# On the metric structure of time in classical and quantum mechanics

E. Cattaruzza  E. Gozzi  D. Mauro  INFN, Section of Trieste,
Via Valerio 2, Trieste, 34100, Italy  Phys. Dept. Theoretical Section, Univ.
of Trieste,
Strada Costiera 11, Miramare, Grignano
Trieste, 34152, Italy

###### Abstract

In this paper we show that, via an extension of time, some metric structures
naturally appear in both classical and quantum mechanics when both are
formulated via path integrals. We calculate the various Ricci scalar and
curvatures associated to these metrics and prove that they can be choosen to
be zero in classical mechanics while this is not possible in quantum
mechanics.

###### keywords:

###### PACS:

31.15.xk , 03.65.-w , 04.20.q

##  1 Introduction

Quantum gravity is for sure one of the most outstanding open problem in
theoretical physics. The usual approach is to take a geometric classical
theory (like Einstein gravity, string or similar) and apply quantum mechanics
to it. People have never reversed the problem, that means first try to
understand if there is some hidden universal geometry in quantum mechanics
and, second, see if this geometry is compatible with the geometry of the
classical model that we want to quantize. In this paper we will concentrate on
the first of the two issues above that means try to understand if there is
some hidden geometry in quantum mechanics. In doing this we will discover some
nice things which may have some application. In a future paper we hope to come
back to the second issue mentioned above.
It was shown in [1] that classical mechanics (CM) and quantum mechanics (QM)
could have a very similar formulation via path-integrals. The generating
function of the first, which we will indicate with the acronym $CPI$ (for
classical path integral) has the form:

$\mathcal{Z}_{CPI}\left[\mathbb{J}\right]=\int\mathcal{D}\Phi^{a}\exp\left[i\int
idtd\theta d\bar{\theta}\,\leavevmode\hbox to14.18pt{\vbox
to14.18pt{\pgfpicture\makeatletter\hbox{\hskip 7.09111pt\lower-7.09111pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{1}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\left(L[\Phi]+\mathbb{J}\,\Phi\right)\right]$
(1)

where $\theta,\bar{\theta}$ are two grassmanian partners of $t$, $\Phi^{a}$
are extensions of the phase space coordinates
$\varphi^{a}\equiv(q^{1},\dots,q^{n};p^{1},\dots,p^{n}),\,a=1,\dots,n$ and $L$
is the usual lagrangian. The generating functional for quantum mechanics,
which we will indicate with $\mathcal{Z}_{QPI}$ (where QPI stands for quantum
path integral), has the form

$\mathcal{Z}_{QPI}\left[\mathbb{J}\right]=\int\mathcal{D}\Phi^{a}\exp\left[i\int
idtd\theta d\bar{\theta}\,\leavevmode\hbox to19.1pt{\vbox
to19.1pt{\pgfpicture\makeatletter\hbox{\hskip 9.54863pt\lower-9.54863pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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}{{}{{{}}}{{}}{}{}{}{}{}{}{}{}{}{\pgfsys@beginscope\pgfsys@invoke{
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}{}\pgfsys@moveto{9.34863pt}{0.0pt}\pgfsys@curveto{9.34863pt}{5.16316pt}{5.16316pt}{9.34863pt}{0.0pt}{9.34863pt}\pgfsys@curveto{-5.16316pt}{9.34863pt}{-9.34863pt}{5.16316pt}{-9.34863pt}{0.0pt}\pgfsys@curveto{-9.34863pt}{-5.16316pt}{-5.16316pt}{-9.34863pt}{0.0pt}{-9.34863pt}\pgfsys@curveto{5.16316pt}{-9.34863pt}{9.34863pt}{-5.16316pt}{9.34863pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-6.03403pt}{-2.83888pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\hbox{{$\cfrac{\theta\bar{\theta}}{\hbar}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\left(L[\Phi]+\mathbb{J}\,\Phi\right)\right]$
(2)

which is very similar to Eq. (1) except that the 1 in Eq. (1) is replaced by
$\frac{\theta\bar{\theta}}{\hbar}$ in Eqs. (2). As these quantities multiply
the measure of integration $\int idtd\theta d\bar{\theta}$, it comes natural
to do the following: let us introduce a general dreinbein $E_{A}^{\,M}$ in the
space $t,\,\theta,\bar{\theta}$ and let us build the following path-integral

$\mathcal{Z}_{GPI}\left[\mathbb{J}\right]=\int\mathcal{D}\Phi^{a}\exp\left[i\int
idtd\theta d\bar{\theta}\,\leavevmode\hbox to16.51pt{\vbox
to16.51pt{\pgfpicture\makeatletter\hbox{\hskip 8.2559pt\lower-8.2559pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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}{}\pgfsys@moveto{8.05591pt}{0.0pt}\pgfsys@curveto{8.05591pt}{4.4492pt}{4.4492pt}{8.05591pt}{0.0pt}{8.05591pt}\pgfsys@curveto{-4.4492pt}{8.05591pt}{-8.05591pt}{4.4492pt}{-8.05591pt}{0.0pt}\pgfsys@curveto{-8.05591pt}{-4.4492pt}{-4.4492pt}{-8.05591pt}{0.0pt}{-8.05591pt}\pgfsys@curveto{4.4492pt}{-8.05591pt}{8.05591pt}{-4.4492pt}{8.05591pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope
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}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-3.97916pt}{-3.41666pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$E$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
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\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\left(L[\Phi]+\mathbb{J}\,\Phi\right)\right]$
(3)

(GPI stands for General Path Integral) and where $E$ is the determinant (or
superdeterminant) of $E_{A}^{M}$. Immediately we notice, comparing Eq. (3)
with Eq. (1) and Eq. (2) that the CPI can be considered a GPI with $E=1$ and
the QPI a GPI with $E=\frac{\theta\bar{\theta}}{\hbar}$. As the GPI has a
general covariance in the $t,\theta,\bar{\theta}$ space we could consider the
CPI and QPI as two “gauge fixed” version of Eq. (3). The reader could object
to this by saying that Eq. (1) and Eq. (2) should contain the Fadeev-Popov
determinant if considered as gauge fixed versions of Eq. (3). The reason we
did not put them in Eq. (1) and Eq. (2) is because we did not path integrate
over the variables $E^{A}_{\,B}$ and the gauge-fixing does not depend on the
matter field $\Phi^{a}$. Somehow we can consider the formulation in Eq. (1),
Eq. (2), Eq. (3) similar to that of a field theory $\Phi$ in a background
gravitational field $E^{A}_{\,B}$. The reader may also object that if both Eq.
(1) and Eq. (2) are different gauge of Eq. (3) then we could turn classical
mecnhainics Eq. (1) into quantum mechanics Eq. (2) via a general covariant
transformation in the extension of time. We will show later on why it is not
possible to turn CM into QM Next we will prove that there are various families
of $E^{A}_{\,B}$ which give the same CPI and the same for the QPI. These
families are parametrized by 4 parameters for the CPI and by 5 for the QPI.
From the $E^{A}_{\,B}$ with the help of Wolfram Mathematica we will build the
metric, the Christoffel symbols, the Ricci curvature tensor $R_{\alpha\beta}$
and the Ricci scalar $R$. All of these depend on the same parameters as the
$E^{A}_{\,B}$. For the CPI we prove that there is a point in parameter space
for which the Ricci scalar and tensors are zero. The same does not happen for
the QPI. This fact may indicate something very profound but we have not been
able to investigate it. We leave to the reader the task of further explore
this last issue. The paper is organized as follows. In Section (2) for
completness we briefly review (1) and (2). In Section (3) we introduce the
vierbein $E^{A}_{\,B}$ and indicate the general strategy. In Section (4) we
show how to obtain the vierbein for both the CPI and the QPI and do the
counting of the free parameters. In Section (5) we calculate we build the
metric for both the CPI and the QPI. In Section (6) we proceed to calculate
the Ricci scalar curvature for both theories. In Section (7) we search for the
point in parameter space where the Ricci scalar and tensors are zero in the
CPI. We also prove that a similar point does not exist for the QPI. In Section
(8) we summarize what we had done and the prospects for the future. In few
appendices we confine some detailed and long calculations.

##  2 Review

In the thirties Koopman and Von Neumann (KVN) proposed [2, 3] an operatorial
and Hilbert space formulation of classical mechanics (CM) on the lines of what
had been done few years before for quantum mechanics (QM). It was then natural
to give [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14] a path-integral version of the
KVN formalism like Feynman had done for the operatorial version of QM [15].
Actually, the path-integral version of classical mechanics (CPI) provided in a
natural way a generalization of the KVN formalism in the sense that it gave
the classical evolution of differential forms and tensors on phase-space [16].
The procedure has been worked out in details in [1] and [4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14] and can be summarized as follows. The KVN postulates for the
Hilbert space and operatorial versioon of CM are the following:

  1. 1.

a state of a classical system, whose phase-space is indicate by $\mathcal{M}$
with coordinates $\varphi^{a}\equiv(q^{1},\dots,q^{n};p^{1},\dots,p^{n})$ is
represented by an elememt $|\psi\rangle$ of the Hilbert space $\mathcal{H}$.

  2. 2.

On this Hilbert space the operators $\hat{p}^{i}$ and $\hat{q}^{j}$, whose
eigenvalues are $p^{i}$ and $q^{j}$, commutes

$[\hat{p}^{i},\hat{q}^{j}]=0$

and their common eigenstates are indicated as $|q,p\rangle$.

  3. 3.

The staes $\langle q,p|\psi\rangle$ are square-integrable and their modulus
squared $|\psi(q,p)|^{2}$ is the probability density $\rho(q,p)$ of finding
the system in $(q,p)$.

  4. 4.

The evolution of $\psi(q,p)$ is given by the Liouville equation

$i\frac{\partial\psi}{\partial t}=\widehat{L}\psi$

where the Liouvillian $\widehat{L}$ is

$\widehat{L}=i\,\left(\frac{\partial H}{\partial
q}\,\frac{\partial\,}{\partial p}-\frac{\partial H}{\partial
p}\,\frac{\partial\,}{\partial q}\right)$

and $H$ is the Hamiltonian of the sistem whose associated equation of motion
are

$\dot{\varphi}^{a}=\omega^{ab}\,\frac{\partial H}{\partial\varphi^{b}}$ (4)

with

$\omega^{ab}=\left(\begin{matrix}0&\mathbb{I}\\\
-\mathbb{I}&0\end{matrix}\right)$

a $2n\times 2n$ matrix called symplectic matrix. It is well-known that the
evolution between some initial point $\varphi_{i}$ and some final point
$\varphi_{j}$ has the following form on the states

$\psi(\varphi_{f},t_{f})=\int\mathcal{K}(\varphi_{f},t_{f}|\varphi_{i},t_{i})\,\psi(\varphi_{i},t_{i})\,d^{2n}\varphi_{i}$
(5)

where

$\mathcal{K}(\varphi_{f},t_{f}|\varphi_{i},t_{i})=\delta[\varphi_{f}-\Phi_{cl}(t_{f};q_{i},t_{i})]$

with $\Phi_{cl}$ the solution of Eq. (4) with initial condition $\varphi_{i}$.
Slicing the time interval $t_{f}-t_{i}$ in $N$ intervals, we can re-write the
kernel $K(\varphi_{f},t_{f}|\varphi_{i},t_{i})$ as follows:

$K(\varphi_{f},t_{f}|\varphi_{i},t_{i})=\displaystyle\lim_{N\to\infty}\left\\{\int\prod^{N-1}_{J=1}d\varphi_{J}\,\delta[\varphi_{J}-\Phi_{cl}(t_{J};\varphi_{i},t_{i}]\right\\}\,\delta[\varphi_{f}-\Phi_{cl}(t_{f};q_{i},t_{i})]$
(6)

where $\varphi_{J}$ are the intermediate points between $\varphi_{i}$ and
$\varphi_{f}$ over which we integrate. The Dirac deltas which appear in Eq.
(5) can be written as

$\delta[\varphi^{J}-\Phi_{cl}(t_{I};\varphi_{i},t_{i})]=\left.\delta\left[\dot{\varphi}^{a}-\omega^{ab}\,\frac{\partial
H}{\partial\varphi^{b}}\right]\right|_{t_{I}}\left.\textrm{det}\left[\delta^{a}_{b}\partial_{t}-\omega^{ac}\frac{\partial^{2}H}{\partial\varphi^{c}\partial\varphi^{b}}\right]\right|_{t_{J}}.$
(7)

Let us now introduce some auxiliary variables $\lambda_{a}$ and let us rewrite
the first term on the RHS of Eq. (7) as

$\delta\left[\dot{\varphi}^{a}-\omega^{ab}\,\frac{\partial
H}{\partial\varphi^{b}}\right]=\int
d\lambda_{a}\,\exp\left[i\,\lambda_{a}\left(\dot{\varphi}^{a}-\omega^{ab}\,\frac{\partial
H}{\partial\varphi^{b}}\right)\right]$ (8)

modulo a normalization factor. Let us also introduce $4n$ grassmanian
variables [17] $c^{a},\bar{c}_{a},a=1,\dots,2n$ so that we can rewrite the det
on the RHS of Eq. (7) as:

$\textrm{det}\left[\delta^{a}_{b}\partial_{t}-\omega^{ac}\frac{\partial^{2}H}{\partial\varphi^{c}\partial\varphi^{b}}\right]=\int
dc^{a}d\bar{c}_{a}\,\exp\left[-\bar{c}_{a}\left(\delta^{a}_{b}\partial_{t}-\omega^{ac}\frac{\partial^{2}H}{\partial\varphi^{c}\partial\varphi^{b}}\right)c^{b}\right].$
(9)

Using Eq. (7), Eq. (8) and Eq. (9) in Eq. (6) we get

$K(\varphi_{f},t_{f}|\varphi_{i},t_{i})=\int_{\varphi_{i}}^{\varphi_{f}}\mathcal{D}^{\prime\prime}\varphi\mathcal{D}\lambda\mathcal{D}c\mathcal{D}\bar{c}\,\exp\left[i\int
dt\widetilde{\mathcal{L}}\right]$ (10)

where

$\widetilde{\mathcal{L}}=\lambda_{a}\left[\dot{\varphi}^{a}-\omega^{ab}\,\frac{\partial
H}{\partial\varphi^{b}}\right]+i\,\bar{c}_{a}\left(\delta^{a}_{b}\partial_{t}-\omega^{ac}\frac{\partial^{2}H}{\partial\varphi^{c}\partial\varphi^{b}}\right)c^{b}$
(11)

and $\mathcal{D}^{\prime\prime}\varphi$ indicates that the integration is done
over all its intermediate points and not on the end points $\varphi_{i}$ and
$\varphi_{f}$. Eq. (10) is basically the path-integral counter-part of the KVN
formalism. Remembering how commutators are obtained from the path-integral
[15] we get

$\displaystyle[\hat{\varphi}^{a},\hat{\varphi}^{b}]$ $\displaystyle=$
$\displaystyle 0$ $\displaystyle[\hat{\varphi}^{a},\hat{\lambda}_{b}]$
$\displaystyle=$ $\displaystyle i\,\delta^{a}_{b}$ (12)
$\displaystyle[\hat{\bar{c}}_{a},\hat{c}^{b}]$ $\displaystyle=$
$\displaystyle\delta^{a}_{b}$

where the last are anticommutators or graded commutators [17]. All the other
commutators are zero. From the second commutators of Eq. (2) we can realize
operatorially the $\hat{\lambda}_{a}$ as

$\hat{\lambda}_{a}=-i\frac{\partial\,}{\partial\varphi^{a}}.$ (13)

Let us now see how the Liouville operator emerges from Eq. (10) from the non-
grassmanian part of $\widetilde{\mathcal{L}}$ which we indicate with
$\tilde{\mathcal{L}}_{B}$ the following quantitity:

$\widetilde{\mathcal{L}}_{B}=\lambda_{a}\,\dot{\varphi}^{a}-\widetilde{\mathcal{H}}_{B}$

where

$\widetilde{\mathcal{H}}_{B}=\lambda_{a}\,\omega^{ab}\frac{\partial
H}{\partial\varphi^{b}}.$ (14)

It is then clear that

$\int\mathcal{D}\varphi\mathcal{D}\lambda\,\exp\left[i\int
dt\widetilde{\mathcal{L}}_{B}\right]\longrightarrow\exp\left[-\widehat{\widetilde{\mathcal{H}}}_{B}\,t\right]$
(15)

where $\widehat{\widetilde{\mathcal{H}}}_{B}$ is the operator associated to
Eq. (14) obtained using Eq. (13)

$\displaystyle{\widetilde{\mathcal{H}}}_{B}$ $\displaystyle=$
$\displaystyle-i\frac{\partial}{\partial\varphi^{a}}\omega^{ab}\frac{\partial
H}{\partial\varphi^{b}}=$ (16) $\displaystyle=$ $\displaystyle-i\frac{\partial
H}{\partial p}\frac{\partial}{\partial q}+i\frac{\partial H}{\partial
q}\frac{\partial}{\partial p}=\widehat{L}$

and this $\widehat{L}$ is the Liouville operator. The reader may ask now which
operator we would get if we had kept also the grassmanian variables. It was
shown in [4] that the $c^{a}$ can be identified with the differential operator
$d\varphi^{a}$. Via these we can build generic differential forms [16]

$\psi(\varphi,d\varphi)$ (17)

and we know that their evolution is given by an operator [16] called the Lie
derivative of the Hamiltonian flow which is simbolically written as
$\mathcal{L}_{(dH)^{\\#}}$. So

$\partial_{t}\psi(\varphi,d\varphi)=\mathcal{L}_{(dH)^{\\#}}\psi(\varphi,d\varphi).$
(18)

This operator is a generalization of the Liouville operator which makes the
evolution of $\psi(\varphi)$ which are called zero forms. As we said in [4, 5,
6, 7, 8, 9, 10, 11, 12, 13, 14] the $c^{a}$ of our path-integral can be
identified with the $d\varphi^{a}$ and so the differential form Eq. (17) can
be turned into a $\psi(\varphi,c)$

$\psi(\varphi,d\varphi)\longrightarrow\psi(\varphi,c).$ (19)

From the path-integral Eq. (10) it is clear that the evolution of
$\psi(\varphi,c)$ is given by

$\partial_{t}\phi(\varphi,c)=\widehat{\widetilde{H}}\psi(\varphi,c),$ (20)

where $\widehat{\widetilde{H}}$ is the operatorial Hamiltonian associated to
the Lagrangian Eq. (11). Comparing Eq. (20) with Eq. (18) we can say that the
Hamiltonian operator of our path-integral is a well-known object [16] in
differential geometry, i.e. it is the Lie derivative of the Hamiltonian flow.
The identification with objects of differential geometry can be also extended
to the exterior derivatives, the inner contractions, the Lie brackets and the
whole Cartan calculus [16]. The details of this important correspondence have
been worked out in [4, 5]. So the auxiliary variables that we introduced
$c^{a},\bar{c}_{a},\lambda_{a}$ are not just tricks to rewrite the path
integral in a simpler form but crucial geometrical objects. Let us now go back
to the commutation relations Eq. (2). We said before that we can realize the
$\hat{\lambda}_{a}$ as a derivative operator (like in Eq. (13) and obviously
the $\varphi^{a}$ as a multiplicative one:

$\hat{\varphi}^{a}|\varphi\rangle=\varphi^{a}\,|\varphi\rangle.$ (21)

The same can be done for the operators $\hat{c}^{a},\hat{\bar{c}}_{a}$. As
they commute with the $\hat{\varphi}^{a}$ and $\hat{\lambda}_{a}$, we can
generalize the states of Eq. (21) to the following ones:

