d^{5}\\!\mu(Zx)\,|\mu|^{2}\tilde{B}_{T}(Zx)=\int\mathcal{D}\mu_{M}\\!\left[dw_{[0,T)}\right]\,|\mu[dw_{[0,T)}]|^{2}\,,$
(316) $\displaystyle
q_{T}=\langle\nu_{T}^{*}\mu_{T}\rangle_{M}=\langle\mu_{T}^{*}\nu_{T}\rangle$
$\displaystyle=\int
d^{5}\\!\mu(Zx)\,\nu^{*}\mu\,\tilde{B}_{T}(Zx)=\int\mathcal{D}\mu_{M}\\!\left[dw_{[0,T)}\right]\,\nu[dw_{[0,T)}]^{*}\mu[dw_{[0,T)}]\,.$
(317)

In this context, that the first moments and all the other second moments of
the phase-space variables are zero follows from the fact that the measure is
invariant under simultaneous rephasing of all the outcome increments; the
reality of $q_{T}$ follows from the fact that the measure is unchanged under
the transformation $dw_{[0,T)}\rightarrow dw^{*}_{[0,T)}$. These properties
come from the fact that $M_{T}$ is real and symmetric.

Plugging in the stochastic-integral solutions for $\nu_{T}$ and $\mu_{T}$ puts
these moments into the following form:

$\displaystyle n_{T}=\langle|\nu_{T}|^{2}\rangle_{M}$
$\displaystyle=\kappa\sum_{k,l=0}^{N-1}\langle
dw_{k}^{*}\,dw_{l}\rangle_{M}\,e^{-2\kappa(T-t_{k})}e^{-2\kappa(T-t_{l})}$
(318)
$\displaystyle=\kappa\,dt\,\sum_{k,l=0}^{N-1}e^{-2\kappa\,dt\,(N-k)}(M_{T}^{\,\text{-}1})_{kl}e^{-2\kappa\,dt\,(N-l)}\,,$
(319) $\displaystyle m_{T}=\langle|\mu_{T}|^{2}\rangle_{M}$
$\displaystyle=\kappa\sum_{k,l=0}^{N-1}\langle
dw_{k}^{*}\,dw_{l}\rangle_{M}e^{-2\kappa t_{k}}e^{-2\kappa t_{l}}$ (320)
$\displaystyle=\kappa\,dt\,\sum_{k,l=0}^{N-1}e^{-2\kappa\,dt\,k}(M_{T}^{\,\text{-}1})_{kl}e^{-2\kappa\,dt\,l}\,,$
(321) $\displaystyle
q_{T}=\langle\nu_{T}^{*}\mu_{T}\rangle_{M}=\langle\mu_{T}^{*}\nu_{T}\rangle_{M}$
$\displaystyle=\kappa\sum_{k,l=0}^{N-1}\langle
dw_{k}^{*}\,dw_{l}\rangle_{M}e^{-2\kappa(T-t_{k})}e^{-2\kappa t_{l}}$ (322)
$\displaystyle=\kappa\,dt\,\sum_{k,l=0}^{N-1}e^{-2\kappa\,dt\,(N-k)}(M_{T}^{\,\text{-}1})_{kl}e^{-2\kappa\,dt\,l}\,.$
(323)

In the final form of $q_{T}$, it is evident that $q_{T}$ is real. Notice that
these expressions satisfy the zero initial conditions.

That $M_{T}$ is a Toeplitz matrix introduces an additional, quite important
symmetry. The inverse matrix does not inherit the Toeplitz property of $M$,
but it does inherit a less restrictive property. That $M_{T}$ is Toeplitz
implies that it is symmetric about the anti-diagonal, that is,
$M_{kl}=M_{N-1-l,N-1-k}$. A matrix that is symmetric about the anti-diagonal
is called _persymmetric_. It is easy to show that the inverse of a
persymmetric matrix is persymmetric, so $M_{T}^{\,\text{-}1}$ satisfies
$(M_{T}^{\,\text{-}1})_{kl}=(M_{T}^{\,\text{-}1})_{N-1-l,N-1-k}$. Persymmetry
has a major consequence for the three moments, which comes from manipulating
$m_{T}$:

$\displaystyle m_{T}$
$\displaystyle=\kappa\,dt\,\sum_{k,l=0}^{N-1}e^{-2\kappa\,dt\,(N-1-k)}(M_{T}^{\,\text{-}1})_{N-1-k,N-1-l}\,e^{-2\kappa\,dt\,(N-1-l)}$
(324)
$\displaystyle=e^{4\kappa\,dt}\,\kappa\,dt\,\sum_{k,l=0}^{N-1}e^{-2\kappa\,dt\,(N-k)}(M_{T}^{\,\text{-}1})_{lk}e^{-2\kappa\,dt\,(N-l)}\,.$
(325)

We can set $e^{4\kappa\,dt}=1$ and thus conclude that

$\displaystyle m_{T}=n_{T}\,.$ (326)

The complementarity in time of $\nu_{T}$ and $\mu_{T}$ was discussed in
section III.2: the post-measurement variable $\nu_{T}$ of equation 148 depends
exponentially on the end of the outcome register, and the POVM variable
$\mu_{T}$ of equation 149 depends exponentially on the beginning of the
register. The persymmetry of $M_{T}$ and $M_{T}^{\,\text{-}1}$ expresses that
the beginning and end of the record look the same statistically, so it is not
surprising that the persymmetry implies that $m_{T}=n_{T}$.

The next step, deriving Ricatti equations for the three moments, involves
incrementing the moments from $T$ to $T+dT$. The tedious part of this task is
determining how $M_{T}^{\,\text{-}1}$ increments, that is, finding
$M_{T+dT}^{\,\text{-}1}$, and that can be done using the Schur complement. We
relegate this entire task to appendix G and here skip directly to the coupled
Ricatti ODEs, taken from equations 535–537:

$\displaystyle\frac{1}{\kappa}\frac{dn_{T}}{dT}$
$\displaystyle=(1-n_{T})^{2}\,,$ (327)
$\displaystyle\frac{1}{\kappa}\frac{dm_{T}}{dT}$
$\displaystyle=\big{(}q_{T}+e^{-2\kappa T}\big{)}^{2}\,,$ (328)
$\displaystyle\frac{1}{\kappa}\frac{dq_{T}}{dT}$
$\displaystyle=-q_{T}(1-n_{T})+e^{-2\kappa T}(1+n_{T})\,.$ (329)

With the zero initial conditions, these have the solutions,

$\displaystyle n_{T}=m_{T}$ $\displaystyle=\frac{\kappa T}{1+\kappa T}\,,$
(330) $\displaystyle q_{T}$ $\displaystyle=\frac{1}{1+\kappa T}-e^{-2\kappa
T}=-\frac{\kappa T}{1+\kappa T}+2e^{-\kappa T}\sinh\kappa T\,.$ (331)

It is quite instructive to notice that the equality $n_{T}=m_{T}$ and the
reality of $q_{T}$ together imply that the sum and difference Harish-Chandra
phase-space variables are uncorrelated,

$\displaystyle\langle(\nu_{T}\pm\mu_{T})^{*}(\nu_{T}\mp\mu_{T}\rangle_{M}=\langle|\nu_{T}|^{2}\rangle_{M}-\langle|\mu_{T}|^{2}\rangle_{M}\mp\langle\nu_{T}^{*}\mu_{T}\rangle_{M}\pm\langle\mu_{T}^{*}\nu_{T}\rangle_{M}=0\,,$
(332)

with second moments,

$\displaystyle\langle|\nu_{T}\pm\mu_{T}|^{2}\rangle_{M}=\langle|\nu_{T}|^{2}\rangle_{M}+\langle|\mu_{T}|^{2}\rangle_{M}\pm\langle\nu_{T}^{*}\mu_{T}\rangle_{M}\pm\langle\mu_{T}^{*}\nu_{T}\rangle_{M}=2(n_{T}\pm
q_{T})\,,$ (333)

where

$\displaystyle n_{T}+q_{T}$ $\displaystyle=2e^{-\kappa T}\sinh\kappa T\,,$
(334) $\displaystyle n_{T}-q_{T}$ $\displaystyle=\frac{-1+\kappa T}{1+\kappa
T}+e^{-2\kappa T}=2e^{-\kappa T}\cosh\kappa T\,\frac{\Sigma_{T}}{1+\kappa
T}\,,$ (335)

The width $\Sigma_{T}=\kappa T-\tanh\kappa T$ was introduced in equation 242.
It is worth noting the early- and late-time behavior of the various moments:

$\displaystyle n_{T}=m_{T}$ $\displaystyle=\begin{cases}\kappa T\,,&\kappa
T\ll 1,\\\ 1\,,&\kappa T\gg 1,\end{cases}$ (336) $\displaystyle q_{T}$
$\displaystyle=\begin{cases}\kappa T\,,&\kappa T\ll 1,\\\ 1/\kappa T\,,&\kappa
T\gg 1,\end{cases}$ (337) $\displaystyle n_{T}+q_{T}$
$\displaystyle=\begin{cases}2\kappa T\,,&\kappa T\ll 1,\\\ 1\,,&\kappa T\gg
1,\end{cases}$ (338) $\displaystyle n_{T}-q_{T}$
$\displaystyle=\begin{cases}\frac{2}{3}(\kappa T)^{3}\,,&\kappa T\ll 1,\\\
1\,,&\kappa T\gg 1.\end{cases}$ (339)

At early times, $\nu$ and $\mu$ are tightly correlated, with their sum
undergoing standard diffusion; at late times, they become uncorrelated, and
their second moments saturate at 1.

The sum and difference variables being uncorrelated, the Gaussian path
integral 313 has the normalized solution,

$\displaystyle\tilde{B}_{T}(Zx)$ $\displaystyle=e^{-4\kappa T}\delta(r-2\kappa
T)\,4\pi\frac{1}{2\pi(n_{T}+q_{T})}\exp\\!\bigg{(}{-}\frac{|\nu+\mu|^{2}}{2(n_{T}+q_{T})}\bigg{)}\,4\pi\frac{1}{2\pi(n_{T}-q_{T})}\exp\\!\bigg{(}{-}\frac{|\nu-\mu|^{2}}{2(n_{T}-q_{T})}\bigg{)}\,.$
(340)

Noting that

$\displaystyle(n_{T}+q_{T})(n_{T}-q_{T})=\sinh 2\kappa
T\,\frac{2\Sigma_{T}}{e^{2\kappa T}(1+\kappa T)}\,,$ (341)

we can put the normalized solution in the form

$\displaystyle\tilde{B}_{T}(Zx)$ $\displaystyle=\frac{1}{\sinh 2\kappa
T}\,\delta(r-2\kappa T)\,\frac{2e^{-2\kappa T}(1+\kappa
T)}{\Sigma_{T}}\exp\\!\bigg{(}{-}|\nu+\mu|^{2}\frac{e^{\kappa T}}{4\sinh\kappa
T}-|\nu-\mu|^{2}\frac{e^{\kappa T}}{4\cosh\kappa T}\frac{1+\kappa
T}{\Sigma_{T}}\bigg{)}\,,$ (342)

which satisfies the $\delta$-function initial condition 302.

