# Role of Dilations in Reversing Physical Processes:
Tabletop Reversibility and Generalized Thermal Operations

Clive Cenxin Aw  Centre for Quantum Technologies, National University of
Singapore, 3 Science Drive 2, Singapore 117543  Lin Htoo Zaw  Centre for
Quantum Technologies, National University of Singapore, 3 Science Drive 2,
Singapore 117543  Maria Balanzó-Juandó  ICFO-Institut de Ciències Fotòniques,
The Barcelona Institute of Science and Technology, Av. Carl Friedrich Gauss 3,
08860 Castelldefels (Barcelona), Spain  Valerio Scarani  Centre for Quantum
Technologies, National University of Singapore, 3 Science Drive 2, Singapore
117543  Department of Physics, National University of Singapore, 2 Science
Drive 3, Singapore 117542

###### Abstract

Irreversibility, crucial in both thermodynamics and information theory, is
naturally studied by comparing the evolution—the (forward) channel—with an
associated reverse—the reverse channel. There are two natural ways to define
this reverse channel. Using logical inference, the reverse channel is the
Bayesian retrodiction (the Petz recovery map in the quantum formalism) of the
original one. Alternatively, we know from physics that every irreversible
process can be modeled as an open system: one can then define the
corresponding closed system by adding a bath (“dilation”), trivially reverse
the global reversible process, and finally remove the bath again. We prove
that the two recipes are strictly identical, both in the classical and in the
quantum formalism, once one accounts for correlations formed between system
and the bath. Having established this, we define and study special classes of
maps: product-preserving maps (including generalized thermal maps), for which
no such system-bath correlations are formed for some states; and tabletop
time-reversible maps, when the reverse channel can be implemented with the
same devices as the original one. We establish several general results
connecting these classes, and a very detailed characterisation when both the
system and the bath are one qubit. In particular, we show that, when reverse
channels are well-defined, product-preservation is a sufficient but not
necessary condition for tabletop reversibility; and that the preservation of
local energy spectra is a necessary and sufficient condition to generalized
thermal operations.

##  I Introduction

Irreversibility is ubiquitous in real life. In science, it was first studied
systematically in the context of thermodynamics: this is captured by the
Second Law, which stipulates the impossibility of putting all the energy to
fruition, leading to the necessary generation of heat—or more generally,
entropy [1, 2, 3, 4, 5, 6, 7]. Eventually, information theory became the
setting in which to study irreversibility: a process is irreversible for an
agent when the agent is unable to retrieve from the output all the information
about the input. In turn, a theory of optimal retrieval of information was
developed, both in classical and in quantum theories [8, 9, 10, 11, 12].

Meanwhile, the field of stochastic thermodynamics developed quantitative
approaches to irreversibility, based on statistical comparisons between the
process under study and its associated reverse process. But how to define the
latter? In the case of fully reversible, deterministic processes, the reverse
process is obviously the dynamics played backwards. For isothermal evolutions
(driven Hamiltonian evolution while the system is in contact with a thermal
bath), a possible and very natural reverse process consists in driving the
evolution backwards in the presence of the same bath [13, 14]. Reverse process
have also been found for more complex processes, through expert control of the
model and its assumptions (see e.g. [15]). A general recipe may be built on
the observation that any irreversible process can be seen as a marginal of a
global, reversible process involving the system and some environment. The
recipe through dilation is then: add a suitable environment (dilation),
trivially reverse the reversible global process, and finally remove the
environment.

Recently, it was proposed to define the reverse process using the Bayesian
recipe for information retrieval, a.k.a. Bayesian inversion or retrodiction
[16, 17]. This recipe through retrodiction requires only choosing a reference
state, which plays the role of a Bayesian prior. The connection between
reverse processes and Bayesian logic had not been noticed in the context of
classical stochastic thermodynamics. In quantum thermodynamics, one the main
tools for information recovery had been used, first occasionally [18, 14],
then systematically [19, 20]: the Petz recovery map [21, 11, 22], which was
also proposed early on as a quantum analog of Bayesian inversion [23, 24, 25,
26].

In this work, after a review of known material on reverse processes (Section
II), we start by proving that the two recipes by dilation and by retrodiction
are identical, both in the classical and the quantum case (Section III). The
fact that the two proposed general recipes coincide, combined with the
knowledge that all the previously known special cases can be recovered with
these recipes, shows that we have the definition of the reverse process under
control. Next we bring up the observation that a process and its associated
reverse process may be very different. It is indeed well known in the quantum
case that implementing the reverse (Petz) of a channel may require very
different resources than those needed to implement the channel itself (see
e.g. [27]). The cases mentioned above of the reversible and the isothermal
processes, whose reverses are “what one would expect” and can be implemented
with the same control and the same environment, seem to be the exception.
Based on this observation, we set to study which processes have a reverse that
can be implemented with the same, or similar, resources (Fig. 1). We shall say
that the latter processes possess tabletop reversibility. This is of interest
for the structure of the theory of the reverse processes, as well as for
possible experimental tests of fluctuation theorems in situations that are not
unitary or isothermal. The mathematical definition of tabletop reversibility
is defined in Section IV, together with the auxiliary notion of product-
preserving channels, which may be of interest in its own right. In Section V,
we present both general results valid and a thorough characterization of two-
qubit channels. In Section VI, we highlight the implications of these results
and scenarios for energetics and reversibility in the quantum regime. Section
VII is a conclusion.

Figure 1:  An illustration of the main goal of this paper. Any channel
$\mathcal{E}$ on a system can be viewed as a larger reversible unitary process
$\mathcal{U}$ involving the system and a bath (top). Applying the recipes by
dilation or by retrodiction (proved identical in Section III), one finds that
the reverse process must in general be implemented with completely different
tools than the original process (bottom left). We set out to characterize the
tabletop reversible situations, in which the reverse process can be
implemented with the same, or similar, tools as the forward process: namely,
by appending an ancilla and inverting the original unitary $\mathcal{U}$. The
exact definitions will be given in Section IV, and the results in the
following Sections.

##  II Reverse Processes

###  II.1 Notations

In classical theory, we consider a discrete state space with $d$ distinct
states. A generic state is represented by a probability distribution $p$:
$p(j)\geq 0$ for $j=1,...,d$; and $\sum_{j=1}^{d}p(j)=1$. It can be
represented by a $d\times 1$ probability vector $\mathbf{p}$, whose entries
are $\mathbf{p}_{j}:=p(j)$. A generic channel is a stochastic map $\varphi$,
defined by $d^{2}$ probabilities $\varphi(j^{\prime}|j)$ of transiting from
the input state $j$ to the output state $j^{\prime}$. These probabilities must
satisfy $\varphi(j^{\prime}|j)\geq 0$ for all $j,j^{\prime}$ and
$\sum_{j^{\prime}=1}^{d}\varphi(j^{\prime}|j)=1$ for all $j$. The channel can
be represented by the $d\times d$ stochastic matrix $\bm{\varphi}$, whose
entries are $\bm{\varphi}_{j^{\prime}j}:=\varphi(j^{\prime}|j)$. In this
representation, the composition of channels is represented by the standard
matrix multiplication: if $\varphi=\varphi_{2}\circ\varphi_{1}$, then
$\bm{\varphi}=\bm{\varphi}_{2}\,\bm{\varphi}_{1}$.

An important remark for what follows: even if the matrix $\bm{\varphi}$ has an
inverse, in general the entries of the matrix $\bm{\varphi}^{-1}$ do not
define a valid stochastic map. Analogously, while every matrix can be
transposed, the map corresponding to $\bm{\varphi}^{\text{T}}$ is a valid map
if and only if the channel is bistochastic, i.e. satisfies the additional
property $\sum_{j=1}^{d}\varphi(j^{\prime}|j)=1$ for all $j^{\prime}$. When
the inverse or the transpose of the matrix do define valid channels, we shall
denote those channels as $\varphi^{-1}$ and $\varphi^{\text{T}}$ respectively.

In quantum theory, we consider a finite-dimensional complex vector space of
dimension $d$. A generic state is described by a semidefinite operator
$\rho\succeq 0$ in this space with $\Tr(\rho)=1$. Channels are represented by
completely positive, trace preserving (CPTP) maps. Given a CPTP map
$\mathcal{E}$, the adjoint $\mathcal{E}^{\dagger}$ is the unique map for which

$\Tr(\mathcal{E}[X]Y)=\Tr(X\mathcal{E}^{\dagger}[Y])$ (1)

for operators $X$ and $Y$. Just as $\varphi^{\text{T}}$ is not a valid
stochastic map in general, $\mathcal{E}^{\dagger}$ is in general not a valid
quantum channel.

Figure 2: (a) Illustrations for standard examples of reversal. (b) Bayesian
inversion or “retrodiction” is a formal recipe that reproduce results of the
standard approach while generalizing the definition of reverse processes for
any characterized process, and any under setting as captured by a reference
state.

###  II.2 Reverse Process: Standard Examples

We first review some classes of processes where the reverse—or at least, a
candidate for it—is known (see Fig. 2(a) for an illustration).

The most obvious class is that of reversible processes, where the map is a
bijection between the space of states. Their reverse processes are naturally
defined as the evolution played backwards, i.e. the inverse map. In classical
theory, such processes are Hamiltonian evolutions $\Phi$ obeying Liouville’s
theorem: reversal is given by inverting the trajectories in the configuration
space. In a discrete state space, the matrix $\bm{\Phi}$ is a permutation
matrix, whose inverse and transpose coincide
($\bm{\Phi}^{-1}=\bm{\Phi}^{\text{T}}$) and define a valid map. Analogously,
in quantum theory, bijective transformations are unitary channels
$\mathcal{U}[\bullet]=U\bullet U^{\dagger}$ where $UU^{\dagger}=\mathbb{1}$:
their inverse exists, coincides with the adjoint, and defines a valid map. For
a reversible process, the inverse map is the only reasonable candidate of the
corresponding reverse process.

Moving to bistochastic/unital processes, their inverse is in general not
defined, but their transpose/adjoint is a valid channel and thus provides an
immediate, natural candidate for the reverse map. Beyond this class, the
transpose/adjoint ceases to define a valid channel. On this basis, it has been
argued that only bistochastic/unital processes are fundamental if one wants
the theory to be fundamentally reversible [28, 29, 30]. We do not take sides
in that discussion: we are not concerned with ultimate constraints on
fundamental theories, but with the description of practical irreversibility.

The most obvious irreversible processes describe the dissipation of
information in an unmonitored environment, or arise from a coarse-graining
over a chaotic dynamics. In both cases, the dynamics is generally not
bistochastic/unital. The reverse of some such processes has been constructed
on a case-by-case basis by invoking physical arguments. For a system
undergoing Hamiltonian evolution while in contact with thermal baths, under
the assumption of detailed balance, the recipe is very sinmple: the reverse
process consists in playing the Hamiltonian evolution backward, while staying
in contact with the same thermal baths [6, 13]. This recipe was used in the
experimental demonstrations [31] of Crooks’ fluctuation theorem with
biological systems [32, 33] and levitating nanospheres [34, 35]. For quantum
channels, the same result holds for the so-called thermal operations
$\mathcal{T}$ [36, 14, 37, 38, 39, 40, 41, 42, 43].

###  II.3 Reverse Processes: General Recipe through Bayesian Retrodiction

####  II.3.1 Generalities

Having reviewed the prime examples of constructed reverse processes, we now
describe a general recipe applicable to every process: Bayesian retrodiction
(sometimes called “Bayesian inversion”) [16, 17]. It can be traced back to the
works of Watanabe [44, 45], ultimately building on the observation that the
laws of physics give us the knowledge of the evolution (the channel, in our
language) and not the initial state.

In classical theory, given the channel $\varphi$, the recipe for Bayesian
inversion is standard: one postulates a reference prior $\gamma$, then applies
Bayes’ rule to the joint probability distribution
$P(a,a^{\prime}):=\varphi(a^{\prime}|a)\gamma(a)$. The resulting reverse map
is

${\hat{\varphi}_{\gamma}}(a|a^{\prime})=\varphi(a^{\prime}|a)\frac{\gamma(a)}{\varphi[\gamma](a^{\prime})}$
(2)

where the distribution $\varphi[\gamma]$ is given by
$\varphi[\gamma](a^{\prime})=\sum_{a}\varphi(a^{\prime}|a)\gamma(a)$, the
output obtained by propagating the reference prior $\gamma$ through the
channel $\varphi$. In matrix notation, Eq. (2) reads

$\displaystyle\bm{\hat{\varphi}}_{\gamma}$ $\displaystyle=$
$\displaystyle\mathbf{D}_{\gamma}\bm{{\varphi}}^{\text{T}}\mathbf{D}^{-1}_{\varphi[\gamma]}.$
(3)

where $\mathbf{D}_{p}$ is the diagonal matrix with entries corresponding to
the distribution $p$.

The extension of Bayesian formalism to the quantum formalism has been the
object of many studies [23, 46, 25, 24, 47]. Here, we don’t need an exhaustive
Bayesian toolbox: only a candidate for the reverse map. Our choice is the Petz
Recovery map ${\hat{\mathcal{E}}_{\alpha}}$ [48, 21, 49]

${\hat{\mathcal{E}}_{\alpha}}[\bullet]=\sqrt{\alpha}\;\mathcal{E}^{\dagger}\\!\left[\frac{1}{\sqrt{\mathcal{E}[\alpha]}}\bullet\frac{1}{\sqrt{\mathcal{E}[\alpha]}}\right]\sqrt{\alpha},$
(4)

where $\alpha$ is a state that plays the role of reference prior. The choice
of the Petz map as the quantum analog of Bayes’ rule is standard [23, 25, 26].
It has recently been shown to fulfill the most crucial intuitions about
reversal [49, 24, 25, 20] and to be suited to recover results in stochastic
quantum thermodynamics [19, 16].

From now onwards, we identify

$\displaystyle\mathcal{R}_{c}\left[\varphi,\gamma\right]$
$\displaystyle={\hat{\varphi}_{\gamma}}$ (5)
$\displaystyle\mathcal{R}_{q}\left[\mathcal{E},\alpha\right]$
$\displaystyle={\hat{\mathcal{E}}_{\alpha}}$ (6)

and will use these notations interchangeably when convenient. A property of
the Bayes/Petz maps that we shall use later is composability [24]:

$\displaystyle\mathcal{R}_{c}[\varphi_{2}\circ\varphi_{1},\gamma]=$
$\displaystyle\mathcal{R}_{c}\left[\varphi_{1},\gamma\right]\circ\mathcal{R}_{c}\left[\varphi_{2},\varphi_{1}[\gamma]\right],$
(7)
$\displaystyle\mathcal{R}_{q}\left[\mathcal{E}_{2}\circ\mathcal{E}_{1},\alpha\right]=$
$\displaystyle\mathcal{R}_{q}\left[\mathcal{E}_{1},\alpha\right]\circ\mathcal{R}_{q}\left[{\mathcal{E}}_{2},\mathcal{E}_{1}[\alpha]\right]\,.$
(8)

One may see Appendix C.1 for proofs. This property holds even when the maps
are not stochastic.

We note in passing that, due to presence of the term
$\mathcal{E}[\alpha]^{-1/2}$ ($\mathbf{D}^{-1}_{\varphi}[\gamma]$),
${\hat{\mathcal{E}}_{\alpha}}$ (${\hat{\varphi}_{\gamma}}$) is ill-defined
when the propagated reference $\mathcal{E}[\alpha]$ ($\varphi[\gamma]$) is
rank-deficient. Of course, one could attempt to side-step this problem several
ways: for example, by defining $\mathcal{E}[\alpha]^{-1/2}$
($\mathbf{D}^{-1}_{\varphi}[\gamma]$) only on its support, or by considering a
neighbourhood of full-rank states around the rank-deficient output and taking
some limit. However, this boils down to a matter of convention, where each
approach gives a different retrodiction channel, as we discuss in Appendix F
in some detail. We do not make a particular choice, and instead leave the
retrodiction channel undefined in this case.

####  II.3.2 Examples revisited

The examples of reverse maps of the previous subsection are recovered, and
possibly clarified, in the retrodictive approach. Reversible maps are the only
ones, for which the Bayes/Petz map does not depend on the reference prior [45,
17, 25]; and, unsurprisingly, coincides with the inverse:

$\displaystyle\mathcal{R}_{c}\left[\Phi\right]=\hat{\Phi}_{\gamma}=\Phi^{\text{T}}={\Phi}^{-1}$
$\displaystyle\quad\forall\gamma\,,$ (9)
$\displaystyle\quad\mathcal{R}_{q}\left[\mathcal{U}\right]=\hat{\mathcal{U}}_{\alpha}=\mathcal{U}^{\dagger}=\mathcal{U}^{-1}$
$\displaystyle\quad\forall\alpha\,.$

For the case of bistochastic/unital maps, the Bayes/Petz does depend on the
reference prior [25, 17]. The reverse described above is obtained for a very
natural choice of reference prior, namely the uniform: $\gamma:=u$ with
$u(j)=1/d$ in the classical case, $\alpha:=\mathbb{1}/d$ in the quantum case.
Indeed, this prior is preserved by these maps ($\varphi[u]=u$,
$\mathcal{E}[\mathbb{1}/d]=\mathbb{1}/d$), and one can immediately see that

$\displaystyle\mathcal{R}_{c}\left[\varphi,u\right]=\varphi^{\text{T}},\quad\mathcal{R}_{q}\left[\mathcal{E},\mathbb{1}/d\right]=\mathcal{E}^{\dagger}$
(10) $\displaystyle\textrm{[bistochastic/unital]}.$

Lastly, let us look at thermal operations in the quantum language. Given a
non-interacting system-bath Hamiltonian
$H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes H_{B}$, one defines

$\mathcal{T}[\bullet]=\Tr_{B}\left\\{U\left[\bullet\otimes\,\tau_{\kappa}({H_{B}})\right]U^{\dagger}\right\\}$
(11)

where $\tau_{\kappa}({H})=e^{\kappa H}/\tr(e^{\kappa H})$ is the thermal (or
Gibbs) state with $\kappa=-1/k_{B}T$ (usually denoted $-\beta$, but later in
the paper we use $\beta$ will denote a state of the bath) and where $U$ is
constrained to satisfy

$[U,H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes H_{B}]=0.$ (12)

A channel thus constructed preserves the thermal, or Gibbs, state of the
system for the same $\kappa$:
$\mathcal{T}[\tau_{\kappa}({H_{A}})]=\tau_{\kappa}({H_{A}})$. The Petz map
with the Gibbs state as reference prior is found to be ([14], see also
Appendix A for the derivation)

$\hat{\mathcal{T}}_{\tau_{\kappa}({H_{A}})}[\bullet]=\Tr_{B}\left\\{U^{\dagger}\left[\bullet\otimes\tau_{\kappa}({H_{B}})\right]U\right\\}\,,$
(13)

which describes indeed a reversal of the unitary dynamics while in contact
with the same thermal bath. Notice how, having adopted the retrodictive
approach, the usual thermodynamical assumptions called “microreversibility”
and “detailed balance” are replaced by the single assumption on the choice of
the reference prior.

##  III Reverse processes and dilations

Before studying tabletop reversibility, we need to introduce the notion of
dilation of a process. The word, common in the language of quantum channels,
describes the extension of a process on system A to include an environment (or
“bath”, or “ancilla”) B.

Typically, a dilation is performed with the goal of making the extended system
AB a closed one, whose dynamics is therefore reversible. Hence, it is natural
to look at defining reverse processes by the following recipe: dilate by
adding the environment, reverse the map of the dilation (trivial if
reversible), then remove the environment. A priori, this recipe is different
from the Bayesian one applied on the system alone: dilations are not unique,
and the reverse process obtained by this recipe might depend on the details of
the chosen dilation. We proceed to prove that the two recipes actually
coincide: given a choice of dilation, only the reference prior chosen on the
system determines the reverse process. This holds both for classical and
quantum systems.