$\left\\{\begin{matrix}\hat{\varphi}^{a}|\varphi,c\rangle&=&\varphi^{a}\,|\varphi,c\rangle\\\
\hat{c}^{a}|\varphi,c\rangle&=&c^{a}\,|\varphi,c\rangle\\\
\end{matrix}\right.$ (22)

and implement $\hat{\bar{c}}_{a}$ as a derivative operator

$\hat{\bar{c}}_{a}=\frac{\partial\,}{\partial c^{a}}.$

There is another manner to realize the Eq. (2). Noting that $\hat{q}^{i}$ and
$\hat{\lambda}_{p_{i}}$ commutes and the same $\hat{c}^{q}$ and $\hat{c}_{p}$
we can diagonalize these operators 111The index $q$ and $p$ on
$\hat{c},\hat{\bar{c}},\hat{\lambda}$ indicates respectively the first and the
last n-indices on $\hat{c}^{a},\hat{\bar{c}}^{a},\hat{\lambda}^{a}$ and obtain
the states

$\left\\{\begin{matrix}\hat{q}\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle&=&q\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle\\\
\hat{\lambda}_{p}\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle&=&\lambda_{p}\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle\\\
\hat{c}^{q}\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle&=&c^{q}\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle\\\
\hat{\bar{c}}_{p}\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle&=&\bar{c}_{p}\,|q,\lambda_{p},c^{q},\bar{c}_{p}\rangle\\\
\end{matrix}\right.$ (23)

while the operators $\hat{p}$ and $\hat{\lambda}_{q}$ are realized as
derivatives operators

$\hat{p}=i\,\frac{\partial\,}{\partial\lambda_{p}},\quad\hat{\lambda}_{q}=-i\,\frac{\partial\,}{\partial
q}.$

The two basis Eq. (22) and Eq. (23) are related by Fourier transformation [1].
The transition amplitudes in the basis of Eq. (23) is a generalization of Eq.
(10) and it has the following path-integral expression

$\langle\varphi_{f},c_{f},t_{f}|\varphi_{i},c_{i},t_{i}\rangle=\int\mathcal{D}^{\prime\prime}\varphi\mathcal{D}\lambda\mathcal{D}^{\prime\prime}c\mathcal{D}\bar{c}\,\exp\left[i\int
dt\widetilde{\mathcal{L}}\right]$ (24)

where the integration over $c$ has been limited to the internal points with
the end-points fixed and indicated with $\mathcal{D}^{\prime\prime}c$. Using
the basis of Eq. (23) the transition amplitudes will have the path-integral
form

$\displaystyle\langle
q_{f},\lambda^{p}_{f},c_{f}^{q},\bar{c}_{f}^{p}|q_{f},\lambda^{p}_{i},c_{i}^{q},\bar{c}_{i}^{p}\rangle=$
$\displaystyle=$
$\displaystyle\int\mathcal{D}^{\prime\prime}q\mathcal{D}p\mathcal{D}^{\prime\prime}\lambda^{p}\mathcal{D}\lambda^{q}\mathcal{D}^{\prime\prime}c^{q}\mathcal{D}c^{p}\mathcal{D}\bar{c}^{q}\mathcal{D}^{\prime\prime}\bar{c}^{p}\,\exp\left[i\int
dt{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mathcal{L}\,}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mathcal{L}\,}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mathcal{L}\,}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mathcal{L}\,}$}}}}\right]$

where $\widetilde{\widetilde{\mathcal{L}}}$ is a Lagrangian which differ from
$\widetilde{\mathcal{L}}$ of Eq. (11) by surface terms. More details can be
found in ref.[1]. At this point we have to introduce two crucial ingredients
which are familiar from the supersymmetry formalism [18]. Let us extend the
variable $t$ via two grassmanian partners $\theta,\bar{\theta}$. The triplet
$(t,\theta,\bar{\theta})$ is often called “supertime”. If we extend $t$ to the
4-dim $x^{\mu}$ then there is an analog extension of super-time called
“superspace” [18]. With this tool we can group-together the various variables
$(\varphi^{a},\lambda_{a},c^{a},\bar{c}_{a})$ into a function of
$(t,\theta,\bar{\theta})$ called superfield and defined as follows:

$\Phi^{a}(t,\theta,\bar{\theta})\equiv\varphi^{a}+\theta\,c^{a}+\bar{\theta}\,\omega^{ab}\,\bar{c}_{b}+i\,\bar{\theta}\theta\,\omega^{ab}\lambda_{b}.$
(26)

We can separate off the $q$ and $p$ part of this superfields as follows

$\Phi^{a}=\left(\begin{matrix}Q_{i}\\\
P_{i}\end{matrix}\right)\equiv\left(\begin{matrix}q_{i}\\\
p_{i}\end{matrix}\right)+\theta\left(\begin{matrix}c^{q_{i}}\\\
c^{p_{i}}\end{matrix}\right)+\bar{\theta}\left(\begin{matrix}\bar{c}_{q_{i}}\\\
-\bar{c}_{p_{i}}\end{matrix}\right)+i\bar{\theta}\theta\left(\begin{matrix}\lambda_{p_{i}}\\\
-\lambda_{q_{i}}\end{matrix}\right).$ (27)

Using the superfield there are some nice identities which we will need later
on. Let us build the Lagrangian associated to $H(\varphi)$ of the original
equations of motion Eq. (4) and let us call it $L(\varphi)$ where we replaced
$\dot{q}$ with $p$. Let us now replace in $L(\varphi)$ the $\varphi$ with the
superfield $\Phi^{a}$ and expand in $\theta,\bar{\theta}$. We get

$L[\Phi]=L(\varphi)+\theta\,\mathcal{M}+\bar{\mathcal{M}}\bar{\theta}-i\bar{\theta}\theta\,{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mathcal{L}\,}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mathcal{L}\,}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mathcal{L}\,}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mathcal{L}\,}$}}}}$
(28)

where
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mathcal{L\,}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mathcal{L\,}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mathcal{L\,}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mathcal{L\,}}$}}}}$
is the Lagrangian which enters in Eq. (24). We will need these identities
later on. Let us drop the indices in Eq. (27):

$\left\\{\begin{matrix}Q(\theta,\bar{\theta})&=&q+\theta\,c^{q}+\bar{\theta}\,c_{p}+i\,\bar{\theta}\theta\,\lambda_{p}\\\
P(\theta,\bar{\theta})&=&p+\theta\,c^{p}-\bar{\theta}\,c_{q}-i\,\bar{\theta}\theta\,\lambda_{q}\end{matrix}\right..$
(29)

As the variables which enter $Q$ they all commute once they are turned into
operators, we could define the following states

$\widehat{Q}|Q\rangle=Q(t,\theta,\bar{\theta})|Q\rangle$ (30)

which clearly satisfy

$\left\\{\begin{matrix}\hat{q}|Q\rangle&=&q\,|Q\rangle\\\
\hat{\lambda}_{p}|Q\rangle&=&\lambda_{p}\,|Q\rangle\\\
\hat{c}^{q}|Q\rangle&=&c^{q}\,|Q\rangle\\\
\hat{\bar{c}}_{p}|Q\rangle&=&\bar{c}_{p}\,|Q\rangle.\\\ \end{matrix}\right.$

So we can identify the states $|Q\rangle$ with those of the basis Eq. (30). We
can now use this fact and Eq. (28) to rewrite Eq. (2) as follows

$\langle
Q_{f},t_{f}|Q_{i},t_{i}\rangle=\int\mathcal{D}^{\prime\prime}Q\mathcal{D}P\,\exp\left[i\int_{t_{0}}^{t}\,i\,dt^{\prime}d\theta
d\bar{\theta}\,L[\Phi]\right]$ (31)

where we have used the standard rule of grassmanian integration

$\int d\theta d\bar{\theta}\,\bar{\theta}\theta=1.$

all the details above are carefully explained in ref-[1]. Let us go back to
the quantum mechanical path-integral [15] which gives the following expression
for the transition amplitude

$\langle
q_{f},t_{f}|q_{i},t_{i}\rangle=\int\mathcal{D}^{\prime\prime}q\mathcal{D}p\,\exp\left[\frac{i}{\hbar}\int
dt^{\prime}\,L[\varphi]\right].$ (32)

Note the great analogy between the classical path-integrac (CPI) Eq. (31) and
the quantum path-integral (QPI) Eq. (32). We pass from one to the other by the
following steps:

$\left\\{\begin{matrix}Q&\longrightarrow&q\\\ P&\longrightarrow&p\\\ i\int
dtd\theta d\bar{\theta}&\longrightarrow&1/\hbar\end{matrix}\right..$ (33)

This is a sort of dimensional reduction which in [1] we proved to be
equivalent to geometric quantization [19]. More details can be found in
ref.[1]. Differently than in [1] in this paper we still exploit the relation
between Eq. (32) and Eq. (31) but following a different route. Let us write
Eq. (32) using the superfield and relation Eq. (28)

$\int dt\,L(\varphi)=\int dtd\theta
d\bar{\theta}\,\bar{\theta}\theta\,L[\Phi]\\\ =\int i\,dtd\theta
d\bar{\theta}\,(-i\,\bar{\theta}\theta)\,L[\Phi]$

so

$\frac{i}{\hbar}\int dt\,L(\varphi)=i\,\int i\,dtd\theta
d\bar{\theta}\,\left(-\frac{i}{\hbar}\,\bar{\theta}\theta\right)\,L[\Phi].$

We can then rewrite Eq. (32) as

$\langle
q_{f},t_{f}|q_{i},t_{i}\rangle=\mathcal{N}\int\mathcal{D}^{\prime\prime}Q\mathcal{D}P\,\exp\left[i\int\,i\,dt^{\prime}d\theta
d\bar{\theta}\,\left(-\frac{i}{\hbar}\,\bar{\theta}\theta\right)\,L[\Phi]\right].$
(34)

where $\mathcal{N}$ is a normalizing factor. On the right hand side of Eq.
(34) the integration over $c,\bar{c},\lambda$ drops off the path-integration
because these variables do not enter the weight. The normalizing factor
$\mathcal{N}$ is there to get $1$ out of those extra intergrations. We could
avoid introducing this normalizing factor if we write the L.H.S. of Eq. (34)
as $\langle Q_{f},t_{f}|Q_{i},t_{i}\rangle$ The pieces
$\langle\lambda_{f},t_{f}|\lambda_{i},t_{i}\rangle$ $\langle
c_{f},t_{f}|c_{i},t_{i}\rangle$
$\langle\bar{c}_{f},t_{f}|\bar{c}_{i},t_{i}\rangle$ turn out to be products of
“1” at each slice in time exactly as on the path-integral on the R.H.S. So we
can summarize the Eq. (34) and Eq. (31) as

$\langle
Q_{f},t_{f}|Q_{i},t_{i}\rangle_{CPI}=\int\mathcal{D}^{\prime\prime}Q\mathcal{D}P\,\exp\left[i\int_{t_{0}}^{t}\,i\,dt^{\prime}d\theta
d\bar{\theta}\,\leavevmode\hbox to13.61pt{\vbox
to13.61pt{\pgfpicture\makeatletter\hbox{\hskip 6.80302pt\lower-6.80302pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setdash{3.0pt,3.0pt}{0.0pt}\pgfsys@invoke{
}{{}{{{}}}{{}}{}{}{}{}{}{}{}{}{}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setdash{3.0pt,3.0pt}{0.0pt}\pgfsys@invoke{
}{}\pgfsys@moveto{6.60303pt}{0.0pt}\pgfsys@curveto{6.60303pt}{3.64679pt}{3.64679pt}{6.60303pt}{0.0pt}{6.60303pt}\pgfsys@curveto{-3.64679pt}{6.60303pt}{-6.60303pt}{3.64679pt}{-6.60303pt}{0.0pt}\pgfsys@curveto{-6.60303pt}{-3.64679pt}{-3.64679pt}{-6.60303pt}{0.0pt}{-6.60303pt}\pgfsys@curveto{3.64679pt}{-6.60303pt}{6.60303pt}{-3.64679pt}{6.60303pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-1.80556pt}{-3.41666pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\mathbb{I}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\,L[\Phi]\right]$ (35)
$\langle
Q_{f},t_{f}|Q_{i},t_{i}\rangle_{QPI}=\int\mathcal{D}^{\prime\prime}Q\mathcal{D}P\,\exp\left[i\int_{t_{0}}^{t}\,i\,dt^{\prime}d\theta
d\bar{\theta}\,\leavevmode\hbox to27.59pt{\vbox
to27.59pt{\pgfpicture\makeatletter\hbox{\hskip
13.79643pt\lower-13.79643pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setdash{3.0pt,3.0pt}{0.0pt}\pgfsys@invoke{
}{{}{{{}}}{{}}{}{}{}{}{}{}{}{}{}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setdash{3.0pt,3.0pt}{0.0pt}\pgfsys@invoke{
}{}\pgfsys@moveto{13.59644pt}{0.0pt}\pgfsys@curveto{13.59644pt}{7.50919pt}{7.50919pt}{13.59644pt}{0.0pt}{13.59644pt}\pgfsys@curveto{-7.50919pt}{13.59644pt}{-13.59644pt}{7.50919pt}{-13.59644pt}{0.0pt}\pgfsys@curveto{-13.59644pt}{-7.50919pt}{-7.50919pt}{-13.59644pt}{0.0pt}{-13.59644pt}\pgfsys@curveto{7.50919pt}{-13.59644pt}{13.59644pt}{-7.50919pt}{13.59644pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-10.53435pt}{-3.29762pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\hbox{{$-i\cfrac{\bar{\theta}\theta}{\hbar}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\,L[\Phi]\right].$
(36)

We have encircled the quantities $\mathbb{I}$ and
$-i\cfrac{\bar{\theta}\theta}{\hbar}$ because they seems to be the only
quantities which are different in QM and CM. They somehow modify the measure
of integration over the superspace $\int dtd\theta d\bar{\theta}$. We can
extend the formalism also to the generating functionals as we have indicated
in the introduction.

##  3 General Strategy

The presence of a factor in the measure, both in Eq. (35) and Eq. (36), is
reminiscent of another factor which appears in the measure of integration.
This happens in Riemannian geometry. There we have distances defined via a
metric $g_{\mu\nu}$ as

$ds^{2}=g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}$

and we require that this distance is invariant under general coordinate
transformations

$x^{\prime\mu}=x^{\prime\mu}(x^{\nu}).$

We also require that the volume of integration is invariant and this happens
only if we multiply the volume by a factor $E$:

$\int\prod_{\nu=1}^{4}\,dx^{v}\,E.$ (37)

The factor $E$ is a determinant which is built in this way. Let us called a
tensor called vierbein $e^{a}_{\,\mu}$ which carries an index $a$ transforming
under Lorentz transformations and a second index $\mu$ transforming under
general coordinate transformations. It is possible to show that the metric
$g_{\mu\nu}$ can be written in terms of the vierbein as follows

$g_{\mu\nu}=\eta_{ab}\,e^{a}_{\,\mu}\,e^{b}_{\,\nu},$ (38)

where $\eta_{ab}$ is a flat-Lorentz metric. For a review the interested reader
can look into [20]. The factor $E$ making the measure invariant is defined as

$E=\textrm{det}\,e^{a}_{\,\mu}.$ (39)

In our case the space on which we would like to introduce the factor $E$ is
not the 4-dim. space time but the 3-dim. space
$z^{A}=(t,\theta,\bar{\theta})$. Riemannian spaces with grassmannian
coordinates have been studied in [21]. We can define flat supertime in many
ways but we choose the following one:

$dz^{A}\,\eta_{AB}\,dz^{B}=dt^{2}-d\theta d\bar{\theta}+d\bar{\theta}d\theta,$
(40)

where $\eta_{AB}$ is

$\eta_{AB}=\begin{pmatrix}1&0&\,0\\\ 0&0&-1\\\ 0&1&\,0\\\ \end{pmatrix}.$ (41)

The analog of the Lorentz transformation in this case is given by the group
$Osp(1|2)$, which is the set of transformations leaving invariant the quantity

$s=t^{2}+\theta\bar{\theta}-\bar{\theta}\theta.$ (42)

The non-flat infinitesimal distance is defined as

$dz^{A}\,g_{AB}\,dz^{B},$ (43)

where $g_{AB}$ is the analog of the metric, but due to the grassmannian
character of some of its elements, is called supermetric. Under a general
superdiffeomorphisms of our coordinates, which we will indicate as:

$z^{A}=z^{A^{\prime}}+\xi^{A}(z),$ (44)

the $g_{AB}$, in order for Eq. (43) to be invariant, must transform as [21]:

$g_{AB}^{\prime}=g_{AB}(z)+\frac{\overset{\rightarrow}{\partial}\xi^{C}}{\partial
z^{A}}\,g_{CB}+g_{AC}\,\frac{\xi^{C}\overset{\leftarrow}{\partial}}{\partial
z^{B}}+g_{AB,C}\,\xi^{C},$ (45)

where the right and left derivatives above are due to the grassmannian
character of some of the $z$ and the fact that in Eq. (43) some infinitesimals
are to the left and some to the right. Like in normal Riemannian geometry also
in super-Riemannian one [21] we can define the super-vierbein which we will
indicate with $E^{A}_{\Lambda}$, where $A$ is the Lorentz analog (Osp(1,2))
index and $\Lambda$ the general covariant (in supertime) one. The relation
between supermetric and super-vierbein is [21]:

$g_{\Lambda\Pi}=E^{A}_{\,\,\Lambda}\,\eta_{AB}\,(-1)^{(1+B)\Pi}E^{B}_{\,\,\Pi}(z).$
(46)

The numbers which are in the exponent of $(-1)$ are $0$ for $t$ and $1$ for
$\theta$ and $\bar{\theta}$. For more details about grassmannian number,
matrices and super-determinant (which are often indicated by
$\textrm{sdet}(\dots)$) the reader can consult [17] or the A of this paper.
The analog of Eq. (37) for the superspace made of $t,\theta,\bar{\theta}$ will
be

$\int i\,dt\,d\theta\,d\bar{\theta}\,E$ (47)

where $E=\textrm{sdet}(E^{A}_{\,M})$. If we compare this with Eq. (35) and Eq.
(36) we can say that the CPI is like a “gauge” fixed version of a “super-
general covariant” formalism in supertime and the “gauge fixing” is such that

$E=\mathbb{I}$ (48)

while for the QPI the “gauge fixing” is such that

$E=-i\frac{\bar{\theta}\theta}{\hbar}.$ (49)

Before going on further we should remember that Eq. (48) and Eq. (49) are not
the only conditions we have to impose in order to obtain respectively the CPI
and the QPI. We should infact remember that in a “general covariant” formalism
also in the kinetic piece of the Lagrangian there is the presence of the
vierbeins. Let us first suppose we integrate out in Eq. (35) and Eq. (36) the
$P$ so that the kinetic piece in both of them is reduced to

$\partial_{t}Q\partial_{t}Q,$ (50)

where we omit the indices on $Q$. An analog “general covariant” piece would be

$D_{t}Q\,D_{t}Q$ (51)

where the general covariant derivative $D_{t}$ would be

$D_{t}=E^{M}_{\,\,\,t}\partial_{M},$ (52)

with $E^{M}_{\,\,\,t}$ components of the inverse of the vierbein matrix
appearing in Eq. (46) and Eq. (47). The $E^{M}_{t}$ should be chosen to be
real because expression Eq. (51) is real. If instead Eq. (51) were of the form

$(D_{t}Q)(D_{t}Q)^{*},$

then we could choose $E^{M}_{t}$ to be complex. We will extend the reality
condition of the $E^{M}_{t}$ to all the components of the vierbein in order to
simplify the treatment.

