#  Quantum communication on the bosonic loss-dephasing channel

Francesco Anna Mele<EMAIL_ADDRESS>NEST, Scuola Normale Superiore and
Istituto Nanoscienze, Consiglio Nazionale delle Ricerche, Piazza dei Cavalieri
7, IT-56126 Pisa, Italy  Farzin Salek<EMAIL_ADDRESS>Department of
Mathematics, Technical University of Munich, Boltzmannstrasse 3, 85748
Garching, Germany.  Vittorio Giovannetti<EMAIL_ADDRESS>NEST,
Scuola Normale Superiore and Istituto Nanoscienze, Consiglio Nazionale delle
Ricerche, Piazza dei Cavalieri 7, IT-56126 Pisa, Italy  Ludovico Lami
<EMAIL_ADDRESS>QuSoft, Science Park 123, 1098 XG Amsterdam, the
Netherlands  Korteweg-de Vries Institute for Mathematics, University of
Amsterdam, Science Park 105-107, 1098 XG Amsterdam, the Netherlands  Institute
for Theoretical Physics, University of Amsterdam, Science Park 904, 1098 XH
Amsterdam, the Netherlands

###### Abstract

Quantum optical systems are typically affected by two types of noise: photon
loss and dephasing. Despite extensive research on each noise process
individually, a comprehensive understanding of their combined effect is still
lacking. A crucial problem lies in determining the values of loss and
dephasing for which the resulting loss-dephasing channel is anti-degradable,
implying the absence of codes capable of correcting its effect or,
alternatively, capable of enabling quantum communication. A conjecture in
[Quantum 6, 821 (2022)] suggested that the bosonic loss-dephasing channel is
anti-degradable if and only if the loss is above $50\%$. In this paper we
refute this conjecture, specifically proving that for any value of the loss,
if the dephasing is above a critical value, then the bosonic loss-dephasing
channel is anti-degradable. While our result identifies a large parameter
region where quantum communication is not possible, we also prove that if two-
way classical communication is available, then quantum communication — and
thus quantum key distribution — is always achievable, even for high values of
loss and dephasing.

> _And the queen gave birth to a child_
> _who was called Asterion._

Quantum optical platforms are key elements of quantum technologies,
contributing significantly to both quantum communication and quantum
computation Hayashi2017 ; BUCCO ; Caves ; weedbrook12 ; hardware-1 ;
hardware-2 ; hardware-3 ; Ofek2016-mu ; WANG20071 . Since the potential
benefits of quantum technologies are hindered by the presence of decoherence
RevModPhys.76.1267 , the investigation of decoherence sources affecting
bosonic systems and the development of bosonic quantum error-correcting codes
have been extensively analysed in recent years PhysRevA.99.032344 ;
Terhal_2020 ; PhysRevA.97.032346 ; Denys2023tqutrittwomode ;
PhysRevX.10.011058 ; PhysRevA.101.012316 ; PhysRevA.106.022431 ;
PhysRevA.107.032423 ; lower-bound ; Die-Hard-2-PRL ; Die-Hard-2-PRA ;
mele2023optical . The primary noise processes in bosonic systems that act as
dominant sources of decoherence are _photon loss_ and _bosonic dephasing_
PhysRevX.6.031006 ; Brito_2008 ; 1992ElL….28…53W , which have been both
extensively analysed PhysRevA.97.032346 ; 8482307 ; Wilde2014-fm ; exact-
solution . Loss affects the system by causing it to dissipate some of its
energy, whereas dephasing works to transform coherent superpositions into
probabilistic mixtures. Although loss and dephasing sources can simultaneously
affect bosonic systems Campagne-Ibarcq2020-cb ; PhysRevX.10.011058 , such as
in superconducting systems PhysRevB.94.014506 ; Rosenblum_2018 , the existing
literature provides only partial results about their combined effect
Loss_dephasing0 . On a technical level, understanding the combined effect of
loss and dephasing is challenging due to the conflicting behaviours they
exhibit: the action of loss takes a simple form when written in the coherent
state basis but is complicated to analyse in the Fock basis BUCCO , whereas
dephasing demonstrates the opposite pattern, making the analysis of their
combined effect quite intricate.

Consider an optical link (e.g. an optical fibre or a free-space link) or a
quantum memory affected by both loss and dephasing, where the link is used for
quantum communication and the memory for quantum computation. A crucial
challenge is to determine the conditions under which there exist protocols
capable of enabling reliable quantum communication across the optical link or
capable of mitigating the combined noise affecting the quantum memory. This
problem is closely related to the anti-degradability condition in quantum
Shannon theory: if a noise channel is _anti-degradable_ MARK ; Sumeet_book ,
there are no quantum communication protocols for reliable information
transmission or quantum error-correcting codes capable of overcoming it.
Consequently, it is crucial to understand whether the combined effect of loss
and dephasing results in an anti-degradable channel. This has been a puzzling
problem, to the point that in Loss_dephasing0  it was conjectured that the
combined loss-dephasing noise results in an anti-degradable channel if and
only if the loss is above $50\%$.

In this paper we refute the above conjecture; specifically, we prove that for
any value of the photon loss there exists a _critical value_ of the dephasing
above which the resulting loss-dephasing channel is anti-degradable. Our
discovery thus identifies a large region of the loss-dephasing parameter space
where correcting the noise and achieving reliable quantum communication is
impossible. On the more positive side, however, we also prove that if the
sender and the receiver are assisted by two-way classical communication, then
reliable quantum communication — and thus quantum key distribution — is always
possible, even in scenarios characterised by arbitrarily high levels of loss
and dephasing. Ours are the first analytical results to characterise the
transmission of quantum information in the presence of both loss and dephasing
noise.

Preliminaries.— The _quantum capacity_ $Q(\pazocal{N})$ of a quantum channel
$\pazocal{N}$ quantifies the efficiency in transmitting qubits reliably across
$\pazocal{N}$ MARK ; Sumeet_book . The condition $Q(\pazocal{N})=0$ implies
that there exist neither reliable quantum communication protocols across
$\pazocal{N}$ nor codes capable of correcting the errors induced by
$\pazocal{N}$. Accordingly, if $\pazocal{N}$ is _anti-degradable_ its quantum
capacity vanishes MARK . This underscores the significance of determining
whether a channel is anti-degradable, as the noise associated with such a
channel cannot be corrected. By definition, a channel $\pazocal{N}$ is anti-
degradable if there exists a channel $\pazocal{A}$ — called the _anti-
degrading map_ — such that
$\pazocal{A}\circ\pazocal{N}^{\text{c}}=\pazocal{N}$, where
$\pazocal{N}^{\text{c}}$ denotes a complementary channel of $\pazocal{N}$ MARK
. Conversely, a channel $\pazocal{N}$ is degradable if there exists a channel
$\pazocal{D}$ such that $\pazocal{D}\circ\pazocal{N}=\pazocal{N}^{\text{c}}$.
Degradable channels are theoretically important because their quantum capacity
can be calculated as the single-letter coherent information of the channel
Devetak-Shor ; MARK ; Sumeet_book .

The phenomenon of photon loss is mathematically modelled by the _pure-loss
channel_ $\pazocal{E}_{\lambda}$ BUCCO ; weedbrook12 , a single-mode
continuous-variable channel that acts on the input state $\rho$ by mixing it
with an environmental vacuum state in a beam splitter of transmissivity
$\lambda\in[0,1]$:

$\displaystyle\pazocal{E}_{\lambda}(\rho)\coloneqq\Tr_{E}\left[U_{\lambda}\left(\rho_{S}\otimes\outerproduct{0}{0}_{E}\right)U_{\lambda}^{\dagger}\right]\,,$
(1)

where
$U_{\lambda}\coloneqq\exp\left[\arccos(\sqrt{\lambda})(\hat{a}^{\dagger}\hat{e}-\hat{a}\,\hat{e}^{\dagger})\right]$
is the beam splitter unitary, $\hat{a}$ and $\hat{e}$ are the annihilation
operators of the input system $S$ and of the environment $E$, and $\Tr_{E}$ is
the partial trace w.r.t. $E$. When a single photon is fed into
$\pazocal{E}_{\lambda}$, it is transmitted to the output with probability
$\lambda$, while it is lost to the environment with probability $1-\lambda$.
More generally, if $n$ photons are fed into the channel, the output is given
by the binomial probability mixture
$\pazocal{E}_{\lambda}(\outerproduct{n}{n})=\sum_{\ell=0}^{n}\binom{n}{l}(1-\lambda)^{\ell}\lambda^{n-\ell}\outerproduct{n-\ell}{n-\ell}$,
where $\ket{n}$ denotes the Fock state with $n$ photons BUCCO . When
$\lambda=1$ the pure-loss channel is noiseless, while when $\lambda=0$ it is
completely noisy — it maps any state into the vacuum. It is known that the
pure-loss channel is anti-degradable for $\lambda\in[0,\frac{1}{2}]$ and
degradable for $\lambda\in[\frac{1}{2}$,1] Caruso_weak ; LossyECEAC1 ;
Caruso2006 ; Wolf2007 .

The phenomenon of bosonic dephasing is mathematically described by the
_bosonic dephasing channel_ $\pazocal{D}_{\gamma}$ Loss_dephasing0 ; exact-
solution ; PhysRevA.102.042413 , which maps the state
$\rho=\sum_{m,n=0}^{\infty}\rho_{mn}\outerproduct{m}{n}$, written in the Fock
basis, to

$\displaystyle\pazocal{D}_{\gamma}(\rho)\coloneqq\sum_{m,n=0}^{\infty}\rho_{mn}e^{-\frac{\gamma}{2}(m-n)^{2}}\outerproduct{m}{n}\,,$
(2)

resulting in a reduction in magnitude of the off-diagonal elements. When
$\gamma=0$, the bosonic dephasing channel is noiseless. In contrast, when
$\gamma\to\infty$, it completely annihilates all off-diagonal components of
the input density matrix, reducing it to an incoherent probabilistic mixture
of Fock states. Moreover, the bosonic dephasing channel is never anti-
degradable and it is always degradable PhysRevA.102.042413 .

Consider an optical system undergoing simultaneous loss and dephasing over a
finite time interval. At each instant, the system is susceptible to both an
infinitesimal pure-loss channel and an infinitesimal bosonic dephasing
channel. Hence, the overall channel, which describes the simultaneous effect
of loss and dephasing, results in a suitable composition of numerous
concatenations between infinitesimal pure-loss and bosonic dephasing channels.
However, given that (i) the pure-loss channel and the bosonic dephasing
channel commute, i.e.
$\pazocal{E}_{\lambda}\circ\pazocal{D}_{\gamma}=\pazocal{D}_{\gamma}\circ\pazocal{E}_{\lambda}$;
(ii) the composition of pure-loss channels is a pure-loss channel,
$\pazocal{E}_{\lambda_{1}}\circ\pazocal{E}_{\lambda_{2}}=\pazocal{E}_{\lambda_{1}\lambda_{2}}$;
and (iii) the composition of bosonic dephasing channels is a bosonic dephasing
channel,
$\pazocal{D}_{\gamma_{1}}\circ\pazocal{D}_{\gamma_{2}}=\pazocal{D}_{\gamma_{1}+\gamma_{2}}$;
it follows that the combined effect of loss and dephasing can be modelled by
the composition

$\displaystyle\pazocal{N}_{\lambda,\gamma}\coloneqq\pazocal{E}_{\lambda}\circ\pazocal{D}_{\gamma}\,,$
(3)

which we will refer to as the _bosonic loss-dephasing channel_ 111The
Supplementary Material provides detailed derivations and additional results
concerning the bosonic loss-dephasing channel for interested readers..

Anti-degradability.— Prior to this work, the only result on the anti-
degradability of the bosonic loss-dephasing channel was that it is anti-
degradable if the transmissivity is below $\frac{1}{2}$ Loss_dephasing0 .
(This result trivially follows from the anti-degradability of the pure-loss
channel for transmissivities below $\frac{1}{2}$, and the fact that the
composition of an anti-degradable channel with another channel inherits the
property of being anti-degradable Note1 ). Notably, in the regime
$\lambda>\frac{1}{2}$, it was an open question to understand whether or not
the bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$ is anti-
degradable for some values of the dephasing $\gamma$, and in Loss_dephasing0
the answer was conjectured to be negative. However, in the forthcoming Theorem
1, we show that the latter conjecture is incorrect, specifically, we prove
that for all $\lambda\in[0,1)$, if $\gamma$ is sufficiently large, then
$\pazocal{N}_{\lambda,\gamma}$ becomes anti-degradable (Note1, , Theorem S27).

######  Theorem 1.

The bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$ is anti-
degradable if the transmissivity $\lambda$ and the dephasing $\gamma$ fall
within one of the following regions: (i) $\lambda\in[0,\frac{1}{2}]$ and
$\gamma\geq 0$; (ii) $\lambda\in(\frac{1}{2},1)$ and $\gamma$ such that
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$,
where $\theta(x,y)\coloneqq\sum_{n=0}^{\infty}x^{n^{2}}y^{n}$. A weaker but
simpler sufficient condition that implies anti-degradability is given by
$\lambda\leq\max\\!\left(\frac{1}{2},\frac{1}{1+9e^{-\gamma}}\right)$.

###### Proof sketch.

Any finite dimensional channel $\pazocal{N}$ is anti-degradable if and only if
its Choi state is _two-extendible_ Myhr2009 , meaning that there exists a
tripartite state $\rho_{AB_{1}B_{2}}$ such that the reduced states on $AB_{1}$
and $AB_{2}$ both coincide with the Choi state:

$\displaystyle\Tr_{B_{2}}\left[\rho_{AB_{1}B_{2}}\right]$
$\displaystyle=C_{AB_{1}}\\!(\pazocal{N})\,,$ (4)
$\displaystyle\Tr_{B_{1}}\left[\rho_{AB_{1}B_{2}}\right]$
$\displaystyle=C_{AB_{2}}\\!(\pazocal{N})\,,$

where the Choi state is defined as
$C_{AB}(\pazocal{N})\coloneqq\operatorname{id}_{A}\otimes\pazocal{N}_{A^{\prime}\to
B}(\Phi_{AA^{\prime}})$, with $\Phi_{AA^{\prime}}$ being the maximally
entangled state. Such a characterisation extends to infinite dimension by
considering the generalised Choi state
$C^{(r)}_{AB}(\pazocal{N})\coloneqq\operatorname{id}_{A}\otimes\pazocal{N}_{A^{\prime}\to
B}\big{(}\Psi^{(r)}_{AA^{\prime}}\big{)}$ extendibility , obtained by
replacing $\Phi_{AA^{\prime}}$ by the two-mode squeezed vacuum state
$\Psi^{(r)}_{AA^{\prime}}$ with squeezing parameter $r>0$ BUCCO . The crux of
our proof is to find a two-extension of
$C^{(r)}_{AB}(\pazocal{N}_{\lambda,\gamma})$ in the region identified by
condition (ii). We do this in two steps.

First, after scrutinising the matrix
$C^{(r)}_{AB}(\pazocal{N}_{\lambda,\gamma})$ written in the Fock basis, we
construct a tripartite state $\tau_{AB_{1}B_{2}}$ such that the reduced states
on $AB_{1}$ and $AB_{2}$ have the same diagonal as
$C^{(r)}_{AB}(\pazocal{N}_{\lambda,\gamma})$, and the same pattern of
vanishing off-diagonal entries. The construction of this tripartite state
involves applying several channels — namely, beam splitter unitaries,
squeezing unitary, partial trace, and a three mode controlled-add-add isometry
— to a 4-mode vacuum state.

The second step consists in transforming $\tau_{AB_{1}B_{2}}$ into a two-
extension of $C^{(r)}_{AB}(\pazocal{N}_{\lambda,\gamma})$ by tweaking its off-
diagonal entries. This is done by using the toolbox of _Hadamard maps_
Sumeet_book . For any matrix $A\coloneqq(a_{mn})_{m,n\in\mathbb{N}}$, the
associated Hadamard map $H^{(A)}$ is defined by
$H^{(A)}(\outerproduct{m}{n})=a_{mn}\outerproduct{m}{n}$ for all $m,n$
Sumeet_book . In practice, $H^{(A)}$ acts on the input density matrix by
multiplying each $(m,n)$ entry by the corresponding coefficient $a_{mn}$.
Importantly, $H^{(A)}$ is a quantum channel if and only if $A$ is Hermitian,
positive semi-definite, and has all $1$’s on the main diagonal Sumeet_book .
The crucial observation is that it is always possible to find an infinite
matrix $A_{\lambda,\gamma}$ (possibly not positive semi-definite), which is
real, symmetric, and all has all $1$’s on the main diagonal, such that the
operator $\operatorname{id}_{A}\otimes H^{(A_{\lambda,\gamma})}_{B_{1}}\otimes
H^{(A_{\lambda,\gamma})}_{B_{2}}\left(\tau_{AB_{1}B_{2}}\right)$ coincides
with $C^{(r)}_{AB}(\pazocal{N}_{\lambda,\gamma})$ when tracing out either
$B_{1}$ or $B_{2}$.

This however does not mean that we have found a two-extension of
$C^{(r)}_{AB}$, because the above operator is not necessarily a state — it may
fail to be positive semi-definite. It _is_ a state, however, whenever
$H^{(A_{\lambda,\gamma})}$ is a quantum channel, i.e. when the infinite matrix
$A_{\lambda,\gamma}$ is positive semi-definite, in formula
$A_{\lambda,\gamma}\geq 0$. Therefore, a sufficient condition on the anti-
degradability of $\pazocal{N}_{\lambda,\gamma}$ is that
$A_{\lambda,\gamma}\geq 0$.