###  III.1 Classical Dilations & Bayes’ Rule

Every classical process $\varphi$ on a system state space $A$ ($|A|=d_{A}$)
can always be expressed as a marginal of a global process $\Phi_{AB}$ on an
extended state space $AB$ ($|B|=d_{B}$), alongside some potentially-correlated
environment $\beta_{B}$. This may be expressed by:

$\displaystyle\varphi(a^{\prime}|a)=\sum_{bb^{\prime}}\Phi(a^{\prime},b^{\prime}|a,b)\beta(b|a)$
(14)

A tuple $(\Phi_{AB},\beta_{B})$ that fulfills (14) will be called a dilation
of $\varphi$. We proceed to prove our claim for classical processes:

######  Result 1.

Given a classical map $\varphi$, the reverse obtained by dilating with an
environment, reversing the dilated map, then removing the environment, is the
same as that obtained directly through the Bayesian recipe (2) on the system.

###### Proof.

Let us construct the reverse with the dilation. Besides the reference prior on
the system A, we have the additional freedom of choosing a dilation
$(\Phi_{AB},\beta_{B})$. The total reference prior is then
$\Gamma(a,b)=\gamma(a)\beta(b|a)$, and we define the reverse of the dilated
map by applying Bayes’ rule to
$P(a,b,a^{\prime},b^{\prime})\equiv\Phi(a^{\prime},b^{\prime}|a,b)\Gamma(a,b)$:

$\Phi(a^{\prime},b^{\prime}|a,b)\Gamma(a,b)=\hat{\Phi}_{\Gamma}(a,b|a^{\prime},b^{\prime})\Phi[\Gamma](a^{\prime},b^{\prime})\,.$
(15)

Finally, we remove the environment. Writing $\eta:=\Phi[\Gamma]$ for
readability, we have

$\displaystyle\underbrace{\sum_{bb^{\prime}}\Phi(a^{\prime},b^{\prime}|a,b)\beta(b|a)}_{\varphi(a^{\prime}|a)}\gamma(a)$
$\displaystyle=$
$\displaystyle\sum_{bb^{\prime}}\hat{\Phi}(a,b|a^{\prime},b^{\prime})\underbrace{\Phi[\Gamma](a^{\prime},b^{\prime})}_{\eta(b^{\prime}|a^{\prime})\eta(a^{\prime})}$
$\displaystyle:=$
$\displaystyle\hat{\varphi}^{\prime}_{\Gamma}(a|a^{\prime})\eta(a^{\prime}).$

where on the left-hand side we have used the fact that $(\Phi_{AB},\beta_{B})$
is a dilation of $\varphi$, and where
$\hat{\varphi}^{\prime}_{\Gamma}(a|a^{\prime})$ is the reverse map obtained
through this recipe. By summing on both sides over $a$, we see that
$\eta(a^{\prime})=\varphi[\gamma](a^{\prime})$: whence
$\hat{\varphi}^{\prime}_{\Gamma}(a|a^{\prime})$ is identical to Eq. (2). In
particular, the only freedom left is indeed that of choosing $\gamma_{A}$. ∎

Notice that we did not have to assume that $\Phi_{AB}$ is reversible: the
proof is valid for any dilation. Also, we did not have to assume that
$\beta(b|a)=\beta(b)$ carries no initial correlations; of course, one can
choose a dilation with this property, if deemed physically important. By
contrast, having chosen the dilation, the posterior $\eta:=\Phi[\Gamma]$ is
what it is: one cannot enforce $\eta(b^{\prime}|a^{\prime})=\eta(b^{\prime})$.

###  III.2 Assignment Maps

Let us now have a more detailed look at the structure of dilations. The first
operation (appending an environment) appears as the natural reverse of the
last operation (tracing out the environment). We are going to show that this
is indeed the case (see Appendix C for supplementary proofs).

Denote the operation of tracing out the environment by $\Sigma_{B}$. The map
of appending an environment B to the system A (assignment map), with reference
state on AB given by $\Lambda$, is given by the Bayesian reverse of
$\Sigma_{B}$:

$\hat{\Sigma}_{B,\Lambda}[\bullet_{A}]=\bullet_{A}\cdot\left(\frac{\Lambda}{\Sigma_{B}[\Lambda]}\right)_{\\!B|A}\,.$
(16)

Explicitly, $\hat{\Sigma}_{B,\Lambda}[p_{A}](a,b)=p(a)\Lambda(b|a)$ has the
form of Jeffrey’s update: given a reference joint distribution $\Lambda(a,b)$,
if one gets the information that the distribution of $A$ is actually given by
$p$, the rational way of updating one’s knowledge is to update $A$’s marginal
while keeping what attains to $B$ unchanged.

In turn, the Bayesian reverse of any classical assignment map, for which
$\Lambda(a,b)$ is product, is the partial trace, for any choice of reference
prior:
$\mathcal{R}_{c}\left[\hat{\Sigma}_{B,\square\otimes\beta},\gamma\right]=\Sigma_{B}$
for all $\beta$ and $\gamma$.

The generic definition (14) of the dilation $(\Phi_{AB},\beta_{B})$ can then
be written as

$\varphi=\Sigma_{B}\circ{\Phi}\circ\hat{\Sigma}_{B,\Lambda}$ (17)

with a choice of $\Lambda$ such that
$\frac{\Lambda(a,b)}{\sum_{a}\Lambda(a,b)}=\beta(b|a)$. By using the
composability property (7), one obtains

$\hat{\varphi}_{\gamma}=\Sigma_{B}\circ\hat{\Phi}_{\Gamma}\circ\hat{\Sigma}_{B,\Phi[\Gamma]}$
(18)

which is what we proved in Result 1. For relevant proofs see Appendix C.2.

Classically, by choosing $\Lambda(b|a)\neq\Lambda(b)$, initial system-bath
correlation are straightforwardly described. The quantum analog, by contrast,
took some discussions to be clarified [50, 51, 52]. The Petz map of the
partial trace satisfies all the properties of a completely positive assignment
map. For a generic reference state $\Omega$ of $AB$, it reads

$\hat{\text{T}}\text{r}_{B,\Omega}[\bullet_{A}]=\sqrt{\Omega}\left[\left(\frac{1}{\sqrt{\Tr_{B}[\Omega]}}\bullet_{A}\frac{1}{\sqrt{\Tr_{B}[\Omega]}}\right)\otimes\mathbb{1}_{B}\right]\sqrt{\Omega}\,.$
(19)

For an uncorrelated $\Omega=\square\otimes\beta$, it takes the form

$\hat{\text{T}}\text{r}_{B,\square\otimes\beta}[\bullet_{A}]=\bullet_{A}\otimes\beta\,.$
(20)

With this definition of the quantum assignment map, we now tackle retrodiction
on dilations in the quantum formalism.

Figure 3: Two routes for Bayesian retrodiction illustrated. One can show that
these two protocols always give the same reverse process, as long as the
propagated correlations formed across the reference prior and the ancillary
environment is accounted for. This is captured by the retrodictive assignment
map (23).

###  III.3 Quantum Dilations & the Petz Recovery Map

For quantum processes, we focus on unitary dilations with an initially
uncorrelated state of the environment. Any quantum channel $\mathcal{E}$ can
be seen as the marginal of a global unitary $\mathcal{U}$ acting on a target
input and an ancillary system prepared in a suitable density operator $\beta$
[22, 53]:

$\mathcal{E}[\bullet]=\Tr_{B}\left[U(\bullet\otimes\beta)U^{\dagger}\right]$
(21)

As in the classical case, it can be seen as the composition of channels

$\mathcal{E}[\bullet]=\Tr_{B}{}\circ\mathcal{U}\circ\hat{\text{T}}\text{r}_{B,\square\otimes\beta}[\bullet]\,.$
(22)

The Petz map of $\mathcal{E}$ with reference state $\alpha$ can be computed
directly, using the fact that the adjoint of (21) is
$\mathcal{E}^{\dagger}[\bullet]=\Tr_{B}[\sqrt{\mathbb{1}\otimes\beta}U^{\dagger}(\bullet\otimes\mathbb{1})U\sqrt{\mathbb{1}\otimes\beta}]$:

$\displaystyle{\hat{\mathcal{E}}_{\alpha}}[\bullet]$
$\displaystyle=\Tr_{B}\Bigg{\\{}\sqrt{\alpha\otimes\beta}\,U^{\dagger}\left[\left(\frac{1}{\sqrt{\Tr_{B}[U(\alpha\otimes\beta)U^{\dagger}]}}\,\bullet\,\frac{1}{\sqrt{\Tr_{B}[U(\alpha\otimes\beta)U^{\dagger}]}}\right)\otimes\mathbb{1}\right]U\,\sqrt{\alpha\otimes\beta}\Bigg{\\}}$
(23)
$\displaystyle=\Tr_{B}\Bigg{\\{}U^{\dagger}\underbrace{\left[\sqrt{U(\alpha\otimes\beta)U^{\dagger}}\left[\left(\frac{1}{\sqrt{\Tr_{B}[U(\alpha\otimes\beta)U^{\dagger}]}}\,\bullet\,\frac{1}{\sqrt{\Tr_{B}[U(\alpha\otimes\beta)U^{\dagger}]}}\right)\otimes\mathbb{1}\right]\sqrt{U(\alpha\otimes\beta)U^{\dagger}}\right]}_{\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}[\bullet]}U\Bigg{\\}}$
$\displaystyle=\Tr_{B}{}\circ\mathcal{U}^{\dagger}\circ\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}[\bullet]\,,$

where between the first and the second line we inserted two identities
$U^{\dagger}U=\mathbb{1}$ and used
$U\sqrt{(\alpha\otimes\beta)}U^{\dagger}=\sqrt{U(\alpha\otimes\beta)U^{\dagger}}$,
and where we used (19).

| Classical | Quantum
---|---|---
Dilation Definition | $\varphi(a^{\prime}|a)=\sum_{bb^{\prime}}\Phi(a^{\prime}b^{\prime}|ab)\beta(b)$ | $\mathcal{E}[\bullet]=\Tr_{B}[U\bullet\otimes\beta U^{\dagger}]$
Dilation as Channels | $\varphi=\Sigma_{B}\circ\Phi\circ\hat{\Sigma}_{B,\square\otimes\beta}$ | $\mathcal{E}=\Tr_{B}{}\circ\mathcal{U}\circ\hat{\text{T}}\text{r}_{B,\square\otimes\beta}$
Bayesian Inversion | ${\hat{\varphi}_{\gamma}}=\Sigma_{B}\circ\Phi^{-1}\circ\hat{\Sigma}_{B,\Phi[\gamma\otimes\beta]}$ | ${\hat{\mathcal{E}}_{\alpha}}=\Tr_{B}{}\circ\mathcal{U}^{\dagger}\circ\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}$
Via decomposability
Assignment Map to Partial Trace | $\mathcal{R}_{c}\left[\hat{\Sigma}_{B,\square\otimes\beta},\gamma\right]=\Sigma_{B}$ | $\mathcal{R}_{q}\left[\hat{\text{T}}\text{r}_{B,\square\otimes\beta},\alpha\right]=\Tr_{B}{}$
Inversion of Global Process | $\mathcal{R}_{c}\left[\Phi,\gamma\otimes\beta\right]=\Phi^{-1}$ | $\mathcal{R}_{q}\left[\mathcal{U},\alpha\otimes\beta\right]=\mathcal{U}^{\dagger}$
Partial Trace to Retrodictive Assignment | $\mathcal{R}_{c}\left[\Sigma_{B},\Phi[\gamma\otimes\beta]\right]=\hat{\Sigma}_{B,\Phi[\gamma\otimes\beta]}$ | $\mathcal{R}_{q}\left[\Tr_{B}{},\mathcal{U}[\alpha\otimes\beta]\right]=\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}$
If no correlations are formed, and reverse map is well-defined $\Rightarrow$
tabletop reversibility
Tabletop Time-Reversibility | ${\hat{\varphi}_{\gamma}}^{\text{TR}}=\Sigma_{B}\circ\Phi^{-1}\circ\hat{\Sigma}_{B,\square\otimes\beta^{\prime}}$ | ${\hat{\mathcal{E}}_{\alpha}}^{\text{TR}}=\Tr_{B}{}\circ\mathcal{U}^{\dagger}\circ\hat{\text{T}}\text{r}_{B,\square\otimes\beta^{\prime}}$
Table 1: Summary table of the relation between retrodiction and dilations. To
stress the comparison between classical and quantum theory, in this table the
classical dilation $\Phi$ is a reversible channels, and the bath is taken as
initially uncorrelated with the system (although these restrictions are not
needed, as proved in the text).

Thus, we verified directly the composition that was expected on formal grounds
(8). We can then state the claimed:

######  Result 2.

Given a quantum map $\mathcal{E}$, the reverse obtained by dilating with an
environment, reversing the dilated map (accounting for propagated correlations
through the assignment map), then removing the environment, is the same as the
Petz map (4) computed directly on the map. In other words, the knowledge of a
dilation of $\mathcal{E}$ does not add any useful information to define the
reverse of $\mathcal{E}$.

###### Proof.

The recipe through the dilation is the composition of

$\displaystyle\mathcal{R}_{q}\left[\hat{\text{T}}\text{r}_{B,\square\otimes\beta},\alpha\right]=\Tr_{B}{}$
(24)
$\displaystyle\mathcal{R}_{q}\left[\mathcal{U},\hat{\text{T}}\text{r}_{B,\square\otimes\beta}[\alpha]\right]=\hat{\mathcal{U}}_{\alpha\otimes\beta}=\mathcal{U}^{\dagger}\;\;\;\because\eqref{eq:deterministic}$
$\displaystyle\mathcal{R}_{q}\left[\Tr_{B}{},\mathcal{U}\circ\hat{\text{T}}\text{r}_{B,\square\otimes\beta}[\alpha]\right]=\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}.$

Proofs for each individual part of the decomposition are found in Appendix
C.3. This composition indeed coincides with the Petz map, as proved in (23). ∎

We summarize the structural symmetries that relate dilation and retrodiction,
for classical and quantum theory, in Table 1. In both regimes, the role of
retrodictive assignment maps
$\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]},\hat{\Sigma}_{B,\Phi[\gamma\otimes\beta]}$
ensures consistency in expressing the reverse process. The last line of the
table anticipates the definition of tabletop reversibility, the central object
of this paper, which we are going to introduce next.

##  IV Definition of tabletop reversibility and related classes of channels

In this Section, we introduce the new classes of channels that are the central
object of this work. From here onward, we work only in the quantum formalism.
When required, we shall highlight whether a result is purely quantum, or is
also true for classical processes.

###  IV.1 Tabletop Reversibility

Our primary concern is the implementation of a Petz map
${\hat{\mathcal{E}}_{\alpha}}$, given the control on the implementation of the
channel $\mathcal{E}[\bullet]=\Tr_{B}[U(\bullet\otimes\beta)U^{\dagger}]$.

Implementing the Petz map is not straightforward, and approximate realisations
have been studied recently [27, 54]. With what we introduced, we can
understand the reason of this difficulty. On the one hand, since the Petz map
is a CPTP map, there exist a unitary $V$ and an ancillary state $\bar{\beta}$
such that
${\hat{\mathcal{E}}_{\alpha}}[\bullet]=\text{Tr}_{\bar{B}}\bqty{V\pqty{\bullet\otimes\bar{\beta}}V^{\dagger}}$.
But in general, $V\neq U^{\dagger}$: we may have to build a dedicated unitary.
On the other hand, we have just seen in Eq. (23) that every Petz map can be
written as
${\hat{\mathcal{E}}_{\alpha}}[\bullet]=\Tr_{B}\bqty{U^{\dagger}\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}[\bullet]U}$.
But in general,
$\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}[\bullet]\neq\bullet\otimes\beta^{\prime}$,
as shown in (19): we may have to do something more complicated than appending
an ancilla.

We want to identify the special cases, in addition to unitary and isothermal
channels, where the reverse channel can be obtained by just appending an
ancilla and inverting the unitary:

######  Definition 1.

A quantum channel $\mathcal{E}$ with a unitary dilation
$\mathcal{E}[\bullet]=\Tr_{B}[U(\bullet\otimes\beta)U^{\dagger}]$ is called
tabletop reversible for the prior $\alpha$ [shorthand
$\textrm{TR}(\alpha,\beta^{\prime}|U,\beta)$] if there exists a state
$\beta^{\prime}=\beta^{\prime}(\alpha)$ of the bath such that the Petz map
with respect to $\alpha$ is

${\hat{\mathcal{E}}_{\alpha}}[\bullet]=\Tr_{B}\bqty{U^{\dagger}\pqty{\bullet\otimes\beta^{\prime}}U}$
(25)

for the same $U$ that enters the dilation of $\mathcal{E}$.

Notice that this definition does not mean that the reverse should be
implementable by acting only on the system, a situation studied by Aberg [55]
and inspired by dynamical decoupling. Even in the generic case of isothermal
processes one may have to invert the system-bath interaction, if the latter is
not constant.

###  IV.2 Product-Preserving Maps & Generalized Thermal Operations

Here we introduce two more definitions that will be used below.

######  Definition 2.

A unitary that acts in a joint Hilbert space
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is product-preserving if

$\exists(\alpha,\beta,\alpha^{\prime},\beta^{\prime}):\,U(\alpha\otimes\beta)U^{\dagger}=\alpha^{\prime}\otimes\beta^{\prime}$
(26)

Here, $\alpha,\alpha^{\prime}\in S(\mathcal{H}_{A})$ and
$\beta,\beta^{\prime}\in S(\mathcal{H}_{B})$ are positive semidefinite
operators with trace one, and at least one of $\alpha$, $\beta$ is not
maximally mixed (to exclude the obvious case
$U(\mathbb{1}\otimes\mathbb{1})U^{\dagger}=\mathbb{1}\otimes\mathbb{1}$). We
shall also call any tuple $(U,\alpha,\beta)$ for which $U$ is product-
preserving with respect to $\alpha$ and $\beta$ a product-preserving tuple.

Contrary to the production of correlations (e.g. in universal entanglers [56,
57]) and the preservation of maximal entanglement (Bell-to-Bell maps [57]),
product-preservation has not been studied systematically prior to this work.
It appears as a natural property of thermal maps, and is at the origin of
several results in entropy production, thermalization and reversibility in the
quantum regime [36, 14, 37, 38]. By relaxing Eq. (12), we enlarge that natural
setting in a way that was already used in some other works in the literature
[14, 18]:

######  Definition 3.

A unitary that acts on a joint Hilbert space
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is a generalized thermal unitary if

$\displaystyle\exists(H_{A},H_{B},H^{\prime}_{A},H^{\prime}_{B}):$ (27)
$\displaystyle\qquad U(H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes
H_{B})U^{\dagger}=H_{A}^{\prime}\otimes\mathbb{1}+\mathbb{1}\otimes
H_{B}^{\prime},$

where either $H_{A}$ or $H_{B}$ is not proportional to the identity. The
corresponding generalized thermal map is given by
$\mathcal{E}[\bullet]=\Tr_{B}\\{U[\bullet\otimes\,\tau_{\kappa}({H_{B}})]U^{\dagger}\\}$.

It is straightforward (see Appendix A) that

$U\tau_{\kappa}({H_{A}})\otimes\tau_{\kappa}({H_{B}})U^{\dagger}=\tau_{\kappa}({H^{\prime}_{A}})\otimes\tau_{\kappa}({H^{\prime}_{B}})\,,$
(28)

that is, $(U,\tau_{\kappa}({H_{A}}),\tau_{\kappa}({H_{B}}))$ are a product-
preserving tuple for all generalized thermal operations.

##  V Results

Having defined tabletop time-reversibility (25), product-preservation (26) and
generalized thermal channels (27), we prove several results connecting them.

In subsection V.1, we prove some general connections between these classes of
processes. At a glance:

  * •

Theorem 1 fully characterizes the generalized thermal as a subset of product-
preserving unitaries.

  * •

Theorem 2 establishes a strong relation between the input pair
$(\alpha,\beta)$ and the output pair $(\alpha^{\prime},\beta^{\prime})$ of any
product-preserving unitary (and Corollary 1 interprets that relation in the
thermal case).

  * •

Theorem 3 is almost obvious, but is central to our work: it proves that
product preservation leads to tabletop reversibility.

  * •

Theorem 4 proves that the converse is not true, by exhibiting examples of
tabletop reversibility that do not arise from product preservation.

In subsection V.2, we present a thorough study for two-qubit unitaries. At a
glance:

  * •

Theorem 5 provides a parametric characterization of two-qubit generalized
thermal unitaries.