For the expression Eq. (51) to be the same as Eq. (50), we will see later on
that we have to make a particular choice for the vierbein. This choice, beside
the Eq. (48) for the CPI and the Eq. (49) for the QPI, is something like a
gauge fixing that we need to impose on the ”general covariant” formalism where
the vierbein are free. The reader may object that we should also insert a
Faddev-Popov (F.P.) determinant in the functional measure. As we already said
earlier, we think this is not necessary in our two cases because the F.P.
would depend only on $E^{A}_{\,M}$ in our two gauge-fixings and we do not have
the integration over $E^{A}_{\,M}$ in the path-integral. Of course what we get
is not a “gauge fixing” independent formalism neither one in which we can pass
from the CPI to the QPI via a “gauge transformation”. So we should be careful
in saying that the CPI and the QPI are something like a “gauge fixing” of a
general covariant formalism. In fact we have used the expression “something
like”. Nevertheless we think that is worth to pursue this analogy and see if
it helps us better undertand the interplay between CM and QM

##  4 Vierbeins

In this section we shall build the vierbein $E^{M}_{\,A}$, which gives the CPI
and the one which gives the QPI. We will bother the reader with several
details which are crucial in order to get the precise form of vierbeins. We
will show that there is a whole family of $E^{A}_{\,B}$, which reproduce the
same CPI and the same for the QPI. The vierbein is a $3\times 3$ super matrix

$E^{M}_{\,A}=\begin{pmatrix}a&\alpha&\beta\\\ \gamma&b&c\\\ \delta&d&e\\\
\end{pmatrix},$ (53)

where the greek letter indicate an odd element while the latin one indicates
an even element. It is easy to see why the elements of $E^{A}_{\,M}$ have the
features indicated above by considering how the supermetric $g_{AB}$ is built
out of the vierbein Eq. (46) and the odd/even characters of the elements of
the $g_{AB}$. The two conditions that we have to satisfy to get the CPI are:

$\left\\{\begin{matrix}E=1\Longrightarrow\textrm{sdet}(E^{M}_{\,A})=1\\\ \\\
D_{t}Q\,D_{f}Q=\partial_{t}Q\,\partial_{t}Q\quad.\end{matrix}\right.$ (54)

It is a long calculation, reported in B, to prove that the vierbein for the
CPI, satisfying the constraints (54) is given by

$E^{M}_{\,A}(CPI)=\begin{pmatrix}\pm 1&0&0\\\ \gamma&b&c\\\ \delta&d&e\\\
\end{pmatrix}.$

where the variables $b,c,d,e$ have to satisfy two constraints reported in B. A
similar but much longer calculation, reported in C, gives the form of the
vierbein for the QPI

$E^{M}_{A}(QPI)=\begin{pmatrix}1+a_{S}\theta\bar{\theta}&\alpha&\beta\\\
\gamma&b&c\\\ \delta&d&e\end{pmatrix},$

where $a_{S}$ is the “soul” (see A or ref.[17] for the definition of soul).
Also the elements of $E^{M}_{A}(QPI)$ are not free but must satisfy two
constraints presented in A. Of course for the QPI also Eq. (54) is different
and it is reported in details in C.

##  5 Metrics

In this section we will calculate the metric from the vierbeins and show that
they depend on a lower number of free parameters than the vierbeins. Also in
this section we will bother the reader with details but they are crucial in
order to build the metric with the least number of parameters. We will skip
the similar details in later sections for the curvatures because most of those
calculations were done using Wolfram Mathematica and using symbols already
defined in this and the previous section.
Let us start from the CPI. The vierbein, before implementing the constraint
Eq. (96), has the form Eq. (95), i.e.

$E^{M}_{\,A}=\begin{pmatrix}\pm 1&0&0\\\ \gamma&b&c\\\ \delta&d&e\\\
\end{pmatrix},$ (55)

and the “super-metric” has the following form as a function of the vierbein:

$g^{\Lambda\Pi}=E^{\Lambda}_{\,\,A}\,\eta^{AB}\,(-1)^{(1+B)\Pi}E^{\Pi}_{\,\,B}.$
(56)

A long but easy calculation gives

$g^{MN}=\begin{pmatrix}1-2\,\gamma\,\delta&\gamma\,d-\delta\,b&\gamma\,e-\delta\,c\\\
\gamma\,d-\delta\,b&0&b\,e-c\,d\\\
\gamma\,e-\delta\,c&-(b\,e-c\,d)&0\end{pmatrix}$ (57)

and implementing the constraint (96) we get:

$g^{MN}=\begin{pmatrix}1-2\,\gamma\,\delta&\gamma\,d-\delta\,b&\gamma\,e-\delta\,c\\\
\gamma\,d-\delta\,b&0&\pm 1\\\ \gamma\,e-\delta\,c&\mp 1&0\end{pmatrix}.$ (58)

Apparently this metric depends on $\gamma,\delta,b,c,d,e$ which means 12
parameters minus the two constraints on $b,c,d,e$ so on 10 parameters.
Actually the combinations of parameters which enter the $g^{MN}$ is less. In
fact let us define the following variables:

$\left\\{\begin{matrix}\pi_{1}&\equiv&\gamma_{\theta}\,e_{B}-\delta_{\theta}\,c_{B}\\\
\pi_{2}&\equiv&\gamma_{\bar{\theta}}\,e_{B}-\delta_{\bar{\theta}}\,c_{B}\\\
\pi_{3}&\equiv&\delta_{\theta}\,b_{B}-\gamma_{\theta}\,d_{B}\\\
\pi_{4}&\equiv&\delta_{\bar{\theta}}\,b_{B}-\delta_{\bar{\theta}}\,d_{B}\end{matrix}\right.$
(59)

and

$\pi_{5}\equiv\gamma_{\bar{\theta}}\,\delta_{\theta}-\gamma_{\theta}\,\delta_{\bar{\theta}}.$
(60)

This $\pi_{5}$ is actually dependent on the other four $\pi_{i}$ of Eq. (59),
in fact it is easy to show that

$\pi_{5}=\pm(\pi_{2}\,\pi_{3}-\pi_{1}\,\pi_{4}).$

It is easy to see that the metric Eq. (57) can be written in term of the
$\pi_{i}$ as follows

$g^{MN}=\begin{pmatrix}1\pm
2\bar{\theta}\theta(\pi_{2}\pi_{3}-\pi_{1}\pi_{4})&-\pi_{3}\theta-\pi_{4}\bar{\theta}&\pi_{1}\theta+\pi_{2}\,\bar{\theta}\\\
-\pi_{3}\theta-\pi_{4}\bar{\theta}&0&\pm 1&\\\
\pi_{1}\theta+\pi_{2}\,\bar{\theta}&\mp 1&0\end{pmatrix}.$ (61)

Later on, in order to build the Christoffel symbols and the various curvatures
tensor, we shall need also the inverse of $g^{MN}$ which turns out to have the
following expression:

$g_{MN}=\begin{pmatrix}1&\mp(\theta\pi_{1}+\bar{\theta}\pi_{2})&\mp(\theta\pi_{3}+\bar{\theta}\pi_{4})\\\
\pm(\theta\pi_{1}+\bar{\theta}\pi_{2})&0&\mp(1+\bar{\theta}\theta(\pi_{2}\,\pi_{3}-\pi_{1}\,\pi_{4}))\\\
\pm(\theta\pi_{3}+\bar{\theta}\pi_{4})&\pm(1+\bar{\theta}\theta(\pi_{2}\,\pi_{3}-\pi_{1}\,\pi_{4}))&0\end{pmatrix}.$
(62)

In both metrics above we have made the choice $a=\pm 1$ which is consistent
with the CPI. The reader may wonder why the metric has less free parameters
than the vierbein. We feel the reason is because of the particular combination
of vierbeins which enters the metric (see Eq. (55)). Moreover the vierbein is
a more general object than the metric; in fact it enters the dynamics of
particles of any spin. One last question the reader may have is if the
$\pi_{i}$ are really free parameters or not. We feel they are free because
they are made (see Eq. (59)) of combinations of odd variables $\gamma,\delta$
and even one $b,c,d,e$ and only these last ones are constrained (see Eq. (96),
while the first ones are totally free. Let us now build the metric for the
quantum case (QPI) or better for the “regularized” quantum case. The very long
details of the calculations are confined in D. The result is anyhow the
following

$g^{MN}=\begin{pmatrix}1-2\,\pi_{5}^{Q}\,\bar{\theta}\theta&-\pi_{3}^{Q}\,\theta-\pi_{4}^{Q}\,\bar{\theta}&\pi_{1}^{Q}\,\theta+\pi_{2}^{Q}\,\bar{\theta}\\\
-\pi_{3}^{Q}\,\theta-\pi_{4}^{Q}\,\bar{\theta}&0&\pi_{7}^{Q}+\pi^{Q}_{6}\bar{\theta}\theta\\\
\pi_{1}^{Q}\theta+\pi_{2}^{Q}\bar{\theta}&-\pi_{7}^{Q}-\pi_{6}^{Q}\,\bar{\theta}\theta&0\end{pmatrix}.$

where the variables $\pi_{1}^{Q},\dots,\pi_{7}^{Q}$ are properly defined in D.

##  6 Curvatures

In this section we will build the curvatures from the metric presented in the
previous section. As the calculations are very long we have made used of a
package of Wolfram Mathematica  dedicated to calculations containing
grassmannian variables [22]. The same package has been used also for
calculating the metric of the previous section and other calculations
presented all through the paper. The first thing we have calculated has been
the Christoffel symbols associated to our varoius metrics. If we work in a
space with odd and even variables the Christoffel symbols $\Gamma^{C}_{AB}$
have the following expression [21]:

$\displaystyle\Gamma_{AB}^{C}$ $\displaystyle=$
$\displaystyle(-)^{BC}\frac{1}{2}\left[(-)^{BD}\,g_{AD,B}+(-)^{A+B+AB+AD}\,g_{BD,A}-g_{AB,D}\right]\,g^{DC}$

where the comma on the metric like $g_{AD,B}$ means the derivative of $g_{AD}$
respect to the variable $B$ . As usual the exponent on the $(-)$ indicate the
even $(0)$ or odd $(1)$ nature of the associated variables. The results of the
calculations of Eq. (LABEL:eq:77-1) for both the CPI metric Eq. (61) and Eq.
(62) and for the QPI one Eq. (149) and Eq. (156) are confined in E. Once we
have the Christoffel symbols we can calculate the various curvatures. The
definition we will use is the one of ref.[21] for SuperRiemannian space:

$\displaystyle R_{ABC}^{D}$ $\displaystyle=$
$\displaystyle-\Gamma_{AC,B}^{D}+(-)^{BC}\,\Gamma_{AB,C}^{D}-(-)^{C(D+E)}\,\Gamma_{AC}^{E}\,\Gamma_{EB}^{D}+$
(64) $\displaystyle(-)$ ${}^{B(C+D+E)}\,\Gamma_{AB}^{E}\,\Gamma_{EC}^{D}.$

From this we can build the Ricci curvature tensor defined as

$R_{AB}\equiv(-)^{C}\,R_{ABC}^{C}.$ (65)

Its expression in term of Christoffel symbols is:

$\displaystyle R_{AB}$ $\displaystyle=$
$\displaystyle(-)^{C+1}\,\Gamma_{AC,B}^{C}+(-)^{C(B+1)}\,\Gamma_{AB,C}^{C}+$
$\displaystyle-$
$\displaystyle(-)^{C(C+E-1)}\,\Gamma_{AC}^{E}\,\Gamma_{EB}^{C}+(-)^{BE+C}\,\Gamma_{AB}^{E}\,\Gamma_{EC}^{C}.$

From the Ricci curvature tensor we can calculate the so-called Ricci scalar
defined in [21] as

$R=(-)^{B}\,g^{BA}\,R_{AB}.$ (67)

The explicit expression of all components of the Ricci tensor and of the Ricci
scalar has been confined to F and G. Its calculation, again, has been made
possible by the use of Wolfram Mathematica [22]. The things to notice for the
curvatures of the QPI (see Eq. (163) and (G.2)) is they are singular for
$\epsilon\to 0$. Infact in many of its component we have the $\pi^{Q}_{7}$ in
the denominator and $\pi^{Q}_{7}$ is proportional to $\epsilon$. Only the
$R_{\theta\theta}$ and $R_{\bar{\theta}\bar{\theta}}$ are not singular because
they are equal to zero. What we should take care of is the Ricci scalar, Eq.
(F.2) and (G.2)), where the singularity cannot be a coordinate artifact
because it is a scalar independent of the coordinates. The way out could came
from the fact that in the true quantum case ($\epsilon\to 0$) se also have
that $\theta$ and $\bar{\theta}$ have to be sent to zero [21]. So for example
in Eq. (F.2) we have that the fourth contribution is proportional to
$\bar{\theta}\theta/\pi_{7}$ that would give a $0/0$, which is an undefined
term. But being $\epsilon$ and $\bar{\theta}\theta$ totally independent, we
could choose that this undefined form is a finite grassman number. The next
term is proportional to $\theta\bar{\theta}$ and it would go to zero. In this
manner the Ricci scalar would not blow up in QM, which seems a natural thing
to require as nothing goes to infinity in QM

##  7 Zeros of the curvature

In this section we will check if there are value of the $\pi_{i}$ for which
the various curvatures turn out to be zero or at least some of them. As the
$\pi_{i}$ are arbitrary we can choose the system to sit on those values and so
conclude that those curvatures are zero. This is what happens in the CPI as we
will check. Surprisingly this does not happens in the QPI. There is no point
where the curvatures vanish. This “may” indicate that there is some sort of
“hidden matter” in QM. Of course we don’t identify this with the so called
“hidden variables” of Einstein [23].

###  7.1 Zeros of the curvature in the CPI

Let us start with the CPI in the case where the $\pi_{i}$ are indipendent of
time. The Ricci scalar was given in Eq. (162) and it had the following
expression:

$\displaystyle R^{CPI}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left(\pi_{2}^{2}-22\,\pi_{2}\,\pi_{3}+\pi_{3}^{2}+20\,\pi_{1}\,\pi_{4}\right)+$
(68) $\displaystyle+$ $\displaystyle
8\,\frac{\bar{\theta}\theta}{a}\,\left(\pi_{2}\,\pi_{3}-\pi_{1}\,\pi_{4}\right)^{2}.$

In order to have $R^{CPI}=0$ we need to have zero both its soul and body,
i.e.:

$\left\\{\begin{aligned} &\pi_{2}\,\pi_{3}-\pi_{1}\,\pi_{4}=0\\\
&\pi_{2}^{2}-22\,\pi_{2}\,\pi_{3}+\pi_{3}^{2}+20\,\pi_{1}\,\pi_{4}=0.\end{aligned}\right.$
(69)

From the first equation in (69) we get

$\pi_{1}=\frac{\pi_{2}\,\pi_{3}}{\pi_{4}}$ (70)

and putting this into the second of Eq. (69) we get

$\pi_{2}=\pi_{3}.$ (71)

In the space described by the four parameters
$\pi_{1},\pi_{2},\pi_{3},\pi_{4}$ the Ricci scalar is zero on a 2-dim surface
descibed by Eq. (70) and Eq. (71). Let us now check if on this surface, or at
least on some points, also the Ricci tensor is zero. Let us look at the
various components of the Ricci tensor presented in Eq. (161). Let us start
with

$R_{tt}=\frac{1}{2}\left(\pi_{2}+\pi_{3}\right)^{2}-2\,\pi_{1}\,\pi_{4}.$

Using Eq. (70) and Eq. (71) it is straightforward to show that $R_{tt}=0$. Let
us now move on to

$R_{t\theta}=-R_{\theta
t}=\frac{(\theta\,\pi_{1}+\bar{\theta}\,\pi_{2})\,(-(\pi_{2}+\pi_{3})^{2}+4\,\pi_{1}\,\pi_{4})}{2\,a}.$

The second factor on the right is zero on Eq. (69) so $R_{t\theta}=R_{\theta
t}=0$. Let us now check

$R_{t\bar{\theta}}=-R_{\bar{\theta}t}=-\frac{(\theta\,\pi_{3}+\bar{\theta}\,\pi_{4})\,((\pi_{2}+\pi_{3})^{2}-4\,\pi_{1}\,\pi_{4})}{2\,a}.$

Again the second factor on the right is zero on Eq. (69) so
$R_{t\bar{\theta}}=R_{\bar{\theta}t}=0$. Next let us check

$\displaystyle R_{\theta\bar{\theta}}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[\bar{\theta}\theta\underbrace{\left(\pi_{1}\,\pi_{4}-\pi_{2}\,\pi_{3}\right)}_{A}\,\left(\pi_{2}^{2}-6\,\pi_{2}\,\pi_{3}+\pi^{2}_{3}+4\,\pi_{1}\,\pi_{4}\right)\right.$
$\displaystyle-$ $\displaystyle
a\,\left.\underbrace{\left(\pi_{2}^{2}-10\,\pi_{2}\,\pi_{3}+\pi^{2}_{3}+8\,\pi_{1}\,\pi_{4}\right)}_{B}\right].$

The term $A=0$ because of Eq. (69) while $B$, using Eq. (69), can be
transformed as follows:

$B=\pi_{2}^{2}-10\,\pi_{2}\,\pi_{3}+\pi^{2}_{3}+8\,\pi_{1}\,\pi_{4}=\pi_{2}^{2}-10\,\pi_{2}^{2}+\pi^{2}_{2}+8\,\pi_{2}^{2}=0.$

So $R_{\theta\bar{\theta}}=-R_{\bar{\theta}\theta}=0$.
The reason why we can choose the values of the $\pi_{i}$ on which our
curvature is zero is because the $\pi_{i}$ do not enter the lagrangian of the
CPI and we can change them without the Lagrangian getting modified. So far in
the CPI we have proved that both the Ricci scalar and the Ricci curvature can
be brought to zero. This is a situation very similar to the Schwarzschild case
where, for points outside the mass region we have both $R$ and $R_{ab}$ equal
to zero. What is not zero there is another scalar built up from the curvature
tensor:

$R_{abcd}R^{abcd}\neq 0.$

This quantity in the Schwarzschild case is proportional to $G/r^{6}$ and it is
the indicator of the presence of matter somewhere. We should calculate the
analog quantity for the CPI. In this case the quantity to calculate is:

$R_{abc}R^{abc}.$

Instead of doing this rather complicated calculation, we should remind
ourselves that our analog of space-time is $(t,\theta,\bar{\theta})$ so we
should just check if our Ricci scalar and tensor are zero for any value of
$(t,\theta,\bar{\theta})$. This would not happen in the Schwarzschild case in
the area where there is matter. The calculation we have done is without $t$
and with $\theta,\bar{\theta}\neq 0$ and it gives zero everywhere. Let us now
generalize it to the case with $t$ present that means when the $\pi_{i}$
depend on $t$ and see if we get zero everywhere in $(t,\theta,\bar{\theta})$.
Let us start by finding the points on which the Ricci scalar is zero for
$\pi_{i}$ depending on time. The $R^{CPI}$ is given by Eq. (G.1)

$\displaystyle R^{CPI}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left[\pi_{2}^{2}-22\,\pi_{2}\pi_{3}+\pi^{2}_{3}+20\,\pi_{1}\pi_{4}+4(\pi_{3}^{\prime}-\pi_{2}^{\prime})\right]$
$\displaystyle+$
$\displaystyle\frac{\bar{\theta}\theta}{a}\left[8\,(\pi_{2}\pi_{3}-\pi_{1}\pi_{4})^{2}+4\,a\,(\pi_{3}^{\prime}-\pi_{2}^{\prime})\,\pi_{5}+7\,a\,\pi_{5}^{\prime}(\pi_{3}-\pi_{2})+2\,a\,\pi_{5}^{\prime\prime}\right].$

Both the soul and the body must be zero, i.e.