To retrieve the unnormalized distribution
$B_{T}(Zx)=\mathcal{N}_{T}\tilde{B}_{T}(Zx)$, one needs $\det
M_{T}=1/\mathcal{N}_{T}$, which according to equation 530 or equation 306,
satisfies the equation

$\displaystyle\frac{1}{\kappa}\frac{d\ln\det
M_{T}}{dT}=-(1+n_{T})=-2+\frac{1}{1+\kappa T}\,,$ (343)

with solution, for initial condition $\det M_{0}=1$,

$\displaystyle\det M_{T}=e^{-2\kappa T}(1+\kappa
T)=\frac{1}{\mathcal{N}_{T}}\,.$ (344)

The unnormalized distribution is therefore

$\displaystyle B_{T}(Zx)$ $\displaystyle=\frac{1}{\sinh 2\kappa
T}\,\delta(r-2\kappa
T)\,\frac{2}{\Sigma_{T}}\exp\\!\bigg{(}{-}|\nu+\mu|^{2}\frac{e^{\kappa
T}}{4\sinh\kappa T}-|\nu-\mu|^{2}\frac{e^{\kappa T}}{4\cosh\kappa
T}\frac{1+\kappa T}{\Sigma_{T}}\bigg{)}$ (345) $\displaystyle=\frac{1}{\sinh
2\kappa T}\,\delta(r-2\kappa
T)\,\frac{2}{\Sigma_{T}}\exp\\!\bigg{(}{-}|\beta+\alpha|^{2}e^{-\kappa
T}\sinh\kappa T-|\beta-\alpha|^{2}e^{-\kappa T}\cosh\kappa T\frac{1+\kappa
T}{\Sigma_{T}}\bigg{)}\,.$ (346)

The second line transforms to Cartan phase-space coördinates. Unnormalizing
changes the Gaussian’s prefactor; transforming to Cartan coördinates changes
the Gaussian. The final step is to undo the gauge transformation to get back
to the Cartan reduced distribution,

$\displaystyle C_{T}(Zx)=e^{f(Zx)}B_{T}(Zx)=\frac{1}{\sinh 2\kappa
T}\,\delta(r-2\kappa
T)\,\frac{2}{\Sigma_{T}}\exp\\!\bigg{(}{-}\frac{|\beta-\alpha|^{2}}{\Sigma_{T}}\bigg{)}\,,$
(347)

which matches the solution 241 obtained from the PDE 235.

It is fair to ask whether the point of this section is just to provide a
different, more complicated route to the solution for $C_{T}(Zx)$? We think it
is more than that, and here’s why. It all comes back to the elephant in the
room, that is, how one handles the normalization or scaling of the Kraus
operators that comes from the center $Z$. The Cartan reduced distribution,
defined in equation 230 and determined from the diffusion equation 235,
succeeds in representing POVM completeness by marginalizing the _ur_
-distribution $D_{T}(x)$ over the Cartan-center normalization,
$e^{-2l}=e^{2f(Zx)}e^{-2s}$; this gives a distribution uniform in the Cartan
sum variable $\beta+\alpha$ and thus spread over all of phase space in a way
that gives POVM completeness. As a consequence, however, the moments of
$C_{T}(Zx)$ are not those of the stochastic integrals for the phase-plane
variables. The Harish-Chandra reduced distribution $B_{T}(Zx)$, defined in
equation 261, marginalizes $D_{T}(x)$ over the Harish-Chandra-center
normalization, $e^{-2s}$. After normalization to unity by the factor
$\mathcal{N}_{T}$, the normalized distribution $\tilde{B}_{T}(Zx)$ is
determined from the diffusion equation 310 or by applying the path-integral
expression 281, with its modified path measure, to the stochastic integrals
for the Harish-Chandra phase-space variables. The route from
$\tilde{B}_{T}(Zx)$ to POVM completeness runs backwards through the
normalization factor $\mathcal{N}_{T}$ and the anti-Gaussian gauge
transformation $e^{2f(Zx)}$ and arrives at $C_{T}(Zx)$. The point of this
section, one might say, is to find and reveal these connections among path
integrals, diffusion equations, and SDEs; discovering these connections,
driven in this paper by a combination of necessity and opportunity, allows us
to re-unite the three faces of the stochastic trinity. Accessing the entire
stochastic trinity by using a positive gauge transformation that is grounded
in a problem’s Lie group—this, we hope, might be generally applicable to
Feynman-Kac formulas for non-normalization-preserving diffusion equations.

##  V Concluding Remarks. The Stochastic Trinity

We set out on the project of analyzing simultaneous measurements of
noncommuting observables [39, 40] with the goal of showing that such
measurements end up with a POVM in the overcomplete coherent-state basis. We
now think that we have uncovered something more ambitious, a distinctive new
window into the space of quantum dynamics. The formulation of the problem of
continual, differential measurements invites one—compels one, really—to think
in terms of the paths of Wiener outcome increments. These outcome paths, as
sample-paths drawn from the Wiener measure, know nothing about a space in
which they are wandering. When they are instantiated in the exponents of Kraus
operators, however, the time-ordered products of the differential Kraus
operators generate a (complex) Lie-group manifold, the instrumental Lie group,
in which the Kraus operators are, in the way of groups, both the
transformations and the moving points. The motion in the instrumental Lie-
group manifold is described by Kraus-operator SDEs or by the diffusion of the
KOD, as embodied in a FPK diffusion equation. The continuous, but not
differentiable paths are handled effortlessly by the Itô calculus of the
outcome increments, with its terms of order $\sqrt{dt}$ and $dt$. This
requires getting just beyond the linear structure of vector fields (right-
invariant derivatives) and one-forms (right-invariant one-forms) on the Lie-
group manifold, as is evident from our discussion of Stratonovich vs. Itô SDEs
and in the derivation of the FPK diffusion equation, where right-invariant
derivatives end up as diffusive second derivatives. We end up in a very
comfortable place, working in all three corners of the stochastic trinity:
path integrals, FPK diffusion equations, and SDEs, all three describing
motion, equivalently, on the instrumental Lie-group manifold.

Why don’t others find the same comfort in all three faces of the trinity?
Field theorists, interested in the propagator of closed-system dynamics, have
a different way of handling the continuous, but not differentiable paths,
coming from a Stratonovich-calculus way of dealing with the temporal
derivatives in the kinetic terms in a Lagrangian. While they usually have an
equivalent Schrödinger-like equation for the propagator, they do not have the
analog of SDEs, even though the problem often undergoes a Wick rotation into
“Euclidean spacetime.”

Open-systems theorists, both in condensed-matter physics and in quantum
optics, generally work with master equations or stochastic master equations
for the evolution of quantum states and sometimes with diffusion equations for
a probability distribution associated with the states. Those who start with
diffusion equations can avail themselves of a Feynman-Kac formula for a path
integral, but the connection to SDEs has generally not been made. The reason
for not using all three faces of the stochastic trinity is, we think, a
failure to identify the appropriate Lie-group manifold; this failure is
connected to the emphasis on the evolution of quantum states, which obscures
nearly entirely the Lie-group manifold on which the open-system dynamics is
occurring.

Suffice it to say that we think we have found something: the home of quantum
dynamics, the Lie-group manifold that supports all three faces of the trinity.
The exhausted reader who has survived to read this concluding sentence of a
very long paper might be pleased—or so we hope—to learn that our ambition is
larger than was evident at the beginning.

###### Acknowledgements.

CSJ thanks Mohan Sarovar for all of the helpful discussions and financial
support through Sandia National Laboratories and thanks CMC for the incredibly
fruitful collaboration. CMC supported himself and is eternally grateful for
the opportunity to work with CSJ. This work was supported in part by the
Center for Quantum Information and Control at the University of New Mexico and
in part by the U.S. Department of Energy, Office of Science, Office of
Advanced Scientific Computing Research, under the Quantum Computing
Application Teams (QCAT) program. Sandia National Laboratories is a
multimission laboratory managed and operated by NTESS, LLC., a wholly owned
subsidiary of Honeywell International, Inc., for the U.S. DOE’s NNSA under
contract DE-NA-0003525. This paper describes objective technical results and
analysis. Any subjective views or opinions that might be expressed in the
paper do not necessarily represent the views of the U.S. Department of Energy
or the United States Government.

##  Appendix A Exterior Derivatives of the Right-Invariant Basis

The Lie algebra of right-invariant derivatives is equivalent to the exterior
algebra of right-invariant one-forms. This can be seen by considering the
second exterior differential of a scalar function,

$\displaystyle 0$ $\displaystyle=ddf$ (348)
$\displaystyle=d\Big{(}\underleftarrow{e_{\mu}}[f]\theta^{\mu}\Big{)}$ (349)
$\displaystyle=\underleftarrow{e_{\mu}}\Big{[}\underleftarrow{e_{\nu}}[f]\Big{]}\theta^{\mu}\\!\wedge\theta^{\nu}+\underleftarrow{e_{\lambda}}[f]d\theta^{\lambda}$
(350)
$\displaystyle=\frac{1}{2}\bigg{(}\underleftarrow{e_{\mu}}\Big{[}\underleftarrow{e_{\nu}}[f]\Big{]}-\underleftarrow{e_{\nu}}\Big{[}\underleftarrow{e_{\mu}}[f]\Big{]}\bigg{)}\theta^{\mu}\\!\wedge\theta^{\nu}+\underleftarrow{e_{\lambda}}[f]d\theta^{\lambda}$
(351)
$\displaystyle=\frac{1}{2}\big{[}\underleftarrow{e_{\mu}},\underleftarrow{e_{\nu}}\big{]}[f]\,\theta^{\mu}\\!\wedge\theta^{\nu}+\underleftarrow{e_{\lambda}}[f]d\theta^{\lambda}$
(352)
$\displaystyle=\underleftarrow{e_{\lambda}}[f]\Big{(}{-}\frac{1}{2}{c_{\mu\nu}}^{\lambda}\,\theta^{\mu}\\!\wedge\theta^{\nu}+d\theta^{\lambda}\Big{)}\,,$
(353)

which by linear independence means that

$[\underleftarrow{e_{\mu}},\underleftarrow{e_{\nu}}]=-{c_{\mu\nu}}^{\lambda}\underleftarrow{e_{\lambda}}\hskip
20.0pt\Longleftrightarrow\hskip
20.0ptd\theta^{\lambda}=\frac{1}{2}{c_{\mu\nu}}^{\lambda}\,\theta^{\mu}\\!\wedge\theta^{\nu}\,.$
(354)