The rest of the proof consists in showing that under condition (ii) one indeed
finds $A_{\lambda,\gamma}\geq 0$. This is not straightforward to check,
because $A_{\lambda,\gamma}$ is an _infinite_ matrix, and it cannot be
diagonalised analytically nor numerically. To by-pass this last hurdle we
employ the theory of diagonally dominant matrices, and in particular the
statement that if a matrix $A$ is such that $a_{nn}-\sum_{m:\,m\neq
n}|a_{mn}|\geq 0$ for all $n$, then necessarily $A\geq 0$ (HJ1, , Chapter 6).
We demonstrate that if $\lambda$ and $\gamma$ satisfy condition (ii), then
$A_{\lambda,\gamma}$ satisfies this condition, which establishes that
$A_{\lambda,\gamma}\geq 0$ and hence concludes the proof Note1 . ∎

Theorem 1 identifies a region of the parameter space $(\lambda,\gamma)$, with
$\lambda$ identifying the transmissivity and $\gamma$ the dephasing, where the
channel is anti-degradable and thus its quantum capacity vanishes, thereby
implying the absence of viable error correcting codes for quantum data
transfer and storage. This region is illustrated in Fig. 1. Interestingly,
Theorem 1 implies that even if $\lambda>\frac{1}{2}$ one can pick $\gamma$
large enough so that there exists an anti-degrading map achieving the
transformation
$\pazocal{N}_{\lambda,\gamma}^{\text{c}}(\outerproduct{n}{n}_{F})\longrightarrow\pazocal{N}_{\lambda,\gamma}(\outerproduct{n}{n}_{F})$,
which can be expressed as (Note1, , SubsectionI.7)

$\displaystyle\pazocal{E}_{1-\lambda}(\outerproduct{n}{n}_{F})\otimes\outerproduct{\sqrt{\gamma}n}{\sqrt{\gamma}n}_{C}\longrightarrow\pazocal{E}_{\lambda}(\outerproduct{n}{n}_{F})\,,$
(5)

where $\ket{n}_{F}$ denotes the $n$th Fock state and
$\ket{\sqrt{\gamma}n}_{C}$ denotes a coherent state BUCCO  (see (Note1, ,
Theorem S29) for an explicit construction of such anti-degrading map). This
entails the following remarkable fact: for $\lambda>1/2$ and large enough
$\gamma$ there exists an _$n$ -independent_ strategy to convert the lossy Fock
state $\pazocal{E}_{1-\lambda}(\outerproduct{n}{n}_{F})$ into the less lossy
Fock state $\pazocal{E}_{\lambda}(\outerproduct{n}{n}_{F})$ using the coherent
state $\ket{\sqrt{\gamma}n}_{C}$ as a resource. In other words, one can undo
part of the loss on $\ket{n}_{F}$ if one has a coherent state that contains
some information on $n$, sufficiently amplified so that that information is
accessible enough. The nontrivial and somewhat surprising nature of this exact
conversion strategy arises from the fact that the coherent states
$\\{\ket{\sqrt{\gamma}n}_{C}\\}_{n\in\mathbb{N}}$ are not orthogonal, meaning
that the strategy that consists in measuring the coherent state, guessing $n$,
and re-preparing $\pazocal{E}_{\lambda}(\outerproduct{n}{n}_{F})$ cannot
succeed with probability $1$.

Theorem 1 does not identify the entire anti-degradability region of
$\pazocal{N}_{\lambda,\gamma}$, but only a subset of it. One way to improve
this approximation is to determine numerically the region where the infinite
matrix $A_{\lambda,\gamma}$ introduced in the proof sketch of Theorem 1 is
positive semi-definite. In Fig. 1 we depict a numerical estimate of this
region (see the crossed red part of the plot).

So far we have been concerned with inner approximations of the anti-
degradability region. To obtain outer approximations, instead, one can start
by observing that the action of $\pazocal{N}_{\lambda,\gamma}$ can only
subtract and never add any photons. Mathematically, if the input state to
$\pazocal{N}_{\lambda,\gamma}$ is supported on the span of the first $d$ Fock
states, so is the output state Note1 . This restriction defines a qu$d$it-to-
qu$d$it channel $\pazocal{N}_{\lambda,\gamma}^{(d)}$, analysing which can
yield some insights into $\pazocal{N}_{\lambda,\gamma}$ itself. First, if
$\pazocal{N}^{(d)}_{\lambda,\gamma}$ is _not_ anti-degradable then the same is
true of $\pazocal{N}_{\lambda,\gamma}$ Note1 ; secondly, as discussed above
the anti-degradability of $\pazocal{N}^{(d)}_{\lambda,\gamma}$ is equivalent
to the two-extendibility of the corresponding Choi state Myhr2009 , and for
moderate values of $d$ this latter condition can be efficiently checked
numerically via _semi-definite programming_ Note1 ; WATROUS ; Sumeet_book .
Interestingly, already the qubit restriction
$\pazocal{N}^{(2)}_{\lambda,\gamma}$, which coincides with the composition
between the amplitude damping channel and the qubit dephasing channel
Sumeet_book , yields the following necessary condition on the anti-
degradability of $\pazocal{N}_{\lambda,\gamma}$ (Note1, , Theorem S35).

######  Theorem 2.

If $\lambda>\frac{1}{1+e^{-\gamma}}$ then $\pazocal{N}_{\lambda,\gamma}$ is
not anti-degradable.

###### Proof sketch.

A qubit channel $\pazocal{N}$ is anti-degradable if and only if

$\displaystyle\frac{1}{4}\Tr\\!\left[\pazocal{N}\\!\left(\mathbb{1}_{2}\right)^{2}\right]\geq\Tr\\!\left[C(\pazocal{N})^{2}\right]-4\sqrt{\det[C(\pazocal{N})]},$
(6)

where $C(\pazocal{N})$ is the Choi state paddock2017characterization ;
Chen_2014 ; Myhr2009 . When focusing on the qubit restriction
$\pazocal{N}^{(2)}_{\lambda,\gamma}$, Eq. (6) is equivalent to the condition
$\lambda\leq\frac{1}{1+e^{-\gamma}}$. ∎

We are interested in the anti-degradability of the loss-dephasing channel
because it implies that the quantum capacity vanishes, entailing the
impossibility of quantum communication. We now look at the complementary
question: when is the quantum capacity $Q(\pazocal{N}_{\lambda,\gamma})$
strictly positive? A simple sufficient condition can be obtained by optimising
the coherent information Sumeet_book ; MARK  of $\pazocal{N}_{\lambda,\gamma}$
over input states of the form $\rho_{p}\coloneqq
p\outerproduct{0}{0}+(1-p)\outerproduct{1}{1}$. By doing so we identify a
region of the $(\lambda,\gamma)$ parameter space where
$Q(\pazocal{N}_{\lambda,\gamma})>0$ (see the crossed green region in Fig. 1).
In this region quantum communication and quantum error correction become
feasible.

Figure 1: Summary of results on the anti-degradability of the bosonic loss-
dephasing channel $\pazocal{N}_{\lambda,\gamma}$. The vertical axis represents
the transmissivity $\lambda$, while the horizontal axis corresponds to
$e^{-\gamma}$, where $\gamma$ is the dephasing parameter. In the green region
$\pazocal{N}_{\lambda,\gamma}$ is not anti-degradable, while in the red region
it is anti-degradable. In the crossed green region, the quantum capacity of
$\pazocal{N}_{\lambda,\gamma}$ is strictly positive. The crossed red region is
a numerical estimate of the region where the infinite matrix
$A_{\lambda,\gamma}$ is positive semi-definite, a condition implying that
$\pazocal{N}_{\lambda,\gamma}$ is anti-degradable, as explained in the proof
sketch of Theorem 1. Such an estimate can be obtained by examining the
positive semi-definiteness of the $d\times d$ top-left corner of
$A_{\lambda,\gamma}$ for large values of $d$ (here we employ $d=30$, but
increasing $d$ already beyond $d\geq 20$ yields no discernible change in the
plot). The restriction $\pazocal{N}_{\lambda,\gamma}^{(6)}$ is anti-degradable
if and only if $\lambda$ and $\gamma$ fall within the green region, and this
is the reason why $\pazocal{N}_{\lambda,\gamma}$ is not anti-degradable in the
green region. Below the curve
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)=\frac{3}{2}$,
the channel $\pazocal{N}_{\lambda,\gamma}$ is anti-degradable, as stated in
Theorem 1. Above the curve $\lambda=\frac{1}{1+e^{-\gamma}}$, the channel is
not anti-degradable, as guaranteed by Theorem 2.

Two-way quantum communication.— As we have just seen, (unassisted) quantum
communication is not possible when the combined effects of loss and dephasing
are too strong. However, in the forthcoming Theorem 3 we show that if Alice
(the sender) and Bob (the receiver) have access to a _two-way_ classical
communication line, then quantum communication, entanglement distribution, and
quantum-key distribution Ekert91  become again achievable for any value of
loss and dephasing, even when Alice’s input signals are constrained to have
limited energy. Hence, the bosonic loss-dephasing channel is never
entanglement breaking MARK ; Sumeet_book .

In this two-way communication setting the relevant notion of capacity is the
_two-way quantum capacity_ $Q_{2}(\pazocal{N})$ MARK ; Sumeet_book , defined
as the maximum achievable rate of qubits that can be reliably transmitted
across $\pazocal{N}$ with the aid of two-way classical communication. Since in
practice Alice has only a limited amount of energy to produce her input
signals, one usually defines the so-called _energy-constrained_ two-way
quantum capacity Davis2018 ; 8541091 , denoted as $Q_{2}(\pazocal{N},N_{s})$.
Here, $N_{s}$ is the energy constraint on the mean photon number of any input
state $\rho$, i.e. $\Tr[\rho\,\hat{a}^{\dagger}\hat{a}]\leq N_{s}$. Now, we
are ready to state the following result (Note1, , Theorem S38).

######  Theorem 3.

For all $N_{s}>0$, $\lambda\in(0,1]$, and $\gamma\geq 0$, the energy-
constrained two-way quantum capacity of the bosonic loss-dephasing channel is
strictly positive, i.e. $Q_{2}(\pazocal{N}_{\lambda,\gamma},N_{s})>0$. In
particular, for $\lambda>0$ $\pazocal{N}_{\lambda,\gamma}$ is not entanglement
breaking.

###### Proof sketch.

Assume, without loss of generality, that $N_{s}<1$. Alice prepares many copies
of the state
$\ket{\Psi_{N_{s}}}_{\\!AA^{\prime}}\coloneqq\sqrt{1-N_{s}}\ket{0}_{\\!A}\\!\otimes\\!\ket{0}_{\\!A^{\prime}}+\sqrt{N_{s}}\ket{1}_{\\!A}\\!\otimes\\!\ket{1}_{\\!A^{\prime}}$,
which satisfies the energy constraint, and sends the subsystems $A^{\prime}$
to Bob via the channel. Now Alice and Bob share many copies of the effectively
two-qubit state
$\rho_{AB}\coloneqq\operatorname{id}_{A}\otimes\pazocal{N}_{\lambda,\gamma}(\outerproduct{\Psi_{N_{s}}}{\Psi_{N_{s}}})$.
By a straightforward calculation one verifies that the ‘partial transpose’
PeresPPT  of $\rho_{AB}$ is not positive, hence its entanglement is
distillable 2-qubit-distillation , implying that Alice and Bob can extract
singlets at a strictly positive rate via LOCC Sumeet_book . This rate is a
lower bound on $Q_{2}(\pazocal{N}_{\lambda,\gamma},N_{s})$, thanks to quantum
teleportation teleportation . ∎

Degradability.— The bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is never degradable, except in the simple cases
when either $\lambda=1$ or $\gamma=0$ and $\lambda\geq 1/2$ Loss_dephasing0 .
This in turn implies that no single-letter formula for its quantum capacity is
known. The failure of degradability has been demonstrated in Loss_dephasing0
through pages-long proof; we are now in position to provide a significantly
simpler proof of this result. The key ideas behind our proof are as follows:
(i) If the qubit restriction $\pazocal{N}_{\lambda,\gamma}^{(2)}$ is not
degradable, then $\pazocal{N}_{\lambda,\gamma}$ is not degradable either Note1
; and (ii) if the rank of the Choi state of a qubit channel is greater or
equal to $3$ than such channel is not degradable (Cubitt2008, , Theorem 4).
The result then follows by observing that the rank of the Choi state of the
qubit channel $\pazocal{N}_{\lambda,\gamma}^{(2)}$ is exactly $3$ (for a
detailed proof see (Note1, , Theorem S34)).

Conclusions.— In this paper we have provided the first analytical
investigation of the quantum communication capabilities of the bosonic
dephasing channel, a much more realistic model of noise than dephasing and
loss treated separately. Refuting a conjecture put forth in Loss_dephasing0 ,
we showed that the bosonic loss-dephasing channel is anti-degradable in a
large region of the loss-dephasing parameter space, entailing that neither
quantum communication nor quantum error correcting codes are possible in this
region. On the positive side, we also showed that if two-way classical
communication is suitably exploited, then quantum communication is always
achievable, even in scenarios characterised by high levels of loss and
dephasing, and even in the presence of stringent energy constraints.

An intriguing open problem is to find the entire anti-degradability region of
the bosonic loss-dephasing channel. Based on the analysis of the qu$d$it
restrictions for small values of $d$ Note1 , we conjecture that for
$e^{-\gamma}\leq\sqrt{2}-1$ the anti-degradability condition is _precisely_
$\lambda\leq\frac{1}{1+e^{-\gamma}}$. Furthermore, although the capacities of
the bosonic dephasing channel and the pure-loss channel have been determined
exact-solution ; PLOB ; holwer ; Caruso2006 ; Wolf2007 ; Mark2012 ; Mark-
energy-constrained ; Noh2019 , the capacities of the bosonic loss-dephasing
remain unknown. Another pressing challenge is to calculate or approximate
these capacities. A lower bound on $Q_{2}(\pazocal{N}_{\lambda,\gamma},N_{s})$
can be determined by calculating and optimising the rate of the protocol
introduced in the proof of Theorem 3, following the approach outlined in
lower-bound .

Acknowledgements.— FAM and VG acknowledge financial support by MUR (Ministero
dell’Istruzione, dell’Università e della Ricerca) through the following
projects: PNRR MUR project PE0000023-NQSTI, PRIN 2017 Taming complexity via
Quantum Strategies: a Hybrid Integrated Photonic approach (QUSHIP) Id.
2017SRN-BRK, and project PRO3 Quantum Pathfinder. FS is supported by Walter
Benjamin Fellowship, DFG project No. 524058134. FAM and LL thank the Freie
Universität Berlin for hospitality. FAM thanks the University of Amsterdam for
hospitality.

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Supplementary Material

###### Contents

  1. I Setting the Stage and Notations
    1. I.1 Quantum states and Channels
    2. I.2 Bosonic systems
    3. I.3 Hadamard maps
    4. I.4 Beam splitter
    5. I.5 Pure-loss channel
    6. I.6 Bosonic dephasing channel
    7. I.7 Bosonic loss-dephasing channel
    8. I.8 Preliminaries on capacities of quantum channels
  2. II Anti-degradability and Degradability of bosonic loss-dephasing channel
    1. II.1 Sufficient condition on anti-degradability
      1. Expanding the anti-degradability region numerically
      2. Anti-degrading maps
    2. II.2 Analysis of the bosonic loss-dephasing channel via its finite-dimensional restrictions
      1. The bosonic loss-dephasing channel is never degradable
      2. Necessary condition on anti-degradability via qubit restriction
      3. Necessary condition on anti-degradability via qudit restrictions
  3. III Coherence preservation of the bosonic loss-dephasing channel
  4. IV Appendix
    1. IV.1 Hadamard maps
    2. IV.2 Miscellaneous Lemmas

##  I Setting the Stage and Notations

###  I.1 Quantum states and Channels

In this subsection, we present a summary of the notation and fundamental
properties used in the paper, drawing from the conventions established in
standard quantum information theory textbooks [60, 33, 45, 34]. Every quantum
system is associated with a separable complex Hilbert space $\pazocal{H}$
whose dimension is denoted by $|\pazocal{H}|$. We use subscripts to denote the
system associated to a Hilbert space and also systems on which the operators
act. The composite quantum systems $A$ and $B$ exist within the tensor product
of their individual Hilbert spaces $\pazocal{H}_{A}\otimes\pazocal{H}_{B}$
which is also denoted by $\pazocal{H}_{AB}$.

We shall use $\mathbb{1}$ to denote the identity operator on $\pazocal{H}$.
The operator norm of a linear operator $\Theta:\pazocal{H}\to\pazocal{H}$ is
defined as

$\displaystyle\|\Theta\|_{\infty}\coloneqq\sup_{\ket{\psi}\in\pazocal{H}:\,\langle{\psi}|\psi\rangle=1}\sqrt{\langle\psi|\Theta^{\dagger}\Theta|\psi\rangle}\,.$
(S1)

An alternative (but equivalent) definition of the operator norm is as follows:

$\displaystyle\|\Theta\|_{\infty}\coloneqq\sup_{\ket{v},\ket{w}\in\pazocal{H},\,\innerproduct{v}{v}=\innerproduct{w}{w}=1}\left|\bra{v}\Theta\ket{w}\right|\,.$
(S2)

An operator is called bounded if its operator norm is bounded, i.e.
$\|\Theta\|_{\infty}<\infty$. A bounded operator $\Theta$ is positive semi-
definite if $\bra{\psi}\Theta\ket{\psi}\geq
0,\forall\,\ket{\psi}\in\pazocal{H}\,$, while it is positive definite if
$\bra{\psi}\Theta\ket{\psi}>0,\,\forall\,\ket{\psi}\in\pazocal{H}\,$. The
trace norm of a linear operator $\Theta:\pazocal{H}\to\pazocal{H}$ is defined
as $\|\Theta\|_{1}\coloneqq\Tr\sqrt{\Theta^{\dagger}\Theta}$. The set of trace
class operators, denoted as $\pazocal{T}(\pazocal{H})$, is the set of all the
linear operators on $\pazocal{H}$ such that their trace norm is finite, i.e.
$\|\Theta\|_{1}<\infty$. The operator and trace norm satisfy
$\|\Theta\|_{\infty}\leq\|\Theta\|_{1}$. The set of quantum states (density
operators), denoted as $\pazocal{P}(\pazocal{H})$, is the set of positive
semi-definite trace class operators on $\pazocal{H}$ with unit trace. The
_fidelity_ between two quantum states $\rho,\sigma\in\pazocal{P}(\pazocal{H})$
is defined as
$F(\rho,\sigma)\coloneqq\Tr[\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}]$.