  * •

Theorem 6 shows that, given any two qubit unitary $U$ and a pure state
$\ket{\beta}$, there always exists $\ket{\alpha}$ such that
$U(\ket{\alpha}\otimes\ket{\beta})$ is product. This is unique to low-
dimensional cases, as the existence of non-product-preserving unitaries have
been proven in higher dimensions [58].

  * •

Theorems 7 and 8 provide two connections between the generalized thermal
character and the product-preservation properties of two-qubit unitaries.

###  V.1 General Results

######  Theorem 1.

$U$ is generalized thermal if and only if it is product-preserving with regard
to full rank $\alpha,\beta$.

###### Proof.

The “only if” direction has already been established in Eq. (28). For the “if”
implication, we note that for any full-rank state $\alpha$, one can always
find a corresponding $H_{A}$ for which
$\alpha=\tau_{\kappa}({H_{A}})=\exp{\kappa(H_{A}-Z_{H_{A}})}$, and likewise
for the bath state $\beta$ with the same inverse temperature $-\kappa$. Then,

$\displaystyle\ln(\alpha\otimes\beta)$ $\displaystyle=$
$\displaystyle\kappa(H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes
H_{B})-(Z_{H_{A}}+Z_{H_{B}})\mathbb{1}\otimes\mathbb{1}.$

The same construction can be done for the logarithm of the output product
states $\ln(\alpha^{\prime}\otimes\beta^{\prime})$. By invoking

$UAU^{\dagger}=A^{\prime}\;\iff\;Uf(A)U^{\dagger}=f(A^{\prime}),$ (29)

the product-preserved behavior of the tuple implies
$U\ln(\alpha\otimes\beta)U^{\dagger}=\ln(\alpha^{\prime}\otimes\beta^{\prime})$,
which is Eq. (27). Thus there will always exist for every product-preserved
tuple with full rank $\alpha,\beta$ some
$H_{A},H_{B},H_{A}^{\prime},H_{B}^{\prime}$ such that $U$ is a generalized
thermal unitary. ∎

Notice that, in the previous theorem, the condition of full rank cannot be
relaxed. Indeed, on the one hand, the logarithm of rank-defective states is
ill-defined. On the other hand, it is simple to find unitaries that preserve
one pure product state, and that are not even close (in any meaningful
distance) to a generalized thermal unitary. One such example is the two-qubit
unitary $U\ket{00}=\ket{00}$,
$U\ket{01}=\frac{1}{\sqrt{3}}(\omega\ket{01}+\omega^{*}\ket{10}+\ket{11})$,
$U\ket{10}=\frac{1}{\sqrt{3}}(\omega^{*}\ket{01}+\omega\ket{10}+\ket{11})$,
$U\ket{11}=\frac{1}{\sqrt{3}}(\ket{01}+\ket{10}+\ket{11})$ with
$\omega=e^{2\pi i/3}$.

Next, in the definition of product-preservation, we have merely required the
input states $\alpha\otimes\beta$ and output states
$\alpha^{\prime}\otimes\beta^{\prime}$ to be uncorrelated. Now we show that
product-preservation implies a stronger relationship between the two:

######  Theorem 2.

If $(U,\alpha,\beta)$ is a product-preserving tuple with output states
$\alpha^{\prime}$ and $\beta^{\prime}$, then there exist local unitaries
$u_{A}$ and $u_{B}$ such that
$\alpha^{\prime}\otimes\beta^{\prime}=(u_{A}\alpha
u_{A}^{\dagger})\otimes(u_{B}\beta u_{B}^{\dagger})$. In the case
$d_{A}=d_{B}$, this may hold up to a swap: i.e. it could be
$\alpha^{\prime}\otimes\beta^{\prime}=(u_{A}\beta
u_{A}^{\dagger})\otimes(u_{B}\alpha u_{B}^{\dagger})$.

###### Proof.

We denote $\sigma[\rho]$ as the eigenspectrum of $\rho$, and write the spectra
of $\alpha$ and $\beta$ in the following way:

  1. 1.

$\sigma[\alpha]=\\{\lambda_{1},\dots,\lambda_{d_{A}}\\}$, where $\forall
i:\lambda_{i}\geq\lambda_{i+1}$;

  2. 2.

$\sigma[\beta]=\\{\mu_{1},\dots,\mu_{d_{B}}\\}$, where $\forall
j:\mu_{j}\geq\mu_{j+1}$.

As such, $\sigma[\alpha\otimes\beta]$ is given by the set of values
$m_{i,j}=\lambda_{i}\mu_{j}$. This implies that $m_{i,j}\geq m_{i+1,j}$ and
$m_{i,j}\geq m_{i,j+1}$ for all $i,j$. Since the input and output products
differ by a unitary transformation,
$\sigma[\alpha^{\prime}\otimes\beta^{\prime}]$ is also given by the same set
of $m_{i,j}$, for which some
$\sigma[\alpha^{\prime}]=\\{\lambda_{1}^{\prime},\dots,\lambda_{d_{A}}^{\prime}\\}$
and $\sigma[\beta^{\prime}]=\\{\mu_{1}^{\prime},\dots,\mu_{d_{B}}^{\prime}\\}$
exist such that $m_{i,j}=\lambda_{i}^{\prime}\mu_{j}^{\prime}$. It is
therefore necessary that $\forall
i:\lambda_{i}^{\prime}\geq\lambda_{i+1}^{\prime}$, and likewise for
$\mu_{j}^{\prime}$.

Now, assume that there is some $i$ for which
$\lambda_{i}^{\prime}=c\lambda_{i}$. This means that for every $j$,
$\mu_{j}^{\prime}=c^{-1}\mu_{j}$. But since $\sum_{j}\mu_{j}=1$ and
$\sum_{j}\mu_{j}^{\prime}=1$, summing over $j$ for both sides of
$\mu_{j}^{\prime}=c^{-1}\mu_{j}$ gives $c=1$. This argument also works for the
values of $\mu_{j}$. Therefore, the spectra must always be conserved.

An “up to a swap” is obtained if we begin by assuming that there is some $i$
for which $\lambda_{i}^{\prime}=c\mu_{i}$. Thus, $(U,\alpha,\beta)$ is a
product-preserving tuple if and only if the spectra of $\alpha$ and $\beta$
are conserved up to a swap:

$\displaystyle\sigma[\alpha^{\prime}]=\sigma[\alpha]\wedge\sigma[\beta^{\prime}]=\sigma[\beta]$
(30) $\displaystyle\vee$
$\displaystyle\sigma[\alpha^{\prime}]=\sigma[\beta]\wedge\sigma[\beta^{\prime}]=\sigma[\alpha].$

Note that the second set of conditions can only be fulfilled if
$|\text{supp}(\alpha)|\leq d_{B}$ and $|\text{supp}(\beta)|\leq d_{A}$.

Since the spectra are conserved, up to a swap, we can always find some local
unitary $u_{A}$ that brings $\alpha$ to $\alpha^{\prime}$, and similarly for
$u_{B}$, which completes the proof. ∎

Notably, Theorem 2 shows that demanding the global unitary to preserve
noncorrelation is enough to ensure that the _local spectra_ of the input
states are preserved, up to a swap between the input and ancilla. This
provides us with a corollary on the level of the Hamiltonian:

######  Corollary 1.

If $(U,H_{A},H_{B})$ is a generalized thermal tuple with output Hamiltonians
$H_{A}^{\prime}$ and $H_{B}^{\prime}$, then
$H_{A}^{\prime}=u_{A}H_{A}u_{A}^{\dagger}$ and
$H_{B}^{\prime}=u_{B}H_{B}u_{B}^{\dagger}$, or
$H_{A}^{\prime}=u_{A}H_{B}u_{A}^{\dagger}$ and
$H_{B}^{\prime}=u_{B}H_{A}u_{B}^{\dagger}$, where $u_{A}$ and $u_{B}$ are some
local unitaries on the system and ancilla respectively.

###### Proof.

To prove this, we use Theorem 2 on Eq. (28), and then apply Eq. (29) to find
$\tau_{c}(H_{A}^{\prime})\otimes\tau_{c}(H_{B}^{\prime})=[u_{A}\tau_{c}(H_{A})u_{A}^{\dagger}]\otimes[u_{B}\tau_{c}(H_{B})u_{B}^{\dagger}]=\tau_{c}(u_{A}H_{A}u_{A}^{\dagger})\otimes\tau_{c}(u_{B}H_{B}u_{B}^{\dagger})$,
up to a swap. Finally, noting that the exponential of a Hermitian operator is
full rank, we can take the logarithm to complete the proof. ∎

Let us now state the following theorem

######  Theorem 3.

If $(U,\alpha,\beta)$ is a product-preserving tuple, the channel
$\mathcal{E}[\bullet]=\Tr_{B}\bqty{U(\bullet\otimes\beta)U^{\dagger}}$ is
tabletop reversible for reference prior $\alpha$.

###### Proof.

Using (20), it is immediate that

$\displaystyle\mathcal{U}[\alpha\otimes\beta]$
$\displaystyle=\alpha^{\prime}\otimes\beta^{\prime}$ (31)
$\displaystyle\Rightarrow{\hat{\mathcal{E}}_{\alpha}}[\bullet]$
$\displaystyle=\Tr_{B}{}\circ\mathcal{U}^{\dagger}\circ\hat{\text{T}}\text{r}_{B,\alpha^{\prime}\otimes\beta^{\prime}}[\bullet]$
$\displaystyle=\Tr_{B}\bqty{U^{\dagger}\pqty{\bullet\otimes\beta^{\prime}}U}$

which is tabletop reversibility (25). ∎

Next, we prove that the converse of Theorem (3) is not true:

######  Theorem 4.

$\textrm{TR}(\alpha,\beta^{\prime}|U,\beta)$ does not imply that
$(U,\alpha,\beta)$ is generalized thermal.

###### Proof.

We prove this with two counterexamples.

For the first example, we look at the two-qubit channel
$\mathcal{E}_{1}[\bullet]:=\Tr_{B}[U_{1}(\bullet\otimes\beta_{1})U_{1}^{\dagger}]$
with
$U_{1}=\outerproduct{0}{0}\otimes\mathbb{1}+\outerproduct{1}{1}\otimes\sigma_{x}$
and $\commutator{\beta_{1}}{\sigma_{x}}\neq 0$.

Given a prior
$\alpha_{1}=a_{0}\outerproduct{0}{0}+(1-a_{0})\outerproduct{1}{1}$, it can be
verified that $\mathcal{E}_{1}[\alpha_{1}]=\alpha_{1}$, while
$U_{1}(\alpha_{1}\otimes\beta_{1})U_{1}^{\dagger}=a_{0}\outerproduct{0}{0}\otimes\beta_{1}+(1-a_{0})\outerproduct{1}{1}\otimes(\sigma_{x}\beta_{1}\sigma_{x})\neq\alpha_{1}\otimes\beta_{1}$.
Though separable, this is not a product state, so
$(U_{1},\alpha_{1},\beta_{1})$ is not generalized thermal.

The retrodiction of $\mathcal{E}_{1}$ with respect to $\alpha_{1}$ can be
found to be
$\hat{\mathcal{E}}_{1,\alpha_{1}}[\bullet]=\sum_{k}\outerproduct{k}{k}\bullet\outerproduct{k}{k}$.
Meanwhile, by setting
$\beta_{1}^{\prime}=(\mathbb{1}+b_{y}^{\prime}\sigma_{y}+b_{z}^{\prime}\sigma_{z})/2$,
we have
$\overline{\mathcal{E}}_{1}[\bullet]:=\Tr_{B}[U_{1}^{\dagger}(\bullet\otimes\beta_{1}^{\prime})U_{1}]=\hat{\mathcal{E}}_{1,\alpha_{1}}[\bullet]$.
Therefore, $\textrm{TR}(\alpha_{1},\beta_{1}^{\prime}|U_{1},\beta_{1})$.

The second example is
$\mathcal{E}_{2}[\bullet]:=\Tr_{B}[U_{2}(\bullet\otimes\beta_{2})U_{2}^{\dagger}]$
with

$\displaystyle U_{2}$ $\displaystyle=\spmqty{undef},$ (32)
$\displaystyle\beta_{2}$
$\displaystyle=\frac{1}{217}\left(\small\begin{array}[]{llll}70&84e^{\frac{i\pi}{3}}&35&30e^{-\frac{2i\pi}{3}}\\\
84e^{-\frac{i\pi}{3}}&112&42e^{-\frac{i\pi}{3}}&-35\\\
35&42e^{\frac{i\pi}{3}}&21&15e^{-\frac{2i\pi}{3}}\\\
30e^{\frac{2i\pi}{3}}&-35&15e^{\frac{2i\pi}{3}}&14\\\ \end{array}\right).$

The retrodiction channel $\hat{\mathcal{E}}_{2,\alpha_{2}}$ with respect to
the prior $\alpha_{2}=\mathbb{1}/2$ is the depolarizing channel
$\hat{\mathcal{E}}_{2,\alpha_{2}}[\bullet]=\text{Tr}[\bullet]\mathbb{1}/2$. At
the same time,
$\overline{\mathcal{E}}_{2}[\bullet]:=\Tr_{B}[U^{\dagger}(\bullet\otimes\beta_{2}^{\prime})U]$
with $\beta_{2}^{\prime}=\mathbb{1}/2$ is also the depolarizing channel, so
$\textrm{TR}(\alpha_{2},\beta_{2}^{\prime}|U_{2},\beta_{2})$.

Meanwhile, the log negativity of
$\Omega_{2}:=U_{2}(\alpha_{2}\otimes\beta_{2})U_{2}^{\dagger}$ is found to be
$\log_{2}\tr\|\Omega_{2}^{T_{B}}\|_{1}\approx 0.9135$, where $\bullet^{T_{B}}$
is a partial transpose on the ancilla and $\|\bullet\|_{1}$ is the trace norm.
Since log negativity is an entanglement monotone [59], and considering that
the maximally-entangled pure state has a log negativity of 1, $\Omega_{2}$ is
in fact very highly entangled. So, $(U_{2},\alpha_{2},\beta_{2})$ is certainly
not a product-preserving tuple, and hence is not generalized thermal either.

These examples demonstrate that channels can be tabletop time-reversible
without being generalized thermal with respect to certain priors. Note that
the first example also holds for the classical case by setting
$b_{y}^{\prime}=0$, while the second example is inherently quantum due to the
presence of entanglement. ∎

Figure 4: Schematic of the order relationships between the tuple sets we have
introduced. Note that the non-tabletop time-reversible but product-preserved
tuples are those for which a Bayesian inversion is not well-defined in general
due to the presence of rank-deficient output states: although it is still
possible to define some retrodiction channel in this case, it will depend on
the chosen convention, and the different approaches are inconsistent with each
other (Section VI.2, Appendix F).

###  V.2 Results for Qubit Channels with Two-Qubit Dilations

We shall now focus our attention to one-qubit generalized thermal channels
with two-qubit dilations, and product-preserving unitaries acting on two
qubits. It is known that every two-qubit unitary permits the Cartan
decomposition [60]

$U=(u_{A}\otimes
u_{B})\exp\left(i\sum_{k=1}^{3}t_{k}\sigma_{k}\otimes\sigma_{k}\right)(v_{A}^{\dagger}\otimes
v_{B}^{\dagger}),$ (33)

where $\\{\sigma_{k}\\}_{k=1}^{3}$ are the usual Pauli operators, while
$v_{A}$, $v_{B}$, $u_{A}$, and $u_{B}$ are single-qubit unitaries. Hence,
every $U$ would be specified by these local unitaries and angles
$\\{t_{k}\\}_{k=1}^{3}$. With reference to this parametrization, we fully
characterize all two-qubit generalized thermal unitaries:

######  Theorem 5.

A two-qubit unitary $U$, parameterized as Eq. (33), is generalized thermal if
and only if $(t_{j}-t_{k})\bmod(\pi/2)=0$ or $(t_{j}+t_{k})\bmod(\pi/2)=0$ for
some $j\neq k$.

###### Proof.

For a given two-qubit unitary $U$, we prove this by characterizing every
possible pair of Hamiltonians $H_{A}$ and $H_{B}$ such that $(U,H_{A},H_{B})$
is a generalized thermal tuple. The proof by direct inspection is done in
Appendix E, divided in three lemmas: the main one covers all the $U$ such that
$t_{k}\bmod(\pi/4)=0$ for at most one $t_{k}$; the other two settle the
remaining special cases. ∎

Due to Theorem 1, the above also fully characterizes all two-qubit unitaries
that are product-preserving with respect to full rank states. There are in
fact many more product-preserving unitaries for two-qubits. Indeed, every two-
qubit unitary is not just product-preserving, but product-preserving with
respect to every pure ancilla:

######  Theorem 6.

For every two-qubit $U$ and ancilla $\ket{\beta}$, there exists an
$\ket{\alpha}$ such that $(U,\ket{\alpha},\ket{\beta})$ is a product-
preserving tuple.

###### Proof.

Let $v_{B}^{\dagger}\ket{\beta}\widehat{=}(b_{0},b_{1})^{T}$,
$v_{A}^{\dagger}\ket{\alpha}\widehat{=}(a_{0},a_{1})^{T}\propto(1,x)^{T}$,
where $v_{A}$ and $v_{B}$ are the same as in Eq. (33), and we have assumed for
now that $a_{0}\neq 0$. Then,

$\displaystyle(u_{A}^{\dagger}\otimes
u_{B}^{\dagger})U(\ket{\alpha}\otimes\ket{\beta})$
$\displaystyle{}\widehat{=}{}\pmqty{undef},$ (34)
$\displaystyle(u_{A}^{\dagger}\ket{\alpha^{\prime}})\otimes(u_{B}^{\dagger}\ket{\beta^{\prime}})$
$\displaystyle{}\widehat{=}{}\pmqty{undef}\otimes\pmqty{undef},$

where $f_{jk}(x)$ are linear functions of $x$. The condition that $U$ is
product-preserving with respect to $\ket{\alpha}\otimes\ket{\beta}$ requires
$(a_{0}^{\prime}b_{0}^{\prime})(a_{1}^{\prime}b_{1}^{\prime})=(a_{0}^{\prime}b_{1}^{\prime})(a_{1}^{\prime}b_{0}^{\prime})$,
which implies

$f_{00}(x)f_{11}(x)=f_{01}(x)f_{10}(x).$ (35)

Since this is at most a quadratic equation in $x$, a solution almost always
exists for any choice of $U$, $b_{0}$, and $b_{1}$. If a solution for $x$
exists, then $U$ is product-preserving with respect to
$\ket{\alpha}=v_{A}\pqty{\ket{0}+x\ket{1}}/\sqrt{1+\absolutevalue{x}^{2}}$ and
the specified $\ket{\beta}$. A solution would not exist for Eq. (35) if it
results in a contradiction of the form $c=0$ for a nonzero constant $c$.
However, in those cases, it is shown in Appendix D that $U$ is product-
preserving with respect to $\ket{\alpha}=v_{A}\ket{1}$ and $\ket{\beta}$. ∎

######  Corollary 2.

Every two-qubit unitary is product-preserving with regard to some states.

This corollary was already known in the context of universal entanglers, where
it has been shown that there is no two-qubit unitary that takes _every_
product state to an entangled state [58].

Finally, we present an characterisation of the generalized thermal two-qubit
unitaries, an alternative to Theorem 5, in terms of the product states that
they preserve:

######  Theorem 7.

A two-qubit unitary $U$ is generalized thermal if and only if it is product-
preserving with respect to two pure states
$\ket{\alpha_{+}}\otimes\ket{\beta_{+}}$ and
$\ket{\alpha_{-}}\otimes\ket{\beta_{-}}$, such that
$\langle{\alpha_{+}}|{\alpha_{-}}\rangle=\langle{\beta_{+}}|{\beta_{-}}\rangle=0$.