Specifically, if we write out all of the different Lie brackets,

$\displaystyle H_{\textrm{o}}$ $\displaystyle=\text{no such brackets}\,,$
(355) $\displaystyle Q$ $\displaystyle=[H_{\textrm{o}},-iP]\,,$ (356)
$\displaystyle P$ $\displaystyle=[H_{\textrm{o}},iQ]\,,$ (357) $\displaystyle-
iP$ $\displaystyle=[H_{\textrm{o}},Q]\,,$ (358) $\displaystyle iQ$
$\displaystyle=[H_{\textrm{o}},P]\,,$ (359) $\displaystyle-\Omega$
$\displaystyle=[Q,iP]=[iQ,P]\,,$ (360) $\displaystyle i\Omega$
$\displaystyle=[Q,P]=[iQ,-iP],$ (361)

we can read off all of the one-form differentials or “curls”,

$\displaystyle d\theta^{i\Omega}$
$\displaystyle=\theta^{Q}\\!\wedge\theta^{P}+\theta^{iQ}\\!\wedge\theta^{-iP}\,,$
(362) $\displaystyle d\theta^{-\Omega}$
$\displaystyle=\theta^{-iP}\\!\wedge\theta^{Q}+\theta^{iQ}\\!\wedge\theta^{P}\,,$
(363) $\displaystyle d\theta^{-iP}$
$\displaystyle=\theta^{H_{\textrm{o}}}\\!\wedge\theta^{Q}\,,$ (364)
$\displaystyle d\theta^{iQ}$
$\displaystyle=\theta^{H_{\textrm{o}}}\\!\wedge\theta^{P}\,,$ (365)
$\displaystyle d\theta^{Q}$
$\displaystyle=\theta^{H_{\textrm{o}}}\\!\wedge\theta^{-iP}\,,$ (366)
$\displaystyle d\theta^{P}$
$\displaystyle=\theta^{H_{\textrm{o}}}\\!\wedge\theta^{iQ}\,,$ (367)
$\displaystyle d\theta^{-H_{\textrm{o}}}$ $\displaystyle=0\,,$ (368)

##  Appendix B Global Coördinate Transformation between Cartan and Harish-
Chandra

The global transformation from Cartan to Harish-Chandra coördinates is most
easily accomplished by inserting the displacement-operator normal ordering of
equation 76 into the Cartan decomposition 156 and then pushing the factors
around until they are in Harish-Chandra form:

$\displaystyle\left(D_{\beta}\,e^{i\Omega\phi}\right)e^{-H_{\textrm{o}}r-\Omega\ell}D_{\alpha}^{\,\text{-}1}$
$\displaystyle=e^{-\Omega[\ell+\frac{1}{2}(|\beta|^{2}+|\alpha|^{2})]}e^{i\Omega\phi}e^{a^{\dagger}\beta}e^{-a\beta^{*}}e^{-H_{\textrm{o}}r}e^{-a^{\dagger}\alpha}e^{a\alpha^{*}}$
(369)
$\displaystyle=e^{-\Omega[\ell+\frac{1}{2}(|\beta|^{2}+|\alpha|^{2})]}e^{i\Omega\phi}e^{a^{\dagger}\beta}e^{-a\beta^{*}}e^{-a^{\dagger}e^{-r}\alpha}e^{-H_{\textrm{o}}r}e^{a\alpha^{*}}$
(370)
$\displaystyle=e^{-\Omega[\ell+\frac{1}{2}(|\beta|^{2}+|\alpha|^{2})]}e^{i\Omega\phi}e^{\Omega\beta^{*}\alpha
e^{-r}}e^{a^{\dagger}\beta}e^{-a^{\dagger}e^{-r}\alpha}e^{-a\beta^{*}}e^{-H_{\textrm{o}}r}e^{a\alpha^{*}}$
(371)
$\displaystyle=e^{-\Omega[\ell+\frac{1}{2}(|\beta|^{2}+|\alpha|^{2})]}e^{i\Omega\phi}e^{\Omega\beta^{*}\alpha
e^{-r}}e^{a^{\dagger}\beta}e^{-a^{\dagger}e^{-r}\alpha}e^{-H_{\textrm{o}}r}e^{-ae^{-r}\beta^{*}}e^{a\alpha^{*}}$
(372)
$\displaystyle=e^{a^{\dagger}(\beta-e^{-r}\alpha)}e^{-H_{\textrm{o}}r-\Omega[\ell+\frac{1}{2}(|\beta|^{2}+\frac{1}{2}|\alpha|^{2})-\beta^{*}\alpha
e^{-r}-i\phi]}e^{a(\alpha-e^{-r}\beta)^{*}}\,.$ (373)

Identifying the parameters of the last line gives the coördinate
transformation,

$\displaystyle\nu$ $\displaystyle=\beta-e^{-r}\alpha\,,$ (374)
$\displaystyle\mu$ $\displaystyle=\alpha-e^{-r}\beta\,,$ (375)
$\displaystyle-s+i\psi=z$ $\displaystyle=-\ell-f+i(\phi+\xi)\,,$ (376)

where

$\displaystyle f$
$\displaystyle\equiv\frac{1}{2}\Big{(}|\beta|^{2}+|\alpha|^{2}-\beta^{*}\alpha
e^{-r}-\beta\alpha^{*}e^{-r}\Big{)}=\frac{1-e^{-r}}{4}|\beta+\alpha|^{2}+\frac{1+e^{-r}}{4}|\beta-\alpha|^{2}\,,$
(377) $\displaystyle\xi$ $\displaystyle\equiv
e^{-r}\frac{\beta^{*}\alpha-\beta\,\alpha^{*}}{2i}=e^{-r}\frac{(\beta-\alpha)^{*}(\beta+\alpha)-(\beta-\alpha)(\beta+\alpha)^{*}}{4i}\,,$
(378)

are functions of the RWIH $G/Z$.

The inverse coördinate transformation is

$\displaystyle\beta$ $\displaystyle=\frac{e^{r}\nu+\mu}{2\sinh r}\,,$ (379)
$\displaystyle\alpha$ $\displaystyle=\frac{e^{r}\mu+\nu}{2\sinh r}\,,$ (380)
$\displaystyle\ell$ $\displaystyle=s-f\,,$ (381) $\displaystyle\phi$
$\displaystyle=\psi-\xi\,.$ (382)

It is quite useful to notice that the sum and difference phase-space
coördinates simply rescale under this transformation,

$\displaystyle\beta+\alpha$
$\displaystyle=\frac{\nu+\mu}{1-e^{-r}}=\frac{e^{r/2}(\nu+\mu)}{2\sinh(r/2)}\,,$
(383) $\displaystyle\beta-\alpha$
$\displaystyle=\frac{\nu-\mu}{1+e^{-r}}=\frac{e^{r/2}(\nu-\mu)}{2\cosh(r/2)}\,.$
(384)

We also have

$\displaystyle f$
$\displaystyle=\frac{|\nu+\mu|^{2}}{4(1-e^{-r})}+\frac{|\nu-\mu|^{2}}{4(1+e^{-r})}=\frac{(|\nu|^{2}+|\mu|^{2})e^{r}+\nu^{*}\mu+\nu\mu^{*}}{4\sinh
r}\,,$ (385) $\displaystyle\xi$
$\displaystyle=\frac{(\nu-\mu)^{*}(\nu+\mu)-(\nu-\mu)(\nu+\mu)^{*}}{8i\sinh
r}=\frac{\nu^{*}\mu-\nu\mu^{*}}{4i\sinh r}\,.$ (386)

It is crucial to appreciate that the transformation from Harish-Chandra to
Cartan coördinates is singular at $r=0$, with the consequence that positive
transformations of the form $e^{Qq+Pp}$ are not represented in Cartan
coördinates. At $r=0$ the Harish-Chandra decomposition becomes

$\displaystyle x_{r=0}=e^{\Omega
z}e^{a^{\dagger}\nu}e^{a\mu^{*}}=e^{\Omega(z-\frac{1}{2}\nu\mu^{*})}e^{a^{\dagger}\nu+a\mu^{*}}=e^{\Omega(z-\frac{1}{2}\nu\mu^{*})}D_{(\nu-\mu)/2}e^{a^{\dagger}\frac{1}{2}(\nu+\mu)+a\frac{1}{2}(\nu+\mu)^{*}}\,.$
(387)

Each displacement operator and each positive transformation of the form
$e^{Qq+Pp}$ is represented uniquely in the 4-plane of $\nu$ and $\mu$
coördinates; in particular, the identity operator has coördinates $r=0$ and
$z=\nu=\mu=0$. At $r=0$ the Cartan decomposition reduces to

$\displaystyle
x_{r=0}=e^{i\Omega\phi}e^{-\Omega\ell}D_{\beta}D_{-\alpha}=e^{i\Omega[\phi+(\beta^{*}\alpha-\beta\alpha^{*})/2i]}e^{-\Omega\ell}D_{\beta-\alpha}\,.$
(388)

The Cartan coördinate singularity at $r=0$ is that $x_{r=0}$ does not depend
on $\beta+\alpha$. A displacement operator $D_{\beta-\alpha}=D_{\tau}$ is
represented by a plane of $\beta+\alpha$ values, specifically, by all the
coördinates satisfying $\beta-\alpha=\tau$, $r=0$, $\ell=0$, and
$\phi=\big{(}\beta+\alpha)^{*}\tau-(\beta+\alpha)\tau^{*}\big{)}/4i$. Most
importantly, the identity $1$ is represented by the plane of Cartan coördinate
values satisfying $\beta=\alpha$ and $r=\phi=\ell=0$. Positive transformations
of the form $e^{Qq+Pp}$ are not represented at all in Cartan coördinates; this
is not a problem because these positive operators lie on a boundary that is
not accessible to the Kraus operators of SPQM.

