# Symmetry-reduced Loop Quantum Gravity: Plane Waves, Flat Space and the
Hamiltonian Constraint

F. Hinterleitner111e-mail<EMAIL_ADDRESS>
Institute of Theoretical Physics and Astrophysics, Masaryk University,
Kotlářská 2, 611 37 Brno, Czech Republic

###### Abstract

Loop quantum gravity methods are applied to a symmetry-reduced model with
homogeneity in two dimensions, derived from a Gowdy model [5, 6]. The
conditions for propagation of unidirectional plane gravitational waves at
exactly the speed of light are set up in form of null Killing equations in
terms of Ashtekar variables and imposed as operators on quantum states of the
system. Due to the effective one-dimensionality, holonomies and holonomy
operators appear as simple phase factors. In correspondence state functions
might be considered as $U(1)$ elements with the usual inner product. Under the
assumption of equal spacing of the eigenvalues of geometrical quantities the
solutions are not normalizable in this sense. With decreasing spacing for
growing eigenvalues, as introduced for example in [11], the situation becomes
worse. Taking over the inner product from the genuine gauge group $SU(2)$ of
LQG renders the obtained states normalizable, nevertheless fluctuations of
geometrical quantities remain divergent. In consequence, the solutions of the
Killing conditions are modified, which means allowing for small fluctuations
of the propagation speed, i. e. dispersion of gravitational waves. Vacuum
fluctuations of Minkowski space are sketched. Finally the same methods are
applied to the Hamiltonian constraint with the same result concerning
normalizability. With such a modification also the constraint is not exactly
satisfied any more, which indicates the necessary presence of some kind of
interacting matter.

##  1 Introduction

The main motivation for this paper is the question about the occurrence of
quantum vacuum dispersion of electromagnetic or gravitational waves. There
exists an extensive literature about this topic, energy-dependent speed of
light, Lorentz invariance violations, modifications of special relativity,…,
see an overview in [1], most of it based on semiclassical models. A more
specific example, dealing with gravitational wave dispersion, is [2], where
generally the effects of inhomogeneous perturbations of homogenous
cosmological models on gravitational and electromagnetic waves are studied. In
the present paper we study a simpler, but fully quantized model, symmetry
reduced by homogeneity in two dimensions, but manifestly inhomogeneous in the
third one, so that the dynamics happens in one spatial dimension. This
restriction to effectively 1+1 dimensions is very convenient for the study of
the speed of waves, so that here we are able to obtain results from a fully
quantized model.

The dispersionfree propagation of plane gravitational waves in classical
theory, briefly introduced in section 2, is translated into the existence of a
null Killing vector field in the direction of propagation. In section 3 the
system is quantized along the rules of loop quantum gravity, defining triad
and connection operators.

In the main part of the paper, beginning with section 4, the Killing
conditions are formulated in terms of loop quantum operators acting on quantum
states on one-dimensional spin networks, effectively on the vertex functions,
whereas the edge functions can be given freely. All these considerations are
locally restricted to a vertex. The solutions turn out to be not normalizable,
at least not in the apparently obvious way. After making the states
normalizable by choosing a more convenient inner product, a further
modification, applied in section 5, is necessary to render also length and
volume fluctuations finite. These modifications violate the Killing conditions
for a propagating at exactly the speed light. The resulting velocity
uncertainties are negatively correlated to the uncertainties of metrical
quantities.

In chapter 6 the formalism is used to indicate fluctuations of the geometry in
Minkowski space, and in 7 the methods developed in the main part lead to
solutions of a certain discrete version the Hamiltonian constraint,
constructed analogously to the Killing conditions. Like the solutions of the
latter ones, they require renormalization.

##  2 Classical background

###  2.1 Unidirectional gravitational waves

To establish a quantum model for unidirectional plane gravitational waves, we
start with a rather restricted sector of full General Relativity, the
polarized cosmological Gowdy model [3] and impose further restrictions to
unidirectionality at the classical level by postulating a null Killing field
in form of conditions on the phase space variables.

For the sake of a better intuition we begin with an even more restricted
solution that describes precisely unidirectional plane waves, according to
Ehlers and Kundt [4], with the following space-time metric

${\rm d}s^{2}={\rm d}t^{2}-L^{2}e^{2\beta}{\rm d}x^{2}-L^{2}e^{-2\beta}{\rm
d}y^{2}-{\rm d}z^{2},$ (1)

where $L$ and $\beta$ depend on $t-z$ or on $t+z$. The function $L$,
describing the transversal area, is called “background factor”, and $\beta$,
describing the deformation of a ring of test particles in the transversal
$(x,y)$ plane by gravitational forces, is called “wave factor”. The one and
only Einstein equation for this system is

$L^{\prime\prime}+(\beta^{\prime})^{2}L=0.$ (2)

It determines the background, when a wave factor is given, or vice versa, so
one function is left free, up to the condition that $L$ and $L^{\prime\prime}$
must have opposite signs. In consequence, a positive $L$ in front of a wave
pulse leads necessarily to $L=0$ somewhere inside or behind the pulse and the
coordinate system becomes invalid there. Gravitational waves of his kind,
traveling exactly at the speed of light along $z$, do not interact with each
other and thus can be superimposed like in a linear theory.

###  2.2 Generalization, Ashtekar variables

The metric components $g_{tt}$ and $g_{zz}$ in (1) being equal to one, this
model does not display any longitudinal fluctuations. In a quantum
description, however, we would expect geometric fluctuations in every
direction, even in flat space without gravitational waves. For this reason we
start from a slightly more general classical point of view, taken from a
polarized Gowdy model and a quantum version thereof [5, 6]. The metric for a
polarized model is given by

${\rm d}s^{2}=-N^{2}{\rm d}t^{2}+{\cal E}\frac{E^{y}}{E^{x}}\,{\rm
d}x^{2}+{\cal E}\frac{E^{x}}{E^{y}}\,{\rm d}y^{2}+\frac{E^{x}E^{y}}{\cal
E}\,{\rm d}z^{2},$ (3)

where $E^{x}$ and $E^{y}$ are radial densitized triad components,

$E^{a}=\sqrt{({E^{a}}_{1})^{2}+({E^{a}}_{2})^{2}},\hskip 56.9055pta=x,y,$ (4)

in the $x$ and $y$ direction, and $\cal E$ is the corresponding quantity in
the $z$ direction. All of them, as well as the lapse function $N$, are assumed
to be functions of $t$ and $z$ (not necessarily only of $t\pm z$ from the
beginning). The model is effectively 1+1-dimensional.

In the case $N^{2}=g_{zz}$ and when the metric functions depend only on $t-z$,
the Einstein equation corresponding to (2) is

$\frac{{\cal E}^{\prime\prime}}{\cal E}+\frac{1}{2}\left(\frac{{\cal
E}^{\prime}}{\cal E}\right)^{2}-(\ln E^{x}E^{y})^{\prime}\,\frac{{\cal
E}^{\prime}}{\cal
E}+\frac{1}{2}\left(\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}\right)^{2}=0.$
(5)

For comparison with (2) we set $\sqrt{\cal E}=L$ and
$\;\ln(E^{y}/E^{x})=2\beta$, then we find

$L^{\prime\prime}-(\ln
E^{x}E^{y})^{\prime}\,L^{2}\,L^{\prime}+(\beta^{\prime})^{2}\,L=0,$ (6)

which reduces to (2), when $E^{x}E^{y}=$ const. In (5) the second derivative
of $L$ does not necessarily have the opposite sign as $L$, in contrast to (2).

The canonically conjugate connection to $E^{x}$, $E^{y}$, and $\cal E$ is
derived from the (diagonal) extrinsic curvature

$K_{aa}=\frac{1}{2N}\,\dot{g}_{aa}.$ (7)

The sign convention is the same as in [6]. With the aid of the triad covectors
${e^{i}}_{b}$ we construct the connection components
${K_{a}}^{i}={e^{i}}_{b}\,{K_{a}}^{b}$, in the notation of [5, 6], and find
the radial components $K_{a}=\sqrt{({K_{a}}^{1})^{2}+({K_{a}}^{2})^{2}}\,$
with $a=x,y$, relevant for the polarized model, and $\cal A$,

$K_{x}=\frac{1}{N}\,\partial_{t}\sqrt{{\cal E}\frac{E^{y}}{E^{x}}},\hskip
22.76219ptK_{y}=\frac{1}{N}\,\partial_{t}\sqrt{{\cal
E}\frac{E^{x}}{E^{y}}},\hskip 22.76219pt{\cal
A}=\frac{1}{N}\,\partial_{t}\sqrt{\frac{E^{x}E^{y}}{\cal E}}$ (8)

with the nonzero Poisson brackets

$\\{K_{a}(z),E^{b}(z^{\prime})\\}=\kappa\,{\delta_{a}}^{b}\,\delta(z-z^{\prime}),\hskip
14.22636pt\\{{\cal A}(z),{\cal
E}(z^{\prime})\\}=\kappa\,\delta(z-z^{\prime}).$ (9)

$\kappa=2G_{\rm N}/\pi$, $G_{\rm N}$ is Newton’s gravitational constant. With
the rotations around the $z$ axis being fixed by the Gauß constraint at an
early stage, we have only the canonical variables introduced above, subject to
the Hamiltonian constraint and one diffeo constraint, generating diffeos along
$z$:

$\displaystyle H=-\frac{1}{\kappa}\frac{1}{\sqrt{{\cal
E}E^{x}E^{y}}}\left[K_{x}E^{x}K_{y}E^{y}+(K_{x}E^{x}+K_{y}E^{y}){\cal
AE}-\rule{0.0pt}{17.07164pt}\right.\hskip 14.22636pt$ (10)
$\displaystyle\;\;\;\;-\left.\frac{1}{4}\,({\cal
E}^{\prime})^{2}-\left(\frac{1}{2}{\cal
E}\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}\right)^{2}+\frac{1}{2}{\cal
EE}^{\prime}(\ln E^{x}E^{y})^{\prime}-{\cal EE}^{\prime\prime}\right]=0,$
$\displaystyle
D=\frac{1}{\kappa}\,(K_{x}^{\prime}E^{x}+K_{y}^{\prime}E^{y}-{\cal
AE}^{\prime})=0.\hskip 128.0374pt$ (11)

###  2.3 One-way Killing conditions

Unidirectional waves at the speed of light can be characterized by a null
Killing vector field $k^{\mu}$ with the corresponding covector

$k_{\mu}=\left(Nk,0,0,\pm\sqrt{\frac{E^{x}E^{y}}{\cal E}}\,k\right),$ (12)

$k$ is a function of $t$ and $z$. The plus/minus sign determines plane waves
into the positive/negative $z$ direction. The Killing equation $k_{0;0}=0$
leads to

$\dot{k}=\pm\sqrt{\frac{\cal E}{E^{x}E^{y}}}\,N^{\prime}k,$ (13)

the equation $k_{3;3}=0$ to

$k^{\prime}=\pm\frac{1}{N}\left(\partial_{t}\sqrt{\frac{E^{x}E^{y}}{\cal
E}}\right)k,$ (14)

where dot and prime mean derivatives w. r. to $t$ and $z$. Equating the $z$
derivative of $(\ln k)\dot{}$ from (13) with the $t$ derivative of $(\ln
k)^{\prime}$ from (14) yields the integrability condition

$\left(\frac{N^{\prime}}{\sqrt{g_{zz}}}\right)^{\prime}=\left(\frac{\dot{\sqrt{g_{zz}}}}{N}\right)\\!\\!\raisebox{11.38109pt}{$\dot{}$}.$
(15)

The equation $k_{0;3}+k_{3;0}=0$ does not give rise to further relations, with
(13) and (14) it is identically fulfilled. Assuming some lapse function $N$
means to choose a time unit, and then (15) fixes the dependence of $g_{zz}$ on
the coordinate time $t$ and on $z$. For the choice $N=\sqrt{g_{zz}}$ it gives

$\left(\ln g_{zz}\right)^{\prime\prime}=\left(\ln g_{zz}\right)\ddot{}$

with the solution

$\ln g_{zz}=f(t+z)+g(t-z),$

leading to the following form of the integrability condition

$g_{zz}=\frac{E^{x}E^{y}}{\cal E}=F(t+z)\cdot G(t-z).$ (16)

This condition is only necessary, not sufficient; for the actual existence of
a null Killing vector field in one direction, either $F$ or $G$ must be
constant.

In the remaining two non-trivial Killing equations $k_{1;1}=0$ and $k_{2;2}=0$
the time derivatives in the Christoffel symbols $\Gamma^{0}_{11}$ and
$\Gamma^{0}_{22}$ are expressed in terms of $K_{x}$ and $K_{y}$ with the help
of (8):

$\Gamma_{11}^{0}=\frac{1}{2N^{2}}\partial_{t}\left({\cal
E}\frac{E^{y}}{E^{x}}\right)=\frac{K_{x}}{N}\sqrt{{\cal
E}\frac{E^{y}}{E^{x}}}$ (17)

and

$\Gamma_{22}^{0}=\frac{1}{2N^{2}}\partial_{t}\left({\cal
E}\frac{E^{x}}{E^{y}}\right)=\frac{K_{y}}{N}\sqrt{{\cal
E}\frac{E^{x}}{E^{y}}}.$ (18)

Inserted into the Killing conditions this gives relations independent of $N$,

$\displaystyle U_{x}:=E^{x}K_{x}+\frac{1}{2}\,{\cal
E}^{\prime}+\frac{1}{2}\,{\cal
E}\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}=0,$ (19) $\displaystyle
U_{y}:=E^{y}K_{y}+\frac{1}{2}\,{\cal E}^{\prime}-\frac{1}{2}\,{\cal
E}\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}=0.$ (20)

With this choice of signs, these are one-way conditions for waves propagating
into the positive $z$ direction ($\partial_{t}=-\partial_{z}$). For waves in
the opposite direction the signs of the derivative terms have to be changed.