###### Proof.

The “if” direction: We shall first consider the case where $H_{A}$ and $H_{B}$
are nondegenerate. Let $\ket{\alpha_{\pm}}$ and $\ket{\beta_{\pm}}$ be the
eigenstates of $H_{A}$ and $H_{B}$ respectively, with the ground states
labelled $\ket{\alpha_{-}}$ and $\ket{\beta_{-}}$. Taking the low temperature
limit $\lim_{c\to\infty}U\left[\tau_{\pm c}(H_{A})\otimes\tau_{\pm
c}(H_{B})\right]U^{\dagger}=\lim_{c\to\infty}\tau_{\pm
c}(H_{A}^{\prime})\otimes\tau_{\pm c}(H_{B}^{\prime})$, we have

$U\left[\outerproduct{\alpha_{\mp}}{\alpha_{\mp}}\otimes\outerproduct{\beta_{\mp}}{\beta_{\mp}}\right]U^{\dagger}=\outerproduct{\alpha^{\prime}_{\mp}}{\alpha^{\prime}_{\mp}}\otimes\outerproduct{\beta^{\prime}_{\mp}}{\beta^{\prime}_{\mp}},$
(36)

so $U$ is product-preserving with respect to
$\ket{\alpha_{\pm}}\otimes\ket{\beta_{\pm}}$.

The “only if” direction: Let
$\ket{\alpha_{\pm}}\otimes\ket{\beta_{\pm}}=:(v_{A}\otimes
v_{B})(\ket{\pm}\otimes\ket{\pm})$ and
$\ket{\alpha^{\prime}_{\pm}}\otimes\ket{\beta^{\prime}_{\pm}}=:(u_{A}\otimes
u_{B})(\ket{\pm}\otimes\ket{\pm})$. Then,

$\displaystyle(u_{A}^{\dagger}\otimes u_{B}^{\dagger})U(v_{A}\otimes v_{B})$
(37) $\displaystyle\quad{}\widehat{=}{}\left(\begin{matrix}1&0&0&0\\\
0&e^{i(a+\phi)}\cos(\frac{\theta}{2})&e^{-i(b-\phi)}\sin(\frac{\theta}{2})&0\\\
0&-e^{i(b+\phi)}\sin(\frac{\theta}{2})&e^{-i(a-\phi)}\cos(\frac{\theta}{2})&0\\\
0&0&0&1\end{matrix}\right),$

where the top-left and bottom-right entries originate from the product-
preserving condition, while the rest of the entries are parameterised to
impose unitarity. Define
$H_{A}:=\omega_{A}\outerproduct{\alpha_{+}}{\alpha_{+}}-\omega_{A}\outerproduct{\alpha_{-}}{\alpha_{-}}$
and
$H_{B}:=\omega_{B}\outerproduct{\beta_{+}}{\beta_{+}}-\omega_{B}\outerproduct{\beta_{-}}{\beta_{-}}$.
From direct computation, $U$ will be found to be generalized thermal with
respect to $H_{A}$ and $H_{B}$ for any $\omega_{A}$ and $\omega_{B}$ if
$\theta=2n\pi$ for some integer $n$, and for $\omega_{A}=\omega_{B}$ if
$\theta\neq 2n\pi$.

To handle the degenerate case, we use the full characterization of two-qubit
generalized thermal unitaries from Appendix E. Specifically, lemmas 1′ & 1′′
states that $U$ is generalized thermal with respect to
$H_{A}\not\propto\mathbb{1}$ and $H_{B}\propto\mathbb{1}$ if and only if it is
generalized thermal with respect to $H_{A}\not\propto\mathbb{1}$ and
$\bar{H}_{B}:=v_{B}v_{A}^{\dagger}H_{A}v_{A}v_{B}^{\dagger}$. Since
$\bar{H}_{B}$ shares the same spectrum as $H_{A}$ and is therefore
nondegenerate, the rest of the proof follows as stated above. ∎

If we have a situation that only one of the states in the product-preserved
tuple is pure, then we can also conclude that the unitary is generalized
thermal:

######  Theorem 8.

For a two-qubit $U$ and full-rank $\alpha$, if
$(U,\alpha,\outerproduct{\beta}{\beta})$ is a product-preserving tuple, then
$U$ is generalized thermal.

###### Proof.

Up to a swap, $U$ is product-preserving with respect to a full-rank $\alpha$
and pure $\outerproduct{\beta}{\beta}$ if

$U\left(\alpha\otimes\outerproduct{\beta}{\beta}\right)U^{\dagger}=\alpha^{\prime}\otimes\outerproduct{\beta^{\prime}}{\beta^{\prime}},$
(38)

where we have used Theorem 2 to conclude that $\alpha^{\prime}$ is full-rank
and $\outerproduct{\beta^{\prime}}{\beta^{\prime}}$ is pure. Consider first
the special case of $\alpha=\mathbb{1}/2$. Let $\ket{\beta_{+}}:=\ket{\beta}$
and $\ket{\beta_{-}}$ to be its orthogonal state. Then,

$\displaystyle
U\pqty{\mathbb{1}\otimes\outerproduct{\beta_{-}}{\beta_{-}}}U^{\dagger}$
$\displaystyle=\mathbb{1}-U\pqty{\mathbb{1}\otimes\outerproduct{\beta_{+}}{\beta_{+}}}U^{\dagger}$
(39)
$\displaystyle=\mathbb{1}\otimes\outerproduct{\beta^{\prime}_{-}}{\beta^{\prime}_{-}}.$

Hence,
$\beta=\tau_{\kappa}\pqty{\outerproduct{\beta_{+}}{\beta_{+}}-\outerproduct{\beta_{-}}{\beta_{-}}}$
for any $\kappa$ will satisfy
$U(\alpha\otimes\beta)U^{\dagger}=\alpha^{\prime}\otimes\beta^{\prime}$, so
$U$ is generalized thermal.

For $\alpha\neq\mathbb{1}/2$, let
$\alpha=p_{+}\outerproduct{\alpha_{+}}{\alpha_{+}}+p_{-}\outerproduct{\alpha_{-}}{\alpha_{-}}$
with $p_{+}>p_{-}$. It is clear that

$\displaystyle
U\pqty{\alpha^{n}\otimes\outerproduct{\beta_{+}}{\beta_{+}}}U^{\dagger}$
$\displaystyle=\bqty{U\pqty{\alpha\otimes\outerproduct{\beta_{+}}{\beta_{+}}}U^{\dagger}}^{n}$
(40) $\displaystyle=\alpha^{\prime
n}\otimes\outerproduct{\beta_{+}}{\beta_{+}}.$

Taking the limit $n\to\infty$ with $\alpha^{n}/\Tr(\alpha^{n})$ leads to
$U\pqty{\ket{\alpha_{+}}\otimes\ket{\beta_{+}}}=\ket{\alpha_{+}^{\prime}}\otimes\ket{\beta_{+}^{\prime}}$.
Then, with $\outerproduct{\alpha_{-}}{\alpha_{-}}\propto\alpha-
p_{+}\outerproduct{\alpha_{+}}{\alpha_{+}}$, where the product form of both
terms on the right are preserved by $U$ with respect to the same ancilla state
$\outerproduct{\beta_{+}}{\beta_{+}}$, we also have
$U\pqty{\ket{\alpha_{-}}\otimes\ket{\beta_{+}}}=\ket{\alpha_{-}^{\prime}}\otimes\ket{\beta_{+}^{\prime}}$.

Therefore, $U$ is product-preserving with respect to
$\outerproduct{\alpha_{+}}{\alpha_{+}}+\outerproduct{\alpha_{-}}{\alpha_{-}}=\mathbb{1}$
and $\outerproduct{\beta}{\beta}$. From the first part of the proof, this
implies that $U$ is generalized thermal. ∎

Note that the converse does not hold. From the proof, for a $U$ to be product-
preserving with respect to some input state of the form
$\alpha\otimes\outerproduct{\beta_{+}}{\beta_{+}}$, it must necessarily be
generalized thermal with respect to
$\mathbb{1}\otimes\tau_{\kappa}(\outerproduct{\beta_{+}}{\beta_{+}}-\outerproduct{\beta_{-}}{\beta_{-}})$
for all $\kappa$. From the proofs in Appendix E, this is only possible when
there exists $j,k\in\\{1,2,3\\}$ and $j\neq k$ such that either
$t_{j},t_{k}\bmod(\pi/2)=0$ or $t_{j},t_{k}\bmod(\pi/2)=\pi/4$. Therefore, a
generalized thermal $U$ is in general not product-preserving with respect to
some $\alpha\otimes\outerproduct{\beta}{\beta}$.

##  VI Further observations

While the preceding sections put emphasis on the mathematical aspects of
product-preserving and generalized thermal unitaries, we now consolidate
notable physical insights that the results elucidate. We devote one subsection
to each of the notions we introduced: generalized thermal operations, product-
preservation, and tabletop reversibility.

###  VI.1 On Generalized Thermal Maps

Generalized thermal operations are the largest class of operations that can be
described by the following procedure [61]: a system (with free Hamiltonian
$H_{A}$) and a bath (with free Hamiltonian $H_{B}$) are brought into contact,
allowed to interact for some time with a unitary $U$, then are decoupled
again. As a result, the final Hamiltonian must be without interaction. A
scattering experiment, where two distinct collection of particles start far
apart, come together to interact in a complicated way, and two (possibly
different) collection of particles leave the interaction region, would be an
example of such a process. That said, we do not make any assumptions about the
interaction unitary $U$ apart from the fact that the initial and final
Hamiltonians are decoupled. This procedure indeed describes our definition of
generalized thermal operations [Eq. (27)].

Corollary 1 shows that every process described by the procedure above must
obey a strong constraint: the _local conservation of energy spectra_ , up to a
possible swap (if the system and the bath have the same dimension, we can
always choose to redefine which is which).

The requirement [Eq. (12)] that the local Hamiltonians be unchanged, which is
the standard definition of thermal operation [61, 62], is strictly stronger.
That being said, given a generalized thermal tuple $(U,H_{A},H_{B})$, up to a
swap there exists local unitaries $v_{A}$ and $v_{B}$, such that
$(U^{\prime},H_{A},H_{B})$ is a thermal tuple, with $U^{\prime}=(v_{A}\otimes
v_{B})U$. Indeed, using Corollary 1:

$\displaystyle U\bqty{H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes
H_{B}}U^{\dagger}=u_{A}H_{A}u_{A}^{\dagger}\otimes\mathbb{1}+\mathbb{1}\otimes
u_{B}H_{B}u_{B}^{\dagger}$ $\displaystyle\underbrace{(u_{A}^{\dagger}\otimes
u_{B}^{\dagger})U}_{=:U^{\prime}}\bqty{H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes
H_{B}}=\bqty{H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes H_{B}}U^{\prime}$
$\displaystyle\implies[U^{\prime},H_{A}\otimes\mathbb{1}+\mathbb{1}\otimes
H_{B}]=0$ (41)

which is the claimed result with $v_{X}=u^{\dagger}_{X}$.

In Appendix E, we have characterized all possible generalized thermal tuples
$(U,H_{A},H_{B})$ for two qubits; whence all generalized thermal maps that
possess a two-qubit dilation can be inferred. This complements the recent
characterization of all thermal maps for one qubit coupled to a bosonic bath
[63] and similar studies in resource theoretic contexts [43]. Such
characterizations are necessary for one of the main goals of a resource
theory, namely, to identify all states reachable from some initial state using
only free operations [64].

| For all $\mathbb{C}^{d}$ channels
---|---
For any $\alpha,\beta$ |  $U,H_{A},H_{B}$ is a generalized thermal tuple to $H_{A}^{\prime},H_{B}^{\prime}$ | $\Rightarrow$ |  $U,\alpha,\beta$ is product preserved to $\alpha^{\prime},\beta^{\prime}$ | Eq.(28)
$U,\alpha,\beta$ is product preserved to $\alpha^{\prime},\beta^{\prime}$ | $\Rightarrow$ |  $\mathcal{E}(U,\beta)$ is tabletop reversible with regard to $\alpha$ | Eq.(31)
For any full-rank $\alpha,\beta$ |  $U,\alpha,\beta$ is product preserved to $\alpha^{\prime},\beta^{\prime}$ | $\Leftrightarrow$ |  $U,H_{A},H_{B}$ is a generalized thermal tuple to $H_{A}^{\prime},H_{B}^{\prime}$ | Thm.1

Table 2: Summary of key results for general channels. For relevant results,
$\alpha=\tau_{\kappa}({H_{A}}),\beta=\tau_{\kappa}({H_{B}}),\alpha^{\prime}=\tau_{\kappa}({H_{A}^{\prime}}),\beta^{\prime}=\tau_{\kappa}({H_{B}^{\prime}})$
and $\mathcal{E}[\bullet]=\Tr_{B}{[U(\bullet\otimes\beta)U^{\dagger}]}$

| For all two-qubits channels
---|---
Full-rank $\alpha,\beta$ |  $U,\alpha,\beta$ is product preserved tuple to $\alpha^{\prime},\beta^{\prime}$ | $\Leftrightarrow$ |  $U$ s.t. $t_{j}-t_{k}\bmod\frac{\pi}{2}$ or $t_{j}+t_{k}\bmod\frac{\pi}{2}=0$ under (33) | Thm.5
Pure $\alpha,\beta$ | $\forall U,\outerproduct{\beta}{\beta}$ | $\Rightarrow$ |  $\exists\outerproduct{\alpha}{\alpha}$ for which $U,\alpha,\beta$ is product preserved | Thm.6
$(U,\ket{\alpha},\ket{\beta})$, $(U,\ket{\alpha_{\perp}},\ket{\beta_{\perp}})$ are product-preserved | $\Leftrightarrow$ |  $\exists H_{A},H_{B}$ for which $(U,H_{A},H_{B})$ is generalized thermal | Thm.7
Full rank $\alpha$, pure $\beta$ |  $U,\alpha,\outerproduct{\beta}{\beta}$ is product preserved tuple to $\alpha^{\prime},\beta^{\prime}$ | $\Rightarrow$ |  $U,H_{A},H_{B}$ is a generalized thermal tuple to $H_{A}^{\prime},H_{B}^{\prime}$ | Thm.8

Table 3: Summary of key results for channels with two-qubit dilations Figure
5: All two-qubit unitaries are equivalent up to local unitaries to the Weyl
chamber [60], as plotted here with coordinates $(t_{1},t_{2},t_{3})$. The
generalized thermal unitaries are marked out in gray, and occur only on the
surfaces
$(0,0,0)\text{--}(\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{4})\text{--}(\frac{\pi}{4},\frac{\pi}{4},0)$
and
$(0,0,0)\text{--}(\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{4})\text{--}(\frac{\pi}{2},0,0)$.

###  VI.2 Product-preservation Involving Rank-Deficient States

The discrepancy between product-preserving and generalized thermal unitaries
boils down to the states they are acting upon. In the preceding sections, we
have given explicit examples of unitaries that are product-preserving, but not
generalized thermal, with respect to pure states.

However, Gibbs states are full rank for finite inverse temperature. So, pure
states, or more generally rank-deficient states, are zero-temperature states
in the thermodynamic picture, and are not physically feasible except in the
limiting sense. Even outside the field of thermodynamics, since real-world
experiments are always susceptible to noise and uncertainty, pure states are
really idealizations that can never be actually prepared in the lab.

Whichever the motivation, a product-preserving unitary with respect to a rank-
deficient state is pragmatically useful only when there is a neighborhood of
full-rank states around it whose product structure is also preserved by the
same unitary. If so, one can choose a full-rank approximation of the target
state whose product structure is also preserved by the same unitary.

For two-qubit unitaries, which are all product-preserving, Theorem 7 therefore
provides a simple check for when that unitary has a product-preserving
property. Its proof also offers an exact construction of the input and ancilla
Hamiltonians for which they are a generalized thermal tuple together with
interaction unitary.

Meanwhile, the Stinespring dilation theorem asserts the uniqueness of the
dilation unitary for every quantum channel when the ancilla state is pure, up
to an isomorphism on the ancilla [65]. Therefore, for channels with dilations
whose input and ancilla are both a single qubit, Theorem 8 connects the
product-preserving property of the Stinespring dilation and the generalized-
thermal property of the dilation unitary.

It must be emphasized here that considering full-rank product-preserving
states in the neighborhood of a rank-deficient state, and taking the former to
be full-rank approximations of the latter, is a convenient choice: one that is
motivated by thermodynamic arguments, but an otherwise arbitrary one. Bayes’
rule and the Petz alike involve the inverse of the propagated reference. When
this state is rank-deficient, an inverse does not exist, rendering such
channels undefined. One might consider a neighborhood of full-rank states
whose inverses exists, and define the retrodiction channel with some limiting
process. However, our prevening discussion and an explicit example in appendix
F.1.3 shows that defining retrodiction this way depends on the chosen
neighborhood of full-rank states.

###  VI.3 Composable Tabletop Time-reversibility

Figure 6: Where the composition of many copies of the same channel involves
the same $\beta$ as the ancilla in each step, it can be desirable for the
reversal of the composite channel to involve the same $\beta^{\prime}$ as the
ancilla in each step in the opposite direction. We call channels that satisfy
Eq. (42) as composable tabletop time-reversible channels.

Apart from the reduction in reversal complexity of a tabletop-time-reversible
channel, one might also desire for this behavior to apply for _compositions_
of the same channel. For example, when the prior and ancilla are thermal
states, the reverse channel of the composition of a thermal operation is the
composition of the reverse channel of a single thermal operation [14].

More generally, it would be convenient if the reverse channel of a composition
of many copies of the same quantum operation can be implemented as illustrated
in Fig. 6. Formally, for a tabletop-time-reversible channel $\mathcal{E}$ with
unitary dilation ($U$, $\beta$) and positive integer $L$, one desires the
reverse channel of
$\mathcal{E}^{L}:=\underbrace{\mathcal{E}\circ\mathcal{E}\circ\dots\circ\mathcal{E}}_{L\text{
times}}$ to be

$\displaystyle\mathcal{R}_{q}\left[\mathcal{E}^{L},\alpha\right]$
$\displaystyle=\hat{\mathcal{E}}_{\alpha}\circ\hat{\mathcal{E}}_{\mathcal{E}[\alpha]}\circ\hat{\mathcal{E}}_{\mathcal{E}^{2}[\alpha]}\circ\dots\circ\hat{\mathcal{E}}_{\mathcal{E}^{L-1}[\alpha]}$
(42)
$\displaystyle\overset{!}{=}\hat{\mathcal{E}}_{\alpha}\circ\hat{\mathcal{E}}_{\alpha}\circ\hat{\mathcal{E}}_{\alpha}\circ\dots\circ\hat{\mathcal{E}}_{\alpha}$
$\displaystyle=\mathcal{R}_{q}\left[\mathcal{E},\alpha\right]^{L}\quad\forall
L\in\mathbb{Z}^{+},\text{where}$ $\displaystyle\hskip
70.0001pt\hat{\mathcal{E}}_{\alpha}[\bullet]=\Tr[U^{\dagger}(\bullet\otimes\beta^{\prime})U].$

For brevity, we shall use $\textrm{TR}_{c}(\alpha,\beta^{\prime}|U,\beta)$ to
denote such _composable_ tabletop-time-reversible
channels—$\textrm{TR}(\alpha,\beta^{\prime}|U,\beta)$ that also satisfy Eq.
(42). Let us provide a few examples of such channels.

_Unitary channels._ The action of both the unitary and its inverse on the
input is independent of the state of the ancilla, so unitary channels are
composably tabletop time-reversible. While this is a trivial case, it aligns
with the intended definition of composable tabletop time-reversible channels
as illustrated in Fig. 6.

