}$$\scriptstyle{\Lambda^{*}\Lambda}$

where the both the embeddings $H_{m}^{1/2}(B)\hookrightarrow H_{m}^{-1/2}(B)$
and $\Lambda$ are compact by Rellich’s theorem, thus also $\Lambda^{*}\Lambda$
is compact.  $\square$

As $\mu_{m}^{-2}=-(\nabla^{2}-m^{2})^{-1}:L^{2}({\mathbb{R}}^{d})\to
L^{2}({\mathbb{R}}^{d})$, we have

$(\mu_{m}^{-2}f)({\bf x})=\int_{{\mathbb{R}}^{d}}{\rm G}_{m}({\bf x}-{\bf
y})f({\bf y})d{\bf y}\,,\quad f\in{\mathcal{S}}\,,$

with ${\rm G}_{m}$ the Green function for the Helmholtz operator

${\rm G}_{m}({\bf
x})=\frac{1}{(2\pi)^{d/2}}\left(\frac{m}{r}\right)^{d/2-1}{\rm
K}_{d/2-1}(mr)\,,$ (54)

where ${\rm K}_{\alpha}$ is a modified Bessel function of the second kind. We
may denote $\mu_{m}^{-2}$ also by $G_{m}$.

Now, if $\Phi_{k}=w_{m}(f_{k},g_{k})\in{\mathcal{T}}_{m}(B)$, we have

$(\Phi_{1},V^{B}_{m}\Phi_{2})=(\Phi_{1},V_{m}\Phi_{2})$

with

$V_{m}=\left[\begin{matrix}0&0\\\ G_{m}&0\end{matrix}\right]\,,$ (55)

therefore

$\Re(\Phi_{1},\imath_{m}V^{B}_{m}\Phi_{2})=-\Im(\Phi_{1},V_{m}\Phi_{2})=(f_{1},G_{m}f_{2})_{L^{2}}=\int_{B\times
B}{\rm G}_{m}({\bf x}-{\bf y})f_{1}({\bf x})f_{2}({\bf y})d{\bf x}d{\bf y}\,.$
(56)

###  5.4 The operators $K_{m}$ and $A_{m}=-\imath_{m}K_{m}$

We now define the complex linear operator $K_{m}$ on ${\mathcal{H}}_{m}$ by
extending $K_{m}^{B}$ by complex linearity, namely

$K_{m}(\Phi+\imath_{m}\Psi)=K_{m}^{B}\Phi+\imath_{m}K_{m}^{B}\Psi\,,\
\Phi,\Psi\in D(K_{m}^{B})\,,$

so

$D(K_{m})=D(K_{m}^{B})+\imath_{m}D(K_{m}^{B})\,.$

We shall see that $A_{m}=\imath_{m}K_{m}$ is proportional to the mass $m$
local modular Hamiltonian $\log\Delta_{B}$.

######  Proposition 5.7.

$A_{m}$ is essentially selfadjoint on $D(K_{m})$. Moreover,

$e^{isA_{m}}H_{m}(B)=H_{m}(B)\,,\quad s\in{\mathbb{R}}\,.$

  * Proof.

Immediate by Corollary 2.3.  $\square$

We now give an explicit formula for the quadratic form on $H_{m}(B)$
associated with $A_{m}$, in terms of the stress-energy tensor and a Green
integral operator with Yukawa kernel.

With $A_{m}=-\imath_{m}K_{m}$, note first that

$(\Phi,A_{m}\Phi)=\Im
i(\Phi,A_{m}\Phi)=\Im(\Phi,\imath_{m}A_{m}\Phi)=\Im(\Phi,K_{m}\Phi)\\\
=\beta(\Phi,K_{m}\Phi)=\beta(\Phi,K^{B}_{m}\Phi)=\beta(\Phi,\widetilde{K}_{m}\Phi)\,,$
(57)

if $\Phi\in H_{m}(B)$ is in the domain of $A_{m}$.

######  Theorem 5.8.

If $\Phi\in{\mathcal{T}}_{m}(B)$, we have

$(\Phi,A_{m}\Phi)=\frac{1}{2}\int_{B}(1-r^{2})\langle
T^{(m)}_{00}\rangle_{\Phi}d{\bf x}+\frac{D}{2}\int_{B}\Phi^{2}d{\bf
x}+\frac{1}{2}m^{2}\int_{B\times B}G_{m}({\bf x}-{\bf y})\Phi({\bf
x})\Phi({\bf y})d{\bf x}$ (58)

($t=0$ integrals) with $\langle
T^{(m)}_{00}\rangle_{\Phi}=\frac{1}{2}\big{(}(\partial_{0}\Phi)^{2}+|\nabla\Phi|^{2}+m^{2}\Phi^{2}\big{)}$
the energy density given by the classical, massive stress-energy tensor
$T^{(m)}$.

  * Proof.