As already mentioned, the integrability condition (16) does not uniquely
determine left- or right-going waves, they must be picked out by hand. In view
of this one might also choose a more elementary, alternative approach:
Postulating that the derivative of the metric components $g_{xx}$ and $g_{yy}$
w. r. to the physical time $Nt$ is equal to minus the derivative w. r. to the
physical length $\sqrt{g_{zz}}\,z$,

$\sqrt{\frac{E^{x}E^{y}}{\cal E}}\,\dot{g}_{aa}=-Ng_{aa}^{\prime},\hskip
28.45274pta=1,2$ (21)

leads to (19) and (20), when the time derivatives are expressed in terms of
$K_{x}$ and $K_{y}$. When $N$ is chosen to be equal to $\sqrt{E^{x}E^{y}/{\cal
E}}$, for a wave traveling into the positive $z$ direction, this gives further

$U_{z}:={\cal AE}-\frac{1}{2}\,{\cal E}^{\prime}+\frac{1}{2}\,{\cal
E}\left(\ln E^{y}E^{x}\right)^{\prime}=0$ (22)

as a consequence of the integrability condition (15), but as this is dependent
on a choice of the lapse function, it is not an independent Killing condition.
In the present approach $U_{z}$ just appears later on to make the relations
(23) and (30), appearing below, more convenient. Namely, when we combine
$U_{x}$, $U_{y}$, their derivatives w. r. to $z$, and $U_{z}$, with the
Hamiltonian and the diffeo constraint, we obtain the following relation for
waves in the positive $z$ direction

$\kappa\left(\sqrt{\frac{E^{x}E^{y}}{\cal E}}\,H+D\right)=-\frac{1}{\cal
E}\left(U_{x}U_{y}+U_{x}U_{z}+U_{y}U_{z}\right)+(U_{x}+U_{y})^{\prime}.$ (23)

This expression has density weight two. $\sqrt{E^{x}E^{y}/{\cal E}}\,H$ is the
evolution generator in the case of the special lapse function
$N=\sqrt{g_{zz}}$. $U_{x}\equiv 0$ and $U_{y}\equiv 0$ are sufficient for the
essential equivalence of the Hamiltonian with the diffeo constraint,
$\sqrt{E^{x}E^{y}/{\cal E}}\,H=\pm D$, i. e. the strict proportionality of the
time and the $z$ derivatives, whereas $U_{z}=0$ is not needed.

###  2.4 Einstein equations, massless fields

When the metric components in (3) are merely functions of $t$ and $z$ and not
only of $t\pm z$, but the lapse function is chosen to be
$N=\sqrt{\frac{E^{x}E^{y}}{\cal E}}$, the Einstein tensor has five independent
non-vanishing components, the diagonal ones and $G_{03}=G_{30}$, giving rise
to five equations for six variables. $G_{00}$ and $G_{33}$ do not contain time
derivatives of the phase space variables, so in the case of vacuum they
represent the Hamiltonian and the diffeo constraint,

$G_{00}=-\frac{1}{\cal E}\sqrt{\frac{E^{x}E^{y}}{\cal E}}\,H,\hskip
28.45274ptG_{03}=-\frac{1}{\cal E}\,D.$ (24)

In the presence of matter they represent the gravitational energy and momentum
density in the chosen reference frame. The remaining Einstein equations are
dynamical, they determine $\dot{K}_{x}$, $\dot{K}_{y}$, and $\dot{\cal A}$ in
terms of the fundamental geometrical (and perhaps matter) variables, thus
leading to a system of three coupled nonlinear first-order PDEs for time
evolution.

When, under the assumption $N=\sqrt{g_{zz}}$, the one-way conditions
(19,20,23) are satisfied, then all metric components depend on $t\pm z$, and
the Einstein tensor reduces to three identical nonzero components,

$G_{00}=G_{33}=-G_{03},$ (25)

(see (5)) of constraint type, no further nontrivial components arise. - With
the above assumptions the time evolution is already fixed. If, in addition,
the vacuum Einstein equation is imposed on (25), one obtains pure
unidirectional gravitational waves and $G_{00}$ becomes the Hamiltonian
constraint, which is in this case equivalent to the diffeo constraint.

In the presence of matter $G_{ab}$ is coupled to the energy-momentum tensor of
a field. As the trace of the Einstein tensor in (25) is zero, with or without
matter, the field must be massless for consistency. Without imposing the
vacuum constraint $G_{00}=0$, the condition “time derivative $=\pm z$
derivative” does not distinguish between pure gravitational waves and the
gravitational background accompanying massless fields.

###  2.5 Alternative set of conditions, physical meaning

The sum and the difference of the two $N$-independent conditions $U_{x}$ and
$U_{y}$ for wave propagation into the positive $z$ direction are

$\displaystyle U_{+}:=K_{x}E^{x}+K_{y}E^{y}+{\cal E}^{\prime}=0,$ (26)
$\displaystyle U_{-}:=K_{x}E^{x}-K_{y}E^{y}+{\cal
E}\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}=0.$ (27)

With $K_{x}$ and $K_{y}$ from (8) inserted, $U_{+}=0$ means

$\sqrt{\frac{E^{x}E^{y}}{\cal E}}\,\dot{\cal E}=-N{\cal E}^{\prime},$ (28)

namely the physical time derivative of $\cal E$ is equal to the true physical
spatial derivative along $z$. This refers to the background factor, the
transversal area $\cal E$. In the same way $U_{-}=0$ means

$\sqrt{\frac{E^{x}E^{y}}{\cal
E}}\,\left(\ln\frac{E^{y}}{E^{x}}\right)\\!\\!\raisebox{8.53581pt}{$\dot{}$}=-N\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime},$
(29)

indicating equivalence of time and space derivatives of the wave factor
$\ln(E^{y}/E^{x})$.

As (23) can also be written in the form

$\kappa\left(\sqrt{\frac{E^{x}E^{y}}{\cal E}}\,H+D\right)=\frac{1}{4\cal
E}\,U_{-}^{2}-\frac{1}{4\cal E}\,U_{+}^{2}-\frac{1}{\cal
E}\,U_{+}U_{z}+U_{+}^{\prime},$ (30)

it follows from the equivalence of $H$ and $D$ (particularly when $H=0$ and
$D=0$) that $U_{-}=0$, whenever $U_{+}=0$, thus the wave factor propagates at
the speed of light, when the background factor does.

The one-way conditions Poisson-commute weakly with the constraints of General
Relativity (see next subsection), $U_{+}$ also with $U_{-}$ and with itself,
$U_{-}$ does not commute with itself in general. In detail the Poisson
brackets between the one-way conditions are the following:

$\displaystyle\\{U_{x}[f],U_{x}[g]\\}=\\{U_{y}[f],U_{y}[g]\\}=\frac{1}{2}\,{\cal
E}[f^{\prime}g-fg^{\prime}],$
$\displaystyle\\{U_{x}[f],U_{y}[g]\\}=\frac{1}{2}\,{\cal
E}[fg^{\prime}-f^{\prime}g],$ (31)
$\displaystyle\\{U_{+}[f],U_{+}[g]\\}=\\{U_{+}[f],U_{-}[g]\\}=0,$
$\displaystyle\\{U_{-}[f],U_{-}[g]\\}=2\,{\cal E}[f^{\prime}g-fg^{\prime}],$

As can be seen, some of the conditions do Poisson-commute only for constant
test functions $f$ and $g$, this means that only their averages commute. As
one-way conditions are not gauge constraints, we are only looking for a
certain class of solutions, not for a theory with additional symmetries, so
the Poisson brackets between the conditions do not have to be imposed as
further conditions. In quantum theory the nonzero commutators will not be
relevant as operators, in other words, we do not have to enlarge the algebra
of constraints by the one-way conditions, this would over-constrain the
system.

###  2.6 Relation to the constraints of GR

$U_{+}$ and $U_{-}$, and even $U_{z}$, as densities of weight one, do weakly
Poisson-commute with the diffeo constraint,

$\displaystyle\\{U_{+}[f],D[g]\\}=U_{+}^{\prime}[fg]+U_{+}[fg^{\prime}]=-U_{+}[f^{\prime}g],$
(32)
$\displaystyle\\{U_{-}[f],D[g]\\}=U_{-}^{\prime}[fg]+U_{-}[fg^{\prime}]=-U_{-}[f^{\prime}g],$
(33) $\displaystyle\\{U_{z}[f],D[g]\\}=-U_{z}[f^{\prime}g].$ (34)

The Poisson brackets with the Hamiltonian constraint are the following

$\\{U_{+}[f],H[g]\\}=U_{+}\left[\sqrt{\frac{\cal
E}{E^{x}E^{y}}}\,f^{\prime}g\right]-H[fg]=\sqrt{\frac{\cal
E}{E^{x}E^{y}}}\>U_{+}[f^{\prime}g]-H[fg],$ (35)
$\\{U_{-}[f],H[g]\\}=\frac{1}{\sqrt{{\cal
E}E^{x}E^{y}}}\left(U_{-}\left[({\cal E}f)^{\prime}g\right]-{\cal
E}\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}U_{+}[fg]\right).$ (36)

$U_{+}$ weakly commutes with $H$, the condition $U_{-}$ for the wave factor
needs also the condition $U_{+}=0$ for the background factor to be satisfied,
in order to commute with $H$. Just for completeness: the coordinate dependent
condition $U_{z}$ does not commute with $H$ with arbitrary lapse functions, it
does with $N=\sqrt{g_{zz}}$,

$\displaystyle\left\\{U_{z}[f],H\left[\sqrt{\frac{E^{x}E^{y}}{\cal
E}}\right]\right\\}=-U_{z}\,[f^{\prime}]+\frac{1}{\cal
E}\,(U_{x}U_{y}+U_{x}U_{z}+U_{y}U_{z})\,[f]-$ (37)
$\displaystyle-\left(\frac{1}{2}\,(\ln
g_{xx})^{\prime}\,U_{x}+\frac{1}{2}\,(\ln g_{yy})^{\prime}\,U_{y}+2\,(\ln{\cal
E})^{\prime}\,U_{z}\right)[f].$

Summarizing the classical part, we state that there are two Killing conditions
for making time evolution equivalent to space translation. In the following
sections we shall investigate in how far these relations carry over to quantum
theory.

##  3 Quantum states and operators

Quantum states of the present model are one-dimensional spin network functions
on a sequence of edges and vertices along the $z$ axis, made from point
holonomies at vertices (labeled by $i$ along the $z$-axis) and the usual path
holonomies along edges. The former ones are

$|m\rangle=e^{i\frac{m}{2}K_{x}(v_{i})}\hskip 14.22636pt\mbox{and}\hskip
14.22636pt|n\rangle=e^{i\frac{n}{2}K_{y}(v_{i})}$ (38)

with integer $m$ and $n$.

In [6] the state functions are elements of the Bohr compactification of the
reals [8], with representation labels $m$ and $n$. But in fact all these
holonomies are phases and in the following sections 4 - 7 we will need only
sequences of integers $m$ and $n$, determined by the raising and lowering
parts of the connection operators, see (40) below. So nothing is lost when we
dispense with ${\bf R}_{\rm Bohr}$ and reduce our interest until further notes
effectively to $U(1)$ representations with the labels $m/2$ and $n/2$.

Edge holonomues are

$|k\rangle=e^{i\frac{k}{2}\int_{e}{\cal A}\,{\rm d}z}$ (39)

where $e$ denotes an edge and $k$ is an integer number, also denoting a
representation of $U(1)$ by its very nature, connected to rotations around the
$z$ axis, which are fixed by the Gauß constraint. $k$ has the meaning of a
quantum number of transversal area, as well as $m$ and $n$ are area quantum
numbers, too.

The connection components are promoted to holonomy operators. As in the
present model all holonomies are mere phase factors, we will, in contrast to
[6], where holonomy operators act as traces of $SU(2)$ operators on states,
introduce $U(1)$ vertex operators by first setting

$K_{a}\rightarrow\hat{K}_{a}(v_{i})=\sin\left(K_{a}\right)=\frac{1}{i}\left(e^{\frac{i}{2}K_{a}}-e^{-\frac{i}{2}K_{a}}\right).$
(40)

When $\hat{K}_{x}$ acts on a holonomy $|m\rangle$, this gives

$\frac{1}{i}(|m+1\rangle-|m-1\rangle),$ (41)

so $K_{x}$ acts simply as a raising and lowering operator. Also here we
content ourselves with a lower mathematical level in favor of clearer and
simpler calculations. (For comparison: in [7] calculations like the following
ones in section 4 were done with traces of $SU(2)$ operators, leading to
qualitatively the same outcomes.) The same is carried out with $K_{y}$, so
that

$\hat{K}_{y}(v_{i})|n\rangle=\frac{1}{i}(|n+1\rangle-|n-1\rangle),$ (42)

if $|n\rangle$ is a holonomy of $K_{y}$.