$\left\\{\begin{aligned}
&\pi_{2}^{2}-22\,\pi_{2}\pi_{3}+\pi^{2}_{3}+20\,\pi_{1}\pi_{4}+4(\pi_{3}^{\prime}-\pi_{2}^{\prime})=0\\\
&8\,(\pi_{2}\pi_{3}-\pi_{1}\pi_{4})^{2}+4\,a\,(\pi_{3}^{\prime}-\pi_{2}^{\prime})\,\pi_{5}++7\,a\,\pi_{5}^{\prime}(\pi_{3}-\pi_{2})+2\,a\,\pi_{5}^{\prime\prime}=0.\end{aligned}\right.$
(73)

If in the time-independent case Eq. (70) and Eq. (71) are choosen, starting
from the relation between $\pi_{5}$ and $\pi_{1},\pi_{2},\pi_{3},\pi_{4}$

$\pi_{5}=\frac{\pi_{2}\,\pi_{3}-\pi_{1}\,\pi_{4}}{a},$

it can be easily seen that

$\pi_{5}=0,$ (74)

from which it follows that $\pi_{5}^{\prime}=\pi_{5}^{\prime\prime}=0$. The
system of Eq. (73) consequently reduces to

$\left\\{\begin{aligned}
&\pi_{2}^{2}-22\,\pi_{2}\pi_{3}+\pi^{2}_{3}+20\,\pi_{1}\pi_{4}=0\\\
&8\,(\pi_{2}\pi_{3}-\pi_{1}\pi_{4})^{2}=0.\end{aligned}\right.$ (75)

which is equivalent to the system of Eq. (69), whose solutions Eq. (70) and
Eq. (71) are the one we started with in the time independent case. It follows
that the constraints on which $R^{CPI}$ is zero, even in the time dependent
case, are Eq. (70) and Eq. (71) like in the case of $\pi_{i}$ independent on
time. Of course there may be other set of points on which it is zero (because
$\pi_{2}=\pi_{3}$ was our choice) but what is important is that we have found
points in which it is zero. Next we should check that also the Ricci tensor is
zero on the same set of points. Their expression is given in (165) of D and it
is easy to check they are all zero on the points where the following condition
are satisfied (70),(71),(74):

$\left\\{\begin{aligned} &\pi_{2}=\pi_{3}\\\
&\pi_{1}=\frac{\pi_{2}\,\pi_{3}}{\pi_{4}}\\\ &\pi_{5}=0.\end{aligned}\right.$
(76)

Let us consider for example $R_{tt}$ of (165):

$\displaystyle R_{tt}$
$\displaystyle=R_{tt}^{CPI}(\pi_{i})+(\pi_{3}^{\prime}-\pi_{2}^{\prime})+\bar{\theta}\theta\left[\pi_{3}^{2}\pi_{2}^{\prime}-\pi_{2}^{2}\pi_{3}^{\prime}+(\pi_{2}-\pi_{3})(\pi_{4}\pi_{1}^{\prime}+\pi_{4}^{\prime}\pi_{1})+\right.$
$\displaystyle+\left.(\pi_{3}^{\prime}-\pi_{2}^{\prime})(a\,\pi_{5}-2\,\pi_{2}\pi_{3}+3\,\pi_{1}\pi_{4})+2\,a\pi_{5}^{\prime}\right].$
(77)

It results that the first term is equal to zero because it has the same
expression as the time independent $R_{tt}^{CPI}$, which was zero on the
contraints (70),(71). As for the other contributions they are trivially zero
because the conditions $\pi_{2}=\pi_{3}$ and $\pi_{5}=0$ imply
$\pi_{2}^{\prime}=\pi_{3}^{\prime}$ and $\pi_{5}^{\prime}=0$. In the same way
it can be shown that all the other contributions are identically equal to
zero. As we explainied in the time independent case we do not calculate

$R_{abc}R^{abc},$

bacause we proved that the Ricci scalar and the tensors are zero over the
whole $(t,\theta,\bar{\theta})$ space. As a consequence we have that there is
no matter anywhere differently than in the Schwarzschild case.

###  7.2 Lack of zeros in the QPI curvature

Let us now turn to the QPI case starting with the Ricci scalar in the case
independent on time. Its expression was given in Eq. (F.2):

$\displaystyle R^{QPI}$ $\displaystyle=$ $\displaystyle
R^{CPI}(\pi^{Q}_{i})+(2\,\sigma_{1}-3\,\pi^{Q}_{6})+$ $\displaystyle+$
$\displaystyle\frac{\bar{\theta}\theta}{\pi_{7}^{Q}}\left[-(\pi_{2}^{Q^{2}}+6\,\pi_{2}^{Q}\pi_{3}^{Q}+\pi_{3}^{Q^{2}}-8\,\pi_{1}^{Q}\pi_{4}^{Q})\pi_{6}^{Q}+4\,\pi_{6}^{Q^{2}}+\right.$
$\displaystyle-$
$\displaystyle\left.4\,\sigma_{1}(\pi_{2}^{Q^{2}}+3\,\pi_{2}^{Q}\pi_{3}^{Q}+\pi_{3}^{Q^{2}}-5\,\pi_{1}^{Q}\pi_{4}^{Q}-\pi_{6}^{Q}+\sigma_{1})\right].$

The $\sigma_{1}$ was defined in (158) of E and the parameters of (7.2) are not
four but six. The various $\pi_{i}^{Q}$ were introduced in (151). Note that
$\pi_{5}^{Q}$ does not appear because it is related to the others. Let us
suppose we stay on the surface where

$R^{CPI}(\pi_{i}^{Q})=0.$ (79)

and this happens when the constraints Eq. (70) and Eq. (71) are satisfied but
with the $\pi_{i}$ replaced by the $\pi_{i}^{Q}$. Next, for the body of Eq.
(7.2) to be zero we need that

$2\,\sigma_{1}-3\,\pi^{Q}_{6}=0$

i.e.

$\sigma_{1}=\frac{3}{2}\,\pi_{6}^{Q}.$ (80)

Like in the CPI, we keep $\pi_{4}^{Q}$ and $\pi_{3}^{Q}$ free and link the
other vabiables to these two via the $R^{CPI}(\pi_{i}^{Q})=0$. Also
$\pi_{6}^{Q}$ seems to be free and the same $\pi_{7}^{Q}$, which does not make
its appearance in Eq. (7.2) but was present in the formalism. So Eqs. (80) and
(79) make the body of $R^{QPI}$ zero; now we have to make the soul zero. Using
the constrain Eq. (79), which leads to $\pi_{2}^{Q}=\pi_{3}^{Q}$, and Eq.
(80), after straightforward calculations we get that the soul of $R^{QPI}$ is
given by

$\textrm{soul}(R^{QPI})=\frac{\pi^{{Q}^{2}}_{6}}{\pi_{7}^{Q}}.$

So to be zero we have to set

$\pi_{6}^{Q}=0.$ (81)

We can summarize the set of constraints which make $R^{QPI}=0$ as:

$\left\\{\begin{aligned} \pi_{2}^{Q}&=\pi_{3}^{Q}\\\
\pi_{1}^{Q}&=\cfrac{\pi_{2}^{Q}\,\pi_{3}^{Q}}{\pi_{4}^{Q}}\\\
\pi_{6}^{Q}&=0\\\ \sigma_{1}&=\cfrac{3}{2}\,\pi_{6}^{Q}\end{aligned}\right.$
(82)

Let us now see if also the Ricci tensor for the QPI, given by Eq. (163) is
zero on the points of Eq. (82). Let us start from $R_{tt}^{QPI}$:

$\displaystyle R_{tt}^{QPI}$ $\displaystyle=$
$\displaystyle\underbrace{R_{tt}^{CPI}(\pi_{i}^{Q})}_{A}+2\,\sigma_{1}+\underbrace{\frac{\bar{\theta}\theta}{\pi_{7}^{Q}}\left[((\pi_{2}^{Q}+\pi_{3}^{Q})^{2}-4\,\pi_{1}^{Q}\pi_{4}^{Q})\pi_{6}^{Q}\right.}_{B}$
$\displaystyle+$
$\displaystyle\left.\underbrace{2\,(\pi_{2}^{Q^{2}}+4\,\pi_{2}^{Q}\pi_{3}^{Q}+\pi_{3}^{Q^{2}}-6\,\pi_{1}^{Q}\pi_{6}^{Q}-\pi_{6}^{Q})\,\sigma_{1}+6\sigma_{1}^{2}}_{C}\right].$

The $A$-term calculated using Eq. (82) gives

$\displaystyle A$ $\displaystyle=$
$\displaystyle\left[\frac{1}{2}(\pi_{2}^{Q}+\pi_{3}^{Q})^{2}-2\,\pi_{1}^{Q}\pi_{4}^{Q}\right]$
$\displaystyle=$
$\displaystyle\left[\frac{1}{2}(2\,\pi_{2}^{Q})^{2}-2\,\pi_{2}^{Q}\pi_{3}^{Q}\right]=2\,\pi_{2}^{Q^{2}}-2\,\pi_{2}^{Q^{2}}=0.$

The term $B$ is zero because is multiplied by $\pi_{6}^{Q}$ which is zero by
Eq. (82). The term $C$ is zero because $\sigma_{1}=0$. After the $A$ piece
there is a $\sigma_{1}$ which is zero. So $R_{tt}^{QPI}=0$. Let us now analyze
the $R_{t\theta}^{QPI}$ which is

$\displaystyle R_{t\theta}^{QPI}$ $\displaystyle=$ $\displaystyle-R_{\theta
t}^{QPI}$ $\displaystyle=$ $\displaystyle
R_{t\theta}^{CPI}(\pi_{i}^{Q})-\frac{\pi_{6}^{Q}+3\,\sigma_{1}}{\pi_{7}^{Q}}\left[\theta\pi_{1}^{Q}+\bar{\theta}(\pi_{2}^{Q}+\pi_{3}^{Q})\right].$

The $R_{t\theta}^{CPI}=0$ on the constraints Eq. (82). The second piece in Eq.
(7.2) is zero because $\pi_{6}^{Q}=\sigma_{1}=0$, so $R_{t\theta}^{QPI}=0$.
Next let us analyze $R_{\theta\bar{\theta}}$ which is

$\displaystyle R_{\theta\bar{\theta}}^{QPI}$ $\displaystyle=$ $\displaystyle-
R_{\bar{\theta}\theta}=R_{\theta\bar{\theta}}^{CPI}(\pi_{i}^{Q})+\frac{\sigma_{1}-3\,\pi_{6}^{Q}}{\pi_{7}^{Q}}$
$\displaystyle+$
$\displaystyle\frac{\bar{\theta}\theta}{\pi_{7}^{Q^{2}}}\left[2\,\pi_{6}^{Q^{2}}+8\,\pi_{1}^{Q}\pi_{4}^{Q}\pi_{6}^{Q}-8\,\pi_{2}^{Q}\pi_{3}^{Q}\pi_{6}^{Q}\sigma_{1}\left((\pi_{2}^{Q}-\pi_{3}^{Q}\right)^{2}+4\,\pi_{6}^{Q}+2\,\sigma_{1})\right].$

Again $R_{\theta\bar{\theta}}^{CPI}=0$ is zero on the constraints Eq. (82) and
all the rest is zero because $\pi_{6}^{Q}=\sigma_{1}=0$.

Now let us check if the constraint Eq. (82) leads to any contradiction. Let us
go back to the definition of $\pi_{6}^{Q}$ given in Eq. (154)

$\pi_{6}^{Q}=\epsilon\,a_{S}-\cfrac{i}{\hbar}\,a_{B}\,(1-\epsilon)+a_{B}(\alpha_{\theta}\,\pi_{2}-\alpha_{\bar{\theta}}\,\pi_{1}+\beta_{\theta}\,\pi_{4}-\beta_{\bar{\theta}}\,\pi_{3})+\alpha_{\bar{\theta}}\,\beta_{\theta}-\alpha_{\theta}\,\beta_{\bar{\theta}}.$

Going to the true-quantum case $\epsilon=0$ we would get

$\pi_{6}^{Q}=-\cfrac{i}{\hbar}\,a_{B}+a_{B}\underbrace{(\alpha_{\theta}\,\pi_{2}-\alpha_{\bar{\theta}}\,\pi_{1}+\beta_{\theta}\,\pi_{4}-\beta_{\bar{\theta}}\,\pi_{3})}_{A}+\underbrace{\alpha_{\bar{\theta}}\,\beta_{\theta}-\alpha_{\theta}\,\beta_{\bar{\theta}}}_{B}.$
(85)

If, as we did before in our calculations, we choose $\alpha=\beta=0$, we get
in the true quantum case:

$\pi_{6}^{Q}=-\cfrac{i}{\hbar}\,a_{B}$

as $a_{B}=\pm 1$ we get

$\pi_{6}^{Q}=\mp\cfrac{i}{\hbar}\,\,.$ (86)

So it is never zero and this contradicts Eq. (82) or, saying it better,
because of Eq. (86) the constraint Eq. (82) is not satisfied. The Ricci scalar
curvature and the associated tensor are never zero in the true quantum case.
Let us suppose we do not make the choice $\alpha=\beta=0$, then in Eq. (154)
we would have three terms: one which is a complex number
$\epsilon\,a_{S}-\cfrac{i}{\hbar}\,a_{B}\,(1-\epsilon)$ and the $A,B$ of Eq.
(85) which are the product of couple of grassmannian odd number like $\pi_{2}$
and $\alpha_{\theta}$ and similar. These $A$ and $B$ are grassmann even and
they will never be equal to a complex number. So $A$ and $B$ cannot cancel the
above mentioned complex number in order to put $\pi_{6}^{Q}$ equal to zero. As
the $A$ and $B$ contain the parameterm $\pi_{i}^{Q}$ which are free they could
be put to zero, but then also the complex number has to be put to zero. This
is possible in the regularized $QPI$ and it would mean

$\epsilon\,a_{S}-\frac{i}{\hbar}a_{B}(1-\epsilon)=0$

which is equivalent to:

$a_{S}=\frac{i}{\hbar}\,a_{B}\,\frac{1-\epsilon}{\epsilon}.$ (87)

But in the true quantum case $\epsilon\to 0$ we would get $a_{S}\to\infty$,
which does not make sense. Let us do the last attempt and see if we can put
$R^{QPI}=0$ without putting $\pi_{6}^{Q}=0$. Let us go back to formula Eq.
(7.2) and let us see if there is a different method to get $R^{QPI}=0$. If in
Eq. (7.2) we put first to zero the body and then the soul we get

$R^{CPI}(\pi^{Q}_{i})+2\sigma_{1}-3\pi_{6}^{Q}=0$

which leads to

$\pi_{6}^{Q}=\frac{1}{3}\,(R^{CPI}(\pi_{i}^{Q})+2\sigma_{1}).$ (88)

From this formula it seems that we do not have to put $\pi_{6}^{Q}=0$. But let
us analize Eq. (88) in detail. On the R.H.S. we have only terms which are
product of grassmann variables like

$\sigma_{1}=\pi_{2}^{Q}\pi_{3}^{Q}-\pi_{1}^{Q}\pi_{4}^{Q}-\pi_{5}^{Q}\pi_{7}^{Q}.$

The same for $R^{CPI}$ which is

$\displaystyle R^{CPI}(\pi_{i}^{Q})$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left(\pi_{2}^{Q^{2}}-22\,\pi_{2}^{Q}\pi^{Q}_{4}+\pi_{3}^{Q^{2}}+20\,\pi_{1}^{Q^{2}}\pi_{4}^{Q}\right).$
$\displaystyle+$
$\displaystyle\bar{\theta}\theta\frac{8}{a}(\pi_{2}^{Q}\pi_{3}^{Q}-\pi_{1}^{Q}\pi_{4}^{Q}).$

So on the R.H.S. of Eq. (88) we have products of 2 grassmannian numbers or
higher terms (like those with $\bar{\theta}\theta$ which anyhow goes to zero
in the true quantum case $\theta,\bar{\theta}\to 0$), while on the L.H.S. we
have $\pi_{6}^{Q}$ which (see Eq. (85)) contains both product of grassmannian
numbers but also complex number which must be put to zero separately and we go
back to the case descibed by Eq. (87). So we can conlude that in the true
quantum case we cannot bring the curvature to zero.

##  8 Conclusions

In this paper we have shown that “intrinsic” vierbein, persent in the CPI
version of CM, gives zero curvature (at least the Ricci one). This seems
natural because there is no external mass generating a curvature in the space
on which our test particle of the CPI would move. In the quantum case there is
an intrinsic curvature. One could immediately ask what the matter, which
produce this curvature, is. We could speculate saying that there are some non-
local hidden variables of the type Bell [23] proposed long ago or it is some
sort of dark matter or dark energy so fashionable these days. We do not know
and we prefer not to speculate. Our goal at the beginning was to see if there
was some “intrinsic” geometry in Q.M and we feel we have found some hints of
it. We also would like to notice that this intrinsic geometry appear when we
look not only at the usual bosonic variables $\varphi^{a}$ of Q.M but also at
their differential forms $c^{a}$, which we would like to call “quantum forms”.
This is a topic which has not been studied deeply except by few mathematicians
and in a language difficult for physicists. We feel a more intense study
should be done of this sector of mathematics.

## Appendix A

###  A.1 Grassmannian algebras

Given a set of $N$-elements $\xi^{a},a=1,\dots,N$ obeying the following
properties

$\xi^{a}\xi^{b}=-\xi^{b}\xi^{a},\,(\xi^{a})^{2}=0,\textrm{for all $a,b$},$

they are called generators of a grassmannian algebra ${\Lambda}_{N}$.
The elements $1,\xi^{a},\xi^{a}\xi^{b},\xi^{a}\xi^{b}\xi^{c},\dots$ form a set
of $2^{N}$ objects, called the basis of the algebra. An addition in this basis
and a multiplication by complex numbers is defined among its elements and so
they form a linear vector space of dimension $2^{N}$.

###  A.2 Super-numbers

Every element $z$ of the vector space above can be written as

$z=z_{B}+z_{S}$

where $z_{B}$ is an ordinary complex number and it is called the “body” of $z$
and $z_{S}$, called the “soul”, is:

$z_{S}=\sum_{n=1}^{2^{N}}\frac{1}{n!}c_{a_{1}\cdots
a_{n}}\xi^{a_{n}}\cdots\xi^{a_{1}},$ (89)

where the $c_{a_{1}\cdots a_{n}}$ are also complex numbers. The
$c_{a_{1}\cdots a_{n}}$ are antisymmetric in the exchange of their indices. It
is easy to prove that

$z_{S}^{N+1}=0.$

###  A.3 Inverse of a super-number

The inverse $z^{-1}$ of a super-number, defined by $z\,z^{-1}=1$, turns out to
be

$z^{-1}=z_{B}^{-1}\sum_{n=0}^{2^{N}}\left(-z_{B}^{-1}\,z_{S}\right),$

so if $z_{B}=0$ the inverse does not exist.

###  A.4 C-number and A-number

Any super-number can be split into its even “e” and odd “o” part as

$\displaystyle z$ $\displaystyle=$ $\displaystyle e+o$ $\displaystyle e$
$\displaystyle\equiv$ $\displaystyle z_{B}+\sum_{n}\frac{1}{2n!}c_{a_{2}\cdots
a_{2n}}\xi^{a_{2n}}\cdots\xi^{a_{2}}$ $\displaystyle o$ $\displaystyle\equiv$
$\displaystyle\sum_{n}\frac{1}{2n+1!}c_{a_{1}\cdots
a_{2n+1}}\xi^{a_{2n+1}}\cdots\xi^{a_{1}}.$

If a super-number has only an “$e$” part is called an even super-number while
if it has only an “$o$” part it is called an odd super-number. The grassmann
index of even or odd numbers is the number $2n$ or $2n+1$ modulo 2 so for even
number is zero and for odd number is one. Usually it is put as exponent of
$(-1)$ and is indicated with square brackets: $[e],[o]$.

###  A.5 Super-vectors and super-matrices

Super-vectors are defined regorously in [17], but basically are rows or
columns of super-numbers. The elements in the basis of these vectors are
arranged in such a manner that the even elements “$e$” come above the odd
“”o”” one, like

$\begin{pmatrix}e\\\ o\end{pmatrix}.$ (90)

If the basis has the form Eq. (90) then a super-matrix $\mathcal{K}$ can
always be arranged in the form

$K=\begin{pmatrix}A&C\\\ D&B\end{pmatrix},$ (91)

where the elements of the super-matrices $A$ and $B$ are made of super-numbers
while $C$ and $D$ are made of odd numbers. More details are given in [17].

###  A.6 Super-trace

The super-trace of the matrix $\mathcal{K}$ is defined as

$\textrm{str}\mathcal{K}=(-1)^{[i]}\mathcal{K}_{i}^{i},$

where $[i]$ is the grassmann index of the “i” elements.