Let $\Phi_{m}=w_{m}(f,g)\in{\mathcal{T}}_{m}(B)$. Since
$K^{B}_{m}=K_{0}^{B}-P_{m}-V^{B}_{m}$ and the symplectic form $\beta$ is mass
independent, by the identities (57) we have

$(\Phi_{m},A_{m}\Phi_{m})=\beta(\Phi_{m},\widetilde{K}_{m}\Phi_{m})=\beta(\Phi_{0},K_{0}\Phi_{0})-\beta(\Phi_{m},P_{m}\Phi_{m})-\beta(\Phi_{m},V_{m}\Phi_{m})\,,$

so the theorem follows by the $m=0$ case, Prop. 4.5, eq. (56) and the identity

$-\beta(\Phi_{m},P_{m}\Phi_{m})=\frac{m^{2}}{2}\int_{B}(1-r^{2})\Phi_{m}^{2}\,d{\bf
x}\,.$

$\square$

###  5.5 The massive modular local Hamiltonian

This section contains our main result Theorem 5.13.

Denote by $P_{m}$ the cutting projection on ${\mathcal{H}}_{m}$ relative to
$H_{m}(B)$.

######  Proposition 5.9.

$P_{m}$ is given by the matrix

$P_{m}=\left[\begin{matrix}P_{+}&0\\\ 0&P_{-}\end{matrix}\right]\,,$ (59)

with $P_{\pm}:D(P_{\pm})\subset H_{m}^{\pm 1/2}\to H_{m}^{\pm 1/2}$ the
operator of multiplication by the characteristic function $\chi_{B}$ of $B$ in
$H_{m}^{\pm 1/2}$.

  * Proof.

By duality (33), $H_{m}(B)^{\prime}=H_{m}(B^{\prime})$, therefore
$D(P_{m})=H_{m}(B)+H_{m}(B^{\prime})$.

By Prop. 2.5, $P_{m}\imath_{m}|_{H_{m}(B)}$ is a real linear, densely defined
operator on $H_{m}(B)$. Clearly $P_{m}$ acts as in (59) on vectors in
$H_{m}(B)+H_{m}(B^{\prime})$ given by smooth functions, hence on all
$D(P_{m})$ as $P_{m}$ is a closed operator [11].  $\square$

By Prop. 2.5, $P_{m}\imath_{m}|_{H_{m}(B)}$ is a real linear, densely defined
operator on $H_{m}(B)$.

######  Proposition 5.10.

$(P_{m}\imath_{m}-P_{0}\imath_{0})|_{H_{m}(B)}$ is a compact operator on
$H_{m}(B)$, $m\geq 0$.

  * Proof.

By (59), we have

$P_{m}\imath_{m}=\left[\begin{matrix}0&P_{+}\mu_{m}^{-1}\\\
-P_{\\!-}\mu_{m}&0\end{matrix}\right]\,.$

As $P_{m}\imath_{m}$ is skew-selfadjoint, it suffices to show that
$P_{-}\mu_{m}-P_{-}\mu_{0}:H_{m}^{1/2}(B)\to H_{m}^{-1/2}(B)$ is compact.
Namely, that

$f\in H_{m}^{1/2}(B)\mapsto(\mu_{m}-\mu_{0})f|_{B}\in H_{m}^{-1/2}(B)$

is compact. Now, in Fourier transform,

$\big{(}(\mu_{m}-\mu_{0})f\big{)}{}^{\widehat{\phantom{x}}}({\bf
p})=\big{(}\sqrt{|{\bf p}|^{2}+m^{2}}-\sqrt{|{\bf p}|^{2}}\big{)}\hat{f}({\bf
p})=\frac{m^{2}}{\sqrt{|{\bf p}|^{2}+m^{2}}+\sqrt{|{\bf p}|^{2}}}\hat{f}({\bf
p})$ (60)

so $(\mu_{m}-\mu_{0}):H_{m}^{1/2}\to H_{m}^{1/2}$ is a bounded operator. We
have the following diagram

${H_{m}^{1/2}(B)}$${H_{m}^{-1/2}(B)}$${H_{m}^{1/2}}$${H_{m}^{0}=L^{2}({\mathbb{R}}^{d})}$${H_{m}^{0}(B)}$$\scriptstyle{P_{-}\mu_{m}\,-\,P_{-}\mu_{0}}$$\scriptstyle{\mu_{m}-\mu_{0}}$$\scriptstyle{\iota_{1}}$$\scriptstyle{\chi_{B}}$$\scriptstyle{\iota_{2}}$
(61)

where $\chi_{B}$ is the multiplication by the characteristic function of $B$
in $L^{2}({\mathbb{R}}^{d})$, i.e. the orthogonal projection
$L^{2}({\mathbb{R}}^{d})\to L^{2}(B)$, and $\iota_{1}$, $\iota_{2}$ are
natural embeddings. Therefore $P_{-}\mu_{m}-P_{-}\mu_{0}:H_{m}^{1/2}\to
H_{m}^{-1/2}$ is compact because all the operators in the diagram are bounded
and the embedding $\iota_{2}:H_{m}^{0}(B)\hookrightarrow H_{m}^{-1/2}(B)$ is
compact by Rellich’s theorem.  $\square$

######  Lemma 5.11.

The point spectrum of $\Delta_{B}=\Delta_{B,m}$ is empty.