A superoperator is a linear map between spaces of linear operators. The
identity superoperator will be denoted as $\operatorname{id}$. Quantum
channels are completely-positive trace-preserving (cptp) superoperators. In
this paper, we will use two different representations of a quantum channel
that are known as Stinespring and Choi–Jamiołkowski representation. A quantum
channel $\pazocal{N}_{A^{\prime}\to B}$ can be represented in Stinespring
representation as

$\displaystyle\pazocal{N}_{A^{\prime}\to
B}(\cdot)=\Tr_{E}\\!\left[U_{A^{\prime}E\to
BE}(\cdot\otimes\outerproduct{0}{0}_{E})U_{A^{\prime}E\to
BE}^{\dagger}\right].$

Here, $E$ is an environment system, $\ket{0}_{E}$ is a pure state of the
environment, and $U_{A^{\prime}E\to BE}$ is an isometry that takes as input
the systems $A^{\prime}$ and $E$ and outputs the systems $B$, $E$. The
associated complementary channel $\pazocal{N}_{A^{\prime}\to B}^{\text{c}}$ is
given by

$\displaystyle\pazocal{N}^{\text{c}}_{A^{\prime}\to
E}(\cdot)=\Tr_{B}\\!\left[U_{A^{\prime}E\to
BE}(\cdot\otimes\outerproduct{0}{0}_{E})U_{A^{\prime}E\to
BE}^{\dagger}\right].$

A channel $\pazocal{N}$ is called degradable if there exist a quantum channel
$\pazocal{J}$, such that when is used after $\pazocal{N}$, we get back to the
complementary channel $\pazocal{N}^{\text{c}}$, i.e.
$\pazocal{J}\circ\pazocal{N}=\pazocal{N}^{\text{c}}$. On the other hand, a
channel is called anti-degradable if there is another quantum channel
$\pazocal{A}$, such that using it after the complementary channel, gives back
the original channel, i.e.
$\pazocal{A}\circ\pazocal{N}^{\text{c}}=\pazocal{N}$. The channels
$\pazocal{J}$ and $\pazocal{A}$ are usually called the degrading map and the
anti-degrading map of $\pazocal{N}$, respectively.

The Choi–Jamiołkowski representation of the channel
$\pazocal{N}_{A^{\prime}\to B}$ is the operator
$C(\pazocal{N})\in\pazocal{T}(\pazocal{H}_{A}\otimes\pazocal{H}_{B})$ that is
defined as

$\displaystyle
C(\pazocal{N})\coloneqq\operatorname{id}_{A}\otimes\pazocal{N}_{A^{\prime}\to
B}(\outerproduct{\Phi}{\Phi}_{AA^{\prime}})\,,$ (S3)

where
$\ket{\Phi}=\frac{1}{\sqrt{|\pazocal{H}_{A}|}}\sum_{i=0}^{|\pazocal{H}_{A}|-1}\ket{i}_{A}\otimes\ket{i}_{A^{\prime}}$
is a maximally entangled state of schmidt rank $|\pazocal{H}_{A}|$, the set of
states $\\{\ket{i}\\}_{i=0,\ldots,|\pazocal{H}_{A}|-1}$ forms a basis for
$\pazocal{H}_{A}$, and $\pazocal{H}_{A}=\pazocal{H}_{A^{\prime}}$. It is a
well-established fact that the superoperator $\pazocal{N}$ is a quantum
channel if and only if $C(\pazocal{N})\geq 0$ and
$\Tr_{B}C(\pazocal{N})=\mathbb{1}_{A}/|\pazocal{H}_{A}|$ [33].

######  Definition S1.

A bipartite state $\rho_{AB}$ is symmetric two-extendible on $B$ if there
exists a tripartite state $\tau_{AB_{1}B_{2}}$ such that

  * •

$F_{B_{1}B_{2}}\tau_{AB_{1}B_{2}}F_{B_{1}B_{2}}^{\dagger}=\tau_{AB_{1}B_{2}}$

  * •

$\Tr_{B_{1}}\tau_{AB_{1}B_{2}}=\rho_{AB}$,

where $B_{1}$ and $B_{2}$ are two copies of the system $B$, the operator
$F_{B_{1}B_{2}}\coloneqq\sum_{i,j}\outerproduct{i}{j}_{B_{1}}\otimes\outerproduct{j}{i}_{B_{2}}$
denotes the swap unitary, and $\\{\ket{i}_{B_{1}}\\}_{i}$ and
$\\{\ket{i}_{B_{2}}\\}_{i}$ form an orthonormal basis. The state
$\tau_{AB_{1}B_{2}}$ is called a symmetric two-extension of $\rho_{AB}$ on
$B$.

######  Definition S2.

A bipartite state $\rho_{AB}$ is called two-extendible on $B$ if there exists
a tripartite state $\tau_{AB_{1}B_{2}}$ such that

$\displaystyle\Tr_{B_{1}}\tau_{AB_{1}B_{2}}=\Tr_{B_{2}}\tau_{AB_{1}B_{2}}=\rho_{AB}\,,$
(S4)

where $B_{1}$ and $B_{2}$ are two copies of the system $B$.

######  Lemma S3 ([61]).

A bipartite state $\rho_{AB}$ is two-extendible on $B$ if and only if it is
symmetric two-extendible on $B$ .

###### Proof.

First, assume $\rho_{AB}$ is symmetric two-extendible on $B$. Since
$F_{B_{1}B_{2}}\tau_{AB_{1}B_{2}}F_{B_{1}B_{2}}^{\dagger}=\tau_{AB_{1}B_{2}}$,
it holds that $\Tr_{B_{2}}\tau_{AB_{1}B_{2}}=\Tr_{B_{1}}\tau_{AB_{1}B_{2}}$.
This implies that $\rho_{AB}$ is two-extendible on $B$. Second, let
$\rho_{AB}$ be two-extendible on $B$. One can easily check that the state
$1/2(\tau_{AB_{1}B_{2}}+F_{B_{1}B_{2}}\tau_{AB_{1}B_{2}}F_{B_{1}B_{2}}^{\dagger})$
is a symmetric two-extension of $\rho_{AB}$. ∎

It has been demonstrated that a quantum channel is anti-degradable if and only
if its Choi state is two-extendible on the output system [42]. This
equivalence leads to a simple necessary and sufficient condition for the anti-
degradability of qubit channels:

######  Lemma S4.

[46, Corollary 4](See also [47, 42]) Any qubit quantum channel $\pazocal{N}$
is anti-degradable if and only if it satisfies

$\displaystyle\Tr\\!\left[\left(\pazocal{N}\left(\frac{\mathbb{1}_{2}}{2}\right)\right)^{2}\right]\geq\Tr\\!\left[(C(\pazocal{N}))^{2}\right]-4\sqrt{\det(C(\pazocal{N}))},$

where $\mathbb{1}_{2}$ denotes the identity operator on the qubit Hilbert
space.

###  I.2 Bosonic systems

In this subsection, we will provide an overview of relevant definitions and
properties concerning quantum information with continuous variable systems;
refer to [2] for detailed explanations. A single-mode of electromagnetic
radiation with definite frequency and polarisation is represented by the
Hilbert space $L^{2}(\mathbb{R})$, which comprises all square-integrable
complex-valued functions over $\mathbb{R}$. Let $\mathbb{N}_{+}$ be the set of
strictly positive integers and let
$\mathbb{N}\coloneqq\\{0\\}\cup\mathbb{N}_{+}$. For any $n\in\mathbb{N}$, the
construction of the _Fock state_ $\ket{n}$ (the quantum state with $n$
photons) involves the application of the bosonic creation operator
$\hat{a}^{\dagger}$ to the _vacuum state_ $\ket{0}$:

$\displaystyle\ket{n}\coloneqq\frac{1}{\sqrt{n!}}(\hat{a}^{\dagger})^{n}\ket{0}\,.$
(S5)

The Fock states $\\{\ket{n}\\}_{n\in\mathbb{N}}$ form an orthonormal basis of
$L^{2}(\mathbb{R})$. The bosonic annihilation operator $\hat{a}$ and creation
operator $\hat{a}^{\dagger}$ satisfy the well-known canonical commutation
relation $[\hat{a},\hat{a}^{\dagger}]=\mathbb{1}$.

Let $\mathbb{C}$ be the set of complex numbers. For any $\alpha\in\mathbb{C}$,
let $D(\alpha)\coloneqq e^{\alpha\hat{a}^{\dagger}-\alpha^{\ast}\hat{a}}$ be
the displacement operator. A coherent state of parameter $\alpha$, denoted by
$\ket{\alpha}$, is defined by applying the displacement operator $D(\alpha)$
to the vacuum state, i.e. $\ket{\alpha}\coloneqq D(\alpha)\ket{0}$. The
overlap between coherent states is given by

$\displaystyle\innerproduct{\alpha}{\beta}=e^{-\frac{1}{2}(|\alpha|^{2}+|\beta|^{2}-2\alpha^{*}\beta)}\,.$
(S6)

Quantum channels acting on bosonic systems are sometimes called bosonic
channels. Similar to finite-dimensional channels, bosonic channels admit a
Choi–Jamiołkowski representation, usually referred to as _generalised_
Choi–Jamiołkowski representation [62, 63]. Consider isomorphic Hilbert spaces
$\pazocal{H}_{A},\pazocal{H}_{A^{\prime}}$ which are possibly infinite
dimensional. Let $\ket{\psi}_{AA^{\prime}}$ be a pure state satisfying
$\Tr_{A^{\prime}}\\!\left[\outerproduct{\psi}{\psi}_{AA^{\prime}}\right]>0$
(See [64, Lemma 26]). The generalised Choi state is constructed by applying
the channel to the subsystem $A^{\prime}$ of $\ket{\psi}_{A^{\prime}A}$ (see
Lemma S48 in the Appendix):
$\operatorname{id}_{A}\otimes\pazocal{N}_{A^{\prime}\to
B}(\outerproduct{\psi}{\psi}_{AA^{\prime}})$. The construction of the
generalised Choi state usually utilises the two-mode squeezed vacuum state
with squeezing parameter $r>0$, defined as follows [2]:

$\displaystyle\ket{\psi(r)}_{AA^{\prime}}\coloneqq\frac{1}{\cosh(r)}\sum_{n=0}^{\infty}\tanh^{n}(r)\ket{n}_{A}\ket{n}_{A^{\prime}}\,.$
(S7)

The equivalence between anti-degradability of a channel and two-extendibility
of its Choi state extends to the infinite-dimensional channels [43]. We
provide a detailed proof of this equivalence in Lemma S49 in the Appendix as
it helps us in developing our intuition in inventing an explicit example of an
anti-degrading map of the bosonic loss-dephasing channel.

###  I.3 Hadamard maps

Let $A=(a_{mn})_{m,n\in\mathbb{N}},a_{mn}\in\mathbb{C}$, be an infinite matrix
of complex numbers. Consider the superoperator $H$ on
$\pazocal{T}\left(L^{2}(\mathbb{R})\right)$, recognised as the _Hadamard map_
, whose action on rank one operator $\outerproduct{m}{n}$ is defined as
follows:

$\displaystyle
H(\outerproduct{m}{n})=a_{mn}\outerproduct{m}{n},\quad\forall\,m,n\in\mathbb{N}\,.$

In Section IV.1 in the Appendix, we provide an overview of relevant properties
of Hadamard maps. In particular, by combining known results about Hadamard
maps and matrix analysis, in Lemma S46 in the Appendix, we show that given an
infinite matrix $A=(a_{mn})_{m,n\in\mathbb{N}},a_{mn}\in\mathbb{C}$, the
associated Hadamard map is a quantum channel if

  * •

$A$ is Hermitian

  * •

$a_{nn}=1$, $\forall n\in\mathbb{N}$

  * •

$A$ is diagonally dominant, i.e. $\sum_{\begin{subarray}{c}m=0\\\ m\neq
n\end{subarray}}^{\infty}|a_{mn}|\leq 1,\,\forall\,n\in\mathbb{N}\,.$

###  I.4 Beam splitter

A beam splitter serves as a linear optical tool employed for creating quantum
entanglement between two modes, referred to as the _system_ mode (denoted as
$S$) and the _environment_ mode (denoted as $E$). A depiction of a beam
splitter is reported in Fig 2.

######  Definition S5.

Let $\pazocal{H}_{S},\pazocal{H}_{E}\coloneqq L^{2}(\mathbb{R})$. Let
$\hat{a}$ and $\hat{e}$ denote the annihilation operator of $\pazocal{H}_{S}$
and $\pazocal{H}_{E}$, respectively. For all $\lambda\in[0,1]$, the beam
splitter unitary of transmissivity $\lambda$ given by

$\displaystyle U_{\lambda}^{SE}$
$\displaystyle\coloneqq\exp\left[\arccos\sqrt{\lambda}\left(\hat{a}^{\dagger}\hat{e}-\hat{a}\,\hat{e}^{\dagger}\right)\right]\,.$
(S8)

######  Lemma S6.

For all $\lambda\in[0,1]$, it holds that

$\displaystyle(U_{\lambda}^{SE})^{\dagger}\hat{a}\,U_{\lambda}^{SE}$
$\displaystyle=\sqrt{\lambda}\,\hat{a}+\sqrt{1-\lambda}\,\hat{e}\,,$ (S9)
$\displaystyle U_{\lambda}^{SE}\hat{a}\,(U_{\lambda}^{SE})^{\dagger}$
$\displaystyle=\sqrt{\lambda}\,\hat{a}-\sqrt{1-\lambda}\,\hat{e}\,,$
$\displaystyle(U_{\lambda}^{SE})^{\dagger}\hat{e}\,U_{\lambda}^{SE}$
$\displaystyle=-\sqrt{1-\lambda}\,\hat{a}+\sqrt{\lambda}\,\hat{e}\,,$
$\displaystyle U_{\lambda}^{SE}\hat{e}\,(U_{\lambda}^{SE})^{\dagger}$
$\displaystyle=\sqrt{1-\lambda}\,\hat{a}+\sqrt{\lambda}\,\hat{e}\,.$

###### Proof.

These identities can be readily proved by applying the Baker-Campbell-
Hausdorff formula. For an alternative proof see [21, Lemma A.2]. ∎

######  Lemma S7.

For all $\lambda\in[0,1]$ and all $n\in\mathbb{N}$, it holds that

$\displaystyle U^{SE}_{\lambda}\ket{n}_{S}\otimes\ket{0}_{E}$
$\displaystyle=\sum_{l=0}^{n}(-1)^{l}\sqrt{\binom{n}{l}}\lambda^{\frac{n-l}{2}}(1-\lambda)^{\frac{l}{2}}\ket{n-l}_{S}\otimes\ket{l}_{E}\,,$
(S10) $\displaystyle U^{SE}_{\lambda}\ket{0}_{S}\otimes\ket{n}_{E}$
$\displaystyle=\sum_{l=0}^{n}\sqrt{\binom{n}{l}}(1-\lambda)^{\frac{l}{2}}\lambda^{\frac{n-l}{2}}\ket{l}_{S}\otimes\ket{n-l}_{E}\,.$
(S11)

###### Proof.

Thanks to Lemma (S6), we have that
$U_{\lambda}^{SE}\hat{a}\left(U_{\lambda}^{SE}\right)^{\dagger}=\sqrt{\lambda}\hat{a}-\sqrt{1-\lambda}\hat{e}$.
Consequently, it holds that

$\displaystyle U_{\lambda}^{SE}\ket{n}_{S}\otimes\ket{0}_{E}$
$\displaystyle=\frac{1}{\sqrt{n!}}U_{\lambda}^{SE}(a^{\dagger})^{n}\ket{0}_{S}\otimes\ket{0}_{E}$
(S12)
$\displaystyle=\frac{1}{\sqrt{n!}}\left(U_{\lambda}^{SE}a^{\dagger}\left(U_{\lambda}^{SE}\right)^{\dagger}\right)^{n}\ket{0}_{S}\otimes\ket{0}_{E}$
$\displaystyle=\frac{1}{\sqrt{n!}}\left(\sqrt{\lambda}a^{\dagger}-\sqrt{1-\lambda}b^{\dagger}\right)^{n}\ket{0}_{S}\otimes\ket{0}_{E}$
$\displaystyle=\frac{1}{\sqrt{n!}}\sum_{l=0}^{n}(-1)^{l}\binom{n}{l}\lambda^{\frac{n-l}{2}}(1-\lambda)^{\frac{l}{2}}(a^{\dagger})^{n-l}\ket{0}_{S}\otimes(b^{\dagger})^{l}\ket{0}_{E}$
$\displaystyle=\sum_{l=0}^{n}(-1)^{l}\sqrt{\binom{n}{l}}\lambda^{\frac{n-l}{2}}(1-\lambda)^{\frac{l}{2}}\ket{n-l}_{S}\otimes\ket{l}_{E}\,.$

Analogously, one can show the validity of (S11) by exploiting
$U_{\lambda}^{SE}\hat{e}\,(U_{\lambda}^{SE})^{\dagger}=\sqrt{1-\lambda}\,\hat{a}+\sqrt{\lambda}\,\hat{e}$.
∎

######  Remark S8.

It is easily seen that Eq. (S10) is equivalent to

$\displaystyle U^{SE}_{\lambda}\ket{n}_{S}\otimes\ket{0}_{E}$
$\displaystyle=(-1)^{n}\sum_{l=0}^{n}(-1)^{l}\sqrt{\binom{n}{l}}\lambda^{\frac{l}{2}}(1-\lambda)^{\frac{n-l}{2}}\ket{l}_{S}\otimes\ket{n-l}_{E}\,.$

###  I.5 Pure-loss channel

In optical platforms, the most common source of noise is photon loss, which is
modelled by the _pure-loss channel_ [2]. For any $\lambda\in[0,1]$, the pure-
loss channel $\pazocal{E}_{\lambda}$ of transmissivity $\lambda$ is a single-
mode bosonic channel which acts on the input system by mixing it in a beam
splitter of transmissivity $\lambda$ with an environmental vacuum state; see
Fig 2. In this model, the input signal is partially transmitted and partially
reflected by the beam splitter, representing the loss of photons (or energy)
in the channel. When $\lambda=1$, the pure-loss channel is noiseless (it
equals the identity superoperator). Conversely, when $\lambda=0$, the pure-
loss channel is completely noisy (specifically, it is a constant channel that
maps any state in $\outerproduct{0}{0}$).