_Reverse channel with $N$-steady state prior._ Consider
$\mathcal{E}[\bullet]=\Tr_{B}[U(\bullet\otimes\beta)U^{\dagger}]$ that is
_not_ a unitary channel, with a set of priors
$\\{\alpha_{0},\alpha_{1},\dots,\alpha_{N-1}\\}$ that satisfy

$U(\alpha_{n}\otimes\beta)U^{\dagger}=\alpha_{(n+1)\bmod
N}\otimes\beta^{\prime}.$ (43)

Then,
$\hat{\mathcal{E}}_{\alpha_{n}}[\bullet]=\Tr[U^{\dagger}(\bullet\otimes\beta^{\prime})U]$
for all $n$, hence $\textrm{TR}_{c}(\alpha_{n},\beta^{\prime}|U,\beta)$. We
call this the “$N$_-steady state_ ” as the set $\\{\alpha_{n}\\}_{n=0}^{N-1}$
is unchanged under the channel. The $N=1$ case is the usual steady state of a
channel, which includes the class of thermal operations as previously studied
[14]. An example of such a channel for any $N$ is an $N$-dit channel
$\mathcal{E}$ given by the ancilla
$\beta=\sum_{k=0}^{N-1}b_{k}\outerproduct{k}{k}$ and the dilation unitary

$U=\sum_{k=0}^{N-1}u_{k}\otimes\outerproduct{\psi_{k}}{k}.$ (44)

Here, $\sigma$ is a permutation of order $N$,
$\\{\ket{\psi_{k}}\\}_{k=0}^{N-1}$ and $\\{\ket{k}\\}_{k=0}^{N-1}$ are
orthonormal bases, and

$u_{k}=\sum_{j=0}^{N-1}e^{i\phi_{j}^{(k)}}\outerproduct{\sigma(j)}{j},$ (45)

with $u_{k}^{\dagger}u_{k^{\prime}}\not\propto\mathbb{1}$ for all $k\neq
k^{\prime}$, which ensures that $\mathcal{E}$ is not a unitary channel.
Meanwhile, the priors are defined as
$\alpha_{n}=\sum_{k=0}^{N-1}a_{k}\outerproduct{\sigma^{n}(k)}{\sigma^{n}(k)}$
where $\alpha_{n}$ is not degenerate, so that
$\alpha_{n}\neq\alpha_{n^{\prime}}$ for all $n\neq n^{\prime}$. Since
$u_{k}\outerproduct{\sigma^{n}(j)}{\sigma^{n}(j)}u_{k}^{\dagger}=\outerproduct{\sigma^{n+1}(j)}{\sigma^{n+1}(j)}$,
we have that

$\displaystyle U(\alpha_{n}\otimes\beta)U^{\dagger}$ (46)
$\displaystyle\quad{}={}\sum_{k=0}^{N-1}\sum_{j=0}^{N-1}a_{j}b_{k}\pqty\Big{u_{k}\outerproduct{\sigma^{n}(j)}{\sigma^{n}(j)}u_{k}^{\dagger}}\otimes\outerproduct{\psi_{k}}{\psi_{k}}$
$\displaystyle\quad{}={}\underbrace{\pqty{\sum_{j=0}^{N-1}a_{j}\outerproduct{\sigma^{n+1}(j)}{\sigma^{n+1}(j)}}}_{=\alpha_{(n+1)\bmod
N}}\otimes\underbrace{\pqty{\sum_{k=0}^{N-1}b_{k}\outerproduct{\psi_{k}}{\psi_{k}}}}_{:=\beta^{\prime}}.$

This satisfies Eq. (43), and hence this channel is composably tabletop time-
reversible with respect to $\alpha_{n}$.

_Channels with idempotent reverse channels._ These are tabletop time-
reversible channels whose reverse channels are idempotent, both with
themselves and subsequent priors: that is, they have the property
$\hat{\mathcal{E}}_{\alpha}\circ(\hat{\mathcal{E}}^{L-1})_{\mathcal{E}[\alpha]}=\hat{\mathcal{E}}_{\alpha}\circ\hat{\mathcal{E}}_{\alpha}=\hat{\mathcal{E}}_{\alpha}$.
If so, Eq. (42) is also satisfied. An example of such a channel is when the
dilation unitary is a swap, as

$\displaystyle U(\alpha\otimes\beta)U^{\dagger}$
$\displaystyle=\beta\otimes\alpha$ (47)
$\displaystyle\implies\hat{\mathcal{E}}_{\alpha}[\bullet]$
$\displaystyle=\Tr_{B}[U^{\dagger}(\bullet\otimes\alpha)U]$

for every $\alpha$. Clearly,
$\hat{\mathcal{E}}_{\alpha}\circ\hat{\mathcal{E}}_{\mathcal{E}[\alpha]}\circ\dots\circ\hat{\mathcal{E}}_{\mathcal{E}^{L-1}[\alpha]}[\bullet]=\alpha=\hat{\mathcal{E}}_{\alpha}[\bullet]=\hat{\mathcal{E}}_{\alpha}\circ\hat{\mathcal{E}}_{\alpha}[\bullet]$,
so the reverse channel with respect to the prior $\alpha$ is idempotent.

These are the classes of composable tabletop time-reversible channels that we
have thus far identified. In our study of generalized thermal channels with
two-qubit dilations, all $\textrm{TR}_{c}$ channels we have found belong to
one of the above classes. It is not known if this is an exhaustive list, as
there might be richer families of $\textrm{TR}_{c}$ channels in higher
dimensions.

##  VII Conclusion

Using a recipe from Bayesian inference, one can associate any physical
channel, however irreversible, to a family of reverse channels indexed by the
choice of a reference prior. We proved that an apparently different recipe,
based on dilating the system to include the bath into which information is
dissipated, leads exactly to the same family of reverse channels.

For thermal channels, it was known that a natural reverse channel consists of
reversing the unitary evolution while in contact with the same thermal bath
(in our framework, this is obtained by choosing the Gibbs state as reference
prior). We ask for which other channels such phenomenon happens: that the
reverse channel can be implemented with the same devices and similar baths as
the original channel (“tabletop reversibility”). These questions further
inspire the related definition of “product-preserving” maps that contains
generalized thermal channels as an important subclass. We then proved several
relations between these classes, both in general and with more detail for two-
qubit unitaries. In particular, we show that when the reverse channel is well-
defined (Section VI.2), product-preservation leads to tabletop reversibility
(Theorem 3). As such, with full-rank states, the latter is strictly more
general (Theorem 4). As a by-product of this work, we found that the
preservation of local energy spectra is a necessary and sufficient
characterization of generalized thermal operations. Characterizing tabletop
reversibility in a necessary and sufficient way remains an open problem.

## Acknowledgments

We thank Francesco Buscemi for many discussions, and in particular for
stressing the role of assignment maps. We acknowledge discussions with Mile
Gu, Matteo Lostaglio, Eric Lutz, and Nelly Ng.

This work is supported by the National Research Foundation, Singapore and
A*STAR under its CQT Bridging Grant; and by the Ministry of Education,
Singapore, under the Tier 2 grant “Bayesian approach to irreversibility”
(Grant No. MOE-000504-01). M.B.J. acknowledges support from the Government of
Spain (Severo Ochoa CEX2019-000910-S and TRANQI), Fundació Cellex, Fundació
Mir-Puig, Generalitat de Catalunya (CERCA program), as well as funding from
the European Union’s Horizon 2020 research and innovation programme under the
Marie Skłodowska-Curie Grant Agreement No. 847517.

## References

  * Gibbs [1878] J. W. Gibbs, On the equilibrium of heterogeneous substances, American Journal of Science s3-16, 441 (1878), https://www.ajsonline.org/content/s3-16/96/441.full.pdf .
  * Onsager [1931] L. Onsager, Reciprocal relations in irreversible processes. ii., Physical review 38, 2265 (1931).
  * Seifert [2012] U. Seifert, Stochastic thermodynamics, fluctuation theorems and molecular machines, Reports on Progress in Physics 75, 126001 (2012).
  * Brandao  _et al._ [2015] F. Brandao, M. Horodecki, N. Ng, J. Oppenheim, and S. Wehner, The second laws of quantum thermodynamics, Proceedings of the National Academy of Sciences 112, 3275 (2015).
  * Evans and Searles [2002] D. J. Evans and D. J. Searles, The fluctuation theorem, Advances in Physics 51, 1529 (2002).
  * Crooks [2008] G. E. Crooks, Quantum operation time reversal, Phys. Rev. A 77, 034101 (2008).
  * Campisi  _et al._ [2011] M. Campisi, P. Hänggi, and P. Talkner, Colloquium: Quantum fluctuation relations: Foundations and applications, Rev. Mod. Phys. 83, 771 (2011).
  * Jaynes [1957] E. T. Jaynes, Information theory and statistical mechanics, Physical review 106, 620 (1957).
  * Shannon [1948] C. E. Shannon, A mathematical theory of communication, The Bell system technical journal 27, 379 (1948).
  * Bennett [1982] C. H. Bennett, The thermodynamics of computation—a review, International Journal of Theoretical Physics 21, 905 (1982).
  * Barnum and Knill [2002] H. Barnum and E. Knill, Reversing quantum dynamics with near-optimal quantum and classical fidelity, Journal of Mathematical Physics 43, 2097 (2002).
  * Hilt  _et al._ [2011] S. Hilt, S. Shabbir, J. Anders, and E. Lutz, Landauer’s principle in the quantum regime, Physical Review E 83, 030102 (2011).
  * Jarzynski [2000] C. Jarzynski, Hamiltonian derivation of a detailed fluctuation theorem, Journal of Statistical Physics 98, 77 (2000).
  * Alhambra  _et al._ [2018] A. M. Alhambra, S. Wehner, M. M. Wilde, and M. P. Woods, Work and reversibility in quantum thermodynamics, Phys. Rev. A 97, 062114 (2018).
  * Manzano  _et al._ [2018] G. Manzano, J. M. Horowitz, and J. M. R. Parrondo, Quantum fluctuation theorems for arbitrary environments: Adiabatic and nonadiabatic entropy production, Phys. Rev. X 8, 031037 (2018).
  * Buscemi and Scarani [2021] F. Buscemi and V. Scarani, Fluctuation theorems from bayesian retrodiction, Phys. Rev. E 103, 052111 (2021).
  * Aw  _et al._ [2021] C. C. Aw, F. Buscemi, and V. Scarani, Fluctuation theorems with retrodiction rather than reverse processes, AVS Quantum Science 3, 045601 (2021), https://doi.org/10.1116/5.0060893 .
  * Alhambra  _et al._ [2016] A. M. Alhambra, L. Masanes, J. Oppenheim, and C. Perry, Fluctuating work: From quantum thermodynamical identities to a second law equality, Phys. Rev. X 6, 041017 (2016).
  * Kwon and Kim [2019] H. Kwon and M. S. Kim, Fluctuation theorems for a quantum channel, Phys. Rev. X 9, 031029 (2019).
  * Kwon  _et al._ [2022] H. Kwon, R. Mukherjee, and M. S. Kim, Reversing lindblad dynamics via continuous petz recovery map, Phys. Rev. Lett. 128, 020403 (2022).
  * Petz [1988] D. Petz, Sufficiency of channels over von Neumann algebras, The Quarterly Journal of Mathematics 39, 97 (1988).
  * Wilde [2013] M. M. Wilde,  _Quantum Information Theory_ (Cambridge University Press, 2013).
  * Leifer and Spekkens [2013] M. S. Leifer and R. W. Spekkens, Towards a formulation of quantum theory as a causally neutral theory of bayesian inference, Phys. Rev. A 88, 052130 (2013).
  * Parzygnat and Buscemi [2023] A. J. Parzygnat and F. Buscemi, Axioms for retrodiction: achieving time-reversal symmetry with a prior, Quantum 7, 1013 (2023).
  * Parzygnat and Fullwood [2023] A. J. Parzygnat and J. Fullwood, From time-reversal symmetry to quantum bayes’ rules, PRX Quantum 4, 020334 (2023).
  * Aw  _et al._ [2023] C. C. Aw, K. Onggadinata, D. Kaszlikowski, and V. Scarani, Quantum bayesian inference in quasiprobability representations, PRX Quantum 4, 020352 (2023).
  * Gilyén  _et al._ [2022] A. Gilyén, S. Lloyd, I. Marvian, Y. Quek, and M. M. Wilde, Quantum Algorithm for Petz Recovery Channels and Pretty Good Measurements, Phys. Rev. Lett. 128, 220502 (2022).
  * Di Biagio  _et al._ [2021] A. Di Biagio, P. Donà, and C. Rovelli, The arrow of time in operational formulations of quantum theory, Quantum 5, 520 (2021).
  * Chiribella and Liu [2022] G. Chiribella and Z. Liu, Quantum operations with indefinite time direction, Communications Physics 5, 190 (2022).
  * Chiribella  _et al._ [2021] G. Chiribella, E. Aurell, and K. Życzkowski, Symmetries of quantum evolutions, Physical Review Research 3, 033028 (2021).
  * Ciliberto  _et al._ [2010] S. Ciliberto, S. Joubaud, and A. Petrosyan, Fluctuations in out-of-equilibrium systems: from theory to experiment, Journal of Statistical Mechanics: Theory and Experiment 2010, P12003 (2010).
  * Collin  _et al._ [2005] D. Collin, F. Ritort, C. Jarzynski, S. B. Smith, I. Tinoco, and C. Bustamante, Verification of the Crooks fluctuation theorem and recovery of RNA folding free energies, Nature 437, 231 (2005).
  * Hayashi [2018] K. Hayashi, Application of the fluctuation theorem to motor proteins: from f1-atpase to axonal cargo transport by kinesin and dynein, Biophysical reviews 10, 1311 (2018).
  * Hoang  _et al._ [2018] T. M. Hoang, R. Pan, J. Ahn, J. Bang, H. Quan, and T. Li, Experimental test of the differential fluctuation theorem and a generalized jarzynski equality for arbitrary initial states, Physical review letters 120, 080602 (2018).
  * Carberry  _et al._ [2004] D. Carberry, J. C. Reid, G. Wang, E. M. Sevick, D. J. Searles, and D. J. Evans, Fluctuations and irreversibility: an experimental demonstration of a second-law-like theorem using a colloidal particle held in an optical trap, Physical review letters 92, 140601 (2004).
  * Horodecki and Oppenheim [2013] M. Horodecki and J. Oppenheim, Fundamental limitations for quantum and nanoscale thermodynamics, Nature Communications 4, 2059 (2013).
  * Santos  _et al._ [2020] J. Santos, A. Timpanaro, and G. Landi, Joint fluctuation theorems for sequential heat exchange, Entropy 22, 10.3390/e22070763 (2020).
  * Landi and Paternostro [2021] G. T. Landi and M. Paternostro, Irreversible entropy production: From classical to quantum, Reviews of Modern Physics 93, 035008 (2021).
  * Horodecki  _et al._ [2003] M. Horodecki, P. Horodecki, and J. Oppenheim, Reversible transformations from pure to mixed states and the unique measure of information, Physical Review A 67, 062104 (2003).
  * Brandao  _et al._ [2013] F. G. Brandao, M. Horodecki, J. Oppenheim, J. M. Renes, and R. W. Spekkens, Resource theory of quantum states out of thermal equilibrium, Physical review letters 111, 250404 (2013).
  * Lostaglio  _et al._ [2018] M. Lostaglio, Á. M. Alhambra, and C. Perry, Elementary thermal operations, Quantum 2, 52 (2018).
  * Faist  _et al._ [2015] P. Faist, J. Oppenheim, and R. Renner, Gibbs-preserving maps outperform thermal operations in the quantum regime, New Journal of Physics 17, 043003 (2015).
  * Hu and Ding [2019] X. Hu and F. Ding, Thermal operations involving a single-mode bosonic bath, Phys. Rev. A 99, 012104 (2019).
  * Watanabe [1955] S. Watanabe, Symmetry of physical laws. part iii. prediction and retrodiction, Rev. Mod. Phys. 27, 179 (1955).
  * Watanabe [1965] S. Watanabe, Conditional probabilities in physics, Progr. Theor. Phys. Suppl. E65, 135 (1965).
  * Parzygnat and Russo [2022] A. J. Parzygnat and B. P. Russo, A non-commutative Bayes’ theorem, Linear Algebra and its Applications 644, 28 (2022).
  * Surace and Scandi [2023] J. Surace and M. Scandi, State retrieval beyond bayes’ retrodiction, Quantum 7, 990 (2023).
  * Petz [1986] D. Petz, Sufficient subalgebras and the relative entropy of states of a von neumann algebra, Comm. Math. Phys. 105, 123 (1986).
  * Wilde [2015] M. Wilde, Recoverability in quantum information theory, Proceedings of the Royal Society A 471, 20150338 (2015).
  * Pechukas [1994] P. Pechukas, Reduced dynamics need not be completely positive, Phys. Rev. Lett. 73, 1060 (1994).
  * Alicki [1995] R. Alicki, Comment on “Reduced dynamics need not be completely positive”, Phys. Rev. Lett. 75, 3020 (1995).
  * Buscemi [2014] F. Buscemi, Complete positivity, markovianity, and the quantum data-processing inequality, in the presence of initial system-environment correlations, Phys. Rev. Lett. 113, 140502 (2014).
  * Nielsen and Chuang [2002] M. A. Nielsen and I. Chuang, Quantum computation and quantum information (2002).
  * Lautenbacher  _et al._ [2022] L. Lautenbacher, F. de Melo, and N. K. Bernardes, Approximating invertible maps by recovery channels: Optimality and an application to non-markovian dynamics, Phys. Rev. A 105, 042421 (2022).
  * Åberg [2018] J. Åberg, Fully quantum fluctuation theorems, Phys. Rev. X 8, 011019 (2018).
  * Puchała  _et al._ [2015] Z. Puchała, L. Rudnicki, K. Chabuda, M. Paraniak, and K. Życzkowski, Certainty relations, mutual entanglement, and nondisplaceable manifolds, Phys. Rev. A 92, 032109 (2015).
  * Brahmachari  _et al._ [2022] S. Brahmachari, R. N. Rajmohan, S. A. Rather, and A. Lakshminarayan, Dual unitaries as maximizers of the distance to local product gates, arXiv preprint arXiv:2210.13307 (2022).
  * Chen  _et al._ [2008] J. Chen, R. Duan, Z. Ji, M. Ying, and J. Yu, Existence of universal entangler, Journal of Mathematical Physics 49, 012103 (2008).
  * Plenio [2005] M. B. Plenio, Logarithmic negativity: A full entanglement monotone that is not convex, Phys. Rev. Lett. 95, 090503 (2005).
  * Zhang  _et al._ [2003] J. Zhang, J. Vala, S. Sastry, and K. B. Whaley, Geometric theory of nonlocal two-qubit operations, Phys. Rev. A 67, 042313 (2003).
  * Korzekwa [2016] K. Korzekwa,  _Coherence, thermodynamics and uncertainty relations_ , Ph.D. thesis, Imperial College London (2016).
  * Lostaglio [2019] M. Lostaglio, An introductory review of the resource theory approach to thermodynamics, Reports on Progress in Physics 82, 114001 (2019).
  * vom Ende [2022] F. vom Ende, Which bath Hamiltonians matter for thermal operations?, Journal of Mathematical Physics 63, 112202 (2022).
  * Korzekwa [2017] K. Korzekwa, Structure of the thermodynamic arrow of time in classical and quantum theories, Phys. Rev. A 95, 052318 (2017).
  * Stinespring [1955] W. F. Stinespring, Positive Functions on C*-Algebras, Proceedings of the American Mathematical Society 6, 211 (1955).
  * Khatri and Wilde [2020] S. Khatri and M. M. Wilde, Principles of Quantum Communication Theory: A Modern Approach (2020), arXiv:2011.04672 [quant-ph] .

##  Appendix A Thermal Operations & Gibbs States

In [14], we have it that $U$ commutes with a global Hamiltonian
$H=H_{A}\otimes\mathbb{1}_{B}+\mathbb{1}_{A}\otimes H_{B}$. We simply that
under this definition of $H$, the Gibbs state of $H$ is none other than the
product of the Gibbs states of the local Hamiltonian:

$\displaystyle\frac{e^{\kappa H}}{Z_{H}}$
$\displaystyle=\frac{1}{Z_{A}Z_{B}}e^{\kappa(H_{A}\otimes\mathbb{1}_{B}+\mathbb{1}_{A}\otimes
H_{B})}$ $\displaystyle=\frac{1}{Z_{A}Z_{B}}\sum_{nm}e^{\kappa
a_{n}}\outerproduct{a_{n}b_{m}}{a_{n}b_{m}}\sum_{ij}e^{\kappa
b_{j}}\outerproduct{a_{i}b_{j}}{a_{i}b_{j}}$
$\displaystyle=\frac{1}{Z_{A}Z_{B}}\sum_{nmij}e^{\kappa a_{n}}e^{\kappa
b_{j}}\delta_{in}\delta_{jm}\outerproduct{a_{n}b_{m}}{a_{i}b_{j}}$ (48)
$\displaystyle=\frac{1}{Z_{A}}\sum_{n}e^{\kappa
a_{n}}\outerproduct{a_{n}}{a_{n}}\otimes\frac{1}{Z_{B}}\sum_{j}e^{\kappa
b_{j}}\outerproduct{b_{j}}{b_{j}}$
$\displaystyle=\tau_{\kappa}({H_{A}})\otimes\tau_{\kappa}({H_{B}})$

Thus, if $[U,H]=0$ then
$[U,\tau_{\kappa}({H_{A}})\otimes\tau_{\kappa}({H_{B}})]=0$. Clearly then this
means $(U,\tau_{\kappa}({H_{A}}),\tau_{\kappa}({H_{B}}))$ is a product-
preserved tuple. For a generalized thermal scenarios, we have that
$U(H_{A}+H_{B})U^{\dagger}=H_{A}^{\prime}+H_{B}^{\prime}$. Invoking (29), it
holds that for generalized thermal maps,
$U[\tau_{\kappa}({H_{A}})\otimes\tau_{\kappa}({H_{B}})]U^{\dagger}=\tau_{\kappa}({H_{A}^{\prime}})\otimes\tau_{\kappa}({H_{B}^{\prime}})$.

