# Zeta-regularized Lattice Field Theory with Lorentzian background metrics

Tobias Hartung  Department of Mathematical Sciences, University of Bath, 4
West, Claverton Down, Bath, BA2 7AY, United Kingdom
and
Computation-based Science and Technology Research Center, The Cyprus
Institute, 20 Kavafi Street, 2121 Nicosia, Cyprus  ,  Karl Jansen  NIC, DESY
Zeuthen, Platanenallee 6, 15738 Zeuthen, Germany  and  Chiara Sarti  Computer
Laboratory, University of Cambridge, William Gates Building, 15 JJ Thomson
Avenue, Cambridge, CB3 0FD, United Kingdom

###### Abstract.

Lattice field theory is a very powerful tool to study Feynman’s path integral
non-perturbatively. However, it usually requires Euclidean background metrics
to be well-defined. On the other hand, a recently developed regularization
scheme based on Fourier integral operator $\zeta$-functions can treat
Feynman’s path integral non-pertubatively in Lorentzian background metrics. In
this article, we formally $\zeta$-regularize lattice theories with Lorentzian
backgrounds and identify conditions for the Fourier integral operator
$\zeta$-function regularization to be applicable. Furthermore, we show that
the classical limit of the $\zeta$-regularized theory is independent of the
regularization. Finally, we consider the harmonic oscillator as an explicit
example. We discuss multiple options for the regularization and analytically
show that they all reproduce the correct ground state energy on the lattice
and in the continuum limit. Additionally, we solve the harmonic oscillator on
the lattice in Minkowski background numerically.

##  1\. Introduction

###  1.1. Lattice Field Theory

Feynman’s path integral feynman; feynman-hibbs-styer approach to quantum
mechanics and quantum field theory has become an indispensable conceptual tool
to evaluate physical observables, in particular in high energy physics. One
very fascinating aspect of the path integral formulation is that it allows in
principle to compute also non-perturbative phenomena and thus reaches out to
problems that can not be addressed in perturbation theory.

The major obstacle to employ the path integral for studying non-perturbative
effects is, however, that it is in general ill defined, if we take it at face
value, i.e., using a Lorentzian background metric. An elegant way out is to
analytically continue to imaginary – Euclidean – time and discretize the
system either on a time lattice in quantum mechanics or on a space-time grid
in quantum field theory. The so introduced lattice spacing acts then as an
ultraviolet regulator, whereas a finite time or box length serves as an
infrared regulator resulting thus in completely regularized version of the
path integral, see Gattringer:2010zz; Rothe:1992nt for introductions to the so
obtained lattice field theories.

Another important advantage of the Euclidean lattice regulated path integral
is its resemblance to systems in statistical physics where it corresponds to
the partition function. This allows in particular to resort to numerical
calculations using Markov Chain Monte Carlo (MCMC) methods to evaluate the
path integral. This approach has been employed very successfully in the last
years to obtain many non-perturbative results mainly for quantum
chromodynamics, the theory of the strong interaction between quarks and
gluons. In particular, it became possible to perform an ab initio calculation
of the low-lying baryon spectrum using only QCD as the underlying theory
Durr:2008zz. Furthermore, lattice QCD calculations led to a detailed insight
into the structure of hadrons Constantinou:2015agp; Cichy:2018mum and they
have provided non-perturbative contributions to electroweak processes
Meyer:2018til and flavor physics Juettner:2016atf, see e.g. Aoki:2019cca;
kronfeld2012twenty for overviews. They were also very successful to determine
thermodynamic properties Ding:2015ona. The remarkable progress in lattice QCD
simulations can be seen by the fact that nowadays calculations are performed
on large lattices – presently of the order of around $100^{3}\times 200$
lattice points – and directly in physical conditions.

Despite this impressive success, such Euclidean time lattice simulations can
not be employed in important and so far open questions such as the matter
anti-matter asymmetry, topological theories, the discrepancy of the observed
and theoretically predicted amount of charge-parity (CP) violation and non-
equilibrium physics. The reason is that in these cases the integrand in the
path integral becomes complex such that MCMC can not be used.

It would therefore be extremely valuable to have an alternative regularization
of the path integral which allows, on one hand, to address non-perturbative
physics and, on the other hand, to use a Minkowski background metric or even
more general Lorentzian metrics. It is precisely at this point where the
proposed $\zeta$-regularization of the path integral hartung-jmp; hartung-
iwota; hartung-jansen; jansen-hartung is filling this so far missing gap. By
“gauging” the path integral (cf. subsection 1.2), it allows for a fully non-
perturbative regularization for very general metrics. In this way, physical
observables can be computed even in cases where the Euclidean lattice
techniques such as MCMC simulations fail and even on quantum computers
hartung-jmp; hartung-iwota; hartung-jansen; jansen-hartung.

In this article, we aim to extend the mathematical formulation of
$\zeta$-regularization within the context of lattice theories and to
illustrate its application at the example of the quantum mechanical harmonic
oscillator. In particular, we will test different ways of gauging the path
integral and demonstrate that all the here employed choices lead to correct
physical results. In addition, we will show that the $\zeta$-regularization
gives the correct classical limit. Our work will therefore not only provide a
rather simple example of the $\zeta$-regularization on the lattice with
Lorentzian background metric, but it also discusses different classes of
gauges providing thus suitable choices when applied to a particular physical
model of interest.

###  1.2. $\zeta$-regularization

Operator $\zeta$-functions are a means of constructing traces on certain
operator algebras. Given an algebra $\mathcal{A}$ of operators, a trace $\tau$
that is defined on a sub-algebra $\mathcal{A}_{0}\subseteq\mathcal{A}$, and an
operator $A\in\mathcal{A}\setminus\mathcal{A}_{0}$, the aim of operator
$\zeta$-functions is to define $\tau(A)$. To this end, we construct a
holomorphic family $\varphi:\ \mathbb{C}\to\mathcal{A}$ such that $\varphi$
maps an open connected subset of $\mathbb{C}$ into $\mathcal{A}_{0}$ and
$\varphi(0)=A$. Given this setup and sufficient regularity of the operator
algebra and construction of $\varphi$, we can study the operator
$\zeta$-function $\zeta(\varphi)$ which is defined via analytic continuation
of $z\mapsto\tau(\varphi(z))$ on the domain that is mapped to
$\mathcal{A}_{0}$ under $\varphi$. If this analytic extension exists in a
neighborhood of $0$ and the value $\zeta(\varphi)(0)$ is independent of the
choices made in the construction of $\varphi$, then we can use
$\zeta(\varphi)(0)$ as our definition of $\tau(\varphi(0))$, i.e., $\tau(A)$.

For example, if we consider the absolute value of the differential operator
$A:=\left\lvert{\partial}\right\lvert$ on $\mathbb{R}/2\pi\mathbb{Z}$, then
$\mathcal{A}$ is the algebra of classical pseudo-differential operators on
$\mathbb{R}/2\pi\mathbb{Z}$ and $\tau$ the trace on trace-class operators,
i.e., the sum of eigenvalues counting multiplicities. Furthermore, we may
construct $\varphi(z):=A^{z+1}$. Formally evaluating the trace on $\varphi(z)$
thus yields
$\tau(\varphi(z))=\sum_{n\in\mathbb{Z}}\left\lvert{n}\right\lvert^{z+1}=2\sum_{n\in\mathbb{N}}n^{z+1}$.
This makes sense for $\Re(z)<-2$ and implies that the corresponding operator
$\zeta$-function is $\zeta(\varphi)(z)=2\zeta_{R}(-z-1)$ where $\zeta_{R}$
denotes the Riemann $\zeta$-function. In particular, the $\zeta$-regularized
trace of $\left\lvert{\partial}\right\lvert$ is given by
$\zeta(\varphi)(0)=2\zeta_{R}(-1)=-\frac{1}{6}$.

Operator $\zeta$-functions were first studied for pseudo-differential
operators by Ray and Singer ray; ray-singer whose work has been extended by
many authors including Seeley seeley, Guillemin guillemin-lagrangian;
guillemin-residue-traces; guillemin-wave, Kontsevich and Vishik kontsevich-
vishik; kontsevich-vishik-geometry, Lesch lesch, Paycha and Scott paycha;
paycha-scott, and Wodzicki wodzicki. This notion of $\zeta$-regularization has
been introduced to physics by Hawking hawking in his work on path integrals
with a curved space-time background, successfully applied in many physical
settings (e.g., the Casimir effect, defining one-loop functional determinants,
the stress-energy tensor, conformal field theory, and string theory
beneventano-santangelo; blau-visser-wipf; bordag-elizalde-kirsten; bytsenko-
et-al; culumovic-et-al; dowker-critchley; elizalde2001; elizalde; elizalde-et-
al; elizalde-vanzo-zerbini; fermi-pizzocchero; hawking; iso-murayama;
marcolli-connes; mckeon-sherry; moretti97; moretti99; moretti00; moretti11;
robles; shiekh; tong-strings), and is related to Hadamard parametrix
renormalization hack-moretti. As such the approach plays a foundational role
for an effective Lagrangian to be defined blau-visser-wipf, for relatively
easy computations of heat kernel coefficients bordag-elizalde-kirsten, and for
non-trivial extensions of the Chowla-Selberg formula elizalde2001.
Furthermore, the residues associated with operator $\zeta$-functions give rise
to the multiplicative anomaly in perturbation theory elizalde-vanzo-zerbini
and contribute to the energy momentum tensor of a black hole hawking.

Although $\zeta$-regularization based on pseudo-differential operators has
been very successful, it cannot be applied in general. Radzikowski
radzikowski92; radzikowski96 showed that, in general, Fourier integral
operator $\zeta$-functions guillemin-lagrangian; guillemin-residue-traces;
guillemin-wave; hartung-phd; hartung-scott are necessary. The application of
Fourier integral operator $\zeta$-functions to quantum field theory has
recently been developed in the context of the Hamiltonian formulation of
quantum field theories hartung-phd; hartung-jmp; hartung-iwota; hartung-
jansen; hartung-jansen-gauge-fields; hartung-scott; jansen-hartung. The
starting point remains fundamentally the same as with the pseudo-differential
application of $\zeta$-regularization. Given an algebra of Fourier integral
operators $\mathcal{A}$ corresponding to a quantum field theory, the time
evolution operator $U$, and an observable $\Omega$, we would like to compute
the vacuum expectation value

($*$)
$\displaystyle\langle\Omega\rangle=\lim_{T\to\infty+i0}\frac{\operatorname{tr}(U(0,T)\Omega)}{\operatorname{tr}(U(0,T))}.$

Both $U(0,T)$ and $U(0,T)\Omega$ are Fourier integral operators in general,
$U(0,T)$ is unitary, and $U(0,T)\Omega$ is usually unbounded. Hence, neither
numerator nor denominator are well-defined. This justifies the Fourier
integral operator $\zeta$-function approach to defining a $\zeta$-regularized
vacuum expectation value which replaces the traces in Equation ($*$ ‣ 1.2) by
operator $\zeta$-functions. To construct these $\zeta$-regularized vacuum
expectation values hartung-jmp; hartung-iwota; hartung-jansen; hartung-jansen-
gauge-fields; jansen-hartung, we construct a suitable holomorphic family of
operators $\mathfrak{G}$ called “gauge”. This gauge has to satisfy a number of
properties. Most importantly, for $\Re(z)\ll 0$ both
$U(0,T)\mathfrak{G}(z)\Omega$ and $U(0,T)\mathfrak{G}(z)$ have to be trace-
class operators. This will imply that both $\zeta(U(0,T)\mathfrak{G}\Omega)$
and $\zeta(U(0,T)\mathfrak{G})$ are well-defined meromorphic functions and
therefore

$\displaystyle\langle\Omega\rangle_{\zeta}(z)=\lim_{T\to\infty+i0}\frac{\zeta(U(0,T)\mathfrak{G}\Omega)}{\zeta(U(0,T)\mathfrak{G})}(z)$

is a well-defined meromorphic function (provided that
$\zeta(U(0,T)\mathfrak{G})$ is not the constant zero-function). In particular,
if $\langle\Omega\rangle_{\zeta}$ is holomorphic in a neighborhood of $0$,
then we can define the $\zeta$-regularized vacuum expectation value as
$\langle\Omega\rangle_{\zeta}(0)$. This construction has been shown to be
generally possible hartung-jmp; hartung-iwota; hartung-jansen-gauge-fields and
to be physically meaningful and accessible using quantum computing hartung-
jansen; jansen-hartung.

Furthermore, the construction is formally applicable to lattice formulations
in Lorentzian space-time as well. Applying a lattice discretization to
$\zeta(U(0,T)\mathfrak{G}\Omega)(z)$ for example yields integrals of the form
$\int\omega(\Phi)\mathfrak{g}(z)(\Phi)e^{iS(\Phi)}\mathcal{D}(\Phi)$, i.e.,
precisely the type of integral we expect from a lattice discretization but
with an additional term $\mathfrak{g}(z)(\Phi)$. This additional term
$\mathfrak{g}(z)(\Phi)$ moreover satisfies the asymptotic bound
$\left\lvert{\mathfrak{g}(z)(\Phi)}\right\lvert\leq
C(1+\left\lVert{\Phi}\right\lVert^{2})^{\delta\Re(z)}$ for some constants
$C,\delta>0$ and $\left\lVert{\Phi}\right\lVert$ sufficiently large. Since
$\omega$ is polynomially bounded for large $\left\lVert{\Phi}\right\lVert$,
this implies that $\Phi\mapsto\omega(\Phi)\mathfrak{g}(z)(\Phi)e^{iS(\Phi)}$
is integrable for $\Re(z)\ll 0$ and the $\zeta$-regularized lattice vacuum
expectation can be obtained analytically continuing to $z=0$.

