# Reproducing Kernel Hilbert Space Approach to Non-Markovian Quantum
Stochastic Models

John E. Gough<EMAIL_ADDRESS>Aberystwyth University, SY23 3BZ, Wales, United
Kingdom  Haijin Ding<EMAIL_ADDRESS>Laboratoire des Signaux
et Systémes (L2S), CNRS-CentraleSupélec-Université Paris-Sud, Université
Paris-Saclay, 3, Rue Joliot Curie, 91190, Gif-sur-Yvette, France.  Nina H.
Amini<EMAIL_ADDRESS>Laboratoire des Signaux et Systémes
(L2S), CNRS-CentraleSupélec-Université Paris-Sud, Université Paris-Saclay, 3,
Rue Joliot Curie, 91190, Gif-sur-Yvette, France.

###### Abstract

We give a derivation of the non-Markovian quantum state diffusion equation of
Diósi and Strunz starting from a model of a quantum mechanical system coupled
to a bosonic bath. We show that the complex trajectories arises as a
consequence of using the Bargmann-Segal (complex wave) representation of the
bath. In particular, we construct a reproducing kernel Hilbert space for the
bath auto-correlation and realize the space of complex trajectories as a
Hilbert subspace. The reproducing kernel naturally arises from a feature space
where the underlying feature space is the one-particle Hilbert space of the
bath quanta. We exploit this to derive the unravelling of the open quantum
system dynamics and show equivalence to the equation of Diósi and Strunz. We
also give an explicit expression for the reduced dynamics of a two-level
system coupled to the bath via a Jaynes-Cummings interaction and show that
this does indeed correspond to an exact solution of the Diósi-Strunz equation.
Finally, we discuss the physical interpretation of the complex trajectories
and show that they are intrinsically unobservable.

##  I Introduction

An ensemble of quantum states of a system $S$ is mathematically described as
follows: the state of the system at time $t$ is given by
$|\Psi_{t}(\xi)\rangle_{S}$ where $\xi$ is information describing external
influence and where $\xi$ belongs to some collection $\mathcal{E}$ which is a
measurable space with a probability measure $\mathbb{P}$. Given an observable
$\hat{X}$, its ensemble average at time $t$ should be

$\displaystyle\langle\hat{X}\rangle_{t}=\int_{\mathcal{E}}\langle\Psi_{t}(\xi)|\hat{X}|\Psi_{t}(\xi)\rangle_{S}\,\mathbb{P}[d\xi].$
(1)

It may be that $\langle\hat{X}\rangle_{t}$ computed this way yields the
correct physical average value of every observable $\hat{X}$ of the system at
any time $t$, in which case we say that the ensemble gives an unraveling of
the systems dynamics. However, two points should be noted: firstly, the
ensemble is here a classical randomization of an otherwise quantum dynamics;
secondly, the unraveling only tells us the one-time average. We have not
obliged the unravelling to provide the correct multi-dimensional correlations
between several observables at different times: only the one-time averages.
Kolmogorov’s results on encapsulating stochastic processes tell us that
unravelings address only a very specific modeling aspect of the noise.

The rationale falls into three (potentially overlapping) camps:

  1. 1.

(Quantum State Diffusion)Pearle ; Gisin ; GRW ; Diosi ; GPR  $\xi$ represents
some universal external influence which leads to a non-unitary dynamics for
the ensemble and may be used to explain the collapse of the wave-function;

  2. 2.

(Quantum Filtering)Belavkin ; BB ; BvHJ  $\xi$ is information obtained through
indirect (weak) measurement of the system and $\langle\hat{X}\rangle_{t}$ is
the state of the system at time $t$ conditioned on the information at that
time;

  3. 3.

(Quantum Monte Carlo)DCM ; Carmichael  the ensemble is just a handy device for
computing the average behaviour of an open system, with $\xi$ being one
simulation out of many which we repeat to get a good estimate.

In each of the three applications above, $\xi$ will typically be a sample path
and $\mathcal{E}$ some space of admissible trajectories which we average over.
The sample path structure may lead to a filtration with
$\xi\mapsto|\Psi_{t}(\xi)\rangle_{S}$ being an adapted process. In the quantum
filtering approach, the $\xi$ give the continuous-in-time measurement readout
from indirect measurements and parallels traditional stochastic filtering
methodology - this is often the motivation in the quantum monte carlo
approach, even if not made explicit. The quantum state diffusion is often more
speculative and vague when it comes to the origins of the external noise. See
the well-known book of Joos et al. on Decoherence for more detailsDecoherence
.

Our interest in this paper will be the non-markovian quantum stochastic
dynamics introduced by Diósi and Strunz DS97  (see also their later work with
GisinSDG )

$\displaystyle\frac{d}{dt}|\Psi_{t}(\zeta)\rangle_{S}=\bigg{(}-i\hat{H}+\hat{L}\zeta_{t}-\hat{L}^{\ast}\int_{0}^{t}d\tau\,K\left(t,\tau\right)\frac{\delta}{\delta\zeta_{\tau}}\bigg{)}|\Psi_{t}(\zeta)\rangle_{S},$
(2)

where $\zeta$ is a complex-valued trajectory. Here $K(t,s)$ is the auto-
correlation function of the bath $B$ which acts as environment for $S$.

Complex trajectories originally appeared in the (markovian) quantum state
diffusion approach where they were motivated by the internal symmetries of the
Lindblad generator: here the $\zeta$ appear as the sample paths of a complex
Wiener process $Z(t)=Z^{\prime}(t)+iZ^{\prime\prime}(t)$ where $Z^{\prime}$
and $Z^{\prime\prime}$ are independent standard Wiener processes. However,
they are atypical of the other approaches where each measurement process is a
real-valued process. However, they are compatible with quantum
filteringPUD2017 ; JG2017 .

In the paper, we show that the equation (2) can be derived from a microscopic
model. The origin of the complex trajectories ultimately comes from the fact
that we are using the Bargmann-Segal (complex-wave) representation for the
bath. Here our approach follows the derivation given in Diósi and Strunz DS97
, but we deal with technical mathematical issues relating to infinite baths -
we also explain the origin of the causal structure of (2): the kernel $K(t,s)$
is not causal and this is not built in to the Bargmann-Segal representation,
but instead comes from the form of the dynamical equations we wish model. In
setting up a rigorous description, we make use of the fact that the kernel
leads to a reproducing kernel Hilbert space (RKHS) and we show that space of
complex trajectories is a Hilbert subspace of this RKHS. In particular, the
mircroscopic model naturally supplies a feature map for the kernel.

As an application, in Section V we construct an explicit unravelling for the
problem of a two-level atom coupled to the bath via Jaynes-Cummings
interaction, and then show that the state $|\Psi_{t}(\zeta)\rangle_{S}$ does
indeed satisfy (2).

In section VI, we show that the complex trajectories have no physical
interpretation as observable processes. They are entirely a mathematical by-
product of the Bargmann-Segal representation.

Care is needed, therefore, in the interpretation. For instance, the equation
(2) is often compared with the quantum Zakai equationSDG . Both of these are
indeed linear stochastic equations for an un-normalized vector state, however,
only the latter has an interpretation as the conditional evolution of a
continuously monitored system.

##  II The Microscopic Model

We consider a system with Hilbert space $\mathfrak{h}_{S}$ coupled to a
bosonic bath with Hilbert space $\mathfrak{H}_{B}=\Gamma(\mathfrak{f}_{B})$,
that is, the (Bose) Fock space over a fixed one-particle space
$\mathfrak{f}_{B}$.

###  II.1 The Bath Fock space

Recall that $\mathfrak{H}_{B}=\Gamma(\mathfrak{f}_{B})$ will be the direct sum
of the symmetrized Hilbert spaces
$\otimes_{\mathrm{symm.}}^{n}\mathfrak{f}_{B}$ for $n=0,1,2,\cdots$, with the
$n=0$ space being spanned by a single vector $|\mathrm{vac}\rangle_{B}$ which
is the Fock vacuum. (See, for instance, Chapter 10 of Gough and Kupsch GK  for
details.)

The creation and annihilation operators for test-function
$g\in\mathfrak{f}_{B}$ are denoted as $\hat{a}(g)^{\ast},\hat{a}(g)$
respectively, and the differential second-quantization of an observable
$\hat{M}$ by $d\Gamma(\hat{M})$.

For example, a bath consisting of a discrete set of oscillators with non-
degenerate frequencies $\Omega$ is represented by taking
$\mathfrak{f}_{B}=\mathbb{C}^{\Omega}$ with inner-product $\langle
f|g\rangle=\sum_{\omega\in\Omega}f(\omega)^{\ast}g(\omega)$ and

$\displaystyle\hat{a}(g)^{\ast}=\sum_{\omega\in\Omega}g(\omega)a_{\omega}^{\ast},\quad\hat{a}(g)=\sum_{\omega\in\Omega}g(\omega)^{\ast}a_{\omega},\quad
d\Gamma(\hat{M})=\sum_{\omega,\omega^{\prime}\in\Omega}M_{\omega,\omega^{\prime}}a^{\ast}_{\omega}a_{\omega^{\prime}},$
(3)

where
$[\hat{a}_{\omega},\hat{a}_{\omega^{\prime}}^{\ast}]=\delta_{\omega,\omega^{\prime}}$
and $\hat{M}$ is a matrix with entries $M_{\omega,\omega^{\prime}}$. The bath
Hamiltonian would then be
$\hat{H}_{B}=d\Gamma(\hat{h}_{B})=\sum_{\omega}\omega\hat{a}^{\ast}_{\omega}\hat{a}_{\omega}$,
with $\hat{h}_{B}$ being the diagonal matrix with entries $\omega\in\Omega$.

For a continuum of oscillators, we could take
$\mathfrak{f}_{B}=L^{2}(\mathbb{R}_{+},d\omega)$ with
$\hat{a}(g)^{\ast}=\int_{0}^{\infty}g(\omega)\hat{a}_{\omega}^{\ast}\,d\omega,\,\hat{a}(g)=\int_{0}^{\infty}g(\omega)^{\ast}\hat{a}_{\omega}\,d\omega$
where now
$[\hat{a}_{\omega},\hat{a}_{\omega^{\prime}}^{\ast}]=\delta(\omega-\omega^{\prime})$.

###  II.2 Bargmann-Segal (Complex Wave) Representation

For a given vector $|f\rangle\in\mathfrak{f}_{B}$, its exponential vector is
defined as

$\displaystyle|e^{f}\rangle_{B}=e^{\hat{a}(f)}|\mathrm{vac}\rangle_{B}\equiv|\mathrm{vac}\rangle\oplus|f\rangle\oplus\big{(}\frac{1}{\sqrt{2!}}|f\rangle\otimes|f\rangle\big{)}\oplus\cdots.$
(4)

In particular the vacuum state corresponds to
$|\mathrm{vac}\rangle_{B}\equiv|e^{f}\rangle_{B}$. The exponential vectors
form a total subset in $\mathfrak{H}_{B}$ and we note that

$\displaystyle\langle e^{f}|e^{g}\rangle_{B}=e^{\langle f|g\rangle}.$ (5)

The Complex Wave (or Bargmann-Segal) representation of a vector
$|\Psi\rangle\in\mathfrak{H}_{B}$ is in terms of the function $\tilde{\Psi}$
given by

$\displaystyle\tilde{\Psi}\left(f\right)=\langle e^{f}\,|\Psi\rangle.$ (6)

The mapping $f\mapsto\langle e^{f}|\Psi\rangle$ is naturally anti-holomorphic.