  * Proof.

It was proved by Figliolini and Guido [15, Prop. 3.5] that $1$ is not
eigenvalue of $(\Delta_{B}+1)(\Delta_{B}-1)^{-1}$, and the same argument in
that proof shows that no $\lambda\in\mathbb{R}$ can be an eigenvalue, implying
the same for $\Delta_{B}$.  $\square$

We mention that Lemma 5.11 implies that the promotion of $\Delta_{B}$ on the
Bose Fock space has no point spectrum but 1 with multiplicity one. Then the
vacuum modular group of acts ergodically on the second quantisation von
Neumann algebra ${\mathcal{A}}_{m}(B)$ the Bose Fock space over
${\mathcal{H}}_{m}$. Therefore, ${\mathcal{A}}_{m}(B)$ is a factor of type
$III_{1}$ [12], see [1, 15].

The massive modular Hamiltonian can be characterised as “minimal” perturbation
of the massless modular Hamiltonian in the sense of Proposition 5.12 or
Corollary 5.14.

######  Proposition 5.12.

$K_{m}=\imath_{m}A_{m}$ is the unique skew-selfadjoint operator
$K=\imath_{m}A$ on ${\mathcal{H}}_{m}$, leaving $H_{m}(B)$ invariant (as in
Prop. (2.2)), such that

$\imath_{m}f_{\lambda}(A)|_{H_{m}(B)}-\imath_{0}f_{\lambda}(A_{0})|_{H_{m}(B)}$
(62)

is a compact operator on $H_{m}(B)$ for all $\lambda>0$, with
$f_{\lambda}(x)=e^{-\lambda x^{2}}\tanh(\pi x)$.

  * Proof.

$K^{B}_{m}$ is a compact perturbation of $K^{B}_{0}$ by Proposition 5.6, and
this implies that

$\imath_{m}f_{\lambda}(A_{m})|_{H_{m}(B)}-\imath_{0}f_{\lambda}(A_{0})|_{H_{m}(B)}$
(63)

is also compact by Prop. 2.4, any $\lambda>0$.

On the other hand, assume now that the operator (62) is compact on $H_{m}(B)$.
Then

$T\equiv\imath_{m}f_{\lambda}(A)|_{H_{m}(B)}-\imath_{m}f_{\lambda}(A_{m})|_{H_{m}(B)}$

is also a compact, by taking the side by side difference of equations (62) and
(63).

Since $H_{m}(B)$ is $K$ and $K_{m}$-invariant, both $K$ and $K_{m}$ commute
with $\Delta^{is}_{B,m}$. Thus $\Delta^{is}_{B,m}|_{H_{m}(B)}$ commutes with
the compact operator $T$, so with $T^{*}T$.

Thus, the finite-dimensional eigen-spaces of $T^{*}T$ are left invariant by
$\Delta_{B,m}^{it}$, so the point spectrum of $\Delta_{B,m}$ would be not
empty if $T\neq 0$. By Lemma 5.11 we must then have $T=0$. Therefore, for all
$\lambda>0$, we have $f_{\lambda}(A)=f_{\lambda}(A_{m})$ on $H_{m}(B)$, thus
on ${\mathcal{H}}_{m}$ because $H_{m}(B)+\imath_{m}H_{m}(B)$ is dense and
$f_{\lambda}$ is bounded. Since $e^{-\lambda x^{2}}\nearrow 1$ as
$\lambda\searrow 0$, we conclude that $\tanh(\pi A)=\tanh(\pi A_{m})$, so
$A=A_{m}$ because the hyperbolic tangent is one-to-one.  $\square$

######  Theorem 5.13.

The modular Hamiltonian $\log\Delta_{B}$ associated with the unit ball $B$ in
the free scalar, mass $m$ quantum field theory is given by

$-2\pi A_{m}=\log\Delta_{B,m}\,.$

Thus, on $H_{m}(B)$, we have

$\log\Delta_{B,m}=2\pi\imath_{m}\left[\begin{matrix}0&\frac{1}{2}(1-r^{2})\\\
\frac{1}{2}(1-r^{2})(\nabla^{2}-m^{2})-r\partial_{r}-D-\frac{1}{2}m^{2}G^{B}_{m}&0\end{matrix}\right]$
(64)

where $G_{m}^{B}$ is the operator (53). $G_{m}^{B}$ is given by the Helmholtz
Green integral operator, with Yukawa potential if $d=3$ , see (56).

  * Proof.