Figure 2:  Stinespring representation of the pure-loss channel
$\pazocal{E}_{\lambda}$. The pure-loss channel $\pazocal{E}_{\lambda}$ acts on
the input state $\rho$ by mixing it in a beam splitter of transmissivity
$\lambda$ (represented by the grey box) with an environmental vacuum state
$\ket{0}$. Moreover, $\hat{a}$ and $\hat{e}$ are the annihilation operators of
the input mode and the environmental mode, respectively. The complementary
channel of the pure-loss channel is given by
$\pazocal{E}^{\text{c}}_{\lambda}(\rho)=\pazocal{E}_{1-\lambda}\big{(}(-1)^{\hat{a}^{\dagger}\hat{a}}\rho(-1)^{\hat{a}^{\dagger}\hat{a}}\big{)}$.

######  Definition S9.

Let $\pazocal{H}_{S},\pazocal{H}_{E}\coloneqq L^{2}(\mathbb{R})$. For all
$\lambda\in[0,1]$, the pure-loss channel of transmissivity $\lambda$ is a
quantum channel
$\pazocal{E}_{\lambda}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{S})$
defined as follows:

$\displaystyle\pazocal{E}_{\lambda}(\rho)$
$\displaystyle\coloneqq\Tr_{E}\left[U_{\lambda}^{SE}\big{(}\rho_{S}\otimes\outerproduct{0}{0}_{E}\big{)}(U_{\lambda}^{SE})^{\dagger}\right],\quad\forall\,\rho\in\pazocal{T}(\pazocal{H}_{S})\,,$

where $\outerproduct{0}{0}_{E}$ denotes the vacuum state of $\pazocal{H}_{E}$
and $U_{\lambda}^{SE}$ denotes the beam splitter unitary defined in (S8).

######  Lemma S10.

For all $\lambda\in[0,1]$ and all $n,m\in\mathbb{N}$, it holds that

$\displaystyle\pazocal{E}_{\lambda}(\outerproduct{m}{n})$
$\displaystyle=\sum_{\ell=0}^{\min(n,m)}\sqrt{\binom{m}{\ell}\binom{n}{\ell}}\,(1-\lambda)^{\ell}\lambda^{\frac{n+m}{2}-\ell}\outerproduct{m-\ell}{n-\ell}\,.$

###### Proof.

This is a direct consequence of (S10) and of the definition of pure-loss
channel. ∎

######  Lemma S11.

For all $\lambda\in[0,1]$, a complementary channel
$\pazocal{E}^{\text{c}}_{\lambda}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{E})$
of the pure-loss channel
$\pazocal{E}_{\lambda}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{S})$
is given by

$\displaystyle\pazocal{E}^{\text{c}}_{\lambda}(\rho)\coloneqq\Tr_{S}\left[U_{\lambda}^{SE}\big{(}\rho_{S}\otimes\outerproduct{0}{0}_{E}\big{)}\left(U_{\lambda}^{SE}\right)^{\dagger}\right]=\pazocal{E}_{1-\lambda}\circ\pazocal{R}(\rho),\qquad\forall\,\rho\in\pazocal{T}(\pazocal{H}_{S})\,,$
(S13)

where $\pazocal{R}(\cdot)\coloneqq V\cdot V^{\dagger}$ with
$V\coloneqq(-1)^{\hat{a}^{\dagger}\hat{a}}$.

###### Proof.

By linearity, it suffices to show the identity in (S13) for rank-one Fock
operators of the form $\outerproduct{m}{n}$ for any $n,m\in\mathbb{N}$, i.e.

$\displaystyle\Tr_{S}\left[U_{\lambda}^{SE}\big{(}\outerproduct{m}{n}_{S}\otimes\outerproduct{0}{0}_{E}\big{)}\left(U_{\lambda}^{SE}\right)^{\dagger}\right]=\pazocal{E}_{1-\lambda}\circ\pazocal{R}(\outerproduct{m}{n})\,.$
(S14)

This follows directly from (S10). ∎

######  Proposition S12.

[39, 38] The pure-loss channel $\pazocal{E}_{\lambda}$ is degradable if and
only if $\lambda\in[\frac{1}{2},1]$, and it is anti-degradable if and only if
$\lambda\in[0,\frac{1}{2}]$.

######  Lemma S13.

[21, Lemma A.7] For all $\lambda_{1},\lambda_{2}\in[0,1]$ it holds that
$\pazocal{E}_{\lambda_{1}}\circ\pazocal{E}_{\lambda_{2}}=\pazocal{E}_{\lambda_{1}\lambda_{2}}$.

###  I.6 Bosonic dephasing channel

Another main source of noise in optical platforms is bosonic dephasing, which
serves as a prominent example of a non-Gaussian channel [40, 28].

######  Definition S14.

Let $\pazocal{H}_{S}\coloneqq L^{2}(\mathbb{R})$ and let $\hat{a}$ be the
corresponding annihilation operator. For all $\gamma\geq 0$, the bosonic
dephasing channel
$\pazocal{D}_{\gamma}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{S})$
is a quantum channel defined as follows:

$\displaystyle\pazocal{D}_{\gamma}(\rho)\coloneqq\frac{1}{\sqrt{2\pi\gamma}}\int_{-\infty}^{\infty}\mathrm{d}\phi\,e^{-\frac{\phi^{2}}{2\gamma}}\,e^{i\phi\hat{a}^{\dagger}\hat{a}}\,\rho\,e^{-i\phi\hat{a}^{\dagger}\hat{a}},\quad\forall\,\rho\in\pazocal{T}(\pazocal{H}_{S})\,.$

In other words, $\pazocal{D}_{\gamma}$ is a convex combination of phase space
rotations $\rho\longmapsto
e^{i\phi\hat{a}^{\dagger}\hat{a}}\,\rho\,e^{-i\phi\hat{a}^{\dagger}\hat{a}}$,
where the random variable $\phi$ follows a centered Gaussian distribution with
a variance of $\gamma$.

######  Lemma S15.

For all $\gamma\geq 0$ and all $n,m\in\mathbb{N}$, it holds that

$\displaystyle\pazocal{D}_{\gamma}(\outerproduct{m}{n})=e^{-\frac{1}{2}\gamma(n-m)^{2}}\outerproduct{m}{n}\,.$

###### Proof.

This result follows from the Fourier transform of the Gaussian function:

$\displaystyle\frac{1}{\sqrt{2\pi\gamma}}\int_{-\infty}^{\infty}\mathrm{d}\phi\,e^{-\frac{\phi^{2}}{2\gamma}}e^{i\phi
k}=e^{-\frac{1}{2}\gamma k^{2}},\quad\forall\,k\in\mathbb{R}\,.$ (S15)

∎

When $\gamma=0$, the bosonic dephasing channel is noiseless. In contrast, when
$\gamma\to\infty$ it annihilates all off-diagonal components of the input
density matrix, reducing it to an incoherent probabilistic mixture of Fock
states. We now present a Stinespring dilation of the bosonic dephasing
channel.

######  Lemma S16.

Let $\pazocal{H}_{S}=\pazocal{H}_{E}\coloneqq L^{2}(\mathbb{R})$ and let
$\hat{a}$ and $\hat{e}$ be annihilation operators on $\pazocal{H}_{S}$ and
$\pazocal{H}_{E}$, respectively. For all $\gamma>0$, the bosonic dephasing
channel
$\pazocal{D}_{\gamma}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{S})$
can be expressed in Stinespring representation as

$\displaystyle\pazocal{D}_{\gamma}(\rho)=\Tr_{E}\left[V_{\gamma}^{SE}\big{(}\rho_{S}\otimes\outerproduct{0}{0}_{E}\big{)}(V_{\gamma}^{SE})^{\dagger}\right],\qquad\forall\,\rho\in\pazocal{T}(\pazocal{H}_{S})\,,$
(S16)

where $V_{\gamma}^{SE}$ denotes the conditional displacement unitary defined
by

$\displaystyle
V_{\gamma}^{SE}\coloneqq\exp\left[\sqrt{\gamma}\,\hat{a}^{\dagger}\hat{a}\otimes\,(\hat{e}^{\dagger}-\hat{e})\right]=\sum_{n=0}^{\infty}\outerproduct{n}{n}_{S}\otimes
D(\sqrt{\gamma}n).$ (S17)

###### Proof.

For any $n,m\in\mathbb{N}$ it holds that

$\displaystyle\Tr_{E}\left[V_{\gamma}^{SE}\big{(}\outerproduct{m}{n}_{S}\otimes\outerproduct{0}{0}_{E}\big{)}(V_{\gamma}^{SE})^{\dagger}\right]$
(S18)
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(i)}}}}{{=}}\Tr_{E}\left[\outerproduct{m}{n}_{S}\otimes\outerproduct{\sqrt{\gamma}n}{\sqrt{\gamma}m}_{E}\right]$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ii)}}}}{{=}}e^{-\frac{\gamma}{2}(n-m)^{2}}\outerproduct{m}{n}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iii)}}}}{{=}}\pazocal{D}_{\gamma}(\outerproduct{m}{n})\,.$

Here, in (i) we used that
$V^{SE}_{\gamma}\ket{n}_{S}\otimes\ket{0}_{E}=\ket{n}_{S}\otimes\ket{\sqrt{\gamma}n}_{E}$,
where $\ket{\sqrt{\gamma}n}_{E}$ denotes the coherent state with parameter
$\sqrt{\gamma}n$. In (ii), we exploited the formula for the overlap between
coherent states provided in (S6), and in (iii) we used Lemma S15. The proof is
completed by linearity. ∎

######  Remark S17.

A different approach to dilating the bosonic dephasing channel, as outlined in
the existing literature [40, 65, 66], is as follows:

$\displaystyle\tilde{V}_{\gamma}^{\text{SE}}=\exp\left[-i\sqrt{\gamma}\,\hat{a}^{\dagger}\hat{a}\,(\hat{e}^{\dagger}+\hat{e})\right].$

This unitary transformation is achieved by rotating the environmental mode of
the unitary operator $V_{\gamma}^{\text{SE}}$ in (S17) by $\frac{\pi}{2}$,
that is
$\tilde{V}_{\gamma}^{\text{SE}}=e^{i\frac{\pi}{2}\hat{e}^{\dagger}\hat{e}}V_{\gamma}^{\text{SE}}e^{-i\frac{\pi}{2}\hat{e}^{\dagger}\hat{e}}.$
These dilations yield the same dephasing channel, as all dilations are
equivalent up to unitary transformations.

######  Proposition S18 ([32]).

The bosonic dephasing channel $\pazocal{D}_{\gamma}$ is degradable for all
$\gamma\geq 0$.

######  Proposition S19 ([32]).

The bosonic dephasing channel $\pazocal{D}_{\gamma}$ is never anti-degradable.

######  Lemma S20.

For all $\gamma_{1},\gamma_{2}\geq 0$, the composition of bosonic dephasing
channels is given by
$\pazocal{D}_{\gamma_{1}}\circ\pazocal{D}_{\gamma_{2}}=\pazocal{D}_{\gamma_{1}+\gamma_{2}}$.

###### Proof.

This can be shown by leveraging Lemma S15. ∎

###  I.7 Bosonic loss-dephasing channel

Consider an optical system undergoing simultaneous loss and dephasing over a
finite time interval. At each instant, the system is susceptible to both an
infinitesimal pure-loss channel and an infinitesimal bosonic dephasing
channel. Hence, the overall channel describing the simultaneous effect of loss
and dephasing results in a suitable composition of numerous concatenations
between infinitesimal pure-loss and bosonic dephasing channels. However, given
that:

  * •

The pure-loss channel and the bosonic dephasing channel commute, i.e.
$\pazocal{E}_{\lambda}\circ\pazocal{D}_{\gamma}=\pazocal{D}_{\gamma}\circ\pazocal{E}_{\lambda}$,
as implied by Lemma S10 and Lemma S15;

  * •

The composition of pure-loss channels is a pure-loss channel,
$\pazocal{E}_{\lambda_{1}}\circ\pazocal{E}_{\lambda_{2}}=\pazocal{E}_{\lambda_{1}\lambda_{2}}$
(Lemma S13);

  * •

The composition of bosonic dephasing channels is a bosonic dephasing channel,
$\pazocal{D}_{\gamma_{1}}\circ\pazocal{D}_{\gamma_{2}}=\pazocal{D}_{\gamma_{1}+\gamma_{2}}$
(Lemma S20);

it follows that the combined effect of loss and dephasing can be modelled by
the composition between pure-loss channel and bosonic dephasing channel,

$\displaystyle\pazocal{N}_{\lambda,\gamma}\coloneqq\pazocal{E}_{\lambda}\circ\pazocal{D}_{\gamma}\,,$
(S19)

dubbed the _bosonic loss-dephasing channel_.

######  Definition S21.

For all $\gamma\geq 0$ and $\lambda\in[0,1]$, the bosonic loss-dephasing
channel $\pazocal{N}_{\lambda,\gamma}$ is a quantum channel defined by the
composition between pure-loss channel and bosonic dephasing channel:
$\pazocal{N}_{\lambda,\gamma}\coloneqq\pazocal{D}_{\gamma}\circ\pazocal{E}_{\lambda}\,.$

######  Lemma S22.

For all $\lambda\in[0,1]$ and $\gamma\geq 0$, it holds that
$\pazocal{N}_{\lambda,\gamma}\coloneqq\pazocal{D}_{\gamma}\circ\pazocal{E}_{\lambda}=\pazocal{E}_{\lambda}\circ\pazocal{D}_{\gamma}$.
Moreover, for all $n,m\in\mathbb{N}$ it holds that

$\displaystyle\pazocal{N}_{\lambda,\gamma}(\outerproduct{m}{n})$
$\displaystyle=e^{-\frac{1}{2}\gamma(n-m)^{2}}\pazocal{E}_{\lambda}(\outerproduct{m}{n})$
$\displaystyle=e^{-\frac{1}{2}\gamma(n-m)^{2}}\sum_{\ell=0}^{\min(n,m)}\sqrt{\binom{n}{\ell}\binom{m}{\ell}}\,(1-\lambda)^{\ell}\lambda^{\frac{n+m}{2}-\ell}\outerproduct{m-\ell}{n-\ell}\,.$

###### Proof.

It follows from Lemma S10 and Lemma S15. ∎

######  Lemma S23.

Let $\pazocal{H}_{S},\pazocal{H}_{E_{1}},\pazocal{H}_{E_{2}}\coloneqq
L^{2}(\mathbb{R})$. For all $\lambda\in[0,1]$ and all $\gamma\geq 0$, the
bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{S})$
admits the following Stinespring representation:

$\displaystyle\pazocal{N}_{\lambda,\gamma}(\rho)=\Tr_{E_{1}E_{2}}\left[U_{\lambda}^{SE_{1}}V_{\gamma}^{SE_{2}}\big{(}\rho_{S}\otimes\outerproduct{0}{0}_{E_{1}}\otimes\outerproduct{0}{0}_{E_{2}}\big{)}\left(U_{\lambda}^{SE_{1}}V_{\gamma}^{SE_{2}}\right)^{\dagger}\right],\qquad\forall\,\rho\in\pazocal{T}(\pazocal{H}_{S}),$

The associated complementary channel
$\pazocal{N}_{\lambda,\gamma}^{\text{c}}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{E_{1}}\otimes\pazocal{H}_{E_{2}})$
is defined as follows:

$\displaystyle\pazocal{N}_{\lambda,\gamma}^{\text{c}}(\rho)\coloneqq\Tr_{S}\\!\left[U_{\lambda}^{SE_{1}}V_{\gamma}^{SE_{2}}\big{(}\rho_{S}\otimes\outerproduct{0}{0}_{E_{1}}\otimes\outerproduct{0}{0}_{E_{2}}\big{)}\left(U_{\lambda}^{SE_{1}}V_{\gamma}^{SE_{2}}\right)^{\dagger}\right],\qquad\forall\,\rho\in\pazocal{T}(\pazocal{H}_{S})\,.$

In particular,

$\displaystyle\pazocal{N}^{\text{c}}_{\lambda,\gamma}(\outerproduct{m}{n})=(-1)^{m-n}\pazocal{E}_{1-\lambda}\left(\outerproduct{m}{n}_{E_{1}}\right)\otimes\outerproduct{\sqrt{\gamma}m}{\sqrt{\gamma}n}_{E_{2}},\qquad\forall\,m,n\in\mathbb{N}\,,$
(S20)

where $\ket{\sqrt{\gamma}n}_{E_{2}}$ denotes the coherent state with parameter
$\sqrt{\gamma}n$.

###### Proof.

The Eq. (S20) is derived by first applying (S13) and subsequently utilising
the dilation of the bosonic dephasing channel. Finally, the non-environment
mode of the bosonic dephasing channel is traced out. ∎

######  Lemma S24.

Consider the Hilbert spaces $\pazocal{H}_{S}$, $\pazocal{H}_{E_{1}}$, and
$\pazocal{H}_{E_{2}}$, all isomorphic to $L^{2}(\mathbb{R})$. Let
$\lambda\in[0,1]$ and $\gamma\geq 0$. The bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}:\pazocal{T}(\pazocal{H}_{S})\rightarrow\pazocal{T}(\pazocal{H}_{S})$
exhibits anti-degradability if and only if there exists a quantum channel
$\pazocal{A}_{\lambda,\gamma}:\pazocal{T}(\pazocal{H}_{E_{\text{out}}})\rightarrow\pazocal{T}(\pazocal{H}_{S})$
satisfying the following condition:

$\displaystyle\pazocal{A}_{\lambda,\gamma}\circ\pazocal{N}^{\text{c}}_{\lambda,\gamma}(\outerproduct{m}{n})=\pazocal{N}_{\lambda,\gamma}(\outerproduct{m}{n}),\quad\forall\,m,n\in\mathbb{N}\,,$
(S21)

where
$\pazocal{N}^{\text{c}}_{\lambda,\gamma}:\pazocal{T}(\pazocal{H}_{S})\to\pazocal{T}(\pazocal{H}_{E_{\text{out}}})$
denotes the complementary channel reported in (S20), and
$\pazocal{H}_{E_{\text{out}}}\subset\pazocal{H}_{E_{1}}\otimes\pazocal{H}_{E_{2}}$
is defined in (S42). Moreover, the condition in (S21) is equivalent to

$\displaystyle\pazocal{A}_{\lambda,\gamma}\left(\pazocal{E}_{1-\lambda}\left(\outerproduct{m}{n}\right)\otimes\outerproduct{\sqrt{\gamma}m}{\sqrt{\gamma}n}\right)=(-1)^{m-n}e^{-\frac{\gamma}{2}(m-n)^{2}}\pazocal{E}_{\lambda}(\outerproduct{m}{n}),\quad\forall\,m,n\in\mathbb{N}\,,$
(S22)

where $\ket{\sqrt{\gamma}n}$ denotes the coherent state with parameter
$\sqrt{\gamma}n$.