###### Definition 1

The pre-measure $\tilde{\mathbb{P}}$ on the one-particle space
$\mathfrak{f}_{B}$ by

$\displaystyle\int_{\mathfrak{f}_{B}}e^{\langle g_{1}|f\rangle}\,e^{\langle
f|g_{2}\rangle}\,\tilde{\mathbb{P}}\left[df\right]=e^{\langle
g_{1}|g_{2}\rangle}.$ (7)

We remark that for a finite assembly of oscillators the one-particle Hilbert
space is $\mathfrak{f}_{B}\cong\mathbb{C}^{\Omega}$ which can be considered as
the measurable space $\mathbb{R}^{2\Omega}$. In this case, $\mathbb{P}$ is a
probability measure and is given explicitly by

$\displaystyle\tilde{\mathbb{P}}\left[df\right]=\prod_{\omega}\left(\frac{1}{\pi}e^{-|f(\omega)|^{2}}df(\omega)^{\prime}df(\omega)^{\prime\prime}\right)$
(8)

where we write each variable $f(\omega)$ in terms of its real and imaginary
parts as $f(\omega)=f(\omega)^{\prime}+if(\omega)^{\prime\prime}$.

As a technical aside, $\tilde{\mathbb{P}}$ will define a Gaussian measure only
for the finite dimensional case. In the infinite-dimensional case, we should
extend $\tilde{\mathbb{P}}$ to a $\sigma$-additive measure over a larger space
$\mathfrak{f}_{B}^{>}$. As before, for each $|\Psi\rangle\in\mathfrak{H}_{B}$
we arrive at functions $:f\mapsto\tilde{\Psi}(f)$ this time defined for
$\alpha\in\mathfrak{f}_{B}^{>}$. These functions are properly square-
integrable with respect to the measure (which we again denote as
$\tilde{\mathbb{P}}$), however, one a dense subset of these functions will now
be continuous and anti-holomorphic. See Gough and Kupsch GK  for details.

In general, we have

$\displaystyle\langle\Phi|\Psi\rangle_{B}=\int_{\mathfrak{f}_{B}}\tilde{\Phi}(f)\tilde{\Psi}(f)\,\tilde{\mathbb{P}}\left[df\right],$
(9)

with (7) being the special case $\Phi=e^{g_{1}}$ and $\Psi=e^{g_{2}}$. We also
have the resolution of identity

$\displaystyle\int_{\mathfrak{f}_{B}}|e^{f}\rangle\langle
e^{f}|\,\,\tilde{\mathbb{P}}\left[df\right]=\hat{I}_{B}.$ (10)

###  II.3 The System-Bath Interaction

The total Hamiltonian for our system and bath will be of the form

$\displaystyle\hat{H}_{\mathrm{Tot.}}=\hat{H}\otimes\hat{I}_{B}+\hat{I}_{S}\otimes\hat{H}_{B}+i\hat{L}\otimes\hat{a}(g)^{\ast}-i\hat{L}^{\ast}\otimes\hat{a}(g).$
(11)

where $\hat{H}=\hat{H}^{\ast},\hat{L}$ are bounded operators of the system,
$|g\rangle\in\mathfrak{f}_{B}$ and $\hat{H}_{B}=d\Gamma(\hat{h}_{B})$ for some
self-adjoint operator $\hat{h}_{B}$ on $\mathfrak{f}_{B}$. We note that the
annihilation operators evolve under the bath Hamiltonian according to

$\displaystyle
e^{it\hat{H}_{B}}\hat{a}(g)e^{-it\hat{H}_{B}}\equiv\hat{a}(g_{t}),\,\text{where}\,|g_{t}\rangle=e^{it\hat{h}_{B}}|g\rangle.$
(12)

The bath correlation function is then given by

$\displaystyle K(t,s)\triangleq\langle g_{t}|g_{s}\rangle=\langle
g|e^{-i(t-s)\hat{h}_{B}}g\rangle.$ (13)

We note that the kernel is sesquilinear, i.e.,
$K\left(t,s\right)^{\ast}=K\left(s,t\right)$, and defines a positive definite
kernel. We note that kernels of the form (13) cannot be both non-trivial and
causal. Causality is the condition that $K(t,\tau)\equiv 0$ for $\tau<t$, but
by the sesquilinear symmetry we would also have $K(t,\tau)=0$ for $t<\tau$
too. The only nontrivial and causal model is the memoriless kernel
$K(t,s)=k_{0}\,\delta(t-s)$.

###### Definition 2

The vector $g\in\mathfrak{f}_{B}$ is said to be faithful if the measure
$\mu_{g}$, with Fourier transform $\int_{-\infty}^{\infty}e^{i\omega
t}\,\mu_{g}[d\omega]=\langle g|e^{it\hat{h}_{B}}|g\rangle$, has support equal
to the spectrum of $\hat{h}_{B}$.

This basically means that all the bath modes enter in the interaction. In
general, $\mathfrak{h}_{B}$ mat be written as the direct sum of a maximal
Hilbert space for which $g$ is faithful (the modes that couple to the system)
and its orthogonal complement. By ignoring this complement we can always
assume our setup is faithful.

###### Remark 3

For instance, the kernel for a discrete assembly of oscillators will be
$K(t,s)=\sum_{\omega\in\Omega}|g(\omega)|^{2}e^{-i(t-s)\omega}$ and
faithfulness would just mean that $g(\omega)\neq 0$ for all $\omega\in\Omega$.

###  II.4 The Open System Dynamics

It is convenient to move to the interaction picture using the unitary family
$U_{t}=e^{+itH_{0}}e^{-itH_{\mathrm{Tot.}}}$ where
$\hat{H}_{0}=\hat{I}_{S}\otimes\hat{H}_{B}$ and here we find the differential
equation

$\displaystyle\frac{d}{dt}\hat{U}_{t}=-i\hat{\Upsilon}_{t}\,\hat{U}_{t},\qquad-i\hat{\Upsilon}_{t}=-i\hat{H}\otimes\hat{I}_{B}+\hat{L}\otimes\hat{Z}(t)-\hat{L}^{\ast}\otimes\hat{Z}(t)^{\ast},$
(14)

where $\hat{Z}(t)=\hat{a}(g_{t})^{\ast}$. It is important to note that
$\hat{Z}(t)$ is not Hermitean and therefore not an observable! Indeed, the
canonical commutation relations imply that

$\displaystyle[\hat{Z}(t)^{\ast},\hat{Z}(s)]=K(t,s)\,\hat{I}_{B}.$ (15)

(The reason for taking $\hat{Z}(t)$ to be the creator rather than the
annihilator is that we will be shortly using the anti-holomorphic Bargmann
representation.) We note the eigen-relation

$\displaystyle\bigg{(}\hat{Z}^{\ast}_{t}-\langle
g_{t}|f\rangle\bigg{)}\,|e^{f}\rangle_{B}=0.$ (16)

The two-point vacuum expectations are

$\displaystyle\langle\mathrm{vac}|\hat{Z}_{t}^{\ast}\hat{Z}_{s}|\mathrm{vac}\rangle=K\left(t,s\right),\quad\langle\mathrm{vac}|\hat{Z}_{t}\hat{Z}_{s}|\mathrm{vac}\rangle=\langle\mathrm{vac}|\hat{Z}_{t}^{\ast}\hat{Z}_{s}^{\ast}|\mathrm{vac}\rangle=\langle\mathrm{vac}|\hat{Z}_{t}\hat{Z}_{s}^{\ast}|\mathrm{vac}\rangle=0.$
(17)

It is convenient to introduce the following operators

$\displaystyle j_{t}(\hat{X})$ $\displaystyle=$
$\displaystyle\hat{U}_{t}^{\ast}(\hat{X}\otimes\hat{I}_{B})\hat{U}_{t},$ (18)
$\displaystyle\hat{Z}_{\mathrm{in}}(t)$ $\displaystyle=$
$\displaystyle\hat{I}_{S}\otimes\hat{Z}(t),$ (19)
$\displaystyle\hat{Z}_{\mathrm{out}}(t)$ $\displaystyle=$
$\displaystyle\hat{U}_{t}^{\ast}\hat{Z}_{\mathrm{in}}(t)\hat{U}_{t}.$ (20)

######  Proposition 4 (Equal time commutation relations)

For each system operator $\hat{X}$, the operator $j_{t}(\hat{X})$ commutes
with both $\hat{Z}_{\mathrm{out}}(t)$ and $\hat{Z}_{\mathrm{out}}(t)^{\ast}$.

To see this, we note that, for instance,

$\displaystyle[j_{t}(\hat{X}),\hat{Z}_{\mathrm{out}}(t)]=\hat{U}_{t}^{\ast}[\hat{X}\otimes\hat{I}_{B},\hat{I}_{S}\otimes\hat{Z}(t)]\hat{U}_{t}=0,\,\mathrm{etc.}$
(21)

We should emphasize that the outputs $\hat{Z}_{\mathrm{out}}(t)$ and
$\hat{Z}_{\mathrm{out}}(t)^{\ast}$ do not commute with $j_{s}(\hat{X})$ for
$s\neq t$! Moreover, while the inputs satisfy
$[\hat{Z}_{\mathrm{in}}(t)^{\ast},\hat{Z}_{\mathrm{in}}(s)]=K(t,s)\hat{I}$,
where $\hat{I}=\hat{I}_{S}\otimes\hat{I_{B}}$, there is no such simple
relation for the outputs.

######  Proposition 5 (Input-Output Relation)

The processes $\hat{Z}_{\mathrm{in}}(t)$ and $\hat{Z}_{\mathrm{out}}(t)$ are
related by the input-output relations

$\displaystyle\hat{Z}_{\mathrm{out}}(t)^{\ast}=\hat{Z}_{\mathrm{in}}(t)^{\ast}+\int_{0}^{t}K(t,\tau)j_{\tau}(\hat{L})d\tau.$
(22)

Proof. Fix $f\in\mathfrak{f}_{B}$, then
$\frac{d}{dt}\,\hat{U}_{t}^{\ast}(\hat{I}_{S}\otimes\hat{a}(f)\big{)}\hat{U}_{t}=\langle
f|g_{t}\rangle\,\hat{U}_{t}^{\ast}\big{(}\hat{L}\otimes\hat{I}_{B}\big{)}U_{t}=\langle
f|g_{t}\rangle\,j_{t}(\hat{L})$, and integrating leads to

$\displaystyle\hat{U}_{t}^{\ast}(\hat{I}_{S}\otimes\hat{a}(f)\big{)}\hat{U}_{t}=\hat{I}_{S}\otimes\hat{a}(f)+\int_{0}^{t}\langle
f|g_{\tau}\rangle\,j_{\tau}(\hat{L})\,d\tau.$ (23)

Setting $f=g_{t}$ gives the desired result.