If this general setup can be obtained from lattice discretizing
$\zeta$-regularized vacuum expectation values starting from the Hamiltonian
formulation, we may also take it as a starting point to an a priori
$\zeta$-regularization of a lattice field theory without needing to construct
the Hamiltonian formulation first. However, this raises a number of
applicability questions. Most importantly:

  1. Q1:

What kind of gauge families $\mathfrak{g}$ can be used to $\zeta$-regularize
lattice field theories in a Lorentzian background?

  2. Q2:

Under what conditions is it possible to construct a $\zeta$-regularized
Hamiltonian theory starting from a “naïvely” $\zeta$-regularized lattice field
theory?

The first question therefore addresses the freedom we have in choosing gauge
families most suitable for a given lattice simulation whereas answering the
second question allows us to prove equivalence with the Fourier integral
operator $\zeta$-function approach to $\zeta$-regularized vacuum expectation
values. This is important because it means that the known applicability and
physicality results hartung-jmp; hartung-iwota; hartung-jansen; hartung-
jansen-gauge-fields; jansen-hartung extend to the thus constructed
$\zeta$-regularized lattice field theory. In particular, it provides an avenue
of using quantum computation to simulate lattice field theories in Lorentzian
backgrounds.

###  1.3. Aims of this article

Since $\zeta$-regularized lattice field theories in Lorentzian backgrounds
have not been studied from this point of view, we aim to gain insights into
questions Q1 and Q2. In section 2 we will begin with a general discussion of
lattice field theory from a distribution theory point of view. This is
important for Q2 because the Hamiltonian theory uses microlocal analysis and
therefore is fundamentally a theory of distributions. Formally introducing the
gauge family $\mathfrak{g}$ in the second half of section 2 we will show that
the $\zeta$-regularized lattice field theory and the original lattice field
theory coincide in the distributional limit $z\to 0$. This is a necessary
first step in the construction of gauged transfer matrices and gauged time
evolution operators in section 3. Thus, section 2 addresses Q1 and section 3
addresses Q2. In section 4 we will show that $\zeta$-regularized lattice field
theories have correct classical limits. In particular, we will explicitly
compute the classical limit for the harmonic oscillator. We will then continue
discussing the harmonic oscillator analytically in section 5 and numerically
in section 6. For the analytic discussion, we will solve the harmonic
oscillator five times, once with Euclidean background and then again with four
different gauge families $\mathfrak{g}$. Each of these gauges will be
representative of a generic property for the gauge that may be useful in
different circumstances. The numerical discussion of the harmonic oscillator
will focus on classical computing rather than quantum computing and highlight
extrapolation and convergence behavior for one of the choices of gauge.

##  2\. $\zeta$-regularizing lattice vacuum expectation values

In this section, we will investigate the connection between lattice vacuum
expectation values with Euclidean and Lorentzian backgrounds. In particular,
we will highlight how additional complications arise in Lorentzian backgrounds
and how a distributional understanding of the lattice integrals and the
$\zeta$-regularization are used to formally understand the lattice integrals
in Lorentzian backgrounds. To this end, approximation on compact subsets will
be instrumental. In Theorem 1 and Theorem 2, we will provide conditions that
ensure the infinite volume limit exists and coincides with the
$\zeta$-regularized vacuum expectation value considered in hartung-iwota;
hartung-jmp; hartung-jansen; hartung-jansen-gauge-fields; jansen-hartung.

In a Euclidean lattice theory with $T$ time slices, we can express the vacuum
expectation value $\langle\Omega\rangle_{E}$ of an observable $\Omega$ as

$\displaystyle\langle\Omega\rangle_{E}=\frac{\int_{X^{T}}\Omega(\Phi)e^{-S(\Phi)}d\mu^{T}(\Phi)}{\int_{X^{T}}e^{-S(\Phi)}d\mu^{T}(\Phi)}$

where $X$ is a locally compact Hausdorff space, $\mu$ a $\sigma$-finite Radon
measure on $X$, and $\mu^{T}$ denotes the $T$-fold product measure
$\mu\times\mu\times\cdots\times\mu$ on $X^{T}$.

  * Example

    1. (i)

For the topological rotor (i.e., a massive particle on a circle of radius
$R$), $X$ is the circle of radius $R$.

    2. (ii)

For quantum mechanics in $\mathbb{R}^{N}$, $X=\mathbb{R}^{N}$.

    3. (iii)

For a quantum field with values in $V$ on $L$ spacial lattice points,
$X=V^{L}$. In particular, for a gauge field on $L$ spacial lattice points, $X$
often is $G^{L}$ for some (locally) compact Hausdorff group and $\mu$ the Haar
measure on $X$.

$\square$

Vacuum expectation values of a lattice theory with Lorentzian background can
be expressed as

$\displaystyle\langle\Omega\rangle_{L}=\frac{\int_{X^{T}}\Omega(\Phi)e^{iS(\Phi)}d\mu^{T}(\Phi)}{\int_{X^{T}}e^{iS(\Phi)}d\mu^{T}(\Phi)}$

where the integrals $\int_{X^{T}}F(\Phi)e^{iS(\Phi)}d\mu^{T}(\Phi)$ with
$F\in\\{1,\Omega\\}$ are defined distributionally if $\mu(X)=\infty$. In terms
of the transfer matrix, we can write

$\displaystyle
Z=\int_{X^{T}}e^{iS(\Phi)}d\mu^{T}(\Phi)=\int_{X^{T}}\prod_{j=1}^{T}e^{iS_{\Delta}(\Phi_{j+1},\Phi_{j})}d\mu^{T}(\Phi)$

where the kernel $\kappa_{\Delta}$ of the transfer matrix111It should be noted
that in complete generality, the transfer matrix may be “time-dependent.” If
that is the case, then we have different $S_{\Delta,j}(\Phi_{j+1},\Phi_{j})$,
$\kappa_{\Delta,j}$, and $U_{\Delta,j}$. For the purpose of legibility, we
will assume $\forall j,k:\ S_{\Delta,j}=S_{\Delta,k}=S_{\Delta}$ throughout
this article. This assumption is not necessary for any of the results reported
here to be true except that various powers, such as $U_{\Delta}^{T}$, would
need to be replaced with the corresponding products, such as
$\prod_{j=1}^{T}U_{\Delta,j}$.  $U_{\Delta}$ is given by
$\kappa_{\Delta}(\Phi^{\prime},\Phi)=e^{iS_{\Delta}(\Phi^{\prime},\Phi)}$. In
this distributional setting, $U_{\Delta}$ is therefore to be understood as the
operator satisfying,

($**$) $\displaystyle\forall\varphi,\psi\in C_{c}^{\infty}(X):\
\langle\varphi,U_{\Delta}\psi\rangle=\int_{X}\int_{X}\varphi(\Phi^{\prime})^{*}e^{iS_{\Delta}(\Phi^{\prime},\Phi)}\psi(\Phi)d\mu(\Phi^{\prime})d\mu(\Phi).$

At this point we can make the connection to one of the necessary assumptions
in the construction of $\zeta$-regularized vacuum expectation values in the
Hamiltonian setting hartung-iwota; hartung-jmp; hartung-jansen; hartung-
jansen-gauge-fields; jansen-hartung. For the approach to be applicable, we
need to ensure that Cauchy surfaces (i.e., time slices) of the spacetime are
compact. Equation ($**$ ‣ 2), however, implies that this assumption in the
$\zeta$-formalism does not restrict generality. Let $K\subseteq X$ be compact.
Then we define $U_{\Delta}|_{K}$ to be the restriction of $U_{\Delta}$ to $K$
which means that the kernel of $U_{\Delta}|_{K}$ is given by $1_{K\times
K}\kappa_{\Delta}$ as a distribution on $K^{2}$ and $U_{\Delta}|_{K}$ is
uniquely determined by

$\displaystyle\forall\varphi,\psi\in C^{\infty}(K):\
\langle\varphi,U_{\Delta}|_{K}\psi\rangle=\int_{K}\int_{K}\varphi(\Phi^{\prime})^{*}e^{iS_{\Delta}(\Phi^{\prime},\Phi)}\psi(\Phi)d\mu|_{K}(\Phi^{\prime})d\mu|_{K}(\Phi).$

The $\zeta$-formalism by construction only works for $U_{\Delta}|_{K}$.
However, since for every $\varphi,\psi\in C_{c}^{\infty}(X)$, there exists a
compact $K_{\varphi,\psi}\subseteq X$ such that
$\mathrm{spt}\varphi\cup\mathrm{spt}\psi\subseteq K_{\varphi,\psi}$, where
$\mathrm{spt}\varphi$ denotes the support of $\varphi$, we obtain that
$\langle\varphi,U_{\Delta}\psi\rangle=\langle\varphi,U_{\Delta}|_{K_{\varphi,\psi}}\psi\rangle$.
Hence, $U_{\Delta}|_{K}$ converges to $U_{\Delta}$ for $K\nearrow X$ in the
sense that the kernel $1_{K\times K}\kappa_{\Delta}$ of $U_{\Delta}|_{K}$
converges to the kernel $\kappa_{\Delta}$ of $U_{\Delta}$ in the space of
distributions.

In the Euclidean setting, we generally don’t have to discuss this aspect
because the integrand $Fe^{-S}$ is in $L_{1}(X^{2})$, and $\mu$ is a
$\sigma$-finite Radon measure. This implies

$\displaystyle\int_{X^{T}}Fe^{-S}d\mu^{T}-\int_{K^{T}}Fe^{-S}d\mu^{T}=$
$\displaystyle\int_{X^{T}}1_{X^{T}\setminus K^{T}}Fe^{-S}d\mu^{T}\to$
$\displaystyle 0\quad(K\nearrow X)$

and since $\int_{X^{T}}Fe^{-S}d\mu^{T}$ is a well-defined integral, the limit
$K\nearrow X$ usually needs not to be considered.

In the $\zeta$-regularized setting, we enforce a similar situation by
enforcing asymptotic bounds of the form
$\left\lvert{\Omega(\Phi)}\right\lvert\leq c\left\lVert{\Phi}\right\lVert^{d}$
for $\left\lVert{\Phi}\right\lVert\to\infty$ and some constants
$c,d\in\mathbb{R}$ on the observable $\Omega$, and by introducing a suitably
chosen holomorphic family of functions $\mathfrak{g}$ with $\mathfrak{g}(0)=1$
and
$\left\lvert{\mathfrak{g}(z)(\Phi)}\right\lvert\leq\alpha\left\lVert{\Phi}\right\lVert^{\beta\Re(z)}$
for constants $\alpha,\beta>0$ and $\left\lVert{\Phi}\right\lVert\gg 1$. This
family $\mathfrak{g}$ is called a “gauge” in the mathematical literature and
ensures that the gauged integrand $Fe^{iS}\mathfrak{g}(z)$ is in
$L_{1}(X^{T})$ for $\Re(z)\ll 0$. We therefore obtain

$\displaystyle\int_{X^{T}}Fe^{iS}\mathfrak{g}(z)d\mu^{T}-\int_{K^{T}}Fe^{iS}\mathfrak{g}(z)d\mu^{T}=$
$\displaystyle\int_{X^{T}}1_{X^{T}\setminus
K^{T}}Fe^{iS}\mathfrak{g}(z)d\mu^{T}\to$ $\displaystyle 0$

whenever $\Re(z)\ll 0$. Thus, considering a sequence
$(K_{n})_{n\in\mathbb{N}}$ of compacta in $X$ with $K_{n}\nearrow X\
(n\to\infty)$, we conclude

$\displaystyle\lim_{n\to\infty}\frac{\int_{K_{n}^{T}}\Omega
e^{iS}\mathfrak{g}(z)d\mu^{T}}{\int_{K_{n}^{T}}e^{iS}\mathfrak{g}(z)d\mu^{T}}=\frac{\int_{X^{T}}\Omega
e^{iS}\mathfrak{g}(z)d\mu^{T}}{\int_{X^{T}}e^{iS}\mathfrak{g}(z)d\mu^{T}}$

whenever $\Re(z)\ll 0$. Since the $\zeta$-regularized vacuum expectation
$\langle\Omega\rangle_{\zeta}$ is defined via analytic continuation of
$z\mapsto\frac{\langle\Omega\mathfrak{g}(z)\rangle_{L}}{\langle\mathfrak{g}(z)\rangle_{L}}$,
the above identity implies pointwise convergence on a half-space with
$\Re(z)\ll 0$.

######  Theorem 1.

Let $(K_{n})_{n\in\mathbb{N}}$ be a sequence of compacta in $X$ with
$K_{n}\nearrow X$ and $\left(\frac{\langle
1_{K_{n}^{T}}\Omega\mathfrak{g}\rangle_{L}}{\langle
1_{K_{n}^{T}}\mathfrak{g}\rangle_{L}}\right)_{n\in\mathbb{N}}$ locally bounded
in $C(D)$ where $D\subseteq\mathbb{C}$ is open and connected containing a
half-space with $\Re(z)\ll 0$. Then, $\left(\frac{\langle
1_{K_{n}^{T}}\Omega\mathfrak{g}\rangle_{L}}{\langle
1_{K_{n}^{T}}\mathfrak{g}\rangle_{L}}\right)_{n\in\mathbb{N}}$ converges
compactly to the analytic continuation of
$\frac{\langle\Omega\mathfrak{g}\rangle_{L}}{\langle\mathfrak{g}\rangle_{L}}$
on $D$. In particular, if $0\in D$, we obtain

$\displaystyle\lim_{n\to\infty}\langle
1_{K_{n}^{T}}\Omega\rangle_{\zeta}=\frac{\langle\Omega\mathfrak{g}\rangle_{L}}{\langle\mathfrak{g}\rangle_{L}}(0)=\langle\Omega\rangle_{\zeta}.$

Theorem 1 follows directly from Vitali’s Theorem since the set of pointwise
convergence contains a half-space with $\Re(z)\ll 0$ and we have asserted
local boundedness.