In $E^{x}$ and $E^{y}$ the functional derivatives $\delta/\delta K_{a}$ are
replaced by the derivatives $\partial/\partial K_{a}(v_{i})$, so that the
densitized triads act as multiplication operators, when integrated over some
interval $\cal I$ containing a vertex

$\int_{\cal I}{\rm d}z\,\hat{E}^{x}(z)|m\rangle=\gamma l_{\rm
P}^{2}\,m|m\rangle,\hskip 28.45274pt\int_{\cal I}{\rm
d}z\,\hat{E}^{y}(z)|n\rangle=\frac{\gamma l_{\rm P}^{2}}{2}\,n|n\rangle.$ (43)

$l_{\rm P}$ is the Planck length, $\gamma$ is the Immirzi parameter.
$\gamma\,l_{\rm P}^{2}$ is the typical scale of discreteness of both
$|m\rangle$ and $|n\rangle$.

Analogously $\cal A$ is promoted to an operator; being a scalar density we
have

${\cal A}\rightarrow 2\sin\left(\frac{1}{2}\int_{e}{\cal A}\,{\rm
d}z\right)=\frac{1}{i}\left(e^{\frac{i}{2}\int_{e}{\cal A}\,{\rm
d}z}-e^{-\frac{i}{2}i\int_{e}{\cal A}\,{\rm d}z}\right).$ (44)

Its action on edge holonomies is

$\int_{e}{\rm d}z\,\hat{\cal
A}(z)\>|k\rangle=\frac{1}{i}\left(|k+1\rangle-|k-1\rangle\right),$ (45)

and the conjugate multiplication operator $\cal E$ acts as

$\hat{\cal E}|k\rangle=\frac{\gamma l_{\rm P}^{2}}{2}\,k\,|k\rangle.$ (46)

So, $\hat{\cal E}$ is essentially the transversal area operator in units of
$\gamma\,l_{\rm P}^{2}/2$, $\hat{V}=\widehat{\sqrt{{\cal E}E^{x}E^{y}}}$ is
the volume operator.

##  4 The one-way conditions as operators on quantum state functions,
solutions

###  4.1 An operator corresponding to $U_{+}$

The expressions $K_{x}E^{x}$ and $K_{y}E^{y}$, containing connection
components, act nontrivially on vertex holonomies, but leave edge holonomies
unchanged, so we can assume the latter ones as arbitrarily given. $U_{+}$,
commuting with itself, is easier to promote to an operator than $U_{-}$, it
can naturally be associated to each vertex. First, we choose a symmetric
factor ordering,

$U_{+}=\sqrt{E^{x}}\,K_{x}\,\sqrt{E^{x}}+\sqrt{E^{y}}\,K_{y}\,\sqrt{E^{y}}+{\cal
E}^{\prime}.$ (47)

The derivative term ${\cal E}^{\prime}$ is naturally interpreted as
$(k_{+}-k_{-})/\epsilon$, where $k_{\pm}$ are the edge labels right and left
from the considered vertex $v_{i}$, and $\epsilon$ is the length of the same
interval $\cal I$ as in the definition of the triad operators. To obtain an
operator from the density expressions in (47), the equation is integrated over
$\cal I$, so ${\cal E}^{\prime}$ gives simply the difference $k_{+}-k_{-}$.
Taking a (double) vertex holonomy

$|m,n\rangle:=|m\rangle\otimes|n\rangle,$ (48)

the action of $\hat{U}_{+}:=\int_{\cal I}{\rm d}z\,U_{+}(z)$, with
$\gamma_{\rm P}^{2}$ set equal to one, on such a holonomy is the following

$\displaystyle\hat{U}_{+}|m,n\rangle=\frac{1}{2i}\left[\sqrt{m(m+1)}\,|m+1,n\rangle-\sqrt{m(m-1)}\,|m-1,n\rangle\right.$
(49) $\displaystyle\hskip
28.45274pt+\left.\sqrt{n(n+1)}\,|m,n+1\rangle-\sqrt{n(n-1)}\,|m,n-1\rangle\right]+(k_{+}-k_{-})\,|m,n\rangle.$

####  4.1.1 The difference equation and solutions

For a vertex state of the form

$|\psi\rangle=\sum_{m,n}a_{m,n}\,|m,n\rangle$ (50)

the condition $\hat{U_{+}}\,|\psi\rangle=0$ leads to the difference equation

$\displaystyle\sqrt{m(m+1)}\,a_{m+1,n}-\sqrt{m(m-1)}\,a_{m-1,n}+\sqrt{n(n+1)}\,a_{m,n+1}-\sqrt{n(n-1)}\,a_{m,n-1}$
$\displaystyle=i\,b|m,n\rangle.$ (51)

$b$, coming from the difference of the cross section area (the background),
left and right, is defined as

$b:=(k_{+}-k_{-}).$ (52)

As an introduction, in the following we consider in detail the simplest case
with $b=0$, indicating the same cross-section area to the left and to the
right of the given vertex (in classical GR exactly constant cross section
means no waves at all).

The partial difference equation (4.1.1) is separable and has a one-parameter
family of solutions. When separated, the parameter is the separation constant.
Here we prefer not to separate and to parametrize the solutions by initial
values $a_{m,m}$ along the diagonal in an $(m,n)$ diagram and the differences
$a_{m,m-1}-a_{m+1,m}$. The advantage is some systematics: the first kind of
initial values give rise to symmetric, the second one to antisymmetric
solutions.

For three different sets of initial values we give three explicit solutions.
Inserting $(m,n)=(0,0)$ gives identically zero, so we can choose the
coefficients $a_{1,0}$ and $a_{0,1}$ (and also $a_{0,0}$) arbitrarily, setting
them equal to zero, we get $a_{0,2}=0$ from $(m,n)=(0,1)$, and so on. In this
way all $a_{m,0}$ and $a_{0,n}$ become zero and we can leave zero-volume
states aside and restrict our attention to the quadrant ($m\geq 1$, $n\geq 1$)
in a discrete coordinate system of pairs $(m,n)$.

From $(m,n)=(1,1)$ we find $a_{1,2}+a_{2,1}=0$, from $(2,2)$ follows
$a_{2,3}+a_{3,2}=0$, and so on. This means that we can conveniently choose the
diagonal elements $a_{m,m}$ as free initial values, with the symmetric parts
$a_{m+1,m}+a_{m,m-1}$ determined to be zero by the difference equation,
whereas the antisymmetric part remains undetermined. So, from initial values
$a_{m,m}$ with $a_{m,m\pm 1}=0$ we can construct symmetric solutions, on the
other hand, from $a_{m,m}=0$ and arbitrary antisymmetric initial values
$a_{m,m-1}=-a_{m-1,m}$ we get antisymmetric ones.

####  4.1.2 Three explicit solutions for $b=0$

Let’s consider the simplest case of a symmetric solution coming from the
initial value $a_{1,1}=1$, with all other diagonal values and the values
beside the diagonal being zero. Inserting $(2,1)$ into (4.1.1) gives

$a_{3,1}=\frac{1}{\sqrt{3}}=\frac{2\sqrt{1\cdot 3}}{1\cdot 2\cdot 3},$ (53)

for $(3,2)$ we get

$a_{4,2}=\frac{2\sqrt{2\cdot 4}}{2\cdot 3\cdot 4}.$ (54)

From these coefficients and a few further ones one can easily read off the
general formula

$a_{n+2,n}=\frac{2\sqrt{n(n+2)}}{n(n+1)(n+2)}.$ (55)

Inserting $(n+2,n)$ gives $a_{n+3,n}=0$ $\forall n$, for $(n+3,n)$ inserted
into (4.1.1) we obtain

$a_{n+4,n}=\frac{4\sqrt{n(n+4)}}{(n+1)(n+2)(n+3)},$ (56)

for $(n+5,n)$ we find

$a_{n+6,n}=\frac{6\,\sqrt{n(n+6)}}{(n+2)(n+3)(n+4)},$ (57)

from which we may read off the general form

$a_{m,n}=\frac{8\,(m-n)\,\sqrt{mn}}{(m+n-2)(m+n)(m+n+2)},$ (58)

where $n$ was set equal to $\frac{m+n}{2}-\frac{m-n}{2}$ in the denominator,
and prove by induction. With the definitions $p=1/2(m+n)$ and $q=1/2(m-n)$,
this becomes

$a_{p+q,p-q}=\frac{2q\,\sqrt{p^{2}-q^{2}}}{(p-1)p(p+1)}.$ (59)

In the same way from the only nonzero initial value $a_{2,2}=1$ one obtains
the slightly more complicated symmetric solution

$a_{p+q,p-q}=\frac{4q\,(p^{2}-3q^{2}-1)\,\sqrt{p^{2}-q^{2}}}{p\,(p^{2}-1)\,(p^{2}-4)}.$
(60)

For large $p$ both these solutions go as $p^{-2}$. In the symmetric solutions
for all nonzero coefficients $a_{m,n}$ the sum $m+n$ is even, in the
antisymmetric solutions only coefficients with odd sum $m+n$ are nonzero, in
both cases there is a chessboard like pattern in the discrete $(m,n)$ diagram.
As an antisymmetric example we present the solution coming from
$a_{2,1}=-a_{1,2}=1$ with all other possible initial values being zero. For
$m>n$ it is

$a_{m,n}=\frac{6\sqrt{2}[(m-n)^{2}-1]\,\sqrt{mn}}{[(m+n)^{2}-1][(m+n)^{2}-9]},$
(61)

or

$a_{p+q,p-q}=\frac{6\sqrt{2}\,(4q^{2}-1)\,\sqrt{p^{2}-q^{2}}}{(4p^{2}-1)(4p^{2}-9)}$
(62)

(for half-integer $q$), respectively. Here the coefficients go as $p^{-3}$ for
large $p$. The nontrivial case with $b\neq 0$, where symmetric and
antisymmetric solutions get inevitably mixed, will be treated together with
$U_{-}$. Before doing this we have a look at the normalizability problem of
the solutions.

####  4.1.3 Normalizability

With the obvious inner product $\langle
m,n|m^{\prime},n^{\prime}\rangle=\delta_{m,m^{\prime}}\,\delta_{n,n^{\prime}}$
of holonomies, which offers itself at first sight, the norm of a solution
$|\psi\rangle$ (50) is equal to the sum of $a_{m,n}^{2}$ over $m>0$ and $n>0$.
Relabeled with $p$ and $q$ and reordered, for $a_{1,1}=1$ the norm square of
solution (58), or in the form (59) respectively, is

$\langle\psi|\psi\rangle=1+\sum_{p=2}^{\infty}\,2\sum_{q=1}^{p-1}a_{p+q,p-q}^{2}.$
(63)

The finite sum over $q$ for fixed $p$ is

$\sum_{q=1}^{p-1}\frac{4q^{2}(p^{2}-q^{2})}{p^{2}(p^{2}-1)^{2}}=\frac{4}{(p^{2}-1)^{2}}\sum_{q=1}^{p-1}q^{2}-\frac{4}{p^{2}(p^{2}-1)^{2}}\sum_{q=1}^{p-1}q^{4}.$
(64)

The leading term of $\sum q^{2}$ being $(p-1)^{3}/3$ and that of $\sum q^{4}$
being $(p-1)^{5}/5$, for large $p$ the sum (64) goes as
$\frac{8}{15}\,p^{-1}$, thus the norm square (63) diverges logarithmically.
The same occurs with the solutions (60) and (62).

To understand the meaning of this, recall that with zero spatial derivative
${\cal E}^{\prime}$ (in the discrete case with zero difference $b=0$),
$\hat{U_{+}}\,|\psi\rangle=0$ determines states with zero space and time
derivatives of the background. This satisfies one condition for quantum
Minkowski space, but without $\hat{U_{-}}\,|\psi\rangle=0$ it is not
restrictive enough to give at least an approximative model of flat space. On
the other hand, with zero difference of cross section areas, it deals only
with longitudinal fluctuations, as can be expected for the present essentially
$1+1$ \- dimensional model.

The non-normalizability (under the assumption of the inner product introduced
above) points out a problem already at this early stage and indicates that
also more realistic solutions must be renormalized in one or another way. This
can be achieved by some kind of cutoff in $p$ (suppression of large
volumes/lengths) or in $q$ (suppression of large wave factors), or in both
directions. As can be seen from (59, 60, 62), the coefficients fall off
sufficiently rapidly in the $p$-direction for fixed $q$, whereas the sums over
$q$ from 1 to $p-1$ seem to be “too large”. Intuitively it seems more
convenient to cut off large wave factors by introducing some decreasing
function of $q$. Physically this could be plausibly interpreted as caused by
interactions of gravity with other fields in the form of particle creation at
the cost of quanta of geometry, mainly of those with large wave factors. By
any modification of the solution the condition $\hat{U_{+}}\,|\psi\rangle=0$
becomes violated, so that the time derivative of the background picks up
fluctuations. Nevertheless, for a cutoff in $p$ or a symmetric cutoff in $q$
the expectation value of $\hat{U_{+}}$ remains zero, there are just
fluctuations in time.

###  4.2 The operator $U_{-}$

####  4.2.1 Difference equation for $U_{-}$

$U_{-}$ contains the derivative of $\ln(E^{y}/E^{x})$, so it cannot be
associated as straightforwardly as $U_{+}$ to one and the same vertex, the
discrete version must involve at least one further vertex. Lets denote the
labels at the considered vertex as $m_{i}$ and $n_{i}$, and those left and
right as $m_{i\pm 1}$, $n_{i\pm 1}$.

In the classical theory, when background and wave factor are differentiable
functions, it does not make a difference whether we take the left or the right
derivative. In the discrete case the derivative of the wave factor can be
interpreted in different ways:

(i) Central difference

$\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}\rightarrow\ln\frac{n_{i+1}}{m_{i+1}}-\ln\frac{n_{i-1}}{m_{i-1}}.$
(65)

This seems to be the most natural interpretation.