As a corollary, we obtain the following version of the Ehrenfest equations for
system operators.

######  Proposition 6 (Ehrenfest equation)

We fix the initial state to be
$|\phi\rangle_{S}\otimes|\mathrm{vac}\rangle_{B}$ and set $\langle
j_{t}(\hat{X})\rangle=\langle\phi\otimes\mathrm{vac}|j_{t}(\hat{X})\,\phi\otimes\mathrm{vac}\rangle$.
Then

$\displaystyle\frac{d}{dt}\langle j_{t}(\hat{X})\rangle=\langle
j_{t}(\frac{1}{i}[\hat{X},\hat{H}])\rangle+\int_{0}^{t}d\tau\,K(t,\tau)\langle
j_{t}([\hat{L}^{\ast},\hat{X}])j_{\tau}(\hat{L})\rangle+\int_{0}^{t}d\tau\,K(t,\tau)^{\ast}\langle
j_{\tau}(\hat{L}^{\ast})j_{t}([\hat{X},\hat{L}])\rangle.$

Proof. We start with the equation of motion
$\frac{d}{dt}j_{t}(\hat{X})=\hat{U}_{t}^{\ast}\frac{1}{i}[\hat{X}\otimes\hat{I}_{B},\hat{\Upsilon}_{t}]\hat{U}_{t}$,
which implies

$\displaystyle\frac{d}{dt}j_{t}(\hat{X})=j_{t}(\frac{1}{i}[\hat{X},\hat{H}])+j_{t}([\hat{X},\hat{L}])\hat{Z}_{\mathrm{out}}(t)+j_{t}([\hat{L}^{\ast},\hat{X}])\hat{Z}_{\mathrm{out}}(t)^{\ast}.$
(24)

Note that the equal time commutation relations allow us to commute either
$\hat{Z}_{\mathrm{out}}(t)$ or $\hat{Z}_{\mathrm{out}}(t)^{\ast}$ with the
terms multiplying them. The desired equation is obtained by replacing the
output processes with the inputs using (22) and using the fact that
$\hat{Z}_{\mathrm{in}}(t)^{\ast}\,|\phi\otimes\mathrm{vac}\rangle\equiv 0$
since $\hat{Z}_{\mathrm{in}}(t)^{\ast}$ is an annihilator.

###  II.5 Caldeira-Leggett Baths

We take a continuum of bath modes. The bath Hamiltonian is given by

$\displaystyle\hat{H}_{B}=\frac{1}{2}\int_{0}^{\infty}d\omega\bigg{\\{}\frac{\omega}{J(\omega)}\hat{\pi}(\omega)^{2}+\omega
J(\omega)\hat{q}(\omega)^{2}\bigg{\\}}$ (25)

where $\hat{q}(\omega)$ and $\hat{\pi}(\omega)$ satisfy canonical commutation
relations
$[\hat{q}(\omega),\hat{\pi}(\omega^{\prime})]=i\,\delta(\omega-\omega^{\prime})$.
Here $J$ is a function known as the $J$ spectral density function. We
introduce the mode operators as

$\displaystyle\hat{a}_{\omega}=\sqrt{\frac{J(\omega)}{2}}\hat{q}(\omega)+i\frac{1}{\sqrt{2J(\omega)}}\hat{\pi}(\omega)$
(26)

in which case
$\hat{H}_{B}=\int_{0}^{\infty}\omega\,\hat{a}(\omega)^{\ast}\hat{a}(\omega)\,d\omega$.

We now couple our system $S$ coupled to the bath through its position
coordinate $\hat{q}$: the total Hamiltonian will be taken to be of Caldeira-
Leggett form CL

$\displaystyle\hat{H}_{S+B}$ $\displaystyle=$
$\displaystyle\hat{H}\otimes\hat{I}_{B}+\frac{1}{2}\int_{0}^{\infty}d\omega\bigg{\\{}\frac{\omega}{J(\omega)}\big{[}\hat{I}_{S}\otimes\hat{\pi}(\omega)-\sqrt{\frac{2}{\pi}}\hat{q}\otimes\hat{I}_{B}\big{]}^{2}+\omega
J(\omega)\hat{I}_{S}\otimes\hat{q}(\omega)^{2}\bigg{\\}}$ (27)
$\displaystyle=$
$\displaystyle\hat{H}\otimes\hat{I}_{B}+\hat{I}_{S}\otimes\hat{H}_{B}+\hat{V}_{SB}.$

where the interaction is
$\hat{V}_{SB}=i\hat{q}\otimes\int_{0}^{\infty}d\omega\sqrt{\frac{J(\omega)}{\pi}}\big{(}\hat{a}_{\omega}-\hat{a}_{\omega}^{\ast}\big{)}$.

This can be put in the standard form (11) with $\hat{L}=-i\hat{q}$ and
$g(\omega)=\sqrt{J(\omega)/\pi}$. At this stage, we identify the complex wave
representation. Our one particle Hilbert space is
$\mathfrak{f}_{B}=L^{2}(\mathbb{R}_{+},d\omega)$ and for every $f=f(\omega)$
in we associate a complex trajectory
$\zeta(t)=\int_{0}^{\infty}\sqrt{J(\omega)/\pi}e^{i\omega
t}f(\omega)^{\ast}\,d\omega$. The corresponding quantum process is

$\displaystyle\hat{Z}(t)=\int_{0}^{\infty}\sqrt{J(\omega)/\pi}e^{i\omega
t}a_{\omega}^{\ast}\,d\omega,$ (28)

and the kernel will be the Fourier transform of the $J$ spectral density:

$\displaystyle
K(t,s)=\frac{1}{\pi}\int_{0}^{\infty}J(\omega)e^{-i\omega(t-s)}\,d\omega.$
(29)

##  III Complex Trajectories

The eigen-relation (16) may be alternatively written as

$\displaystyle\bigg{(}\hat{Z}^{\ast}_{t}-\zeta_{t}(f)^{\ast}\bigg{)}\,|e^{f}\rangle_{B}=0.$
(30)

and we refer to $t\to\zeta_{t}(f)$ is the complex trajectory associated with
$f\in\mathfrak{f}_{B}$.

###  III.1 Reproducing Kernel Hilbert Space Formalism

In the following, $\mathbb{T}$ will denote an interval in $\mathbb{R}$. We fix
a positive definite kernel $K:\mathbb{T}\times\mathbb{T}\mapsto\mathbb{C}$.
For each $t\in\mathbb{T}$, a function
$\mathbb{k}_{t}:\mathbb{T}\mapsto\mathbb{C}$ is then defined by

$\displaystyle\mathbb{k}_{t}(\cdot)\triangleq K(t,\cdot).$ (31)

We recall that a Hilbert space $\mathscr{H}$ of complex-valued functions on
$\mathbb{T}$ forms a reproducing kernel Hilbert space (RKHS) for the kernel
$K$ if $\mathbb{k}_{t}\in\mathscr{H}$ for each $t\in\mathbb{T}$ and we have
the reproducing propertyRKHS

$\displaystyle\langle\mathbb{k}_{t},f\rangle_{\mathscr{H}}=f(t)$ (32)

for all $t\in\mathbb{T}$ and all $f\in\mathscr{H}$.

In this situation, the maps $\mathbb{k}_{t}$ are called representers, or
evaluation maps, and they play the role of Dirac delta-functions for the test
functions $\mathscr{H}$. Crucially the representers are bounded maps in the
RKHS setting. We also note the identity

$\displaystyle\langle\mathbb{k}_{t},\mathbb{k}_{s}\rangle_{\mathscr{H}}=\mathbb{k}_{s}(t)=K(s,t).$
(33)

A kernel is said to be derivable from a feature mapRKHS  if there exists a
Hilbert space $\mathfrak{f}$ (called the feature space) and a function
$g:\mathbb{T}\mapsto\mathbb{C}$ (called the feature map) such that

$\displaystyle K(t,s)\equiv\langle g_{t},g_{s}\rangle_{\mathfrak{f}}.$ (34)

###### Remark 7

The bath correlation kernel introduced in (13) is clearly derivable from a
feature map with $\mathfrak{f}_{B}$ as feature space. The set $\mathbb{T}$ may
generally be taken to be some interval of the time in $\mathbb{R}$ and we
shall fix $dt$ as Lebesgue measure. In order to construct an RKHS we need
further assumptions (Mercer conditions) which essentially allow us to write
the kernel in the form

$\displaystyle K(t,s)=\sum_{n}\lambda_{n}\phi_{n}(t)\phi_{n}(s)^{\ast}$ (35)

where the $\lambda_{n}$ are positive numbers and the $\phi_{n}$ form a
complete orthonormal basis for $L^{2}(\mathbb{T},dt)$. Here we require that
the integral operator $T_{K}$ given by
$\big{(}T_{K}u\big{)}(t)=\int_{\mathbb{T}}K\big{(}t,s\big{)}u(s)\,ds$ is
Hilbert-Schmidt in which case the $\lambda_{n}$ are the eigenvalues and the
$\phi_{n}$ are the eigen-functions normalized in $L^{2}(\mathbb{T},dt)$. Here
one solves the homogeneous Fredholm integral equation

$\displaystyle\int_{\mathbb{T}}K(t,s)\phi_{n}(d)\,ds=\lambda_{n}\,\phi_{n}(t).$
(36)

This is automatically satisfied when $\mathbb{T}$ is compact but otherwise
will require that the kernel has a finite-trace property.RKHS ; DUV

With the assumption that the kernel admits a Mercer expansion (35), we may
construct the RKHS $\mathscr{H}_{K}(\mathbb{T},dt)$ \- which we write as
$\mathscr{H}$ as shorthand when no confusion arise - as follows: (step 1) we
define $\mathscr{H}$ to be the set of all functions $u$ of the form
$u(t)=\sum_{n}u_{n}\,\phi_{n}(t)$ with
$\sum_{n}\frac{1}{\lambda_{n}}|u_{n}|^{2}<\infty$ and we endow this with the
inner-product

$\displaystyle\langle
u,v\rangle_{\mathscr{H}}\triangleq\sum_{n}\frac{1}{\lambda_{n}}u_{n}^{\ast}v_{n};$
(37)

(step 2) we see that the representers take the form

$\displaystyle\mathbb{k}_{t}(\cdot)=\sum_{n}\lambda_{n}\phi_{n}(t)^{\ast}\,\phi_{n}(\cdot),$
(38)

so that the coefficients of $\mathbb{k}_{t}$ are
$\lambda_{n}\phi_{n}(t)^{\ast}$; (step 3) for $u\in\mathscr{H}$ we obtain the
reproducing property

$\displaystyle\langle\mathbb{k}_{t},u\rangle_{\mathscr{H}}=\sum_{n}\frac{1}{\lambda_{n}}\big{(}\lambda_{n}\phi_{n}(t)^{\ast}\big{)}^{\ast}u_{n}=\sum_{n}u_{n}\phi_{n}(t)\equiv
u(t).$ (39)

As a check, we note that

$\displaystyle\langle\mathbb{k}_{t},\mathbb{k}_{s}\rangle_{\mathscr{H}}=\sum_{n}\frac{1}{\lambda_{n}}\big{(}\lambda_{n}\phi_{n}(t)^{\ast}\big{)}^{\ast}\lambda_{n}\phi_{n}(s)^{\ast}=\sum_{n}\lambda_{n}\phi_{n}(t)\phi_{n}(s)^{\ast}\equiv
K(t,s).$ (40)

Note that the “delta-function” on $\mathbb{T}$ is given by

$\displaystyle\delta_{\mathbb{T}}(t-s)\equiv\sum_{n}\phi_{n}(t)\phi_{n}(s)^{\ast}.$
(41)

###  III.2 Hilbert Space of Complex Trajectories

###### Definition 8

For each $f\in\mathfrak{f}$, we define its associated complex trajectory over
the time interval $\mathbb{T}$ to be the function
$\zeta(f):\mathbb{T}\mapsto\mathbb{C}:t\mapsto\zeta_{t}(f)$ where

$\displaystyle\zeta_{t}(f)=\langle f|g_{t}\rangle_{\mathfrak{f}}.$ (42)

The space of complex trajectories will be denoted as
$\mathscr{C}_{K}(\mathbb{T},dt)$.