  * Remark

The local boundedness assumption in Theorem 1 means that for every $z_{0}$ in
the domain $D$, there exists a neighborhood $U\subseteq D$ of $z_{0}$ such
that the sequence of functions $\frac{\langle
1_{K_{n}^{T}}\Omega\mathfrak{g}(z)\rangle_{L}}{\langle
1_{K_{n}^{T}}\mathfrak{g}(z)\rangle_{L}}$ with $z\in U$ is uniformly bounded
as $K_{n}\nearrow X$, that is,

$\displaystyle\sup_{n\in\mathbb{N}}\ \sup_{z\in U}\ \left\lvert{\frac{\langle
1_{K_{n}^{T}}\Omega\mathfrak{g}(z)\rangle_{L}}{\langle
1_{K_{n}^{T}}\mathfrak{g}(z)\rangle_{L}}}\right\lvert<\infty.$

From a physical point of view, this means that any finite volume effects
arising from the restriction to $K_{n}$ remain bounded under small variation
of the gauge parameter $z$. In this sense, the local boundedness assumption
requires the physical theory to have locally bounded finite volume effects if
the volume is sent to infinity and the observable is subject to holomorphic
deformation. In particular, Theorem 1 then implies that locally bounded finite
volume effects already imply vanishing finite volume effects in the infinite
volume limit. $\square$

Since the $\zeta$-regularization requires compactification of $X$ and Theorem
1 ensures that we can reconstruct the non-compact case from compact
restrictions, we will from now on only consider the case in which $X$ is
compact.

  * Example

For quantum mechanics in $\mathbb{R}^{N}$, we can generally consider $X$ to be
the flat torus $(\mathbb{R}/L\mathbb{Z})^{N}$ of side length
$L\in\mathbb{R}_{>0}$. In this case the limit $K\nearrow\mathbb{R}^{N}$
corresponds to $L\nearrow\infty$. $\square$

Returning to the transfer matrix $U_{\Delta}$ on compact $X$, we observe that
the kernel

$\displaystyle\kappa_{U_{\Delta}^{T}}(\Phi_{T+1},\Phi_{1})=\int_{X^{T-1}}\prod_{j=1}^{T}e^{iS_{\Delta}(\Phi_{j+1},\Phi_{j})}d\mu^{T-1}(\Phi_{2},\ldots,\Phi_{T})$

of $U_{\Delta}^{T}$ is integrable along the diagonal which implies that the
nuclear trace

$\displaystyle\operatorname{tr}(U_{\Delta}^{T})=\langle\kappa_{U_{\Delta}^{T}},\delta_{\mathrm{diag}}\rangle=\int_{X}\kappa_{U_{\Delta}^{T}}(\Phi,\Phi)d\mu(\Phi)$

of $U_{\Delta}^{T}$ is well-defined. Introducing an observable $\Omega\in
L_{\infty}(X)$ and gauge $\mathfrak{g}$, we thus obtain (using
$F\in\\{1,\Omega\\}$) that

$\displaystyle\langle\kappa_{U_{\Delta}^{T}}F\mathfrak{g}(z),\delta_{\mathrm{diag}}\rangle=$
$\displaystyle\int_{X}\kappa_{U_{\Delta}^{T}}(\Phi,\Phi)F(\Phi)\mathfrak{g}(z)(\Phi)d\mu(\Phi)$

is a holomorphic function in $z$ which directly implies Theorem 2.

######  Theorem 2.

If $X$ is compact, then

$\displaystyle\langle\Omega\rangle_{L}=\frac{\int_{X^{T}}\Omega(\Phi)e^{iS(\Phi)}d\mu^{T}(\Phi)}{\int_{X^{T}}e^{iS(\Phi)}d\mu^{T}(\Phi)}=\langle\Omega\rangle_{\zeta}.$

Furthermore, if $X$ is not compact, the assumptions of Theorem 1 are
satisfied, and $(K_{n})_{n\in\mathbb{N}}$ is as in Theorem 1, then

$\displaystyle\lim_{n\to\infty}\frac{\int_{K_{n}^{T}}\Omega(\Phi)e^{iS(\Phi)}d\mu^{T}(\Phi)}{\int_{K_{n}^{T}}e^{iS(\Phi)}d\mu^{T}(\Phi)}=\langle\Omega\rangle_{\zeta}.$

##  3\. Gauging the transfer matrix and time evolution operator

In section 2, we have discussed how the $\zeta$-regularization can make sense
of the formal integrals arising in a Lorentzian lattice theory through a
process of gauging. However, the gauging process as discussed in section 2 can
be understood as an operation on the observable. More precisely, as was shown
in hartung-jansen; jansen-hartung, the entire construction of
$\zeta$-regularized vacuum expectation values is equivalent to the
construction of an analytic continuation of a quotient $\frac{\langle
O_{1}(z)\rangle}{\langle O_{2}(z)\rangle}$ of expectation values $\langle
O_{j}(z)\rangle$ for suitably constructed observables $O_{j}(z)$. In order to
make the connection to the general $\zeta$-regularization theory of vacuum
expectation values hartung-jmp; hartung-iwota; hartung-jansen-gauge-fields, on
the other hand, we need to consider the gauge as part of the time evolution
operator. Furthermore, for the $\zeta$-regularization point of view to be as
versatile as methods used in Euclidean lattice theories, it would be
advantageous to find gauges that can be attributed to the transfer matrix as
opposed to the entire time evolution operator only. In this section, we will
therefore discuss the relation between the point of view in section 2 and the
formalism of hartung-iwota; hartung-jmp; hartung-jansen; hartung-jansen-gauge-
fields; jansen-hartung. As such, we will be able to conclude that all
properties proven in hartung-iwota; hartung-jmp; hartung-jansen; hartung-
jansen-gauge-fields; jansen-hartung also hold for the $\zeta$-regularization
of lattice field theories. In particular, we will obtain that the
$\zeta$-regularized lattice vacuum expectation values are independent of the
choice of gauge $\mathfrak{g}$.

The transfer matrix $U_{\Delta}$ is given by its kernel
$\kappa_{\Delta}(\Phi^{\prime},\Phi)=e^{iS_{\Delta}(\Phi^{\prime},\Phi)}$,
i.e.,

$\displaystyle\forall\varphi,\psi\in C_{c}^{\infty}(X):\
\langle\varphi,U_{\Delta}\psi\rangle=\int_{X}\int_{X}\varphi(\Phi^{\prime})^{*}e^{iS_{\Delta}(\Phi^{\prime},\Phi)}\psi(\Phi)d\mu(\Phi^{\prime})d\mu(\Phi),$

and its adjoint $U_{\Delta}^{*}$ has kernel
$\kappa_{U_{\Delta}^{*}}(\Phi^{\prime},\Phi)=\kappa_{\Delta}(\Phi,\Phi^{\prime})^{*}=e^{-iS_{\Delta}(\Phi,\Phi^{\prime})}$.
Since the transfer matrix connects neighboring values $\Phi_{j}$ and
$\Phi_{j+1}$ in terms of the global action
$e^{iS(\Phi)}=\prod_{j=1}^{T}e^{iS_{\Delta}(\Phi_{j+1},\Phi_{j})}$, it can
often be advantageous to mirror this structure in the choice of gauge
$\mathfrak{g}$. Introducing a gauge
$\mathfrak{g}(z)(\Phi)=\prod_{j=1}^{T}\mathfrak{g}_{\Delta}(z)(\Phi_{j+1},\Phi_{j})$,
we obtain the gauged transfer matrix $\tilde{U}_{\Delta}$ defined as

$\displaystyle\left(\tilde{U}_{\Delta}(z)\psi\right)(\Phi^{\prime})=\int_{X}e^{iS_{\Delta}(\Phi^{\prime},\Phi)}\mathfrak{g}_{\Delta}(z)(\Phi^{\prime},\Phi)\psi(\Phi)d\mu(\Phi).$

This directly yields the gauge family $\mathfrak{G}_{\Delta}$ in the operator
point of view via

$\displaystyle\left(\mathfrak{G}_{\Delta}(z)\psi\right)(\Phi^{\prime\prime})=$
$\displaystyle\left(U_{\Delta}^{*}\tilde{U}_{\Delta}(z)\psi\right)(\Phi^{\prime\prime})$
$\displaystyle=$
$\displaystyle\int_{X}\int_{X}e^{-iS_{\Delta}(\Phi^{\prime},\Phi^{\prime\prime})}e^{iS_{\Delta}(\Phi^{\prime},\Phi)}\mathfrak{g}_{\Delta}(z)(\Phi^{\prime},\Phi)\psi(\Phi)d\mu(\Phi)d\mu(\Phi^{\prime})$

and thus $\tilde{U}_{\Delta}=U_{\Delta}\mathfrak{G}_{\Delta}$. In other words,
$\mathfrak{G}_{\Delta}(z)$ is an integral operator with kernel

$\displaystyle\kappa_{\mathfrak{G}_{\Delta}(z)}(\Phi^{\prime},\Phi)=\int_{X}e^{-iS_{\Delta}(y,\Phi^{\prime})}e^{iS_{\Delta}(y,\Phi)}\mathfrak{g}_{\Delta}(z)(y,\Phi)d\mu(y).$

######  Theorem 3.

Let $U:=U_{\Delta}^{T}$ be the time evolution operator defined by a transfer
matrix $U_{\Delta}$ corresponding to an action $S_{\Delta}$,
$\mathfrak{g}(z)(\Phi)=\prod_{j=1}^{T}\mathfrak{g}_{\Delta}(z)(\Phi_{j+1},\Phi_{j})$,
$\mathfrak{G}_{\Delta}(z)$ the gauge family with the kernel

$\displaystyle\kappa_{\mathfrak{G}_{\Delta}(z)}(\Phi^{\prime},\Phi)=\int_{X}e^{-iS_{\Delta}(y,\Phi^{\prime})}e^{iS_{\Delta}(y,\Phi)}\mathfrak{g}_{\Delta}(z)(y,\Phi)d\mu(y),$

and $\omega$ an observable with corresponding operator $\Omega$. Then,

$\displaystyle\langle\omega\rangle(z)=\frac{\int_{X^{T}}\omega(\Phi)e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)d\mu^{T}(\Phi)}{\int_{X^{T}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)d\mu^{T}(\Phi)}$

is the $\zeta$-regularized lattice vacuum expectation value which, in the
operator formulation, corresponds to the $\zeta$-regularized vacuum
expectation value

$\displaystyle\langle\Omega\rangle(z)=\frac{\zeta\left((U_{\Delta}\mathfrak{G}_{\Delta})^{T}\Omega\right)(z)}{\zeta\left((U_{\Delta}\mathfrak{G}_{\Delta})^{T}\right)(z)}.$

Since $(U_{\Delta}\mathfrak{G}_{\Delta})^{T}$ is a gauge of $U$ (more
precisely, $(U_{\Delta}\mathfrak{G}_{\Delta})^{T}=U\mathfrak{G}$ using the
“global” gauge family
$\mathfrak{G}=U^{*}(U_{\Delta}\mathfrak{G}_{\Delta})^{T}$), we obtain gauge
independence of $\langle\omega\rangle(0)$ in the sense of hartung-iwota;
hartung-jmp; hartung-jansen; hartung-jansen-gauge-fields; jansen-hartung. In
particular, if $U_{\Delta}$ and $\mathfrak{G}_{\Delta}$ commute, then
$\mathfrak{G}=(\mathfrak{G}_{\Delta})^{T}$.

We note that the assumptions of Theorem 3 permit gauge families $\mathfrak{g}$
of the form
$\mathfrak{g}(z)(\Phi)=\left(\prod_{j=1}^{T}\left\lvert{\Phi_{j}}\right\lvert\right)^{\frac{z}{T}}$
and
$\mathfrak{g}(z)(\Phi)=\left(\prod_{j=1}^{T}\left\lvert{\Phi_{j+1}-\Phi_{j}}\right\lvert\right)^{\frac{z}{T}}$
but not $\mathfrak{g}(z)(\Phi)=\left\lVert{\Phi}\right\lVert^{z}$. The latter
is a gauge for $U=U_{\Delta}^{T}$ rather than $U_{\Delta}$ itself. The kernel
$\kappa_{U}$ of $U$ is given by

$\displaystyle\kappa_{U}(\Phi_{T+1},\Phi_{1})=\int_{X^{T-1}}e^{iS(\Phi_{T+1},\ldots,\Phi_{1})}d\mu^{T-1}(\Phi_{T},\ldots,\Phi_{2}).$

Here, we do not identify $\Phi_{T+1}$ with $\Phi_{1}$ yet, since that step is
eventually part of taking the trace. To reduce notation, we will introduce
$\hat{\Phi}:=(\Phi_{T},\ldots,\Phi_{2})$ and $\hat{\mu}:=\mu^{T-1}$. Then, the
action $S$ is a function of $(\Phi_{T+1},\hat{\Phi},\Phi_{1})$ due to the open
boundary conditions and the gauge family $\mathfrak{g}$ is a family of
functions of $(\hat{\Phi},\Phi_{1})$. Furthermore, $U^{*}$ has kernel

$\displaystyle\kappa_{U^{*}}(x,y)=\kappa_{U}(y,x)^{*}=\int_{X^{T-1}}e^{-iS(y,\hat{\Phi},x)}d\hat{\mu}(\hat{\Phi}).$

Hence, the gauged time evolution $\tilde{U}$ with kernel

$\displaystyle\kappa_{\tilde{U}(z)}(\Phi_{T+1},\Phi_{1})=\int_{X^{T-1}}e^{iS(\Phi_{T+1},\hat{\Phi},\Phi_{1})}\mathfrak{g}(z)(\hat{\Phi},\Phi_{1})d\hat{\mu}(\hat{\Phi})$

can be expressed as $\tilde{U}=U\mathfrak{G}$ with
$\mathfrak{G}=U^{*}\tilde{U}$, i.e., $\mathfrak{G}(z)$ has kernel

$\displaystyle\kappa_{\mathfrak{G}(z)}(x,y)=$
$\displaystyle\int_{X}\kappa_{U^{*}}(x,w)\kappa_{\tilde{U}(z)}(w,y)d\mu(w)$
$\displaystyle=$
$\displaystyle\int_{X}\int_{X^{T-1}}e^{-iS(w,\hat{\Phi},x)}d\hat{\mu}(\hat{\Phi})\int_{X^{T-1}}e^{iS(w,\hat{\Psi},y)}\mathfrak{g}(z)(\hat{\Psi},y)d\hat{\mu}(\hat{\Psi})d\mu(w).$

This directly implies Theorem 4 which is the extension of Theorem 3 to gauges
$\mathfrak{g}$ that cannot be split into gauges of the transfer matrix.