###  A.7 Super-determinant and its inverse

For a standard matrix $X$ we know that the following relation holds between
the variation “$\delta$” of parameters entering the determinant and the ones
entering the trace:

$\delta[\textrm{ln det}X]=\textrm{tr}[X^{-1}\delta X].$

We use this relation to define the super-determinant in case of a super-matrix
$\mathcal{K}$ which has the form $\mathcal{K}$ of Eq. (91). The result [17]
is:

$\textrm{sdet}\begin{pmatrix}A&C\\\
D&B\end{pmatrix}=\textrm{det}(A-CB^{-1}D)(\textrm{det}B)^{-1}$ (92)

where the symbol “det” has the same meaning as if the entries were complex
numbers. It is also possible to define the inverse of the supermatrix $X$ as:

$X^{-1}=\begin{pmatrix}\widetilde{A}&\widetilde{C}\\\
\widetilde{D}&\widetilde{B}\end{pmatrix}$ (93)

where

$\displaystyle\widetilde{A}$
$\displaystyle=(\mathbb{I}-A^{-1}CB^{-1}D)^{-1}A^{-1}$
$\displaystyle\widetilde{C}$
$\displaystyle=-(\mathbb{I}-A^{-1}CB^{-1}D)^{-1}A^{-1}CB^{-1}$
$\displaystyle\widetilde{D}$
$\displaystyle=-(\mathbb{I}-B^{-1}DA^{-1}C)^{-1}B^{-1}DA^{-1}$
$\displaystyle\widetilde{B}$
$\displaystyle=(\mathbb{I}-B^{-1}DA^{-1}C)^{-1}B^{-1}.$

Note that this inverse exists if only $A$ and $B$ are not singular. It is also
easy to calculate the determinant of the inverse:

$\textrm{sdet}\begin{pmatrix}A&C\\\
D&B\end{pmatrix}^{-1}=(\textrm{det}A)^{-1}\textrm{det}(B-DA^{-1}C).$

###  A.8 Left and right derivatives

The rigourous definition for these operations are given in [17]. Here we will
give only an example. Let us give a function $f(\xi_{1},\xi_{2})$ of two
grassmannian odd variables $\xi_{1},\,\xi_{2}$ of the form

$f(\xi_{1},\xi_{2})=\xi_{1}\,\xi_{2}.$

Let us define the right or right derivative of $f$ with respect to $\xi_{1}$:

$\displaystyle\frac{\overset{\rightarrow}{\partial}f}{\partial\xi_{1}}$
$\displaystyle=$ $\displaystyle\xi_{2}\quad\textrm{\quad left derivative}$
$\displaystyle\frac{\overset{\leftarrow}{\partial}f}{\partial\xi_{1}}$
$\displaystyle=$ $\displaystyle-\xi_{2}\quad\textrm{\quad right derivative}$

“Somehow” roughly speaking in the right derivative it is as if we had put
$\overset{\leftarrow}{\partial}f/\partial\xi_{1}$ to the right of the function
so that $\overset{\leftarrow}{\partial}f/\partial\xi_{1}$ has to pass through
$\xi_{2}$ in order to act on $\xi_{1}$. In going through $\xi_{2}$ it acquires
a minus sign because $\xi_{1}$ and $\xi_{2}$ anticommute. On grassmannian
spaces we can also define the concept of integration. All the details are
given in [17]. The few things we need in this paper were already indicated in
the body of the paper and will not be repeated in this appendix.

## Appendix B

In this appendix we will give details of the calculations of the vierbein in
the CPI case.
Using the expression Eq. (53) for the $E^{M}_{\,A}$ we get

$D_{t}=\partial_{M}E^{M}_{\,\,t}=a\,\partial_{t}+\alpha\,\partial_{\theta}+\beta\,\partial_{\bar{\theta}},$

so

$(D_{t}Q)(D_{t}Q)=a^{2}\,\partial_{t}Q\,\partial_{t}Q+2\,a\,\alpha\,\partial_{t}Q\,\partial_{\theta}Q+2\,a\,\beta,\partial_{t}Q\,\partial_{\bar{\theta}}Q+\alpha\,\beta\,\partial_{\theta}Q\,\partial_{\bar{\theta}}Q.$

Using the expression above, it is easy to prove that a choice of parameters
for which the second of Eq. (54) holds is the following one:

$a=\pm 1,\quad\alpha=\beta=0.$ (94)

So the supervierbein in Eq. (53) takes the form

$E^{M}_{\,A}=\begin{pmatrix}\pm 1&0&0\\\ \gamma&b&c\\\ \delta&d&e\\\
\end{pmatrix}.$ (95)

Next we have to impose the first of the condition Eq. (54) using the
definition of superdeterminant given in [17] or A of this paper and applied to
Eq. (95):

$\textrm{sdet}E^{M}_{\,A}=1\Longrightarrow\textrm{det}\begin{pmatrix}b&c\\\
d&c\end{pmatrix}\Longrightarrow b\,e-c\,d=\pm 1.$ (96)

The quantity $b,c,d,e$ are even so they have the form

$\begin{matrix}b\equiv b_{B}+b_{S}\,\bar{\theta}\theta\\\ c\equiv
c_{B}+c_{S}\,\bar{\theta}\theta\\\ d\equiv d_{B}+d_{S}\,\bar{\theta}\theta\\\
e\equiv e_{B}+e_{S}\,\bar{\theta}\theta,\end{matrix}$ (97)

where $b_{B},c_{B},d_{B},e_{B}$ are the “bodies” of the numbers while
$b_{S},c_{S},d_{S},e_{S}$ are called the “souls” of the numbers. For details
about these numbers see A of this paper and consult ref.[17]. Using Eq. (97)
the relation (96) gives the two equations

$\left\\{\begin{matrix}b_{B}\,e_{B}-c_{B}\,d_{B}=\pm 1\\\
b_{S}\,e_{B}+b_{B}\,e_{S}-c_{S}\,d_{B}-c_{B}\,d_{S}=0\end{matrix}\right..$
(98)

A set of two solutions has the form

$\displaystyle(1)$
$\displaystyle\,e_{B}=0,\,c_{B}=\mp\frac{1}{d_{B}},\,c_{S}=\frac{\mp
d_{S}+d_{B}\,b_{B}\,e_{S}}{d_{B}^{2}}$ $\displaystyle(2)$
$\displaystyle\,b_{B}=\frac{\pm 1+c_{B}\,d_{B}}{e_{B}},\,b_{S}=\frac{\mp
e_{S}-c_{B}\,d_{B}\,e_{S}}{e_{B}^{2}}+\frac{c_{B}\,d_{S}\,e_{B}+c_{S}\,d_{B}\,e_{B}}{e_{B}^{2}}$

It is a long but easy calculation to build the inverse [17] of the matrix
$E^{M}_{\,A}$, the result is

$\begin{pmatrix}\pm 1&0&0\\\ \pm c\,\delta\mp e\,\gamma&\pm e&\mp c\\\ \pm
d\,\gamma\mp b\,\delta&\mp d&\pm b\end{pmatrix}.$ (100)

So now we have all the matrix elements to build the kinetic term and the
superdeterminat. The number of free elements that we know in Eq. (95) is $12$
because each $b,c,d,e,\gamma,\delta$ is made of two numbers either the body
and the soul of the even ones or for the odd elements, like
$\gamma=\gamma_{\theta}\,\theta+\gamma_{\bar{\theta}}\,\bar{\theta}$, they are
the coefficient of $\theta$ and of $\bar{\theta}$. We have $2$ constraints in
Eq. (98), so the number of free variables is $10$ and we choose them to be
real. Considering the gauge freedom the careful reader may envision the
following problem. We said that building the CPI or the QPI is “like a gauge
fixing”. For the CPI this “gauge fixing ” is given by the constraints of Eq.
(54). One important thing to check is that the “ gauge fixing” Eq. (54) does
not fix more parameters than those allowed by the gauge freedom. This is not
so and we prove it below. Our diffeomorphism, Eq. (44), can be explicitly
written as:

$\displaystyle\delta t$ $\displaystyle=$ $\displaystyle
A(t)+\widetilde{\alpha}(t)\theta+\widetilde{\beta}(t)\bar{\theta}+\beta(t)\theta\bar{\theta}$
$\displaystyle\delta\theta$ $\displaystyle=$
$\displaystyle\widetilde{\gamma}(t)+C(t)\theta+D(t)\bar{\theta}+\epsilon(t)\theta\bar{\theta}$
(101) $\displaystyle\delta\bar{\theta}$ $\displaystyle=$
$\displaystyle\widetilde{\delta}(t)+F(t)\theta+G(t)\bar{\theta}+\xi(t)\theta\bar{\theta},$

where the latin symbols are real even numbers and the greek ones are odd
number functions only of $t$ and not of $\theta$ and $\bar{\theta}$. So in (B)
we have 12 parameters. The general vierbein has the form

$E^{M}_{A}(z)=\begin{pmatrix}a&\alpha&\beta\\\ \gamma&b&c\\\
\delta&d&e\end{pmatrix}$ (102)

and it contains 18 variables because each $a,\alpha,\dots,e$ are made of two
entries. This vierbein transforms in the following manner under (B) or (44):

$E^{\prime
M}_{A}(z)=\frac{\overset{\rightarrow}{\partial}z^{\prime}_{B}}{\partial
z_{A}}E^{M}_{B}(z^{\prime}).$ (103)

If we were able to fully exploit the 12-parameter gauge freedom of (B), we
could reduce the 18-variables of $E^{M}_{A}$ to just six. In the CPI the
vierbein that we use is:

$E^{M}_{A}(CPI)=\begin{pmatrix}\pm 1&0&0\\\ \gamma&b&c\\\
\delta&d&e\end{pmatrix}$ (104)

so we have 12 parameters minus the 2 constraints (98) and this brings down to
10 parameters that are more than 6. So we have done only a partial gauge
fixing.

## Appendix C

In this appendix we will give details of the calculation of the vierbein for
the QPI.
For the QPI we will get a vierbein of the form

$E^{M}_{A}(QPI)=\begin{pmatrix}1+a_{s}\theta\bar{\theta}&\alpha&\beta\\\
\gamma&b&c\\\ \delta&d&e\end{pmatrix},$ (105)

which has $17$ parameters minus $2$ constraints (that we will see later on)
bringing the total free parameters down to $15$. Moreover we will choose
$\alpha=\beta=0$ like in the CPI, so we will come down to $11$ parameters,
which again is more than $6$. This is consistent with considering our
procedure as a partial gauge fixing. This would not be so if the procedure
would bring the number of free parameters to less than $6$ both in the CPI and
in the QPI.

Let us now build the vierbein for the quantum case that is the QPI of Eq. (2).
In this case the determinant of the vierbein $E=\textrm{sdet}(E^{A}_{\,M})$
has to be

$E=-i\frac{\bar{\theta}\theta}{\hbar}.$ (106)

This number has a body equal to zero and, as explained in [17], and in A, it
does not admit an inverse $E^{-1}$. The inverse would be the determinant of
the elements $E^{M}_{\,A}$ which eneter the kinetic piece of the action like
it happened in the CPI (see Eq. (52)). So these elements cannot be built if we
stick to the condition Eq. (106). The way out is is to add a small
“regulating” body $\epsilon$ to Eq. (106) so that the determinant can be
inverted. This “regularized” determinant is

$E^{\textrm{reg}}=\epsilon-i\frac{\bar{\theta}\theta}{\hbar}.$ (107)

The inverse can now be built [17] and it is

$E^{-1}=\frac{1}{\epsilon}+\frac{1}{\epsilon^{2}}\,\frac{\bar{\theta}\theta}{\hbar}.$
(108)

We will now go on to find for the QPI the analog of the two constraints of Eq.
(54). Let us insert the regularized $E^{\textrm{reg}}$ of Eq. (107) into the
action of the QPI written in Eq. (36) once we have integrated out the $P$.
Moreover let us keep only the kinetic piece:

$S^{\textrm{reg}}_{QPI}=i\int dtd\theta
d\bar{\theta}\left(\epsilon-i\frac{\bar{\theta}\theta}{\hbar}\right)\left(\frac{1}{2}D_{t}Q\,D_{t}Q\right).$
(109)

Performing the products above,we get:

$S^{\textrm{reg}}_{QPI}=i\,\epsilon\int dtd\theta
d\bar{\theta}\left(\frac{1}{2}D_{t}Q\,D_{t}Q\right)+\frac{i}{\hbar}\int
dtd\theta
d\bar{\theta}\,\bar{\theta}\theta\left(\frac{1}{2}D_{t}Q\,D_{t}Q\right).$
(110)

The first term goes to zero in the true quantum-case because in this case
$\epsilon\to 0$. So we will work out only the second term in Eq. (110) using
the general form of the vierbein written in Eq. (53) and using the expression
Eq. (52) for the covariant derivative. The second term in Eq. (110) turns out
to be

$\frac{1}{\hbar}\int
dt\,d\theta\,d\bar{\theta}\,\frac{1}{2}\left(a\,\partial_{t}Q+d\,\partial_{t}Q+\beta\,\partial_{\bar{\theta}}Q\right)^{2}\,\bar{\theta}\theta..$
(111)

As the $\alpha$ and $\beta$ are odd and get bultiplied by $\bar{\theta}\theta$
the only term which survives is

$\frac{1}{\hbar}\int
dt\,d\theta\,d\bar{\theta}\,\frac{1}{2}\,a^{2}\,(\partial_{t}Q)^{2}\,\bar{\theta}\theta.$
(112)

Differently than in the classical case of Eq. (94), note that in the quantum
case $a$ is an even element made of a body $a_{B}$ and a soul $a_{S}$. So Eq.
(112) turns out to be

$\frac{1}{\hbar}\int
dt\,d\theta\,d\bar{\theta}\cfrac{1}{2}\,\left(a_{B}^{2}+2\,a_{B}\,a_{S}\,\bar{\theta}\theta\right)\bar{\theta}\theta\partial_{t}Q\partial_{t}Q=\frac{1}{2\,\hbar}\int\,a_{B}^{2}\,(\partial_{t}q)(\partial_{t}q).$
(113)

where $q$ is the first component of $Q$ like in Eq. (27) and we have omitted
the indices for simplicity. In order to get the usual kinetic piece of quantum
mechanics in Eq. (113) we need to have:

$a_{B}=\pm 1,$ (114)

while $a_{S}$ is free. Next we have to impose the conditions Eq. (108) and Eq.
(114) on the determinant, i.e.

$\textrm{sdet}\begin{pmatrix}\pm 1+a_{S}\,\bar{\theta}\theta&\alpha&\beta\\\
\gamma&b&c\\\
\delta&d&e\end{pmatrix}=\frac{1}{\epsilon}+\frac{i}{\epsilon}\frac{\bar{\theta}\theta}{\hbar}.$
(115)

Working out the sdet on the L.H.S. of Eq. (115) using the usual rules given in
[17], we get

sdet $\displaystyle\begin{pmatrix}\pm
1+a_{S}\,\bar{\theta}\theta&\alpha&\beta\\\ \gamma&b&c\\\
\delta&d&e\end{pmatrix}$ $\displaystyle=$ $\displaystyle\left[\pm
1+a_{S}\,\bar{\theta}\theta-\begin{pmatrix}\alpha&\beta\end{pmatrix}\begin{pmatrix}b&c\\\
d&e\end{pmatrix}^{-1}\begin{pmatrix}\gamma\\\
\delta\end{pmatrix}\right]\cdot\textrm{det}^{-1}\begin{pmatrix}b&c\\\
d&e\end{pmatrix}.$

Let us now simplify things by introducing some new symbols $p,q,r$ defined as:

$\displaystyle p\,\bar{\theta}\theta$ $\displaystyle\equiv$
$\displaystyle\begin{pmatrix}\alpha&\beta\end{pmatrix}\begin{pmatrix}b&c\\\
d&e\end{pmatrix}^{-1}\begin{pmatrix}\gamma\\\ \delta\end{pmatrix}$
$\displaystyle q+r\,\bar{\theta}\theta$ $\displaystyle\equiv$
$\displaystyle\textrm{det}^{-1}\begin{pmatrix}b&c\\\ d&e\end{pmatrix}.$ (117)

The powers of $\theta,\bar{\theta}$ present on the L.H.S.of Eq. (117) can be
easily understood by remembering the powers of $\theta,\bar{\theta}$ present
in the even and odd elements. Using Eq. (117) the relation Eq. (C) can be
written as

$\textrm{sdet}\begin{pmatrix}\pm 1+a_{S}\,\bar{\theta}\theta&\alpha&\beta\\\
\gamma&b&c\\\ \delta&d&e\end{pmatrix}=\pm q+(a_{S}\,q-p\,q\pm
r)\bar{\theta}\theta.$ (118)

Combining Eq. (118) with Eq. (115) we get

$\left\\{\begin{matrix}&&q=\pm\cfrac{1}{\epsilon}\\\ &&a_{s}q-p\,q\pm
r=\cfrac{i}{\epsilon^{2}\,\hbar}\end{matrix}\right..$ (119)

which can be combined to give

$\pm\left(\frac{a_{S}}{\epsilon}-\frac{p}{\epsilon}+r\right)=\frac{1}{\epsilon^{2}\,\hbar}.$
(120)

From the second equation in (117) we get that the matrix

$D\equiv\begin{pmatrix}b&c\\\ d&e\end{pmatrix}$ (121)

must be invertible and the determinant of the inverse must be equal to the
L.H.S. of the following equation:

$q+r\,\bar{\theta}\theta=\pm\frac{1}{\epsilon}+r\,\bar{\theta}\theta.$ (122)

The R.H.S. of Eq. (122) is otained from the first of Eq. (119). From the
determinant of the inverse we can get the determinant of $D$ which from Eq.
(122) turns out to be

$\textrm{det}D=\pm\epsilon-\epsilon^{2}\,r\,\bar{\theta}\theta.$ (123)

The determinant of $D$ is equal to $(b\,e-c\,d)$ so Eq. (123) becomes

$b\,e-c\,d=\pm\epsilon-\epsilon^{2}\,r\,\bar{\theta}\theta,$ (124)

which is equal to

$(b_{B}+b_{S}\,\bar{\theta}\theta)(e_{B}+e_{S}\,\bar{\theta}\theta)-(c_{B}+c_{S}\,\bar{\theta}\theta)(d_{B}+d_{S}\,\bar{\theta}\theta)=\pm\epsilon-\epsilon^{2}\,r\,\bar{\theta}\theta$
(125)

and comparing equal powers of $\theta$ and $\bar{\theta}$ we get that Eq.
(125) is equivalent to the following two equations

$\displaystyle\left\\{\begin{matrix}b_{B}\,e_{B}-c_{B}\,d_{B}=\pm\epsilon\\\
b_{S}\,e_{B}+b_{B}\,e_{S}-c_{S}\,d_{B}-c_{B}\,d_{S}=-\epsilon^{2}\,r.\end{matrix}\right.$
(126)

The first equation is a true constraint equation while the second one relates
the parameter $r$ to the variables $b,c,d,e$. From Eq. (117) we can also
obtain the detail expression of $p$ in terms of the entries of the vierbein. A
long calculation leads to the following equation

$\displaystyle
p=\pm\frac{1}{\epsilon}\left(\alpha_{\bar{\theta}}\,\gamma_{\theta}\,e_{B}-\gamma_{\bar{\theta}}\,\alpha_{\theta}\,e_{B}-\alpha_{\bar{\theta}}\,\delta_{\theta}\,c_{B}+\alpha_{\theta}\,\delta_{\bar{\theta}}\,c_{B}\right.$
$\displaystyle\left.+\beta_{\theta}\,\gamma_{\bar{\theta}}\,d_{B}-\beta_{\bar{\theta}}\,\gamma_{\theta}\,d_{B}+\beta_{\bar{\theta}}\,\delta_{\theta}\,b_{B}-\beta_{\theta}\,\delta_{\bar{\theta}}\,b_{B}\right).$
(127)

Inserting Eq. (127) and the second of Eq. (126) into the second of Eq. (119)
we get a constraint among the $a,b,c,d,\alpha,\beta,\gamma,\delta$. This
constraint together with the first of Eq. (126) provides the two QPI
constraints analog to the two of the CPI of Eq. (98) but much more
complicated. In order to simplify things let us choose $\alpha=\beta=0$ like
in the CPI case. This choice, once inserted in Eq. (127), gives $p=0$. Using
this inside Eq. (120) we get