(ii) Left and right difference, involving the local variables

$\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}\rightarrow\ln\frac{n_{i}}{m_{i}}-\ln\frac{n_{i-1}}{m_{i-1}},\hskip
28.45274pt\mbox{or}\hskip
28.45274pt\rightarrow\ln\frac{n_{i+1}}{m_{i+1}}-\ln\frac{n_{i}}{m_{i}}.$ (66)

(iii) Difference left or right from the considered vertex, for example

$\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}\rightarrow\ln\frac{n_{i-1}}{m_{i-1}}-\ln\frac{n_{i-2}}{m_{i-2}}.$
(67)

For the choice (i) or (iii) at each vertex the appertaining variables $m_{i}$
and $n_{i}$ do not occur, $\hat{U_{+}}$ and $\hat{U_{-}}$ commute and have a
common solution for each pair of neighboring vertex holonomies and each pair
$(k_{-},k_{+})$ of adjacent edge labels. In (ii) there is no common solution.
The non-commutativity of the Poisson brackets of the classical $U_{-}$ with
itself is reflected in the non-commutativity of the corresponding operators at
neighboring vertices. For the construction of a global solution at (some
extended part of) the whole SNW choice (i) is not very well-suited, whereas
for (ii) and (iii) one could construct a global solution going from vertex to
vertex. However, it will turn out that already the local solution of the
Killing conditions has to be modified for physical reasons anyway, a global
construction of an exact solution is a secondary problem.

For the expression $\cal E$ at a vertex we take the mean value of the edge
labels left and right:

${\cal E}\rightarrow\frac{1}{2}\,(k_{-}+k_{+})=:k.$ (68)

The difference equation: For the choice (i) for
$\left(\ln(E^{y}/E^{x})\right)^{\prime}$ and (68) for $\cal E$ the difference
equation derived from $U_{-}$ is the following

$\displaystyle\sqrt{m(m+1)}\,a_{m+1,n}-\sqrt{m(m-1)}\,a_{m-1,n}-\sqrt{n(n+1)}\,a_{m,n+1}+\sqrt{n(n-1)}\,a_{m,n-1}$
$\displaystyle=iwa_{m,n},$ (69)

where $m\equiv m_{i}$, $n\equiv n_{i}$, and

$w:=\frac{1}{2}(k_{+}+k_{-})\left(\ln\frac{n_{i+1}}{m_{i+1}}-\ln\frac{n_{i-1}}{m_{i-1}}\right)$
(70)

stands for wave factor and is constant at the vertex under consideration. It
depends only on neighboring vertex and edge labels. (For the choice (ii) we
would have to write $w(m,n)$ instead of simply $w$.)

####  4.2.2 Common solutions

With the decision for $w$ in the form (70), $\hat{U_{+}}$ and $\hat{U_{-}}$
are commuting Hermitian operators, so there is a nontrivial one-parameter
solution. The partial difference equations (4.1.1) and (4.2.1) are in this
case separable with the product ansatz $a_{m,n}=\alpha_{m}\beta_{n}$,
resulting in

$\hat{U_{x}}\,\sum_{m}\alpha_{m}|m\rangle=0\hskip 28.45274pt\mbox{and}\hskip
28.45274pt\hat{U_{y}}\sum_{n}\beta_{n}|n\rangle=0$ (71)

in terms of operators corresponding to the original expressions $U_{x}$ and
$U_{y}$, as $U_{\pm}=U_{x}\pm U_{y}$. The separated difference equations are

$\displaystyle\sqrt{m(m+1)}\,\alpha_{m+1}-\sqrt{m(m-1)}\,\alpha_{m-1}=i(b+w)\alpha_{m},$
(72)
$\displaystyle\sqrt{n(n+1)}\,\beta_{n+1}-\sqrt{n(n-1)}\,\beta_{n-1}=i(b-w)\beta_{n}.$
(73)

With $\alpha_{1}=\beta_{1}=1$ the first few coefficients in a solution of (72)
are

$\displaystyle\alpha_{2}=i\,\frac{b+w}{\sqrt{2}},\hskip
28.45274pt\displaystyle\alpha_{3}=\frac{1}{\sqrt{3}}\,\left(1-\frac{1}{2}\,(b+w)^{2}\right),\hskip
28.45274pta_{4}=i\frac{b+w}{3}\left(2-\frac{1}{2}\,(b+w)^{2}\right),$
$\displaystyle\alpha_{5}=\frac{1}{\sqrt{5}}\left(1-\frac{5}{6}\,(b+w)^{2}+\frac{1}{12}\,(b+w)^{4}\right),\ldots.$
(74)

$\beta_{n}$ is of the same form as $\alpha_{n}$ with $b+w$ replaced by $b-w$.
The resulting coefficients $a_{m,n}=\alpha_{m}\,\beta_{n}$ are symmetric in
$b$ and antisymmetric in $w$.

To see the asymptotic behavior of the solution of (72) and (73) we replace
$\alpha_{m}$ by a differentiable function $\alpha(m)$ and approximate the
difference equation by a differential equation:

$\sqrt{m}\,\frac{\rm d}{{\rm
d}m}\left(\sqrt{m}\,\alpha(m)\right)=i(b+w)\,\alpha(m)$ (75)

with the solution

$\alpha(m)=\frac{1}{\sqrt{m}}\,e^{i(b+w)\,\ln m}.$ (76)

Doing the same with the second equation gives

$\beta(n)=\frac{1}{\sqrt{n}}\,e^{i(b-w)\,\ln n},$ (77)

so that the asymptotic behavior is

$a_{m,n}\rightarrow\frac{1}{\sqrt{mn}}\,e^{i\left(b\,\ln(mn)+w\ln\frac{m}{n}\right)}.$
(78)

As expected, this is not normalizable in the $U(1)$ inner product.

##  5 Reormalization, expectation values

###  5.1 Norm

The solution with arbitrary $b$ and $w$, presented in the foregoing section,
contains the function $1/\sqrt{mn}$ plus powers of $b\pm w$. First we will
consider the simpler state with $b=w=0$, introduced below as contribution to
the vacuum state, (117) as representative and then show that in the general
case the normalization problem is the same.

So far we had assumed the obvious inner product $\langle
m,n|m^{\prime},n^{\prime}\rangle=\delta_{m,m^{\prime}}\delta_{n,n^{\prime}}$
of holonomies (=phase factors) for vertex states. However, basic states in
full LQG are $SU(2)$ holonomies, in the present context they are artificially
squeezed to $U(1)$ phase factors by symmetry reduction.

We might prefer an inner product of the basic holonomies close to that of full
LQG: There the basic SNW functions are the $SU(2)$ matrix elements
${D^{j}(g)^{\mu}}_{\nu}$ of a group element $g$ in a representation labeled by
some half-integer $j$. According to the Peter-Weyl theorem their inner product
is

$\langle{{D^{j}(g)}^{\mu}}_{\nu},{{D^{j^{\prime}}(g)}^{\rho}}_{\sigma}\rangle=\frac{1}{2j+1}\int_{SU(2)}\\!\\!{{D^{j}}(g)^{\mu}}_{\nu}\,{{D^{j^{\prime}}(g)}^{\rho}}_{\sigma}\,{\rm
d}_{\rm
H}(g)=\frac{1}{2j+1}\,\delta^{jj^{\prime}}\,\delta^{\mu\rho}\,\delta_{\nu\sigma},$
(79)

derived from the Haar measure on $SU(2)$. $2j+1$ is the dimension of the
representation $(j)$. Taking this over, we have

$\langle m,m^{\prime}\rangle=\frac{1}{2m+1}\,\delta_{m,m^{\prime}}\hskip
14.22636pt\mbox{and}\hskip 14.22636pt\langle
n,n^{\prime}\rangle=\frac{1}{2n+1}\,\delta_{n,n^{\prime}},$ (80)

thus

$\langle
m,n|m^{\prime},n^{\prime}\rangle=\frac{1}{(2m+1)(2n+1)}\,\delta_{m,m^{\prime}}\,\delta_{n,n^{\prime}},$
(81)

in spite of the fact that the states appear as mere phase factors as an
artifact of symmetry reduction. The label spacing of $m$ and $n$ by $1/2$ has
also been chosen in accordance with $SU(2)$.

Applying this to the simple state with $a_{m,n}=1/\sqrt{mn}$ gives the
following contribution of the state $|m,n\rangle$ to the norm square

$\frac{a_{m,n}^{2}}{(2m+1)(2n+1)}=\frac{1}{m(2m+1)n(2n+1)},$ (82)

going to $1/4m^{2}n^{2}$ for large $m$ and $n$, so in the sense of this scalar
product the state is normalizable.

There is, however, still a problem: $\sqrt{mnk}$, with
$k=\frac{1}{2}(k_{+}+k_{-})$ is the volume eigenvalue of a vertex with
holonomy $|m,n\rangle$ with $k_{\pm}$ denoting the labels of the adjacent edge
functions, which can be chosen arbitrarily without fluctuation. $k$ denoting
the cross section of the cell containing the considered vertex. $\sqrt{mn/k}$
is the length eigenvalue. With the inner product borrowed from $SU(2)$ and
given edge holonomies, the length expectation value
$\frac{1}{\sqrt{k}}\langle\sqrt{mn}\rangle$ of a solution is finite, but the
root-mean-square deviation $\Delta\left(\sqrt{mn}\right)$, containing the
expectation value $\langle mn\rangle$ is not: Going along the $m$ axis, the
sum

$\sum_{m}\frac{|a_{m,1}|^{2}\,m}{(2m+1)\cdot 3}\sim\sum_{m}\frac{1}{6m}$ (83)

diverges logarithmically, as well as the sum along the $n$ axis. On the other
hand, the sums in the direction of the diagonal and parallel to it, like

$\sum_{m}\frac{|a_{m,m}|^{2}\,m^{2}}{(2m+1)^{2}}\sim\sum_{m}\frac{1}{4m^{2}},$
(84)

is finite.

In conclusion, the Killing conditions in the present model do not give
physically acceptable states, at least the states away from the diagonal with
large wave factors $\ln\frac{n}{m}$ have to be suppressed, thus violating the
Killing conditions. Consider $U_{+}$: The solutions are states with $\langle
U_{+}\rangle=0$ without fluctuations, so it is not a big surprise that
something goes wrong when quantized.

A possible modification of the conditions, leading to a desirable result, is
modifying the Killing conditions by raising the powers of the triads. Writing

$\left(E^{a}\right)^{\frac{1}{2}+\lambda}\,K_{a}\,\left(E^{a}\right)^{\frac{1}{2}+\lambda}\hskip
28.45274pt\mbox{for}\hskip 28.45274pta=x,y$ (85)

with some same small positive parameter $\lambda$ instead of
$\sqrt{E^{a}}\,K_{a}\,\sqrt{E^{a}}$. In the simple case of zero
$z$-derivatives, $b=w=0$, (which is the subject of the next section) and
$a_{1,1}=1$ (72) is modified to

$(m+1)^{\frac{1}{2}+\lambda}\,\alpha_{m+1}-(m-1)^{\frac{1}{2}+\lambda}\,\alpha_{m-1}=0$
(86)

with the solution

$\alpha_{m}=\frac{1}{m^{\frac{1}{2}+\lambda}},$ (87)

(73) analogously. In the result the coefficients $a_{m,n}$ fall off
sufficiently rapidly in all directions in $m$ and $n$, and all of them are
modified, including the diagonal ones, which do not make problems, as soon as
the $SU(2)$ measure is introduced.

Here we concentrate on another approach, leading to a result closer to the
solutions of the Killing conditions. Instead of modifying the operators
$\hat{U_{\pm}}$ in a symmetric way, we multiply
$\sqrt{E^{a}}\,K_{a}\,\sqrt{E^{a}}$ from the right by

$e^{\lambda\left|\ln\frac{E^{y}}{E^{x}}\right|}.$ (88)

In this way the equations remain unmodified for $m=n$, but the coefficients
fall off more rapidly in the directions of growing $m$ and $n$. For $m>n$,
$\hat{U_{+}}|\psi\rangle=0$ is replaced by

$\begin{array}[]{l}\displaystyle
m^{\frac{1}{2}}(m+1)^{\frac{1}{2}+\lambda}n^{-\lambda}\,a_{m+1,n}-m^{\frac{1}{2}}(m-1)^{\frac{1}{2}+\lambda}n^{-\lambda}\,a_{m-1,n}+\\\\[8.53581pt]
\displaystyle
n^{\frac{1}{2}}(n+1)^{\frac{1}{2}-\lambda}m^{\lambda}\,a_{m,n+1}-n^{\frac{1}{2}}(n-1)^{\frac{1}{2}-\lambda}m^{\lambda}\,a_{m,n-1}=ib\,a_{m,n}\end{array}$
(89)

and $\hat{U_{-}}|\psi\rangle=0$ by

$\begin{array}[]{l}\displaystyle
m^{\frac{1}{2}}(m+1)^{\frac{1}{2}+\lambda}n^{-\lambda}\,a_{m+1,n}-m^{\frac{1}{2}}(m-1)^{\frac{1}{2}+\lambda}n^{-\lambda}\,a_{m-1,n}-\\\\[8.53581pt]
\displaystyle
n^{\frac{1}{2}}(n+1)^{\frac{1}{2}-\lambda}m^{\lambda}\,a_{m,n+1}+n^{\frac{1}{2}}(n-1)^{\frac{1}{2}-\lambda}m^{\lambda}\,a_{m,n-1}=iw\,a_{m,n}.\end{array}$
(90)