######  Theorem 4.

Let $U$ be the time evolution operator corresponding to an action $S$,
$\mathfrak{g}(z)$ a family of gauge functions, $\mathfrak{G}(z)$ the gauge
family with the kernel

$\displaystyle\kappa_{\mathfrak{G}(z)}(x,y)=\int_{X}\int_{X^{T-1}}\int_{X^{T-1}}e^{-iS(w,\hat{\Phi},x)}e^{iS(w,\hat{\Psi},y)}\mathfrak{g}(z)(\hat{\Psi},y)d\hat{\mu}(\hat{\Phi})d\hat{\mu}(\hat{\Psi})d\mu(w),$

and $\omega$ an observable with corresponding operator $\Omega$. Then,

$\displaystyle\langle\omega\rangle(z)=\frac{\int_{X^{T}}\omega(\Phi)e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)d\mu^{T}(\Phi)}{\int_{X^{T}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)d\mu^{T}(\Phi)}$

is the $\zeta$-regularized lattice vacuum expectation value which, in the
operator formulation, corresponds to the $\zeta$-regularized vacuum
expectation value

$\displaystyle\langle\Omega\rangle(z)=\frac{\zeta\left(U\mathfrak{G}\Omega\right)(z)}{\zeta\left(U\mathfrak{G}\right)(z)}.$

In particular, $\langle\omega\rangle(0)$ is independent of the gauge
$\mathfrak{g}$ in the sense of hartung-iwota; hartung-jmp; hartung-jansen;
hartung-jansen-gauge-fields; jansen-hartung.

This identifies sufficient conditions on the choice of gauge
$\mathfrak{g}_{\Delta}$. Primarily, we need to ensure that $\mathfrak{G}(z)$
is a Fourier integral operator of order $\delta\Re(z)$ for some $\delta>0$ for
the construction of $\zeta$-regularized vacuum expectation values to make
sense in the operator picture hartung-jmp; hartung-iwota; hartung-jansen;
jansen-hartung. A sufficient condition for this to be true is for
$\mathfrak{g}(z)$ to be in the Hörmander class $S^{\delta\Re(z)}$ hoermander-
books, that is,
$\left\lvert{\partial_{\Phi}^{\alpha}\mathfrak{g}(z)(\Phi)}\right\lvert\leq
c_{\alpha}(1+\left\lVert{\Phi}\right\lVert)^{\delta\Re(z)-\left\lvert{\alpha}\right\lvert}$
has to hold for all $\Phi$, $z$, multiindices $\alpha$, and some constant
$c_{\alpha}$. This condition can also be weakened by assuming that the
inequality holds asymptotically for $\left\lVert{\Phi}\right\lVert\to\infty$
hartung-phd; hartung-scott; hartung-jmp; hartung-iwota. In doing so, we can
allow for homogeneous singularities at $\Phi=0$, such as,
$\mathfrak{g}(z)(\Phi)=\left\lVert{\Phi}\right\lVert^{z}$ or even consider
$\mathfrak{g}(z)$ to be polyhomogeneous $\mathfrak{g}(z)\sim\sum_{\iota\in
I}a_{\iota}(z)(\Phi)$ where $a_{\iota}(z)$ is homogeneous of degree
$d_{\iota}+z$ and the $\Re(d_{\iota})$ are bounded from above hartung-phd;
hartung-scott; hartung-jmp; hartung-iwota.

##  4\. The classical limit

In the construction of $\zeta$-regularized vacuum expectation values, we
consider a quotient of $\zeta$-functions
$\frac{\zeta(U\mathfrak{G}\Omega)}{\zeta(U\mathfrak{G})}$ which in the lattice
setting takes the form $\frac{\int e^{iS}\mathfrak{g}\omega d\mu}{\int
e^{iS}\mathfrak{g}d\mu}$. Although the general theory of operator
$\zeta$-functions would allow for different choices of $\mathfrak{G}$ (and
thus different choices of $\mathfrak{g}$) in the numerator and denominator,
the definition of $\zeta$-regularized vacuum expectation values hartung-jmp
implicitly requires us to make the same choice. This was introduced to handle
the problem that
$\frac{\zeta(U\mathfrak{G}_{1}\Omega)}{\zeta(U\mathfrak{G}_{2})}$ can depend
on the choice of $\mathfrak{G}_{1}$ and $\mathfrak{G}_{2}$ if
$\zeta(U\mathfrak{G}_{1}\Omega)(0)=\zeta(U\mathfrak{G}_{2})(0)=0$. While
enforcing $\mathfrak{G}_{1}=\mathfrak{G}_{2}=\mathfrak{G}$ does not obviously
solve the problem of
$\frac{\zeta(U\mathfrak{G}\Omega)}{\zeta(U\mathfrak{G})}(0)$ being dependent
on $\mathfrak{G}$ if $\zeta(U\mathfrak{G}\Omega)(0)=\zeta(U\mathfrak{G})(0)=0$
– in fact, independence of $\mathfrak{G}$ is a consequence of the physicality
proof for $\zeta$-regularized vacuum expectation values hartung-jansen – there
is a physical reason to hope for this to be the correct requirement. In a very
loose sense, we can interpret the denominator $\zeta(U\mathfrak{G})(z)$ to be
the partition function of a quantum field theory $\mathrm{QFT}(z)$ with “time
evolution” $U\mathfrak{G}(z)$. For $\Re(z)\ll 0$ the vacuum expectation
$\langle\Omega\rangle(z)$ of $\Omega$ is then well-defined in
$\mathrm{QFT}(z)$ using Feynman’s construction and satisfies
$\langle\Omega\rangle(z)=\frac{\zeta(U\mathfrak{G}\Omega)(z)}{\zeta(U\mathfrak{G})(z)}$.
The $\zeta$-regularized vacuum expectation value $\langle\Omega\rangle(0)$ can
therefore be loosely understood as the analytic continuation of well-defined
vacuum expectation values in Feynman’s sense with respect to the holomorphic
family of quantum field theories $\mathrm{QFT}$.

Of course, this interpretation requires many philosophical concessions because
in the lattice picture, $\mathrm{QFT}(z)$ corresponds to the “action”
$S(z)=S(0)-i\ln(\mathfrak{g}(z))$ where $S(0)$ is the action of
$\mathrm{QFT}(0)$, i.e., the action of the QFT we wish to study. However, if
we accept this interpretation, then we can ask the question regarding the
classical limit. In general, interpreting $S(z)$ as a classical action is
likely to be difficult due to $-i\ln(\mathfrak{g}(z))$ being complex. The
closest classical action we can extract from $S(z)$ is the original action
$S(0)$ which leads to the hypothesis that the classical limit should always be
the classical limit of $\mathrm{QFT}(0)$ independent of the value of $z$.
Theorem 5 is precisely this surprising result.

######  Theorem 5.

Let $S:\ \mathbb{R}^{N}\to\mathbb{R}$ be a non-degenerate action with a unique
minimum, that is, there is exactly one point $C\in\mathbb{R}^{N}$ with
$S^{\prime}(C)=0$ which furthermore satisfies $\det S^{\prime\prime}(C)\neq
0$. Let $\omega:\ \mathbb{R}^{N}\to\mathbb{R}$ be an observable and
$\mathfrak{g}(z)$ a gauge function with $\forall z\in\mathbb{C}:\
\mathfrak{g}(z)(C)\neq 0$. Then,

$\displaystyle\forall z\in\mathbb{C}:\ \lim_{\hbar\searrow
0}\langle\omega\rangle(z)=\lim_{\hbar\searrow
0}\frac{\int_{\mathbb{R}^{N}}e^{\frac{i}{\hbar}S(\Phi)}\mathfrak{g}(z)(\Phi)\omega(\Phi)d\Phi}{\int_{\mathbb{R}^{N}}e^{\frac{i}{\hbar}S(\Phi)}\mathfrak{g}(z)(\Phi)d\Phi}=\omega(C).$

In particular, $\lim_{\hbar\searrow 0}\langle\omega\rangle(0)=\omega(C)$

###### Proof.

Since we are interested in the classical limit $\hbar\searrow 0$, we will
expand the integrals in

$\displaystyle\langle\omega\rangle(z)=\frac{\int_{\mathbb{R}^{N}}e^{\frac{i}{\hbar}S(\Phi)}\mathfrak{g}(z)(\Phi)\omega(\Phi)d\Phi}{\int_{\mathbb{R}^{N}}e^{\frac{i}{\hbar}S(\Phi)}\mathfrak{g}(z)(\Phi)d\Phi}.$

using stationary phase approximation and obtain

$\displaystyle\langle\omega\rangle(z)=$
$\displaystyle\frac{e^{\frac{i}{\hbar}S(C)}\left\lvert{\det
S^{\prime\prime}(C)}\right\lvert^{-\frac{1}{2}}e^{\frac{i\pi}{4}\mathrm{sgn}S^{\prime\prime}(C)}\mathfrak{g}(z)(C)\omega(C)\left(2\pi\hbar\right)^{\frac{N}{2}}+o\left(\hbar^{\frac{N}{2}}\right)}{e^{\frac{i}{\hbar}S(C)}\left\lvert{\det
S^{\prime\prime}(C)}\right\lvert^{-\frac{1}{2}}e^{\frac{i\pi}{4}\mathrm{sgn}S^{\prime\prime}(C)}\mathfrak{g}(z)(C)\left(2\pi\hbar\right)^{\frac{N}{2}}+o\left(\hbar^{\frac{N}{2}}\right)}$
$\displaystyle=$ $\displaystyle\frac{\omega(C)+o(1)}{1+o(1)}.$

Hence, we observe

$\displaystyle\forall z\in\mathbb{C}:\ \lim_{\hbar\searrow
0}\langle\omega\rangle(z)=\omega(C).$

∎

  * Example

Consider the harmonic oscillator on the lattice with action

$\displaystyle S:\ \mathbb{R}^{\tau}\to\mathbb{R};\ x\mapsto
a\sum_{j=0}^{\tau-1}\frac{m}{2}\frac{(x_{j+1}-x_{j})^{2}}{a^{2}}+\mu^{2}x_{j}^{2}.$

Then

$\displaystyle\partial_{k}S(x)=$ $\displaystyle
m\frac{x_{k}-x_{k-1}}{a}-m\frac{x_{k+1}-x_{k}}{a}+2a\mu^{2}x_{k}$
$\displaystyle=$ $\displaystyle
m\frac{2x_{k}-x_{k-1}-x_{k+1}}{a}+2a\mu^{2}x_{k}$ $\displaystyle=$
$\displaystyle(Mx)_{k}$

with
$M_{j,k}=\left(\frac{2m}{a}+2a\mu^{2}\right)\delta_{j,k}-\frac{m}{a}\delta_{j+1,k}-\frac{m}{a}\delta_{j-1,k}$.
Thus, $M$ is a circulant matrix and setting
$\Sigma_{0}=(\delta_{j+1,k})_{j,k}$ implies
$M=\frac{2m}{a}+2a\mu^{2}-\frac{m}{a}\Sigma_{0}-\frac{m}{a}\Sigma_{0}^{\tau-1}=p(\Sigma_{0})$.
Since the eigenvalues of $\Sigma_{0}$ are $\lambda_{k}=e^{\frac{2\pi
ik}{\tau}}$, the eigenvalues of $M$ are

$\displaystyle p(\lambda_{k})=\frac{2m}{a}+2a\mu^{2}-\frac{m}{a}e^{\frac{2\pi
ik}{\tau}}-\frac{m}{a}e^{-\frac{2\pi
ik}{\tau}}=2a\mu^{2}+\frac{2m}{a}\left(1-\cos\left(\frac{2\pi
k}{\tau}\right)\right).$

Hence, $S^{\prime}(x)=0$ fails to have a unique solution ($C=0$) if and only
if $\exists k:\ p(\lambda_{k})=0$. The observation $p(\lambda_{k})\geq
2a\mu^{2}>0$ therefore implies that $C=0$ is the unique critical point, and we
furthermore observe $S^{\prime\prime}(C)=M\in GL(N)$. Thus, the classical
limit $\hbar\searrow 0$ of $\langle\omega\rangle(z)$ is the observable
$\omega$ evaluated on the classical vacuum $C=0$. $\square$

  * Remark

The proof of Theorem 5 can be extended to the case of multiple critical points
$C$ with $S^{\prime}(C)=0$ and $\det S^{\prime\prime}(C)\neq 0$ such as in the
case of a double well potential. Let $\Gamma$ be the set of critical points
and $\omega$ and $\mathfrak{g}$ as in Theorem 5. Then, stationary phase
approximation yields