$\frac{a_{S}}{\epsilon}+r=\pm\frac{i}{\epsilon^{2}\,\hbar}$ (128)

and using for $r$ the expression in the second equation of Eq. (126) we get
from Eq. (128) the following constraint

$b_{S}\,e_{B}+b_{B}\,e_{S}-c_{S}\,d_{B}-c_{B}\,d_{S}=\mp\frac{i}{\hbar}+\epsilon\,a_{S}.$
(129)

This together with the first relation in Eq. (126) are the two constraints for
the QPI analog to the two for the CPI in Eq. (98). Let us write together those
of the QPI

$\left\\{\begin{aligned} &b_{B}\,e_{B}-c_{B}\,d_{B}=\pm\epsilon\hskip
85.35826pt\\\
&b_{S}\,e_{B}+b_{B}\,e_{S}-c_{S}\,d_{B}-c_{B}\,d_{S}=\mp\cfrac{i}{\hbar}+\epsilon\,a_{S}.\end{aligned}\right.$
(130)

We can find some solutions of these equations like for example

$\displaystyle(1)$ $\displaystyle\quad e_{B}=0,\quad
c_{B}=\mp\frac{1}{d_{B}},\quad
c_{S}=\frac{\pm\epsilon\,d_{S}+d_{B}\,b_{B}\,a_{S}\pm\frac{i}{\hbar}a_{S}\,d_{B}-\epsilon\,a_{S}\,d_{B}}{d_{B}^{2}}$
(131) $\displaystyle(2)$
$\displaystyle\,b_{B}=\frac{\pm\epsilon+c_{B}\,d_{B}}{e_{B}},\quad
b_{S}=\frac{\mp\epsilon\,e_{S}-c_{B}\,d_{B}\,b_{S}+c_{B}\,d_{S}\,e_{B}}{e_{B}^{2}}$
$\displaystyle+\frac{c_{B}\,d_{B}\,e_{B}\mp\frac{i}{\hbar}e_{B}+\epsilon\,a_{S}\,a_{B}}{e_{B}^{2}}.$

## Appendix D

Here we will give details for the construction of the metric in the QPI case.
Let us start with the most general vierbein:

$E_{A}^{\,\,M}=\begin{pmatrix}a&\alpha&\beta\\\ \gamma&b&c\\\
\delta&d&e\end{pmatrix}$ (132)

and later on we will insert the quantum constraints Eq. (126) in the
associated matrix. Using the relation Eq. (56) between vierbein and metric we
get that the metric associated to the general vierbein Eq. (132) has the form:

$\begin{pmatrix}1-2\gamma\,\delta+2\,\bar{\theta}\theta\,a_{B}a_{S}&d\,\gamma-b\delta+\alpha
a_{B}&e\,\gamma-c\,\delta+\beta\,a_{B}\\\
d\,\gamma-b\,\delta+\alpha\,a_{B}&0&be-cd+\alpha\,\beta\\\
e\,\gamma-c\,\delta+\beta\,a_{B}&cd-be-\alpha\,\beta&0\end{pmatrix}.$ (133)

Let us now rewrite the metric using the $\pi_{i}$ introduced in Eq. (59) and
Eq. (60). From the definition of $\pi_{5}$ in Eq. (60) it is easy to prove
that

$\gamma\,\delta=\pi_{5}\,\bar{\theta}\theta$

so the element $g^{11}$ of Eq. (133) can be written as

$\displaystyle 1-2\gamma\,\delta+2\,\bar{\theta}\theta\,a_{B}a_{S}$
$\displaystyle=$ $\displaystyle 1-2(\pi_{5}-a_{B}a_{S})\bar{\theta}\theta$
(134) $\displaystyle\equiv$ $\displaystyle 1-\pi_{5}^{Q}\,\bar{\theta}\theta$

where we have defined a new quantity $\pi^{Q}_{5}$ as

$\pi^{Q}_{5}\equiv\pi_{5}-a_{B}a_{S}.$ (135)

The index “Q” is to indicate that these are objects related to the QPI. The
element $g^{12}$ of Eq. (133) can be written as

$\displaystyle d\,\gamma-b\,\delta+\alpha\,a_{B}$ $\displaystyle=$
$\displaystyle-\pi_{3}\,\theta-\pi_{4}\,\bar{\theta}+\alpha_{\theta}\,a_{B}\,\theta+\alpha_{\bar{\theta}}\,a_{B}\,\bar{\theta}$
(136) $\displaystyle=$
$\displaystyle-(\pi_{3}-\alpha_{\theta}a_{B})\theta-(\pi_{4}-\alpha_{\bar{\theta}}a_{B})\bar{\theta}$
$\displaystyle\equiv$
$\displaystyle-\pi_{3}^{Q}\,\theta-\pi^{Q}_{4}\,\bar{\theta}$

where

$\displaystyle\pi_{3}^{Q}$ $\displaystyle\equiv$
$\displaystyle\pi_{3}-\alpha_{\theta}a_{B}$ $\displaystyle\pi_{4}^{Q}$
$\displaystyle\equiv$ $\displaystyle\pi_{4}-\alpha_{\bar{\theta}}a_{B}.$ (137)

The element $g^{13}$ of Eq. (133) can be written as

$\displaystyle e\,\gamma-c\,\delta+\beta\,a_{B}$ $\displaystyle=$
$\displaystyle\pi_{1}\,\theta+\pi_{2}\,\bar{\theta}+\beta\,a_{B}$ (138)
$\displaystyle=$
$\displaystyle\pi_{1}\,\theta+\pi_{2}\,\bar{\theta}+\beta_{\theta}\,\theta\,a_{B}+\beta_{\bar{\theta}}\,\bar{\theta}\,a_{B}$
$\displaystyle=$
$\displaystyle(\pi_{1}+\beta_{\theta}a_{B})\theta+(\pi_{2}+\alpha_{\bar{\theta}}a_{B})\bar{\theta}$
$\displaystyle\equiv$
$\displaystyle\pi^{Q}_{1}\,\theta+\pi^{Q}_{2}\,\bar{\theta}$

where

$\displaystyle\pi_{1}^{Q}$ $\displaystyle\equiv$
$\displaystyle\pi_{1}+\beta_{\theta}a_{B}$ $\displaystyle\pi_{2}^{Q}$
$\displaystyle\equiv$ $\displaystyle\pi_{2}+\beta_{\bar{\theta}}a_{B}.$ (139)

Next let now examine the term $g^{23}$ of Eq. (133)

$\displaystyle be-cd+\alpha\,\beta$ $\displaystyle=$
$\displaystyle(b_{B}\,e_{B}-c_{B}\,d_{B})+$ $\displaystyle+$
$\displaystyle(b_{S}\,e_{B}+b_{B}\,e_{S}-c_{S}\,d_{B}-c_{B}\,d_{S}+\alpha_{\bar{\theta}}\,\beta_{\theta}-\beta_{\bar{\theta}}\,\alpha_{\theta})\,\bar{\theta}\theta.$

When $a_{B}\neq\pm 1$ the first of relation (126) and (120) will turn into the
following two relations:

$\left\\{\begin{matrix}b_{B}\,e_{B}-e_{B}\,d_{B}&=&a_{B}\,\epsilon\\\ &&\\\
\cfrac{a_{B}a_{S}}{\epsilon}-\cfrac{a_{B}p}{\epsilon}+a_{B}r&=&\cfrac{i}{\epsilon^{2}\,\hbar}\end{matrix}\right..$
(141)

The first of relations (119) will turn into

$q=\frac{a_{B}}{\epsilon}$

while the relation Eq. (127) for $p$ becomes

$\displaystyle p$ $\displaystyle=$
$\displaystyle\frac{a_{B}}{\epsilon}\left[\alpha_{\bar{\theta}}(\gamma_{\theta}\,e_{B}-\delta_{\theta}\,c_{B})+\alpha_{\theta}(\delta_{\bar{\theta}}\,c_{B}-\gamma_{\bar{\theta}}\,e_{B})+\right.$
(142)
$\displaystyle\left.+\beta_{\theta}(\gamma_{\bar{\theta}}\,d_{B}-\delta_{\bar{\theta}}\,b_{B})+\beta_{\bar{\theta}}(\delta_{\theta}\,b_{B}-\gamma_{\theta}\,d_{B})\right]$
$\displaystyle=$
$\displaystyle\frac{a_{B}}{\epsilon}\left(\alpha_{\bar{\theta}}\,\pi_{1}-\alpha_{\theta}\,\pi_{2}-\beta_{\theta}\,\pi_{4}+\beta_{\bar{\theta}}\,\pi_{3}\right)$

Multiplying the second equation of (141) by $\epsilon^{2}/a_{B}$ we get

$\epsilon\,a_{S}-\epsilon\,p+\epsilon^{2}\,r=\cfrac{i}{\hbar\,a_{B}}$

which, using Eq. (142), implies

$\displaystyle-\epsilon^{2}\,r$ $\displaystyle=$
$\displaystyle\epsilon\,a_{S}-\epsilon\,p-\frac{i}{\hbar\,a_{B}}=$
$\displaystyle=$
$\displaystyle\epsilon\,a_{S}-\frac{i}{\hbar\,a_{B}}-a_{B}(\alpha_{\bar{\theta}}\,\pi_{1}-\alpha_{\theta}\,\pi_{2}-\beta_{\theta}\,\pi_{4}+\beta_{\bar{\theta}}\,\pi_{3}).$

Let us now remember the second relation of Eq. (126)

$b_{S}\,e_{B}+b_{B}\,e_{S}-c_{S}\,d_{B}-c_{B}\,d_{S}=-\epsilon^{2}\,r.$ (144)

Note that the L.H.S. of this equation are exactly the first four terms of the
soul of $b\,e-c\,d+\alpha\,\beta$ in Eq. (D). Replacing them with the
expression of $-\epsilon^{2}\,r$, which appear on the L.H.S. of Eq. (D), we
get that the soul of $b\,e-c\,d+\alpha\,\beta$ is equal to

$\displaystyle\epsilon\,a_{S}$ $\displaystyle-$
$\displaystyle\frac{i}{\hbar\,a_{B}}+a_{B}(\alpha_{{\theta}}\,\pi_{2}-\alpha_{\bar{\theta}}\,\pi_{1}+\beta_{\theta}\,\pi_{4}-\beta_{\bar{\theta}}\,\pi_{3})+$
(145) $\displaystyle+$
$\displaystyle\alpha_{\bar{\theta}}\,\beta_{\theta}-\alpha_{\theta}\,\beta_{\bar{\theta}}\equiv\pi_{6}^{Q}.$

In the equation above the soul of $b\,e-c\,d+\alpha\,\beta$ has been set equal
to $\pi_{6}^{Q}$.

Going now back to Eq. (D) and using for its body the first constraint of Eq.
(141) we get

$b\,e-c\,d+\alpha\,\beta=a_{B}\,\epsilon+\bar{\theta}\theta\,\pi_{6}^{Q}.$
(146)

We have now all elements to write down the metric with all constraints
implemented

$g^{MN}=\begin{pmatrix}1-2\,\pi_{5}^{Q}\,\bar{\theta}\theta&-\pi_{3}^{Q}\,\theta-\pi_{4}^{Q}\,\bar{\theta}&\pi_{1}^{Q}\,\theta+\pi_{2}^{Q}\,\bar{\theta}\\\
-\pi_{3}^{Q}\,\theta-\pi_{4}^{Q}\,\bar{\theta}&0&a_{B}\,\epsilon+\pi^{Q}_{6}\bar{\theta}\theta\\\
\pi_{1}^{Q}\theta+\pi_{2}^{Q}\bar{\theta}&-a_{B}\,\epsilon-\pi_{6}^{Q}\,\bar{\theta}\theta&0\end{pmatrix}..$
(147)

If we define a new variable $\pi_{7}^{Q}$ as

$\pi_{7}^{Q}=a_{B}\,\epsilon,$ (148)

the metric Eq. (147) can be written using only $\pi_{i}^{Q}$ variables as
follows:

$g^{MN}=\begin{pmatrix}1-2\,\pi_{5}^{Q}\,\bar{\theta}\theta&-\pi_{3}^{Q}\,\theta-\pi_{4}^{Q}\,\bar{\theta}&\pi_{1}^{Q}\,\theta+\pi_{2}^{Q}\,\bar{\theta}\\\
-\pi_{3}^{Q}\,\theta-\pi_{4}^{Q}\,\bar{\theta}&0&\pi_{7}^{Q}+\pi^{Q}_{6}\bar{\theta}\theta\\\
\pi_{1}^{Q}\theta+\pi_{2}^{Q}\bar{\theta}&-\pi_{7}^{Q}-\pi_{6}^{Q}\,\bar{\theta}\theta&0\end{pmatrix}.$
(149)

Let us summarize the various quantities we have introduced:

$\left\\{\begin{aligned} \pi_{1}^{Q}&\equiv\pi_{1}+\beta_{\theta}\,a_{B}\\\
\pi_{2}^{Q}&\equiv\pi_{2}+\beta_{\theta}\,a_{B}\\\
\pi_{3}^{Q}&\equiv\pi_{3}-\alpha_{\theta}\,a_{B}\\\
\pi_{4}^{Q}&\equiv\pi_{4}-\alpha_{\bar{\theta}}\,a_{B}\\\
\pi_{5}^{Q}&\equiv\pi_{5}-a_{B}\,a_{S}\\\
\pi_{6}^{Q}&\equiv\epsilon\,a_{S}-\cfrac{i}{\hbar\,a_{B}}+a_{B}(\alpha_{\theta}\,\pi_{2}-\alpha_{\bar{\theta}}\pi_{1}+\beta_{\theta}\pi_{4}-\beta_{\bar{\theta}}\pi_{3})+\alpha_{\bar{\theta}}\beta_{\theta}-\alpha_{\theta}\beta_{\bar{\theta}}\\\
\pi_{7}^{Q}&\equiv a_{B}\epsilon\end{aligned}\right.$ (150)

Let us now count the number of free parameters. $\alpha$ and $\beta$ do not
enter any of the constraints in Eq. (126) so they are free. In order to
simplify things we can put them by hand equal to zero like in the CPI and we
suggested this already after Eq. (127). Moreover we should remember that
$a_{B}=\pm 1$ as proved in Eq. (114) but differently than the classical case
$a_{S}$ is a free parameter. So with this choice Eq. (150), becomes

$\left\\{\begin{aligned} \pi_{1}^{Q}&=\pi_{1}\\\ \pi_{2}^{Q}&=\pi_{2}\\\
\pi_{3}^{Q}&=\pi_{3}\\\ \pi_{4}^{Q}&=\pi_{4}\\\ \pi_{5}^{Q}&=\pi_{5}\mp
a_{S}\\\ \pi_{6}^{Q}&=\epsilon\,a_{S}\mp\cfrac{i}{\hbar}\\\
\pi_{7}^{Q}&=\pm\epsilon\end{aligned}\right.$ (151)

So the “quantum” metric depend on $5$ parameters
$\pi_{1}^{Q},\pi_{2}^{Q},\pi_{3}^{Q},\pi_{4}^{Q}$ and $a_{S}$ while the
classical one only on $4$. The reader may object that also $\alpha$ and
$\beta$ were free and should be counted. He is right. Anyhow as we put them
equal to zero both in the CPI and the QPI and so the difference in the numbers
of free parameters remains one between QM and CM Let us look at the vierbein.
For the CPI we have $10$ free parameters, while in the QPI will be $11$
because we have $a_{S}$ as extra variable. If we had not put $\alpha=\beta=0$
we would have $14$ parameters for the vierbein of the CPI and $15$ for the
QPI. For the metrics instead, as the $\alpha$ and $\beta$ get incorporated
into the $\pi^{Q}_{i}$ (see Eq. (150)) the number of free parameters is $6$
for the QPI, while for the CPI we do not know because we should re-derive the
metric keeping the $\alpha$ and $\beta$ different from zero.To finish this
section let us explore the issue of wether we can recover the classical case
from the “regulated quantum” one without setting $\hbar\to 0$ but manipolating
the $\epsilon$ parameter and the others. For sure we have to require that
$\alpha=\beta=a_{S}=0$ which are the values set previously in the CPI.
Moreover in the CPI we had the constraint

$b\,e-c\,d=\pm 1$ (152)

while in the QPI we had (with $\alpha=\beta=0$) Eq. (146):

$b\,e-c\,d=a_{B}\,\epsilon+\bar{\theta}\theta\,\pi_{6}^{Q}.$ (153)

For Eq. (153) to be equal to Eq. (152), as we know that $a_{B}=\pm 1$, we have
to require that $\epsilon\to 1$ and $\pi^{Q}_{6}=0$. Actually, remembering the
form of $\pi^{Q}_{6}$ present in Eq. (150), we see that the following other
form of $\pi_{6}^{Q}$

$\displaystyle\pi_{6}^{Q}$ $\displaystyle=$
$\displaystyle\epsilon\,a_{S}-\cfrac{i}{\hbar}\,a_{B}\,(1-\epsilon)+a_{B}(\alpha_{\theta}\,\pi_{2}-\alpha_{\bar{\theta}}\,\pi_{1}+\beta_{\theta}\,\pi_{4}+$
(154) $\displaystyle-$
$\displaystyle\beta_{\bar{\theta}}\,\pi_{3})+\alpha_{\bar{\theta}}\,\beta_{\theta}-\alpha_{\theta}\,\beta_{\bar{\theta}}$

has the same quantum limit ($\epsilon\to 0$) as the one conained in Eq. (150).
So we can use Eq. (154) in order to reproduce QM This new $\pi_{6}^{Q}$ has
the feature that it goes to zero for $\epsilon=1$ (of course this has to be
combined with the other things we require for CM: $a_{S}=\alpha=\beta=0$). So
in the limit $\epsilon\to 0$ we would get QM and in the limit $\epsilon\to 1$
we would get CM This is equivalent of having required that the determinant of
the vierbein had the form

$E=\epsilon-i\,(1-\epsilon)\,\frac{\bar{\theta}\theta}{\hbar}.$ (155)

For $\epsilon\to 1$ we would get from Eq. (155)

$E=\mathbb{I}$

which is the $C.P.I$ and for $\epsilon\to 0$ we would get

$E=-i\,\frac{\bar{\theta}\theta}{\hbar},$

which is QM For $\epsilon$ in between 0 and 1 we would get a family of models
which are betwwen CM and QM and could interpolate all the mesoscopic physics.
Before concluding this section let us provide the inverse of $g^{MN}$ of Eq.
(149). This quantity will be useful for the calculations provided in the next
section.

$g_{MN}=\begin{pmatrix}1-\frac{2\,\bar{\theta}\theta(\varphi^{Q}-\pi_{5}^{Q}\pi_{7}^{Q})}{\pi_{7}^{Q}}&-\frac{\theta\,\pi_{1}^{Q}+\bar{\theta}\,\pi_{2}^{Q}}{\pi_{7}^{Q}}&-\frac{\theta\,\pi_{3}^{Q}+\bar{\theta}\,\pi_{4}^{Q}}{\pi_{7}^{Q}}\\\
\frac{\theta\,\pi_{1}^{Q}+\bar{\theta}\,\pi_{2}^{Q}}{\pi_{7}^{Q}}&0&-\frac{\pi_{7}^{Q}+(\varphi^{Q}-\pi_{6}^{Q})}{\pi_{7}^{Q^{2}}}\\\
\frac{\theta\,\pi_{3}^{Q}+\bar{\theta}\,\pi_{4}^{Q}}{\pi_{7}^{Q}}&\frac{\pi_{7}^{Q}+\bar{\theta}\theta(\varphi^{Q}-\pi_{6}^{Q})}{\pi_{7}^{Q^{2}}}&0\end{pmatrix}.$
(156)

In the expression above to be compact we have defined

$\varphi^{Q}\equiv\pi_{2}^{Q}\,\pi_{3}^{Q}-\pi_{1}^{Q}\,\pi_{4}^{Q}.$

Note that in the true quantum limit $\epsilon\to 0$ this metric is singular
because $\pi_{7}^{Q}\to 0$.