We note that $\zeta(f)$ is defined to be anti-holonomic in $f$ and this is the
natural choice for the Bargmann representation which we will use in the next
section. Unfortunately, it makes the following result slightly awkward to
state.

###### Theorem 9

The set of complex trajectories, $\mathscr{C}_{K}(\mathbb{T},dt)$, forms a
Hilbert subspace of the RKHS $\mathscr{H}_{K}(\mathbb{T},dt)$ (inheriting the
same inner product) and the map $\zeta$ is conjugate-linear isometry into the
feature space $\mathfrak{f}_{B}$.

$\displaystyle\langle\zeta(f_{1})^{\ast},\zeta(f_{2})^{\ast}\rangle_{\mathscr{H}}=\langle
f_{1}|f_{2}\rangle_{\mathfrak{f}}.$ (43)

The space $\mathscr{C}_{K}(\mathbb{T},dt)$ is conjugate-isomorphic to
$\mathfrak{f}_{B}$ when the set up is faithful.

Proof. Let us introduce the following vectors in the feature space:

$\displaystyle|\pi_{n}\rangle_{\mathfrak{f}}\triangleq\int_{\mathbb{T}}\phi_{n}(t)|g_{t}\rangle_{\mathfrak{f}}\,dt,$
(44)

where the integral is understood as a Gel’fand-Pettis integral. We see that

$\displaystyle\langle\pi_{n}|\pi_{m}\rangle_{\mathfrak{f}}=\int_{\mathbb{T}}dt\int_{\mathbb{T}}ds\,\phi_{n}(t)^{\ast}\phi_{n}(s)\,\langle
g_{t}|g_{s}\rangle_{\mathfrak{f}}=\int_{\mathbb{T}}dt\int_{\mathbb{T}}ds\,\phi_{n}(t)^{\ast}K(t,s)\phi_{n}(s)=\lambda_{n}\,\delta_{nm}.$
(45)

In particular, this leads to the resolution of identity
$\sum_{n}\frac{1}{\lambda_{n}}|\pi_{n}\rangle\langle\pi_{n}|=I_{\mathfrak{f}}$.

The complex trajectories then have the expansion

$\displaystyle\zeta_{t}(f)^{\ast}=\sum_{n}\langle\phi_{n},\zeta(f)^{\ast}\rangle_{L^{2}}\,\phi_{n}(t)=\sum_{n}\bigg{(}\int_{\mathbb{T}}\phi_{n}(s)^{\ast}\langle
g_{t}|f\rangle_{\mathfrak{f}}\bigg{)}\phi_{n}(t)=\sum_{n}\langle\pi_{n}|f\rangle_{\mathfrak{f}}\,\phi_{n}(t)$
(46)

and therefore

$\displaystyle\langle\zeta(f_{1})^{\ast},\zeta(f_{2})^{\ast}\rangle_{\mathscr{H}}=\sum_{n}\frac{1}{\lambda_{n}}\langle
f_{1}|\pi_{n}\rangle_{\mathfrak{f}}\langle\pi_{n}|f_{2}\rangle_{\mathfrak{f}}=\langle
f_{1}|f_{2}\rangle_{\mathfrak{f}}.$ (47)

Generally speaking, the map $\zeta$ from the feature space to the RKHS may be
many-to-one.

###  III.3 Quantum Karhunen-Loève Theorem

Under the condition that the kernel is a Mercer kernel, the inputs can be
given the expansion

$\displaystyle\hat{Z}_{t}^{\ast}=\sum_{n}\hat{z}_{n}^{\ast}\phi_{n}(t),$ (48)

where the coefficient operators satisfy the commutation relations

$\displaystyle[\hat{z}_{n}^{\ast},\hat{z}_{m}]=\lambda_{n}\,\delta_{nm}.$ (49)

Specifically, these are given
$\hat{z}_{n}=\int_{\mathbb{T}}\phi_{n}(t)\hat{Z}_{t}\,dt$. One checks that

$\displaystyle[\hat{z}_{n}^{\ast},\hat{z}_{m}]=\int_{\mathbb{T}}dt\int_{\mathbb{T}}ds\,\phi_{n}(t)^{\ast}[\hat{Z}_{t}^{\ast},\hat{Z}_{s}]\phi_{n}(s)=\int_{\mathbb{T}}dt\int_{\mathbb{T}}ds\,\phi_{n}(t)^{\ast}\,K(t,s)\phi_{n}(s)=\lambda_{n}\,\delta_{nm}.$

The convergence will be uniform in $t$ in the $L^{2}$-norm. This is
essentially a quantum analogue of the classical Karhunen-Loève Theorem. It is
natural to introduce the bath annihilators

$\displaystyle\hat{b}_{n}\triangleq\frac{1}{\sqrt{\lambda}_{n}}\hat{z}_{n}^{\ast},$
(50)

and we have the canonical commutation relations
$[\hat{b}_{n},\hat{b}_{m}^{\ast}]=\delta_{nm}$.

The interaction picture Hamiltonian $\hat{\Upsilon}_{t}$ may then be written
as

$\displaystyle-i\hat{\Upsilon}_{t}=-i\hat{H}\otimes\hat{I}_{B}+\sum_{n}\sqrt{\lambda_{n}}\bigg{\\{}\hat{L}\otimes\hat{b}_{n}^{\ast}\,\phi_{n}(t)^{\ast}-\hat{L}^{\ast}\otimes\hat{b}_{n}\,\phi_{n}(t)\bigg{\\}},$
(51)

###### Remark 10

What the Karhunen-Loève expansion does is to decorrelate the process
$\hat{Z}_{t}$. Specifically, the auto-correlation was described by the kernel
$K(t,s)$ however this is replaced by the modes $\hat{b}_{n}$ which are
independent oscillators in their ground state. Note that for a finite bath, we
go from a finite number of modes $\hat{a}_{\omega}$ to an infinite number of
the $\hat{b}_{n}$.

###### Remark 11

The annihilation and creation processes $\hat{A}_{t},\hat{A}_{t}^{\ast}$ of
the Hudson-Parthasarathy theory satisfy
$[\hat{A}_{t},\hat{A}_{s}^{\ast}]=t\wedge s$. restricting to the interval
$\mathbb{T}=[0,T]$, the kernel $K(t,s)=t\wedge s$ has eigen-functions
$\phi_{n}(t)=\sqrt{\frac{2}{T}}\sin(\omega_{n}t)$ for $n=1,2,3,\cdots$, where
$\omega_{n}=\frac{\pi}{T}(n-\frac{1}{2})$, and corresponding eigenvalues
$\lambda_{n}=1/\omega_{n}^{2}$. We therefore find that

$\displaystyle\hat{A}_{t}\equiv\sum_{n=1}^{\infty}\frac{\sqrt{2T}}{\pi(n-\frac{1}{2})}\sin\bigg{(}\frac{\pi(n-\frac{1}{2})t}{T}\bigg{)}\,\hat{b}_{n}.$
(52)

This expansion was essentially given in QKL . Note however that, for the
markovian dynamics, the kernel is a $\delta$-function and the driving input
processes are quantum white noises, that is, the formal derivatives of these
processes, see section IV.3.

###  III.4 Causal Form

Let us set the time interval to be $\mathbb{T}=[0,T]$ where $T$ is a finite
time horizon. We shall write the eigenvalues as $\lambda_{n}(T)$ and the
eigen-functions as $\phi_{n}(\cdot,T)$ to emphasize the dependence on $T$. We
then have the expansion

$\displaystyle\hat{Z}^{\ast}_{t}\equiv\sum_{n}\sqrt{\lambda_{n}(T)}\phi_{n}(t;T)\,\hat{b}_{n}(T),$
(53)

for each $0\leq t\leq T$. One of the problems with this expansion is that
$z_{n}^{\ast}(T)=\lambda_{n}(T)\,\hat{b}_{n}(T)$ will be given by
$\int_{0}^{T}\phi_{n}(s)\hat{Z}_{s}\,ds$, so we are effectively computing
$\hat{Z}_{t}^{\ast}$ using both its past values $0\leq s<t$ and its future
values $s<t\leq T$.

This may be avoided by using a running time horizon (set equal to the time
parameter $t$) rather than a fixed future time $T$. We therefore have the
causal form

$\displaystyle\hat{Z}^{\ast}_{t}\equiv\sum_{n}\sqrt{\lambda_{n}(t)}\phi_{n}(t;t)\,\hat{b}_{n}(t).$
(54)

This has the effect of replacing the kernel with its causal version and we
have

$\displaystyle
K(t,s)\,\theta(t-s)=\sum_{n}\sqrt{\lambda_{n}(t)}\,\phi_{n}(t;t)\phi_{n}(s;t)^{\ast},$
(55)

where $\theta(\cdot)$ is the Heaviside step function. The support of each
$\phi_{n}(\cdot;t)$ being contained in $[0,t]$.