$\displaystyle\langle\omega\rangle(z)=$
$\displaystyle\frac{\sum_{C\in\Gamma}e^{\frac{i}{\hbar}S(C)}\left\lvert{\det
S^{\prime\prime}(C)}\right\lvert^{-\frac{1}{2}}e^{\frac{i\pi}{4}\mathrm{sgn}S^{\prime\prime}(C)}\mathfrak{g}(z)(C)\omega(C)+o\left(1\right)}{\sum_{C\in\Gamma}e^{\frac{i}{\hbar}S(C)}\left\lvert{\det
S^{\prime\prime}(C)}\right\lvert^{-\frac{1}{2}}e^{\frac{i\pi}{4}\mathrm{sgn}S^{\prime\prime}(C)}\mathfrak{g}(z)(C)+o\left(1\right)}$

and thus

$\displaystyle\langle\omega\rangle(0)=$
$\displaystyle\frac{\sum_{C\in\Gamma}e^{\frac{i}{\hbar}S(C)}\left\lvert{\det
S^{\prime\prime}(C)}\right\lvert^{-\frac{1}{2}}e^{\frac{i\pi}{4}\mathrm{sgn}S^{\prime\prime}(C)}\omega(C)+o\left(1\right)}{\sum_{C\in\Gamma}e^{\frac{i}{\hbar}S(C)}\left\lvert{\det
S^{\prime\prime}(C)}\right\lvert^{-\frac{1}{2}}e^{\frac{i\pi}{4}\mathrm{sgn}S^{\prime\prime}(C)}+o\left(1\right)}.$

In the case of the double well potential, where we have exactly two critical
points $C_{1}$ and $C_{2}$ with $S(C_{1})=S(C_{2})$ and
$S^{\prime\prime}(C_{1})=S^{\prime\prime}(C_{2})$, we can further simplify to

$\displaystyle\langle\omega\rangle(z)=$
$\displaystyle\frac{\mathfrak{g}(z)(C_{1})\omega(C_{1})+\mathfrak{g}(z)(C_{2})\omega(C_{2})+o\left(1\right)}{\mathfrak{g}(z)(C_{1})+\mathfrak{g}(z)(C_{2})+o\left(1\right)},$
$\displaystyle\langle\omega\rangle(0)=$
$\displaystyle\frac{\omega(C_{1})+\omega(C_{2})+o\left(1\right)}{2+o\left(1\right)},$

and thus

$\displaystyle\lim_{\hbar\to 0}\langle\omega\rangle(0)=$
$\displaystyle\frac{\omega(C_{1})+\omega(C_{2})}{2}.$

$\square$

##  5\. The harmonic oscillator: four gauges and a Wick rotation

In this section we will consider the harmonic oscillator and analytically
compute the ground state energy with five different approaches. First (in
subsection 5.1), we will compute it using a Wick rotation; thus reproducing
well-known results which will serve as a basis for comparison with the four
different choices of gauge $\mathfrak{g}$. These four choices of gauge arise
from two binary choices. The first choice is whether we wish our gauge to be
local in time or global, i.e., whether or not it is possible to split the
gauge function $\mathfrak{g}(z)(\Phi)$ into a product
$\prod_{j}\mathfrak{g}_{\Delta}(z)(\Phi_{j+1},\Phi_{j})$ and thus gauging the
transfer matrix $U_{\Delta}$. In applications, this choice will likely be a
matter of convenience. The more important binary decision is related to the
behavior of the gauge near zero. Since the regularization is introduced
through the asymptotic behavior of the gauge for large
$\left\lVert{\Phi}\right\lVert$, there is a considerable degree of freedom in
any fixed compact set. In particular, the functional analysis of a
$\zeta$-lattice field theory is easier if the gauge is of the form
$\mathfrak{g}(z)(\Phi)=\left\lVert{\Phi}\right\lVert^{z}$ but this introduces
a pole in zero. The corresponding integrals are still well-defined in terms of
homogeneous distributions hartung-iwota; hartung-jmp; hartung-jansen; hartung-
jansen-gauge-fields; jansen-hartung but not absolutely integrable. Thus, for
numerical purposes, absolutely integrable gauges like
$\mathfrak{g}(z)(\Phi)=(1+\left\lVert{\Phi}\right\lVert)^{z}$ are preferable.
In subsection 5.2, subsection 5.3, subsection 5.4, and subsection 5.5 we will
consider each of these cases (cf., Table 1).

Table 1. Different choices of gauge for the $\zeta$-regularized hamonic oscillator on the lattice with Lorentzian background. | regularization
---|---
| distributional | abs. integrable
gauging |  |  |
global in time | subsection 5.2 | subsection 5.3
local in time | subsection 5.4 | subsection 5.5

In our treatment of the harmonic oscillator, we will absorb $\hbar$ into the
action

$\displaystyle S:\ \mathbb{R}^{\tau}\to\mathbb{R};\
\Phi\mapsto\frac{a}{\hbar}\sum_{j=0}^{\tau-1}\frac{m}{2}\frac{(\Phi_{j+1}-\Phi_{j})^{2}}{a^{2}}+\frac{m\omega^{2}}{2}\Phi_{j}^{2}.$

Given an observable $\Omega$ and a gauge $\mathfrak{g}$, we hence want to
solve

$\displaystyle\langle\Omega\rangle(z)=\frac{\int_{\mathbb{R}^{\tau}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)\Omega(\Phi)d\Phi}{\int_{\mathbb{R}^{\tau}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)d\Phi}$

in the $\zeta$-regularized cases and

$\displaystyle\langle\Omega\rangle_{E}=\frac{\int_{\mathbb{R}^{\tau}}e^{-S(\Phi)}\Omega(\Phi)d\Phi}{\int_{\mathbb{R}^{\tau}}e^{-S(\Phi)}d\Phi}$

in the Wick rotated case. For the calculations in this section, we will
consider observables $\Omega$ of the form

$\displaystyle\Phi^{\alpha}=\frac{1}{\tau}\sum_{j=0}^{\tau-1}\Phi_{j}^{\alpha}.$

Many interesting observables are related to these $\Phi^{\alpha}$. For
instance, the ground state energy is given by
$iE_{0}=m\omega^{2}\langle\Phi^{2}\rangle(0)$ or
$E_{0}=m\omega^{2}\langle\Phi^{2}\rangle_{E}$ where we have lost a factor of
$i$ due to the Wick rotation.

We will compute the continuum limit
$E_{0}^{\mathrm{continuum}}=\frac{\hbar\omega}{2}$ in the Wick rotated case
explicitly. For all other cases, we will show that the ground state energy
$E_{0}$ as computed with the $\zeta$-regularized lattice theory coincides with
$E_{0}$ as computed with the Wick rotated lattice theory. To achieve this, we
will define the matrix

$\displaystyle\Sigma:=\frac{1}{\hbar}\left(\left(\frac{m}{a}+\frac{am\omega^{2}}{2}\right)\delta_{j,k}-\frac{m}{2a}(\delta_{j,k+1}+\delta_{j+1,k})\right)_{j,k}$

because

$\displaystyle S(\Phi)=\langle\Phi,\Sigma\Phi\rangle.$

$\Sigma$ is a circulant matrix and setting $\Sigma_{0}=(\delta_{j+1,k})_{j,k}$
implies

$\displaystyle\Sigma=\frac{m}{a\hbar}+\frac{am\omega^{2}}{2\hbar}-\frac{m}{2a\hbar}\Sigma_{0}-\frac{m}{2a\hbar}\Sigma_{0}^{\tau-1}=p(\Sigma_{0}).$

Since the eigenvalues of $\Sigma_{0}$ are $\mu_{k}=e^{\frac{2\pi ik}{\tau}}$,
the eigenvalues of $\Sigma$ are

$\displaystyle
p(\mu_{k})=\frac{m}{a\hbar}+\frac{am\omega^{2}}{2\hbar}-\frac{m}{2a\hbar}e^{\frac{2\pi
ik}{\tau}}-\frac{m}{2a\hbar}e^{-\frac{2\pi
ik}{\tau}}=\frac{am\omega^{2}}{2\hbar}+\frac{m}{a\hbar}\left(1-\cos\left(\frac{2\pi
k}{\tau}\right)\right).$

Hence, $\Sigma$ is strictly positive and

$\displaystyle S(\Phi)=\left\lVert{\sqrt{\Sigma}\Phi}\right\lVert^{2}.$

In other words, the change of coordinates $\Psi:=\sqrt{\Sigma}\Phi$ implies

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{\int_{\mathbb{R}^{\tau}}e^{i\left\lVert{\Psi}\right\lVert^{2}}\mathfrak{g}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}{\int_{\mathbb{R}^{\tau}}e^{i\left\lVert{\Psi}\right\lVert^{2}}\mathfrak{g}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}$
$\displaystyle=$ $\displaystyle\frac{\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{ir^{2}}\mathfrak{g}(z)\left(r\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(r\sqrt{\Sigma}^{-1}\Psi\right)r^{\tau-1}drd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{ir^{2}}\mathfrak{g}(z)\left(r\sqrt{\Sigma}^{-1}\Psi\right)r^{\tau-1}drd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}$ $\displaystyle=$
$\displaystyle\frac{\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{is}\mathfrak{g}(z)\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)s^{\frac{\tau}{2}-1}dsd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{is}\mathfrak{g}(z)\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)s^{\frac{\tau}{2}-1}dsd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}$

and

$\displaystyle\langle\Omega\rangle_{E}=\frac{\int_{\mathbb{R}^{\tau}}e^{-\left\lVert{\Psi}\right\lVert^{2}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}{\int_{\mathbb{R}^{\tau}}e^{-\left\lVert{\Psi}\right\lVert^{2}}d\Psi}$

where $\sqrt{\Sigma}^{-1}$ has the eigenvalues

$\displaystyle\lambda_{k}=\left(\frac{am\omega^{2}}{2\hbar}+\frac{m}{a\hbar}\left(1-\cos\left(\frac{2\pi
k}{\tau}\right)\right)\right)^{-\frac{1}{2}}.$

###  5.1. The Wick rotation

We can compute $\langle\Omega\rangle_{E}$ by choosing an eigenbasis
$(e_{j})_{j}$ of $\sqrt{\Sigma}^{-1}$ such that $\Psi=\sum_{j}\Psi_{j}e_{j}$
and $\sqrt{\Sigma}^{-1}e_{j}=\lambda_{j}e_{j}$. Then, we obtain

$\displaystyle\langle\Omega\rangle_{E}=$
$\displaystyle\frac{\int_{\mathbb{R}^{\tau}}e^{-\left\lVert{\Psi}\right\lVert^{2}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}{\int_{\mathbb{R}^{\tau}}e^{-\left\lVert{\Psi}\right\lVert^{2}}d\Psi}$
$\displaystyle=$
$\displaystyle\frac{\int_{\mathbb{R}^{\tau}}\prod_{j=0}^{\tau-1}e^{-\Psi_{j}^{2}}\frac{1}{\tau}\sum_{k=0}^{\tau-1}(\lambda_{k}\Psi_{k})^{\alpha}d\Psi}{\int_{\mathbb{R}^{\tau}}\prod_{j=0}^{\tau-1}e^{-\Psi_{j}^{2}}d\Psi}$
$\displaystyle=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\frac{\int_{\mathbb{R}}e^{-\Psi_{k}^{2}}(\lambda_{k}\Psi_{k})^{\alpha}d\Psi_{k}}{\int_{\mathbb{R}}e^{-\Psi_{k}^{2}}d\Psi_{k}}$
$\displaystyle=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{\left((-1)^{\alpha}+1\right)\Gamma\left(\frac{\alpha+1}{2}\right)}{2\Gamma\left(\frac{1}{2}\right)}.$

For $\alpha=2$, we conclude

$\displaystyle\langle\Phi^{2}\rangle_{E}=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{2}\frac{\Gamma\left(\frac{1}{2}+1\right)}{\Gamma\left(\frac{1}{2}\right)}=\frac{1}{2\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{2}.$

To compute $\lim_{\tau\to\infty}\langle\Phi^{2}\rangle_{E}$, we use
$\lim_{\tau\to\infty}\frac{1}{\tau}\sum_{k=0}^{\tau-1}f\left(\frac{2\pi
k}{\tau}\right)=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)dx$ to obtain222This
$\tau\to\infty$ result coincides with the $N\to\infty$ limit of equation
(C.29) in Creutz-Freedman.

$\displaystyle\lim_{\tau\to\infty}\langle\Phi^{2}\rangle_{E}=$
$\displaystyle\frac{1}{4\pi}\int_{0}^{2\pi}\left(\frac{am\omega^{2}}{2\hbar}+\frac{m}{a\hbar}\left(1-\cos\left(x\right)\right)\right)^{-1}dx$
$\displaystyle=$
$\displaystyle\left.\frac{\arctan\left(\sqrt{\frac{2m}{a\hbar}\frac{2\hbar}{am\omega^{2}}+1}\tan\left(\frac{x}{2}\right)\right)}{2\pi\sqrt{\frac{am\omega^{2}}{2\hbar}\left(\frac{am\omega^{2}}{2\hbar}+\frac{2m}{a\hbar}\right)}}\right|_{x=0}^{\pi}$
$\displaystyle+\left.\frac{\arctan\left(\sqrt{\frac{2m}{a\hbar}\frac{2\hbar}{am\omega^{2}}+1}\tan\left(\frac{x}{2}\right)\right)}{2\pi\sqrt{\frac{am\omega^{2}}{2\hbar}\left(\frac{am\omega^{2}}{2\hbar}+\frac{2m}{a\hbar}\right)}}\right|_{x=\pi}^{2\pi}$
$\displaystyle=$
$\displaystyle\frac{\hbar}{2m\omega\sqrt{\frac{a^{2}\omega^{2}}{4}+1}}$

and thus

$\displaystyle E_{0}^{\mathrm{continuum}}=\lim_{a\to
0}\lim_{\tau\to\infty}m\omega^{2}\langle\Phi^{2}\rangle_{E}=$
$\displaystyle\frac{\hbar\omega}{2}.$

For the $\zeta$-regularized computations to be consistent and independent of
the choice of gauge, it suffices to show
$\langle\Phi^{2}\rangle(0)=\frac{i}{2\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{2}$.
If that is the case, we not only obtain the correct continuum limit for the
ground state energy of the harmonic oscillator on the lattice, but the ground
state energy coincides with the Wick rotated ground state energy result for
all choices of lattice spacing $a$ and number of lattice sites $\tau$.