###### Proof.

$\pazocal{N}_{\lambda,\gamma}$ is anti-degradable if and only if there exists
a quantum channel
$\pazocal{A}_{\lambda,\gamma}:\pazocal{T}(\pazocal{H}_{E_{\text{out}}})\to\pazocal{T}(\pazocal{H}_{S})$
such that

$\displaystyle\pazocal{A}_{\lambda,\gamma}\circ\pazocal{N}^{\text{c}}_{\lambda,\gamma}(\rho)=\pazocal{N}_{\lambda,\gamma}(\rho),\qquad\forall\,\rho\in\pazocal{T}(\pazocal{H}_{S})\,.$
(S23)

By linearity, it suffices to show the condition in (S21), i.e. which
corresponds to the condition in (S23) restricted to rank-one Fock operators of
the form $\rho=\outerproduct{m}{n}$ with $m,n\in\mathbb{N}$. Moreover, by
exploiting Lemma S22 and S20, the condition in (S21) is equivalent to (S22). ∎

######  Lemma S25.

For all $\gamma_{1},\gamma_{2}\geq 0$ and all $\lambda_{1},\lambda_{2}$ it
holds that
$\pazocal{N}_{\lambda_{1},\gamma_{1}}\circ\pazocal{N}_{\lambda_{2},\gamma_{2}}=\pazocal{N}_{\lambda_{1}\lambda_{2}\,,\,\gamma_{1}+\gamma_{2}}$.

###### Proof.

This can be shown by exploiting Lemma S13 and Lemma S20. ∎

###  I.8 Preliminaries on capacities of quantum channels

Quantum channels can be suitably exploited in order to transfer information
from their input port to a possibly distant output port, a crucial task in
quantum information theory [60, 45]. In particular, quantum Shannon theory
[33, 34] primarily helps us understand the fundamental limits of quantum
communication using a quantum channel $\pazocal{N}$. These limits are called
_capacities_ and tell us the ultimate amount of information we can send
through the channel when we use it many times [33, 34]. Different capacities
are defined based on the type of information that is being sent down the
channel. For example, classical and quantum capacities of a quantum channel
correspond to its ultimate capability of transmission of classical and quantum
information, respectively. A channel can also be used to generate secret bits
and the relevant capacity in this context is the so-called secret-key
capacity. Each of the above-mentioned capacities might be endowed with other
resources such as initial shared entanglement between the sender and the
receiver or (possibly interactive) classical communication over a noiseless
but public channel. This latter scenario gives rise to the notion of two-way
capacities.

Specifically, the _quantum capacity_ $Q(\pazocal{N})$ of a quantum channel
$\pazocal{N}$ is the maximum rate at which qubits can be reliably transmitted
through $\pazocal{N}$ [34]. We can further assume that both the sender, Alice,
and the receiver, Bob, have free access to a public, noiseless two-way
classical channel. In this two-way communication setting, the relevant notion
of capacities are the _two-way quantum capacity_ $Q_{2}(\pazocal{N})$ and the
_secret-key capacity_ $K(\pazocal{N})$ [33, 34], defined as the maximum
achievable rate of qubits and secret-key bits, respectively, that can be
reliably transmitted across $\pazocal{N}$ with the aid of two-way classical
communication. Since an ebit (i.e. a maximally entangled state of Schmidt rank
$2$) can always be used to generate one bit of secret key [48], a trivial
bound relates these capacities: $Q_{2}(\pazocal{N})\leq K(\pazocal{N})$.

In practical scenarios, it is important to consider that the input state
prepared by Alice can not have unlimited energy and it adheres to specific
energy constraints. In bosonic systems, it is common to limit the average
photon number of any input state $\rho$ as
$\Tr[\hat{a}^{\dagger}\hat{a}\rho]\leq N_{s}$, where $N_{s}>0$ is a given
energy constraint. For any $N_{s}>0$, the energy-constrained (EC) two-way
capacities for transmitting qubits and secret-key bits, denoted as
$Q_{2}(\pazocal{N},N_{s})$ and $K(\pazocal{N},N_{s})$ respectively, are
defined similarly to the unconstrained capacities with the difference that now
the optimisation is performed over those strategies that exploit input states
that adhere to the specified energy constraint. As in the unconstrained
scenario, the relation between EC two-way capacities continue to hold, i.e.
$Q_{2}(\pazocal{N},N_{s})\leq K(\pazocal{N},N_{s})$. Moreover, the
unconstrained capacities are upper bounds for the corresponding energy
constrained capacities and they become equal in the limit $N_{s}\to\infty$.

Two-way capacities of a quantum channel are closely related to another
important information-processing task, namely, entanglement distillation over
a quantum channel. Suppose Alice generates $n$ copies of a state
$\rho_{A^{\prime}A}$ and sends the $A^{\prime}$ subsystems to Bob using the
channel $\pazocal{N}$ for $n$ times. Now, Alice and Bob share $n$ copies of
the state
$\rho^{\prime}_{AB}\coloneqq\operatorname{id}_{A}\otimes\pazocal{N}_{A^{\prime}\to
B}(\rho_{AA^{\prime}})$. The task of an entanglement distillation protocol
concerns identifying the largest number $m$ of ebits that can be extracted
using $n$ copies of $\rho^{\prime}_{AB}$ via LOCC (Local Operations and
Classical Communication) operations. The rate of an entanglement distillation
protocol is defined by the ratio $m/n$. The distillable entanglement
$E_{d}(\rho_{AB}^{\prime})\,$ of $\rho_{AB}^{\prime}$ is defined as the
maximum rate over all the possible entanglement distillation protocols [67]
[34, Chapter 8]. Note that without extra classical communication, entanglement
distillation is not possible [68]. The following lemma establishes a link
between the two-way quantum capacity, secret-key capacity, and distillable
entanglement.

######  Lemma S26.

Let $\pazocal{H}_{A},\pazocal{H}_{A^{\prime}},\pazocal{H}_{B}\coloneqq
L^{2}(\mathbb{R})$. Let
$\pazocal{N}:\pazocal{H}_{A^{\prime}}\rightarrow\pazocal{H}_{B}$ be a quantum
channel. Let $N_{s}>0$ be the energy constaint, and let
$\rho_{A^{\prime}A}\in\pazocal{P}(\pazocal{H}_{A}\otimes\pazocal{H}_{A^{\prime}})$
be a two-mode state satisfying the energy constraint
$\Tr[(\hat{a}^{\dagger}\hat{a}\otimes\operatorname{id}_{A})\rho_{A^{\prime}A}]\leq
N_{s}$, where $\hat{a}$ denotes the annihilation operator on
$\pazocal{H}_{A^{\prime}}$. Then, it holds that

$\displaystyle K(\pazocal{N},N_{s})\geq Q_{2}(\pazocal{N},N_{s})\geq
E_{d}\\!\left(\operatorname{id}_{A}\otimes\pazocal{N}(\rho_{AA^{\prime}})\right)\,,$
(S24)

where $K(\pazocal{N},N_{s})$ denotes the energy-constrained secret-key
capacity of $\pazocal{N}$, $Q_{2}(\pazocal{N},N_{s})$ denotes the energy-
constrained two-way quantum capacity of $\pazocal{N}$, and
$E_{d}\\!\left(\operatorname{id}_{A}\otimes\pazocal{N}(\rho_{AA^{\prime}})\right)$
denotes the distillable entanglement of the state
$\operatorname{id}_{A}\otimes\pazocal{N}(\rho_{AA^{\prime}})$.

The proof idea of the above lemma is the following. Suppose that Alice
produces $n$ copies of a state $\rho_{AA^{\prime}}$ such that the energy
constraint is satisfied. Then, she can use the channel $n$ times to send all
subsystems $A^{\prime}$ to Bob. Then, Alice and Bob share $n$ copies of
$\operatorname{id}_{A}\otimes\pazocal{N}(\rho_{AA^{\prime}})$, which can now
be used to generate $\approx
n\,E_{d}\\!\left(\operatorname{id}_{A}\otimes\pazocal{N}(\rho_{AA^{\prime}})\right)$
ebits by means of a suitable entanglement distillation protocol. The ebit rate
of this protocol is thus
$E_{d}\\!\left(\operatorname{id}_{A}\otimes\pazocal{N}(\rho_{AA^{\prime}})\right)$,
which provides a lower bound on $Q_{2}(\pazocal{N},N_{s})$ thanks to quantum
teleportation [53]. In addition, it holds that $K(\pazocal{N},N_{s})\geq
Q_{2}(\pazocal{N},N_{s})$ because an ebit can generate a secret-key bit [48].
Consequently, (S24) holds.

##  II Anti-degradability and Degradability of bosonic loss-dephasing channel

This section is split into two parts based on the observation that if the
input state to the bosonic loss-dephasing channel is chosen from a finite-
dimensional subspace, the bosonic loss-dephasing channel effectively becomes a
finite-dimensional channel, a fact we show in Lemma S33. This property allows
us to apply established insights about the finite-dimensional channels to the
bosonic loss-dephasing channel. In subsection II.1, we present our study of
the bosonic loss-dephasing channel when the input resides in the entire
infinite-dimensional space, while subsection II.2 is dedicated to the findings
resulting from analysis of finite-dimensional restrictions of the bosonic
loss-dephasing channel.

###  II.1 Sufficient condition on anti-degradability

It is known that the bosonic dephasing channel $\pazocal{D}_{\gamma}$ is
degradable across all dephasing parameter range $\gamma\geq 0$ and also it is
never anti-degradable [40]. The pure-loss channel $\pazocal{E}_{\lambda}$
displays the peculiar characteristic of being anti-degradable for
transmissivity values within the range $\lambda\in[0,\frac{1}{2}]$ and
degradable for $\lambda\in[\frac{1}{2},1]$ [39, 38]. It turns out that when an
anti-degradable channel is concatenated with another channel, the resulting
channel inherits the property of being anti-degradable (see Lemma S50). This
implies the following: if $\lambda\in[0,\frac{1}{2}]$ and $\gamma\geq 0$, then
the bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$ is anti-
degradable [32]. The authors of [32] left as an open question to understand
whether or not the bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is anti-degradable in the region
$\lambda\in(\frac{1}{2},1]$. In particular, they conjecture that
$\pazocal{N}_{\lambda,\gamma}$ is not anti-degradable for all transmissivity
values $\lambda\in(\frac{1}{2},1]$ and for all $\gamma\geq 0$. In the
following theorem, we refute this conjecture by explicitly finding values of
$\lambda\in(\frac{1}{2},1]$ and $\gamma\geq 0$ where the channel is anti-
degradable. Our approach also yields an explicit expression for an anti-
degrading map of the bosonic loss-dephasing channel.

######  Theorem S27.

Each of the following is a sufficient condition for the bosonic loss-dephasing
channel $\pazocal{N}_{\lambda,\gamma}$ to exhibit anti-degradability:

  1. (i)

$\lambda\in[0,\frac{1}{2}]$ and $\gamma\geq 0$.

  2. (ii)

$\lambda\in(\frac{1}{2},1)$ and
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$,
where $\theta$ is defined as
$\theta(x,y)\coloneqq\sum_{n=0}^{\infty}x^{n^{2}}y^{n},\,\forall\,x,y\in[0,1).$

In particular, $\pazocal{N}_{\lambda,\gamma}$ is anti-degradable if
$\lambda\leq\max\\!\left(\frac{1}{2},\frac{1}{1+9e^{-\gamma}}\right)$.

###### Proof.

The proof of the sufficient condition (i) follows directly from the
observation that the composition of a pure-loss channel with transmissivity
$\lambda\in[0,\frac{1}{2}]$ with any other channel inherits the anti-
degradability from the pure-loss channel (see Lemma S50).

The proof of the sufficient condition (ii) is more involved, and it is the
main technical contribution of our work. We rely on the equivalence between
anti-degradability of a quantum channel and the two-extendibility of its Choi
state [42, 43]. To provide a comprehensive and intuitive understanding of this
idea and to aid in the construction of an anti-degrading map of the bosonic
loss-dephasing channel, we present this equivalence in Lemma S49 within the
Appendix.

Assume $\lambda\in(\frac{1}{2},1)$ and
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$.
Let
$\pazocal{H}_{A},\pazocal{H}_{B},\pazocal{H}_{B_{1}},\pazocal{H}_{B_{2}}=L^{2}(\mathbb{R})$
and suppose that the bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is a quantum channel from the system
$A^{\prime}$ to $B$. We want to show that the generalised Choi state of
$\pazocal{N}_{\lambda,\gamma}$ is two-extendible on $B$. In other words, we
want to show that there exists a tripartite state $\rho_{AB_{1}B_{2}}$ such
that

$\displaystyle\Tr_{B_{2}}\left[\rho_{AB_{1}B_{2}}\right]$
$\displaystyle=\tau_{AB_{1}}\,,$ (S25)
$\displaystyle\Tr_{B_{1}}\left[\rho_{AB_{1}B_{2}}\right]$
$\displaystyle=\tau_{AB_{2}}\,,$

where
$\tau_{AB}\coloneqq\operatorname{id}_{A}\otimes\pazocal{N}_{\lambda,\gamma}(\outerproduct{\psi(r)}{\psi(r)}_{AA^{\prime}})\,$
is the generalised Choi state of $\pazocal{N}_{\lambda,\gamma}$, and
$\ket{\psi(r)}_{AA^{\prime}}$ is the two-mode squeezed vacuum state with
squeezing parameter $r>0$ defined in (S7). By Lemma (S22), the generalised
Choi state of the bosonic loss-dephasing channel can be expressed as follows:

$\displaystyle\tau_{AB}=\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell=0}^{\min(m,n)}(\tanh(r))^{m+n}e^{-\frac{\gamma}{2}(m-n)^{2}}\sqrt{\pazocal{B}_{\ell}(m,1-\lambda)\,\pazocal{B}_{\ell}(n,1-\lambda)}\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell}{n-\ell}_{B}\,,$
(S26)

where
$\pazocal{B}_{\ell}\\!\left(n,\lambda\right)\coloneqq\binom{n}{\ell}\lambda^{\ell}(1-\lambda)^{n-\ell}\,.$
We observe that $\tau_{AB}$ is a linear combination of
$\outerproduct{m}{n}\otimes\outerproduct{j_{1}}{j_{2}}$, where
$m,n,j_{1},j_{2}\in\mathbb{N}$, $j_{1}\leq n_{1}$, $j_{2}\leq n$, and
$m-n=j_{1}-j_{2}$. This insight leads to the following educated guess about
the structure of a potential two-extension:

$\displaystyle\tilde{\rho}_{AB_{1}B_{2}}=\sum_{m,n=0}^{\infty}\sum_{\ell_{1}=0}^{m}\sum_{\ell_{2}=0}^{n}\sum_{k=0}^{\min(m-\ell_{1},n-\ell_{2})}c(m,n,\ell_{1},\ell_{2},k)\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell_{1}}{n-\ell_{2}}_{B_{1}}\otimes\outerproduct{\ell_{1}+k}{\ell_{2}+k}_{B_{2}}\,,$
(S27)

where $\\{c(m,n,\ell_{1},\ell_{2},k)\\}_{m,n,\ell_{1},\ell_{2},k}$ are some
suitable coefficients. The soundness of this guess is confirmed by the fact
that both $\Tr_{B_{2}}\left[\tilde{\rho}_{AB_{1}B_{2}}\right]$ and
$\Tr_{B_{1}}\left[\tilde{\rho}_{AB_{1}B_{2}}\right]$ are linear combinations
of $\outerproduct{m}{n}\otimes\outerproduct{j_{1}}{j_{2}}$ with
$m,n,j_{1},j_{2}\in\mathbb{N}$, $j_{1}\leq m$, $j_{2}\leq n$, and
$m-n=j_{1}-j_{2}$, similar to the generalised Choi state $\tau_{AB}$ in (S26).

At this point, one could try to define the coefficients
$\\{c(m,n,\ell_{1},\ell_{2},k)\\}_{m,n,\ell_{1},\ell_{2},k}$ so as to satisfy
the required conditions of the extendibility. However, the resulting
tripartite operator may not qualify as a quantum state. In order to ensure
that we obtain a quantum state, our approach consists in producing the
operator $\tilde{\rho}_{AB_{1}B_{2}}$ via a physical process that consists in
applying a sequence of quantum channels to a quantum state.

We begin by constructing a quantum state that has the same operator structure
as the operator $\tilde{\rho}_{AB_{1}B_{2}}$ in (S27). This means that at this
initial stage, we only aim to build a tripartite state consisting of a linear
combination of operators
$\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell_{1}}{n-\ell_{2}}_{B_{1}}\otimes\outerproduct{\ell_{1}+k}{\ell_{2}+k}_{B_{2}}$
with the summation limits identical to those in (S27). This ensures that the
coefficients equal to zero coincide in the two operators.