For $n>m$ $m$ and $n$ exchange their roles. The solutions do not correspond to
hermitian operators, moreover, there is no nontrivial common solution at all.
What is possible, is to find solutions that have at least the diagonal and
near-diagonal coefficients $a_{m,m}$, $a_{m,m\pm 1}$ and $a_{m,m\pm 2}$ in
common. In search of them we begin by inserting $(1,1)$, $(2,2)$, $(3,3)$,…for
$(m,n)$ into both of them and find

$a_{2,1}=\frac{i(b+w)}{2\sqrt{2}}\,a_{1,1},\hskip
56.9055pta_{1,2}=\frac{i(b-w)}{2\sqrt{2}}\,a_{1,1},$ (91)
$a_{3,2}=\frac{i}{2\sqrt{2\cdot
3}}\left[(b-w)\,a_{1,1}+(b+w)\,a_{2,2}\right],$ (92)
$a_{4,3}=\frac{i}{2\sqrt{3\cdot
4}}\left[(b+w)\,a_{1,1}+(b-w)\,a_{2,2}+(b+w)\,a_{3,3}\right],\ldots$ (93)

with so far free diagonal elements. $a_{2,3}$ and $a_{3,4}$ are obtained from
$a_{3,2}$ and $a_{4,3}$ by replacing $w$ by $-w$. Inserting $(m,n)=(2,1)$ into
(89) gives

$a_{3,1}^{(+)}=\frac{1}{3^{\frac{1}{2}+\lambda}}\left[\left(1-\frac{b(b+w)}{4}\right)a_{1,1}-a_{2,2}\right],$
(94)

whereas inserting into (90) gives

$a_{3,1}^{(-)}=\frac{1}{3^{\frac{1}{2}+\lambda}}\left[\left(1+\frac{w(b-w)}{4}\right)a_{1,1}+a_{2,2}\right].$
(95)

Equating these two expressions determines $a_{2,2}$:

$a_{2,2}=\frac{1}{8}\,(w^{2}-b^{2})\,a_{1,1},$ (96)

so $a_{3,1}$ becomes finally

$a_{3,1}=\frac{1}{3^{\frac{1}{2}+\lambda}}\left(1-\frac{(b+w)^{2}}{8}\right)a_{1,1}.$
(97)

Repeating the same procedure for $(m,n)=(3,2)$ gives

$a_{3,3}=\frac{1}{3}\left(1-\frac{b^{2}+w^{2}}{4}+\frac{(b^{2}-w^{2})^{2}}{64}\right)a_{1,1}$
(98)

and

$a_{4,2}=\frac{w^{2}-b^{2}}{3\cdot
2^{\frac{3}{2}+\lambda}}\left(1-\frac{(b+w)^{2}}{32}\right)a_{1,1},$ (99)

and so on. The coefficients $a_{m,m-1}$ contain the parameter $\lambda$ and
their uniqueness determines $a_{2,2}$, $a_{3,3}$, …. In relation to the
unmodified difference equations, diagonal coefficients are not modified at
all. Again the coefficients $a_{m,n}$ and $a_{n,m}$ are symmetric in $b$ and
antisymmetric in $w$. Farther away from the diagonal, in the next two lines,
containing $a_{m,m\pm 3}$, the solutions of the modified equations do not
coincide, unlike the solutions of the original conditions. In other words,
“background factor” and “wave factor” do not propagate at exactly the same
speed any more. This is not a big loss, as the Killing property is violated
anyway. Each of the two solutions, their average value, or some linear
combination of them, is a state with finite length fluctuations that fulfils
approximately the Killing conditions. And, of course, the exact equivalence of
the Hamiltonian and the diffeo constraints gets lost, such states are not
exact solutions of the Hamiltonian constraint, thus not states of pure
gravity.

Concerning the asymptotic behavior, consider (89,90) for large $m$ or large
$n$. Then, in these asymptotic cases, the right-hand sides can be neglected,
the equations become separable. For $m>n$ (89) becomes

$m^{\frac{1}{2}-\lambda}\,\frac{\partial}{\partial
m}\left[m^{\frac{1}{2}+\lambda}\,a(m,n)\right]+n^{\frac{1}{2}-\lambda}\,\frac{\partial}{\partial
n}\left[n^{\frac{1}{2}-\lambda}\,a(m,n)\right]\approx 0,$ (100)

(90) differs from this by a minus sign between the two terms, so that both of
them can be set approximately to zero. After a product ansatz
$a(m,n)=\alpha(m)\,\beta(n)$ and for $m\gg n$ the second term becomes
negligible and the relevant $m$-dependent part becomes simply

$\frac{\rm d}{{\rm d}m}\,\left[m^{\frac{1}{2}+\lambda}\alpha(m)\right]\approx
0$ (101)

with solution

$\alpha(m)\approx\frac{1}{m^{\frac{1}{2}+\lambda}},$ (102)

analogously

$\beta(n)\approx\frac{1}{n^{\frac{1}{2}+\lambda}}$ (103)

for $n\gg m$. So all solutions of the modified equations fall off sufficiently
rapidly to make the root-mean-square deviation of length finite.

###  5.2 Speed and length uncertainties

Obviously, the larger the parameter $\lambda$, the larger is the departure of
a vertex state from the Killing conditions and the smaller is the
length/volume uncertainty of a segment containing one vertex. This qualitative
result can be shown more explicitly for the simple state with zero spatial
derivatives ($b=w-0$), but stationary fluctuations, described in then next
section in more detail (117) as a contribution to the Minkowski vacuum state.
The modified equation (75) is (86) with the solution (87), and the common
solution with the corresponding equation for $\beta_{n}$ is

$a_{m,n}=\frac{1}{(mn)^{\frac{1}{2}+\lambda}}$ (104)

for even $m+n$; for odd $m+n$ the coefficients $a_{m,n}$ iare zero.

Now consider the expectation value of $mn$, the essential part of the square
of length/volume. It is

$\langle mn\rangle=\sum_{m,n}\frac{1}{(mn)^{2\lambda}(2m+1)(2n+1)},$ (105)

which is finite in all directions in the $(m,n)$ plane. (Here and in the
following $\langle\ldots\rangle$ means an expectation value without the
normalizing factor $a_{1,1}^{-2}$.) To estimate the contribution of large $m$
and $n$, we approximate

$\langle mn\rangle\approx\frac{1}{4}\sum_{m,n}\frac{1}{(mn)^{1+2\lambda}}$
(106)

and estimate the sums by integrals from some constant (say 1, for simplicity)
to infinity

$\langle mn\rangle\approx\frac{1}{4}\int_{1}^{\infty}\frac{{\rm
d}m}{m^{1+2\lambda}}\,\int_{1}^{\infty}\frac{{\rm
d}n}{n^{1+2\lambda}}=\frac{1}{16}\frac{1}{\lambda^{2}}.$ (107)

The corresponding part $\langle\sqrt{mn}\rangle$ of the length/volume
expectation value is

$\langle\sqrt{mn}\rangle\approx\frac{1}{4}\int_{1}^{\infty}\frac{{\rm
d}m}{m^{\frac{3}{2}+2\lambda}}\,\int_{1}^{\infty}\frac{{\rm
d}n}{n^{\frac{3}{2}+2\lambda}}=\frac{1}{4(2\lambda+\frac{1}{2})}$ (108)

and the root-mean-square deviation of the length of an elementary cell is
estimated by

$\Delta(\ell)\approx\frac{1}{\sqrt{k}}\frac{1}{\sqrt{8}\,\lambda}.$ (109)

To calculate the fluctuation of the Killing conditions we consider first the
action of $K_{x}E^{x}$ on the state (104). Acting on $a_{m-1,n}|m-1,n\rangle$
and $a_{m+1,n}|m+1,n\rangle$ it gives a nonzero contribution to $K_{x}E^{x}$
at the position $(m,n)$:

$\left(\frac{\sqrt{m(m-1)}}{(m-1)^{\frac{1}{2}+\lambda}\,n^{\frac{1}{2}+\lambda}}-\frac{\sqrt{m(m+1)}}{(m+1)^{\frac{1}{2}+\lambda}\,n^{\frac{1}{2}+\lambda}}\right)|m,n\rangle.$
(110)

In the limit of large $m$ this gives

$\frac{2\,\lambda}{(mn)^{\frac{1}{2}+\lambda}}\,|m,n\rangle,$ (111)

a function looking like the original one (104), but nonzero when $m+n$ is odd.
In the limit of large $n$ the action of $K_{y}E^{y}$ gives the same
contribution, so that the contribution of $K_{x}E^{x}+K_{y}E^{y}$ is

$\frac{4\,\lambda}{(mn)^{\frac{1}{2}+\lambda}}\,|m,n\rangle.$ (112)

(Up to the normalization factor) the square of this gives the expectation
value

$\left\langle(K_{x}E^{x}+K_{y}E^{y})^{2}\right\rangle=16\lambda^{2}\sum_{m,n}\frac{\langle
m,n|m,n\rangle}{(mn)^{1+2\lambda}}=16\lambda^{2}\sum_{m,n}\frac{1}{(mn)^{1+2\lambda}(2m+1)(2n+1)}.$
(113)

Estimating again the sums by integrals from 1 to $\infty$, we get

$\left\langle(K_{x}E^{x}+K_{y}E^{y})^{2}\right\rangle\approx\frac{4\lambda^{2}}{1+4\lambda+4\lambda^{2}}\approx
4\lambda^{2}$ (114)

for $\lambda\ll 1$. As in this simple example the expectation value is of
order $\lambda$ (for the unmodified equation it is zero), the square root of
the above expression gives the root-mean-square deviation of $U_{+}$, so that
in lowest order

$\Delta(U_{+})\cdot\Delta\left(\sqrt{mn}\right)={\rm const.}$ (115)

with the constant containing the normalization factor and the edge labels
$k_{\pm}$. In the analogous case with $w=0$ in $U_{-}$ the contributions of
$K_{x}E^{x}$ and $K_{y}E^{y}$ of the type (111) cancel, so that in this case
with zero derivative of the wave factor all the contributions to the departure
from the Killing conditions come from the background.

For the physical meaning of the modification of the Killing conditions write
down the corresponding modification of the classical expression of $U_{+}$,

$(K_{x}E^{x}+K_{y}E^{y})e^{\lambda\left|\frac{E^{x}}{E^{y}}\right|}={\cal
E}^{\prime},$ (116)

wherefrom we can see the following:
$e^{\lambda\left|\frac{E^{x}}{E^{y}}\right|}$ being equal to one for
$E^{y}=E^{x}$ and larger than one in all other cases, the above equation
indicates that the time derivative $K_{x}E^{x}+K_{y}E^{y}$ is by a factor
$e^{-\lambda\left|\frac{E^{x}}{E^{y}}\right|}$, an amount growing with growing
wave factor, smaller than the $z$ derivative. In other words, the background
of gravitational waves with large wave factors is slowed down. The same
happens in (4.2.1) for the wave factor.

Remark. With the non-symmetric interpretation of the derivative of the wave
factor as $\ln\frac{n}{m}-\ln\frac{n_{-}}{m_{-}}$ (66) the non-commutativity
of $\hat{U}_{+}$ and $\hat{U}_{-}$ is shifted from between the vertices
directly to each vertex. For this reason then there are no common solutions.
At the diagonal $m=n$ (66) is equal to $-\ln\frac{n_{-}}{m_{-}}$, and so does
not contain the variables $m$ and $n$ at the considered vertex. In view of the
fact that exact common solutions of both Killing conditions are not physically
acceptable anyway, one could choose the diagonal elements from (96,98) and so
on, and calculate the coefficients $a_{m,m\pm 1}$ beside the diagonal in a
unique way. Farther away from the diagonal the coefficients obtained from
$U_{+}$ and $U_{-}$ do not coincide, like in the case of the modified
equations. Even with this non-compatibility, such an approach need not be
worse than that followed in this paper, although the differences between the
coefficients obtained with the one or the other condition is not of the order
of a small parameter. The advantage of such an approach would be the
possibility to calculate directly $a_{m,n}$ for each single state
$a_{m_{-},n_{-}}$ given at a neighboring vertex and thus go straightforwardly
from vertex to vertex.

Finally one could also replace the derivative of $\ln(E^{y}/E^{x})$ by the
difference completely left or right from the vertex $v_{i}$, say by
$\ln\frac{n_{i-1}}{m_{i-1}}-\ln\frac{n_{i-2}}{m_{i-2}}$. With this
interpretation one can directly construct even global common solutions of the
Killing conditions.

##  6 Flat space

Flat space, as a state without waves in either direction, is classically given
by $U_{+}=U_{-}=0$, see (26,27), with both signs of the derivative terms.
Classically this means $K_{x}E^{x}=K_{y}E^{y}=0$ as well as also $b=w=0$
(background and wave factors are constant).

In quantum theory the common solution of
$\hat{U}_{+}|\psi\rangle=\hat{U}_{-}|\psi\rangle=0$ with $b=w=0$ (and
$a_{1,1}=1$) is simply

$a_{m,n}=\frac{1}{\sqrt{mn}},$ (117)

nonzero only for odd $m$ and $n$. Such a state describes one-dimensional
quantum length fluctuations in the $z$ direction in absence of cross section
fluctuations and with one and the same holonomy at two neighboring vertices
left and right (or also both to the left or to the right) of the considered
vertex. As a state without waves, this is a local contribution to the
Minkowski vacuum \- up to the normalization issue, which is a general feature
of this model.