###  5.2. The global distributional gauge

The first type of gauge will be global in time and distributionally
regularizing. These are generally of the form

$\displaystyle\forall r\in\mathbb{R}_{>0}\ \forall\Phi\in\partial
B_{\mathbb{R}^{\tau}}:\
\mathfrak{g}(z)(r,\Phi)=r^{z}\mathfrak{g}_{\partial}(z)(\Phi)=r^{z}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right).$

Using such a gauge, we can evaluate the integral with respect to $s$ in

$\displaystyle\langle\Omega\rangle(z)=$ $\displaystyle\frac{\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{is}\mathfrak{g}(z)\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)s^{\frac{\tau}{2}-1}dsd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{is}\mathfrak{g}(z)\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)s^{\frac{\tau}{2}-1}dsd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}$

via analytic continuation of the Laplace transform $\mathcal{L}$ and obtain

$\displaystyle I(z):=$ $\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{is}\mathfrak{g}(z)\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)s^{\frac{\tau}{2}-1}dsd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)$ $\displaystyle=$ $\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{is}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)s^{\frac{\tau+z+\alpha}{2}-1}dsd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)$ $\displaystyle=$ $\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)\mathcal{L}\left(s\mapsto
s^{\frac{\tau+z+\alpha}{2}-1}\right)(-i)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)$ $\displaystyle=$ $\displaystyle
i^{\frac{\tau+z+\alpha}{2}}\Gamma\left(\frac{\tau+z+\alpha}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi).$

Hence,

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{i^{\frac{\tau+z+\alpha}{2}}\Gamma\left(\frac{\tau+z+\alpha}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{i^{\frac{\tau+z}{2}}\Gamma\left(\frac{\tau+z}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}$ $\displaystyle=$
$\displaystyle\frac{i^{\frac{\alpha}{2}}\Gamma\left(\frac{\tau+z+\alpha}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\Gamma\left(\frac{\tau+z}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}$

and

$\displaystyle\langle\Omega\rangle(0)=$
$\displaystyle\frac{i^{\frac{\alpha}{2}}\Gamma\left(\frac{\tau+\alpha}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\Gamma\left(\frac{\tau}{2}\right)\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\partial B_{\mathbb{R}^{\tau}})}.$

If we choose the eigenbasis of $\Sigma$ to parametrize $\mathbb{R}^{\tau}$
again, then

$\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)=$
$\displaystyle\frac{1}{\tau}\sum_{j=0}^{\tau-1}\lambda_{j}^{\alpha}\int_{\partial
B_{\mathbb{R}^{\tau}}}\Psi_{j}^{\alpha}d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)$

needs to be evaluated. In this case, we can perform the calculation
analytically

$\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\Psi_{j}^{\alpha}d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)=$
$\displaystyle\int_{0}^{\pi}\ldots\int_{0}^{\pi}\int_{0}^{2\pi}\cos(\varphi_{1})^{\alpha}\prod_{j=1}^{\tau-2}\sin(\varphi_{j})^{\tau-1-j}d\varphi_{\tau-1}\ldots
d\varphi_{1}$ $\displaystyle=$ $\displaystyle\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau-1}}}\left(\partial
B_{\mathbb{R}^{\tau-1}}\right)\int_{0}^{\pi}\cos(\varphi)^{\alpha}\sin(\varphi)^{\tau-2}d\varphi$
$\displaystyle=$ $\displaystyle\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau-1}}}\left(\partial
B_{\mathbb{R}^{\tau-1}}\right)\frac{\left((-1)^{\alpha}+1\right)\Gamma\left(\frac{\alpha+3}{2}\right)\Gamma\left(\frac{\tau-1}{2}\right)}{(\alpha+1)\Gamma\left(\frac{\tau+\alpha}{2}\right)}$

which yields

$\displaystyle\langle\Omega\rangle(0)=$
$\displaystyle\frac{1}{\tau}\sum_{j=0}^{\tau-1}\lambda_{j}^{\alpha}\frac{i^{\frac{\alpha}{2}}\Gamma\left(\frac{\tau+\alpha}{2}\right)\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau-1}}}\left(\partial
B_{\mathbb{R}^{\tau-1}}\right)\left((-1)^{\alpha}+1\right)\Gamma\left(\frac{\alpha+3}{2}\right)\Gamma\left(\frac{\tau-1}{2}\right)}{\Gamma\left(\frac{\tau}{2}\right)\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\partial
B_{\mathbb{R}^{\tau}})(\alpha+1)\Gamma\left(\frac{\tau+\alpha}{2}\right)}$
$\displaystyle=$
$\displaystyle\frac{i^{\frac{\alpha}{2}}}{(\alpha+1)\tau}\sum_{j=0}^{\tau-1}\lambda_{j}^{\alpha}\frac{\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau-1}}}\left(\partial
B_{\mathbb{R}^{\tau-1}}\right)\left((-1)^{\alpha}+1\right)\Gamma\left(\frac{\alpha+3}{2}\right)\Gamma\left(\frac{\tau-1}{2}\right)}{\Gamma\left(\frac{\tau}{2}\right)\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\partial B_{\mathbb{R}^{\tau}})}$ $\displaystyle=$
$\displaystyle\frac{i^{\frac{\alpha}{2}}\left((-1)^{\alpha}+1\right)\Gamma\left(\frac{\alpha+3}{2}\right)}{(\alpha+1)\sqrt{\pi}\tau}\sum_{j=0}^{\tau-1}\lambda_{j}^{\alpha}$
$\displaystyle=$
$\displaystyle\frac{i^{\frac{\alpha}{2}}\left((-1)^{\alpha}+1\right)\Gamma\left(\frac{\alpha+1}{2}\right)}{2\sqrt{\pi}\tau}\sum_{j=0}^{\tau-1}\lambda_{j}^{\alpha}.$

In particular, for $\alpha=2$ we obtain

$\displaystyle\langle\Phi^{2}\rangle(0)=$
$\displaystyle\frac{i\Gamma\left(\frac{3}{2}\right)}{\sqrt{\pi}\tau}\sum_{j=0}^{\tau-1}\lambda_{j}^{2}=\frac{i}{2\tau}\sum_{j=0}^{\tau-1}\lambda_{j}^{2}$

and thus the correct ground state energy.

###  5.3. The global absolutely integrable gauge

For the gauge to yield an absolutely integrable regularization, we need to
remove the pole in $0$. For instance, a gauge of the form

$\displaystyle\forall r\in\mathbb{R}_{>0}\ \forall\Phi\in\partial
B_{\mathbb{R}^{\tau}}:\ \mathfrak{g}(z)(r,\Phi)=\begin{cases}1&,\ r\leq 1\\\
\mathfrak{g}_{\partial}(z)(\Phi)r^{z}&,\ r>1\end{cases}$

may be chosen. Then

$\displaystyle I(z):=$ $\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\int_{\mathbb{R}_{>0}}e^{is}\mathfrak{g}(z)\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{s}\sqrt{\Sigma}^{-1}\Psi\right)s^{\frac{\tau}{2}-1}dsd\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)$ $\displaystyle=$ $\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)\int_{0}^{1}e^{is}s^{\frac{\tau+\alpha}{2}-1}ds$
$\displaystyle+\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)\int_{\mathbb{R}_{>1}}e^{is}s^{\frac{\tau+z+\alpha}{2}-1}ds$
$\displaystyle=$ $\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)i^{\frac{\tau+\alpha}{2}}\gamma\left(\frac{\tau+\alpha}{2},-i\right)$
$\displaystyle+\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)i^{\frac{\tau+z+\alpha}{2}}\Gamma\left(\frac{\tau+z+\alpha}{2},-i\right).$

In particular, for $\mathfrak{g}_{\partial}(z)=1$ we obtain

$\displaystyle I(z)=$ $\displaystyle\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)i^{\frac{\tau+\alpha}{2}}\left(\gamma\left(\frac{\tau+\alpha}{2},-i\right)+\Gamma\left(\frac{\tau+z+\alpha}{2},-i\right)i^{\frac{z}{2}}\right)$

and thus

$\displaystyle\langle\Omega\rangle(z)=\frac{\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\partial
B_{\mathbb{R}^{\tau}})}\frac{i^{\frac{\alpha}{2}}\left(\gamma\left(\frac{\tau+\alpha}{2},-i\right)+\Gamma\left(\frac{\tau+z+\alpha}{2},-i\right)i^{\frac{z}{2}}\right)}{\left(\gamma\left(\frac{\tau}{2},-i\right)+\Gamma\left(\frac{\tau+z}{2},-i\right)i^{\frac{z}{2}}\right)}.$

At $z=0$, we may use

$\displaystyle\frac{i^{\frac{\alpha}{2}}\left(\gamma\left(\frac{\tau+\alpha}{2},-i\right)+\Gamma\left(\frac{\tau+\alpha}{2},-i\right)\right)}{\left(\gamma\left(\frac{\tau}{2},-i\right)+\Gamma\left(\frac{\tau}{2},-i\right)\right)}=\frac{i^{\frac{\alpha}{2}}\Gamma\left(\frac{\tau+\alpha}{2}\right)}{\Gamma\left(\frac{\tau}{2}\right)}$

and comparing the remaining integrals to the results of subsection 5.2, we
observe that the two choices of gauge give the same result (as expected since
we have gauge invariance). At the same time this appears a lot more
complicated than the computation in subsection 5.2. However, there is a major
advantage in that the initial $\mathbb{R}^{\tau}$ integrals can be evaluated
numerically for $\Re(z)\ll 0$ without using the Laplace transform. Hence, we
can compute $\frac{\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\partial B_{\mathbb{R}^{\tau}})}$, which requires
integration over high dimensional spheres, by computing

$\displaystyle\frac{\gamma\left(\frac{\tau}{2},-i\right)+\Gamma\left(\frac{\tau+z}{2},-i\right)i^{\frac{z}{2}}}{i^{\frac{\alpha}{2}}\left(\gamma\left(\frac{\tau+\alpha}{2},-i\right)+\Gamma\left(\frac{\tau+z+\alpha}{2},-i\right)i^{\frac{z}{2}}\right)}\frac{\int_{\mathbb{R}^{\tau}}e^{i\left\lVert{\Psi}\right\lVert^{2}}\mathfrak{g}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}{\int_{\mathbb{R}^{\tau}}e^{i\left\lVert{\Psi}\right\lVert^{2}}\mathfrak{g}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}$

for values of $z$ with $\Re(z)\ll 0$. This may be easier to implement as more
numerical techniques have been developed for integrating over
$\mathbb{R}^{\tau}$ than $\partial B_{\mathbb{R}^{\tau}}$.

###  5.4. The local distributional gauge

With the third approach we want to gauge the transfer matrix (as opposed to
“only” gauging the entire time evolution operator). This is often useful
because it retains locality in time – a property frequently exploited in many
lattice methods; e.g., when computing correlation functions. In terms of the
following computation, this choice of gauge allows us to use Fubini as in the
Wick rotated computation. Let us start with a generic
$\mathfrak{g}(z)(\Phi)=\prod_{j=0}^{\tau-1}\mathfrak{g}_{\Delta}(\Phi_{j})$.
Then we observe

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{\int_{\mathbb{R}^{\tau}}e^{i\left\lVert{\Psi}\right\lVert^{2}}\mathfrak{g}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}{\int_{\mathbb{R}^{\tau}}e^{i\left\lVert{\Psi}\right\lVert^{2}}\mathfrak{g}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)d\Psi}$
$\displaystyle=$
$\displaystyle\frac{\int_{\mathbb{R}^{\tau}}\prod_{j=0}^{\tau-1}e^{i\Psi_{j}^{2}}\mathfrak{g}_{\Delta}(z)\left(\lambda_{j}\Psi_{j}\right)\frac{1}{\tau}\sum_{k=0}^{\tau-1}(\lambda_{k}\Psi_{k})^{\alpha}d\Psi}{\int_{\mathbb{R}^{\tau}}\prod_{j=0}^{\tau-1}e^{i\Psi_{j}^{2}}\mathfrak{g}_{\Delta}(z)\left(\lambda_{j}\Psi_{j}\right)d\Psi}$
$\displaystyle=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\frac{\int_{\mathbb{R}}e^{i\Psi_{k}^{2}}\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\Psi_{k}\right)(\lambda_{k}\Psi_{k})^{\alpha}d\Psi_{k}}{\int_{\mathbb{R}}e^{i\Psi_{k}^{2}}\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\Psi_{k}\right)d\Psi_{k}}$

where

$\displaystyle\lambda_{j}=\left(\frac{am\omega^{2}}{2}+\frac{m}{a}\left(1-\cos\left(\frac{2\pi
j}{\tau}\right)\right)\right)^{-\frac{1}{2}}.$

If we choose
$\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\Psi_{k}\right)=\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\left\lvert{\Psi_{k}}\right\lvert\right)$,
we further obtain