## Appendix E

We will now calculate the Christofel symbols for the CPI leaving the body
“$a$” of the vierbein undetermined. The results, obtained assuming the
$\pi_{i}$ independent on $t$ and using Mathematica, are the following ones:

$\displaystyle\Gamma_{t\theta}^{t}$ $\displaystyle=$
$\displaystyle\cfrac{\theta\pi_{1}(\pi_{2}-\pi_{3})+\bar{\theta}(\pi_{2}(\pi_{2}+\pi_{3})-2\pi_{1}\pi_{4})}{2\,a}=-\Gamma_{\theta
t}^{t}$ $\displaystyle\Gamma_{t\theta}^{\theta}$ $\displaystyle=$
$\displaystyle\cfrac{\pi_{2}+\pi_{3}}{2}=\Gamma_{\theta
t}^{\theta}=-\Gamma_{t\bar{\theta}}^{\bar{\theta}}=-\Gamma_{\bar{\theta}t}^{\bar{\theta}}$
$\displaystyle\Gamma_{t\theta}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle-\pi_{1}=\Gamma_{\theta t}^{\bar{\theta}}$
$\displaystyle\Gamma_{t\bar{\theta}}^{t}$ $\displaystyle=$
$\displaystyle\cfrac{\theta(-\pi_{3}(\pi_{2}+\pi_{3})+2\pi_{1}\pi_{4})+\bar{\theta}\,\pi_{4}(\pi_{2}-\pi_{3})}{2\,a}=-\Gamma_{\bar{\theta}t}^{t}$
(157) $\displaystyle\Gamma_{t\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\pi_{4}=\Gamma_{\bar{\theta}t}^{\theta}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{t}$ $\displaystyle=$
$\displaystyle\cfrac{\pi_{2}-\pi_{3}}{2\,a}-2\,\bar{\theta}\theta(\pi_{2}-\pi_{3})(\pi_{2}\pi_{3}-\pi_{1}\pi_{4})=-\Gamma_{\bar{\theta}\theta}^{t}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\cfrac{\theta(\pi_{3}(3\pi_{2}-\pi_{3})-2\pi_{1}\pi_{4})+\bar{\theta}\pi_{4}(\pi_{2}-\pi_{3})}{2\,a}=-\Gamma_{\bar{\theta}\theta}^{\theta}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\cfrac{\theta(\pi_{1}(\pi_{3}-\pi_{2})+\bar{\theta}(\pi_{2}(3\pi_{3}-\pi_{2})-2\pi_{1}\pi_{4}))}{2\,a}=-\Gamma_{\bar{\theta}\theta}^{\bar{\theta}}.$

All the other Christofel symbols are equal to zero. Similarly we can calculate
the Christofel symbols for the QPI using the metric (149) and (156). In order
to simplify the expression for the Christofel symbols and curvatures, we need
to introduce the following quantity:

$\sigma_{1}\equiv\pi_{2}^{Q}\pi_{3}^{Q}-\pi_{1}^{Q}\pi_{4}^{Q}-\pi_{5}^{Q}\pi_{7}^{Q}.$
(158)

Note that in the classical limit $\pi_{i}^{Q}\to\pi_{i},\,i=1,\dots,4$ since
$\alpha_{\theta},\alpha_{\bar{\theta}},\beta_{\theta},\beta_{\bar{\theta}}\to
0$ and $\pi_{7}^{Q}\to a$ being $\epsilon\to 1$. Therefore
$\sigma_{1}\to(\pi_{2}\pi_{3}-\pi_{1}\pi_{4}-\pi_{5}\,a_{B})$, that in the
classical limit is equal to zero. The result, via Mathematica [22], for
parameters $\pi_{i}$ independent on $t$, turns out to be:

$\displaystyle\Gamma_{tt}^{t}$ $\displaystyle=$
$\displaystyle\bar{\theta}\theta\cfrac{(\pi_{3}^{Q}-\pi_{2}^{Q})}{\pi_{7}^{Q}}\,\sigma_{1}$
$\displaystyle\Gamma_{tt}^{\theta}$ $\displaystyle=$
$\displaystyle-\bar{\theta}\,\sigma_{1}$ $\displaystyle\Gamma_{t\theta}^{t}$
$\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{t\;CPI}(\pi_{i}^{Q})+\cfrac{\sigma_{1}}{\pi_{7}^{Q}}=-\Gamma_{\theta
t}^{t}$ $\displaystyle\Gamma_{t\theta}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{\theta\;CPI}(\pi_{i}^{Q})+\bar{\theta}\theta\,\cfrac{\pi_{6}^{Q}(\pi_{2}^{Q}+\pi_{3}^{Q})+2\pi_{3}^{Q}\,\sigma_{1}}{2\,\pi_{7}^{Q}}=\Gamma_{\theta
t}^{\theta}$ $\displaystyle\Gamma_{t\theta}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{\bar{\theta}\;CPI}(\pi_{i}^{Q})-\bar{\theta}\theta\,\cfrac{\pi_{1}^{Q}\,(\sigma_{1}+\pi_{6}^{Q})}{\pi_{7}^{Q}}=-\Gamma_{\theta
t}^{\bar{\theta}}$ $\displaystyle\Gamma_{t\bar{\theta}}^{t}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{t\;CPI}(\pi_{i}^{Q})-\cfrac{\sigma_{1}}{\pi_{7}^{Q}}=-\Gamma_{\bar{\theta}t}^{t}$
$\displaystyle\Gamma_{t\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{\theta\;CPI}(\pi_{i}^{Q})+\bar{\theta}\theta\,\cfrac{\pi_{4}^{Q}\,(\sigma_{1}+\pi_{6}^{Q})}{\pi_{7}^{Q}}=\Gamma_{\bar{\theta}t}^{\theta}$
(159) $\displaystyle\Gamma_{t\bar{\theta}}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{\bar{\theta}\;CPI}(\pi_{i}^{Q})-\bar{\theta}\theta\,\cfrac{\pi_{6}^{Q}\,(\pi_{2}^{Q}+\pi_{3}^{Q})+2\,\pi_{2}^{Q}\,\sigma_{1}}{2\,\pi_{7}^{Q}}=\Gamma_{\bar{\theta}t}^{\bar{\theta}}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{t}$ $\displaystyle=$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{t\;CPI}(\pi_{i}^{Q})-\bar{\theta}\theta\,\cfrac{(\pi_{2}^{Q}-\pi_{3}^{Q})(\sigma_{1}+\pi_{6}^{Q}+2\,\pi_{5}^{Q}\pi_{7}^{Q})}{2\,\pi_{7}^{Q^{2}}}$
$\displaystyle=$ $\displaystyle-\Gamma_{\bar{\theta}\theta}^{t}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\theta\;CPI}(\pi_{i}^{Q})-\theta\,\cfrac{\pi_{6}^{Q}}{\pi_{7}^{Q}}=-\Gamma_{\bar{\theta}\theta}^{\theta}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\bar{\theta}\;CPI}(\pi_{i}^{Q})-\bar{\theta}\,\cfrac{\pi_{6}^{Q}}{\pi_{7}^{Q}}=-\Gamma_{\bar{\theta}\theta}^{\bar{\theta}},$

where

$\Gamma_{AB}^{C\;CPI}(\pi_{i}^{Q})\equiv\Gamma_{AB}^{C\;CPI}(\pi_{i}\to\pi_{i}^{Q},a\to\pi_{7}^{Q}).$

All the other Christofel symbols are equal to zero.

If we choose the $\pi_{i}$ dependent on time for the CPI Christofel symbols we
get the following expressions:

$\displaystyle\Gamma_{tt}^{t}$ $\displaystyle=$
$\displaystyle\bar{\theta}\theta\,\pi_{5}^{\prime}$
$\displaystyle\Gamma_{tt}^{\theta}$ $\displaystyle=$
$\displaystyle\theta\,\pi_{3}^{\prime}+\bar{\theta}\,\pi_{4}^{\prime}$
$\displaystyle\Gamma_{tt}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle-\theta\,\pi_{1}^{\prime}-\bar{\theta}\,\pi_{2}^{\prime}$
$\displaystyle\Gamma_{t\theta}^{t}$ $\displaystyle=$
$\displaystyle-\theta\,\frac{\pi_{1}\,(\pi_{2}-\pi_{3})}{2\,a}+\bar{\theta}\,\frac{(\pi_{2}+\pi_{3})\pi_{2}-2\,\pi_{1}\,\pi_{4}}{2\,a}$
$\displaystyle\Gamma_{t\theta}^{\theta}$ $\displaystyle=$
$\displaystyle\frac{\pi_{2}+\pi_{3}}{2}+\bar{\theta}\theta\,\frac{\pi_{5}^{\prime}}{2}$
$\displaystyle\Gamma_{t\theta}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle-\pi_{1}$ $\displaystyle\Gamma_{t\bar{\theta}}^{t}$
$\displaystyle=$
$\displaystyle\bar{\theta}\,\frac{(\pi_{2}-\pi_{3})\pi_{4}}{2\,a}+\theta\,\frac{2\,\pi_{1}\,\pi_{4}-\pi_{3}\,(\pi_{2}+\pi_{3})}{2\,a}$
$\displaystyle\Gamma_{t\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\pi_{4}$ $\displaystyle\Gamma_{t\bar{\theta}}^{\bar{\theta}}$
$\displaystyle=$
$\displaystyle-\frac{\pi_{7}}{2}+\bar{\theta}\theta\,\frac{\pi_{5}^{\prime}}{2}$
$\displaystyle\Gamma_{\theta t}^{t}$ $\displaystyle=$
$\displaystyle-\Gamma_{t\theta}^{t}$ $\displaystyle\Gamma_{\theta t}^{\theta}$
$\displaystyle=$ $\displaystyle\Gamma_{t\theta}^{\theta}$
$\displaystyle\Gamma_{\theta t}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{\bar{\theta}}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{t}$ $\displaystyle=$
$\displaystyle\frac{\pi_{2}-\pi_{3}}{2\,a}-\bar{\theta}\theta\,\frac{4\,\pi_{2}\,\pi_{5}-\pi_{5}^{\prime}}{2\,a}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\bar{\theta}\,\frac{(\pi_{2}-\pi_{3})\,\pi_{4}}{2\,a}+\theta\,\frac{(\pi_{2}-\pi_{3})\,\pi_{3}+2\,a\,\pi_{5}}{2\,a}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle-\bar{\theta}\,\frac{(\pi_{2}-\pi_{3})\,\pi_{1}}{2\,a}+\bar{\theta}\,\frac{2\,a\,\pi_{5}-(\pi_{2}-\pi_{3})\,\pi_{2}}{2\,a}$
$\displaystyle\Gamma_{\bar{\theta}t}^{t}$ $\displaystyle=$
$\displaystyle-\Gamma_{t\bar{\theta}}^{t}$
$\displaystyle\Gamma_{\bar{\theta}t}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{\theta}$
$\displaystyle\Gamma_{\bar{\theta}t}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle-\frac{\pi_{2}+\pi_{3}}{2}+\bar{\theta}\theta\,\frac{\pi_{2}^{\prime}+\pi_{3}^{\prime}}{2}$
$\displaystyle\Gamma_{\bar{\theta}\theta}^{t}$ $\displaystyle=$
$\displaystyle-\Gamma_{\theta\bar{\theta}}^{t}$
$\displaystyle\Gamma_{\bar{\theta}\theta}^{\theta}$ $\displaystyle=$
$\displaystyle-\Gamma_{\theta\bar{\theta}}^{\theta}.$
$\displaystyle\Gamma_{t\theta}^{\theta}$ $\displaystyle=$
$\displaystyle-\frac{\pi_{2}+\pi_{3}}{2}+\bar{\theta}\theta\,\frac{\pi_{5}^{\prime}}{2}$

For the QPI case, when the coefficient $\pi_{i}$ depend on time the Christofel
symbols turn out to be:

$\displaystyle\Gamma_{tt}^{t}$ $\displaystyle=$
$\displaystyle\Gamma_{tt}^{t\;CPI}-\bar{\theta}\theta\,\frac{(\pi_{2}-\pi_{3})\,\sigma_{1}}{\pi_{7}}$
$\displaystyle\Gamma_{tt}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{tt}^{\theta\;CPI}-\theta\,\sigma_{1}$
$\displaystyle\Gamma_{tt}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{tt}^{\bar{\theta}\;CPI}-\bar{\theta}\,\sigma_{1}$
$\displaystyle\Gamma_{t\theta}^{t}$ $\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{t\;CPI}+\bar{\theta}\,\frac{\sigma_{1}}{\pi_{7}}$
$\displaystyle\Gamma_{t\theta}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{\theta\;CPI}+\bar{\theta}\theta\,\frac{\sigma_{1}^{\prime}+\pi_{6}\,(\pi_{2}+\pi_{3})+2\,\pi_{3}\,\sigma_{1}-\pi_{6}^{\prime}}{2\,\pi_{7}}$
$\displaystyle\Gamma_{t\theta}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{\bar{\theta}\;CPI}-\bar{\theta}\theta\,\frac{\pi_{1}\,(\sigma_{1}+\pi_{6})}{\pi_{7}}$
$\displaystyle\Gamma_{t\bar{\theta}}^{t}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{t\;CPI}-\theta\,\frac{\sigma_{1}}{\pi_{7}}$
$\displaystyle\Gamma_{t\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{\theta\;CPI}+\bar{\theta}\theta\,\frac{\pi_{4}\,(\sigma_{1}+\pi_{6})}{\pi_{7}}$
$\displaystyle\Gamma_{t\bar{\theta}}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{\bar{\theta}\;CPI}+\bar{\theta}\theta\,\frac{\sigma_{1}^{\prime}-\pi_{6}\,(\pi_{2}+\pi_{3})-2\,\pi_{2}\,\sigma_{1}-\pi_{6}^{\prime}}{\pi_{7}}$
$\displaystyle\Gamma_{\theta t}^{t}$ $\displaystyle=$
$\displaystyle-\Gamma_{t\theta}^{t}$ $\displaystyle\Gamma_{\theta t}^{\theta}$
$\displaystyle=$ $\displaystyle\Gamma_{t\theta}^{\theta}$
$\displaystyle\Gamma_{\theta t}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{t\theta}^{\bar{\theta}}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{t}$ $\displaystyle=$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{t\;CPI}-\bar{\theta}\theta\,\frac{\pi_{6}^{\prime}-\sigma_{1}^{\prime}+2\,(\pi_{2}-\pi_{3})(\sigma_{1}-\pi_{6})}{2\,\pi_{7}^{2}}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\theta\;CPI}+\theta\,\frac{\sigma_{1}-\pi_{6}}{\pi_{7}}$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{\theta\bar{\theta}}^{\bar{\theta}\;CPI}+\bar{\theta}\,\frac{\sigma_{1}-\pi_{6}}{\pi_{7}}$
$\displaystyle\Gamma_{\bar{\theta}t}^{t}$ $\displaystyle=$
$\displaystyle-\Gamma_{t\bar{\theta}}^{t}$
$\displaystyle\Gamma_{\bar{\theta}t}^{\theta}$ $\displaystyle=$
$\displaystyle\Gamma_{t\bar{\theta}}^{\theta}$
$\displaystyle\Gamma_{\bar{\theta}t}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle\Gamma_{\bar{\theta}t}^{\bar{\theta}}$
$\displaystyle\Gamma_{\bar{\theta}\theta}^{t}$ $\displaystyle=$
$\displaystyle-\Gamma_{\theta\bar{\theta}}^{t}$
$\displaystyle\Gamma_{\bar{\theta}\theta}^{\theta}$ $\displaystyle=$
$\displaystyle-\Gamma_{\theta\bar{\theta}}^{\theta}$
$\displaystyle\Gamma_{\bar{\theta}\theta}^{\bar{\theta}}$ $\displaystyle=$
$\displaystyle-\Gamma_{\theta\bar{\theta}}^{\bar{\theta}}.$

The other symbols are zero.

## Appendix F

###  F.1 Time Independent CPI Ricci tensor and scalar

In the case of $\pi_{i}$ independent of $t$ the CPI Ricci curvature tensor
turns out to be:

$\displaystyle R_{\theta\theta}$ $\displaystyle=$ $\displaystyle
R_{\bar{\theta}\bar{\theta}}=0$ (161) $\displaystyle R_{tt}$ $\displaystyle=$
$\displaystyle\cfrac{(\pi_{2}+\pi_{3})^{2}-4\,\pi_{1}\pi_{4}}{2}$
$\displaystyle R_{t\theta}$ $\displaystyle=$
$\displaystyle-\cfrac{(\theta\,\pi_{1}+\bar{\theta}\,\pi_{2})((\pi_{2}+\pi_{3})^{2}-4\,\pi_{1}\pi_{4})}{2\,a}=-R_{\theta
t}$ $\displaystyle R_{t\bar{\theta}}$ $\displaystyle=$
$\displaystyle-\cfrac{(\theta\,\pi_{3}+\bar{\theta}\,\pi_{4})((\pi_{2}+\pi_{3})^{2}-4\,\pi_{1}\pi_{4})}{2\,a}=-R_{\bar{\theta}t}$
$\displaystyle R_{\theta\bar{\theta}}$ $\displaystyle=$
$\displaystyle\cfrac{\bar{\theta}\theta}{2}(\pi_{1}\pi_{4}-\pi_{2}\pi_{3})(\pi_{2}^{2}-6\,\pi_{2}\pi_{3}+\pi_{3}^{2}+4\,\pi_{1}\,\pi_{4})$
$\displaystyle-$
$\displaystyle\cfrac{a}{2}\,(\pi_{2}^{2}-10\,\pi_{2}\pi_{3}+\pi_{3}^{2}+8\,\pi_{1}\,\pi_{4})=-R_{\bar{\theta}\theta}$

From the components we can build the Ricci scalar

$R^{CPI}=-\cfrac{1}{2}\,(\pi_{2}^{2}-22\,\pi_{2}\pi_{3}+\pi_{3}^{2}+20\,\pi_{1}\,\pi_{4})+8\,\cfrac{\bar{\theta}\theta}{a}\,(\pi_{2}\pi_{3}-\pi_{1}\pi_{4})^{2}$
(162)

###  F.2 Time Independent QPI Ricci tensor and scalar

For the QPI, the components of the Ricci tensor can be written as:

$\displaystyle R_{\theta\theta}$ $\displaystyle=$ $\displaystyle
R_{\bar{\theta}\bar{\theta}}=0$ (163) $\displaystyle R_{tt}$ $\displaystyle=$
$\displaystyle
R_{tt}^{CPI}(\pi_{i}^{Q})+2\sigma_{1}+\cfrac{\bar{\theta}\theta}{\pi_{7}^{Q}}\left[\pi_{6}^{Q}((\pi_{2}^{Q}+\pi_{3}^{Q})^{2}-4\,\pi_{1}^{Q}\pi_{4}^{Q})+\right.$
$\displaystyle+$
$\displaystyle\left.2\,\sigma_{1}\,(\pi_{2}^{Q^{2}}+4\,\pi_{2}^{Q}\pi_{3}^{Q}+\pi_{3}^{Q^{2}}-6\,\pi_{1}^{Q}\pi_{4}^{Q}-\pi_{6}^{Q})+6\,\sigma_{1}^{2}\right]$
$\displaystyle R_{t\theta}$ $\displaystyle=$ $\displaystyle
R_{t\theta}^{CPI}(\pi_{i}^{Q})-\cfrac{\pi_{6}^{Q}+3\,\sigma_{1}}{\pi_{7}^{Q}}\left[\theta\pi_{1}^{Q}+\bar{\theta}(\pi_{2}^{Q}+\pi_{3}^{Q})\right]=-R_{\theta
t}$ $\displaystyle R_{t\bar{\theta}}$ $\displaystyle=$ $\displaystyle
R_{t\bar{\theta}}^{CPI}(\pi_{i}^{Q})-\cfrac{\pi_{6}^{Q}+3\,\sigma_{1}}{\pi_{7}^{Q}}\left[\bar{\theta}\pi_{4}^{Q}+\theta(\pi_{2}^{Q}+\pi_{3}^{Q})\right]=-R_{\bar{\theta}t}$
$\displaystyle R_{\theta\bar{\theta}}$ $\displaystyle=$ $\displaystyle
R_{\theta\bar{\theta}}^{CPI}(\pi_{i}^{Q})-\cfrac{\sigma_{1}-3\,\pi_{6}^{Q}}{\pi_{7}^{Q}}+\cfrac{\bar{\theta}\theta}{2\,\pi_{7}^{Q^{2}}}\left[2\,\pi_{6}^{Q^{2}}+8\,\pi_{1}^{Q}\pi_{4}^{Q}\pi_{6}^{Q}\right.$
$\displaystyle-$
$\displaystyle\left.8\,\pi_{2}^{Q}\pi_{3}^{Q}\pi_{6}^{Q}-\sigma_{1}((\pi_{2}^{Q}-\pi_{3}^{Q})^{2}+4\,\pi_{6}^{Q}+2\,\sigma_{1})\right]=-R_{\bar{\theta}\theta},$

where

$R_{AB}^{CPI}(\pi_{i}^{Q})\equiv
R_{AB}^{CPI}(\pi_{i}\to\pi_{i}^{Q},a\to\pi_{7}^{Q}).$

The Ricci scalar turns out to be

$\displaystyle R^{QPI}$ $\displaystyle=$ $\displaystyle
R^{CPI}(\pi_{i}^{Q})+2\,\sigma_{1}-3\,\pi_{6}^{Q}+$ $\displaystyle+$
$\displaystyle\cfrac{\bar{\theta}\theta}{\pi_{7}^{Q}}\left[-\pi_{6}^{Q}(\pi_{2}^{Q^{2}}+6\,\pi_{2}^{Q}\pi_{3}^{Q}+\pi_{3}^{Q^{2}}-8\,\pi_{1}^{Q}\pi_{4}^{Q})+4\,\pi_{6}^{Q^{2}}\right.$
$\displaystyle-$
$\displaystyle\left.4\,\sigma_{1}(\pi_{2}^{Q^{2}}+3\,\pi_{2}^{Q}\pi_{3}^{Q}+\pi_{3}^{Q^{2}}-5\,\pi_{1}^{Q}\pi_{4}^{Q}\\-\pi_{6}^{Q}+\sigma_{1})\right]$

where

$R^{CPI}(\pi_{i}^{Q})=R^{CPI}(\pi_{i}\to\pi_{i}^{Q},a\to\pi_{7}^{Q}).$

## Appendix G

###  G.1 Time Dependent CPI Ricci tensor and scalar

In the case of $\pi_{i}$ dependent on $t$ we get for the $CPI$ the expressions
below where $\pi_{i}^{\prime}$ and $\pi_{i}^{\prime\prime}$ indicate the first
and second derivative of $\pi_{i}$ with respect to $t$.