###  III.5 Complex Trajectories from The Bargmann Representation

For technical simplicity, we first consider the case where
$\mathfrak{f}_{B}=\mathbb{C}^{\Omega}$ with $\Omega$ the finite set of (non-
degenerate) frequencies. The canonical operators are given in the Bargmann
representation by

$\displaystyle\hat{a}_{\omega}^{\ast}\tilde{\Psi}\left(f\right)=f(\omega)^{\ast}\,\tilde{\Psi}\left(f\right),\qquad\hat{a}_{\omega}\tilde{\Psi}\left(f\right)=\frac{\partial}{\partial
f_{\omega}^{\ast}}\tilde{\Psi}\left(f\right).$ (56)

Note that

$\displaystyle\hat{a}(g)^{\ast}\,\tilde{\Psi}\left(f\right)\equiv\langle
f|g\rangle\,\tilde{\Psi}\left(f\right).$ (57)

We shall assume that $g(\omega)\neq 0$ for each $\omega\in\Omega$. Under the
free evolution of the bath Hamiltonian $H_{B}$, we find that
$\hat{a}_{\omega}\to e^{-i\omega t}\hat{a}_{\omega}$ and so

$\displaystyle\hat{Z}_{t}=\sum_{\omega}g(\omega)e^{i\omega
t}\hat{a}_{\omega}^{\ast}\equiv\hat{a}(g_{t})^{\ast}.$ (58)

with $g_{t}(\omega)=e^{i\omega t}g(\omega)$. The eigen-relation (16) may be
written as

$\displaystyle\bigg{(}\hat{Z}_{t}^{\ast}-\zeta_{t}(f)^{\ast}\bigg{)}\,|e^{f}\rangle_{B}=0,$
(59)

where we recall the complex trajectory from Definition 8

$\displaystyle\zeta_{t}(f)=\langle
f|g_{t}\rangle=\sum_{\omega}g(\omega)e^{i\omega t}f(\omega)^{\ast}.$ (60)

We see that, in the Bargmann representation, $\hat{Z}(t)$ is the operator of
multiplication by $\zeta_{t}(\cdot)$:

$\displaystyle\hat{Z}(t)\,\tilde{\Psi}(f)=\zeta_{t}(f)\,\tilde{\Psi}(f).$ (61)

This implies to the more general case where the feature space
$\mathfrak{f}_{B}$ is not finite dimensional.

###### Definition 12

Suppose that the set up is faithful so that the mapping
$\zeta:\mathfrak{f}_{B}\mapsto\mathscr{C}_{K}(\mathbb{T},dt)$ is invertible.
Then, for each Bargamann function $\tilde{\Psi}$, we define the functions
$\Psi$ by

$\displaystyle\Psi(\cdot)=\tilde{\Psi}\circ\zeta^{-1}.$ (62)

Additionally, we can endow $\mathscr{C}_{K}(\mathbb{T},dt)$ with the pull-back
pro-measure

$\displaystyle\mathbb{P}=\tilde{\mathbb{P}}\circ\zeta^{-1}.$ (63)

We may extend $\mathbb{P}$ to a probability measure as outlined earlier.

In detail, for each complex trajectory $\zeta\equiv\zeta(f)$, we have the
change of variable $\Psi(\zeta)\equiv\tilde{\Psi}(f)$.

In this way, $\zeta_{t}=\zeta_{t}\left(\cdot\right)$ is the coordinate for
trajectories $\mathscr{C}_{K}(\mathbb{T},dt)$. in and $\hat{Z}(t)$ may be
thought of as the operator corresponding to multiplication by
$\zeta_{t}(\cdot)$. Indeed, for the case where $\mathbb{T}=\mathbb{R}$, the
adjoint operator $\hat{Z}(t)^{\ast}$ may be represented as a functional
differential operator:

$\displaystyle\hat{Z}(t)^{\ast}\equiv\int_{-\infty}^{\infty}d\tau\,K\left(t,\tau\right)\frac{\delta}{\delta\zeta_{\tau}}.$
(64)

###### Remark 13

It is easy to see that the correct commutation relations (15) follow from the
representation of $\hat{Z}(t)$ with multiplication by $\zeta_{t}$ and
$\hat{Z}(t)^{\ast}$ with 64. However, it is instructive to repeat here the
argument from DS97  for a finite dimensional feature space. We have

$\displaystyle\hat{Z}(t)^{\ast}\tilde{\Psi}\left(f^{\ast}\right)=\sum_{\omega}g(\omega)^{\ast}e^{-i\omega
t}\,\frac{\partial}{\partial
f(\omega)^{\ast}}\tilde{\Psi}\left(f^{\ast}\right).$

The transform may be inverted to give
$f(\omega)^{\ast}=\frac{1}{2\pi}\frac{1}{g(\omega)}\int_{-\infty}^{\infty}\zeta_{\tau}(t)e^{-i\omega\tau}d\tau$
and so $\frac{\delta
f(\omega)^{\ast}}{\delta\zeta_{\tau}}\equiv\frac{1}{2\pi}\frac{1}{g(\omega)}e^{-i\omega\tau}$.
It follows that

$\displaystyle\int_{-\infty}^{\infty}d\tau\,K\left(t,\tau\right)\frac{\delta}{\delta\zeta_{\tau}}=\int_{-\infty}^{\infty}d\tau\,\sum_{\omega^{\prime}}|g(\omega^{\prime})|^{2}e^{-i\omega^{\prime}(t-\tau)}\sum_{\omega}\frac{e^{-i\omega\tau}}{2\pi\,g(\omega)}\frac{\partial}{\partial
f(\omega)^{\ast}}\equiv\sum_{\omega}g(\omega)^{\ast}e^{i\omega
t}\,\frac{\partial}{\partial f(\omega)^{\ast}}.$

In the special case here, where $\Omega$ is a finite set of frequencies and
$g$ is faithful, we may also construct an inverse kernel.

###### Proposition 14

For a finite dimensional feature space in the faithful case, we can construct
the inverse kernel

$\displaystyle
G\left(t,s\right)=\frac{1}{2\pi}\sum_{\omega\in\Omega}\frac{1}{|g(\omega)|^{2}}e^{-i\omega(t-s)},$
(65)

that is,
$\int_{-\infty}^{\infty}G\left(t,\tau\right)K\left(\tau,s\right)=\delta\left(t-s\right)$.
Moreover, we have the identity

$\displaystyle\sum_{\omega}|f_{\omega}|^{2}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\zeta_{t}(f)^{\ast}G\left(t,s\right)\zeta_{s}(f)\,dtds.$
(66)

##  IV Complex Trajectory Unravellings

###### Definition 15

Let $\mathcal{E}$ be a measurable space with measure $\mu$, then a family of
maps $\mathcal{E}\mapsto\mathfrak{h}_{S}:\xi\mapsto|\Psi_{t}(\xi)\rangle_{S}$,
parametrized by time $t$, is an unravelling of an open quantum system if we
have

$\displaystyle\langle
j_{t}(\hat{X})\rangle=\int_{\mathcal{E}}\langle\Psi_{t}(\xi)|\hat{X}|\Psi_{t}(\xi)\rangle_{S}\,\mu[d\xi].$
(67)

We will now construct an unravelling for the open system dynamics encountered
in subsection II.3 using the Bargmann representation.

###  IV.1 Complex Wave Representation for the Open System

We now consider a system with Hilbert space $\mathfrak{h}_{S}$ coupled to a
the assembly of oscillators which acts as a bath. Any state
$|\Psi\rangle\in\mathfrak{h}_{S}\otimes\mathfrak{H}_{B}$ can be given a hybrid
complex wave representation as a map
$:\mathbb{C}^{\Omega}\mapsto\mathfrak{h}_{S}$ taking
$f\mapsto|\tilde{\Psi}(f)\rangle_{S}$ where the complex wave representation is
used for the bath.

It is convenient to parametrize using complex trajectories $\zeta$ rather than
the amplitudes $f$: for $\zeta\in\mathscr{C}_{K}(\mathbb{T},dt)$ we shall use
the form $|\Psi(\zeta)\rangle_{S}\equiv|\tilde{\Psi}(f_{\zeta})\rangle_{S}$.
The inner product is then

$\displaystyle\langle\Phi|\Psi\rangle=\int_{\mathbb{C}^{\Omega}}\langle\tilde{\Phi}(f)|\tilde{\Psi}(f)\rangle_{S}\,\tilde{\mathbb{P}}\left[df\right]=\int_{\mathscr{C}_{K}(\mathbb{T},dt)}\langle\Phi(\zeta)|\Psi(\zeta)\rangle_{S}\,\mathbb{P}\left[d\zeta\right].$
(68)

Generally speaking, the vector states $|\Psi(\zeta)\rangle_{S}$ are not
normalized. Instead we have (switching to the distribution over the complex
trajectories $t\mapsto\zeta_{t}$ rather than the complex amplitudes
$\omega\mapsto f(\omega)$)

$\displaystyle\int_{\mathscr{C}_{K}(\mathbb{T},dt)}\|\Psi(\zeta)\|^{2}_{S}\,\,\mathbb{P}[d\zeta]=1.$
(69)

More generally, for a system observable $\hat{X}$,

$\displaystyle\langle\Phi|\hat{X}\otimes\hat{I}_{B}|\Psi\rangle=\int_{\mathscr{C}_{K}(\mathbb{T},dt)}\langle\Phi(\zeta)|\hat{X}|\Psi(\zeta)\rangle_{S}\,\mathbb{P}\left[d\zeta\right].$
(70)

###### Proposition 16

Let $|\Psi_{t}(\zeta)\rangle_{S}$ be the hybrid complex wave representation of
the the vector
$|\Psi_{t}\rangle=\hat{U}_{t}\,|\phi\otimes\mathrm{vac}\rangle$. We have that

$\displaystyle\frac{\delta}{\delta\zeta_{\tau}}|\Psi_{t}(\zeta)\rangle_{S}=0,\quad\mathrm{whenever}\quad\tau\notin[0,t].$
(71)

Proof. Set $\zeta=\zeta(f)$ and fix $\psi\in\mathfrak{h}_{S}$, then

$\displaystyle\langle\psi|\Psi_{t}(\zeta)\rangle_{S}=\langle\psi|\tilde{\Psi}_{t}(f)\rangle_{S}=\langle\psi\otimes
e^{f}|\vec{T}e^{-i\int_{0}^{t}\hat{\Upsilon}(s)ds}|\phi\otimes\mathrm{vac}\rangle=\langle\psi\otimes
e^{f}|\vec{T}e^{-i\int_{0}^{t}[\hat{\Upsilon}-\hat{L}\otimes\zeta_{s}]\,ds}|\phi\otimes\mathrm{vac}\rangle.$
(72)

Here we use the fact that $\langle e^{f}|=\langle\mathrm{vac}|e^{\hat{a}(f)}$
and that we may move the operator $e^{\hat{a}(f)}$ to the right side of
$\hat{U}_{t}$ with the effect of shifting the creator fields as
$\hat{a}(g_{s})^{\ast}\to\hat{a}(g_{s})^{\ast}+\langle
f|g_{t}\rangle\equiv\hat{a}(g_{s})^{\ast}+\zeta_{s}(f)$. It is then clear that
$\langle\psi|\Psi_{t}(\zeta)\rangle_{S}$ depends on $\zeta_{s}$ for $s\leq t$
only.

Collecting all our results so far together, we arrive at the following
conclusion.

###### Theorem 17

For the model encountered in subsection II.3, an unravelling is given on the
space $\mathscr{C}_{K}(\mathbb{T},dt)$ of complex trajectories with measure
$\mathbb{P}$. The unravelling $\zeta\mapsto|\Psi_{t}(\zeta)\rangle_{S}$
satisfies the Diói-Strunz differential equation (2) with initial condition
$|\Psi_{0}(\zeta)\rangle_{S}=|\phi\rangle_{S}$ for all $\zeta$.