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{\int_{\mathbb{R}_{>0}}e^{i\psi}\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\sqrt{\Psi_{k}}\right)\Psi_{k}^{\frac{\alpha-1}{2}}d\Psi_{k}}{\int_{\mathbb{R}_{>0}}e^{i\psi}\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\sqrt{\Psi_{k}}\right)\Psi_{k}^{-\frac{1}{2}}d\Psi_{k}}.$

Again, we have the choice between a distributionally regularizing gauge or an
absolutely integrable gauge. In this subsection, we will proceed with a
distributionally regularizing gauge
$\mathfrak{g}_{\Delta}(z)(\psi)=\left\lvert{\psi}\right\lvert^{z}$ which
implies

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{\int_{\mathbb{R}_{>0}}e^{i\psi}\psi^{\frac{\alpha+z-1}{2}}d\psi}{\int_{\mathbb{R}_{>0}}e^{i\psi}\psi^{\frac{z-1}{2}}d\psi}=\frac{i^{\frac{\alpha}{2}}\Gamma\left(\frac{\alpha+z+1}{2}\right)}{\Gamma\left(\frac{z+1}{2}\right)\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}$

and hence the required

$\displaystyle\langle\Phi^{2}\rangle(0)=$ $\displaystyle\lim_{z\to
0}\frac{i\Gamma\left(\frac{z+1}{2}+1\right)}{\Gamma\left(\frac{z+1}{2}\right)\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{2}=\frac{i}{2\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{2}.$

###  5.5. The local absolutely integrable gauge

The fourth and final choice of gauge would be an absolutely integrable gauge
for the transfer matrix. For example, we may consider

$\displaystyle\mathfrak{g}_{\Delta}(z)(\psi)=\begin{cases}1&,\
\left\lvert{\psi}\right\lvert<1\\\ \left\lvert{\psi}\right\lvert^{z}&,\
\left\lvert{\psi}\right\lvert\geq 1\end{cases}.$

In this case, we obtain

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{\int_{\mathbb{R}_{>0}}e^{i\psi}\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\sqrt{\psi}\right)\psi^{\frac{\alpha-1}{2}}d\psi}{\int_{\mathbb{R}_{>0}}e^{i\psi}\mathfrak{g}_{\Delta}(z)\left(\lambda_{k}\sqrt{\psi}\right)\psi^{-\frac{1}{2}}d\psi}$
$\displaystyle=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{\int_{0}^{\lambda_{k}^{-1}}e^{i\psi}\psi^{\frac{\alpha-1}{2}}d\psi+\lambda_{k}^{z}\int_{\mathbb{R}_{>{\lambda_{k}^{-1}}}}e^{i\psi}\psi^{\frac{\alpha+z-1}{2}}d\psi}{\int_{0}^{\lambda_{k}^{-1}}e^{i\psi}\psi^{-\frac{1}{2}}d\psi+\lambda_{k}^{z}\int_{\mathbb{R}_{>{\lambda_{k}^{-1}}}}e^{i\psi}\psi^{\frac{z-1}{2}}d\psi}$
$\displaystyle=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{i^{\frac{\alpha+1}{2}}\gamma\left(\frac{\alpha+1}{2},\lambda_{k}^{-1}\right)+\lambda_{k}^{z}i^{\frac{\alpha+z+1}{2}}\Gamma\left(\frac{\alpha+z+1}{2},\lambda_{k}^{-1}\right)}{i^{\frac{1}{2}}\gamma\left(\frac{1}{2},\lambda_{k}^{-1}\right)+\lambda_{k}^{z}i^{\frac{z+1}{2}}\Gamma\left(\frac{z+1}{2},\lambda_{k}^{-1}\right)}$

which in the limit $z\to 0$ yields

$\displaystyle\langle\Omega\rangle(0)=$
$\displaystyle\frac{1}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{i^{\frac{\alpha}{2}}\gamma\left(\frac{\alpha+1}{2},\lambda_{k}^{-1}\right)+i^{\frac{\alpha}{2}}\Gamma\left(\frac{\alpha+1}{2},\lambda_{k}^{-1}\right)}{\gamma\left(\frac{1}{2},\lambda_{k}^{-1}\right)+\Gamma\left(\frac{1}{2},\lambda_{k}^{-1}\right)}=\frac{i^{\frac{\alpha}{2}}}{\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{\alpha}\frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}$

and thus

$\displaystyle\langle\Phi^{2}\rangle(0)=$
$\displaystyle\frac{i}{2\tau}\sum_{k=0}^{\tau-1}\lambda_{k}^{2}.$

##  6\. The harmonic oscillator numerically

While the harmonic oscillator can be solved analytically, as seen in section
5, more complicated lattice theories will need to be solved numerically. To
this end, we may choose between a quantum simulation of the lattice theory
hartung-jansen; jansen-hartung or a classical simulation. In this section, we
will therefore consider the numerical approach to the $\zeta$-regularized
harmonic oscillator using classical computation as an example for
$\zeta$-regularized lattice theories in detail.

If we want to treat the $\zeta$-regularized harmonic oscillator numerically,
we have in principle two options. The first option is to choose a gauge
$\mathfrak{g}$ and compute the dependence on the gauge parameter $z$
explicitly. In this case, we may choose a distributionally regularizing gauge
$\mathfrak{g}(z)(r,\Phi)=r^{z}\mathfrak{g}_{\partial}(z)(\Phi)$ and obtain an
expression of the form

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{i^{\frac{\alpha}{2}}\Gamma\left(\frac{\tau+z+\alpha}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\Gamma\left(\frac{\tau+z}{2}\right)\int_{\partial
B_{\mathbb{R}^{\tau}}}\mathfrak{g}_{\partial}(z)\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}.$

This enables us to take the $z\to 0$ limit explicitly and we are left with the
numerical problem of integrating

$\displaystyle E_{0}=$
$\displaystyle\frac{m\omega^{2}\tau}{2}\frac{\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\partial B_{\mathbb{R}^{\tau}})}$

where $\Omega$ is the observable $\Phi^{2}$. Alternatively, if the analytic
continuation cannot be computed explicitly, then we can use the fact that
$E_{0}(z)=-im\omega^{2}\langle\Omega\rangle(z)$ is comprised of integrals that
are well-posed for $\Re(z)\ll 0$. Hence, $E_{0}(z)$ can be computed for a
sufficiently large number of values $z$ with $\Re(z)\ll 0$. The $z=0$ value
can then be obtained by fitting the $z$-dependence (which is known up to some
parameters) to the thus generated data and extrapolating the fit to $z=0$. In
either case, we need to evaluate high-dimensional spherical integrals.

###  6.1. Solving the spherical integrals

The numerical difficulty of computing

$\displaystyle E_{0}=$
$\displaystyle\frac{m\omega^{2}\tau}{2}\frac{\int_{\partial
B_{\mathbb{R}^{\tau}}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)d\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\Psi)}{\mathrm{vol}_{\partial
B_{\mathbb{R}^{\tau}}}(\partial B_{\mathbb{R}^{\tau}})}$

is the quadrature of the high-dimensional spherical integrals. We are using
“Sobol’ points on the sphere $\partial B_{\mathbb{R}^{\tau}}$” which can be
generated from a “standard” Sobol’ sequence sobol in three steps.

  1. Step 1.

Generate a $\tau$-dimensional Sobol’ sequence (uniform distribution on
$(0,1)^{\tau}$).

  2. Step 2.

Apply the standard normal ($\mathcal{N}(0,1)$) percent point function
$\mathrm{ppf}$ in each coordinate (standard normal distribution on
$\mathbb{R}^{\tau}$).

  3. Step 3.

Normalize each sample (uniform distribution on $\partial
B_{\mathbb{R}^{\tau}}$).

In order to make full use of the properties of the Sobol’ sequence, we
generate $2^{n}$ Sobol’ points for some $n\in\mathbb{N}$ and skip the first
point since the first (non-randomized) Sobol’ point is always
$\left(\frac{1}{2},\ldots,\frac{1}{2}\right)^{T}$ which gets mapped to the
origin under $\mathrm{ppf}$ and thus cannot be normalized.

Let $\mathcal{S}$ be the set of generated Sobol’ points. Then we can
approximate $E_{0}$ via

$\displaystyle
E_{0}\approx\frac{m\omega^{2}\tau}{2}\frac{1}{\\#\mathcal{S}}\sum_{\Psi\in\mathcal{S}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right).$

The convergence in terms of the number of (randomized) Sobol’ samples is shown
in Figure 1. The standard deviations, which were extracted from $100$
simulations each, indicate an error scaling proportional to
$\frac{1}{\sqrt{\\#\mathcal{S}}}$. In other words, the $\zeta$-regularized
lattice simulation of the harmonic oscillator with Lorentzian background is
comparable to a Markov Chain Monte-Carlo simulation in Euclidean background in
terms of numerical effort.

Figure 1. This figure shows the standard deviation of
$\frac{1}{\\#\mathcal{S}}\sum_{\Psi\in\mathcal{S}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right)$
using $N=\\#\mathcal{S}=2^{n}-1$ sample points. The standard deviation is
estimated using $100$ simulations with randomized Sobol’ points. Furthermore,
a least square error fit $C(N+1)^{\alpha}$ to the simulation data is shown.
The exponent $\alpha\approx-0.575$ indicates an error scaling
$\mathrm{error}\propto\frac{1}{\sqrt{\\#\mathcal{S}}}$. The simulation
parameters are $m=\omega=\hbar=1$, $\tau=128$, and $a=0.01$.

Having a reasonable quadrature to estimate

$\displaystyle
E_{0}\approx\frac{m\omega^{2}\tau}{2}\frac{1}{\\#\mathcal{S}}\sum_{\Psi\in\mathcal{S}}\Omega\left(\sqrt{\Sigma}^{-1}\Psi\right),$

we can now consider the behavior of $E_{0}(m,\omega,\hbar)$ as a function of
the lattice spacing $a$ and the number of lattice points $\tau$. The physical
volume is then given by $T=a\tau$. In order to extract the continuum limit, we
are interested in the limit $a\searrow 0$ at constant physical volume $T$ and
the limit $T\nearrow\infty$. Figure 2 (top) shows simulation results of
$E_{0}(m=1,\omega=1,\hbar=1)$ with lattice spacing
$a\in\\{2^{-2},2^{-3},2^{-4},2^{-5}\\}$ at physical volume $T=32$. Similarly,
Figure 2 (bottom) shows simulation results of $E_{0}(m=1,\omega=1,\hbar=1)$
with lattice spacing $a=2^{-5}$ at physical volume $T\in\\{4,8,16,32\\}$. The
error bars are the standard deviations extracted from $100$ simulations using
$2^{10}-1$ randomized Sobol’ points and the horizontal lines are the analytic
continuum limit $E_{0}(m=1,\omega=1,\hbar=1)=\frac{1}{2}$. In particular, we
note that the results for $a=2^{-5}$ and $T\in\\{16,32\\}$ are already
comparable to the continuum limit at the given level of accuracy.

---
Figure 2. This figure shows physical simulations indicating the lattice
spacing $a\to 0$ limit at constant physical volume $T=32$ (top) and the
physical volume $T\to\infty$ limit (bottom) at constant lattice spacing
$a=2^{-5}$. The horizontal lines are the continuum ground state energy
$E_{0}=\frac{\hbar\omega}{2}$. In the simulations, $m=\omega=\hbar=1$ was used
and error bars denote standard deviations extracted from $100$ independent
simulations. Each simulation uses $2^{10}-1$ randomized Sobol’ points. It
should be noted that the vertical axis range is within $1\%$ of $E_{0}$ at the
top and within $4\%$ of $E_{0}$ at the bottom.

###  6.2. Extrapolation from $\Re(z)\ll 0$

In more complicated situations, it may not always be possible to compute the
dependence on the gauge parameter $z$ explicitly, or computing the spherical
integrals may be more difficult than computing the $\mathbb{R}^{\tau}$
integrals

$\displaystyle\langle\Omega\rangle(z)=\frac{\int_{\mathbb{R}^{\tau}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)\Omega(\Phi)d\Phi}{\int_{\mathbb{R}^{\tau}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)d\Phi}.$

In such cases, it may be advantageous to compute $\langle\Omega\rangle(z)$ at
$\Re(z)\ll 0$ where the integrals are well-defined and numerically well-posed.
In order to guide the extrapolation to $z=0$, we need to account for the
unknown $z$-dependence $\Delta(z)$. However, if the gauge is chosen well, then
we can use the fact that the $z$-dependence comes from analytic continuations
of Laplace transforms of poly-($\log$)-homogeneous distributions, for which
the general form of these Laplace transforms in terms of the variable $z$ are
known. Thus, by combining those constituent terms, we obtain a parametric
expression $\Delta(z,p)$ such that $\Delta(z,p^{*})=\langle\Omega\rangle(z)$
is satisfied for some set of parameters $p^{*}$. Once this general dependence
$\Delta(z,p)$ is identified by analyzing the poly-($\log$)-homogeneity
structure of the integrands, the parameters $p^{*}$ can be obtained from
fitting $\Delta(z,p)$ against the numerical data $\langle\Omega\rangle(z)$.
Finally we obtain $\langle\Omega\rangle(0)$ by extrapolation, i.e., by
evaluating $\Delta(z,p^{*})$ at $z=0$.