$\displaystyle R_{\theta\theta}$
$\displaystyle=R_{\bar{\theta}\bar{\theta}}=0$ (165) $\displaystyle R_{tt}$
$\displaystyle=R_{tt}^{CPI}(\pi_{i})+(\pi_{3}^{\prime}-\pi_{2}^{\prime})+\bar{\theta}\theta\left[\pi_{3}^{2}\pi_{2}^{\prime}-\pi_{2}^{2}\pi_{3}^{\prime}+\right.$
$\displaystyle+\left.(\pi_{3}^{\prime}-\pi_{2}^{\prime})(a\,\pi_{5}-2\,\pi_{2}\pi_{3}+3\,\pi_{1}\pi_{4})+2\,a\pi_{5}^{\prime}+\right.$
$\displaystyle+\left.(\pi_{2}-\pi_{3})(\pi_{4}\pi_{1}^{\prime}+\pi_{4}^{\prime}\pi_{1})\right]$
$\displaystyle R_{t\theta}$
$\displaystyle=R_{t\theta}^{CPI}(\pi_{i})+\cfrac{\theta\pi_{1}+\bar{\theta}\pi_{2}}{2\,a}(\pi_{3}^{\prime}-\pi_{2}^{\prime})+\cfrac{\bar{\theta}\pi_{5}^{\prime}}{2}$
$\displaystyle R_{t\bar{\theta}}$
$\displaystyle=R_{t\bar{\theta}}^{CPI}(\pi_{i})+\cfrac{\theta\pi_{3}+\bar{\theta}\pi_{4}}{2\,a}(\pi_{2}^{\prime}-\pi_{3}^{\prime})-\cfrac{\theta\pi_{5}^{\prime}}{2}$
$\displaystyle R_{\theta t}$ $\displaystyle=R_{\theta
t}^{CPI}(\pi_{i})-\cfrac{\theta\pi_{3}+\bar{\theta}\pi_{4}}{2\,a}(\pi_{2}^{\prime}-\pi_{3}^{\prime})-\cfrac{3\,\theta\pi_{5}^{\prime}}{2}$
$\displaystyle R_{\theta\bar{\theta}}$
$\displaystyle=R_{\theta\bar{\theta}}^{CPI}(\pi_{i})+\cfrac{a\,(\pi_{2}^{\prime}-\pi_{3}^{\prime})}{2}+\cfrac{\bar{\theta}\theta}{2}\left[5\,a\,\pi_{5}^{\prime}(\pi_{3}-\pi_{2})+\right.$
$\displaystyle+\left.4\,a\,\pi_{5}(\pi_{3}^{\prime}-\pi_{2}^{\prime})+a\pi_{5}^{\prime\prime}\right]$
$\displaystyle R_{\bar{\theta}t}$
$\displaystyle=R_{\bar{\theta}t}^{CPI}(\pi_{i})-\cfrac{\theta\pi_{3}+\bar{\theta}\pi_{4}}{2\,a}(\pi_{2}^{\prime}-\pi_{3}^{\prime})-\cfrac{3\,\theta\pi_{5}^{\prime}}{2}$
$\displaystyle R_{\bar{\theta}\theta}$
$\displaystyle=R_{\bar{\theta}\theta}^{CPI}(\pi_{i})-\cfrac{a\,(\pi_{2}^{\prime}-\pi_{3}^{\prime})}{2}-\cfrac{\bar{\theta}\theta}{2}\left[5\,a\,\pi_{5}^{\prime}(\pi_{3}-\pi_{2})+\right.$
$\displaystyle+\left.4\,a\,\pi_{5}(\pi_{3}^{\prime}-\pi_{2}^{\prime})+a\pi_{5}^{\prime\prime}\right].$

$R_{AB}^{CPI}(\pi_{i})$ are the expressions previously presented in C for the
time independent CPI Ricci tensor. The other tensor components are all equal
to zero. The CPI Ricci scalar, when the parameters are explicit functions of
time, turns out to be:

$\displaystyle R^{CPI}$ $\displaystyle=$ $\displaystyle
R^{CPI}(\pi_{i})+2(\pi_{2}^{\prime}-\pi_{3}^{\prime})+$ $\displaystyle+$
$\displaystyle\bar{\theta}\theta\left[4\,\pi_{5}\,(\pi_{3}^{\prime}-\pi_{2}^{\prime})+7\,(\pi_{3}-\pi_{2})\,\pi_{5}^{\prime}+2\,\pi_{5}^{\prime\prime}\right],$

where, once again, $R^{CPI}(\pi_{i})$ is the CPI Ricci scalar given in C.

###  G.2 Time Dependent QPI Ricci Tensor and Curvature

In the case of $\pi_{i}$ dependent on $t$ we get for the $QPI$ Ricci tensor:

$\displaystyle R_{tt}$ $\displaystyle=$ $\displaystyle
R_{tt}^{QPI}(\pi_{i})+\pi_{2}^{\prime}(t)-\pi_{3}^{\prime}(t)+$
$\displaystyle-$
$\displaystyle\cfrac{\bar{\theta}\theta}{\pi_{7}}\left[-\pi_{2}(t)\left(2\pi_{4}(t)\pi_{1}^{\prime}(t)+4\pi_{3}(t)\left(\pi_{2}^{\prime}(t)-\pi_{3}^{\prime}(t)\right)+2\pi_{1}(t)\pi_{4}^{\prime}(t)+\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.\pi_{7}\pi_{5}^{\prime}(t)+\pi_{3}^{\prime\prime}(t)\right)+\pi_{3}(t)\left(2\pi_{4}(t)\pi_{1}^{\prime}(t)+2\pi_{1}(t)\pi_{4}^{\prime}(t)+\pi_{7}\pi_{5}^{\prime}(t)+\right.\right.$
$\displaystyle-$
$\displaystyle\left.\left.\left.\pi_{2}^{\prime\prime}(t)\right)+2\pi_{2}(t){}^{2}\pi_{3}^{\prime}(t)-2\pi_{3}(t){}^{2}\pi_{2}^{\prime}(t)+6\pi_{1}(t)\pi_{4}(t)\pi_{2}^{\prime}(t)+\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.2\pi_{7}\pi_{5}(t)\pi_{2}^{\prime}(t)+\pi_{6}(t)\pi_{2}^{\prime}(t)-6\pi_{1}(t)\pi_{4}(t)\pi_{3}^{\prime}(t)-2\pi_{7}\pi_{5}(t)\pi_{3}^{\prime}(t)+\right.\right.$
$\displaystyle-$
$\displaystyle\left.\left.\pi_{6}(t)\pi_{3}^{\prime}(t)-2\pi_{2}^{\prime}(t)\pi_{3}^{\prime}(t)+2\pi_{1}^{\prime}(t)\pi_{4}^{\prime}(t)+\pi_{4}(t)\pi_{1}^{\prime\prime}(t)+\pi_{1}(t)\pi_{4}^{\prime\prime}(t)+\right.\right.$
$\displaystyle+$ $\displaystyle\left.\pi_{6}^{\prime\prime}(t)\right]$
$\displaystyle R_{t\theta}$ $\displaystyle=$ $\displaystyle
R_{t\theta}^{QPI}(\pi_{i})+\cfrac{\theta}{2\pi_{7}}\,\left(\pi_{1}(t)\pi_{2}^{\prime}(t)-\pi_{1}(t)\pi_{3}^{\prime}(t)\right)+\frac{\bar{\theta}}{2\pi_{7}}\,\left[-\pi_{4}(t)\pi_{1}^{\prime}(t)+\right.$
$\displaystyle+$
$\displaystyle\left.\left(\pi_{2}(t)+\pi_{3}(t)\right)\pi_{2}^{\prime}(t)-\pi_{1}(t)\pi_{4}^{\prime}(t)+3\pi_{6}^{\prime}(t)\right]$
$\displaystyle R_{t\bar{\theta}}$ $\displaystyle=$ $\displaystyle
R_{t\bar{\theta}}^{QPI}(\pi_{i})+\frac{\theta}{2\pi_{7}}\left[\pi_{4}(t)\pi_{1}^{\prime}(t)-\left(\pi_{2}(t)+\pi_{3}(t)\right)\pi_{3}^{\prime}(t)+\pi_{1}(t)\pi_{4}^{\prime}(t)+\right.$
$\displaystyle+$
$\displaystyle\left.-3\pi_{6}^{\prime}(t)\right]+\frac{\bar{\theta}}{2\,\pi_{7}}\,\left(\pi_{4}(t)\left(\pi_{2}^{\prime}(t)-\pi_{3}^{\prime}(t)\right)\right)$
$\displaystyle R_{\theta t}$ $\displaystyle=$ $\displaystyle R_{\theta
t}^{QPI}(\pi_{i})+\frac{\theta}{2\pi_{7}}\,\pi_{1}(t)\left(\pi_{3}^{\prime}(t)-\pi_{2}^{\prime}(t)\right)+\frac{\bar{\theta}}{2\pi_{7}}\left[-3\pi_{4}(t)\pi_{1}^{\prime}(t)+\right.$
$\displaystyle+$
$\displaystyle\left.-\left(\pi_{2}(t)-3\pi_{3}(t)\right)\pi_{2}^{\prime}(t)+4\pi_{2}(t)\pi_{3}^{\prime}(t)-3\left(\pi_{1}(t)\pi_{4}^{\prime}(t)+\pi_{6}^{\prime}(t)\right)\right]$
$\displaystyle R_{\theta\bar{\theta}}$ $\displaystyle=$ $\displaystyle
R_{\theta\bar{\theta}}^{QPI}(\pi_{i})+\frac{\pi_{2}^{\prime}(t)-\pi_{3}^{\prime}(t)}{2\pi_{7}}-\frac{\bar{\theta}\theta}{2\,\pi_{7}^{2}}\left[-\pi_{2}(t)\left(2\pi_{4}(t)\pi_{1}^{\prime}(t)+\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.4\pi_{3}(t)\left(\pi_{3}^{\prime}(t)-\pi_{2}^{\prime}(t)\right)+2\pi_{1}(t)\pi_{4}^{\prime}(t)-3\pi_{7}\pi_{5}^{\prime}(t)+2\pi_{6}^{\prime}(t)+\pi_{3}^{\prime\prime}(t)\right)+\right.$
$\displaystyle+$
$\displaystyle\left.\pi_{3}(t)\left(2\pi_{4}(t)\pi_{1}^{\prime}(t)+2\pi_{1}(t)\pi_{4}^{\prime}(t)-3\pi_{7}\pi_{5}^{\prime}(t)+2\pi_{6}^{\prime}(t)-\pi_{2}^{\prime\prime}(t)\right)+\right.$
$\displaystyle+$
$\displaystyle\left.2\pi_{2}(t){}^{2}\pi_{3}^{\prime}(t)-2\pi_{3}(t){}^{2}\pi_{2}^{\prime}(t)-2\pi_{1}(t)\pi_{4}(t)\pi_{2}^{\prime}(t)+2\pi_{7}\pi_{5}(t)\pi_{2}^{\prime}(t)+\right.$
$\displaystyle-$
$\displaystyle\left.2\pi_{6}(t)\pi_{2}^{\prime}(t)+2\pi_{1}(t)\pi_{4}(t)\pi_{3}^{\prime}(t)-2\pi_{7}\pi_{5}(t)\pi_{3}^{\prime}(t)+2\pi_{6}(t)\pi_{3}^{\prime}(t)+\right.$
$\displaystyle-$
$\displaystyle\left.2\pi_{2}^{\prime}(t)\pi_{3}^{\prime}(t)+2\pi_{1}^{\prime}(t)\pi_{4}^{\prime}(t)+\pi_{4}(t)\pi_{1}^{\prime\prime}(t)+\pi_{1}(t)\pi_{4}^{\prime\prime}(t)+\pi_{6}^{\prime\prime}(t)\right]$
$\displaystyle R_{\bar{\theta}t}$ $\displaystyle=$ $\displaystyle
R_{\bar{\theta}t}^{QPI}(\pi_{i})+\frac{\theta}{2\,\pi_{7}}\left[3\pi_{4}(t)\pi_{1}^{\prime}(t)+\pi_{3}(t)\left(\pi_{3}^{\prime}(t)-4\pi_{2}^{\prime}(t)\right)+3\left(-\pi_{2}(t)\pi_{3}^{\prime}(t)\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.+\pi_{1}(t)\pi_{4}^{\prime}(t)+\pi_{6}^{\prime}(t)\right)\right]+\frac{\bar{\theta}}{2\pi_{7}}\pi_{4}(t)\left[\pi_{3}^{\prime}(t)-\pi_{2}^{\prime}(t)\right]$
$\displaystyle R_{\bar{\theta}\theta}$ $\displaystyle=$ $\displaystyle
R_{\bar{\theta}\theta}^{QPI}(\pi_{i})+\frac{\pi_{3}^{\prime}(t)-\pi_{2}^{\prime}(t)}{2\pi_{7}}-\frac{\bar{\theta}\theta}{\pi_{7}^{2}}\left[\pi_{2}(t)\left(2\pi_{4}(t)\pi_{1}^{\prime}(t)+4\pi_{3}(t)\left(\pi_{3}^{\prime}(t)-\pi_{2}^{\prime}(t)\right)\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.2\pi_{1}(t)\pi_{4}^{\prime}(t)-3\pi_{7}\pi_{5}^{\prime}(t)+2\pi_{6}^{\prime}(t)+\pi_{3}^{\prime\prime}(t)\right)-\pi_{3}(t)\left(2\pi_{4}(t)\pi_{1}^{\prime}(t)+\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.2\pi_{1}(t)\pi_{4}^{\prime}(t)-3\pi_{7}\pi_{5}^{\prime}(t)+2\pi_{6}^{\prime}(t)-\pi_{2}^{\prime\prime}(t)\right)-2\pi_{2}(t){}^{2}\pi_{3}^{\prime}(t)2\pi_{3}(t){}^{2}\pi_{2}^{\prime}(t)\right.+$
$\displaystyle+$
$\displaystyle\left.\left.2\pi_{1}(t)\pi_{4}(t)\pi_{2}^{\prime}(t)-2\pi_{7}\pi_{5}(t)\pi_{2}^{\prime}(t)+2\pi_{6}(t)\pi_{2}^{\prime}(t)-2\pi_{1}(t)\pi_{4}(t)\pi_{3}^{\prime}(t)+\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.2\pi_{7}\pi_{5}(t)\pi_{3}^{\prime}(t)-2\pi_{6}(t)\pi_{3}^{\prime}(t)+2\pi_{2}^{\prime}(t)\pi_{3}^{\prime}(t)-2\pi_{1}^{\prime}(t)\pi_{4}^{\prime}(t)-\pi_{4}(t)\pi_{1}^{\prime\prime}(t)+\right.\right.$
$\displaystyle-$
$\displaystyle\left.\left.\pi_{1}(t)\pi_{4}^{\prime\prime}(t)-\pi_{6}^{\prime\prime}(t)\right]\right.$

where $R_{AB}^{CPI}(\pi_{i})$ are the expression previously presented in C for
the QPI Ricci tensor. The other tensor components are all equal to zero. The
QPI Ricci scalar, when the parameters are explicit functions of time, turns
out:

$\displaystyle R^{QPI}$ $\displaystyle=$ $\displaystyle
R^{QPI}(\pi_{i})+2(\pi_{2}^{\prime}-\pi_{3}^{\prime})+$ $\displaystyle+$
$\displaystyle\cfrac{\bar{\theta}\theta}{\pi_{7}}\left[-\pi_{2}(t)\left(5\pi_{4}(t)\pi_{1}^{\prime}(t)+3\pi_{3}(t)\left(\pi_{3}^{\prime}(t)-\pi_{2}^{\prime}(t)\right)+5\pi_{1}(t)\pi_{4}^{\prime}(t)\right.\right.$
$\displaystyle-$
$\displaystyle\left.\left.2\pi_{7}\pi_{5}^{\prime}(t)+5\pi_{6}^{\prime}(t)+2\pi_{3}^{\prime\prime}(t)\right)+\pi_{3}(t)\left(5\pi_{4}(t)\pi_{1}^{\prime}(t)+5\pi_{1}(t)\pi_{4}^{\prime}(t)\right.\right.$
$\displaystyle-$
$\displaystyle\left.\left.2\pi_{7}\pi_{5}^{\prime}(t)+5\pi_{6}^{\prime}(t)-2\pi_{2}^{\prime\prime}(t)\right)+2\left(\pi_{1}(t)\left(\pi_{4}(t)\left(\pi_{2}^{\prime}(t)-\pi_{3}^{\prime}(t)\right)+\pi_{4}^{\prime\prime}(t)\right)\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.\left(3\pi_{7}\pi_{5}(t)-\pi_{6}(t)\right)\pi_{2}^{\prime}(t)+\left(-2\pi_{2}^{\prime}(t)-3\pi_{7}\pi_{5}(t)+\pi_{6}(t)\right)\pi_{3}^{\prime}(t)\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.2\pi_{1}^{\prime}(t)\pi_{4}^{\prime}(t)+\pi_{4}(t)\pi_{1}^{\prime\prime}(t)+\pi_{6}^{\prime\prime}(t)\right)+5\pi_{2}(t){}^{2}\pi_{3}^{\prime}(t)-5\pi_{3}(t){}^{2}\pi_{2}^{\prime}(t)\right]$

where, once again, $R^{QPI}(\pi_{i})$ is the QPI Ricci scalar given in C.

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