On a global level for flat space we may in principle require the same local
state at every vertex (i. e. the coefficients $a_{m,n}$ are the same at every
vertex) and the same superposition of edge holonomies at every edge. For the
possible edge labels 1, 2, 3,…this means that the differences $b=k_{+}-k_{-}$
acquire all integers, and with the same amplitudes for every $|k\rangle$ left
and right of every vertex, the expectation value of $b$ is zero. This is a
quantum relaxation of the constant cross section area in classical theory. In
the following, however, we consider only the simpler case $b=0$.

Assume states of the form $\sum_{m,n}a_{m,n}|m,n\rangle$ at two neighboring
vertices $v_{i\pm 1}$ of $v\equiv v_{i}$ and for each pair
$|m_{i-1},n_{i-1}\rangle\equiv|i,j\rangle$ at $v_{i-1}$ and
$|m_{i+1},n_{i+1}\rangle\equiv|p,l\rangle$ at $v_{i+1}$ and an amplitude
$\alpha_{ijpl}$ for the state $|1,1\rangle$ at $v$. Then for each
$|i,j\rangle$ and $|p,l\rangle$ we calculate, according to (4.2.2), the state
at $v$ with coefficients $\alpha_{m,n}(i,j,p,l)$ in dependence of the initial
value $\alpha_{ijpl}$ representing the coefficient $a_{1,1}$ at $v$. These
numbers are multiplied with the probability amplitudes $a_{i,j}$ and $a_{p,l}$
from the neighboring vertices, and a superposition is taken. Then the
requirement of equal states, which is a quantum version of homogeneity in $z$,
is formulated as

$\sum_{ijpl}a_{i,j}a_{p,l}\alpha_{ijpl}\>\alpha_{m,n}(i,j,p,l)=a_{m,n}.$ (118)

The above sum gives the coefficient $a_{m,n}$ of $|m,n\rangle$ at $v$, which,
in turn is equal to the corresponding coefficient at $v_{i\pm 1}$). As this is
an infinite system of equations, the requirement of equal states, as
formulated here, is only a principle that can be applied in practice only in
an approximative form.

As a little illustration, we choose an approximation restricted to $m,n\leq
3$, where neighboring states have the nonzero coefficients $a_{1,1}$,
$a_{2,2}$, $a_{3,3}$ and $a_{1,3}=a_{3,1}$. (It turns out that for consistency
$a_{1,2}=a_{2,1}=0$, as well as $a_{1,3}=a_{3,1}$.) The edge label in $U_{-}$
is taken to be equal to one.

Further we may assume that $(|i,j\rangle,\>|p,l\rangle)$ and
$(|p,l\rangle,\>|i,j\rangle)$ at $v_{i-1}$ and $v_{i+1}$, respectively, give
the same contribution, thus $\alpha_{ijpl}=\alpha_{plij}$, these quantities
are totally symmetric and there are 10 non vanishing ones:

$\alpha_{1111},\;\alpha_{1113},\;\alpha_{1122},\;\alpha_{1133},\;\alpha_{1313},\;\alpha_{1322},\;\alpha_{1333},\;\alpha_{2222},\;\alpha_{2233},\;\alpha_{3333}.$
(119)

For the state $|1,1\rangle$ at $v$ we have simply the superposition

$\sum_{ijpl}\alpha_{ijpl}\,a_{i,j}\,a_{p,l}=a_{1,1}.$ (120)

AS $a_{2,2}=w^{2}/2$ with $b=0$ in (4.2.2), the corresponding relation for
$|2,2\rangle$ is

$\left(2\alpha_{1113}\,a_{1,1}\,a_{1,3}+2\alpha_{1333}\,a_{1,3}\,a_{3,3}+2\alpha_{1322}\,a_{1,3}\,a_{2,2}+4\alpha_{1313}\,a_{1,3}^{2}\right)\ln^{2}3=a_{2,2}.$
(121)

Under the assumption $k=1$ everywhere, $w$ can assume the values $\pm\ln 3$ or
zero. Further, from $a_{1,3}=\frac{1}{\sqrt{3}}\left(1-\frac{w^{2}}{2}\right)$
($b=0$ in (4.2.2)) we see directly

$a_{1,3}=\frac{1}{\sqrt{3}}\,(a_{1,1}-a_{2,2}),$ (122)

as the contributions of its two terms were calculated already in (120) and
(121).

For the same reason we get from
$a_{3,3}=\frac{1}{3}\left(1-w^{2}+\frac{w^{4}}{4}\right)$

$\displaystyle a_{3,3}=\frac{1}{3}\,a_{1,1}-\frac{2}{3}\,a_{2,2}+$ (123)
$\displaystyle\hskip
28.45274pt+\left(\alpha_{1113}\,a_{1,1}\,a_{1,3}+\alpha_{1333}\,a_{1,3}\,a_{3,3}+\alpha_{1322}\,a_{1,3}\,a_{2,2}+8\alpha_{1313}\,a_{1,3}^{2}\right)\frac{\ln^{4}3}{3},$

with the second line coming from $w^{4}/4$.

From (121) and (123) we can express

$\alpha_{1313}\,a_{1,3}^{2}=\frac{6a_{3,3}-2a_{1,1}+(4+\ln^{2}3)a_{2,2}}{8\ln^{4}3}$
(124)

and insert it and into (120). This results in the quadratic equation

$\displaystyle\alpha_{1111}\,a_{1,1}^{2}+2\alpha_{1122}\,a_{1,1}\,a_{2,2}+\alpha_{2222}\,a_{2,2}^{2}+2\alpha_{1133}\,a_{1,1}\,a_{3,3}+\alpha_{3333}\,a_{3,3}^{2}+2\alpha_{2233}\,a_{2,2}\,a_{3,3}$
$\displaystyle+\frac{1}{\ln^{4}3}\left[(3-\ln^{4}3)\,a_{1,1}+(7\ln^{2}3-6)\,a_{2,2}-9\,a_{3,3}\right]=0$
(125)

for $a_{1,1}$, $a_{2,2}$, and $a_{3,3}$ with 6 parameters $\alpha_{ijpl}$,
having a trivial solution $a_{1,1}=a_{2,2}=a_{3,3}=0$. Geometrically this is a
quadric containing the origin $(0,0,0)$. Due to the cut-off, this kind of
approximation can be normalized even in the sense of the $U(1)$ scalar
product, unless a full solution. With (122) the normalization condition is

$4a_{1,1}^{2}-2a_{1,1}\,a_{2,2}+4a_{2,2}^{2}+3a_{3,3}^{2}=3,$ (126)

describing an ellipsoid around the origin with axes smaller than one. From the
quadric in (6) we may pick out the point $(0,0,a_{3,3})$. The possible values
of $a_{3,3}$ are zero and $9/(\ln^{4}3\;\alpha_{3333})$, this point lies
clearly outside the normalization ellipsoid, as all $\alpha$ are less than
one. From this we see that, whatever kind of quadric is determined by (6), it
intersects the ellipsoid (126) in a closed curve, and there is a one-parameter
solution for every set of $\alpha_{ijpl}$. The above considerations do
obviously not determine a unique approximated state for flat space. It may be
interesting that the above construction goes through in the same way, whether
the neighboring vertices are assumed both to the left or right or one left,
one right.

Due to the symmetry of $a_{m,n}$ also the expectation value of the wave factor
$\ln(n/m)$ becomes zero, though with deviations. So far, by restricting to
$k\equiv 1$ for all edges, we have considered only one-dimensional
fluctuations in the $z$ direction. With $m$ and $n$ going from 1 to 3 it is
natural to vary also the edge label. Taking also $k_{\pm}=1,2,3$, this gives 9
combinations of $k_{+}$ and $k_{-}$, so the above approximation is
supplemented with 8 further versions, which are more complicated for
$k_{+}\neq k_{-}$. Amplitudes for edge functions are not determined by
$U_{\pm}$, but in view of (117) it seems natural to associate an amplitude
$1/\sqrt{k}$ to each edge holonomy $|k\rangle$. For large $k$ this is
confirmed by (160) below.

Physically the situation described in this section is the following:
$K_{x}E^{x}$ and $K_{y}E^{y}$ represent time derivatives, in states that are
annihilated by the corresponding operators the expectation values are exactly
zero without fluctuations. This means that the obtained spatial fluctuations
of spatial derivatives $\Delta b$ and $\Delta w$ are, so to say, frozen in
time. A renormalization procedure as in the foregoing section, allowing for
fluctuations of $K_{x}E^{x}$ and $K^{y}E^{y}$, too, would “unfreeze” the
stationary vacuum fluctuations.

If we did the same with $m$ and $n$ restricted to 1 and 2, we would get only
$a_{1,1}$ as nonzero coefficient and $a_{1,2}=a_{2,1}=a_{2,2}=0$. If we go to
higher-order approximations, particularly with nontrivial superpositions of
edge holonomies, the system becomes very complicated. The above example is
rather a demonstration of only academic value than a serious approximation.

##  7 The Hamiltonian constraint

###  7.1 The operator

In the Killing operators in the foregoing sections the connection component
$\cal A$ does not occur and for this reason we had to do only with vertex
functions, the edge holonomies could be given arbitrarily, according to the
classical prototype. The dynamics generated by the Hamiltonian constraint, in
contrary, involves all three types of holonomies on an equal footing, so that
the cross section area stands in a dynamical relation to the vertices.

In accordance with [6] we denote the first two terms of the Hamiltonian
constraint in (10), which contain connection components, as “kinetic part”
$H_{K}$, and the rest as “potential part” $H_{P}$.The expression for $H$
contains an inverse volume factor

$\frac{1}{V}=\frac{1}{\sqrt{{\cal E}E^{x}E^{y}}},$ (127)

which does not give rise to a densely defined operator in general and is
usually regularized by replacing it by a commutator of positive powers of
connection variables [9]. But this turns out as unnecessary, as states with
$m$, $n$ or $k$ equal to zero do not play a role.

The Hamiltonian constraint operator can be regularized by letting it act on
small intervals on the $z$ axis, containing at most one vertex, see [6]. In
this way commutativity between operators acting on different vertices is
assured. In view of the derivatives occurring in $H_{P}$, this interpretation
has some disadvantages. Derivatives of functions are reformulated as
differences at different values of $z$. In [6] the derivative of a vertex
function $f(m,n)$ is interpreted as the difference at two endpoints of an
interval, containing only one vertex (or none). As there is no vertex at one
of the endpoints, the derivative of a function $f(m,n)$ at a vertex $v$
results simply in (plus or minus) the value at $v$, when $v$ is an endpoint.
The first derivative of $\cal E$ at $v$ has a natural interpretation as the
difference of the edge labels left and right from $v$, but the second
derivative collapses to the first one in this approach. Such a lowering of the
degree of derivatives in comparison with the classical theory does not seem
very natural, it might be “too local”.

For this reason we prefer to associate the Hamiltonian constraint to up to
four pairs $(v,e)$ of a vertex and an edge. States at vertex-edge pairs are
superpositions of vertex and edge holonomies in the form

$|\psi\rangle=\sum_{m,n,k}a_{m,n,k}|m,n,k\rangle.$ (128)

$a_{m,n,k}$ is the product of the coefficients $a_{m,n}$ as before and the
probability amplitude of an edge holonomy.

The potential part is written as

$H_{P}=\frac{1}{\sqrt{{\cal E}E^{x}E^{y}}}\,P$ (129)

with $P$ containing the $z$-derivatives, and, in addition, the functions $\cal
E$ and ${\cal E}^{2}$. As in the case of the Killing conditions, derivatives
can be replaced by differences in different ways, either containing the
variables $m$, $n$, and $k$ at the considered vertex and edge, or only the
labels from neighboring ones. In the following, in the first case we write
$P(m,n,k)$, in the second one only $P$ and assume pairs $(v,e)$ with the edge
right from the vertex. As an example, in addition to $m$, $n$, $k$, belonging
to $(v,e)$, the quantity $P(m,n,k)$ may be chosen to involve the labels from
the vertex left to $v$ and those of the two edges left from $e$, indexed by
$i-1$ and $i-2$, respectively:

$\displaystyle
P(m,n,k)=\frac{1}{4}\,(k-k_{i-1})^{2}+\frac{1}{4}\left[k\left(\ln\frac{n}{m}-\ln\frac{n_{i-1}}{m_{i-1}}\right)\right]^{2}$
(130) $\displaystyle\hskip
56.9055pt-\frac{1}{2}\,k\,(k-k_{i-1})\left[\ln(nm)-\ln(n_{i-1}m_{i-1})\right]+k\,(k+k_{i-2}-2k_{i-1}).$

In any case $P(m,n,k)$ results in a diagonal operator, and for the difference
equation below it does not make a difference, whatever interpretation of the
differences is chosen.

With the above choice, for every state

$|k_{i-2}\rangle\otimes|m_{i-1},n_{i-1},k_{i-1}\rangle$ (131)

left from $v$, one can formulate and solve a difference equation for the pair
$(v,e)$, and so on, going step by step from one vertex-edge pair to the next
one, from the left to the right. In general, there will be a superposition of
states (131) at the chosen sequence with the probability amplitudes

$\sum_{m_{i-2},\,n_{i-2}}a_{m-2,\,n-2,\,k-2}\cdot
a_{m_{i-1},\,n_{i-1},\,k_{i-1}},$ (132)

and the obtained states at $(v,e)$ have to be multiplied with (132) for every
state (131). And so on from left to right…Starting from such an “initial-
value” state at some sequence “edge-vertex-edge”, one can of course, do
everything also from right to left for pairs $(e,v)$ with the edge left from
the vertex.