  * Example

Consider an action $S$ for which there exists a closed compact manifold $M$ in
$\mathbb{R}^{\tau}$ such that $\forall\Phi\in\mathbb{R}^{\tau}\
\exists!\varphi\in M\ \exists!r\in\mathbb{R}_{>0}:\ \Phi=r\varphi$ and
$S(\Phi)=cr^{\sigma}$ for some constant $c\in\mathbb{R}$. Furthermore, let
$\Omega(\Phi)=\sum_{\iota\in I}r^{\delta_{\iota}}\omega_{\iota}(\varphi)$ be a
polyhomogeneous observable and choose the gauge
$\mathfrak{g}(z)(\Phi)=(1+r)^{z}$. Then,

$\displaystyle\langle\Omega\rangle(z)=$
$\displaystyle\frac{\int_{\mathbb{R}^{\tau}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)\Omega(\Phi)d\Phi}{\int_{\mathbb{R}^{\tau}}e^{iS(\Phi)}\mathfrak{g}(z)(\Phi)d\Phi}$
$\displaystyle=$ $\displaystyle\frac{\sum_{\iota\in
I}\int_{\mathbb{R}_{>0}}e^{icr^{\sigma}}(1+r)^{z}r^{\delta_{\iota}+\tau-1}dr\int_{M}\omega_{\iota}(\varphi)d\mathrm{vol}_{M}(\varphi)}{\int_{\mathbb{R}_{>0}}e^{icr^{\sigma}}(1+r)^{z}r^{\tau-1}dr\mathrm{vol}_{M}(M)}.$

Thus,

$\displaystyle I_{\delta}(z):=$
$\displaystyle\int_{\mathbb{R}_{>0}}e^{icr^{\sigma}}(1+r)^{z}r^{\delta}dr$
$\displaystyle=$
$\displaystyle\int_{0}^{1}e^{icr^{\sigma}}(1+r)^{z}r^{\delta}dr+\int_{\mathbb{R}_{>1}}e^{icr^{\sigma}}(1+r)^{z}r^{\delta}dr$
$\displaystyle=$ $\displaystyle\sum_{k\in\mathbb{N}_{0}}{z\choose
k}\left(\int_{0}^{1}e^{icr^{\sigma}}r^{\delta+k}dr+\int_{\mathbb{R}_{>1}}e^{icr^{\sigma}}r^{\delta+z-k}dr\right)$
$\displaystyle=$ $\displaystyle\sum_{k\in\mathbb{N}_{0}}{z\choose
k}\frac{1}{\sigma}\left(\int_{0}^{1}e^{icr}r^{\frac{\delta+k+1-\sigma}{\sigma}}dr+\int_{\mathbb{R}_{>1}}e^{icr}r^{\frac{\delta+z-k+1-\sigma}{\sigma}}dr\right)$
$\displaystyle=$ $\displaystyle\sum_{k\in\mathbb{N}_{0}}{z\choose
k}\frac{1}{\sigma}\left(\frac{\gamma\left(\frac{\delta+k+1}{\sigma},-ic\right)}{(-ic)^{\frac{\delta+k+1}{\sigma}}}+\frac{\Gamma\left(\frac{\delta+z-k+1}{\sigma},-ic\right)}{(-ic)^{\frac{\delta+z-k+1}{\sigma}}}\right)$

implies

($***$) $\displaystyle\langle\Omega\rangle(z)=$ $\displaystyle\sum_{\iota\in
I}C_{\iota}\frac{\sum_{k\in\mathbb{N}_{0}}{z\choose
k}\frac{1}{\sigma}\left(\frac{\gamma\left(\frac{\delta_{\iota}+\tau+k}{\sigma},-ic\right)}{(-ic)^{\frac{\delta_{\iota}+\tau+k}{\sigma}}}+\frac{\Gamma\left(\frac{\delta_{\iota}+\tau+z-k}{\sigma},-ic\right)}{(-ic)^{\frac{\delta_{\iota}+\tau+z-k}{\sigma}}}\right)}{\sum_{k\in\mathbb{N}_{0}}{z\choose
k}\frac{1}{\sigma}\left(\frac{\gamma\left(\frac{\tau+k}{\sigma},-ic\right)}{(-ic)^{\frac{\tau+k}{\sigma}}}+\frac{\Gamma\left(\frac{\tau+z-k}{\sigma},-ic\right)}{(-ic)^{\frac{\tau+z-k}{\sigma}}}\right)}.$

In this expression, everything is known a priori with the exception of the
$C_{\iota}$. In other words, we can fit Equation ($***$ ‣ Example) against
simulation data to extract the $C_{\iota}$ and obtain

$\displaystyle\langle\Omega\rangle(0)=$ $\displaystyle\sum_{\iota\in
I}C_{\iota}\frac{\Gamma\left(\frac{\delta_{\iota}+\tau}{\sigma}\right)}{(-ic)^{\frac{\delta_{\iota}}{\sigma}}\Gamma\left(\frac{\tau}{\sigma}\right)}.$

$\square$

---
Figure 3.  This figure shows the continuum extrapolation of $1000$ simulations
with randomly chosen $z\in[-2\tau-1,-\tau-1]$, $\tau=40$, $a=0.1$, and
$m=\omega=\hbar=1$. The top plot shows the simulation data, the fit, and the
$z=0$ value $E_{0}(a,\tau)$. The middle plot shows the extrapolated $z=0$
values with standard deviation as error using $50,100,\ldots,950,1000$
simulations and compares these to the $z=0$ value $E_{0}(a,\tau)$. The bottom
plot shows the standard deviations of the extrapolated $z=0$ values as well as
a least square error fit $CN^{\alpha}$ where $N$ is the number of simulations
used for the $z=0$ extrapolation. The exponent $\alpha\approx-0.490$ indicates
an error scaling
$\mathrm{error}\propto\frac{1}{\sqrt{\\#\mathrm{simulations}}}$.

With the choice of gauge $g(z)(r,\varphi)=r^{z}$, the $z$-dependence
$\Delta(z,p)$ is much simpler; namely affine linear
$\Delta(z,p)=p_{0}+p_{1}z$. Using up to $1000$ simulations at randomly chosen
$z\in[-2\tau-1,-\tau-1]$ with $\tau=40$, $a=0.1$, and $m=\omega=\hbar=1$, we
can fit $\Delta(z,p)$ and obtain the $z=0$ extrapolation $p_{0}$. Figure 3
(top) shows the least square error fit of $\Delta(z,p)$ to the data of $1000$
simulations as well as the $z=0$ value $E_{0}(a,\tau)$. In Figure 3 (middle)
the extrapolated value $p_{0}$ extracted from $50,100,\ldots,950,1000$
simulations including the standard deviation as error is shown. The comparison
to the $z=0$ value $E_{0}(a,\tau)$, which we can compute analytically as in
section 5, shows compatibility of the extrapolated value $p_{0}$ with
$E_{0}(a,\tau)$ already for a relatively small number of simulations. In
Figure 3 (bottom), we compare the standard deviation of $p_{0}$ extracted from
$50,100,\ldots,950,1000$ simulations to a least square error fit of the form
$C\cdot(\\#\text{simulations})^{\alpha}$. The fitted value of
$\alpha\approx-0.490$ indicates an error scaling proportional to
$\frac{1}{\sqrt{\\#\text{simulations}}}$.

##  7\. Conclusion

In this article, we have studied the $\zeta$-regularization of lattice field
theories with Lorentzian background metrics. In particular, we have discussed
the connection between lattice field theories with Euclidean backgrounds and
Lorentzian backgrounds in section 2, in order to understand how the
$\zeta$-regularization arises as a naïve means of regularizing lattice
expectation values when it is not possible to analytically continue the theory
to Euclidean time. In section 3, we have then made the connection from the
naïvely $\zeta$-regularized lattice theory to the Hamiltonian formulation so
far discussed in the literature hartung-iwota; hartung-jmp; hartung-jansen;
hartung-jansen-gauge-fields; jansen-hartung. This allowed us to gain insights
into the two main questions that need considering when extending the
$\zeta$-regularization in the Hamiltonian formulation to the Lagrangian
formulation commonly used to describe lattice field theories.

  1. Q1:

What kind of gauge families $\mathfrak{g}$ can be used to $\zeta$-regularize
lattice field theories in a Lorentzian background?

  2. Q2:

Under what conditions is it possible to construct a $\zeta$-regularized
Hamiltonian theory starting from a “naïvely” $\zeta$-regularized lattice field
theory?

Question Q1 is at least partially answered by Theorem 1 and Theorem 2. In
order to $\zeta$-regularize a lattice field theory in a Lorentzian background,
$\mathfrak{g}(z)$ must be chosen in such a way, that the corresponding
integrals are well-defined for $\Re(z)\ll 0$. This is generally satisfied if
$\mathfrak{g}(z)$ has an asymptotic behavior of
$\left\lvert{\mathfrak{g}(z)(\Phi)}\right\lvert\leq
c(1+\left\lVert{\Phi}\right\lVert^{2})^{\delta\Re(z)}$ or exhibits even faster
convergence to zero as $\left\lVert{\Phi}\right\lVert\to\infty$. Furthermore,
for an increasing, exhaustive sequence of compacta $(K_{n})_{n\in\mathbb{N}}$,
the compactified vacuum expectation values $\left(\frac{\langle
1_{K_{n}^{T}}\Omega\mathfrak{g}\rangle_{L}}{\langle
1_{K_{n}^{T}}\mathfrak{g}\rangle_{L}}\right)_{n\in\mathbb{N}}$ need to be
locally bounded in $C(D)$ where $D\subseteq\mathbb{C}$ is open, connected, and
contains $0$ as well as a half-space with $\Re(z)\ll 0$. This local
boundedness condition is likely to be the most intricate assumption to
satisfy, as this is the likely point at which renormalization and modeling
decisions can have a major impact.

It is important to note that the answer to Q1 merely implies the existence of
a $\zeta$-regularized lattice theory. Without further information, there is no
reason to assume that this $\zeta$-regularized lattice theory is at all
physical. In order to conclude physicality of the expectation values extracted
from the $\zeta$-regularized lattice theory, we need to connect the Lagrangian
formulation of the $\zeta$-regularized lattice theory to the Hamiltonian
formulation studied to date hartung-iwota; hartung-jmp; hartung-jansen;
hartung-jansen-gauge-fields; jansen-hartung which includes a proof of
physicality. As such, answering Q2 allows us to make physical sense of the
$\zeta$-regularized lattice vacuum expectation values.

Question Q2 is partially answered by Theorem 3 and Theorem 4. In essence, we
need to ensure that the resulting gauged transfer matrix or gauged time
evolution operator is a suitable family of Fourier integral operators; namely
a gauged family in the sense of hartung-iwota; hartung-jmp; hartung-jansen;
hartung-jansen-gauge-fields; jansen-hartung. This can be ensured by choosing
$\mathfrak{g}(z)$ or $\mathfrak{g}_{\Delta}(z)$ to be in a Hörmander class
$S^{\delta\Re(z)}$ or (poly-)homogeneous. While this restricts the class of
gauge families $\mathfrak{g}$ sufficient to construct the $\zeta$-regularized
lattice theory as per question Q1, computationally this is not a bad class of
functions as many analytic results are known and, thus, allowing for
computational simplifications in numerical simulations of $\zeta$-regularized
lattice theories.

In section 4, we then considered the classical limit of $\zeta$-regularized
lattice theories and observed the remarkable fact that the classical limit is
independent of the value for $z$. If we consider the impact of the gauge on a
classical action, this can be rationalized since the gauge adds an imaginary
component to a classical action. Thus, if we expect for only the real-part of
the gauged action to be retained in the classical limit, then Theorem 5
confirms this point of view.

Finally, we illustrated the theory of $\zeta$-regularized lattice theories
developed in this article at the example of the harmonic oscillator. First, we
treated it analytically in section 5; second, we considered it numerically in
section 6. Both the analytic and numeric studies of the harmonic oscillator
focus on the ground state energy.

In section 5, we computed the ground state energy of the harmonic oscillator
on the lattice five times. Initially, we computed the ground state energy by
using a Wick rotation, i.e., using the canonical approach of analytic
continuation to imaginary (Euclidean) time. This served as a comparison for
the four choices of gauge families that follow, as well as to show that the
lattice theory has the correct continuum limit. The four gauge families that
were used to compute the $\zeta$-regularized harmonic oscillator on the
lattice with Minkowski background highlight the four cases arising from two
binary choices we can make. On one hand, we have the choice of gauging local
in time vs. global in time, that is, gauging the transfer matrix or the entire
time evolution operator. On the other hand, we can choose between absolutely
integrable gauge families and distributionally regularizing gauge families,
where the former is more directly applicable to numerical simulations but the
latter allows for analytic shortcuts via the Laplace transform. Of course, all
four gauge families reproduced the correct ground state energy, as is to be
expected since we have gauge-independence.

The numerical study of the $\zeta$-regularized harmonic oscillator on the
lattice with Minkowski background in section 6 focused on the numerical
obstacles arising in classical simulations of $\zeta$-regularized lattice
theories as opposed to quantum simulations which are a special case of the
quantum simulations discussed in hartung-jansen; jansen-hartung. In
particular, we noted that distributionally regularizing gauge families allow
for numerical shortcuts via Laplace transform and can make for simpler
functional dependence on the complex parameter $z$, which needs to be fitted
against simulation data for extrapolation if the extrapolation cannot be done
analytically to begin with. However, distributionally regularizing gauge
families require numerical quadratures over high-dimensional manifolds (a
sphere for the harmonic oscillator). Absolutely integrable gauge families, on
the other hand, “only” require quadratures in high-dimensional vector spaces
$\mathbb{R}^{\tau}$. As more powerful integration techniques exist for
integrating over $\mathbb{R}^{\tau}$ than codimension $1$ submanifolds of
$\mathbb{R}^{\tau}$, the absolutely integrable gauge families might be
preferable from a numerical point of view. This question may be exacerbated
further when considering gauge fields in lattice field theories.
Unfortunately, the cost of choosing absolutely integrable gauge families is a
more complicated dependence on the complex parameter $z$ which may make
extrapolation to $z=0$ more difficult.

## References