##  Appendix C Solution for Harish-Chandra Center Coördinate

To solve the SDE 146 for the Harish-Chandra center coördinate $z$, it is best
to work with the sums over Wiener increments that underlie the Itô stochastic
integrals. Thus we begin by writing the solution 148 for the post-measurement
Harish-Chandra coördinate $\nu$ as

$\displaystyle\nu_{N}=\sum_{k=0}^{T/dt-1}\sqrt{\kappa}\,dw_{kdt}\,e^{-2\kappa(T-t_{k})}=\sum_{k=0}^{N-1}\sqrt{\kappa}\,dw_{k}\,e^{-2\kappa(N-k)dt}\,,$
(389)

where $dw_{k}=dw_{kdt}$ and $t_{k}=k\,dt$ ($t_{N}=T=N\,dt$). Notice that the
initial condition $\nu_{0}=0$ is enforced by having no terms in the sum for
$N=0$. The solution for $z$ is

$\displaystyle z_{N}$
$\displaystyle=\frac{1}{2}\sum_{k=0}^{N-1}\kappa|dw_{k}|^{2}+\sum_{k=1}^{N-1}\nu_{k}\sqrt{\kappa}\,dw_{k}^{*}$
(390)
$\displaystyle=\frac{1}{2}\sum_{k=0}^{N-1}\kappa|dw_{k}|^{2}+\sum_{k=1}^{N-1}\sum_{l=0}^{k-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\,,$
(391)

where we omit the $k=0$ term on the second line since $\nu_{0}=0$ and where we
can insert the absolute value because $k>l$. Notice that, as for $\nu$, the
way the initial condition $z_{0}=0$ is enforced is that there are no terms in
the sums when $N=0$. Now define

$\displaystyle
y=\sum_{k=1}^{N-1}\sum_{l=0}^{k-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\,,$
(392)

and manipulate $y^{*}$ by specifying the summing range in an equivalent way
and then switching $k$ and $l$,

$\displaystyle
y^{*}=\sum_{l=0}^{N-2}\sum_{k=l+1}^{N-1}\kappa\,dw_{k}\,dw_{l}^{*}\,e^{-2\kappa|k-l|dt}=\sum_{k=0}^{N-2}\sum_{l=k+1}^{N-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\,.$
(393)

With this result we an write our sums as including the entire range of values,
$k,l=1\,\ldots,N-1$. Converting to the real and imaginary parts of
$z=-s+i\psi$, we have

$\displaystyle-s_{N}=\mathrm{Re}(z_{N})$
$\displaystyle=\frac{1}{2}\Bigg{(}\sum_{k=0}^{N-1}\kappa|dw_{k}|^{2}+\sum_{k=1}^{N-1}\sum_{l=0}^{k-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}+\sum_{k=0}^{N-2}\sum_{l=k+1}^{N-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\Bigg{)}$
(394)
$\displaystyle=\frac{1}{2}\sum_{k=0}^{N-1}\sum_{l=0}^{N-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\,,$
(395) $\displaystyle i\psi_{N}=i\mathrm{Im}(z_{N})$
$\displaystyle=\frac{1}{2}\Bigg{(}\sum_{k=1}^{N-1}\sum_{l=0}^{k-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}-\sum_{k=0}^{N-2}\sum_{l=k+1}^{N-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\Bigg{)}$
(396)
$\displaystyle=\frac{1}{2}\sum_{k=0}^{N-1}\sum_{l=0}^{N-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\textrm{sgn}(k-l)\,,$
(397)

where the sign function is defined as

$\displaystyle\textrm{sgn}(u)=\begin{cases}1,&u>0,\\\ 0,&u=0,\\\
-1,&u<0.\end{cases}$ (398)

The center variable 390 is therefore

$\displaystyle
z_{N}=-s_{N}+i\psi_{N}=\sum_{k=0}^{N-1}\sum_{l=0}^{N-1}\kappa\,dw_{k}^{*}\,dw_{l}\,e^{-2\kappa|k-l|dt}\frac{1}{2}\big{(}1+\textrm{sgn}(k-l)\big{)}\,,$
(399)

where one recognizes

$\displaystyle\frac{1}{2}\big{(}1+\textrm{sgn}(u)\big{)}=H(u)=\begin{cases}1,&u>0,\\\
\frac{1}{2},&u=0,\\\ 0&u<0.\end{cases}$ (400)

as the Heaviside step function with the choice $H(0)=\frac{1}{2}$. This
appendix is really an exercise in relating the real and imaginary parts of $z$
to its symmetric and antisymmetric parts and, in the process, getting the
weighting of the diagonal ($k=l$) term right—equivalently, making the right
choice for the $u=0$ value of the sign and Heaviside functions. The
incremental Itô calculus, with its quite explicit diagonal terms
$\frac{1}{2}\kappa|dw_{k}|^{2}$, leaves no doubt about the right choice.

Converting back to integrals gives

$\displaystyle-s_{T}$
$\displaystyle=\frac{1}{2}\int_{0}^{T_{-}}\\!\\!\int_{0}^{T_{-}}\\!\\!\kappa\,dw_{t}^{*}\,dw_{s}\,e^{-2\kappa|t-s|}\,,$
(401) $\displaystyle i\psi_{T}$
$\displaystyle=\frac{1}{2}\int_{0}^{T_{-}}\\!\\!\int_{0}^{T_{-}}\\!\\!\kappa\,dw_{t}^{*}\,dw_{s}\,e^{-2\kappa|t-s|}\textrm{sgn}(t-s)\,,$
(402)

which lead to the formula for $z_{T}=-s_{T}+i\psi_{T}$ in equation 150. To use
these integrals, one should return to the incremental sums. If one started
with the Stratonovich-form SDE 204 for $z$, one could convert it to an
ordinary temporal differential equation with the Wiener increments replaced by
$\delta$-functions. Integrating that equation would give the same results as
here, but only after careful attention to how to weight the diagonal ($t=s$)
$\delta$-functions in the integral. Thinking in terms of the Stratonovich-form
SDE for $z$, it is clear that solving the equation for $z$—what we have done
in this appendix—is identical to deriving a fluctuation-dissipation theorem
for the correlation of $\nu$ and $\mu$.

##  Appendix D Local Transformations Between Cartan-Coördinate and Right-
Invariant Frames

The Cartan decomposition of a position $x\in\mathrm{IWH}$ is

$x=\left(D_{\beta}\,e^{i\Omega\phi}\right)e^{-H_{\textrm{o}}r-\Omega\ell}D_{\alpha}^{\,\text{-}1}\,.$
(403)

Before differentiating $x$ with respect to the Cartan coördinates, it is
useful to differentiate the displacement operator with respect to its
arguments using the ordered forms in equation 75:

$\displaystyle\partial_{\alpha_{1}}D_{\alpha}$
$\displaystyle=\big{(}{-}iP+{\textstyle{\frac{1}{2}}}\alpha_{2}\,i\Omega\big{)}D_{\alpha}\,,$
(404) $\displaystyle\partial_{\alpha_{2}}D_{\alpha}$
$\displaystyle=\big{(}iQ-{\textstyle{\frac{1}{2}}}\alpha_{1}\,i\Omega\big{)}D_{\alpha}\,.$
(405)

We will also need the conjugations

$\displaystyle\frac{1}{\sqrt{2}}D_{\alpha}(Q+iP)D_{\alpha}^{\dagger}=D_{\alpha}aD_{\alpha}^{\dagger}$
$\displaystyle=a-\alpha=\frac{1}{\sqrt{2}}\big{(}Q-\alpha_{1}+i(P-\alpha_{2})\big{)}\,,$
(406)

and

$\displaystyle e^{-rH_{\textrm{o}}}ae^{rH_{\textrm{o}}}$
$\displaystyle=ae^{r}\,,$ (407) $\displaystyle
e^{-rH_{\textrm{o}}}a^{\dagger}e^{rH_{\textrm{o}}}$
$\displaystyle=a^{\dagger}e^{-r}\,,$ (408) $\displaystyle
e^{-rH_{\textrm{o}}}Qe^{rH_{\textrm{o}}}$ $\displaystyle=Q\cosh r+iP\sinh
r\,,$ (409) $\displaystyle e^{-rH_{\textrm{o}}}Pe^{rH_{\textrm{o}}}$
$\displaystyle=P\cosh r-iQ\sinh r\,.$ (410)

Differentiating the position $x$ gives

$\displaystyle\begin{split}\partial_{\phi}x&=i\Omega\,x\,,\end{split}$ (411)
$\displaystyle\begin{split}-\partial_{\ell}x&=\Omega\,x\,,\end{split}$ (412)
$\displaystyle\begin{split}\partial_{\beta_{1}}x&=-iP\,x+{\textstyle{\frac{1}{2}}}\beta_{2}\,i\Omega\,x\,,\end{split}$
(413)
$\displaystyle\begin{split}\partial_{\beta_{2}}x&=iQ\,x-{\textstyle{\frac{1}{2}}}\beta_{1}\,i\Omega\,x\,,\end{split}$
(414)
$\displaystyle\begin{split}{-}\partial_{r}x&=D_{\beta}H_{\textrm{o}}D_{\beta}^{\dagger}\,x\\\
&={\textstyle{\frac{1}{2}}}\big{(}(Q-\beta_{1})^{2}+(P-\beta_{2})^{2}\big{)}x\\\
&=H_{\textrm{o}}\,x-\beta_{1}\,Q\,x-\beta_{2}\,P\,x+{\textstyle{\frac{1}{2}}}(\beta_{1}^{2}+\beta_{2}^{2})\,\Omega\,x\,,\end{split}$
(415)
$\displaystyle\begin{split}\partial_{\alpha_{1}}x&=D_{\beta}\,e^{-rH_{\textrm{o}}}\big{(}iP+{\textstyle{\frac{1}{2}}}\alpha_{2}\,i\Omega\big{)}e^{rH_{\textrm{o}}}D_{\beta}^{\dagger}\,x\\\
&=D_{\beta}\big{(}iP\cosh r+Q\sinh
r+{\textstyle{\frac{1}{2}}}\alpha_{2}\,i\Omega\,\big{)}D_{\beta}^{\dagger}\,x\\\
&=\cosh r\,iP\,x+\sinh r\,Q\,x-\beta_{1}\sinh r\,\Omega\,x-(\beta_{2}\cosh
r-{\textstyle{\frac{1}{2}}}\alpha_{2})\,i\Omega\,x\,,\end{split}$ (416)
$\displaystyle\begin{split}\partial_{\alpha_{2}}x&=D_{\beta}\,e^{-rH_{\textrm{o}}}\big{(}{-}iQ-{\textstyle{\frac{1}{2}}}\alpha_{1}\,i\Omega\big{)}e^{rH_{\textrm{o}}}D_{\beta}^{\dagger}\,x\\\
&=D_{\beta}\big{(}{-}iQ\cosh r+P\sinh
r-{\textstyle{\frac{1}{2}}}\alpha_{1}\,i\Omega\,\big{)}D_{\beta}^{\dagger}\,x\\\
&=-\cosh r\,iQ\,x+\sinh r\,P\,x-\beta_{2}\sinh r\,\Omega\,x+(\beta_{1}\cosh
r-{\textstyle{\frac{1}{2}}}\alpha_{1})\,i\Omega\,x\,.\end{split}$ (417)