For the kinetic part $H_{K}$ (with $\kappa=1$ and the inverse volume factor
included) we choose again the ordering
$K_{a}E^{a}\rightarrow\sqrt{E^{a}}\,K_{a}\,\sqrt{E^{a}}$ and write the inverse
volume term to the left, and we obtain again the factor ordering “connection
left from triad” (and $\kappa=1$)

$H_{K}=\frac{1}{\sqrt{\cal
E}}\,K_{x}\sqrt{E^{x}}\,K_{y}\sqrt{E^{y}}-\frac{1}{\sqrt{E^{y}}}\,K_{x}\sqrt{E^{x}}\,{\cal
A}\,\sqrt{\cal E}-\frac{1}{\sqrt{E^{x}}}\,K_{y}\sqrt{E^{y}}\,{\cal
A}\,\sqrt{\cal E}.$ (133)

The action of the corresponding operator on a basic state $|m,n,k\rangle$ is
the following

$\displaystyle
H_{K}|m,n,k\rangle=\frac{1}{4}\left[\sqrt{\frac{mn}{k}}\left(|m+1,n+1,k\rangle-|m-1,n+1,k\rangle\right.\right.$
$\displaystyle\hskip
113.81102pt-\left.|m+1,n-1,k\rangle+|m-1,n-1,k\rangle\right)$
$\displaystyle\hskip
56.9055pt+\sqrt{\frac{mk}{n}}\left(|m+1,n,k+1\rangle-|m-1,n,k+1\rangle\right.$
(134) $\displaystyle\hskip
113.81102pt-\left.|m+1,n,k-1\rangle+|m-1,n,k-1\rangle\right)$
$\displaystyle\hskip
56.9055pt+\sqrt{\frac{nk}{m}}\left(|m,n+1,k+1\rangle-|m,n-1,k+1\rangle\right.$
$\displaystyle\hskip
113.81102pt-\left.\left.|m,n+1,k-1\rangle+|m,n-1,k-1\rangle\right)\rule{0.0pt}{14.22636pt}\right].$

Zero $m$, $n$, or $k$: For zero $k$, for example, the second and the third
prefactor simply vanish, in the first term, for $k=0$ we have only states of
the form $|m\pm 1,n\pm 1,0\rangle$ on the right-hand side. So this term leaves
the $k=0$ states among themselves, as well as $H_{P}$, being diagonal, does
not mix states with $n=0$ and $m=0$ with other ones. The second kinetic term
acts analogously to the first one on $|m\rangle\otimes|k\rangle$ and the third
one on $|n\rangle\otimes|k\rangle$. Due to this property we can consistently
avoid zero-volume states. Furthermore, $H_{K}$ does not lead to relations
between states with positive and negative labels and we can restrict ourselves
to $m,n,k>0$. For this reason we also do not need absolute values in (7.1).

###  7.2 The difference equation

When the Hamilton operator is imposed to annihilate physical states of the
form (128),

$H_{K}|\psi\rangle=-\frac{P}{\sqrt{{\cal E}E^{x}E^{y}}}\,|\psi\rangle,$ (135)

for each holonomy $|m,n,k\rangle$ on the right-hand side, the inverse volume
factor gives $1/\sqrt{mnk}$, as $P$ is a diagonal operator. After
multiplication with $4\sqrt{mnk}$, we obtain the Hamiltonian constraint
difference equation

$\displaystyle\left[\sqrt{mn}\left(\sqrt{(m-1)(n-1)}\,a_{m-1,n-1,k}-\sqrt{(m+1)(n-1)}\,a_{m+1,n-1,k}\right.\right.$
$\displaystyle\hskip
34.1433pt\left.-\sqrt{(m-1)(n+1)}\,a_{m-1,n+1,k}+\sqrt{(m+1)(n+1)}\,a_{m+1,n+1,k}\right)$
$\displaystyle+\sqrt{mk}\left(\sqrt{(m-1)(k-1)}\,a_{m-1,n,k-1}-\sqrt{(m+1)(k-1)}\,a_{m+1,n,k-1}\right.$
$\displaystyle\hskip
34.1433pt-\left.\sqrt{(m-1)(k+1)}\,a_{m-1,n,k+1}+\sqrt{(m+1)(k+1)}\,a_{m+1,n,k+1}\right)$
(136)
$\displaystyle+\sqrt{nk}\left(\sqrt{(n-1)(k-1)}\,a_{m,n-1,k-1}-\sqrt{(n+1)(k-1)}\,a_{m,n+1,k-1}\right.$
$\displaystyle\hskip
34.1433pt-\left.\left.\sqrt{(n-1)(k+1)}\,a_{m,n-1,k+1}+\sqrt{(n+1)(k+1)}\,a_{m,n+1,k+1}\right)\right]$
$\displaystyle+4P(m,n,k)\,a_{m,n,k}=0.$

Connecting in general 13 coefficients to each other, this equation looks quite
complicated. Lets begin with $(m,n,k)=(1,1,1)$:

$a_{1,2,2}+a_{2,1,2}+a_{2,2,1}=-2\,P(1,1,1)\,a_{1,1,1},$ (137)

the next simplest one, with $(m,n,k)=(1,1,2)$, $(1,2,1)$, or $(2,1,1)$, leads
to the three relations

$\displaystyle\sqrt{3}\,a_{1,2,3}+\sqrt{3}\,a_{2,1,3}+a_{2,2,2}=a_{1,2,1}+a_{2,1,1}-2\,P(1,1,2)\,a_{1,1,2},$
$\displaystyle\sqrt{3}\,a_{1,3,2}+\sqrt{3}\,a_{2,3,1}+a_{2,2,2}=a_{1,1,2}+a_{2,1,1}-2\,P(1,2,1)\,a_{1,2,1},$
(138)
$\displaystyle\sqrt{3}\,a_{3,1,2}+\sqrt{3}\,a_{3,2,1}+a_{2,2,2}=a_{1,1,2}+a_{1,2,1}-2\,P(2,1,1)\,a_{2,1,1}.$

For a systematic look at the difference equation it is convenient to order the
coefficients according to the sum

$p=m+n+k.$ (139)

Equation (137) relates 3 of the 6 coefficients with $p=5$ to an initial value
$a_{1,1,1}$ (with $p=3$), the three equations (7.2) relate 7 of the 10
coefficients with $p=6$ on the left-hand side to three initial values
$a_{1,1,2}$, $a_{1,2,1}$ and $a_{2,1,1}$ with $p=4$ on the right-hand side.

In this way, when we visualize the coefficients in a 3-dimensional grid with
axes $m$, $n$, and $k$, (see the figure) the coefficients $a_{m,1,1}$,
$a_{1,n,1}$, and $a_{1,1,k}$ on the axes are free initial or boundary values
for the solutions, but they do not determine them uniquely. Graphically for
each $p$ the coefficients are represented by points on a triangle, spanned by
$(p-2,1,1)$, $(1,p-2,1)$, and $(1,1,p-2)$ in an $(m,n,k)$ coordinate system.

In the next step the three coefficients $a_{1,2,2}$, $a_{2,1,2}$, and
$a_{2,2,1}$ in (137) plus the free coefficients $a_{1,1,3}$, $a_{1,3,1}$, and
$a_{3,1,1}$ with $p=5$ are related to 12 of 15 coefficients with $p=7$,
written on the left-hand side in the following way

$\displaystyle\sqrt{3}\,a_{2,2,3}+\sqrt{3}\,a_{2,3,2}+3\,a_{1,3,3}=a_{2,1,2}+a_{2,2,1}+\sqrt{3}\,a_{1,3,1}+\sqrt{3}\,a_{1,1,3}$
$\displaystyle\hskip 156.49014pt-a_{1,1,1}-2\,P(1,2,2)\,a_{1,2,2},$
$\displaystyle\sqrt{3}\,a_{2,2,3}+\sqrt{3}\,a_{3,2,2}+3\,a_{3,1,3}=a_{1,2,2}+a_{2,2,1}+\sqrt{3}\,a_{1,1,3}+\sqrt{3}\,a_{3,1,1}$
$\displaystyle\hskip 156.49014pt-a_{1,1,1}-2\,P(2,1,2)\,a_{2,1,2},$
$\displaystyle\sqrt{3}\,a_{2,3,2}+\sqrt{3}\,a_{3,2,2}+3\,a_{3,3,1}=a_{1,2,1}+a_{2,1,2}+\sqrt{3}\,a_{1,3,1}+\sqrt{3}\,a_{3,1,1}$
(140) $\displaystyle\hskip 156.49014pt-a_{1,1,1}-2\,P(2,2,1)\,a_{2,2,1},$
$\displaystyle\sqrt{6}\,a_{1,2,4}+\sqrt{6}\,a_{2,1,4}+a_{2,2,3}=\sqrt{3}\,a_{1,2,2}+\sqrt{3}\,a_{2,1,2}-2\,P(1,1,3)\,a_{1,1,3},$
$\displaystyle\sqrt{6}\,a_{1,4,2}+\sqrt{6}\,a_{2,4,1}+a_{2,3,2}=\sqrt{3}\,a_{1,2,2}+\sqrt{3}\,a_{2,2,1}-2\,P(1,3,1)\,a_{1,3,1}$
$\displaystyle\sqrt{6}\,a_{4,1,2}+\sqrt{6}\,a_{4,2,1}+a_{3,2,2}=\sqrt{3}\,a_{2,1,2}+\sqrt{3}\,a_{2,2,1}-2\,P(3,1,1)\,a_{3,1,1}.$

Having calculated these coefficients for certain holonomies of the form (131)
at the vertex and the two edges left from $(v,e)$, for a general state (128)
at $(v_{i-1},e_{i-1})$ and $(v_{i-2},e_{i-2})$, one has to multiply the
results by the product

$\sum_{m_{i-2},\,n_{i-2},\,k_{i-2}}a_{m_{i-2},\,n_{i-2},\,k_{i-2}}\,a_{m_{i-1},\,n_{i-1},\,k_{i-1}}$
(141)

and form superpositions.

###  7.3 Number of free coefficients

In each further step, functions at $e_{i-1}$ and at $(v,e)$ give rise to a
solution at $(v_{i+1},e_{i+1})$, but do not determine it uniquely, as we have
seen above, at every step there remain undetermined coefficients. In each
further step $\frac{p(p+1)}{2}-3$ coefficients with $m+n+k=p$ are related to
the coefficients with $m+n+k=p-2$, where 3 of the former ones remain
undetermined and serve as initial values for the next group with $m+n+k=p+2$.
Generally, for $m+n+k=p$ we have

$\frac{1}{2}(p-2)(p-1)$ (142)

coefficients. Inserting all combinations with $m+n+k=p-2$ into the difference
equation (7.2) we obtain $\frac{1}{2}(p-4)(p-3)$ linear equations, relating
the coefficients with $p$ to those with $p-2$ and $p-4$. Calculating
coefficients going step by step from $p$ to $p+2$ we can construct special
solutions, either with only even, or with only odd $p$’s. At every step with a
certain value of $p$ we have

$\frac{1}{2}\,(p-3)(p-4)$ (143)

coefficients determined by the Hamiltonian constraint and

$\frac{1}{2}\,(p-1)(p-2)-\frac{1}{2}\,(p-3)(p-4)=2p-5$ (144)

free ones.

Summing up the number of coefficients with even and odd $m+n+k=3$ from 3 to
$p$, we have

$\frac{1}{6}\,p\,(p-1)(p-2)$ (145)

coefficients, where

$(p-2)^{2}$ (146)

of them can be chosen freely, so that

$\frac{1}{6}\,(p-2)(p-3)(p-4)$ (147)

are determined.

$m$$n$$k$Figure: The coefficients for $p=3$ (at $(1,1,1)$,taken here as origin
of the axes),$p=4$(smaller triangle), and $p=5$(larger triangle) in $(m,n,k)$
coordinates.

Had we chosen a different factor ordering, e. g. “connection left from triad”
including the inverse volume factor, as it is often advocated for the
Hamiltonian constraint operator, the square root-coefficients of the
$a_{m,n,k}$ in (7.2) would become different, but the structure of the system
of equations would remain the same, particularly the numbers of the free and
the determined $a_{m,n,k}$’s.

As mentioned before, in the above considerations the diagonal functions
$P(m,n,k)$ are chosen in a different way than in [6]. There the operator acts
individually on every vertex, with a dependence between vertex functions
mediated only via edge labels, and create new vertices. For the choices
advocated here, they depend also directly on the neighboring functions and do
not change the graph.