Proof. The state $|\Psi_{t}\rangle=\hat{U}_{t}|\phi\otimes\mathrm{vac}\rangle$
satisfies the differential equation

$\displaystyle\frac{d}{dt}|\Psi_{t}\rangle=\big{(}-i\hat{H}\otimes\hat{I}_{B}+\hat{L}\otimes\hat{Z}(t)-\hat{L}^{\ast}\otimes\hat{Z}(t)^{\ast}\big{)}\,|\Psi_{t}\rangle.$
(73)

Writing this in the hybrid complex trajectory representation leads to

$\displaystyle\frac{d}{dt}|\Psi_{t}(\zeta)\rangle_{S}=\bigg{(}-i\hat{H}+\hat{L}\zeta_{t}-\hat{L}^{\ast}\int_{-\infty}^{\infty}d\tau\,K\left(t,\tau\right)\frac{\delta}{\delta\zeta_{\tau}}\bigg{)}|\Psi_{t}(\zeta)\rangle_{S},$
(74)

however, by Proposition 16 we may restrict the integration to get (2). As we
have seen, $\mathbb{P}$ is a Gaussian probability measure on the space
$\mathscr{C}_{K}(\mathbb{T},dt)$ of admissible complex trajectories, while
(70) ensures that we have an unravelling.

We may also derive an alternate form.

###### Theorem 18

The differential equation for the unravelling
$\zeta\mapsto|\Psi_{t}(\zeta)\rangle_{S}$ in Theorem 17 can alternatively be
written in the form

$\displaystyle\frac{d}{dt}|\Psi_{t}(\zeta)\rangle_{S}=\big{(}-i\hat{H}+\hat{L}\zeta_{t}\big{)}|\Psi_{t}(\zeta)\rangle_{S}-\int_{\mathscr{C}_{K}(\mathbb{T},dt)}\,\xi_{t}^{\ast}\,\hat{L}^{\ast}|\Psi_{t}(\zeta+\xi)\rangle_{S}\,\mathbb{P}[d\xi],$
(75)

again with the initial condition
$|\Psi_{0}(\zeta)\rangle_{S}=|\phi\rangle_{S}$ for all $\zeta$.

Proof. Let us fix an orthonormal basis $\\{e_{n}\\}$ for $\mathfrak{h}_{S}$.
We take $f\in\mathfrak{f}_{B}$ to be the one-particle state such that
$\zeta_{t}=\zeta_{t}(f)$ and we then have that

$\displaystyle\langle e_{n}|\Psi_{t}(\zeta)\rangle_{S}=\langle e_{n}\otimes
e^{f}|\hat{U}_{t}|\phi\otimes\mathrm{vac}\rangle=\langle
e_{n}\otimes\mathrm{vac}|\hat{U}_{t}^{(f)}|\phi\otimes\mathrm{vac}\rangle$
(76)

where $\hat{U}_{t}^{(f)}=e^{\hat{a}(f)}\hat{U}_{t}e^{-\hat{a}(f)}$. We see
that
$\frac{d}{dt}\hat{U}_{t}^{(f)}=\big{(}-i\hat{\Upsilon}_{t}+\zeta_{t}(f)\,\hat{L}\otimes\hat{I}_{B}\big{)}\hat{U}_{t}^{(f)}$,
where we recall that $\zeta_{t}(f)\equiv\langle f|g_{t}\rangle$, and so

$\displaystyle\frac{d}{dt}\langle e_{n}|\Psi_{t}(\zeta)\rangle_{S}=-i\langle
e_{n}\otimes\mathrm{vac}|\hat{\Upsilon}_{t}\,\hat{U}_{t}^{(f)}|\phi\otimes\mathrm{vac}\rangle+\langle
e_{n}|\hat{L}|\Psi_{t}(\zeta)\rangle_{S}\,\zeta_{t}.$ (77)

It remains to express the first term on the right hand side in a more
convenient form. Here we insert a resolution of identity

$\displaystyle-i\langle
e_{n}\otimes\mathrm{vac}|\hat{\Upsilon}_{t}\,\hat{U}_{t}^{(f)}|\phi\otimes\mathrm{vac}\rangle=-i\sum_{m}\int_{\mathfrak{f}_{B}}\tilde{\mathbb{P}}[dk]\langle
e_{n}\otimes\mathrm{vac}|\hat{\Upsilon}_{t}\,|e_{m}\otimes e^{k}\rangle\langle
e_{m}\otimes e^{k}|\hat{U}_{t}^{(f)}|\phi\otimes\mathrm{vac}\rangle$ (78)

and we see that

$\displaystyle-i\langle
e_{n}\otimes\mathrm{vac}|\hat{\Upsilon}_{t}\,|e_{m}\otimes
e^{k}\rangle=-i\langle e_{n}|\hat{H}|e_{m}\rangle_{S}-\langle
e_{n}|\hat{L}^{\ast}|e_{m}\rangle_{S}\,\langle g_{t}|k\rangle.$ (79)

The $\hat{H}$-term contribution to (78) is then

$\displaystyle-i\sum_{m}\int_{\mathfrak{f}_{B}}\tilde{\mathbb{P}}[dk]\langle
e_{n}\otimes\mathrm{vac}|\hat{H}|e_{m}\otimes e^{k}\rangle\langle e_{m}\otimes
e^{k}|\hat{U}_{t}^{(f)}|\phi\otimes\mathrm{vac}\rangle$ (80)

which reduces to
$-i\langle\hat{H}e_{n}|\tilde{\Psi}_{t}(f)\rangle_{S}=-i\langle
e_{n}|\hat{H}|\Psi_{t}(\zeta)\rangle_{S}$.

The $\hat{L}^{\ast}$-term contribution to (78) is then

$\displaystyle-\sum_{m}\int_{\mathfrak{f}_{B}}\tilde{\mathbb{P}}[dk]\langle
e_{n}|\hat{L}^{\ast}|e_{m}\rangle_{S}\,\langle g_{t}|k\rangle\langle
e_{m}\otimes e^{k}|\hat{U}_{t}^{(f)}|\phi\otimes\mathrm{vac}\rangle$
$\displaystyle=-\sum_{m}\int_{\mathfrak{f}_{B}}\tilde{\mathbb{P}}[dk]\langle
e_{n}|\hat{L}^{\ast}|e_{m}\rangle_{S}\,\zeta_{t}(k)^{\ast}\langle e_{m}\otimes
e^{f+k}|\hat{U}_{t}|\phi\otimes\mathrm{vac}\rangle$
$\displaystyle=-\sum_{m}\int_{\mathfrak{f}_{B}}\tilde{\mathbb{P}}[dk]\langle
e_{n}|\hat{L}^{\ast}|e_{m}\rangle_{S}\,\zeta_{t}(k)^{\ast}\langle
e_{m}|\tilde{\Psi}_{t}(f+k)\rangle_{S}$
$\displaystyle=-\int_{\mathfrak{f}_{B}}\tilde{\mathbb{P}}[dk]\zeta_{t}(k)^{\ast}\langle
e_{n}|\hat{L}^{\ast}\tilde{\Psi}_{t}(f+k)\rangle_{S}.$ (81)

The final step is the change of functional variable from
$k\in\mathfrak{f}_{B}$ to $\xi_{t}=\zeta_{t}(k)$. We note that $\langle
f+k|g_{t}\rangle=\zeta_{t}(f)+\zeta_{t}(k)=\zeta_{t}+\xi_{t}$. This results in
(81) being
$-\int_{\mathscr{C}_{K}(\mathbb{T},dt)}\mathbb{P}[d\xi]\xi_{t}^{\ast}\langle
e_{n}|\hat{L}^{\ast}\Psi_{t}(\zeta+\xi)\rangle_{S}$.

Collecting the terms, and noting that $e_{n}$ was arbitrary, yields the
desired result.

###  IV.2 Thermal Baths

So far we have taken the bath to be in the vacuum state, though the
generalization to thermal baths is straightforward and uses the standard
Araki-Woods representationGK . Here one should replace the bath with two baths
$B_{1}$ and $B_{2}$ to describe the spontaneous and stimulated effects,
respectively. Here one would set

$\displaystyle\hat{a}_{\omega}=\sqrt{n_{\omega}(T)+1}\,\hat{a}_{\omega,1}\otimes\hat{I}_{B_{2}}+\sqrt{n_{\omega}(T)}\hat{I}_{B_{1}}\otimes\hat{a}_{\omega,1}$
(82)

where the
$[\hat{a}_{\omega,j},\hat{a}_{\omega^{\prime},j}^{\ast}]=\delta_{\omega,\omega^{\prime}}\delta_{j,j^{\prime}}$,
for $j=1,2$, and $n_{\omega}(T)=\frac{1}{e^{\omega/kT}-1}$. The thermal state
then corresponds to the joint vacuum for $B_{1}$ and $B_{2}$.

Set $g_{1}(\omega)=\sqrt{n_{\omega}(T)+1}\,g(\omega)$ and
$g_{2}(\omega)=\sqrt{n_{\omega}(T)}\,g(\omega)^{\ast}$, then from complex
amplitudes $f_{j}(\omega)$ we construct complex trajectories $\zeta_{t,j}$
corresponding to $\sum_{\omega}g_{j}(\omega)e^{-i\omega
t}f_{j}(\omega)^{\ast}$. The process $\hat{Z}(t)$ is represented as

$\displaystyle\hat{Z}_{1}(t)\otimes\hat{I}_{B_{2}}+\hat{I}_{B_{1}}\otimes\hat{Z}_{2}(t)^{\ast}$
(83)

where
$\hat{Z}_{j}(t)=\frac{1}{\sqrt{2\pi}}\sum_{\omega,j}\lambda_{\omega}e^{-i\omega
t}\hat{a}_{\omega,j}^{\ast}$.

The appropriate form of the wave representation now involves a pair of complex
trajectories $(\zeta_{1},\zeta_{2})$ and the corresponding Schrödinger
equation will be

$\displaystyle\frac{\partial}{\partial
t}|\Psi_{t}(\zeta_{1},\zeta_{2})\rangle_{S}=\bigg{(}-i\hat{H}+\hat{L}\,\zeta_{t,1}-\hat{L}^{\ast}\int_{0}^{t}d\tau\,K_{1}\left(t,\tau\right)\frac{\delta}{\delta\zeta_{\tau,1}}-\hat{L}^{\ast}\zeta_{t,2}+\hat{L}\int_{0}^{t}d\tau\,K_{2}\left(t,\tau\right)\frac{\delta}{\delta\zeta_{\tau,2}}\bigg{)}|\Psi_{t}(\zeta_{1},\zeta_{2})\rangle_{S},$
(84)

where $K_{j}$ denote the kernels associated to the $\lambda_{j}$, $j=1,2$
respectively.

###  IV.3 The Markovian Case

however this is best handled by the quantum Ito calculus. Here we should
describe proceedings using the Hudson-Parthasarathy quantum stochastic
differential equation (QSDE)

$\displaystyle
d\hat{U}_{t}=\bigg{(}(-i\hat{H}-\frac{\gamma}{2}\,\hat{L}^{\ast}\hat{L})\otimes
dt+\sqrt{\gamma}\,\hat{L}\otimes
d\hat{A}^{\ast}_{t}-\sqrt{\gamma}\hat{L}^{\ast}\otimes
d\hat{A}_{t}\bigg{)}\hat{U}_{t},$ (85)

where $\hat{A}_{t}^{\ast}$ and $\hat{A}_{t}$ are creation and annihilation
processes and the non-trivial component of the quantum Ito table is
$d\hat{A}_{t}\,d\hat{A}_{t}^{\ast}=dt$. as is well-known, the QSDE admits a
unique adapted unitary solution for $\hat{H}=\hat{H}^{\ast}$ and $\hat{L}$
bounded and $\hat{U}_{0}=\hat{I}$. This time, note that the bath is described
in terms of the Fock space with one-particle space $L^{2}(\mathbb{R}_{+},dt)$.

As before, we set

$\displaystyle\langle\psi|\Psi_{t}(\zeta)\rangle_{S}=\langle\psi\otimes
e^{f}|\hat{U}_{t}|\phi\otimes\mathrm{vac}\rangle$ (86)

and we readily obtain

$\displaystyle\frac{d}{dt}|\Psi_{t}(\zeta)\rangle_{S}=\bigg{(}-i\hat{H}-\frac{\gamma}{2}\hat{L}^{\ast}\hat{L}+\hat{L}\zeta_{t}\bigg{)}|\Psi_{t}(\zeta)\rangle_{S},$
(87)

with $\zeta_{t}=\sqrt{\gamma}f(t)$.

##  V Exact Solution: The Jaynes-Cummings Model

We shall now give an exact solution for the state
$|\Psi_{t}(\zeta)\rangle_{S}$ for the Jaynes-Cummings model which is the
Hamiltonian model (11) where our system is a two-level atom with Hilbert space
$\mathfrak{h}_{S}\cong\mathbb{C}^{2}$ spanned by a ground state
$|\mathtt{g}\rangle_{S}$ and an excited state $|\mathtt{e}\rangle_{S}$,
respectively. In this case, we take the system Hamiltonian to be
$\hat{H}=\omega_{0}|\mathtt{e}\rangle\langle\mathtt{e}|$ and the coupling
operator to be the lower operator
$\hat{L}=|\mathtt{g}\rangle\langle\mathtt{e}|$.

In this case, we move to the interaction picture with respect to the
unperturbed Hamiltonian
$\hat{H}_{0}=\hat{H}\otimes\hat{I}_{B}+\hat{I}_{S}\otimes\hat{H}_{B}$. In this
case, the interaction picture is described by

$\displaystyle-i\hat{\Upsilon}_{t}=|\mathtt{g}\rangle\langle\mathtt{e}|\otimes
a(g_{t})^{\ast}-|\mathtt{e}\rangle\langle\mathtt{g}|\otimes a(g_{t}),$ (88)

where this time $g_{t}=e^{it(\hat{h}_{B}-\omega_{0})}g$. The unitary
$\hat{U}_{t}$ admits the formal Dyson series expansion

$\displaystyle\hat{U}_{t}=\hat{I}+\sum_{n=1}^{\infty}(-i)^{n}\int_{t\geq\tau_{n}>\tau_{n-1}>\cdots>\tau_{1}\geq
0}d\tau_{n}\cdots
d\tau_{1}\,\hat{\Upsilon}_{\tau_{n}}\cdots\hat{\Upsilon}_{\tau_{1}}.$ (89)

###### Theorem 19

The state $|\Psi_{t}(\zeta)\rangle_{S}$ for the Jaynes-Cummings model with
initial system state $|\phi\rangle_{S}$ is given by

$\displaystyle|\Psi_{t}(\zeta)\rangle_{S}=|\phi\rangle_{S}+\langle\mathtt{e}|\phi\rangle_{S}\bigg{(}(\lambda(t)-1)|\mathtt{e}\rangle_{S}+\int_{0}^{t}\zeta_{\tau}\lambda(\tau)\,d\tau\,|\mathtt{g}\rangle_{S}\bigg{)},$
(90)

where $\lambda(t)=\sum_{k=0}^{\infty}(-1)^{k}I_{k}(t)$ with $I_{0}(t)=1$ and,
for $k\geq 1$,

$\displaystyle I_{k}(t)=\int_{t\geq\tau_{2k}>\tau_{2k-1}>\cdots>\tau_{1}\geq
0}d\tau_{2k}\cdots d\tau_{1}\,K(\tau_{2k},\tau_{2k-1})\cdots
K(\tau_{2},\tau_{1}).$ (91)

Proof. For a given state vector $|\psi\rangle\in\mathfrak{h}_{S}$ and for
$\zeta_{t}=\zeta_{t}(f)$, we have that
$\langle\psi|\Psi_{t}(\zeta)\rangle_{S}=\langle\psi\otimes
e^{f}|\hat{U}_{t}|\phi\otimes\mathrm{vac}\rangle$.

When computing the terms appearing in the Dyson series, we find that
$\hat{\Upsilon}_{\tau_{n}}\cdots\hat{\Upsilon}_{\tau_{1}}$ will be greatly
simplified by noting that $\hat{L}^{2}=\hat{L}^{*2}=0$. In fact, we have that

$\displaystyle\cdots\hat{\Upsilon}_{\tau_{4}}\hat{\Upsilon}_{\tau_{3}}\hat{\Upsilon}_{\tau_{2}}\hat{\Upsilon}_{\tau_{1}}|\phi\otimes\mathrm{vac}\rangle$
$\displaystyle\equiv$
$\displaystyle\cdots\hat{L}^{\ast}\hat{L}\hat{L}^{\ast}\hat{L}|\phi\rangle_{S}\otimes\cdots
a(g_{\tau_{4}})a(g_{\tau_{3}})^{\ast}a(g_{\tau_{2}})a(g_{\tau_{1}})^{\ast}|\mathrm{vac}\rangle_{B}.$
(92)

Here $(\hat{L}^{\ast}\hat{L})^{k}=|\mathtt{e}\rangle\langle\mathtt{e}|$ for
each integer $k>0$.

The development of the Dyson series will begin as follows:

$\displaystyle\langle\psi\otimes
e^{f}|\hat{U}_{t}|\phi\otimes\mathrm{vac}\rangle$ $\displaystyle=$
$\displaystyle\langle\psi|\phi\rangle_{S}+\langle\psi|\mathtt{g}\rangle_{S}\langle\mathtt{e}|\phi\rangle_{S}\int_{0}^{t}d\tau_{1}\,\langle
f|g_{\tau_{1}}t\rangle-\langle\psi|\mathtt{e}\rangle_{S}\langle\mathtt{e}|\phi\rangle_{S}\int_{0}^{t}d\tau_{2}\int_{0}^{\tau_{2}}d\tau_{1}\,\langle
g_{\tau_{2}}|g_{\tau_{1}}\rangle+\cdots.$ (93)

The general term is easily deduced.

For $n=2k$ (even), we have

$\displaystyle\langle\psi\otimes
e^{f}|\hat{\Upsilon}_{\tau_{2k}}\cdots\hat{\Upsilon}_{\tau_{1}}|\phi\otimes\mathrm{vac}\rangle$
$\displaystyle\qquad=(-1)^{k}\langle\psi|\mathtt{e}\rangle_{S}\langle\mathtt{e}|\phi\rangle_{S}\,\langle
e^{f}|a(g_{\tau_{2k}})a(g_{\tau{2k-1}})^{\ast}\cdots
a(g_{\tau_{2}})a(g_{\tau_{1}})^{\ast}|\mathrm{vac}\rangle_{B}$
$\displaystyle\qquad=(-1)^{k}\langle\psi|\mathtt{e}\rangle_{S}\langle\mathtt{e}|\phi\rangle_{S}\,K(\tau_{2k},\tau_{2k-1})\cdots
K(\tau_{2},\tau_{1}),$ (94)

while for $n=2k+1$ (odd)

$\displaystyle\langle\psi\otimes
e^{f}|\hat{\Upsilon}_{\tau_{2k+1}}\cdots\hat{\Upsilon}_{\tau_{1}}|\phi\otimes\mathrm{vac}\rangle$
$\displaystyle\qquad=(-1)^{k}\langle\psi|\mathtt{g}\rangle_{S}\langle\mathtt{e}|\phi\rangle_{S}\,\langle
e^{f}|a(g_{\tau_{2k+1}})^{\ast}a(g_{\tau_{2k}})\cdots
a(g_{\tau_{2}})a(g_{\tau_{1}})^{\ast}|\mathrm{vac}\rangle_{B}$
$\displaystyle\qquad=(-1)^{k}\langle\psi|\mathtt{g}\rangle_{S}\langle\mathtt{e}|\phi\rangle_{S}\,\zeta_{\tau_{2k+1}}\,K(\tau_{2k},\tau_{2k-1})\cdots
K(\tau_{2},\tau_{1}).$ (95)

Summing the series yields

$\displaystyle\langle\psi\otimes
e^{f}|\hat{U}_{t}|\phi\otimes\mathrm{vac}\rangle=\langle\psi|\phi\rangle_{S}+\langle\mathtt{e}|\phi\rangle_{S}\big{(}\lambda(t)\langle\psi|\mathtt{e}\rangle_{S}+\int_{0}^{t}\zeta_{\tau}\lambda(\tau)\,d\tau\,\langle\psi|\mathtt{g}\rangle_{S}\big{)}$
(96)

and this gives the result.