By the chain rule, we have the frame transformation,

$\displaystyle\begin{array}[]{rclllllll}\vphantom{\Big{(}}\partial_{\phi}&=&+\underleftarrow{i\Omega}\\\
\vphantom{\Big{(}}\partial_{\ell}&=&&{-}\underleftarrow{\Omega}\\\
\vphantom{\Big{(}}\partial_{\beta_{1}}&=&+{\textstyle{\frac{1}{2}}}\beta_{2}\underleftarrow{i\Omega}&&-\underleftarrow{iP}\\\
\vphantom{\Big{(}}\partial_{\beta_{2}}&=&-{\textstyle{\frac{1}{2}}}\beta_{1}\underleftarrow{i\Omega}&&&+\underleftarrow{iQ}\\\
\vphantom{\Big{(}}\partial_{\alpha_{1}}&=&-(\beta_{2}\cosh
r-{\textstyle{\frac{1}{2}}}\alpha_{2})\underleftarrow{i\Omega}&-\beta_{1}\sinh
r\,\underleftarrow{\Omega}&+\cosh r\,\underleftarrow{iP}&&+\sinh
r\underleftarrow{Q}\\\
\vphantom{\Big{(}}\partial_{\alpha_{2}}&=&+(\beta_{1}\cosh
r-{\textstyle{\frac{1}{2}}}\alpha_{1})\underleftarrow{i\Omega}&-\beta_{2}\sinh
r\,\underleftarrow{\Omega}&&-\cosh r\,\underleftarrow{iQ}&&+\sinh
r\underleftarrow{P}\\\
\vphantom{\Big{(}}\partial_{r}&=&&-{\textstyle{\frac{1}{2}}}(\beta_{1}^{2}+\beta_{2}^{2})\underleftarrow{\Omega}&&&+\beta_{1}\underleftarrow{Q}&+\beta_{2}\underleftarrow{P}&-\underleftarrow{H_{\textrm{o}}}\end{array}\,.$
(425)

Inverting the transformation gives

$\displaystyle\underleftarrow{i\Omega}$ $\displaystyle=\partial_{\phi}\,,$
(426) $\displaystyle-\underleftarrow{\Omega}$
$\displaystyle=\partial_{\ell}\,,$ (427) $\displaystyle-\underleftarrow{iP}$
$\displaystyle=\partial_{\beta_{1}}-{\textstyle{\frac{1}{2}}}\beta_{2}\partial_{\phi}\,,$
(428) $\displaystyle\underleftarrow{iQ}$
$\displaystyle=\partial_{\beta_{2}}+{\textstyle{\frac{1}{2}}}\beta_{1}\partial_{\phi}\,,$
(429) $\displaystyle\underleftarrow{Q}$
$\displaystyle=\nabla_{1}-\beta_{1}\partial_{\ell}+\frac{\beta_{2}\cosh
r-\alpha_{2}}{2\sinh r}\partial_{\phi}\,,$ (430)
$\displaystyle\underleftarrow{P}$
$\displaystyle=\nabla_{2}-\beta_{2}\partial_{\ell}-\frac{\beta_{1}\cosh
r-\alpha_{1}}{2\sinh r}\partial_{\phi}\,,$ (431)
$\displaystyle\begin{split}-\underleftarrow{H_{\textrm{o}}}&=\partial_{r}-\beta_{1}\underleftarrow{Q}-\beta_{2}\underleftarrow{P}+\frac{\beta_{1}^{2}+\beta_{2}^{2}}{2}\underleftarrow{\Omega}\\\
&=\partial_{r}-\beta_{1}\nabla_{1}-\beta_{2}\nabla_{2}+\frac{\beta_{1}^{2}+\beta_{2}^{2}}{2}\partial_{\ell}+\frac{\beta_{1}\alpha_{2}-\beta_{2}\alpha_{1}}{2\sinh
r}\partial_{\phi}\,,\end{split}$ (432)

where

$\nabla_{j}\equiv\frac{1}{\sinh r}\left(\partial_{\alpha_{j}}+\cosh
r\,\partial_{\beta_{j}}\right)\,.$ (433)

Transposing the transformation gives

$\displaystyle\begin{split}\theta^{i\Omega}&=d\phi+{\textstyle{\frac{1}{2}}}\beta_{2}d\beta_{1}-{\textstyle{\frac{1}{2}}}\beta_{1}d\beta_{2}-(\beta_{2}\cosh
r-{\textstyle{\frac{1}{2}}}\alpha_{2})\,d\alpha_{1}+(\beta_{1}\cosh
r-{\textstyle{\frac{1}{2}}}\alpha_{1})d\alpha_{2}\\\
&=d\phi+{\textstyle{\frac{1}{2}}}(\beta_{2}d\beta_{1}-\beta_{1}d\beta_{2})+{\textstyle{\frac{1}{2}}}(\alpha_{2}d\alpha_{1}-\alpha_{1}d\alpha_{2})+\cosh
r\,(\beta_{1}d\alpha_{2}-\beta_{2}d\alpha_{1})\,,\end{split}$ (434)
$\displaystyle\theta^{-\Omega}$
$\displaystyle=d\ell+{\textstyle{\frac{1}{2}}}(\beta_{1}^{2}+\beta_{2}^{2})dr+\sinh
r\,(\beta_{1}d\alpha_{1}+\beta_{2}d\alpha_{2})\,,$ (435)
$\displaystyle\theta^{-iP}$ $\displaystyle=d\beta_{1}-\cosh r\,d\alpha_{1}\,,$
(436) $\displaystyle\theta^{iQ}$ $\displaystyle=d\beta_{2}-\cosh
r\,d\alpha_{2}\,,$ (437) $\displaystyle\theta^{Q}$
$\displaystyle=\beta_{1}dr+\sinh r\,d\alpha_{1}\,,$ (438)
$\displaystyle\theta^{P}$ $\displaystyle=\beta_{2}dr+\sinh r\,d\alpha_{2}\,,$
(439) $\displaystyle\theta^{-H_{\textrm{o}}}$ $\displaystyle=dr\,.$ (440)

With the one-form transformations in hand, the Haar measure 110 in Cartan
coördinates is

$d^{7}\\!\mu(x)=d\phi\,d\ell\,\frac{d^{2}\beta}{\pi}\,dr\,\sinh^{2}\\!r\,\frac{d^{2}\alpha}{\pi}\,.$
(441)

The phase-plane measures are $d^{2}\beta=\frac{1}{2}d\beta_{1}d\beta_{2}$ and
$d^{2}\beta=\frac{1}{2}d\beta_{1}d\beta_{2}$. The factors of $1/\pi$ are
conventional in quantum optics and ultimately come from the coherent-state
completeness relation 155.

It is easy to show that the left-invariant derivatives and one-forms can be
obtained from the right-invariant quantities by changing the sign of the
quantities associated with anti-Hermitian operators and transforming the
coördinates according to $\phi\leftrightarrow-\phi$ and
$\beta\leftrightarrow\alpha$. The sign changes don’t change the Haar measure,
and the measure is unchanged by the coördinate transformation, which shows
that the Haar measure is both right- and left-invariant.

##  Appendix E Local Transformations Between Harish-Chandra-Coördinate and
Right-Invariant Frames

The Harish-Chandra decomposition of a position $x\in\mathrm{IWH}$ is

$x=e^{a^{\dagger}\nu}e^{-H_{\textrm{o}}r+\Omega z}e^{a\mu^{*}}\,.$ (442)

Break the complex coördinates into real and imaginary parts, as in equation
134. Differentiating the position gives

$\displaystyle\partial_{\nu_{1}}x$
$\displaystyle=\frac{1}{\sqrt{2}}\,a^{\dagger}\,x\,,$ (443)
$\displaystyle\partial_{\nu_{2}}x$
$\displaystyle=\frac{1}{\sqrt{2}}\,ia^{\dagger}x\,,$ (444)
$\displaystyle-\partial_{s}x$ $\displaystyle=\Omega\,x\,,$ (445)
$\displaystyle\partial_{\psi}x$ $\displaystyle=\,i\Omega\,x\,,$ (446)
$\displaystyle\begin{split}-\partial_{r}x&=e^{a^{\dagger}\nu}H_{\textrm{o}}e^{-a^{\dagger}\nu}\,x\\\
&=H_{\textrm{o}}\,x-\nu\,a^{\dagger}x\\\
&=H_{\textrm{o}}\,x-\nu_{1}\,\frac{1}{\sqrt{2}}a^{\dagger}\,x-\nu_{2}\,\frac{1}{\sqrt{2}}ia^{\dagger}x\,,\end{split}$
(447)
$\displaystyle\begin{split}\partial_{\mu_{1}}x&=\frac{1}{\sqrt{2}}e^{a^{\dagger}\nu}e^{-H_{\textrm{o}}r}ae^{H_{\textrm{o}}r}e^{-a^{\dagger}\nu}x\\\
&=\frac{1}{\sqrt{2}}e^{a^{\dagger}\nu}(e^{r}a)e^{-a^{\dagger}\nu}x\\\
&=\frac{1}{\sqrt{2}}e^{r}(a-\nu\Omega)x\\\
&=e^{r}\left(\frac{1}{\sqrt{2}}a\,x-\frac{1}{2}\nu_{1}\,\Omega\,x-\frac{1}{2}\nu_{2}\,i\Omega\,x\right)\,,\end{split}$
(448)
$\displaystyle\begin{split}\partial_{\mu_{2}}x&=-i\frac{1}{\sqrt{2}}e^{a^{\dagger}\nu}e^{-H_{\textrm{o}}r}ae^{H_{\textrm{o}}r}e^{-a^{\dagger}\nu}x\\\
&=e^{r}\\!\left(\\!{-}\frac{1}{\sqrt{2}}\,ia\,x-\frac{1}{2}\nu_{2}\,\Omega\,x+\frac{1}{2}\nu_{1}\,i\Omega\,x\right)\,.\end{split}$
(449)