###  7.4 Asymptotic behavior

To estimate the asymptotic behavior of solutions for large $m$, $n$, and $k$
we replace the coefficients $a_{m,n,k}$ by differentiable functions $a(m,n,k)$
and approximate the differences in (7.2) by partial derivatives. The resulting
differential equation is separable, when the derivatives of $\ln(E^{y}/E^{x})$
and $\ln(E^{x}E^{y})$ in $P(m,n,k)$ are formulated in terms of only
neighboring labels like in (65) or (67). This is sufficient for separability,
but it is possible, and suitable for simplifying the equation, and for a more
uniform asymptotic behavior in $m$, $n$, and $k$, to reformulate $P$
completely in terms of derivatives,

$\displaystyle P=\frac{1}{2}\left({\cal
E}^{2}\right)^{\prime\prime}-\frac{3}{4}\left({\cal
E}^{\prime}\right)^{2}-\frac{1}{4}({\cal E}^{\prime})^{2}(\ln
E^{x}E^{y})^{\prime}$ (148) $\displaystyle\hskip
28.45274pt+\frac{1}{4}\left({\cal
E}^{2}\,\ln\frac{E^{y}}{E^{x}}\right)^{\prime}\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}-\frac{1}{8}\left({\cal
E}^{2}\right)^{\prime}\left[\left(\ln\frac{E^{y}}{E^{x}}\right)^{2}\right]^{\prime}.$

If we now formulate also $\left({\cal E}^{2}\right)^{\prime}$ and $\left({\cal
E}^{2}\right)^{\prime\prime}$ in terms of neighboring edge labels,

$\left({\cal E}^{2}\right)^{\prime}\rightarrow k_{i+1}^{2}-k_{i-1}^{2},\hskip
28.45274pt\left({\cal E}\right)^{\prime\prime}\rightarrow
k_{i+2}+k_{i-2}-k_{i+1}-k_{i-1}$ (149)

to the left and the right (or on one side), also $k$ disappears from $P$ and
the asymptotic differential equation becomes

$\displaystyle\sqrt{mn}\frac{\partial^{2}}{\partial m\,\partial
n}\left(\sqrt{mn}\,a(m,n,k)\right)+\sqrt{mk}\frac{\partial^{2}}{\partial
m\,\partial k}\left(\sqrt{mk}\,a(m,n,k)\right)$
$\displaystyle+\sqrt{nk}\frac{\partial^{2}}{\partial n\,\partial
k}\left(\sqrt{nk}\,a(m,n,k)\right)=-P\,a(m,n,k).$ (150)

With the product ansatz $a(m,n,k)=M(m)\,N(n)\,K(k)$ this becomes

$\displaystyle\sqrt{mn}\frac{\rm d}{{\rm d}m}\left(\sqrt{m}\,M\right)\frac{\rm
d}{{\rm d}n}\left(\sqrt{n}\,N\right)K+\sqrt{mk}\frac{\rm d}{{\rm
d}m}\left(\sqrt{m}\,M\right)N\frac{\rm d}{{\rm d}k}\left(\sqrt{k}\,K\right)$
$\displaystyle+\sqrt{nk}\,M\frac{\rm d}{{\rm
d}n}\left(\sqrt{n}\,N\right)\frac{\rm d}{{\rm
d}k}\left(\sqrt{k}\,K\right)=-P\,M\,N\,K.$ (151)

Introducing ${\cal M}=\sqrt{m}\,M$, ${\cal N}=\sqrt{n}\,N$, and ${\cal
K}=\sqrt{k}\,K$, the derivatives written in form of primes, and dividing by
$M\,N\,K$ gives

$\sqrt{mn}\,\frac{{\cal M}^{\prime}}{M}\frac{{\cal
N}^{\prime}}{N}+\sqrt{mk}\,\frac{{\cal M}^{\prime}}{M}\frac{{\cal
K}^{\prime}}{K}+\sqrt{nk}\,\frac{{\cal N}^{\prime}}{N}\frac{{\cal
K}^{\prime}}{K}=-P,$ (152)

finally

$\left(m\frac{{\cal M}^{\prime}}{\cal M}\right)\left(n\frac{{\cal
N}^{\prime}}{\cal N}\right)+\left(m\frac{{\cal M}^{\prime}}{\cal
M}\right)\left(k\frac{{\cal K}^{\prime}}{\cal K}\right)+\left(n\frac{{\cal
N}^{\prime}}{\cal N}\right)\left(k\frac{{\cal K}^{\prime}}{\cal K}\right)=-P.$
(153)

When $P=0$, i. e. all derivatives in (10) are zero, the simplest solution to
(153) is given by $\frac{{\cal M}^{\prime}}{\cal M}=\frac{{\cal
N}^{\prime}}{\cal N}=\frac{{\cal K}^{\prime}}{\cal K}=0$, which leads to

$a_{m,n,k}\rightarrow\frac{\rm const.}{\sqrt{mnk}}.$ (154)

When in (10) only the derivatives of $\cal E$ (or the differences of $k$ in
(130)) are zero,

$P=\frac{1}{4}\left[{\cal
E}\left(\ln\frac{E^{y}}{E^{x}}\right)^{\prime}\right]^{2},\hskip
28.45274pt\mbox{or}\hskip
28.45274ptP=\frac{1}{4}\left[k\,\left(\ln\frac{n}{m}-\ln\frac{n_{i-1}}{m_{i-1}}\right)\right]^{2},$
(155)

respectively, is positive. Then for states, where $m\frac{{\cal
M}^{\prime}}{\cal M}$, $n\frac{{\cal N}^{\prime}}{\cal N}$, and $k\frac{{\cal
K}^{\prime}}{\cal K}$ are (approximately) equal, these expressions are
imaginary.

More generally, in standard macroscopic situations the cross section area is
much larger than its derivative, i. e. ${\cal E}\gg{\cal E}^{\prime}$
(particularly in the linear approximation for small waves the cross section
area $\cal E$ is constant along $z$), so that the positive definite term in
(130), coming from ${\cal E}^{2}\left((\ln(E^{y}/E^{x}))^{\prime}\right)^{2}$,
will be dominant, so that $P(m,n,k)>0$ and also $P>0$.

With this in mind, we separate off the part dependent on $m$ and set both
sides equal to an imaginary)constant

$m\frac{{\cal M}^{\prime}}{\cal M}=-\frac{P+n\frac{{\cal N}^{\prime}}{\cal
N}\>k\frac{{\cal K}^{\prime}}{\cal K}}{n\frac{{\cal N}^{\prime}}{\cal
N}+k\frac{{\cal K}^{\prime}}{\cal K}}=i\mu.$ (156)

For $M$ we find a function proportional to $m^{i\mu}$, which means that

$M(m)\propto\frac{1}{\sqrt{m}}\>e^{i\mu\ln m}.$ (157)

Continuing separation gives analogously

$N(n)\propto\frac{1}{\sqrt{n}}\>e^{i\nu\ln n}\hskip 28.45274pt\mbox{and}\hskip
28.45274ptK(k)\propto\frac{1}{\sqrt{k}}\>e^{i\rho\ln k}$ (158)

with

$\mu\nu+\mu\rho+\nu\rho=P.$ (159)

(The equation is separable also for the choice (130), where $P$ is a linear
function of $k$. Without the reformulation (148), $P$ would be a quadratic
function in $k$ and in the exponent in $K(k)$ also an (imaginary) quadratic
function in $k$ would appear, making $K(k)$ oscillate much faster in $k$ than
$M$ and $N$ in $m$ and $n$.)

For the chosen factor ordering in (133) the factors $1/\sqrt{m}$ and
$1/\sqrt{n}$ in (157) and (158) match with the corresponding factors (76) and
(77) in the unrenormalized solutions of the Killing conditions. This would not
be the case, had we chosen a symmetric factor ordering in $H_{K}$. In the case
of $m$, $n$, and $k$ growing at the same order of magnitude, a symmetric
choice for the separation constants w. r. to the three kinds of holonomies,
$\mu=\nu=\rho=\sqrt{P/3}$, will occur. Then also the oscillation frequencies
of $M$, $N$, and $K$ are the same and the asymptotic behavior of the
coefficients of such a state is

$a_{m,n,k}\rightarrow\frac{1}{\sqrt{mnk}}\,e^{i\sqrt{\frac{P}{3}}\,\ln(mnk)}.$
(160)

With the inner product induced from $SU(2)$ the solutions become normalizable,
although with the same problem of diverging length/volume uncertainties as the
solutions of the Killing conditions. An ad hoc renormalization of the type
(85), say by multiplying the three terms of $H_{K}$ by
$e^{\lambda\left|\ln\frac{m}{n}\right|}$,
$e^{\lambda\left|\ln\frac{m}{k}\right|}$,
$e^{\lambda\left|\ln\frac{n}{k}\right|}$, respectively, results in a state
that does not exactly solve the Hamiltonian constraint. If we trust the
considered model, this might mean that in quantum gravity there are small
energy fluctuations as in other QFTs, or that gravity has to be necessarily
coupled to something “coming from outside”. This also alters the relation
between the Hamiltonian and the diffeo constraint, and leads also to
departures from a constant speed of gravitational waves, in accordance with
the results of sections 4 and 5.

As the expression (160) is symmetric in the three dimensions, a normalizable
version thereof with $P=0$ gives a more complete picture of 3 dimensional
fluctuations in flat space than (117), displaying only longitudinal
fluctuations. The behavior like $1/\sqrt{k}$ for large $k$, conjectured at the
end of section 6, is confirmed by (160).

##  8 Summary

Thanks to its effective one-dimensionality, the symmetry-reduced model
considered in this paper facilitates investigations about possible dispersion
of gravitational waves. In the quantum regime the two independent Killing
equations, guaranteeing dispersion-free propagation, act as conditions on
physical states. The relative simplicity of the model does not require too
much mathematical sophistication and in all of this paper we tried generally
to keep the mathematical apparatus as modest as possible, in order to
facilitate explicit calculations, or to make them possible at all. So, in
comparison with [2], the issue of the inverse volume can be circumvented, (as
the crucial issues happen in the asymptotic limit for large volume,
qualitatively nothing important would happen, had we reformulated the inverse
volume as in [9]).

The holonomies being effectively mere phase factors, we have chosen holonomy
operators looking formally like $U(1)$ operators. In reward, the difference
equations become simpler, allowing even for explicit solutions. In the
relatively simple case of $U_{+}$ three of them are given. On the other hand,
concerning the derivatives of $\cal E$, we stuck to the classical prototype
and represented first/second derivatives by first/second differences. In this
pronounced non-local approach this involves necessarily more than one edge, or
vertex, respectively, and nonzero Poisson brackets of local expressions
containing derivatives are reflected in form of the non-commutativity of
operators at neighboring vertices in the discrete regime. But $U_{\pm}$ are
not constraints of the theory, they just pick out certain states, and our
interpretation of derivatives does not prevent us from the construction of
local or and even global solutions.

The obtained solutions are not normalizable in the obvious $U(1)$ inner
product, and even with the inner product borrowed from $SU(2)$, the true gauge
group of quantum gravity, which falls off more rapidly for large spins, length
uncertainties diverge, whereas there is no fluctuation of the wave velocity. A
violation of the Killing condition means dispersion, a departure from an
exactly constant propagation speed of gravitational waves, this speed and
metric quantities appear as mutually complementary - a typical situation in
quantum theory. Problems of this kind are usually overcome by continuous wave
packets. If the quantities $b$ and $w$ in $U_{\pm}$ and the potential part $P$
in $H$ were continuous, one could construct well-behaved wave packets without
invoking anything “from outside”, but they are discrete. Particularly the
difference $k_{+}-k_{-}$ in (49) is integer, so superpositions of asymptotic
solutions are periodic. This makes some renormalization necessary. In section
5 we have shown an ad-hoc example that manages to suppress the influence of
large spins, and estimated the violation of the Killing condition in relation
to the length uncertainty. When the considered kinds of renormalization are
ascribed to the interaction with matter fields, both the Hamiltonian and the
diffeo constraint would have to be combined with the corresponding matter
quantities.

Reduction to a state without waves at all leads to a sketch of fluctuations in
Minkowski space. In section 7 we have applied the described methods to the
Hamiltonian constraint in a non-graph changing version. Representing
derivatives by differences including variables coming from left or right from
a considered vertex or edge, as before, is a choice that disregards the
generally assumed commutativity of the operators at neighboring vertices. It
is traded for a faithful mapping of derivatives to corresponding differences
and thus avoids ultra-locality. Such an approach may be unorthodox, but it is
possible to construct solutions going step by step from vertex to vertex and
edge to edge. What is sacrificed is the off-shell commutativity on all of the
space of unconstrained states, what remains is on-shell commutativity, which
is trivial on states with $H|\psi\rangle=0$. As a reward we gained quantum
solutions for the Hamiltonian constraint as well as for the Killing
conditions. Reflecting the degree of derivatives is by the corresponding
degree of differences yields a stronger bond between quantities at neighboring
vertices. As can be seen from the (infinite) number of free parameters in
(144), there is a large solution space.

Interestingly, the applied methods lead to the same normalizability problem
for the Killing conditions and for the Hamiltonian constraint, illustrated by
the same asymptotic behavior. This indicates some close relationship between
gravitational wave dispersion and slight departures from the Hamiltonian
constraint, and also the exact equivalence of the Hamiltonian and the diffeo
constraint gets perturbed. As the Hamiltonian constraint describes the
symmetry of pure gravity under changes of the local units of time, a
modification indicates the presence of some other fields, acting as a
regulator for quantum gravity states. Some small fluctuation around the
expectation value 0 of the Hamiltonian would be a common feature with other
QFTs.

As further possibilities, infrared divergences of large spins would in
principle be cut off in a finite universe, or one could at least argue with
the relation between the diameter of the visible universe and the Planck
length. At first sight this looks quite inept and ridiculous, but numerically
the logarithm of this ratio (recall that the length uncertainty diverges
logarithmically) is of the order of $10^{2}$, so the root mean square
deviation of length and volume is of order 10. This is much, if the
expectation value amounts to a few Planck lengths, but for macroscopic lengths
it becomes reasonably tiny, but even in view of the not completely absurd
numerics it is hard to accept.

In an approach in [13] the cosmological constant appears generally as
regulator for LQG, this might avoid invoking matter fields. More recently
there are also attempts to promote the Immirzi parameter to a scalar field
[14], or to add topological terms as fields to the action of GR, see [15] and
citations therein, related directly, but not minimally coupled, to gravity.

The main conclusion that can be drawn from our calculations in a fully quantum
system is that in the scope of the present model at least a small dispersion
of gravitational waves, by whatever physical reasons, has to be expected.

Acknowledgement: The author thanks Seth Major for valuable help and
discussions.

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