We now illustrate on each step of this construction. For a fixed
$n\in\mathbb{N}$, consider the state $\ket{n}_{A}\otimes\ket{n}_{B_{1}}$. We
introduce two auxiliary single-mode systems $B_{2}$ and $C$ initially in
vacuum states:
$\ket{n}_{A}\otimes\ket{n}_{B_{1}}\otimes\ket{0}_{B_{2}}\otimes\ket{0}_{C}$.
We next send the systems $B_{2}$ and $B_{1}$ through the ports of a beam
splitter, resulting in a superposition of
$\ket{n}_{A}\otimes\ket{n-\ell}_{B_{1}}\otimes\ket{\ell}_{B_{2}}\otimes\ket{0}_{C}$
for $\ell=0,1,\ldots,n$, as implied by Lemma (S7). We repeat this for systems
$B_{1}$ and $C$, thus obtaining a superposition of
$\ket{n}_{A}\otimes\ket{n-\ell-k}_{B_{1}}\otimes\ket{\ell}_{B_{2}}\otimes\ket{k}_{C}$,
with $k=0,1,\ldots,n-\ell$ and $\ell=0,1,\ldots,n$. Consider now the isometry
$W^{CB_{1}B_{2}}$ defined by

$\displaystyle
W^{CB_{1}B_{2}}\ket{n}_{B_{1}}\otimes\ket{m}_{B_{2}}\otimes\ket{k}_{C}=\ket{n+k}_{B_{1}}\otimes\ket{m+k}_{B_{2}}\otimes\ket{k}_{C},\quad\forall\,n,m,k\in\mathbb{N}\,,$
(S28)

dubbed _controlled-add-add isometry_ (mode $C$ is the control mode). By
applying this isometry to the superposition we created by using beam
splitters, we obtain a superposition of
$\ket{n}_{A}\otimes\ket{n-\ell}_{B_{1}}\otimes\ket{\ell+k}_{B_{2}}\otimes\ket{k}_{C}$
with $k=0,1,\ldots,n-\ell$ and $\ell=0,1,\ldots,n$. Finally, by tracing out
system $C$, we obtain the same operator structure of the operator
$\tilde{\rho}_{AB_{1}B_{2}}$ in (S27). Having focused on the operator
structure, we have not considered the transmissivities of the two beam
splitters so far. We will see that these transmissivities can be chosen
carefully such that the diagonal elements of the operator at hand becomes
equal to those of the Choi state.

We now apply the outlined construction to the two-mode squeezed vacuum state
with two vacuum states appended to it, i.e.
$\ket{\psi(r)}_{AB_{1}}\otimes\ket{0}_{B_{2}}\otimes\ket{0}_{C}$. For reasons
that will become clear in a moment, we choose the two beam splitter
transmissivities to be $\lambda$ (for the beam splitter acting on
$B_{2}B_{1}$) and $\frac{1-\lambda}{\lambda}$ (for the one acting on
$CB_{1}$). By doing so we obtain the state

$\displaystyle\ket{\phi}_{AB_{1}B_{2}C}$ $\displaystyle\coloneqq
W^{C,B_{1}B_{2}}U_{\frac{1-\lambda}{\lambda}}^{CB_{1}}U_{\lambda}^{B_{2}B_{1}}\ket{\psi(r)}_{AB_{1}}\otimes\ket{0}_{B_{2}}\otimes\ket{0}_{C}$
(S29)
$\displaystyle=\frac{1}{\cosh(r)}\sum_{n=0}^{\infty}\sum_{\ell=0}^{n}\sum_{k=0}^{n-\ell}\tanh^{n}(r)\sqrt{\pazocal{B}_{\ell}(n,1-\lambda)\,\pazocal{B}_{k}\\!\left(n-\ell,\frac{2\lambda-1}{\lambda}\right)}\ket{n}_{A}\otimes\ket{n-\ell}_{B_{1}}\otimes\ket{\ell+k}_{B_{2}}\otimes\ket{k}_{C}\,,$

where we used (S7) and Lemma S7. Note that the transmissivities of the beam
splitters $U_{\frac{1-\lambda}{\lambda}}^{CB_{1}}$ and
$U_{\lambda}^{B_{2}B_{1}}$ are chosen such that the diagonal elements of
$\Tr_{B_{2}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}$ and
$\Tr_{B_{1}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}$ both
coincide with those of the Choi state $\tau_{AB}$ in (S26). To verify this,
let us calculate the state
$\Tr_{C}\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}$:

$\displaystyle\Tr_{C}\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}=\frac{1}{\cosh^{2}(r)}$
$\displaystyle\sum_{m,n=0}^{\infty}\sum_{\ell_{1}=0}^{m}\sum_{\ell_{2}=0}^{n}\sum_{k=0}^{\min(m-\ell_{1},n-\ell_{2})}(\tanh(r))^{m+n}\sqrt{\pazocal{B}_{\ell_{1}}(m,1-\lambda)\,\pazocal{B}_{\ell_{2}}(n,1-\lambda)\,}$
$\displaystyle\left[\sqrt{\pazocal{B}_{k}\\!\left(m-\ell_{1},\frac{2\lambda-1}{\lambda}\right)\,\pazocal{B}_{k}\\!\left(n-\ell_{2},\frac{2\lambda-1}{\lambda}\right)}\right]\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell_{1}}{n-\ell_{2}}_{B_{1}}\otimes\outerproduct{\ell_{1}+k}{\ell_{2}+k}_{B_{2}}.$

Notably, the structure of the state
$\Tr_{C}\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}$ mirrors that
of (S27) with specific coefficients $c(m,n,\ell_{1},\ell_{2},k)$. Moreover, it
holds that

$\displaystyle\Tr_{B_{2}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}=\frac{1}{\cosh^{2}(r)}$
$\displaystyle\sum_{m,n=0}^{\infty}\sum_{\ell=0}^{\min(m,n)}(\tanh(r))^{m+n}\sqrt{\pazocal{B}_{\ell}(m,1-\lambda)\,\pazocal{B}_{\ell}(n,1-\lambda)}$
(S30)
$\displaystyle\qquad\left[\sum_{k=0}^{\min(m-\ell,n-\ell)}\sqrt{\,\pazocal{B}_{k}\\!\left(m-\ell,\frac{2\lambda-1}{\lambda}\right)\,\pazocal{B}_{k}\\!\left(n-\ell,\frac{2\lambda-1}{\lambda}\right)}\right]\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell}{n-\ell}_{B_{1}}$

and that

$\displaystyle\Tr_{B_{1}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}=\Tr_{B_{2}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}\,.$
(S31)

In order to prove (S31), observe that

$\displaystyle\Tr_{B_{1}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}$
$\displaystyle=\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell_{1}=\max(m-n,0)}^{m}\sum_{k=0}^{m-\ell_{1}}(\tanh(r))^{m+n}\sqrt{\pazocal{B}_{\ell_{1}}(m,1-\lambda)\,\pazocal{B}_{n-m+\ell_{1}}(n,1-\lambda)}$
(S32)
$\displaystyle\qquad\qquad\pazocal{B}_{k}\\!\left(m-\ell_{1},\frac{2\lambda-1}{\lambda}\right)\outerproduct{m}{n}_{A}\otimes\outerproduct{\ell_{1}+k}{n-m+\ell_{1}+k}_{B_{2}}$
$\displaystyle=\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell_{1}=\max(m-n,0)}^{m}\sum_{\ell=0}^{m-\ell_{1}}(\tanh(r))^{m+n}\sqrt{\pazocal{B}_{\ell_{1}}(m,1-\lambda)\,\pazocal{B}_{n-m+\ell_{1}}(n,1-\lambda)}$
$\displaystyle\qquad\qquad\pazocal{B}_{m-\ell_{1}-\ell}\\!\left(m-\ell_{1},\frac{2\lambda-1}{\lambda}\right)\outerproduct{m}{n}_{A}\otimes\outerproduct{\ell_{1}+k}{n-m+\ell_{1}+k}_{B_{2}}$
$\displaystyle=\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell=0}^{\min(m,n)}\sum_{\ell_{1}=\max(m-n,0)}^{m-\ell}(\tanh(r))^{m+n}\sqrt{\pazocal{B}_{\ell_{1}}(m,1-\lambda)\,\pazocal{B}_{n-m+\ell_{1}}(n,1-\lambda)}$
$\displaystyle\qquad\qquad\pazocal{B}_{m-\ell_{1}-\ell}\\!\left(m-\ell_{1},\frac{2\lambda-1}{\lambda}\right)\outerproduct{m}{n}_{A}\otimes\outerproduct{\ell_{1}+k}{n-m+\ell_{1}+k}_{B_{2}}$
$\displaystyle=\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell=0}^{\min(m,n)}(\tanh(r))^{m+n}\sum_{k=0}^{\min(n-\ell,m-\ell)}\sqrt{\pazocal{B}_{m-\ell-k}(m,1-\lambda)\,\pazocal{B}_{n-\ell-k}(n,1-\lambda)}$
$\displaystyle\qquad\qquad\pazocal{B}_{k}\\!\left(k+\ell,\frac{2\lambda-1}{\lambda}\right)\outerproduct{m}{n}_{A}\otimes\outerproduct{\ell_{1}+k}{n-m+\ell_{1}+k}_{B_{2}}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(i)}}}}{{=}}\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell=0}^{\min(m,n)}(\tanh(r))^{m+n}\sqrt{\pazocal{B}_{\ell}(m,1-\lambda)\,\pazocal{B}_{\ell}(n,1-\lambda)}$
$\displaystyle\qquad\left[\sum_{k=0}^{\min(m-\ell,n-\ell)}\sqrt{\,\pazocal{B}_{k}\\!\left(m-\ell,\frac{2\lambda-1}{\lambda}\right)\,\pazocal{B}_{k}\\!\left(n-\ell,\frac{2\lambda-1}{\lambda}\right)}\right]\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell}{n-\ell}_{B_{2}}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ii)}}}}{{=}}\Tr_{B_{2}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}$

Here, in (i), we used the identity

$\displaystyle\sqrt{\pazocal{B}_{m-\ell-k}(m,1-\lambda)\,\pazocal{B}_{n-\ell-k}(n,1-\lambda)}\pazocal{B}_{k}\\!\left(k+\ell,\frac{2\lambda-1}{\lambda}\right)$
(S33)
$\displaystyle\qquad=\sqrt{\pazocal{B}_{\ell}(m,1-\lambda)\,\pazocal{B}_{\ell}(n,1-\lambda)}\sqrt{\,\pazocal{B}_{k}\\!\left(m-\ell,\frac{2\lambda-1}{\lambda}\right)\,\pazocal{B}_{k}\\!\left(n-\ell,\frac{2\lambda-1}{\lambda}\right)}\,,$

which can be easily proved by substituting the definition
$\pazocal{B}_{\ell}\\!\left(n,\lambda\right)\coloneqq\binom{n}{\ell}\lambda^{\ell}(1-\lambda)^{n-\ell}$
and by leveraging the binomial identity

$\displaystyle\binom{n}{l+k}\binom{l+k}{k}=\binom{n}{l}\binom{n-l}{k}\,.$
(S34)

Moreover, in (ii), we exploited (S30).

Note that the off-diagonal terms of the state in (S30) are not equal to those
of the Choi state $\tau_{AB}$ in (S26). Specifically, the points of difference
with the Choi state $\tau_{AB}$ are the presence of the term inside the square
brackets and the absence of the dephasing exponent. To address these
additional terms, let us use the toolbox of _Hadamard maps_. Let $H$ be the
Hadamard map, introduced in Sec. I.3, associated with the infinite matrix
$A\coloneqq(a_{mn})_{m,n\in\mathbb{N}}$ defined as follows:

$\displaystyle
a_{mn}\coloneqq\frac{e^{-\frac{\gamma}{2}(n-m)^{2}}}{\sum_{j=0}^{\min(n,m)}\sqrt{\,\pazocal{B}_{j}\\!\left(n,\frac{2\lambda-1}{\lambda}\right)\,\pazocal{B}_{j}\\!\left(m,\frac{2\lambda-1}{\lambda}\right)}},\quad\forall\,n,m\in\mathbb{N}\,.$
(S35)

By construction, we have that

$\displaystyle\operatorname{id}_{A}\otimes
H_{B_{1}}\left(\Tr_{B_{2}C}\\!\big{[}\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\big{]}\right)$
(S36)
$\displaystyle\qquad=\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell=0}^{\min(m,n)}(\tanh(r))^{m+n}\sqrt{\pazocal{B}_{\ell}(m,1-\lambda)\,\pazocal{B}_{\ell}(n,1-\lambda)}$
$\displaystyle\qquad\qquad\left[\sum_{k=0}^{\min(m-\ell,n-\ell)}\sqrt{\,\pazocal{B}_{k}\\!\left(m-\ell,\frac{2\lambda-1}{\lambda}\right)\,\pazocal{B}_{k}\\!\left(n-\ell,\frac{2\lambda-1}{\lambda}\right)}\right]a_{m-\ell,\,n-\ell}\,\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell}{n-\ell}_{B_{1}}$
$\displaystyle\qquad=\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell=0}^{\min(m,n)}(\tanh(r))^{m+n}e^{-\frac{\gamma}{2}(n-m)^{2}}\,\sqrt{\pazocal{B}_{\ell}(m,1-\lambda)\,\pazocal{B}_{\ell}(n,1-\lambda)}\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell}{n-\ell}_{B_{1}}$
$\displaystyle\qquad=\tau_{AB_{1}}\,.$

This means that the operator

$\displaystyle\rho_{AB_{1}B_{2}}\coloneqq\operatorname{id}_{A}\otimes
H_{B_{1}}\otimes
H_{B_{2}}\left(\Tr_{C}\left[\outerproduct{\phi}{\phi}_{AB_{1}B_{2}C}\right]\right)$
(S37)

satisfies the extendibility conditions in (S25). All that remains to prove is
that $\rho_{AB_{1}B_{2}}$ is in fact a quantum state. We will do this by
showing that the superoperator $H$ is a quantum channel. In Sec. I.3, we
establish that a Hadamard map is a quantum channel if its defining infinite
matrix is Hermitian, has diagonal elements equal to one, and is diagonally
dominant. The first two properties are trivially satisfied by the infinite
matrix $A$ defined in (S35). We only need to demonstrate that for the
parameter region $\lambda>\frac{1}{2}$ and
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$,
the infinite matrix $A$ is diagonally dominant. We recall that, by definition,
$A$ is diagonally dominant if it holds that $\sum_{\begin{subarray}{c}m=0\\\
m\neq n\end{subarray}}^{\infty}|a_{mn}|\leq 1,\,\forall\,n\in\mathbb{N}$. Note
that for any $n,m\in\mathbb{N}$ we have that

$\displaystyle|a_{nm}|$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(n-m)^{2}}}{\sum_{j=0}^{\min(n,m)}\sqrt{\binom{n}{j}\binom{m}{j}\left(\frac{1-\lambda}{\lambda}\right)^{n+m-2j}\left(\frac{2\lambda-1}{\lambda}\right)^{2j}}}$
(S38)
$\displaystyle\leq\frac{e^{-\frac{\gamma}{2}(n-m)^{2}}}{\sum_{j=0}^{\min(n,m)}\sqrt{\binom{\min(n,m)}{j}^{2}\left(\frac{1-\lambda}{\lambda}\right)^{n+m-2j}\left(\frac{2\lambda-1}{\lambda}\right)^{2j}}}$
$\displaystyle=e^{-\frac{\gamma}{2}(n-m)^{2}}\left(\frac{\lambda}{1-\lambda}\right)^{\frac{|n-m|}{2}}\,.$

Consequently, if $\lambda$ and $\gamma$ are such that $\lambda>\frac{1}{2}$
and
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$,
for any $n\in\mathbb{N}$ we have that

$\displaystyle\sum_{\begin{subarray}{c}m=0\\\ m\neq
n\end{subarray}}^{\infty}|a_{mn}|$
$\displaystyle\leq\sum_{\begin{subarray}{c}m=0\\\ m\neq
n\end{subarray}}^{\infty}e^{-\frac{\gamma}{2}(m-n)^{2}}\left(\frac{\lambda}{1-\lambda}\right)^{\frac{|m-n|}{2}}$
(S39)
$\displaystyle=\sum_{k=1}^{\infty}e^{-\frac{\gamma}{2}k^{2}}\left(\frac{\lambda}{1-\lambda}\right)^{\frac{k}{2}}+\sum_{k=1}^{n}e^{-\frac{\gamma}{2}k^{2}}\left(\frac{\lambda}{1-\lambda}\right)^{\frac{k}{2}}$
$\displaystyle\leq
2\sum_{k=1}^{\infty}e^{-\frac{\gamma}{2}k^{2}}\left(\frac{\lambda}{1-\lambda}\right)^{\frac{k}{2}}$
$\displaystyle=2\,\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)-2$
$\displaystyle\leq 1\,.$

Therefore, the infinite matrix $A$ is diagonally dominant if
$\lambda>\frac{1}{2}$ and
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$.
This establishes that $H_{B_{1}}$ and $H_{B_{2}}$ are valid quantum channels
in this parameter range, implying that $\rho_{AB_{1}B_{2}}$ is a valid two-
extension of the Choi state of $\pazocal{N}_{\lambda,\gamma}$, and in turn
entailing that $\pazocal{N}_{\lambda,\gamma}$ is anti-degradable. Finally,
note that if $\lambda>\frac{1}{2}$ the condition

$\displaystyle\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$
(S40)

is implied by $\lambda\leq\frac{1}{1+9e^{-\gamma}}$. Indeed,

$\displaystyle\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\coloneqq\sum_{k=0}^{n}e^{-\frac{\gamma}{2}k^{2}}\left(\sqrt{\frac{\lambda}{1-\lambda}}\right)^{k}\leq\sum_{k=0}^{\infty}\left(\frac{e^{-\gamma}\lambda}{1-\lambda}\right)^{k/2}=\frac{1}{1-\sqrt{\frac{e^{-\gamma}\lambda}{1-\lambda}}}\leq\frac{3}{2}\,,$

where the last inequality follows from
$\frac{e^{-\gamma}\lambda}{1-\lambda}\leq\frac{1}{9}$, which is implied by
$\lambda\leq\frac{1}{1+9e^{-\gamma}}$. ∎

#### Expanding the anti-degradability region numerically

Note that Theorem S27 does not identify the entire anti-degradability region
of the bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$. In fact,
from the above argument it becomes clear that a way to obtain a better inner
approximation of this region is to check for which values of the parameters
the infinite matrix $A$ defined by (S35) is positive semi-definite. This is
established in the following theorem.