Due to Prop. 5.12, it suffices to show that

$\imath_{m}f_{\lambda}\Big{(}\frac{1}{2}\log\Delta_{B,m}\Big{)}\Big{|}_{H_{m}(B)}-\imath_{0}f_{\lambda}\Big{(}\frac{1}{2}\log\Delta_{B,0}\Big{)}\Big{|}_{H_{0}(B)}$
(65)

is compact (where $H_{0}(B)$ and $H_{m}(B)$ are identified), all $\lambda>0$.

Now, formula (16) for the cutting projection and Prop. 5.10 entail that

$\imath_{m}\coth\Big{(}\frac{1}{2}\log\Delta_{B,m}\Big{)}\Big{|}_{H_{m}(B)}-\imath_{0}\coth\Big{(}\frac{1}{2}\log\Delta_{B,0}\Big{)}\Big{|}_{H_{0}(B)}$

is densely defined and compact, thus

$\imath_{m}\tanh\Big{(}\frac{1}{2}\log\Delta_{B,m}\Big{)}\Big{|}_{H_{m}(B)}-\imath_{0}\tanh\Big{(}\frac{1}{2}\log\Delta_{B,0}\Big{)}\Big{|}_{H_{0}(B)}$

is compact by equation (11). Therefore also the operator (65) is compact by
applying Prop. 2.4, because

$e^{-\lambda x^{2}}\\!\tanh(x/2)=g\big{(}\tanh(x/2)\big{)}$

with
$g(x)=ix\exp\big{(}-\frac{\lambda}{4\pi^{2}}\log^{2}\frac{1-x}{1+x}\big{)}$.
$\square$

######  Corollary 5.14.

$A_{m}=-\frac{1}{2\pi}\log\Delta_{B,m}$ is the unique selfadjoint operator $A$
on ${\mathcal{H}}_{m}$, leaving $H_{m}(B)$ invariant, such that
$(\imath_{m}A-\imath_{0}A_{0})|_{H_{m}(B)}$ is densely defined and compact on
$H_{m}(B)$.

  * Proof.

$K^{B}_{m}$ is a compact perturbation of $K^{B}_{0}$ by Prop. 5.6. Conversely,
assume that $K|_{H_{m}(B)}$ is a compact perturbation of $K^{B}_{0}$ as above,
with $K=\imath_{m}A$. Then
$\big{(}\imath_{m}f_{\lambda}(A)|_{H_{m}(B)}-\imath_{0}f_{\lambda}(A_{0})|_{H_{m}(B)}\big{)}$
is compact by Prop. 2.4. So $K=K_{m}$ by Prop. 5.13.  $\square$

It would be interesting to have a direct proof of the following corollary,
that manifests the thermodynamical nature of the modular Hamiltonian, cf.
[31].

######  Corollary 5.15.

$A_{m}$ is completely active with respect to $H_{m}(B)$.

  * Proof.

Immediate by Theorem 2.10.  $\square$

Let’s now consider balls of different radius. If $F$ is a function on
$\mathbb{R}^{n}$ and $\lambda>0$, we set $F_{\lambda}(x)=F(\lambda x)$. If
$\Phi\in{\mathcal{T}}_{m}$, we have

$\big{(}(\square+\lambda^{2}m^{2})\Phi_{\lambda}\big{)}(x)=\lambda^{2}\big{(}(\square\Phi)(\lambda
x)+m^{2}\Phi^{2}(\lambda x)\big{)}=0\,,$

thus $\Phi_{\lambda}\in{\mathcal{T}}_{\lambda m}$ and we have a real linear
map bijection $\delta_{\lambda}:{\mathcal{T}}_{m}\to{\mathcal{T}}_{\lambda m}$

$\delta_{\lambda}:\Phi\in{\mathcal{T}}_{m}\to\lambda^{D}\Phi_{\lambda}\in{\mathcal{T}}_{\lambda
m}\ .$ (66)

In terms of the Cauchy data, we have $\delta_{\lambda}:w_{m}(f,g)\to
w_{\lambda m}(\lambda^{D}f_{\lambda},\lambda^{D+1}g_{\lambda})$.

If $\Phi,\Psi\in{\mathcal{T}}_{m}(B)$

$\beta(\delta_{\lambda}\Phi,\delta_{\lambda}\Psi)=\frac{\lambda^{d}}{2}\int_{x_{0}=0}\Big{(}\Psi(\lambda
x)\big{(}\partial_{0}\Phi\big{)}(\lambda x)-\Phi(\lambda
x)\big{(}\partial_{0}\Psi\big{)}(\lambda x)\Big{)}dx=\beta(\Phi,\Psi)\ ,$

namely $\delta_{\lambda}$ preserves the symplectic form $\beta$. One can see
that $\delta_{\lambda}$ intertwines the complex structure on $\imath_{m}$ and
$\imath_{\lambda m}$, so $\delta_{\lambda}$ is a unitary