##  Appendix B Simple Examples of Dilations & Priors That Are Not Tabletop
Time-reversible

Here we include simple classical dilations and priors for which the
retrodiction map does not fulfill tabletop time-reversibility (25). We begin
first with a classical example. For a dilation

$\mathbf{\Phi}=\pmqty{undef},\quad\bm{\beta}=\pmqty{undef},$ (49)

the resultant bit-channel is given by

$\bm{\varphi}=\pmqty{undef}.\vspace{1em}$ (50)

The retrodiction channel
${\hat{\varphi}_{\gamma}}=\Sigma_{B}\circ\Phi^{-1}\circ\hat{\Sigma}_{B,\Phi[\gamma\otimes\beta]}$,
with an arbitrary reference
$\bm{\gamma}^{\text{T}}=(\gamma_{0}\;\;1-\gamma_{0})$, is thus

$\displaystyle\bm{{\hat{\varphi}_{\gamma}}}$
$\displaystyle=\left(\begin{array}[]{cc}\frac{\gamma_{0}\beta_{0}}{(\gamma_{0}-1)(\beta_{0}-1)+\gamma_{0}\beta_{0}}&\frac{\gamma_{0}-\gamma_{0}\beta_{0}}{-2\gamma_{0}\beta_{0}+\gamma_{0}+\beta_{0}}\\\
\frac{(\gamma_{0}-1)(\beta_{0}-1)}{\gamma_{0}(2\beta_{0}-1)-\beta_{0}+1}&\frac{\beta_{0}-\gamma_{0}\beta_{0}}{-2\gamma_{0}\beta_{0}+\gamma_{0}+\beta_{0}}\\\
\end{array}\right)$ (51)

Now, every tabletop-reversible map,
${\hat{\varphi}_{\gamma}}^{\text{TR}}=\Sigma_{B}\circ\Phi^{-1}\circ\hat{\Sigma}_{B,\square\otimes\eta}$
where $\bm{\eta}^{\text{T}}=(\eta_{0}\;\;1-\eta_{0})$, for this dilation gives

$\bm{\hat{\varphi}}^{\text{TR}}_{\bm{\gamma}}=\left(\begin{array}[]{cc}1-\eta_{0}&\eta_{0}\\\
\eta_{0}&1-\eta_{0}\\\ \end{array}\right)$ (52)

This means that unless $\gamma_{0}=1/2$ and $\eta_{0}=1-\beta_{0}$, or
$\beta_{0}=0$ and $\eta_{0}=1$, or $\beta_{0}=1$ and $\eta_{0}=0$,
${\hat{\varphi}_{\gamma}}\neq{\hat{\varphi}_{\gamma}}^{\text{TR}}$ always. For
instance, if $\gamma_{0}=\beta_{0}=1/4$, then the two maps will not be equal.
This simple example highlights how common non-tabletop reversible tuples (of
dilations and priors) are, and that these occur easily even when confined to
classical correlations, formed by a two-bit channel.

Meanwhile, for a quantum channel $\mathcal{E}$ with dilation
$U=\cos\theta\mathbb{1}+\sin\theta\operatorname{SWAP}$, ancilla $\beta$, and
prior $\alpha\neq\beta$, we have used semidefinite programming to verify that
$\hat{\mathcal{E}}_{\alpha}[\bullet]\neq\Tr_{B}\\{U^{\dagger}(\bullet\otimes\beta^{\prime})U\\}$
for all $\beta^{\prime}$ when $\theta\bmod(\pi/2)\neq 0$.

##  Appendix C Proofs Regarding Dilation & Retrodiction

###  C.1 Composability of Bayes’ Rule & The Petz Map

Here we provide some simple proofs for (7) and (8). For any concatenation
$\varphi=\varphi_{2}\circ\varphi_{1}$, applying (3) on $\bm{\varphi}$ gives:

$\displaystyle\mathcal{R}_{c}\left[\varphi_{2}\circ\varphi_{1},\gamma\right]$
$\displaystyle=\mathbf{D}_{\gamma}(\bm{\varphi}_{2}\bm{\varphi}_{1})^{\text{T}}\mathbf{D}_{\varphi_{2}\circ\varphi_{1}[\gamma]}^{-1}$
(53)
$\displaystyle=\mathbf{D}_{\gamma}\,\bm{\varphi}_{1}^{\text{T}}\,\bm{\varphi}_{2}^{\text{T}}\,\mathbf{D}_{\varphi_{2}\circ\varphi_{1}[\gamma]}^{-1}$
$\displaystyle=\mathbf{D}_{\gamma}\,\bm{\varphi}_{1}^{\text{T}}\,\mathbf{D}_{\varphi_{1}[\gamma]}^{-1}\mathbf{D}_{\varphi_{1}[\gamma]}\bm{\varphi}_{2}^{\text{T}}\mathbf{D}_{\varphi_{2}\circ\varphi_{1}[\gamma]}^{-1}$
$\displaystyle=\bm{\hat{\varphi}}_{1,\gamma}\bm{\hat{\varphi}}_{2,\varphi_{1}[\gamma]},$

with slight abuse of notation: on the right-hand side we have matrices instead
of maps. We may write this more generally:

$\displaystyle\mathcal{R}_{c}[\varphi_{L}\circ\dots\circ\varphi_{1},\gamma]$
$\displaystyle=\mathcal{R}_{c}\left[\varphi_{1},\gamma\right]\circ\mathcal{R}_{c}\left[\varphi_{2},\varphi_{1}[\gamma]\right]\circ\cdots$
(54)
$\displaystyle\qquad\cdots\circ\mathcal{R}_{c}\left[\varphi_{L},\varphi_{L-1}\circ\dots\circ\varphi_{1}[\gamma]\right]\,.$

A similar proof is available for the Petz map:

$\displaystyle\mathcal{R}_{q}\left[\mathcal{E}_{2}\circ\mathcal{E}_{1},\alpha\right][\bullet]$
$\displaystyle=\sqrt{\alpha}(\mathcal{E}_{2}\circ\mathcal{E}_{1})^{\dagger}\left(\frac{1}{\sqrt{\mathcal{E}_{2}\circ\mathcal{E}_{1}[\alpha]}}\bullet\frac{1}{\sqrt{\mathcal{E}_{2}\circ\mathcal{E}_{1}[\alpha]}}\right)\sqrt{\alpha}$
(55)
$\displaystyle=\sqrt{\alpha}(\mathcal{E}_{1}^{\dagger}\circ\mathcal{E}_{2}^{\dagger})\left(\frac{1}{\sqrt{\mathcal{E}_{2}\circ\mathcal{E}_{1}[\alpha]}}\bullet\frac{1}{\sqrt{\mathcal{E}_{2}\circ\mathcal{E}_{1}[\alpha]}}\right)\sqrt{\alpha}$
$\displaystyle=\sqrt{\alpha}\;\mathcal{E}_{1}^{\dagger}\Bigg{(}\frac{1}{\sqrt{\mathcal{E}_{1}[\alpha]}}\underbrace{\sqrt{\mathcal{E}_{1}[\alpha]}\;\mathcal{E}_{2}^{\dagger}\left(\frac{1}{\sqrt{\mathcal{E}_{2}\circ\mathcal{E}_{1}[\alpha]}}\bullet\frac{1}{\sqrt{\mathcal{E}_{2}\circ\mathcal{E}_{1}[\alpha]}}\right)\sqrt{\mathcal{E}_{1}[\alpha]}}_{\hat{\mathcal{E}}_{2,{\mathcal{E}_{1}[\alpha]}}}\frac{1}{\sqrt{\mathcal{E}_{1}[\alpha]}}\Bigg{)}\sqrt{\alpha}$
$\displaystyle=\mathcal{R}_{q}\left[\mathcal{E}_{1},\alpha\right]\circ\mathcal{R}_{q}\left[{\mathcal{E}}_{2},\mathcal{E}_{1}[\alpha]\right].$

###  C.2 Bayes’ Rule on Classical Decomposition

In this Appendix, we lay out an explicit proof for Result 1 on the level of
matrices. This elucidates how retrodiction occurs on every step of the
decomposition and how Bayesian inversion applies on physically valid channels,
including those that are not stochastic (that is, even channels that do not
correspond to a Markov matrix, such as the marginalization of a subspace and
the assigning of an environment). Note that here, for the sake of symmetry
with the quantum case, we will assume that the global process $\Phi$ is
bijective and that the environment $\beta$ is uncorrelated with the input.
Nevertheless, general insights also apply when these assumptions are removed.

Firstly, we write Eq. (14) in terms of matrices:

$\bm{\varphi}=\mathbf{\Sigma}_{B}\bm{\Phi}\hat{\mathbf{\Sigma}}_{B,\square\otimes\beta}.$
(56)

These matrices can be formalized in the following way. $\mathbf{\Sigma}_{B}$
is a $d_{A}$ by $d_{A}d_{B}$ matrix while
$\hat{\mathbf{\Sigma}}_{B,\Lambda_{AB}}$ is $d_{A}d_{B}$ by $d_{A}$, defined
as follows:

$\displaystyle\mathbf{\Sigma}_{B}$ $\displaystyle=\mathbb{1}_{A}\otimes
v_{\textbf{1}}^{\text{T}}$ (57)
$\displaystyle\hat{\mathbf{\Sigma}}_{B,\Lambda_{AB}}$
$\displaystyle=\pmqty{undef}.$ (58)

Here, $v_{\textbf{1}}$ and $v_{\textbf{0}}$ are $d_{B}$-dimensional column
vectors of ones and zeros respectively, and
$v({\Lambda_{aB}})^{\text{T}}=\pmqty{undef}$, which comes from Eq. (16). Now,
$\square$ indicates any choice of state in $A$. The matrix
$\hat{\Sigma}_{B,\square\otimes\beta}$ is independent of this choice, since

$v(({\square\otimes\beta})_{aB})=\pmqty{undef}=\pmqty{undef}=\bm{\beta}.$ (59)

Thus, it follows that
$\hat{\Sigma}_{B,\square\otimes\beta}[\bullet_{A}]=\bullet_{A}\otimes\beta_{B}$.
Hence, each matrix performs the role of its corresponding channel. Turning now
to retrodiction, we recall that Bayes’ rule is composable [24]. This simply
captures the time-reverse ordering and propagation of the reference prior,
expected of Bayesian inversion. When composability (7) is applied to Eq. (56),
insights are yielded:

$\displaystyle{\hat{\varphi}_{\gamma}}=$
$\displaystyle\mathcal{R}_{c}\left[\hat{\Sigma}_{B,\square\otimes\beta},\gamma\right]\mathcal{R}_{c}\left[\Phi,\hat{\Sigma}_{B,\square\otimes\beta}[\gamma]\right]$
(60)
$\displaystyle\mathcal{R}_{c}\left[\Sigma_{B},\Phi\circ\hat{\Sigma}_{B,\square\otimes\beta}[\gamma]\right]$
$\displaystyle=$
$\displaystyle\mathcal{R}_{c}\left[\hat{\Sigma}_{B,\square\otimes\beta},\gamma\right]\mathcal{R}_{c}\left[\Phi,\gamma\otimes\beta\right]$
$\displaystyle\mathcal{R}_{c}\left[\Sigma_{B},\Phi[\gamma\otimes\beta]\right].$

Now, for both
$\mathcal{R}_{c}\left[\hat{\Sigma}_{B,\square\otimes\beta},\gamma\right]$ and
$\mathcal{R}_{c}\left[\Sigma_{B},\Phi[\gamma\otimes\beta]\right]$ in matrix
form, applying Eq. (3), we get

$\displaystyle\mathbf{D}_{\gamma}\hat{\mathbf{\Sigma}}_{B,\square\otimes\beta}\mathbf{D}_{\gamma\otimes\beta}^{-1}$
$\displaystyle=\mathbf{D}_{\gamma}\pmqty{undef}\\!\pmqty{undef}^{-1}$ (61)
$\displaystyle=\mathbf{D}_{\gamma}\pmqty{undef}$ $\displaystyle=\pmqty{undef}$
$\displaystyle=\mathbf{\Sigma}_{B}.$

$\displaystyle\mathbf{D}_{\Phi[\gamma\otimes\beta]}(\Sigma_{B})^{\text{T}}\mathbf{D}_{\Sigma_{B}\Phi[\gamma\otimes\beta]}^{-1}$
$\displaystyle=\mathbf{D}_{\Phi[\gamma\otimes\beta]}\pmqty{undef}\mathbf{D}_{\Sigma_{B}\Phi[\gamma\otimes\beta]}^{-1}$
(62) $\displaystyle=\mathbf{D}_{\Phi[\gamma\otimes\beta]}\pmqty{undef}$
$\displaystyle=\pmqty{undef}$
$\displaystyle=\hat{\mathbf{\Sigma}}_{B,\Phi[\gamma\otimes\beta]},$

where the $d_{B}$-length vector
$v_{a}(p(\tilde{a}))^{\text{T}}=\pmqty{undef}$.

Thus it is shown that, given the propagated reference priors, the Bayesian
inversion of the assignment map gives a marginalizing channel and vice versa:

$\displaystyle\mathcal{R}_{c}\left[\hat{\Sigma}_{B,\square\otimes\beta},\gamma\right]=\Sigma_{B},$
(63)
$\displaystyle\mathcal{R}_{c}\left[\Phi,\gamma\otimes\beta\right]=\hat{\Phi}_{\gamma\otimes\beta}=\Phi^{-1},\;\;\because\eqref{eq:deterministic}$
(64)
$\displaystyle\mathcal{R}_{c}\left[\Sigma_{B},\Phi[\gamma\otimes\beta]\right]=\hat{\Sigma}_{B,\Phi[\gamma\otimes\beta]}.$
(65)

Together, we can write:

${\hat{\varphi}_{\gamma}}=\Sigma_{B}\circ\Phi^{-1}\circ\hat{\Sigma}_{B,\Phi[\gamma\otimes\beta]}\,.$
(66)

###  C.3 The Petz Map on Quantum Dilations

Here we provide some explicit proofs for each step of Eq.
(LABEL:eq:qdildecomp). Firstly, we know
$\mathcal{R}_{q}\left[\mathcal{U},\hat{\text{T}}\text{r}_{B,\square\otimes\beta}[\alpha]\right]=\hat{\mathcal{U}}_{\alpha\otimes\beta}=\mathcal{U}^{\dagger}$
because of Eqs. (20) and (9). Secondly, one can easily verify that the adjoint
(1) of the quantum assignment map is

$\left(\hat{\text{T}}\text{r}_{B,\Gamma_{AB}}\right)^{\dagger}[\bullet]=\Tr_{B}\left(\frac{1}{\sqrt{\Tr_{B}(\Gamma_{AB})}}\otimes\mathbb{1}_{B}\sqrt{\Gamma_{AB}}\bullet\sqrt{\Gamma_{AB}}\frac{1}{\sqrt{\Tr_{B}(\Gamma_{AB})}}\otimes\mathbb{1}_{B}\right).$
(67)

Hence,

$\displaystyle\left(\hat{\text{T}}\text{r}_{B,\square\otimes\beta}\right)^{\dagger}[\bullet]$
$\displaystyle=\Tr_{B}\left(\frac{1}{\sqrt{\square}}\otimes\mathbb{1}_{B}\sqrt{\square\otimes\beta}\bullet\sqrt{\square\otimes\beta}\frac{1}{\sqrt{\square}}\otimes\mathbb{1}_{B}\right)$
(68)
$\displaystyle\Rightarrow\mathcal{R}_{q}\left[\hat{\text{T}}\text{r}_{B,\square\otimes\beta},\alpha\right]$
$\displaystyle=\sqrt{\alpha}\left(\hat{\text{T}}\text{r}_{B,\square\otimes\beta}\right)^{\dagger}\left(\frac{1}{\sqrt{\alpha\otimes\beta}}\bullet\frac{1}{\sqrt{\alpha\otimes\beta}}\right)\sqrt{\alpha}$
$\displaystyle=\sqrt{\alpha}\Tr_{B}\left(\frac{1}{\sqrt{\square}}\otimes\mathbb{1}_{B}\sqrt{\square\otimes\beta}\frac{1}{\sqrt{\alpha\otimes\beta}}\bullet\frac{1}{\sqrt{\alpha\otimes\beta}}\sqrt{\square\otimes\beta}\frac{1}{\sqrt{\square}}\otimes\mathbb{1}_{B}\right)\sqrt{\alpha}$
$\displaystyle=\sqrt{\alpha}\Tr_{B}\left(\frac{1}{\sqrt{\alpha}}\otimes\mathbb{1}_{B}\bullet\frac{1}{\sqrt{\alpha}}\otimes\mathbb{1}_{B}\right)\sqrt{\alpha}=\Tr_{B}(\bullet).$

Thirdly, one can easily show that the adjoint of the partial trace is given by

$\Tr_{B}^{\dagger}[\bullet_{A}]=\bullet_{A}\otimes\mathbb{1}_{B}.$ (69)

Which implies,

$\displaystyle\mathcal{R}_{q}\left[\Tr_{B}{},\Gamma_{AB}\right]$
$\displaystyle=\sqrt{\Gamma_{AB}}\Tr_{B}^{\dagger}\left(\frac{1}{\sqrt{\Tr_{B}(\Gamma_{AB})}}\bullet\frac{1}{\sqrt{\Tr_{B}(\Gamma_{AB})}}\right)\sqrt{\Gamma_{AB}}$
(70)
$\displaystyle=\sqrt{\Gamma_{AB}}\left(\frac{1}{\sqrt{\Tr_{B}(\Gamma_{AB})}}\bullet\frac{1}{\sqrt{\Tr_{B}(\Gamma_{AB})}}\otimes\mathbb{1}_{B}\right)\sqrt{\Gamma_{AB}}=\hat{\text{T}}\text{r}_{B,\Gamma_{AB}}$
$\displaystyle\Rightarrow\mathcal{R}_{q}\left[\Tr_{B}{},\mathcal{U}[\alpha\otimes\beta]\right]=\hat{\text{T}}\text{r}_{B,\,\mathcal{U}[\alpha\otimes\beta]}.$

Thus, every step of Eq. (LABEL:eq:qdildecomp) is proven, thus giving an
alternative route to Eq. (23). These derivations highlight that Bayesian
inference can be applied in a logically consistent and physically insightful
way to any valid channel, including that of marginalization and the assignment
of environments and so on.

##  Appendix D Special Case of Pure Product-preserving Tuple for Two-qubit
Unitaries

For $a_{0}\neq 0$, the explicit form of Eq. (34) is

$\displaystyle(u_{A}^{\dagger}\otimes
u_{B}^{\dagger})U(\ket{\alpha}\otimes\ket{\beta})$
$\displaystyle\widehat{=}\pmqty{undef}\bqty{\pmqty{undef}\otimes\pmqty{undef}}$
(71) $\displaystyle=\pmqty{undef}$

The condition for $(U,\ket{\alpha},\ket{\beta})$ to be product-preserving is
given in Eq. (35), which can be expressed as a quadratic equation
$ax^{2}+bx+c=0$, with

$\displaystyle a$
$\displaystyle=\frac{i}{2}\bqty{b_{0}^{2}e^{-i2t_{3}}\sin(2t_{1}+2t_{2})-b_{1}^{2}e^{i2t_{3}}\sin(2t_{1}-2t_{2})},$
(72) $\displaystyle b$
$\displaystyle=b_{0}b_{1}\bqty{e^{-i2t_{3}}\cos(2t_{1}+2t_{2})-e^{i2t_{3}}\cos(2t_{1}-2t_{2})},$
$\displaystyle c$
$\displaystyle=-\frac{i}{2}\bqty{b_{0}^{2}e^{i2t_{3}}\sin(2t_{1}-2t_{2})-b_{1}^{2}e^{-i2t_{3}}\sin(2t_{1}+2t_{2})}.$

Unless $a=b=0$ and $c\neq 0$, a solution always exists for $x$, and the
corresponding $\ket{\alpha}$ for the given $U$ and $\ket{\beta}$ to make the
tuple product-preserving.

However, if $a=b=0$ and $c\neq 0$, the condition simplifies to $c=0$, which is
a contradiction. In that case, setting $\ket{\alpha}=\ket{1}$ gives

$(u_{A}^{\dagger}\otimes
u_{B}^{\dagger})U(\ket{1}\otimes\ket{\beta})\widehat{=}\pmqty{undef},$ (73)

and it can be verified that Eq. (35) is satisfied with $a=0$.

In summary, for a given $U$ and $\ket{\beta}$, if $a$ as defined in Eq. (72)
is nonzero, $ax^{2}+bx+c=0$ is solved for $x$ with the coefficients given in
Eq. (72), and we set
$\ket{\alpha}=v_{A}\pqty{\ket{0}+x\ket{1}}/\sqrt{1+\absolutevalue{x}^{2}}$.
Otherwise, if $a=0$, we set $\ket{\alpha}=\ket{1}$. In either case,
$(U,\ket{\alpha},\ket{\beta})$ forms a product-preserving tuple.

##  Appendix E Lemmas About Two-qubit Generalized-thermal Unitaries

######  Lemma 1.

For $U$ such that $t_{k}\bmod(\pi/4)=0$ for at most one $k\in\\{1,2,3\\}$, $U$
is generalized thermal with respect to $H_{A}$, $H_{B}$, $H_{A}^{\prime}$, and
$H_{B}^{\prime}$ if and only if

$\displaystyle H_{A}$ $\displaystyle=v_{A}Hv_{A}^{\dagger},$ $\displaystyle
H_{B}$
$\displaystyle=v_{B}\tilde{\sigma}H\tilde{\sigma}^{\dagger}v_{B}^{\dagger},$
(74) $\displaystyle H_{A}^{\prime}$
$\displaystyle=u_{A}\tilde{\varsigma}H\tilde{\varsigma}^{\dagger}u_{A}^{\dagger},$
$\displaystyle H_{B}^{\prime}$
$\displaystyle=u_{B}\tilde{\varsigma}\tilde{\sigma}H\tilde{\sigma}^{\dagger}\tilde{\varsigma}^{\dagger}u_{B}^{\dagger},$

where $\tilde{\sigma}:=\prod_{j=1}^{3}\sigma_{j}^{m_{-j}}$ and
$\tilde{\varsigma}:=\prod_{j=1}^{3}\sigma_{j}^{m_{+j}}$ are “bit flips” with
$m_{\pm j}$ defined below, and $H$ is any Hamiltonian such that
$\Tr(H\sigma_{j})\neq 0$ only when $(t_{j}-t_{k})\bmod(\pi/2)=0$ or
$(t_{j}+t_{k})\bmod(\pi/2)=0$ for all permutations $(j,k,l)$ of $(1,2,3)$.

The definition of $m_{\pm j}$ is as follows: if $(t_{k}-t_{l})\bmod(\pi/2)=0$
or $(t_{k}+t_{l})\bmod(\pi/2)=0$, where $j\neq k\neq l$, then

$\displaystyle(-1)^{m_{+j}}$
$\displaystyle=\operatorname{sgn}\\!\left[\cos(2t_{k})\cos(2t_{l})\right],$
(75) $\displaystyle(-1)^{m_{-j}}$
$\displaystyle=\operatorname{sgn}\\!\left[\tan(2t_{k})\tan(2t_{l})\right].$

Otherwise, if $(t_{j}\pm t_{k})\bmod(\pi/2)\neq 0$, $m_{\pm j}$ should be
chosen so that $(-1)^{m_{\pm 1}+m_{\pm 2}+m_{\pm 3}}=1$.

###### Proof.

Let us first consider the unitary $\widetilde{U}=e^{i\widetilde{H}}$ where
$\widetilde{H}=\sum_{k=1}^{3}t_{k}\sigma_{k}\otimes\sigma_{k}$, with traceless
Hamiltonians $\widetilde{H}_{A}$, $\widetilde{H}_{B}$,
$\widetilde{H}_{A}^{\prime}$, and $\widetilde{H}_{B}^{\prime}$ with the
decomposition $\widetilde{H}_{A}=\sum_{k=1}^{3}h_{A,k}\sigma_{k}$ (and
similarly for $\widetilde{H}_{B}$, $\widetilde{H}_{A/B}^{\prime}$).

The key step is to work out that for $\mu\in\\{0,1,2,3\\}$, with
$\sigma_{0}=\mathbb{1}$, and cyclic permutations $(j,k,l)$ of $(1,2,3)$,

$\displaystyle=0$ (76)
$\displaystyle[i\widetilde{H},\mathbb{1}\otimes\sigma_{j}\pm\sigma_{j}\otimes\mathbb{1}]$
$\displaystyle=2(t_{k}\mp
t_{l})(\sigma_{k}\otimes\sigma_{l}\pm\sigma_{l}\otimes\sigma_{k})$
$\displaystyle[i\widetilde{H},\sigma_{k}\otimes\sigma_{l}\pm\sigma_{l}\otimes\sigma_{k}]$
$\displaystyle=-2(t_{k}\mp
t_{l})(\mathbb{1}\otimes\sigma_{j}\pm\sigma_{j}\otimes\mathbb{1}).$

Together with the Baker–Campbell–Hausdorff lemma, Eq. (76) can be used to find
$\widetilde{U}(\sigma_{\mu}\otimes\sigma_{\nu})\widetilde{U}^{\dagger}$ for
all $\mu,\nu\in\\{0,1,2,3\\}$.

Having these transformations, we notice that the condition of generalized
thermal channel

$\widetilde{U}(\widetilde{H}_{A}\otimes\mathbb{1}+\mathbb{1}\otimes\widetilde{H}_{B})\widetilde{U}^{\dagger}=\widetilde{H}_{A}^{\prime}\otimes\mathbb{1}+\mathbb{1}\otimes\widetilde{H}_{B}^{\prime},$
(77)

implies that there are no products $\sigma_{j}\otimes\sigma_{k}$ on the right-
hand side. Setting the same components to zero on the left leads to

$\displaystyle h_{A,j}\sin(2t_{k})\cos(2t_{l})$
$\displaystyle=h_{B,j}\sin(2t_{l})\cos(2t_{k}),$ (78) $\displaystyle
h_{A,j}\sin(2t_{l})\cos(2t_{k})$
$\displaystyle=h_{B,j}\sin(2t_{k})\cos(2t_{l}).$

Let us first take $t_{k}\bmod(\pi/4)\neq 0$ for all $k$, so that
$\\{\tan(2t_{k})\\}_{k=1}^{3}$ are all nonzero and finite.

To satisfy Eq. (78), the following must hold:

  * •

if $h_{A,j},h_{B,j}\neq 0$, then $\tan^{2}(2t_{k})=\tan^{2}(2t_{l})$, which
further implies $h_{A,j}=\pm h_{B,j}$,

  * •

if $\tan^{2}(2t_{k})\neq\tan^{2}(2t_{l})$, then $h_{A,j}=h_{B,j}=0$.

So, it is always true that $h_{B,j}=\pm h_{A,j}$, and they are nonzero only if
$\lvert\tan(2t_{k})\rvert=\lvert\tan(2t_{l})\rvert$, which in turn implies
that $(t_{j}-t_{k})\bmod(\pi/2)=0$ or $(t_{j}+t_{k})\bmod(\pi/2)=0$. The sign
ambiguity is due to the absolute value in the latter expression.

Now, turning to the $\mathbb{1}\otimes\sigma_{j}$ and
$\sigma_{j}\otimes\mathbb{1}$ components of Eq. (77), we have

$\displaystyle h^{\prime}_{A,j}$
$\displaystyle=h_{A,j}\cos(2t_{k})\cos(2t_{l})+h_{B,j}\sin(2t_{k})\sin(2t_{l})$
(79) $\displaystyle=$ $\displaystyle\pm h_{A,j},$ $\displaystyle
h^{\prime}_{B,j}$
$\displaystyle=h_{B,j}\cos(2t_{k})\cos(2t_{l})+h_{A,j}\sin(2t_{k})\sin(2t_{l})$
$\displaystyle=\pm h_{B,j}.$

In the end, this means that $h_{B,j}=\pm h_{A,j}=\pm h_{A,j}^{\prime}=\pm
h_{B,j}^{\prime}$. Therefore, after defining $H:=\widetilde{H}_{A}$, we have
$H_{B}=\sigma_{\mu}H\sigma_{\mu}$, $H_{A}^{\prime}=\sigma_{\nu}H\sigma_{\nu}$,
and $H_{B}^{\prime}=\sigma_{\gamma}H\sigma_{\gamma}$ for some “bit-flips” or
identities $\sigma_{\mu}$, $\sigma_{\nu}$, and $\sigma_{\gamma}$ that correct
the sign ambiguities.

Finally, we arrive at Eq. (74) by substituting $\widetilde{U}$ into Eq. (33).
The form of $\tilde{\sigma}$ and $\tilde{\varsigma}$ is found by keeping track
of the signs of the trigonometric functions.

The proof also holds when $t_{k}\bmod(\pi/4)=0$ for at most one $k$:
substituting $t_{k}$ into Eq. (78) gives $h_{A,j}=h_{B,j}=0$ for all $j\neq
k$, after which the rest of the proof follows as stated. ∎

######  Lemma 1′.

For $U$ such that $t_{j},t_{k}\bmod(\pi/4)=0$ and $t_{l}\bmod(\pi/4)\neq 0$,
$U$ is generalized thermal with respect to $H_{A}$, $H_{B}$, $H_{A}^{\prime}$,
and $H_{B}^{\prime}$ if and only if

  * •

$t_{j}\bmod(\pi/2)=t_{k}\bmod(\pi/2)=0$ and, up to a constant offset,

$\displaystyle H_{A}$ $\displaystyle=h_{A}v_{A}\sigma_{l}v_{A}^{\dagger},$
$\displaystyle H_{B}$ $\displaystyle=h_{B}v_{B}\sigma_{l}v_{B}^{\dagger},$
(80) $\displaystyle H_{A}^{\prime}$
$\displaystyle=h_{A}u_{A}\sigma_{l}u_{A}^{\dagger},$ $\displaystyle
H_{B}^{\prime}$ $\displaystyle=h_{B}u_{B}\sigma_{l}u_{B}^{\dagger},$

or;

  * •

$t_{j}\bmod(\pi/2)=t_{k}\bmod(\pi/2)=\pi/4$ and, up to a constant offset,

$\displaystyle H_{A}$ $\displaystyle=h_{A}v_{A}\sigma_{l}v_{A}^{\dagger},$
$\displaystyle H_{B}$ $\displaystyle=h_{B}v_{B}\sigma_{l}v_{B}^{\dagger},$
(81) $\displaystyle H_{A}^{\prime}$
$\displaystyle=h_{B}u_{A}\sigma_{l}u_{A}^{\dagger},$ $\displaystyle
H_{B}^{\prime}$ $\displaystyle=h_{A}u_{B}\sigma_{l}u_{B}^{\dagger}.$

###### Proof.

First, let $2t_{k}=n_{k}\pi/2$ for integer $n_{k}$, which gives
$2\cos(2t_{k})=(-1)^{\lfloor n_{k}/2\rfloor}(1+(-1)^{n_{k}})$ and
$2\sin(2t_{k})=(-1)^{\lfloor n_{k}/2\rfloor}(1-(-1)^{n_{k}})$. Substituting it
into Eq. (78),

$\displaystyle h_{A,j}(1-(-1)^{n_{k}})\cos(2t_{l})$
$\displaystyle=h_{B,j}\sin(2t_{l})(1+(-1)^{n_{k}}),$ (82) $\displaystyle
h_{B,j}\sin(2t_{l})(1+(-1)^{n_{k}})$
$\displaystyle=h_{A,j}(1-(-1)^{n_{k}})\cos(2t_{l}).$

Since one of $1+(-1)^{n_{k}}$ or $1-(-1)^{n_{k}}$ must be zero, it is either
the case that both right columns are zero, or both left columns are zero. As
such, $h_{A,j}=h_{B,j}=0$.

Similarly letting $2t_{j}=n_{j}\pi/2$, repeating the above steps give
$h_{A,k}=h_{B,k}=0$.

Meanwhile, for the $h_{A,l}$ and $h_{B,l}$ case,

$\displaystyle h_{A,l}(1-(-1)^{n_{k}})(1+(-1)^{n_{j}})$ (83)
$\displaystyle\qquad{}={}h_{B,l}(1+(-1)^{n_{k}})(1-(-1)^{n_{j}}),$
$\displaystyle h_{B,l}(1+(-1)^{n_{k}})(1-(-1)^{n_{j}})$
$\displaystyle\qquad{}={}h_{A,l}(1-(-1)^{n_{k}})(1+(-1)^{n_{j}}).$

When the parities of $n_{j}$ and $n_{k}$ do not match, $h_{A,l}=h_{B,l}=0$. In
other words, if $t_{j}\bmod(\pi/2)\neq t_{k}\bmod(\pi/2)$, then $U$ is not
generalized thermal.

Otherwise, the equations are trivially satisfied. Then, Eq. (79) gives

$h_{A,l}^{\prime}=\begin{cases}h_{A,l}&\text{if }t_{j}\bmod\frac{\pi}{2}=0\\\
h_{B,l}&\text{if }t_{j}\bmod\frac{\pi}{2}=\frac{\pi}{4},\\\ \end{cases}$ (84)

with analogous expressions for $h_{B,l}^{\prime}$. Placing these expressions
back into the definitions of the Hamiltonians end the proof. ∎

######  Lemma 1′′.

For $U$ such that $t_{k}\bmod(\pi/4)=0$ for all $k\in\\{1,2,3\\}$, $U$ is
generalized thermal with respect to $H_{A}$, $H_{B}$, $H_{A}^{\prime}$, and
$H_{B}^{\prime}$ if and only if

  * •

$t_{1}\bmod\frac{\pi}{2}=t_{2}\bmod\frac{\pi}{2}=t_{3}\bmod\frac{\pi}{2}=0$,
$H_{A}$ and $H_{B}$ are arbitrary, and

$\displaystyle H_{A}^{\prime}$
$\displaystyle=u_{A}v_{A}^{\dagger}H_{A}v_{A}u_{A}^{\dagger},$ (85)
$\displaystyle H_{B}^{\prime}$
$\displaystyle=u_{B}v_{B}^{\dagger}H_{B}v_{B}u_{B}^{\dagger},$

or;

  * •

$t_{1}\bmod\frac{\pi}{2}=t_{2}\bmod\frac{\pi}{2}=t_{3}\bmod\frac{\pi}{2}=\frac{\pi}{4}$,
$H_{A}$ and $H_{B}$ are arbitrary, and

$\displaystyle H_{A}^{\prime}$
$\displaystyle=u_{A}v_{B}^{\dagger}H_{B}v_{B}u_{A}^{\dagger},$ (86)
$\displaystyle H_{B}^{\prime}$
$\displaystyle=u_{B}v_{A}^{\dagger}H_{A}v_{A}u_{B}^{\dagger}.$

###### Proof.

The proof is almost identical to that of Lemma 1′, so we shall only provide a
sketch. Substituting $2t_{k}=n_{k}\pi/2$ for all $k\in\\{1,2,3\\}$ into Eq.
(78), we will find that $h_{A,k}=h_{B,k}=0$ for all $k$ if the parities of
$n_{1}$, $n_{2}$, and $n_{3}$ do not all match. This means that $U$ is not
generalized thermal if $t_{j}\bmod\frac{\pi}{2}\neq t_{k}\bmod\frac{\pi}{2}$
for any two $j\neq k$.

Otherwise, Eq. (79) gives, for all $k$,

$h_{A,k}^{\prime}=\begin{cases}h_{A,k}&\text{if }t_{j}\bmod\frac{\pi}{2}=0\\\
h_{B,k}&\text{if }t_{j}\bmod\frac{\pi}{2}=\frac{\pi}{4},\\\ \end{cases}$ (87)

with similar expressions for $h_{B,k}$. Substituting these back into the
Hamiltonians concludes the proof. ∎

##  Appendix F Retrodiction with Rank-deficient Outputs is Ill-defined in
General

The Petz recovery map in Eq. (4) involves the operator
$(\mathcal{E}[\alpha])^{-\frac{1}{2}}$, which does not exist when
$\mathcal{E}[\alpha]$ is rank-deficient. One might attempt to circumvent this
problem in in several ways:

(1) _Define $\hat{\mathcal{E}}_{\alpha,1}$ using the pseudoinverse_. The
inverse can be defined only on the support of $\mathcal{E}[\alpha]$, resulting
in the so-called pseudoinverse of $\mathcal{E}[\alpha]$, as is the convention
in quantum information [66].

(2) _Define $\hat{\mathcal{E}}_{\alpha,2}$ in a full-rank neighbourhood_.
Another possibility is to perturb the prior or the channel so that the output
state is perturbed as
$\mathcal{E}[\alpha]\to\mathcal{E}^{(\varepsilon)}[\alpha]$, such that
$\lim_{\varepsilon\to
0}\mathcal{E}^{(\varepsilon)}[\alpha]=\mathcal{E}[\alpha]$ and
$\mathcal{E}^{(\varepsilon)}[\alpha]$ is full rank for every $\varepsilon>0$.
Then, the Petz map $\hat{\mathcal{E}}_{\alpha}^{(\varepsilon)}$ is well-
defined for every $\varepsilon>0$, and the retrodiction channel is defined as
the limit $\hat{\mathcal{E}}_{\alpha,2}:=\lim_{\varepsilon\to
0}\hat{\mathcal{E}}_{\alpha}^{(\varepsilon)}$.

(3) _Naïvely define $\hat{\mathcal{E}}_{\alpha,3}$ with product-preserving
tuples_. As Eq. (31) gives a simple form for the retrodiction channel when
$(U,\alpha,\beta)$ is a product-preserving tuple for full rank $\alpha$ and
$\beta$, one might extend it to the rank-deficient case by imposing that the
retrodiction channel should also take the same form

$\hat{\mathcal{E}}_{\alpha,3}[\bullet]:=\Tr_{B}\bqty{U^{\dagger}\pqty{\bullet\otimes\beta^{\prime}}U}$
(88)

even when $\alpha^{\prime}$ is not full rank.

In the following section, we show that the retrodiction channels as defined by
the above approaches do not agree in general. As such, the retrodiction
channel for a rank-deficient output state depends on the convention chosen,
and cannot be consistently defined when the output state is rank-deficient.