######  Theorem S28.

Let $\ell^{2}(\mathbb{N})$ be the space of square-summable complex-valued
sequences (defined by (S65) below). For any $\lambda\in(\frac{1}{2},1)$ and
$\gamma>0$, let $A=(a_{mn})_{m,n\in\mathbb{N}}$ be the infinite matrix whose
components are defined by (S35). If $A\geq 0$ is positive semi-definite as an
operator on $\ell^{2}(\mathbb{N})$, then the bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is anti-degradable.

###### Proof.

In the proof of Theorem S27 we have seen that the bosonic loss-dephasing
channel is anti-degradable if the Hadamard map associated with the infinite
matrix $A$ (given in (S35)) is a quantum channel. Since the diagonal elements
of $A$ are equal to one, from Lemma S45 we deduce that the Hadamard map
associated to $A$ is a quantum channel if and only if $A$ is positive semi-
definite. This concludes the proof. ∎

In Theorem S27, we showed that the above-mentioned infinite matrix $A$ is
positive semi-definite if
$\theta\\!\left(e^{-\gamma/2},\sqrt{\lambda/(1-\lambda)}\right)\leq\frac{3}{2}$,
where $\theta(x,y)\coloneqq\sum_{n=0}^{\infty}x^{n^{2}}y^{n}$. This identifies
just a portion of the full region of parameters of $\lambda$ and $\gamma$
where the infinite matrix $A$ is positive semi-definite.

To analyse the positive semi-definiteness of the infinite matrix $A$ further,
let $A^{(d)}$ denote its $d\times d$ top-left corner. Note that it is well-
known that an infinite matrix is positive semi-definite if and only if its
$d\times d$ top-left corner is positive semi-definite for all
$d\in\mathbb{N}$. For modest values of $d$, we can numerically determine the
parameter region where $A^{(d)}$ is positive semi-definite. To achieve this,
we plot in Fig. 3 the quantity

$\displaystyle\eta_{d}(\gamma)\coloneqq\max\left(\lambda\in\left(\frac{1}{2},1\right]:\,\,A^{(d)}\text{
is positive semi-definite}\right)\,,$ (S41)

with respect to $e^{-\gamma}$ for various values of $d$. This quantity is
relevant because $A^{(d)}$ is positive semi-definite if and only if
$\lambda\leq\eta_{d}(\gamma)$. Moreover, the quantity $\eta_{d}(\gamma)$
monotonically decreases in $d$ and converges to some $\bar{\eta}(\gamma)$ as
$d\to\infty$. Notably, the condition $\lambda\leq\bar{\eta}(\gamma)$ is
necessary and sufficient for positive semi-definiteness of the infinite matrix
$A$, and also a sufficient condition for the anti-degradability of the the
bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$. Our numerical
investigation seems to suggest that when $d$ is approximately $20$, the
quantity $\eta_{d}(\gamma)$ has already reached its limiting value
$\bar{\eta}(\gamma)$, which can be approximated, for instance, by considering,
the curve $\eta_{30}(\gamma)$.

Figure 3: Numerical estimation of the anti-degradability of the loss-dephasing
channel. The horizontal axis shows the quantity $e^{-\gamma}$, varying from
$0$ to $1$ as dephasing parameter $\gamma$ decreases from $\infty$ to $0$,
while the vertical axis corresponds to the transmissivity $\lambda$. Theorem
S27 establishes that below the region defined by the blue curve, corresponding
to the condition $\theta(e^{-\gamma/2},\sqrt{\lambda/(1-\lambda)})=3/2$, the
bosonic loss-dephasing channel is anti-degradable. Moreover, Theorem S35
establishes that above the region defined by the purple curve, corresponding
to $\lambda=\frac{1}{1+e^{-\gamma}}$, the bosonic loss-dephasing channel is
not anti-degradable. The other curves depict the quantity $\eta_{d}(\gamma)$,
which is defined in (S41) as the maximum value of the transmissivity where
$A^{(d)}$ is positive semi-definite, for various values of $d$. Our numerical
analysis seems to indicate that in the region below the red curve,
corresponding to $\lambda\leq\eta_{30}(\gamma)$, the bosonic loss-dephasing
channel is anti-degradable. Here, we employ $d=30$, as increasing $d$ beyond
$d\geq 20$ yields no discernible change in the plot.

#### Anti-degrading maps

Theorem S27 discovers parameter regions of transmissivity $\lambda$ and
dephasing $\gamma$ in which the bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is anti-degradable. Although the proof of this
theorem ensures the existence of anti-degrading maps for
$\pazocal{N}_{\lambda,\gamma}$ within these parameter regions, it does not
offer explicit constructions of such anti-degrading maps. In the forthcoming
Theorem S29, we present such explicit constructions. Note that, thanks to
Lemma S23, the output operators of the complementary channel
$\pazocal{N}^{\text{c}}_{\lambda,\gamma}$ reside within the space
$\pazocal{T}(\pazocal{H}_{E_{\text{out}}})$, where
$\pazocal{H}_{E_{\text{out}}}$ is the following subspace of the two-mode
Hilbert space
$\pazocal{H}_{E_{1}}\otimes\pazocal{H}_{E_{2}}=L^{2}(\mathbb{R})\otimes
L^{2}(\mathbb{R})$:

$\displaystyle\pazocal{H}_{E_{\text{out}}}\coloneqq\text{Span}\left\\{\ket{\ell}_{E_{1}}\otimes\ket{\sqrt{\gamma}n}_{E_{2}}:\,\,\ell\leq
n\text{ with }\ell,n\in\mathbb{N}\right\\},$ (S42)

where $\ket{n}$ represents the $n$th Fock state, and $\ket{\sqrt{\gamma}n}$
denotes the coherent state with a parameter of $\sqrt{\gamma}n$. These states
correspond to the environmental modes of the pure-loss and dephasing channels,
respectively (we shall maintain this notation throughout).

######  Theorem S29.

Anti-degrading maps corresponding to each parameter region in Theorem S27 can
be defined as follows. In the region (i), i.e.
$\lambda\in\left[0,\frac{1}{2}\right]$ and $\gamma\geq 0$, an anti-degrading
map is given by

$\displaystyle\pazocal{A}_{\lambda,\gamma}=\left(\pazocal{E}_{\frac{\lambda}{1-\lambda}}\circ\pazocal{R}_{E_{1}}\right)\otimes\Tr_{E_{2}},$
(S43)

where
$\pazocal{R}_{E_{1}}(\cdot)\coloneqq(-1)^{\hat{e}_{1}^{\dagger}\hat{e}_{1}}\cdot(-1)^{\hat{e}_{1}^{\dagger}\hat{e}_{1}}$,
with $\hat{e}_{1}$ as the annihilation operator of the output mode of the
pure-loss channel $E_{1}$.

In the region (ii), i.e. $\lambda\in(\frac{1}{2},1)$ and $\gamma$ such that
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$,
an anti-degrading map
$\pazocal{A}_{\lambda,\gamma}:\pazocal{T}\left(\pazocal{H}_{E_{\text{out}}}\right)\to\pazocal{T}\left(\pazocal{H}_{\text{B}}\right)$,
with $\pazocal{H}_{E_{\text{out}}}$ given by (S42), is defined as follows. For
all $\ell_{1},\ell_{2},n_{1},n_{2}\in\mathbb{N}$ with $\ell_{1}\leq n_{1}$ and
$\ell_{2}\leq n_{2}$, it holds that

$\displaystyle{\pazocal{A}}_{\lambda,\gamma}\\!\left(\outerproduct{\ell_{1}}{\ell_{2}}\otimes\outerproduct{\sqrt{\gamma}n_{1}}{\sqrt{\gamma}n_{2}}\right)\coloneqq\sum_{k=0}^{\min(n_{1}-\ell_{1},n_{2}-\ell_{2})}c_{k}^{(\ell_{1},\ell_{2},n_{1},n_{2})}\outerproduct{k+\ell_{1}}{k+\ell_{2}}\,,$
(S44)

where for all $k\in\\{0,1,\ldots,\min(n_{1}-\ell_{1},n_{2}-\ell_{2})\\}$ the
coefficients $c_{k}^{(\ell_{1},\ell_{2},n_{1},n_{2})}$ are defined as

$\displaystyle c_{k}^{(\ell_{1},\ell_{2},n_{1},n_{2})}$
$\displaystyle\coloneqq(-1)^{\ell_{1}-\ell_{2}}\sqrt{\pazocal{B}_{k}\\!\left(n_{1}-\ell_{1},\frac{2\lambda-1}{\lambda}\right)\,\pazocal{B}_{k}\\!\left(n_{2}-\ell_{2},\frac{2\lambda-1}{\lambda}\right)}\,a_{n_{1}-\ell_{1}n_{2}-\ell_{2}}\,a_{k+\ell_{1}k+\ell_{2}}\,,$
(S45)

where
$\pazocal{B}_{l}\\!\left(n,\lambda\right)\coloneqq\binom{n}{l}\lambda^{l}(1-\lambda)^{n-l}$,
and $a_{mn}$ is defined in (S35).

###### Proof of Theorem S29.

Let us suppose that $\lambda$ and $\gamma$ fall within the parameter region
(i). A complementary channel of the pure-loss channel $\pazocal{E}_{\lambda}$
is given by
$\pazocal{E}_{\lambda}^{\text{c}}=\pazocal{E}_{1-\lambda}\circ\pazocal{R}$.
Consequently, Lemma S13 implies that
$\left(\pazocal{E}_{\frac{\lambda}{1-\lambda}}\circ\pazocal{R}\right)\circ\pazocal{E}_{\lambda}^{\text{c}}=\pazocal{E}_{\lambda}$,
i.e. the channel $\pazocal{E}_{\frac{\lambda}{1-\lambda}}\circ\pazocal{R}$ is
an anti-degrading map of the pure-loss channel. The general construction
detailed in the proof of Lemma S50 for the anti-degrading map of the
composition between an anti-degradable channel and another channel
demonstrates that the map given in (S43) is an anti-degrading map of
$\pazocal{N}_{\lambda,\gamma}$.

Let us now suppose that $\lambda$ and $\gamma$ fall within the parameter
region (ii). To come up with the anti-degrading map defined in (S44), we drew
intuition from the proof of Lemma S49, which demonstrates the equivalence
between two-extendibility of the Choi state and the existence of an anti-
degrading map, while also considering the two-extension of the Choi state of
$\pazocal{N}_{\lambda,\gamma}$ explicitly found in (S37). In order to show
that the map ${\pazocal{A}}_{\lambda,\gamma}$ in (S44) is an anti-degrading
map of $\pazocal{N}_{\lambda,\gamma}$, we need to show that it is a quantum
channel satisfying
${\pazocal{A}}_{\lambda,\gamma}\circ\pazocal{N}^{\text{c}}_{\lambda,\gamma}=\pazocal{N}_{\lambda,\gamma}$.
We begin by proving that ${\pazocal{A}}_{\lambda,\gamma}$ is trace preserving.
By linearity, it suffices to show that for any
$\ell_{1},\ell_{2},n_{1},n_{2}\in\mathbb{N}$ with $\ell_{1}\leq n_{1}$ and
$\ell_{2}\leq n_{2}$ it holds that

$\displaystyle\Tr\\!\left[{\pazocal{A}}_{\lambda,\gamma}\\!\left(\outerproduct{\ell_{1}}{\ell_{2}}_{E_{1}}\otimes\outerproduct{\sqrt{\gamma}n_{1}}{\sqrt{\gamma}n_{2}}_{E_{2}}\right)\right]=\Tr\\!\left[\outerproduct{\ell_{1}}{\ell_{2}}_{E_{1}}\otimes\outerproduct{\sqrt{\gamma}n_{1}}{\sqrt{\gamma}n_{2}}_{E_{2}}\right]\,.$
(S46)

Indeed, we obtain

$\displaystyle\Tr\\!\left[{\pazocal{A}}_{\lambda,\gamma}\\!\left(\outerproduct{\ell_{1}}{\ell_{2}}_{E_{1}}\otimes\outerproduct{\sqrt{\gamma}m}{\sqrt{\gamma}n}_{E_{2}}\right)\right]$
$\displaystyle=\,\delta_{\ell_{1},\ell_{2}}\sum_{k=0}^{\min(m-\ell_{1},n-\ell_{1})}c_{k}^{(\ell_{1},\ell_{1},m,n)}$
$\displaystyle=\,\delta_{\ell_{1},\ell_{2}}\sum_{k=0}^{\min(m-\ell_{1},n-\ell_{1})}\sqrt{\pazocal{B}_{k}\\!\left(m-\ell_{1},\frac{2\lambda-1}{\lambda}\right)\pazocal{B}_{k}\\!\left(n-\ell_{1},\frac{2\lambda-1}{\lambda}\right)}a_{m-\ell_{1},n-\ell_{1}}\,a_{k+\ell_{1},k+\ell_{1}}$
$\displaystyle=\,\delta_{\ell_{1},\ell_{2}}e^{-\frac{\gamma}{2}(m-n)^{2}}$
$\displaystyle=\Tr\\!\left[\outerproduct{\ell_{1}}{\ell_{2}}_{E_{1}}\otimes\outerproduct{\sqrt{\gamma}n_{1}}{\sqrt{\gamma}n_{2}}_{E_{2}}\right]\,,$

where $\delta_{\ell_{1},\ell_{2}}$ denotes the kronecker delta and where we
have exploited the formula for the overlap between coherent states provided in
(S6). Now, let us show that ${\pazocal{A}}_{\lambda,\gamma}$ is completely
positive. To achieve this, we need to find a pure state $\Psi$ on
$\pazocal{H}_{\text{anc}}\otimes\pazocal{H}_{E_{\text{out}}}$, where
$\pazocal{H}_{\text{anc}}$ in an auxiliary reference, such that
$\Tr_{\text{anc}}[\outerproduct{\Psi}{\Psi}]>0$ and
$(\operatorname{id}_{\text{anc}}\otimes{\pazocal{A}}_{\lambda,\gamma})(\outerproduct{\Psi}{\Psi})\geq
0$ [69, 62, 63]. Let
$\pazocal{H}_{\text{anc}}\coloneqq\pazocal{H}_{A}\otimes\pazocal{H}_{B_{1}}=L^{2}(\mathbb{R})\otimes
L^{2}(\mathbb{R})$ and let us construct the pure state
$\ket{\Psi}_{AB_{1}E_{1}E_{2}}\in\pazocal{H}_{\text{anc}}\otimes\pazocal{H}_{E_{\text{out}}}$
as follows:

$\displaystyle\ket{\Psi}_{AB_{1}E_{1}E_{2}}$ $\displaystyle\coloneqq
U_{\lambda}^{B_{1}E_{1}}V_{\gamma}^{B_{1}E_{2}}\ket{\psi(r)}_{AB_{1}}\ket{0}_{E_{1}}\ket{0}_{E_{2}}$
(S47)
$\displaystyle=\frac{1}{\cosh(r)}\sum_{n=0}^{\infty}\sum_{\ell=0}^{n}(-1)^{\ell}\tanh^{n}(r)\sqrt{\pazocal{B}_{l}(n,1-\lambda)}\ket{n}_{A}\ket{n-\ell}_{B_{1}}\ket{\ell}_{E_{1}}\ket{\sqrt{\gamma}n}_{E_{2}}\,,$

where $U_{\lambda}^{B_{1}E_{1}}$ is the beam splitter unitary,
$V_{\gamma}^{B_{1}E_{2}}$ is the conditional displacement unitary, and
$\ket{\psi(r)}_{AB_{1}}$ is the two-mode squeezed vacuum state with squeezing
$r>0$. Let us now show that
$\Tr_{AB_{1}}\left[\outerproduct{\Psi}{\Psi}_{AB_{1}E_{1}E_{2}}\right]$ is
positive definite on $\pazocal{H}_{E_{\text{out}}}$. Let
$\ket{\phi}_{E_{1}E_{2}}\in\pazocal{H}_{E_{\text{out}}}$. Since there exists
$\bar{\ell},\bar{n}\in\mathbb{N}$ with $\bar{l}\leq\bar{n}$ such that
$\bra{\phi}_{E_{1}E_{2}}\ket{\bar{\ell}}_{E_{1}}\otimes\ket{\sqrt{\gamma}\bar{n}}_{E_{2}}\neq
0$, (S47) implies that

$\displaystyle\bra{\phi}_{E_{1}E_{2}}\Tr_{AB_{1}}\left[\outerproduct{\Psi}{\Psi}_{AB_{1}E_{1}E_{2}}\right]\ket{\phi}_{E_{1}E_{2}}$
$\displaystyle=\frac{1}{\cosh^{2}(r)}\sum_{n=0}^{\infty}\sum_{l=0}^{n}\tanh^{2n}(r)\,\pazocal{B}_{\ell}\left(n,1-\lambda\right)\left|\bra{\phi}_{E_{1}E_{2}}\ket{\ell}_{E_{1}}\otimes\ket{\sqrt{\gamma}n}_{E_{2}}\right|^{2}$
$\displaystyle\geq\frac{1}{\cosh^{2}(r)}\tanh^{2\bar{n}}(r)\,\pazocal{B}_{\bar{l}}\left(\bar{n},1-\lambda\right)\left|\bra{\phi}_{E_{1}E_{2}}\ket{\bar{\ell}}_{E_{1}}\otimes\ket{\sqrt{\gamma}\bar{n}}_{E_{2}}\right|^{2}$
$\displaystyle>0\,.$

We next show that
$\operatorname{id}_{AB_{1}}\otimes{\pazocal{A}}_{\lambda,\gamma}(\outerproduct{\Psi}{\Psi})$
is positive semi-definite. Let $B_{2}$ denote the output system of
${\pazocal{A}}_{\lambda,\gamma}$. Note that