$\delta_{\lambda}:{\mathcal{H}}_{m}\to{\mathcal{H}}_{\lambda m}\,.$

Let $B_{R}$ the ball in $\mathbb{R}^{d}$ with center at the origin and radius
$R>0$. With $H_{m}(B_{R})$ the standard subspace of ${\mathcal{H}}_{m}$
associated with $B_{R}$, we have

$\delta_{\lambda}:H_{m}(B_{R})\to H_{\lambda m}(B_{\lambda^{-1}R})\,.$

With $\Delta_{R,m}$ the modular operator of $H_{m}(B_{R})$, we then have
$\Delta_{{\lambda^{-1}R},\lambda
m}=\delta_{\lambda}\Delta_{R,m}\delta_{\lambda^{-1}}$, so

$\log\Delta_{m,R}=\delta_{R}\log\Delta_{Rm,1}\delta_{R^{-1}}\,.$ (67)

Now, ${\rm G}_{m}$ is given by (54), so we have

$G_{\lambda m}({\bf x})=\frac{1}{(2\pi)^{d/2}}\left(\frac{\lambda
m}{r}\right)^{d/2-1}\\!\\!\cdot{\rm K}_{d/2-1}(m\lambda
r)=\lambda^{\left(\frac{d-1}{2}\right)^{2}}G_{m}(\lambda{\bf x})\,.$

In particular, if $d=3$, we have ${\rm
K}_{1/2}(r)=\left(\frac{\pi}{2r}\right)^{1/2}e^{-r}$ and

${\rm G}_{m}({\bf x})=\frac{1}{4\pi}\frac{e^{-mr}}{r}$

see e.g. [36, Chapter 3]. ${\rm G}_{m}$ is the Yukawa potential.

The above discussion yields the following formula for the matrix elements of
the modular Hamiltonian.

######  Theorem 5.16.

With $\log\Delta_{R,m}$ the mass $m$ modular Hamiltonian for the radius $R$
ball $B_{R}$, we have

$-\Re(\Phi,\log\Delta_{R,m}\Psi)=\\\
2\pi\int_{B_{R}}\frac{R^{2}-r^{2}}{2R}\langle
T^{(m)}_{00}\rangle_{\Phi,\Psi}d{\bf x}+2\pi\frac{d-1}{R}\int_{B_{R}}\Phi\Psi
d{\bf x}+m^{2}\frac{\pi}{R}\int_{B_{R}\times B_{R}}G_{m}({\bf x}-{\bf
y})\Phi({\bf y})\Psi({\bf x})d{\bf x}d{\bf y}\,,$

where $\Phi,\Psi\in{\mathcal{T}}_{m}(B_{R})$.

  * Proof.

Immediate by the above discussion.  $\square$

######  Remark 5.17.

We make a comment on the locality of the massive Hamiltonian. The skew-
selfadjoint modular Hamiltonian on $H_{m}(B)$ is proportional to $K_{m}^{B}$
and the only non-local term in $K_{m}^{B}$ is $V_{m}$. Thus $K_{m}^{B}$ acts
locally of $\langle f,g\rangle$ iff $G^{B}_{m}$ acts locally on $f$.

Now, $\nabla^{2}-m^{2}$ is a local operator, thus $G_{m}^{B}$ acts locally on
$(\nabla^{2}-m^{2})C_{0}^{\infty}(B)$. Call $T$ the closure of the operator
$-\nabla^{2}+m^{2}|_{C_{0}^{\infty}(B)}$ as an operator on $L^{2}(B)\to
L^{2}(B)$). Then $T$ is a closed, positive Hermitian operator, with lower
bound $m^{2}$, and $T$ is local on its domain. The range $R(T)$ of $T$ is
closed in $L^{2}(B)$, thus $R(T)={\rm ker}(T^{*})^{\perp}$.

Now $T^{*}:D(T^{*})\subset L^{2}(B)\to L^{2}(B)$ is the operator
$-\nabla^{2}+m^{2}$ with domain $D(T^{*})=\\{f\in
L^{2}(B):(-\nabla^{2}+m^{2})f\in L^{2}(B)\\}$.

With $\\{\varphi_{k}\\}$ a basis for ${\rm ker}(T^{*})$, we conclude that
$G^{B}_{m}$ is local on the orthogonal complement of
$\varphi_{1},\varphi_{2},\dots$ in $L^{2}(B)$. That is, if $\langle
f,g\rangle\in D(K^{B}_{m})\cap H_{m}(\widetilde{B})$, with $\widetilde{B}$ a
closed ball contained in $B$, we have

$K^{B}_{m}\langle f,g\rangle\in
H_{m}(\widetilde{B})\Longleftrightarrow\int_{B}f({\bf x})\varphi_{k}({\bf
x})d{\bf x}=0\,,\quad\forall k\,,$

(massive case) namely iff $f$ has zero mean on $B$ with respect to every
$\varphi_{k}$.

##  6 Entropy density of a wave packet

With ${\mathcal{H}}_{m}$ our wave Hilbert space, the following proposition
gives an explicit formula for the action of the cutting projection $P_{m}$ on
waves.

######  Proposition 6.1.

Let $\Phi\in{\mathcal{T}}_{m}$ and set $\Psi=K_{m}\Phi$. Then
$\Psi=w_{m}(f,g)\in H_{m}(B)+H_{m}(B^{\prime})$ and
$P_{m}\Psi=w_{m}(\chi_{B}f,\chi_{B}g)$, with $\chi_{B}$ the characteristic
function of $B$.

  * Proof.

As $\Phi$ belongs to $D(\log\Delta_{B})=D(K_{m})$, it follows that
$K_{m}\Phi\in D(P_{m})$, thus $P_{m}\Psi=w_{m}(\chi_{B}f,\chi_{B}g)$ by Prop.
5.9.  $\square$

We can now compute the local entropy of a wave packet $\Phi$.

######  Proposition 6.2.

Let $\Phi\in{\mathcal{T}}_{m}$. The entropy $S_{\Phi}$ of $\Phi$ with respect
to $H_{m}(B)$ is given by

$S_{\Phi}=\pi\int_{B}\big{(}\Psi\Phi^{\prime}-\Phi\Psi^{\prime}\big{)}d{\bf
x}\ ,$ (68)

with $\Psi=K_{m}\Phi$ and the prime denotes the time derivative.

  * Proof.

The proof now follows from Proposition 6.1 in analogy with the the proof given
in [11] for the wedge region case.  $\square$

######  Theorem 6.3.

The entropy $S_{\Phi}$ of the Klein-Gordon wave $\Phi\in{\mathcal{T}}_{m}$ in
the unit ball $B$, i.e. with respect to $H_{m}(B)$, is given by

$S_{\Phi}=2\pi\int_{B}\frac{1-r^{2}}{2}\langle
T^{(m)}_{00}\rangle_{\Phi}\,d{\bf x}+\pi D\int_{B}\Phi^{2}d{\bf x}+\pi
m^{2}\int_{B\times B}G_{m}({\bf x}-{\bf y})\Phi({\bf y})\Phi({\bf x})d{\bf
x}d{\bf y}\,.$ (69)

In particular, $S_{\Phi}$ is finite if $\Phi\in{\mathcal{T}}_{m}$.

  * Proof.

With $\Phi,\Psi\in{\mathcal{T}}_{m}$, define the real linear quadratic form

$q_{B}(\Phi,\Psi)=2\pi\int_{B}\frac{1-r^{2}}{2}\langle
T^{(m)}_{00}\rangle_{\Phi,\Psi}\,d{\bf x}+\pi D\int_{B}\Phi\Psi d{\bf x}+\pi
m^{2}\int_{B\times B}G_{m}({\bf x}-{\bf y})\Phi({\bf y})\Psi({\bf x})d{\bf
x}d{\bf y}\,.$

Clearly, $q_{B}(\Phi,\Phi)<\infty$ if $\Phi\in{\mathcal{T}}_{m}$. We want to
show that

$S(\Phi,\Phi)=q_{B}(\Phi,\Phi)\,,\quad\Phi\in{\mathcal{T}}_{m}\,,$ (70)

where $S$ is the entropy form with respect to $H_{m}(B)$. As $S$ is a closed
by Prop. 2.7, it suffices to show that $q_{B}$ is closable on
${\mathcal{T}}_{m}$ and that (70) holds on a form core for $S$.

Now, $q_{B}$ is given by

$q_{B}(\Phi,\Psi)=\Re(\Phi,P^{*}_{m}A_{m}P_{m}\Psi)\,,\quad\Phi,\Psi\in{\mathcal{T}}_{m}\,,$

similarly as in [11, Thm. 3.5]. The real linear operator $P^{*}_{m}A_{m}P_{m}$
is Hermitian, thus closable. As $q_{B}$ is positive, $q_{B}$ is closable, cf.
the proof of [20, Thm. 1.27].

On the other hand, (70) holds if $\Phi\in{\mathcal{T}}_{m}(B)$ by Theorem 5.8.
Then it holds if $\Phi\in D(\log\Delta_{B})$ by the same argument, using Prop.
6.2. As $D(\log\Delta_{B})$ is a form core for $S$ (see Sect. 2.4), we
conclude that $S=q_{B}$ on ${\mathcal{T}}_{m}$.  $\square$

Denote by $B_{R}(\bar{\bf x})$ the radius $R$ space ball around the point
$\bar{\bf x}\in\mathbb{R}^{d}$.