###  F.1 Comparison of different approaches to retrodicting rank-deficient
outputs

####  F.1.1 Unitary channels

Consider a unitary channel $\mathcal{U}[\bullet]:=U\bullet U^{\dagger}$ with
the dilation $\mathcal{U}[\bullet]=\Tr_{B}[(U\otimes
V)(\bullet\otimes\beta)(U^{\dagger}\otimes V^{\dagger})]$ for arbitrary
unitary $V$ and ancilla $\beta$. $(U\otimes V,\alpha,\beta)$ is clearly a
product-preserving tuple, since

$(U\otimes V)(\alpha\otimes\beta)(U^{\dagger}\otimes V)=(U\alpha
U^{\dagger})\otimes(V\beta
V^{\dagger})=:\alpha^{\prime}\otimes\beta^{\prime}.$ (89)

(1) _Define $\hat{\mathcal{U}}_{\alpha,1}$ using the pseudoinverse_. Given
$\alpha^{\prime}=\sum_{k=1}^{|\text{supp}(\alpha^{\prime})|}a^{\prime}_{k}\outerproduct{a^{\prime}_{k}}{a^{\prime}_{k}}$
with $a^{\prime}_{k}>0$ for all
$k\in\\{1,2,\dots,|\text{supp}(\alpha^{\prime})|\\}$, the pseudoinverse of
$\sqrt{\alpha^{\prime}}$ is defined as

${\sqrt{\alpha^{\prime}}}^{\,\bm{\pmb{+}}}:=\sum_{k=1}^{|\text{supp}(\alpha^{\prime})|}\frac{1}{\sqrt{a^{\prime}_{k}}}\outerproduct{a^{\prime}_{k}}{a^{\prime}_{k}}.$
(90)

The pseudoinverse has the property
$\sqrt{\alpha^{\prime}}{\sqrt{\alpha^{\prime}}}^{\,\bm{\pmb{+}}}={\sqrt{\alpha^{\prime}}}^{\,\bm{\pmb{+}}}\sqrt{\alpha^{\prime}}=\Pi_{\alpha^{\prime}}$,
where
$\Pi_{\alpha^{\prime}}:=\sum_{k=1}^{|\text{supp}(\alpha^{\prime})|}\outerproduct{a^{\prime}_{k}}{a^{\prime}_{k}}$
is the projector onto the support of $\alpha^{\prime}$. Therefore, with the
replacement
$\alpha^{\prime-\frac{1}{2}}\to{\sqrt{\alpha^{\prime}}}^{\,\bm{\pmb{+}}}$, the
retrodiction of a unitary channel can be found to be

$\hat{\mathcal{U}}_{\alpha,1}[\bullet]=U^{\dagger}\Pi_{\alpha^{\prime}}\bullet\Pi_{\alpha^{\prime}}U.$
(91)

This channel is completely positive but not trace-preserving in general.

(2) _Define $\hat{\mathcal{E}}_{\alpha,2}$ in a full-rank neighbourhood_.
Given any state $\alpha_{\varepsilon}$ in the full-rank neighbourhood of
$\alpha$, that is, $\alpha_{\varepsilon}$ is full rank and
$\lim_{\varepsilon\to 0}\alpha_{\varepsilon}=\alpha$, we find the retrodiction
defined on prior $\alpha_{\varepsilon}$ to be
$\hat{\mathcal{U}}_{\alpha}^{(\varepsilon)}[\bullet]=U^{\dagger}\bullet U$.
Taking the limit gives

$\hat{\mathcal{U}}_{\alpha,2}[\bullet]=\lim_{\varepsilon\to
0}\hat{\mathcal{U}}_{\alpha}^{(\varepsilon)}[\bullet]=U^{\dagger}\bullet U,$
(92)

which does not depend on how we chose to perturb $\alpha$.

(3) _Naïvely define $\hat{\mathcal{E}}_{\alpha,3}$ with product-preserving
tuples_. Substituting the product-preserving tuple $(U,\alpha,\beta)$ into Eq.
(88), we have

$\hat{\mathcal{U}}_{\alpha,3}[\bullet]=U^{\dagger}\bullet U.$ (93)

Therefore, we find that
$\hat{\mathcal{U}}_{\alpha,2}=\hat{\mathcal{U}}_{\alpha,3}$ but
$\hat{\mathcal{U}}_{\alpha,1}\neq\hat{\mathcal{U}}_{\alpha,2}$: choosing the
pseudoinverse results in a different retrodiction channel in comparison to the
other two choices.

####  F.1.2 Erasure channels

Consider the erasure channel
$\mathcal{E}[\bullet]:=\Tr(\bullet)\outerproduct{0}{0}$ with input and output
dimensions $d_{A}$. Let its dilation be

$\mathcal{E}[\bullet]=\Tr_{B}\bqty{U\pqty{\bullet\otimes\outerproduct{0}{0}\otimes\outerproduct{0}{0}\otimes\outerproduct{0}{0}}U^{\dagger}},$
(94)

where the ancilla space is
$\mathcal{H}_{B}=\mathcal{H}_{B_{1}}\otimes\mathcal{H}_{B_{2}}\otimes\mathcal{H}_{B_{3}}=\mathcal{C}^{d_{A}}\otimes\mathcal{C}^{d_{A}}\otimes\mathcal{C}^{2}$,
and

$U=\operatorname{SWAP}{}\otimes{}\mathbb{1}\otimes\outerproduct{0}{0}+V\otimes\outerproduct{1}{1},$
(95)

with $\operatorname{SWAP}$ acting on
$\mathcal{H}_{A}\otimes\mathcal{H}_{B_{1}}$ and $V$ acting on
$\mathcal{H}_{A}\otimes\mathcal{H}_{B_{1}}\otimes\mathcal{H}_{B_{2}}$. Since
$U(\alpha\otimes\outerproduct{000}{000})U^{\dagger}=\outerproduct{0}{0}\otimes\alpha\otimes\outerproduct{00}{00}$,
we again have a product-preserving tuple.

(1) _Define $\hat{\mathcal{E}}_{\alpha,1}$ using the pseudoinverse_. This
gives
$\hat{\mathcal{E}}_{\alpha,1}[\bullet]=\bra{0}\\!\bullet\\!\ket{0}\alpha$.

(2) _Define $\hat{\mathcal{E}}_{\alpha,2}$ in a full-rank neighbourhood_.
Since $\mathcal{E}[\alpha]=\outerproduct{0}{0}$ is rank-deficient for all
$\alpha$, it is not enough to perturb just the prior. Instead, we need to
perturb the ancilla as

$\beta_{\varepsilon}:=(1-\varepsilon)\outerproduct{000}{000}+\varepsilon\Gamma_{12}\otimes\outerproduct{1}{1},$
(96)

with $\Gamma_{12}\in\mathcal{H}_{B_{1}}\otimes\mathcal{H}_{B_{2}}$. This
effectively perturbs the channel as

$\mathcal{E}^{(\varepsilon)}[\bullet]:=\Tr_{B}\bqty{U\pqty{\bullet\otimes\beta_{\varepsilon}}}=(1-\varepsilon)\mathcal{E}[\bullet]+\varepsilon\Phi[\bullet],$
(97)

where
$\Phi[\bullet]:=\Tr_{B_{1},B_{2}}[V(\bullet\otimes\Gamma_{12})V^{\dagger}]$
with $V$ and $\Gamma_{12}$ chosen such that $\Phi[\alpha]$ is full rank. This
ensures that the inverse of $\mathcal{E}^{(\varepsilon)}[\alpha]$ exists for
every $\varepsilon>0$. While obviously $\lim_{\varepsilon\to
0}\mathcal{E}^{(\varepsilon)}=\mathcal{E}$, we are going to show that
$\lim_{\varepsilon\to 0}\hat{\mathcal{E}}_{\alpha}^{(\varepsilon)}$ still
depends on $\Phi$ in general. As such, if we are given the rank-deficient
erasure channel, there is no unique consistent definition of
$\hat{\mathcal{E}}_{\alpha}$.

For simplicity, let us take $\Phi$ to be a classical channel. That is,

$\Phi[\outerproduct{j}{j^{\prime}}]=\delta_{j,j^{\prime}}\sum_{k=0}^{d-1}\varphi(k|j)\outerproduct{k}{k},$
(98)

where $0\leq\varphi(k|j)\leq 1$ is a conditional probability distribution,
normalized as $\sum_{k=0}^{d-1}\varphi(k|j)=1$ but otherwise arbitrary.
Likewise, we take the classical state
$\alpha=\sum_{j=0}^{d-1}a_{j}\outerproduct{j}{j}$ as the prior. By direct
computation of Eq. (4), the action of
$\hat{\mathcal{E}}_{\alpha}^{(\varepsilon)}$ on the basis operators
$\Bqty{\outerproduct{k}{k^{\prime}}}_{k,k^{\prime}=0}^{d-1}$ is found to be

$\hat{\mathcal{E}}_{\alpha}^{(\varepsilon)}[\outerproduct{k}{k^{\prime}}]=\frac{(1-\varepsilon)\delta_{k,k^{\prime}}\delta_{k,0}\alpha+\varepsilon\expectationvalue{\Phi[\alpha]}{k}\hat{\Phi}_{\alpha}[\outerproduct{k}{k^{\prime}}]}{(1-\varepsilon)\delta_{k,0}+\varepsilon\expectationvalue{\Phi[\alpha]}{k}},$
(99)

where

$\hat{\Phi}_{\alpha}[\bullet]=\sum_{j=0}^{d-1}\pqty{\sum_{k=0}^{d-1}\frac{\expectationvalue{\bullet}{k}\varphi(k|j)a_{j}}{\expectationvalue{\Phi[\alpha]}{k}}}\outerproduct{j}{j}$
(100)

is the retrodiction channel of $\Phi$ with respect to the prior $\alpha$,
similarly computed using Eq. (4).

All this is well defined, but taking the limit of Eq. (99), we find

$\hat{\mathcal{E}}_{\alpha,2}[\bullet]=\lim_{\varepsilon\to
0}\hat{\mathcal{E}}_{\alpha}^{(\varepsilon)}[\bullet]=\expectationvalue{\bullet}{0}\alpha+\hat{\Phi}_{\alpha}\bqty\big{\bullet-\outerproduct{0}{0}\\!\bullet\\!\outerproduct{0}{0}}.$
(101)

We would have liked this expression to be independent of $\Phi$ as long as
$\Phi[\alpha]$ is full rank, but this is clearly not the case.

(3) _Naïvely define $\hat{\mathcal{E}}_{\alpha,3}$ with product-preserving
tuples_. A simple substitution gives

$\hat{\mathcal{E}}_{\alpha,3}[\bullet]=\Tr(\bullet)\alpha.$ (102)

We see here that all three approaches result in different retrodiction
channels in general, with
$\hat{\mathcal{E}}_{\alpha,2}=\hat{\mathcal{E}}_{\alpha,3}$ only for the
specific choice $\hat{\Phi}_{\alpha}[\bullet]=\Tr(\bullet)\alpha$.

Since the Petz map is equivalent to classical Bayesian inversion when all
channels and states involved are diagonal in a chosen basis [24], we have
demonstrated that retrodiction channels cannot be consistently defined in
general when the output state is rank-deficient, even in classical theory.
This is of course the case a fortiori in quantum theory.

####  F.1.3 Product-preserving two-qubit dilations

Finally, let us investigate the difference between the three approaches in
more detail for the two-qubit case. Consider the product-preserving tuple
$(U,\outerproduct{\alpha_{+}}{\alpha_{+}},\outerproduct{\beta_{+}}{\beta_{+}})$,
with
$U(\ket{\alpha_{+}}\otimes\ket{\beta_{+}})=\ket{\alpha_{+}^{\prime}}\otimes\ket{\beta_{+}^{\prime}}$.
Let $\ket{\alpha_{-}}$ be the state orthogonal to $\ket{\alpha_{+}}$, with
$\ket{\beta_{-}}$, $\ket{\alpha_{-}^{\prime}}$, and $\ket{\beta_{-}^{\prime}}$
defined analogously.

Meanwhile, define $u_{A}$ as the local unitary such that
$u_{A}\ket{\alpha_{\pm}}=\ket{\alpha^{\prime}_{\pm}}$ and $u_{B}$ as the local
unitary such that $u_{A}\ket{\beta_{\pm}}=\ket{\beta^{\prime}_{\pm}}$.

Then, in the $\\{\ket{\alpha_{\pm}}\\}\otimes\\{\ket{\beta_{\pm}}\\}$ basis,
$U$ takes the form

$\pqty{u_{A}^{\dagger}\otimes u_{B}^{\dagger}}U\;\widehat{=}\pmqty{undef}.$
(103)

The first row and column is fixed by the product-preserving property, while
$u_{jk}$ might take on any value, so long as unitarity is satisfied.

(1) _Define $\hat{\mathcal{E}}_{\outerproduct{\alpha_{+}}{\alpha_{+}},1}$
using the pseudoinverse_. This has the action on the basis operators

$\displaystyle\hat{\mathcal{E}}_{\outerproduct{\alpha_{+}}{\alpha_{+}},1}\bqty{\outerproduct{\alpha^{\prime}_{+}}{\alpha^{\prime}_{+}}}$
$\displaystyle=\outerproduct{\alpha_{+}}{\alpha_{+}},$ (104)
$\displaystyle\hat{\mathcal{E}}_{\outerproduct{\alpha_{+}}{\alpha_{+}},1}\bqty{\outerproduct{\alpha^{\prime}_{-}}{\alpha^{\prime}_{-}}}$
$\displaystyle=0,$
$\displaystyle\hat{\mathcal{E}}_{\outerproduct{\alpha_{+}}{\alpha_{+}},1}\bqty{\outerproduct{\alpha^{\prime}_{+}}{\alpha^{\prime}_{-}}}$
$\displaystyle=0.$

(2) _Define $\hat{\mathcal{E}}_{\outerproduct{\alpha_{+}}{\alpha_{+}},2}$ in a
full-rank neighbourhood_. We perturb the prior as
$\alpha_{\varepsilon}:=(1-\varepsilon)\outerproduct{\alpha_{+}}{\alpha_{+}}+\varepsilon\mathbb{1}/2=(1-\varepsilon/2)\outerproduct{\alpha_{+}}{\alpha_{+}}+(\varepsilon/2)\outerproduct{\alpha_{-}}{\alpha_{-}}$.
If $\mathcal{E}[\alpha_{\varepsilon}]$ is invertible, Eq. (23) gives

$\mathcal{E}_{\alpha_{\varepsilon}}[\bullet]=\Tr_{B}\Bqty{U^{\dagger}\sqrt{\omega}\bqty{\pqty{\frac{1}{\sqrt{\Tr_{B}\omega}}\bullet\frac{1}{\sqrt{\Tr_{B}\omega}}}\otimes\mathbbm{1}}\sqrt{\omega}U},$
(105)

where we have defined
$\omega:=U(\alpha_{\varepsilon}\otimes\outerproduct{\beta_{+}}{\beta_{+}})U^{\dagger}$
for brevity. With this, we have

$\omega=\pqty{1-\frac{\varepsilon}{2}}\outerproduct{\alpha_{+}^{\prime},\beta_{+}^{\prime}}{\alpha_{+}^{\prime},\beta_{+}^{\prime}}+\frac{\varepsilon}{2}\underbrace{U\outerproduct{\alpha_{-},\beta_{+}}{\alpha_{-},\beta_{+}}U^{\dagger}}_{=:\outerproduct{\omega_{\perp}}{\omega_{\perp}}}.$
(106)

Here, we have further defined

$\ket{\omega_{\perp}}:=u_{12}\ket{\alpha^{\prime}_{+},\beta^{\prime}_{-}}+u_{22}\ket{\alpha^{\prime}_{-},\beta^{\prime}_{+}}+u_{32}\ket{\alpha^{\prime}_{-},\beta^{\prime}_{-}}.$
(107)

From this, we also have the reduced state

$\displaystyle\Tr_{B}\omega$
$\displaystyle=\frac{1}{2}\mathbbm{1}+\frac{1}{2}\pqty{1-\varepsilon\pqty{1-\absolutevalue{u_{12}}^{2}}}\pqty{\outerproduct{\alpha_{+}^{\prime}}{\alpha_{+}^{\prime}}-\outerproduct{\alpha^{\prime}_{-}}{\alpha^{\prime}_{-}}}$
(108) $\displaystyle\qquad{}+{}\frac{1}{2}\varepsilon
u_{12}u_{22}^{*}\outerproduct{\alpha^{\prime}_{+}}{\alpha^{\prime}_{-}}+\frac{1}{2}\varepsilon
u_{12}^{*}u_{22}\outerproduct{\alpha^{\prime}_{-}}{\alpha^{\prime}_{+}}$
$\displaystyle=:\frac{1}{2}\mathbbm{1}+\frac{1}{2}\vec{r}\cdot\vec{\sigma},$

where $\vec{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$ are the usual Pauli
oeprators defined in the $\Bqty{\ket{\alpha^{\prime}_{\pm}}}$ basis, and
$\vec{r}$ is a three-dimensional vector given in spherical coordinates
$(r,\theta,\phi)$ with

$\displaystyle r\cos\theta$
$\displaystyle=1-\varepsilon\pqty{1-\absolutevalue{u_{12}}^{2}},$ (109)
$\displaystyle r\sin\theta e^{i\phi}$ $\displaystyle=\varepsilon
u_{12}^{*}u_{22}$ $\displaystyle r$
$\displaystyle=\sqrt{\pqty{1-\varepsilon\pqty{1-\absolutevalue{u_{12}}^{2}}}^{2}+\varepsilon^{2}\absolutevalue{u_{12}}^{2}\absolutevalue{u_{22}}^{2}}.$

The eigenvalues of $\Tr_{B}\omega$ can be verified to be $(1+r)/2$ and
$(1-r)/2$, with eigenstates
$\cos(\frac{\theta}{2})\ket{\alpha_{+}^{\prime}}+e^{i\phi}\sin(\frac{\theta}{2})\ket{\alpha_{-}^{\prime}}$
and
$\sin(\frac{\theta}{2})\ket{\alpha_{+}^{\prime}}-e^{i\phi}\cos(\frac{\theta}{2})\ket{\alpha_{-}^{\prime}}$
respectively. Then, $\Tr_{B}\omega$ can be written in its diagonal form and
$(\Tr_{B}\omega)^{-\frac{1}{2}}$ can be worked out. Since we are considering
the cases where $\mathcal{E}_{\alpha_{\varepsilon}}$ is well-defined, we
require $\absolutevalue{u_{12}}^{2}\neq 1$, so that the inverse of
$\Tr_{B}\omega$ exists. With this,

$\displaystyle\frac{1}{\sqrt{\Tr_{B}\omega}}$
$\displaystyle=\frac{1}{\sqrt{2}r}\Bigg{\\{}\bqty{\frac{r+r\cos\theta}{\sqrt{1+r}}-\sqrt{1-r}+\sqrt{\varepsilon}\times\sqrt{\frac{{\varepsilon}}{{1-r^{2}}}}\times\sqrt{1+r}\times(1-\absolutevalue{u_{12}}^{2})}\outerproduct{\alpha^{\prime}_{+}}{\alpha^{\prime}_{+}}$
(110)
$\displaystyle\qquad{}+{}\bqty\bigg{\frac{r-r\cos\theta}{\sqrt{1+r}}+\underbrace{\frac{1}{\sqrt{\varepsilon}}}_{\text{\footnotesize\emph{possibly
problematic}}}\times\sqrt{\frac{\varepsilon}{1-r^{2}}}\times\sqrt{1+r}\times(r+r\cos\theta)}\outerproduct{\alpha^{\prime}_{-}}{\alpha^{\prime}_{-}}$
$\displaystyle\qquad{}+{}\bqty{\frac{r\sin\theta}{\sqrt{1+r}}-\sqrt{\frac{\varepsilon}{1-r^{2}}}\times\frac{r\sin\theta}{\sqrt{\varepsilon}}\times\sqrt{1+r}}\pqty\Big{\outerproduct{\alpha^{\prime}_{+}}{\alpha^{\prime}_{-}}e^{-i\phi}+\outerproduct{\alpha^{\prime}_{-}}{\alpha^{\prime}_{+}}e^{i\phi}}\Bigg{\\}}.$

From Eq. (109), we have the limits

$\begin{gathered}\lim_{\varepsilon\to
0}r\cos\theta=1,\qquad\lim_{\varepsilon\to
0}\frac{r\sin\theta}{\sqrt{\varepsilon}}=0,\qquad\lim_{\varepsilon\to
0}r=1,\\\\[8.61108pt] \lim_{\varepsilon\to
0}\frac{\varepsilon}{1-r^{2}}=\lim_{\varepsilon\to
0}\frac{\varepsilon}{2\varepsilon\pqty{1-\absolutevalue{u_{12}}^{2}}-\varepsilon^{2}\pqty{(1-\absolutevalue{u_{12}}^{2})^{2}+\absolutevalue{u_{12}}^{2}\absolutevalue{u_{22}}^{2}}}=\frac{1}{2\pqty{1-\absolutevalue{u_{12}}^{2}}}.\end{gathered}$
(111)

Keeping these limits in mind, we have singled out the only _possibly
problematic_ term in Eq. (110), which would diverge if the limit of
$(\Tr_{B}\omega)^{-\frac{1}{2}}$ is considered on its own. Nonetheless, the
following limits hold:

$\lim_{\varepsilon\to