By the chain rule, we have the frame transformation,

$\displaystyle\begin{array}[]{rclllllll}\vphantom{\Big{(}}\partial_{\psi}&=&\underleftarrow{i\Omega}\\\
\vphantom{\Big{(}}-\partial_{s}&=&&\underleftarrow{\Omega}\\\
\vphantom{\Big{(}}\partial_{\nu_{1}}&=&&&\frac{1}{\sqrt{2}}\underleftarrow{a^{\dagger}}\\\
\vphantom{\Big{(}}\partial_{\nu_{2}}&=&&&&\frac{1}{\sqrt{2}}\underleftarrow{ia^{\dagger}}\\\
\vphantom{\Big{(}}e^{-r}\partial_{\mu_{1}}&=&-\frac{1}{2}\nu_{2}\underleftarrow{i\Omega}&-\frac{1}{2}\nu_{1}\underleftarrow{\Omega}&&&\frac{1}{\sqrt{2}}\underleftarrow{a}\\\
\vphantom{\Big{(}}e^{-r}\partial_{\mu_{2}}&=&\frac{1}{2}\nu_{1}\underleftarrow{i\Omega}&-\frac{1}{2}\nu_{2}\underleftarrow{\Omega}&&&&-\frac{1}{\sqrt{2}}\underleftarrow{ia}\\\
\vphantom{\Big{(}}-\partial_{r}&=&&&-\nu_{1}\frac{1}{\sqrt{2}}\underleftarrow{a^{\dagger}}&-\nu_{2}\frac{1}{\sqrt{2}}\underleftarrow{ia^{\dagger}}&&&\underleftarrow{H_{\textrm{o}}}\end{array}\,.$
(457)

Inverting the transformation gives

$\displaystyle\underleftarrow{i\Omega}$ $\displaystyle=\partial_{\psi}\,,$
(458) $\displaystyle-\underleftarrow{\Omega}$ $\displaystyle=\partial_{s}\,,$
(459) $\displaystyle\frac{1}{\sqrt{2}}\underleftarrow{a^{\dagger}}$
$\displaystyle=\partial_{\nu_{1}}\,,$ (460)
$\displaystyle\frac{1}{\sqrt{2}}\underleftarrow{ia^{\dagger}}$
$\displaystyle=\partial_{\nu_{2}}\,,$ (461)
$\displaystyle\frac{1}{\sqrt{2}}\underleftarrow{a}$
$\displaystyle=e^{-r}\partial_{\mu_{1}}-\frac{1}{2}\nu_{1}\partial_{s}+\frac{1}{2}\nu_{2}\partial_{\psi}\,,$
(462) $\displaystyle-\frac{1}{\sqrt{2}}\underleftarrow{ia}$
$\displaystyle=e^{-r}\partial_{\mu_{2}}-\frac{1}{2}\nu_{2}\partial_{s}-\frac{1}{2}\nu_{1}\partial_{\psi}\,,$
(463) $\displaystyle-\underleftarrow{H_{\textrm{o}}}$
$\displaystyle=\partial_{r}-\nu_{1}\partial_{\nu_{1}}-\nu_{2}\partial_{\nu_{2}}\,.$
(464)

Transposing the transformation gives

$\displaystyle\theta^{i\Omega}$
$\displaystyle=d\psi+{\textstyle{\frac{1}{2}}}e^{r}(\nu_{1}d\mu_{2}-\nu_{2}d\mu_{1})\,,$
(465) $\displaystyle\theta^{-\Omega}$
$\displaystyle=ds+{\textstyle{\frac{1}{2}}}e^{r}(\nu_{1}d\mu_{1}+\nu_{2}d\mu_{2})\,,$
(466) $\displaystyle{\sqrt{2}}\,\theta^{a^{\dagger}}$
$\displaystyle=d\nu_{1}+\nu_{1}dr\,,$ (467)
$\displaystyle{\sqrt{2}}\,\theta^{ia^{\dagger}}$
$\displaystyle=d\nu_{2}+\nu_{2}dr\,,$ (468)
$\displaystyle{\sqrt{2}}\,\theta^{a}$ $\displaystyle=e^{r}d\mu_{1}\,,$ (469)
$\displaystyle{\sqrt{2}}\,\theta^{-ia}$ $\displaystyle=e^{r}d\mu_{2}\,,$ (470)
$\displaystyle\theta^{-H_{\textrm{o}}}$ $\displaystyle=dr\,.$ (471)

It is instructive to convert the right-invariant derivatives and forms to
position and momentum,

$\displaystyle-\underleftarrow{iP}$
$\displaystyle=\frac{1}{\sqrt{2}}(\underleftarrow{a^{\dagger}}-\underleftarrow{a})=\partial_{\nu_{1}}-e^{-r}\partial_{\mu_{1}}+\frac{1}{2}\nu_{1}\partial_{s}-\frac{1}{2}\nu_{2}\partial_{\psi}\,,$
(472) $\displaystyle\underleftarrow{iQ}$
$\displaystyle=\frac{1}{\sqrt{2}}(\underleftarrow{ia^{\dagger}}+\underleftarrow{ia})=\partial_{\nu_{2}}-e^{-r}\partial_{\mu_{2}}+\frac{1}{2}\nu_{2}\partial_{s}+\frac{1}{2}\nu_{1}\partial_{\psi}\,,$
(473) $\displaystyle\underleftarrow{Q}$
$\displaystyle=\frac{1}{\sqrt{2}}(\underleftarrow{a^{\dagger}}+\underleftarrow{a})=\nabla_{1}-\frac{1}{2}\nu_{1}\partial_{s}+\frac{1}{2}\nu_{2}\partial_{\psi}\,,$
(474) $\displaystyle\underleftarrow{P}$
$\displaystyle=\frac{1}{\sqrt{2}}(\underleftarrow{ia^{\dagger}}-\underleftarrow{ia})=\nabla_{2}-\frac{1}{2}\nu_{2}\partial_{s}-\frac{1}{2}\nu_{1}\partial_{\psi}\,,$
(475)

where the derivatives

$\nabla_{j}=\partial_{\nu_{j}}+e^{-r}\partial_{\mu_{j}}$ (476)

are equal to the derivatives of equation 433 and

$\displaystyle\theta^{-iP}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\theta^{a^{\dagger}}-\theta^{a}\big{)}=\frac{1}{2}\big{(}d\nu_{1}-e^{r}d\mu_{1}+\nu_{1}dr\big{)}\,,$
(477) $\displaystyle\theta^{iQ}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\theta^{ia^{\dagger}}+\theta^{ia}\big{)}=\frac{1}{2}\big{(}d\nu_{2}-e^{r}d\mu_{2}+\nu_{2}dr\big{)}\,,$
(478) $\displaystyle\theta^{Q}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\theta^{a^{\dagger}}+\theta^{a}\big{)}=\frac{1}{2}\big{(}d\nu_{1}+e^{r}d\mu_{1}+\nu_{1}dr\big{)}\,,$
(479) $\displaystyle\theta^{P}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\theta^{ia^{\dagger}}-\theta^{ia}\big{)}=\frac{1}{2}\big{(}d\nu_{2}+e^{r}d\mu_{2}+\nu_{2}dr\big{)}\,.$
(480)

Comparing the expressions for right-invariant derivatives and one-forms in the
Harish-Chandra coördinates of this appendix with the expressions in Cartan
coördinates from appendix D, one sees that these expressions relate the
coördinate partial derivatives and one-forms in the two coördinate systems.
The relations between coördinate partial derivatives and one-forms can, of
course, be derived straighforwardly from the global coördinate transformations
in appendix B. We do not record these relations because the expressions in
terms of the right-invariant quantities are more useful for our purposes. We
remind the reader that even though the ruler coördinate $r$ is shared between
Harish-Chandra and Cartan coördinates, the partial derivative $\partial_{r}$
is different between the two systems because partial derivatives are defined
by holding the other coördinates constant.

The Haar measure 110 in Harish-Chandra coördinates is

$d^{7}\\!\mu(x)=d\psi\,ds\,\frac{d^{2}\nu}{2\pi}\,dr\,e^{2r}\frac{d^{2}\mu}{2\pi}\,.$
(481)

The factors of $1/2\pi$ in the Harish-Chandra phase-plane measures follow from
transforming the Cartan Haar measure to Harish-Chandra coördinates. Just as
for Cartan coördinates, it is easy to show that the left-invariant one-forms
yield the same measure.

##  Appendix F Delta-Functions and the Singularity in Cartan Coördinates

In this appendix we find the Cartan form of the $\delta$-function
$\delta(x,1)$ in two ways. The first uses that $D_{0}(x)=\delta(x,1)$, so we
can step slightly away from $T=0$, to $T=dt$, and find $D_{dt}(x)$ as an
expression that limits to $\delta(x,1)$ as $dt\rightarrow 0$. The second way
is perhaps more straightforward: show that the expression for $\delta(x,1)$ in
Cartan coördinates transforms to the known form 189 in Harish-Chandra
coördinates. That said, since we need to know $D_{dt}(x)$ in section III.5, we
start with the first way.

Starting from the general path-integral expression 88 for the KOD and setting
$T=dt$,

$\displaystyle D_{dt}(x)=\int
d\mu(dw_{0})\,\delta\big{(}x,\gamma(dw_{0})\big{)}\,,$ (482)

and plugging in the Cartan-coördinate expression 185 for the
$\delta$-function, we have

$\displaystyle D_{dt}(x)=\int
d\mu(dw_{0})\,\delta(\phi-\phi_{dt})\,\delta(\ell-\ell_{dt})\,\frac{1}{\sinh^{2}\\!r}\delta(r-r_{dt})\,\pi\delta^{2}(\beta-\beta_{dt})\,\pi\delta^{2}(\alpha-\alpha_{dt})\,.$
(483)

It is useful to work in terms of sum and difference coördinates for the phase-
space variables, so we note that

$\displaystyle\pi\delta^{2}(\beta-\beta_{dt})\,\pi\delta^{2}(\alpha-\alpha_{dt})=2\pi\delta^{2}\big{(}\beta+\alpha-(\beta_{dt}+\alpha_{dt})\big{)}\,2\pi\delta^{2}\big{(}\beta-\alpha-(\beta_{dt}-\alpha_{dt})\big{)}\,.$
(484)

We now specialize the stochastic-integral solutions for the coördinates to the
first increment. At $T=dt$, $r_{dt}=2\kappa\,dt$ and the phase-space
coördinates are

$\displaystyle\nu_{dt}=\sqrt{\kappa}\,dw_{0}\,e^{-2\kappa
dt}=\sqrt{\kappa}\,dw_{0}\,,\qquad$
$\displaystyle\qquad\mu_{dt}=d\mu_{0}=\sqrt{\kappa}\,dw_{0}\,,$ (485)
$\displaystyle\nu_{dt}+\mu_{dt}=\sqrt{\kappa}\,dw_{0}(e^{-2\kappa
dt}+1)=2\sqrt{\kappa}\,dw_{0}\,,\qquad$
$\displaystyle\qquad\nu_{dt}-\mu_{dt}=\sqrt{\kappa}\,dw_{0}(e^{-2\kappa
dt}-1)=-2\kappa dt\sqrt{\kappa}\,dw_{0}\,,$ (486)
$\displaystyle\beta_{dt}=\sqrt{\kappa}\,dw_{0}\,\mathrm{csch}\,2\kappa
dt=\frac{\sqrt{\kappa}\,dw_{0}}{2\kappa dt}\,,\qquad$
$\displaystyle\qquad\alpha_{dt}=\sqrt{\kappa}\,dw_{0}\coth 2\kappa
dt=\frac{\sqrt{\kappa}\,dw_{0}}{2\kappa dt}\,,$ (487) $\displaystyle\
\beta_{dt}+\alpha_{dt}=\sqrt{\kappa}\,dw_{0}\coth\kappa
dt=\frac{\sqrt{\kappa}\,dw_{0}}{\kappa dt}\,,\qquad$
$\displaystyle\qquad\beta_{dt}-\alpha_{dt}=-\sqrt{\kappa}\,dw_{0}\tanh\kappa
dt=-\kappa dt\sqrt{\kappa}\,dw_{0}\,.$ (488)

In each case, the first expression is exact, and the second is the leading-
order contribution. What should give any reader pause are the divergences in
the single-increment Cartan phase variables. These divergences are an
expression of the $r=0$ singularity in Cartan coördinates, and they make the
decisive contribution to our determination of $D_{dt}(x)$. It is important to
understand that there are subtleties in trying to extract the single-increment
behavior directly from the SDEs, instead of from the stochastic integrals. For
example, the SDEs 144 and 145 say that
$\nu_{dt}=d\nu_{0}=\sqrt{\kappa}\,dw_{0}$ and
$\mu_{dt}=d\mu_{0}=\sqrt{\kappa}\,dw_{0}$, so one misses the leading-order
contribution to $\nu_{dt}-\mu_{dt}$ because that leading-order contribution is
zero from the perspective of the SDEs. More tellingly, the $r=0$ singularity
appears directly in equations 197 and 198 as singularities in $d\beta_{0}$ and
$d\alpha_{0}$, preventing one from determining the single-increment values
$\beta_{dt}$ and $\alpha_{dt}$ from these equations. The single-increment
values can be read off the SDEs 199 and 200, however, and this is because
these are effectively equations for the Harish-Chandra coördinates.

We also need the leading-order single-increment behavior of the several center
coördinates:

$\displaystyle z_{dt}$
$\displaystyle=-s_{dt}+i\psi_{dt}={\textstyle{\frac{1}{2}}}\kappa|dw_{0}|^{2}\,,$
(489) $\displaystyle-\ell_{dt}$
$\displaystyle=f_{dt}-s_{dt}=f_{dt}=2\kappa\,dt\,\frac{|\beta_{dt}+\alpha_{dt}|^{2}}{4}=\frac{1}{2\kappa\,dt}\frac{|\nu_{dt}+\mu_{dt}|^{2}}{4}=\frac{\kappa|dw_{0}|^{2}}{2\kappa\,dt}\,,$
(490) $\displaystyle\phi_{dt}$ $\displaystyle=\psi_{dt}-\xi_{dt}=0\,.$ (491)

Please appreciate that, as indicated, the leading-order contribution to
$\ell_{dt}$ comes entirely from $f_{dt}$.

Putting all this into equation 482, one finds

$\displaystyle\begin{split}D_{dt}(x)&=\delta(\phi)\int\frac{d^{2}(\sqrt{\kappa}\,dw_{0})}{\pi\kappa\,dt}e^{-\kappa|dw_{0}|^{2}/\kappa\,dt}\,\delta\bigg{(}\ell+\frac{\kappa|dw_{0}|^{2}}{2\kappa\,dt}\bigg{)}\,\frac{1}{\sinh^{2}\\!r}\delta(r-2\kappa\,dt)\\\
&\qquad\qquad\qquad\times
2\pi\delta^{2}\bigg{(}\beta+\alpha-\frac{\sqrt{\kappa}\,dw_{0}}{\kappa\,dt}\bigg{)}\,2\pi\delta^{2}(\beta-\alpha)\,.\end{split}$
(492)

Now writing

$\displaystyle\delta^{2}\bigg{(}\beta+\alpha-\frac{\sqrt{\kappa}\,dw_{0}}{\kappa\,dt}\bigg{)}=(\kappa\,dt)^{2}\delta^{2}\big{(}\sqrt{\kappa}\,dw_{0}-\kappa\,dt(\beta+\alpha)\big{)}\,,$
(493)

we do the integral over the initial Wiener increment $dw_{0}$,

$\displaystyle
D_{dt}(x)=\delta(\phi)\,\delta\Big{(}\ell+{\textstyle{\frac{1}{2}}}\kappa\,dt|\beta+\alpha|^{2}\Big{)}\,\frac{1}{\sinh^{2}\\!r}\delta(r-2\kappa\,dt)\,2\pi\frac{\kappa\,dt}{\pi}e^{-\kappa\,dt|\beta+\alpha|^{2}}\,2\pi\delta^{2}(\beta-\alpha)\,.$
(494)

It is easy to see that this integrates to 1 over the Haar measure 183.

It is understood that in defining $\delta(x,1)$, we take the limit
$dt\rightarrow 0$. We might as well take these limits, thus writing

$\displaystyle\delta(x,1)$
$\displaystyle=\delta(\phi)\,\delta(\ell)\,\frac{1}{\sinh^{2}\\!r}\delta(r)\bigg{(}2\pi\lim_{dt\to
0}\frac{\kappa\,dt}{\pi}e^{-\kappa\,dt|\beta+\alpha|^{2}}\bigg{)}\,2\pi\delta^{2}(\beta-\alpha)$
(495)
$\displaystyle=\delta(\phi)\,\delta(\ell)\,\frac{1}{\sinh^{2}\\!r}\delta(r)\bigg{(}\pi\lim_{dt\to
0}\frac{4\kappa\,dt}{\pi}e^{-4\kappa\,dt|\alpha|^{2}}\bigg{)}\,\pi\delta^{2}(\beta-\alpha)\,.$
(496)

This has singular behavior in the ruler $r$, but this singularity goes away
when integrating against the Haar measure. More important is the Gaussian in
$|\beta+\alpha|^{2}$, which limits to an infinitely wide, normalized Gaussian
as $dt\rightarrow 0$. As discussed in appendix B, the identity 1 is
represented by $\phi=\ell=r=0$ and $\beta=\alpha$, with $\beta+\alpha$ free to
take on any complex value. The infinitely wide Gaussian in $\beta+\alpha$ and
the $\delta$-function $\delta(\beta-\alpha)$ express this in $\delta(x,1)$.

Now let’s transform the expression 494 to Harish-Chandra coördinates. With
$r=2\kappa\,dt$, the sum and difference phase-space variables are related by

$\displaystyle\beta+\alpha$ $\displaystyle=\frac{\nu+\mu}{2\kappa\,dt}\,,$
(497) $\displaystyle\beta-\alpha$ $\displaystyle=\frac{\nu-\mu}{2}\,,$ (498)

and the Harish-Chandra center variables are given by, when $\alpha=\beta$,

$\displaystyle s$
$\displaystyle=\ell+f=\ell+{\textstyle{\frac{1}{2}}}\kappa\,dt|\beta+\alpha|^{2}\,,$
(499) $\displaystyle\psi$ $\displaystyle=\phi+\xi=\phi\,.$ (500)

Substituting into equation 494 gives

$\displaystyle
D_{dt}(x)=\delta(\phi)\,\delta(s)\,\delta(r-2\kappa\,dt)\,4\pi\frac{1}{\pi
4\kappa\,dt}e^{-|\nu+\mu|^{2}/4\kappa\,dt}\,4\pi\delta^{2}(\nu-\mu)\,.$ (501)

The wide Gaussian in $\beta+\alpha$ becomes a narrow Gaussian in $\nu+\mu$, so
when we take the limit $dt\rightarrow 0$, we get

$\displaystyle\delta(x,1)=\delta(\phi)\,\delta(s)\,\delta(r)\,4\pi\delta^{2}(\nu+\mu)\,4\pi\delta^{2}(\nu-\mu)=\delta(\phi)\,\delta(s)\,\delta(r)\,2\pi\delta^{2}(\nu)\,2\pi\delta^{2}(\mu)\,,$
(502)

in agreement with equation 189 and consistent with the fact that in Harish-
Chandra coördinates, the identity is represented uniquely by $\phi=s=r=0$ and
$\nu=\mu=0$.

It is now straightforward to integrate over the center variables in either
coördinate system to find the $\delta$-function on $G/Z$,

$\displaystyle\delta(Zx,Z1)$
$\displaystyle=\frac{1}{\sinh^{2}\\!r}\delta(r)\,\bigg{(}\pi\lim_{dt\to
0}\frac{4\kappa\,dt}{\pi}e^{-4\kappa\,dt|\alpha|^{2}}\bigg{)}\,\pi\delta^{2}(\beta-\alpha)$
(503) $\displaystyle=\delta(r)\,2\pi\delta^{2}(\nu)\,2\pi\delta^{2}(\mu)\,.$
(504)

##  Appendix G Riccati Equations for the Three Moments

In this appendix we derive th Ricatti equations 327–329 for the three second
moments, $n_{T}$, $m_{T}$, and $q_{T}$, from their expressions in equations
319, 321, and 323. When we work in terms of the modified path measure
$\mathcal{D}\mu_{M}\\!\left[dw_{[0,T)}\right]$ of equation 274, the
correlation of two outcome increments, at times $t$ and $s$ within the
interval $[0,T)$, depends on the overall time $T$, as noted in equation 275.
Hence it seems natural here to reserve $t$ to denote times within the interval
$[0,T)$ and to increment the overall time from $T$ to $T+dT$. As a
consequence, unlike anywhere else in the paper, we denote the duration of the
increments by $dT$ instead of $dt$.