######  Corollary 6.4.

The entropy $S_{\Phi}(R)=S_{\Phi}(R,t,\bar{\bf x})$ of the wave packet
$\Phi\in{\mathcal{T}}_{m}$ in the space region $B_{R}(\bar{\bf x})$ at time
$t$ is given by

$S_{\Phi}(R)=\pi\int_{B_{R}(\bar{\bf x})}\frac{R^{2}-r^{2}}{R}\langle
T^{(m)}_{00}\rangle_{\Phi}\,d{\bf x}\\\ +\frac{\pi D}{R}\int_{B_{R}(\bar{\bf
x})}\Phi^{2}d{\bf x}+\frac{\pi m^{2}}{R}\int_{{\bf x},{\bf y}\in
B_{R}(\bar{\bf x})}G_{m}({\bf x}-{\bf y})\Phi({\bf y})\Phi({\bf x})d{\bf
x}d{\bf y}\,,$

with $r=|{\bf x}-\bar{\bf x}|$, ($x_{0}=t$ integral).

  * Proof.

In view of Theorem 5.16 and the formula for the cutting projection in
Proposition 6.1, the theorem follows, in the time zero case, by the entropy
formula (19). By translation covariance, we get the formula at arbitrary time.
$\square$

$S$$\Phi$$\Phi$$r$$\frac{1}{r}$$0$$1$ Figure 1: The spherical massless wave
$\Phi(r)=\frac{\sin(r-t)}{r}$, $d=3$, at time $t=0$ and $t=1$ ($\Phi$ is not
everywhere defined). In this example, the entropy densities per area $S(r)$
are close at different times.

For large $R$, $S_{\Phi}(R)/R$ gets proportional to the total local energy
$E=\int_{B_{R}(\bar{\bf x})}\langle T_{00}\rangle_{\Phi}(t,{\bf x})d{\bf x}$:

$\frac{S_{\Phi}(R)}{R}\sim\pi E\,,$

as $R\to\infty$ like in [24, (41)], see also [28].

It follows that, at fixed time $t$, the local entropy $S_{\Phi}(R,{\bf
x})=S_{\Phi}(R,t,{\bf x})$ has an expansion

$\displaystyle S_{\Phi}(R,{\bf x})$ $\displaystyle=\pi\left(\langle
T_{00}\rangle_{\Phi}(t,{\bf x})+D\Phi^{2}(t,{\bf x})\right)V_{d}\,R^{d-1}+...$
$\displaystyle=\frac{\pi}{d}\left(\langle T_{00}\rangle_{\Phi}(t,{\bf
x})+D\Phi^{2}(t,{\bf x})\right)A_{d-1}(R)+...$

as $R\to 0$; here $A_{d-1}(R)=2\frac{\pi^{d/2}}{\Gamma(d/2)}R^{d-1}$ is the
area of the $(d-1)$-dimensional sphere $\partial B_{R}$ and
$V_{d}=A_{d-1}(1)/d$ is the volume of $B$. So the energy density of the wave
packet $\Phi$ around a point gets proportional to the area of the sphere
boundary of $B_{R}$, as expected by the holographic area theorems for the
entropy, in black hole and other contexts, see [5].

Note that the leading coefficient of entropy density
$S_{\Phi}(R,x)/A_{d-1}(R)$ is the sum of two terms, that are proportional one
to the energy $\langle T_{00}\rangle_{\Phi}(x)$ and the other to the square of
the height $\Phi^{2}(x)$ of the wave packet at the point $x$. In other words,
it depends quadratically on the height and on the space and time slopes of
$\Phi$.

As $R\to\infty$, we then have the asymptotic bound for the local entropy

$S_{\Phi}(R)\leq\pi ER$

with $E$ the total energy in the ball $B_{R}$, namely the Bekenstein bound
holds, see [28] and references therein.

###  6.1 Quantum Field Theory

By the analysis in [25, 11], we have an immediate corollary in Quantum Field
Theory concerning the local vacuum relative entropy of a coherent state.

######  Corollary 6.5.

Let ${\mathcal{A}}_{m}(O_{R})$ be the von Neumann algebra associated with the
double cone $O_{R}$ (the causal envelope of $B_{R}$) by the free, neutral
quantum field theory with mass $m\geq 0$. The relative entropy
$S(\varphi_{\Phi}|\\!|\varphi)$ (see [2]) between the vacuum state $\varphi$
and the coherent state $\varphi_{\Phi}$ associated with the one-particle wave
$\Phi\in{\mathcal{H}}_{m}$ is given by $S_{\Phi}(R)$ by Corollary 6.4 (with
$B_{R}$ centred at the origin).

  * Proof.

As shown in [11], $S(\varphi_{\Phi}|\\!|\varphi)$ is equal to the entropy of
the vector $\Phi\in{\mathcal{H}}_{m}$ with respect to the standard subspace
$H_{m}(O_{R})$. So Corollary 6.4 applies.  $\square$

##  7 Appendix. Elementary relations

For the reader’s convenience, we collect a couple of elementary identities
that are used in the text. Note first that $\partial_{r}$ is the partial
derivative in the direction $\vec{r}=(\frac{x_{1}}{r},\dots\frac{x_{d}}{r})$,
thus

$r\partial_{r}f={\bm{x}}\cdot{\nabla}f\,,$ (71)

for any $f\in{\cal S}$.

######  Identity 7.1.

Let $f\in{\mathcal{S}}$. We have

$\int_{\mathbb{R}^{d}}\frac{1}{2}(1-r^{2})|\nabla f|^{2}d{\bf
x}=-\int_{\mathbb{R}^{d}}\frac{1}{2}(1-r^{2})f\nabla^{2}fd{\bf
x}+\int_{\mathbb{R}^{d}}rf\partial_{r}f\,d{\bf x}\,.$

  * Proof.

The identity follows immediately by the following two relations:

$-\int_{\mathbb{R}^{d}}\frac{1}{2}(1-r^{2})f\nabla^{2}fd{\bf
x}=\int_{\mathbb{R}^{d}}\frac{1}{2}(1-r^{2})|\nabla f|^{2}d{\bf
x}+\frac{d}{2}\int_{\mathbb{R}^{d}}f^{2}d{\bf x}\ ,$ (72)
$\int_{\mathbb{R}^{d}}rf\partial_{r}f\,d{\bf
x}=-\frac{d}{2}\int_{\mathbb{R}^{d}}f^{2}d{\bf x}\ .$ (73)

Concerning the second relation, by (71) we have

$\int_{\mathbb{R}^{d}}rf\partial_{r}f\,d{\bf
x}=\sum_{k}\int_{\mathbb{R}^{d}}x_{k}f\partial_{k}f\,d{\bf
x}=\frac{1}{2}\sum_{k}\int_{\mathbb{R}^{d}}x_{k}\partial_{k}(f^{2})d{\bf x}\\\
=-\frac{1}{2}\sum_{k}\int_{\mathbb{R}^{d}}f^{2}d{\bf
x}=-\frac{d}{2}\int_{\mathbb{R}^{d}}f^{2}d{\bf x}\ .$

Then the first relation follows by

$\displaystyle\int_{\mathbb{R}^{d}}(1-r^{2})f\nabla^{2}fd{\bf x}$
$\displaystyle=-\sum_{k}\int_{\mathbb{R}^{d}}\partial_{k}\big{(}(1-r^{2})f\big{)}\partial_{k}f\,d{\bf
x}$ $\displaystyle=2\sum_{k}\int_{\mathbb{R}^{d}}x_{k}f\partial_{k}f\,d{\bf
x}-\sum_{k}\int_{\mathbb{R}^{d}}(1-r^{2})(\partial_{k}f)^{2}d{\bf x}$
$\displaystyle=-d\int_{\mathbb{R}^{d}}f^{2}d{\bf
x}-\int_{\mathbb{R}^{d}}(1-r^{2})|\nabla f|^{2}d{\bf x}\ .$

$\square$

######  Identity 7.2.

With $Z$ and $\gamma$ as in Section 3, we have

$Z^{\prime}(z,s)|_{s=0}=(1-z^{2})/2\,,$
$\gamma^{\prime}(u,v;0)=-\frac{D}{2}(u+v)=-D\,x_{0}\,.$

  * Proof.

Denoting by a prime the derivative with respect to the $s$-parameter, we have

$f(z,0)=1,\quad
f^{\prime}(z,0)=-\frac{e^{-s}(1-z)}{2}\big{|}_{s=0}=-\frac{1-z}{2}\,,$
$g(z,0)=z,\quad g^{\prime}(z,s)=\frac{e^{-s}(1-z)}{2}|_{s=0}=\frac{1-z}{2}\,.$

Since $Z=g/f$, we get

$Z^{\prime}(z,s)|_{s=0}=(1-z^{2})/2\,.$

We have

$\gamma^{\prime}(u,v;s)\big{|}_{s=0}=\big{(}F(u,s)F(-v,-s)\big{)}^{\prime}\big{|}_{s=0}=F^{\prime}(u,s)\big{|}_{s=0}-F^{\prime}(-v,-s)\big{|}_{s=0}\
.$

Since

$F^{\prime}(z,s)\big{|}_{s=0}=-d\big{(}f(z,s)\big{)}^{-D-1}f^{\prime}(z,s)\big{|}_{s=0}=\frac{D}{2}(1-z)\
,$

we have

$\gamma^{\prime}(u,v;0)=-\frac{D}{2}(u+v)=-D\,x_{0}\,.$

$\square$

Acknowledgements. An initial part of this work was done in June 2019 during
program “Operator Algebras and Quantum Physics” at the Simons Center for
Geometry and Physics at Stony Brook. We thank D. Buchholz for conversations.

We acknowledge the MIUR Excellence Department Project awarded to the
Department of Mathematics, University of Rome Tor Vergata, CUP
E83C18000100006.

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