###### Remark 20

We note that we may write
$|\Psi_{t}(\zeta)\rangle_{S}=\hat{\mathtt{U}}_{t}(\zeta)\,|\phi\rangle_{S}$
where

$\displaystyle\hat{\mathtt{U}}_{t}(\zeta)=|\mathtt{g}\rangle\langle\mathtt{g}|+\lambda(t)\,|\mathtt{e}\rangle\langle\mathtt{e}|+\int_{0}^{t}\zeta_{\tau}\lambda(\tau)d\tau\,|\mathtt{g}\rangle\langle\mathtt{e}|.$
(97)

###### Remark 21

We note that the integrals (91) satisfy
$\dot{I}_{k+1}(t)=\int_{0}^{t}K(t,\tau)I_{k}(\tau)\,d\tau$ and consequently

$\displaystyle\dot{\lambda}(t)=-\int_{0}^{t}K(t,\tau)\lambda(\tau)\,d\tau.$
(98)

###### Remark 22

It is worth noting that the solution (90) is indeed a solution to (2) from
Theorem 17 when specified to the Jaynes-Cummings model. Indeed, (2) for this
problem reads as

$\displaystyle\frac{d}{dt}|\Psi_{t}(\zeta)\rangle_{S}=\zeta_{t}\langle\mathtt{e}|\Psi_{t}(\zeta)\rangle_{S}\,|\mathtt{g}\rangle_{S}-\int_{-\infty}^{\infty}d\tau\,K\left(t,\tau\right)\langle\mathtt{g}|\frac{\delta}{\delta\zeta_{\tau}}\Psi_{t}(\zeta)\rangle_{S}\,|\mathtt{e}\rangle_{S}.$
(99)

We likewise see that (90) satisfies

$\displaystyle\frac{d}{dt}|\Psi_{t}(\zeta)\rangle_{S}$ $\displaystyle=$
$\displaystyle\zeta_{t}\langle\mathtt{e}|\Psi_{t}(\zeta)\rangle_{S}\,|\mathtt{g}\rangle_{S}+\dot{\lambda}(t)\langle\mathtt{e}|\phi\rangle_{S}\,|\mathtt{e}\rangle_{S}$
(100) $\displaystyle=$
$\displaystyle\zeta_{t}\langle\mathtt{e}|\Psi_{t}(\zeta)\rangle_{S}\,|\mathtt{g}\rangle_{S}-\int_{0}^{t}K(t,\tau)\lambda(\tau)\,d\tau\,\langle\mathtt{e}|\phi\rangle_{S}\,|\mathtt{e}\rangle_{S},$

where we used identity (98) in the last line. Comparison of the two
differential equations shows that the coincide if

$\displaystyle\langle\mathtt{g}|\frac{\delta}{\delta\zeta_{\tau}}\Psi_{t}(\zeta)\rangle_{S}=\lambda(\tau)\,\langle\mathtt{e}|\phi\rangle_{S}.$
(101)

Again, for the solution (2) we find that

$\displaystyle\langle\mathtt{g}|\frac{\delta}{\delta\zeta_{\tau}}\Psi_{t}(\zeta)\rangle_{S}=\frac{\delta}{\delta\zeta_{\tau}}\bigg{(}\int_{0}^{t}\zeta_{t}\lambda(\tau)d\tau\,\bigg{)}\langle\mathtt{e}|\phi\rangle_{S}=\bigg{(}\int_{0}^{t}\delta(t-\tau)\lambda(\tau)\,d\tau\bigg{)}\langle\mathtt{e}|\phi\rangle_{S}=\lambda(t)\,\langle\mathtt{e}|\phi\rangle_{S}.$
(102)

###### Remark 23

The Markovian case corresponds to $K(t,s)=\gamma\,\delta(t-s)$. Here we find
that the integrals (91) reduce to $k$-simplicial integals and yield
$I_{k}(t)=\frac{1}{k!}(\gamma/2)^{k}$, so that $\lambda(t)=e^{-\gamma t/2}$
giving an exponential relaxation.

It is instructive to see that this solution is indeed a solution of Markovian
equation (87) for the Jaynes-Cumming model: here
$\mathfrak{h}_{S}=\mathbb{C}^{2}$ and we set $\hat{H}=0$ and
$\hat{L}=|\mathtt{g}\rangle\langle\mathtt{e}|$. The markovian complex
trajectory equation is then

$\displaystyle\frac{d}{dt}|\Psi_{t}(\zeta)\rangle_{S}=\bigg{(}-\frac{\gamma}{2}|\mathtt{e}\rangle\langle\mathtt{e}|+\zeta_{t}|\mathtt{g}\rangle\langle\mathtt{e}|\bigg{)}|\Psi_{t}(\zeta)\rangle_{S}$
(103)

and setting

$\displaystyle|\Psi_{t}(\zeta)\rangle_{S}=\left[\begin{array}[]{c}c_{e}(t)\\\
c_{g}(t)\end{array}\right],$ (106)

we find $\dot{c}_{e}(t)=\frac{\gamma}{2}\,c_{e}(t)$ and
$\dot{c}_{g}(t)=\zeta_{t}\,c_{e}(t)$ with solutions
$c_{e}(t)=c_{e}(0)e^{-\gamma t/2}$ and
$c_{g}(t)=c_{g}(0)+c_{e}(0)\int_{0}^{t}\zeta_{\tau}e^{-\gamma\tau/2}d\tau$
with $c_{e}(0)=\langle\mathtt{e}|\phi\rangle_{S}$ and
$c_{g}(0)=\langle\mathtt{g}|\phi\rangle_{S}$. This solution coincides with
that given by (90) when we take the exponential relaxation
$\lambda(t)=e^{-\gamma t/2}$.

##  VI Can We Observe Complex Trajectories?

Quantum filtering theory is based on the requirement that we measure a
commuting family of observable processes
$\\{\hat{Y}_{k}(\cdot):k=1,\cdots,n\\}$ parameterized by time. Commutativity
is essential because otherwise the most recent measurement demolishes previous
ones: this is referred to as the non-self-demolition property of the
observations. Likewise, an observable can be estimated in terms of these
measurements only if it commutes with them - this is the non-demolition
principle.

At this stage it is useful to recall the basics of von Neumann algebras. A
collection of operators is said to form an algebra if it is closed under
addition, scalar multiplication and operator multiplication. It is a *-algebra
if, whenever $\hat{X}$ is in the algebra, $\hat{X}^{\ast}$ is as well. The
commutant of an algebra $\mathfrak{A}$ which is a subalgebra of a larger
algebra $\mathfrak{B}$ is the set of all elements in $\mathfrak{B}$ which
commute with every element of $\mathfrak{A}$: we denote the commutant by
$\mathfrak{A}^{\prime}$. A von Neumann algebra is then as *-algebra which is
its own bicommutant, that is,
$\mathfrak{A}^{\prime\prime}=(\mathfrak{A}^{\prime})^{\prime}=\mathfrak{A}$.
Von Neumann algebras play the role of $\sigma$-algebras in noncommutative
measure theory. Given a collection of operators $\mathfrak{A}_{0}$, we can
construct the smallest von Neummann algebra containing all the elements of the
collection - this will be $\mathfrak{A_{0}}^{\prime\prime}$ and we refer to
this as the von Neumann algebra generated by $\mathfrak{A}_{0}$.

We may now formulate the quantum filtering problem. The von Neumann algebra
$\mathfrak{Y}_{t}$ generated by the measured observables
$\\{\hat{Y}_{k}(\tau):0\leq\tau\leq t,k=1,\cdots,n\\}$ will be commutative, by
virtue of the self-nondemolition property. We say that the measured processes
are essentially classical stochastic processes meaning that if we fix a state
and restrict to just these processes then the classical theory of stochastic
processes is sufficient to describe proceedings.

The nondemolition property then tells us that we should restrict our interests
at each time $t$ to those observables that lie in the commutant of measurement
algebra up to that times, $\mathfrak{Y}_{t}^{\prime}$. If we fix an overall
state, then we can construct the conditional expectation from
$\mathfrak{Y}_{t}^{\prime}$ onto $\mathfrak{Y}_{t}$. In the case of markovian
models, $\mathfrak{Y}_{t}^{\prime}$ consists of all observables of the system
(in the Heisenberg picture) at time $t$ or later. In such cases, the
conditional expectation for an observable $\hat{A}$ of the system
($\hat{A}_{t}$ in the Heisenberg picture) is denoted by $\pi_{t}(\hat{A})$ and
must, of course, belong to $\mathfrak{Y}_{t}$.

As $\mathfrak{Y}_{t}$ is commutative, we may view it as a classical algebra -
effectively treating the observed processes as standard classical stochastic
processes. We may construct a conditioned state $|\Psi_{t}\rangle_{S}$ (a
random variable depending on the observations up to time $t$) such that
$\langle\Psi_{0}|\pi_{t}(\hat{A})|\Psi_{0}\rangle_{S}=\langle\Psi_{t}|\hat{A}|\Psi_{t}\rangle_{S}$.
In general $|\Psi_{t}\rangle_{S}$ satisfies a nonlinear stochastic
differential equation but we may also write

$\displaystyle\langle\chi_{0}|\pi_{t}(\hat{A})|\chi_{0}\rangle_{S}=\frac{\langle\chi_{t}|\hat{A}|\chi_{t}\rangle_{S}}{\langle\chi_{t}|\chi_{t}\rangle_{S}}$
(107)

where $|\chi_{t}\rangle_{S}$ is not normalized but satisfies a linear
stochastic differential equation (the quantum Zakai equation). A typical Zakai
equation takes the form

$\displaystyle
d|\chi_{t}\rangle_{S}=-(\frac{1}{2}\sum_{j}\hat{L}_{j}^{\ast}\hat{L}_{j}+i\hat{H})\,|\chi_{t}\rangle_{S}\,dt+\sum_{j}\hat{L}_{j}|\chi_{t}\rangle_{S}\,dY_{j}(t)$
(108)

where the $\hat{L}_{j}$ are coupling operators.

Returning to the complex trajectory problem, the first issue we encounter is
that the process $\hat{Z}(t)$ not self-adjoint. In principle, we may
decomposed $\hat{Z}(t)=\hat{X}(t)+i\hat{Y}(t)$ where $\hat{X}$ and $\hat{Y}$
are self-adjoint processes which we may refer to as the quadrature processes.
We note the following commutation relations

$\displaystyle[\hat{X}(t),\hat{X}(s)]=[\hat{Y}(t),\hat{Y}(s)]=\frac{i}{2}\,\textrm{Im}\,K(t,s),$
$\displaystyle[\hat{X}(t),\hat{Y}(s)]=\frac{1}{2i}\,\textrm{Re}\,K(t,s).$
(109)

We may construct the algebra $\mathfrak{Z}_{t}$ of functions of the operators
$\\{\hat{Z}(\tau):0\leq\tau\leq t\\}$ and this will indeed be a commutative
algebra. However, it is not a *-algebra and in particular will not be a von
Neumann algebra. In fact, we can construct the adjoint algebra
$\mathfrak{Z}_{t}^{\ast}$ and this does not coincide with $\mathfrak{Z}_{t}$.
Indeed, $\mathfrak{Z}_{t}\cap\mathfrak{Z}_{t}^{\ast}$ consists of just
multiples of the identity.

The von Neumann algebra generated by $\mathfrak{Z}_{t}$ is
$\mathfrak{W}_{t}=(\mathfrak{Z}_{t}\cup\mathfrak{Z}_{t}^{\ast})^{\prime\prime}$
and this will be noncommutative since the process $Z_{t}$ does not commute
with its adjoint. We may form the von Neumann algebras $\mathfrak{X}_{t}$ and
$\mathfrak{Y}_{t}$ generated by the quadrature processes but, again, these two
processes do not commute. Worse still, neither of the quadrature algebras are
commutative - both are self-demolishing! This means that we cannot even do
filtering on just one of these quadratures. (Exceptionally, the quadrature
algebras will be commutative if the kernel was real-valued. This would require
that for each frequency $\omega\in\Omega$ we also have $-\omega\in\Omega$ with
$|g(\omega)|=|g(-\omega)|$.)

In short, $\hat{Z}_{t}$ is not self-adjoint and therefore not observable!

###### Acknowledgements.

This work is supported by the ANR project “Estimation et controle des systémes
quantiques ouverts” QCOAST Projet ANR-19-CE48-0003, the ANR project QUACO
ANR-17-CE40-0007, and the ANR project IGNITION ANR-21-CE47-0015.

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