It is quite convenient to introduce a bra-ket notation and to define vectors

$\displaystyle|{\mathord{+}}_{T}\rangle$
$\displaystyle\equiv\sqrt{dT}\sum_{k=0}^{N-1}e^{-2\kappa\,dT\,(N-k)}|k\rangle\,,$
(505) $\displaystyle|{\mathord{-}}_{T}\rangle$
$\displaystyle\equiv\sqrt{dT}\sum_{k=0}^{N-1}e^{-2\kappa\,dT\,k}|k\rangle\,,$
(506)

where the kets $|k\rangle$ are orthonormal. The matrix elements of $M_{T}$ are

$\displaystyle(M_{T})_{kl}=\langle
k|M_{T}|l\rangle=\delta_{kl}-\kappa\,dT\,e^{-2\kappa\,dT\,|k-l|}\,,\quad
k,l=0,\ldots,N-1.$ (507)

The utility of the bra-ket notation becomes apparent when we write the three
moments as matrix elements of $M_{T}$,

$\displaystyle n_{T}$
$\displaystyle=\kappa{}\langle{\mathord{+}}_{T}|M_{T}^{-1}|{\mathord{+}}_{T}\rangle\,,$
(508) $\displaystyle m_{T}$
$\displaystyle=\kappa\langle{\mathord{-}}_{T}|M_{T}^{-1}|{\mathord{-}}_{T}\rangle\,,$
(509) $\displaystyle q_{T}$
$\displaystyle=\kappa\langle{\mathord{+}}_{T}|M_{T}^{-1}|{\mathord{-}}_{T}\rangle=\kappa\langle{\mathord{-}}_{T}|M_{T}^{-1}|{\mathord{+}}_{T}\rangle\,.$
(510)

Replacing $M_{T}$ and $M_{t}^{\,\text{-}1}$ with the unit matrix gives the
moments with respect to the original Wiener measure; these moments are the
inner products of the kets $|\pm_{T}\rangle$.

Now increment $M_{T}$ forward one step in time, from $T$ to $T+dT$. What
happens is really nothing,

$\displaystyle\langle
k|M_{T+dT}|l\rangle=\delta_{kl}-\kappa\,dT\,e^{-2\kappa\,dT\,|k-l|}\,,\quad
k,l=0,\ldots,N,$ (511)

except that an outside row and column, labeled by the index $N$, are added,

$\displaystyle\vphantom{\Big{(}}k,l=0,\ldots,N-1:\quad$ $\displaystyle\langle
k|M_{T+dT}|l\rangle=\langle
k|M_{T}|l\rangle=\delta_{kl}-\kappa\,dT\,e^{-2\kappa\,dT\,|k-l|}\,,$ (512)
$\displaystyle\vphantom{\Big{(}}k=0,\ldots,N-1:\quad$ $\displaystyle\langle
k|M_{T+dT}|N\rangle=-\kappa\,dT\,e^{-2\kappa\,dT\,(N-k)}=-\kappa\sqrt{dT}\,\langle
k|{\mathord{+}}_{T}\rangle\,,$ (513)
$\displaystyle\vphantom{\Big{(}}l=0,\ldots,N-1:\quad$ $\displaystyle\langle
N|M_{T+dT}|l\rangle=-\kappa\,dT\,e^{-2\kappa\,dT\,(N-k)}=-\kappa\sqrt{dT}\,\langle{\mathord{+}}_{T}|l\rangle\,,$
(514) $\displaystyle\langle N|M_{T+dT}|N\rangle=1-\kappa\,dT\,.$ (515)

The incremented matrix can then be written in block form relative to the outer
row and column,

$\displaystyle
M_{T+dT}=M_{T}-\kappa\sqrt{dT}\big{(}|{\mathord{+}}_{T}\rangle\langle
N|+|N\rangle\langle{\mathord{+}}_{T}|\big{)}+(1-\kappa\,dT)|N\rangle\langle
N|\,.$ (516)

We also need to increment the kets $|\pm_{T}\rangle$:

$\displaystyle|{\mathord{+}}_{T+dT}\rangle$
$\displaystyle=\sqrt{dT}\,\sum_{k=0}^{N}e^{-2\kappa\,dT\,(N+1-k)}|k\rangle=e^{-2\kappa\,dT}\big{(}|{\mathord{+}}_{T}\rangle+\sqrt{dT}\,|N\rangle\big{)}\,,$
(517) $\displaystyle|{\mathord{-}}_{T+dT}\rangle$
$\displaystyle=\sqrt{dT}\,\sum_{k=0}^{N}e^{-2\kappa\,dT\,k}|k\rangle=|{\mathord{-}}_{T}\rangle+\sqrt{dT}\,e^{-2\kappa
T}|N\rangle\,.$ (518)

The harder task, incrementing $M_{T}^{\,\text{-}1}$, is done using the Schur
complement. Generally, for any matrix written in block form,

$\displaystyle M=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}\,,$ (519)

deriving a formal expression for the inverse begins by a block Gaussian
elimination that writes the matrix as

$\displaystyle M=\begin{pmatrix}I&0\\\
CA^{\,\text{-}1}&I\end{pmatrix}\begin{pmatrix}A&0\\\
0&M/A\end{pmatrix}\begin{pmatrix}I&A^{\,\text{-}1}B\\\ 0&I\end{pmatrix}\,,$
(520)

where

$\displaystyle M/A\equiv D-CA^{\,\text{-}1}B$ (521)

is called the Schur complement. A first thing to notice is that

$\displaystyle\det M=\det A\,\det(M/A)\,,$ (522)

and the second thing is that the inverse of $M$ is

$\displaystyle
M^{\,\text{-}1}=\begin{pmatrix}\vphantom{\Big{(}}A^{\,\text{-}1}+A^{\,\text{-}1}B(M/A)^{\,\text{-}1}CA^{\,\text{-}1}&\quad&-A^{\,\text{-}1}B(M/A)^{\,\text{-}1}\\\
\vphantom{\Big{(}}-(M/A)^{\,\text{-}1}CA^{\,\text{-}1}&\quad&(M/A)^{\,\text{-}1}\end{pmatrix}\,.$
(523)

In the case at hand, the block form 516 of $M_{T+dT}$ relative to the outer
row and column can be written in the matrix form 519 by identifying

$\displaystyle A$ $\displaystyle=M_{T}\,,$ (524) $\displaystyle B$
$\displaystyle=-\kappa\sqrt{dT}\,|{\mathord{+}}_{T}\rangle\langle N|\,,$ (525)
$\displaystyle C$
$\displaystyle=-\kappa\sqrt{dT}\,|N\rangle\langle{\mathord{+}}_{T}|\,,$ (526)
$\displaystyle D$ $\displaystyle=(1-\kappa\,dT)|N\rangle\langle N|\,.$ (527)

The Schur complement is the $1\times 1$ matrix

$\displaystyle
M_{T+dT}/M_{T}=\big{(}1-\kappa\,dT-\kappa^{2}\,dT\,\langle{\mathord{+}}_{T}|M_{T}^{-1}|{\mathord{+}}_{T}\rangle\big{)}|N\rangle\langle
N|=\big{[}1-\kappa\,dT(1+n_{T})\big{]}|N\rangle\langle N|\,.$ (528)

One has immediately how the determinant advances by one increment,

$\displaystyle\det M_{T+dT}=\det M_{T}\big{(}1-\kappa\,dT(1+n_{T})\big{)}\,,$
(529)

and thus the ODE

$\displaystyle\frac{1}{\kappa}\frac{d\ln\det M_{T}}{dT}=-(1+n_{T})\,,$ (530)

which agrees with the differential equation 306 for the normalization factor
$\mathcal{N}_{T}=1/\det M_{T}$. Further, one can work out from equation 523
how the inverse increments,

$\displaystyle\begin{split}M_{T+dT}^{\,\text{-}1}&=M_{T}^{\,\text{-}1}+\kappa^{2}\,dT\,M_{T}^{\,\text{-}1}|{\mathord{+}}_{T}\rangle\langle{\mathord{+}}_{T}|M_{T}^{\,\text{-}1}\\\
&\qquad+\kappa\sqrt{dT}\,\big{(}M_{T}^{\,\text{-}1}|{\mathord{+}}_{T}\rangle\langle
N|+|N\rangle\langle{\mathord{+}}_{T}|M_{T}^{\,\text{-}1}\big{)}\\\
&\qquad+\big{[}1+\kappa\,dT\big{(}1+\kappa\langle{\mathord{+}}_{T}|M_{T}^{\,\text{-}1}|{\mathord{+}}_{T}\rangle\big{)}\big{]}|N\rangle\langle
N|\,.\end{split}$ (531)

where we keep only the leading-order term in $dT$ in each block.

Everything is set now to determine how the three moments increment,

$\displaystyle n_{T+dT}$
$\displaystyle=\kappa\langle{\mathord{+}}_{T+dT}|M_{T+dT}^{\,\text{-}1}|{\mathord{+}}_{T+dT}\rangle=n_{T}+\kappa\,dT\,(1-n_{T})^{2}\,,$
(532) $\displaystyle m_{T+dT}$
$\displaystyle=\kappa\langle{\mathord{-}}_{T+dT}|M_{T+dT}^{\,\text{-}1}|{\mathord{-}}_{T+dT}\rangle=m_{T}+\kappa\,dT\,\big{(}q_{T}+e^{-2\kappa
T}\big{)}^{2}\,,$ (533) $\displaystyle q_{T+dT}$
$\displaystyle=\kappa\langle{\mathord{-}}_{T+dT}|M_{T+dT}^{\,\text{-}1}|{\mathord{+}}_{T+dT}\rangle=q_{T}+\kappa\,dT\,\big{(}{-}q_{T}(1-n_{T})+e^{-2\kappa
T}(1+n_{T})\big{)}\,,$ (534)

and these lead immediately to the three (coupled) Riccati equations,

$\displaystyle\frac{1}{\kappa}\frac{dn_{T}}{dT}$
$\displaystyle=(1-n_{T})^{2}\,,$ (535)
$\displaystyle\frac{1}{\kappa}\frac{dm_{T}}{dT}$
$\displaystyle=\big{(}q_{T}+e^{-2\kappa T}\big{)}^{2}\,,$ (536)
$\displaystyle\frac{1}{\kappa}\frac{dq_{T}}{dT}$
$\displaystyle=-q_{T}(1-n_{T})+e^{-2\kappa T}(1+n_{T})\,,$ (537)

which are repeated in the main text as equations 327–329.

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