$\displaystyle\operatorname{id}_{AB_{1}}\otimes{\pazocal{A}}_{\lambda,\gamma}(\outerproduct{\Psi}{\Psi}_{AB_{1}E_{1}E_{2}})$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(i)}}}}{{=}}\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell_{1}=0}^{m}\sum_{\ell_{2}=0}^{n}(-1)^{\ell_{1}+\ell_{2}}[\tanh(r)]^{m+n}\sqrt{\pazocal{B}_{\ell_{1}}(m,1-\lambda)\pazocal{B}_{\ell_{2}}(n,1-\lambda)}$
(S48) $\displaystyle\hskip
102.43008pt\outerproduct{m}{n}_{A}\otimes\outerproduct{m-\ell_{1}}{n-\ell_{2}}_{B_{1}}\otimes{\pazocal{A}}_{\lambda,\gamma}\\!\left(\outerproduct{\ell_{1}}{\ell_{2}}_{E_{1}}\otimes\outerproduct{\sqrt{\gamma}m}{\sqrt{\gamma}n}_{E_{2}}\right)$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ii)}}}}{{=}}\frac{1}{\cosh^{2}(r)}\sum_{m,n=0}^{\infty}\sum_{\ell_{1}=0}^{m}\sum_{\ell_{2}=0}^{n}\sum_{k=0}^{\min(m-\ell_{1},n-\ell_{2})}[\tanh(r)]^{m+n}$
$\displaystyle\hskip
56.9055pt\sqrt{\pazocal{B}_{\ell_{1}}(m,1-\lambda)\,\pazocal{B}_{\ell_{2}}(n,1-\lambda)\,\pazocal{B}_{k}\left(m-\ell_{1},\frac{2\lambda-1}{\lambda}\right)\pazocal{B}_{k}\left(n-\ell_{2},\frac{2\lambda-1}{\lambda}\right)}$
$\displaystyle\hskip
96.73918pta_{m-\ell_{1},n-\ell_{2}}\,a_{k+\ell_{1},k+\ell_{2}}\outerproduct{n_{1}}{n}_{A}\otimes\outerproduct{m-\ell_{1}}{n-\ell_{2}}_{B_{1}}\otimes\outerproduct{k+\ell_{1}}{k+\ell_{2}}_{B_{2}}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iii)}}}}{{=}}\rho_{AB_{1}B_{2}}\,.$

Here, in (i) we used the definition of $\ket{\Psi}_{AB_{1}E_{1}E_{2}}$ given
in (S47); in (ii) we utilised the definition of the map
${\pazocal{A}}_{\lambda,\gamma}$ from (S44); and in (iii) we recognised the
tripartite operator $\rho_{AB_{1}B_{2}}$ defined in (S37), which is a quantum
state provided that $\lambda$ and $\gamma$ satisfy
$\theta\\!\left(e^{-\gamma/2},\sqrt{\frac{\lambda}{1-\lambda}}\right)\leq\frac{3}{2}$.
Therefore, in such a parameter region,
$\operatorname{id}_{AB_{1}}\otimes{\pazocal{A}}_{\lambda,\gamma}(\outerproduct{\Psi}{\Psi}_{AB_{1}E_{1}E_{2}})$
is positive semi-definite, and thus ${\pazocal{A}}_{\lambda,\gamma}$ is a
quantum channel. Let us now verify that
${\pazocal{A}}_{\lambda,\gamma}\circ\pazocal{N}^{\text{c}}_{\lambda,\gamma}=\pazocal{N}_{\lambda,\gamma}$.
To show this, note that

$\displaystyle\operatorname{id}_{A}\otimes\left({\pazocal{A}}_{\lambda,\gamma}\circ\pazocal{N}^{\text{c}}_{\lambda,\gamma}\right)(\outerproduct{\psi(r)}{\psi(r)}_{AB_{1}})$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iv)}}}}{{=}}\Tr_{B_{1}}\left[\operatorname{id}_{AB_{1}}\otimes{\pazocal{A}}_{\lambda,\gamma}(\outerproduct{\Psi}{\Psi}_{AB_{1}E_{1}E_{2}})\right]$
(S49)
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(v)}}}}{{=}}\Tr_{B_{1}}\left[\rho_{AB_{1}B_{2}}\right]$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(vi)}}}}{{=}}\operatorname{id}_{A}\otimes\pazocal{N}_{\lambda,\gamma}\left(\outerproduct{\psi(r)}{\psi(r)}_{AB_{2}}\right)\,,$

Here, in (iv) we employed (S47); in (v) we exploited (S48); and in (vi) we
used that $\rho_{AB_{1}B_{2}}$ is a two-extension of
$\operatorname{id}_{A}\otimes\pazocal{N}_{\lambda,\gamma}\left(\outerproduct{\psi(r)}{\psi(r)}\right)$,
as established in the proof of Theorem S27. Finally, since the two-mode
squeezed vacuum state $\ket{\psi(r)}_{AB}$ satisfies
$\Tr_{B}[\outerproduct{\psi(r)}{\psi(r)}_{AB}]>0$, we conclude
${\pazocal{A}}_{\lambda,\gamma}\circ\pazocal{N}^{\text{c}}_{\lambda,\gamma}=\pazocal{N}_{\lambda,\gamma}$.
∎

###  II.2 Analysis of the bosonic loss-dephasing channel via its finite-
dimensional restrictions

######  Definition S30.

Let $d\in\mathbb{N}$ and let
$\pazocal{H}_{d}\coloneqq\operatorname{Span}(\\{\ket{n}\\}_{n=0,\ldots,d-1})$
be the subspace spanned by the first $d$ Fock states. The qudit restriction of
the bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}^{(d)}$ is a
quantum channel defined by

$\displaystyle\pazocal{N}_{\lambda,\gamma}^{(d)}(\Theta)\coloneqq\pazocal{N}_{\lambda,\gamma}(\Theta)\qquad\forall\,\Theta\in\pazocal{T}(\pazocal{H}_{d})$
(S50)

######  Lemma S31.

Let $\pazocal{H}_{A},\pazocal{H}_{B}\coloneqq L^{2}(\mathbb{R})$. Let
$\pazocal{N}:\pazocal{T}(\pazocal{H}_{A})\to\pazocal{T}(\pazocal{H}_{B})$ be a
quantum channel and let $\pazocal{N}^{(d)}$ be its qudit restriction, defined
by

$\displaystyle\pazocal{N}^{(d)}(\Theta)\coloneqq\pazocal{N}(\Theta)\qquad\forall\,\Theta\in\pazocal{T}(\pazocal{H}_{d})\,.$
(S51)

If $\pazocal{N}$ is (anti-)degradable, then its qudit restriction
$\pazocal{N}^{(d)}$ is (anti-)degradable.

###### Proof.

Let $\pazocal{N}(\cdot)=\Tr_{E}[V_{A\to BE}(\cdot)V_{A\to BE}^{\dagger}]$ be a
Stinespring representation of $\pazocal{N}$, and let
$\pazocal{N}^{\text{c}}(\cdot)=\Tr_{B}[V_{A\to BE}(\cdot)V_{A\to
BE}^{\dagger}]$ be the associated complementary channel. Note that the
isometry $V_{A\to BE}$ provides a Stinespring representation also for the
qudit restriction $\pazocal{N}^{(d)}$. Hence, the qudit restriction of the
complementary channel $\pazocal{N}^{\text{c}}$ is a complementary channel of
the qudit restriction $\pazocal{N}^{(d)}$, i.e.

$\displaystyle(\pazocal{N}^{(d)})^{\text{c}}(\Theta)=\pazocal{N}^{\text{c}}(\Theta)\qquad\forall\,\Theta\in\pazocal{T}(\pazocal{H}_{d})\,.$
(S52)

Moreover, note that any degrading or anti-degrading map of $\pazocal{N}$ is
effective for all input states, including those restricted to
$\pazocal{H}_{d}$. Consequently, an (anti-)degrading map of $\pazocal{N}$ is
also an (anti-)degrading map of its qudit restriction $\pazocal{N}^{(d)}$. ∎

######  Corollary S32.

If the qudit restriction $\pazocal{N}_{\lambda,\gamma}^{(d)}$ is not
(anti)-degradable, then the bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is also not (anti)-degradable.

The following lemma shows that qudit restriction
$\pazocal{N}_{\lambda,\gamma}^{(d)}$ is a qu$d$it-to-qu$d$it channel, mapping
the space spanned by $\\{\ket{n}\\}_{n=0,\ldots,d-1}$ into itself.

######  Lemma S33.

If the input state to the bosonic loss-dephasing channel is confined into the
finite-dimensional subspace spanned by $\\{\ket{n}\\}_{n=0,\ldots,d-1}$, the
resulting output state will similarly be confined to this subspace.

###### Proof.

Examining Lemma S22 reveals that the operator $\outerproduct{m}{n}$, when
acted on by the bosonic loss-dephasing channel, is transformed into linear
combinations of operators $\\{\outerproduct{\ell}{k}\\}_{\ell\leq m,k\leq n}$.
This means that if the input state to the bosonic loss-dephasing channel is
restricted to the $d$-dimensional subspace $\\{\ket{n}\\}_{n=0,\ldots,d-1}$,
the output of the channel will reside within the same subspace. ∎

The qubit restriction $\pazocal{N}_{\lambda,\gamma}^{(2)}$ of the bosonic
loss-dephasing channel coincides with the composition between the amplitude
damping channel and the qubit dephasing channel [34, 33], which we dub
_amplitude-phase damping channel_. Theorems S34 and S35 utilise the amplitude-
phase damping channel $\pazocal{N}_{\lambda,\gamma}^{(2)}$ to find that the
bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$ is never
degradable and, additionally, it is not anti-degradabile for
$\lambda>\frac{1}{1+e^{-\gamma}}$, respectively.

#### The bosonic loss-dephasing channel is never degradable

The bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$ is never
degradable, except when it coincides with either the bosonic dephasing channel
($\gamma>0$ and $\lambda=1$) or the degradable pure-loss channel ($\gamma=0$
and $\lambda\geq\frac{1}{2}$), thereby complicating the derivation of its
quantum capacity [32]. This result has been previously demonstrated in [32]
through pages-long proof; however, here we provide a significantly simpler
proof of this result.

######  Theorem S34.

Let $\lambda\in[0,1]$ and $\gamma\geq 0$. The bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is degradable if and only if one of the
following conditions is satisfied:

  * •

$\gamma=0$ and $\lambda\in[\frac{1}{2},1]$

  * •

$\gamma\geq 0$ and $\lambda=1$

###### Proof.

Thanks to Corollary S32, a necessary condition for
$\pazocal{N}_{\lambda,\gamma}$ to be degradable is the degradability of the
amplitude-phase damping channel $\pazocal{N}_{\lambda,\gamma}^{(2)}$. We now
apply [54, Theorem 4], which establishes a necessary condition on the
degradability of any qubit channel. Specifically, the rank of the Choi state
of a degradable qubit channel is necessarily less or equal to $2$. By using
the notation used in (S3), the matrix associated with the Choi state of the
amplitude-phase damping channel
$C\\!\left(\pazocal{N}_{\lambda,\gamma}^{(2)}\right)$ in the computational
basis $\\{\ket{00},\ket{01},\ket{10},\ket{11}\\}$ is given by:

$\displaystyle\frac{1}{2}\left(\begin{matrix}1\quad&0&0&\sqrt{e^{-\gamma}\lambda}\\\
0&0\quad&0\quad&0\\\ 0&0\quad&1-\lambda\quad&0\\\
\sqrt{e^{-\gamma}\lambda}&0&0&\lambda\quad\end{matrix}\right)\,.$ (S53)

One can easily see that its rank is equal to $3$ for $\gamma>0$ and
$\lambda\in(0,1)$. In addition, for $\lambda=1$ the bosonic loss-dephasing
channel coincides with the bosonic dephasing channel,
$\pazocal{N}_{1,\gamma}=\pazocal{D}_{\gamma}$, which is degradable for any
value of $\gamma\geq 0$ [40]. Finally, for $\gamma=0$ the bosonic loss-
dephasing channel coincides with the pure-loss channel,
$\pazocal{N}_{\lambda,0}=\pazocal{E}_{\lambda}$, which is degradable if and
only if $\lambda\in[\frac{1}{2},1]$ [39, 38]. ∎

#### Necessary condition on anti-degradability via qubit restriction

The next theorem establishes the parameter range where the bosonic loss-
dephasing channel is not anti-degradable. We provide three different proofs
for this theorem.

######  Theorem S35.

Let $\gamma\geq 0$. If $\lambda>\frac{1}{1+e^{-\gamma}}$, then the bosonic
loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$ is not anti-degradable.

###### Proof 1.

Assume that $\pazocal{N}_{\lambda,\gamma}$ is anti-degradable; then
substituting $m=0$ and $n=1$ in (S22) yields

$\displaystyle\pazocal{A}_{\lambda,\gamma}\left(\pazocal{E}_{1-\lambda}(\outerproduct{0}{1})\otimes\outerproduct{0}{\sqrt{\gamma}}\right)=-e^{-\frac{1}{2}\gamma}\pazocal{E}_{\lambda}(\outerproduct{0}{1})\,.$
(S54)

By exploiting
$\pazocal{E}_{\lambda}(\outerproduct{0}{1})=\sqrt{\lambda}\outerproduct{0}{1}$,
we have

$\displaystyle\pazocal{A}_{\lambda,\gamma}\left(\outerproduct{0}{1}\otimes\outerproduct{0}{\sqrt{\gamma}}\right)=-\sqrt{\frac{e^{-\gamma}\lambda}{1-\lambda}}\outerproduct{0}{1}\,.$

Using Lemma S47 in the Appendix, we find
$\sqrt{\frac{e^{-\gamma}\lambda}{1-\lambda}}\leq 1$, or
$\lambda\leq\frac{1}{1+e^{-\gamma}}$. ∎

###### Proof 2.

Assume that $\pazocal{N}_{\lambda,\gamma}$ is anti-degradable. As a
consequence of (S22) and of the data-processing inequality for the fidelity
[60], we find

$\displaystyle
F\big{(}\,\pazocal{E}_{\lambda}(\outerproduct{0}{0})\,,\,\pazocal{E}_{\lambda}(\outerproduct{1}{1})\,\big{)}\geq
F\big{(}\,\pazocal{E}_{1-\lambda}(\outerproduct{0}{0})\otimes\outerproduct{0}{0}\,,\,\pazocal{E}_{1-\lambda}(\outerproduct{1}{1})\otimes\outerproduct{\sqrt{\gamma}}{\sqrt{\gamma}}\,\big{)}\,.$
(S55)

Furthermore, we obtain

$\displaystyle\sqrt{1-\lambda}$
$\displaystyle=F\big{(}\outerproduct{0}{0},\lambda\outerproduct{1}{1}+(1-\lambda)\outerproduct{0}{0}\big{)}$
(S56)
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(i)}}}}{{=}}F\big{(}\pazocal{E}_{\lambda}(\outerproduct{0}{0}),\pazocal{E}_{\lambda}(\outerproduct{1}{1})\big{)}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ii)}}}}{{\geq}}F\big{(}\outerproduct{0}{0},\outerproduct{\sqrt{\gamma}}{\sqrt{\gamma}}\big{)}\,F\big{(}\pazocal{E}_{1-\lambda}(\outerproduct{0}{0}),\pazocal{E}_{1-\lambda}(\outerproduct{1}{1})\big{)}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iii)}}}}{{\geq}}F\big{(}\outerproduct{0}{0},\outerproduct{\sqrt{\gamma}}{\sqrt{\gamma}}\big{)}\,F\big{(}\outerproduct{0}{0},\lambda\outerproduct{0}{0}+(1-\lambda)\outerproduct{1}{1}\big{)}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iv)}}}}{{=}}\sqrt{e^{-\gamma}\lambda}\,.$

Here, (i) follows from Lemma S10, (ii) follows from (S55) and from the fact
that the fidelity is multiplicative under tensor product [60], (iii) uses
Lemma S10 again, and in (iv) we exploited that
$|\innerproduct{0}{\sqrt{\gamma}}|=\sqrt{e^{-\gamma}}$. This yields
$\sqrt{1-\lambda}\geq\sqrt{e^{-\gamma}\lambda}$, or
$\lambda\leq\frac{1}{1+e^{-\gamma}}$. ∎

###### Proof 3.

By exploiting Lemma S4 and the Choi matrix of the qubit channel
$\pazocal{N}^{(2)}_{\lambda,\gamma}$ reported in (S53), one can easily obtain
that $\pazocal{N}^{(2)}_{\lambda,\gamma}$ is anti-degradable if and only if
$\lambda\leq\frac{1}{1+e^{-\gamma}}$. Consequently, thanks to Corollary S32,
the bosonic loss-dephasing channel $\pazocal{N}_{\lambda,\gamma}$ is not anti-
degradable for $\lambda>\frac{1}{1+e^{-\gamma}}$. ∎

#### Necessary condition on anti-degradability via qudit restrictions

Let us introduce the following quantity for any $d\in\mathbb{N}$ and
$\gamma\geq 0$:

$\displaystyle\lambda_{d}(\gamma)\coloneqq\max\left(\lambda\in[0,1]:\,\,\pazocal{N}^{(d)}_{\lambda,\gamma}\text{
is anti-degradable}\right)\,.$ (S57)

This quantity is relevant since it allows us to find parameter region where
the bosonic loss-dephasing channel is not anti-degradable. Specifically, for
$\lambda>\lambda_{d}(\gamma)$ the bosonic loss-dephasing channel
$\pazocal{N}_{\lambda,\gamma}$ is not anti degradable, as established by
Corollary S32. Thanks to the Proof 3 of Theorem S35, it follows that for $d=2$
we have that $\lambda_{2}(\gamma)=\frac{1}{1+e^{-\gamma}}$, thereby
establishing that $\pazocal{N}_{\lambda,\gamma}$ is not anti degradable for
$\lambda>\frac{1}{1+e^{-\gamma}}$. Through an examination of larger values of
$d$, we aim to identify an extended parameter region where the channel is not
anti-degradable (see Fig. 4). We begin by proving some useful properties of
the quantity $\lambda_{d}(\gamma)$.

######  Lemma S36.

For any $\gamma\geq 0$ and $d\in\mathbb{N},d\geq 2$, the following facts hold: