# Thermodynamics of quantum switch information capacity activation

Xiangjing Liu<EMAIL_ADDRESS>Shenzhen Institute for Quantum Science
and Engineering and Department of Physics, Southern University of Science and
Technology, Shenzhen 518055, China  Daniel Ebler<EMAIL_ADDRESS>Theory Lab, Central Research Institute, 2012 Labs, Huawei Technology Co. Ltd.,
Hong Kong Science Park, Hong Kong SAR  Department of Computer Science, The
University of Hong Kong, Pokfulam Road, Hong Kong SAR  Oscar Dahlsten
<EMAIL_ADDRESS>Shenzhen Institute for Quantum Science and Engineering
and Department of Physics, Southern University of Science and Technology,
Shenzhen 518055, China

###### Abstract

We address a new setting where the second law is under question:
thermalizations in a quantum superposition of causal orders, enacted by the
so-called quantum switch. This superposition has been shown to be associated
with an increase in the communication capacity of the channels, yielding an
apparent violation of the data-processing inequality and a possibility to
separate hot from cold. We analyze the thermodynamics of this information
capacity increasing process. We show how the information capacity increase is
compatible with thermodynamics. We show that there may indeed be an
information capacity increase for consecutive thermalizations obeying the
first and second laws of thermodynamics if these are placed in an indefinite
order and moreover that only a significantly bounded increase is possible. The
increase comes at the cost of consuming a thermodynamic resource, the free
energy of coherence associated with the switch.

General Introduction—The study of the Laws of thermodynamics, including the
First Law of energy conservation and the Second Law of free energy non-
increase, are key to the foundations of physics [1, 2]. Thermalizing
evolutions (thermal channels) are central to thermodynamics, representing the
impact of a heat bath on the system of interest [3, 2, 4, 5, 6]. Thermalizing
evolutions will in general remove information about the system’s earlier
interactions, in line with the second law of thermodynamics. One may think of
them as noisy communication channels from the past to the future. How much
information can be conveyed is captured by the information capacity of the
channel, a central concept in the Shannon information theory [7].

Recent results [8, 9, 10, 11, 12, 13] suggest the information capacity of two
consecutive channels may be increased if the order of the channels is in a
quantum superposition of the two possible orders, an operation termed applying
a quantum switch to the two channels [14]. The quantum switch extends the
allowed operations in quantum communication theory [42, 15], and is an example
of how to create indefinite causal order [16], an intriguing area of possible
new physics [17, 18, 19].

Because of the strong connection between information theory and
thermodynamics, e.g. via the central role played by entropy, this phenomenon
may have significance for thermodynamics. The setting wherein information
capacity is increased via the quantum switch was shown to give an apparent
violation of the data-processing inequality [9], which is closely connected to
the second law [20], as well as enable the separation between hot and cold
[21, 22, 23]. A more general study of superpositions of thermodynamic
processes is beginning [24, 25, 26, 27]. These intriguing results raise the
question of whether the laws of thermodynamics and the information capacity
increase of the switch are compatible, and if so how. Is the increased
communication capacity allowed within thermodynamics, if so is there a limit
to how much is allowed? Another concrete open question, posed in [21] concerns
whether the switch is a thermodynamic resource or should be regarded as
thermodynamically free.

In this work we accordingly investigate the interplay between the laws of
thermodynamics and this information capacity increase. We find that switched
thermal channels may have a small increase in information capacity, and no
more, or the laws of thermodynamics would be violated. To show this we derive
an upper-bound on the mutual information (a measure of communication amount)
that can be established in this setting, depicted in Figure 1. The bound
captures how activation is only possible when the switch is ON. We argue that
this increase in capacity comes at a thermodynamical cost. The switch can,
when a careful accounting is done, create free energy. Rather than
interpreting our results as showing that the switch violates the second law of
thermodynamics, we argue it should not be viewed as being in the domain of
applicability of the second law: the quantum switch is a thermodynamical
resource.

Figure 1: The set-up. $M$ carries information about the system $A$, $E$ is
from the heat bath environment. There is a unitary interaction between $M$ and
$E$. If the switch is ON there is a superposition of the black and green
curved-path orders for how the Kraus operators of the channel are applied to
$M$, else there is a fixed order as per the black path.

Laws of thermodynamics and thermal operations—We begin with describing
thermodynamical concepts and quantities that are key to our results. A central
concept is the (Gibb’s) thermal state density matrix $\tau=\exp(-H/kT)/Z$
where $H$ is the system’s Hamiltonian, $T$ the temperature, $k$ Boltzmann’s
constant and $Z=\textrm{Tr}(\exp(-H/kT))$ a normalization constant. A heat
bath is normally assumed to be in the state $\tau$ as well as being so large
that any system interacting with it will eventually reach a thermal state at
the temperature of the bath.

The Second Law for a system interacting with a heat bath can be presented as
demanding that the relative entropy
$S(\rho||\sigma):=\textrm{Tr}\left(\rho(\log\rho-\log\sigma)\right)$ between
the state $\rho$ and the thermal state $\tau$ of the given system cannot
increase under the heat bath’s influence $\varepsilon(\rho)$, i.e.,

$\displaystyle S(\varepsilon(\rho)||\tau)-S(\rho||\tau)\leq 0.$ (1)

The free energy is a crucial related quantity:
$F(\rho,H,T):=U(\rho)-kTS(\rho)=\textrm{Tr}(\rho H)-kTS(\rho)$, for a system
with state $\rho$, Hamiltonian $H$, temperature $T$ and the von Neumann
entropy $S(\rho):=-\textrm{Tr}(\rho\ln\rho)$. The relative entropy (with
natural logarithm) to the thermal state can be equivalently written in terms
of the free energy:

$kTS(\rho||\tau)=F(\rho)-F(\tau).$ (2)

Thus the second law of Eq.(1) can be restated as the free energy gap to the
thermal state not increasing: $\Delta F\leq 0$. Moreover, the thermal state
$\tau$ has minimal free energy, so the free energy cannot increase (whereas
entropy alone can decrease e.g. in cooling scenarios).

The long-established terminology of calling Eq. (1) or similar conditions the
second Law can be considered to be confusing. There are maps, physically and
mathematically, which violate Eq. (1) such as
$\varepsilon(\rho)=\ket{1}\bra{1}$ with $\ket{1}$ not being the ground state
(and thus not a thermal state for any temperature). A safe minimal
interpretation of Eq. (1) is that it is a property that only some dynamical
maps have. We demand that interactions with systems deemed to be heat baths
should have this property. Then, for logical consistency, a map violating Eq.
(1) should not correspond physically to an interaction with a heat bath.

We will moreover refer to the idea of free operations and resources (which are
not free) [1]. A key paradigm in thermodynamics is that systems at some
ambient temperature $T$ are free (e.g. not requiring the burning of fuel) and
anything that is not free is a resource. $kTS(\rho||\tau)=F(\rho)-F(\tau)$ may
be thought of as the currency of thermodynamics, quantifying the resource
value of a given system’s state $\rho$. Accordingly, it is natural to require
Eq. (1) to be respected in order for a map to be termed thermodynamically
free.

Quantum channels (completely positive and trace-preserving maps)
$\varepsilon(\cdot)$ which leave the thermal (Gibbs) state invariant,
$\varepsilon(\tau)=\tau$, are called thermal or Gibbs-preserving. The Gibbs-
preserving condition is equivalent to Eq. (1) (see Appendix A1). Quantum
channels, including Gibbs-preserving channels, are moreover mathematically
equivalent to unitarily interacting the system with an environment $E$ and
then tracing over $E$ [28]. To model a Gibbs-preserving channel this way, one
may take $E$ to be in a thermal state initially, and apply an energy-
preserving unitary $U_{ME}$ (meaning $[U_{ME},H_{M}+H_{E}]=0$) followed by
tracing over $E$. This is commonly called a thermal operation [3, 6, 4, 2, 29,
5]. (We shall later consider thermal operations in superposed orders).

Quantum Switch—The quantum switch $\mathcal{S}$ is a supermap [30], which
takes two quantum channels $\mathcal{C}_{1},\mathcal{C}_{2}$ as input and
outputs another quantum channel
$\mathcal{S}(\mathcal{C}_{1},\mathcal{C}_{2})$. Denoting the Kraus operators
[28] of the channel $\mathcal{C}_{1}$ as $\\{K^{(1)}_{i}\\}$ and
$\mathcal{C}_{2}$ as $\\{K^{(2)}_{j}\\}$, the switched channels
$\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})$ act as

$\displaystyle\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})(\rho)=\sum_{ij}S_{ij}(\sigma_{C}\otimes\rho)S_{ij}^{\dagger},$
(3)

where $\sigma_{C}$ is the state of the control system (also called the
trajectory in the literature [42, 15]) and $S_{ij}$ denotes the Kraus
operators for $\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})$,
namely,

$\displaystyle S_{ij}=|0\rangle_{C}\langle 0|\otimes
K^{(2)}_{j}K^{(1)}_{i}+|1\rangle_{C}\langle 1|\otimes K^{(1)}_{i}K^{(2)}_{j}\
.$ (4)

The quantum switch superposes the two temporal orders of the channels – that
is $\mathcal{C}_{2}\circ\mathcal{C}_{1}$ and
$\mathcal{C}_{1}\circ\mathcal{C}_{2}$ – with amplitudes determined by
$\sigma_{C}$ [14].

The state $\sigma_{C}$ is very important. If $\sigma_{C}=\ket{+}\bra{+}$ there
is a superposition of causal orders but e.g. if $\sigma_{C}=\ket{0}\bra{0}$,
there is a well-defined order. To interpolate between these two extremes, we
shall allow for $\sigma_{C}=\lambda\ket{+}\bra{+}+(1-\lambda)\ket{0}\bra{0},$
in which the free parameter $\lambda\in[0,1]$ describes to what extent the
switch is ON. The alternative choice of
$\ket{\psi}_{C}=\sqrt{\frac{\lambda}{2}}\ket{0}+\sqrt{1-\frac{\lambda}{2}}\ket{1}$
yields qualitatively similar results (Appendix A9).

Model—We now describe the setting, depicted in Figure 1. There is a two-level
system $M$ which carries information about a system $A$. $M$ will be
thermalized via unitary interactions with the $E_{i}$ qubits from the heat
bath. We take the Hamiltonians to be $H_{E_{j}}=-\sigma_{z}\,\forall j$, and
$H_{M}=-\vec{n}\cdot\vec{\sigma}$, where $\vec{n}$ is a three-dimensional unit
vector and $\sigma$ indicates Pauli-matrices. The environmental qubits $E_{i}$
are initially in their thermal states
$\tau_{E_{i}}=q\ket{0}_{E_{i}}\bra{0}+(1-q)\ket{1}_{E_{i}}\bra{1}$ with
$q:=e^{\frac{1}{kT}}/Z\in[1/2,1]$. The compound system $AM$ is initially
uncorrelated with $E$ and is, for simplicity, in the state:

$\rho^{\textrm{in}}_{AM}=p\rho_{A}^{(0)}\otimes\ket{\phi}_{M}\bra{\phi}+(1-p)\rho_{A}^{(1)}\otimes\ket{\phi^{\perp}}_{M}\bra{\phi^{\perp}},$
(5)

where $\ket{\phi},\ket{\phi^{\perp}}$ denote the eigenstates of $H_{M}$.

To determine the dynamics, we prove (see Appendix A2), inspired by Ref. [31],
that the energy conservation condition

$\displaystyle[U_{ME_{j}},H_{M}+H_{E_{j}}]=0,$ (6)

gives a specific class of unitary evolutions for qubit collisions:

$\displaystyle
U_{ME_{j}}(\theta_{1},\theta_{2})=(\tilde{U}\otimes\mathbb{I})e^{i\theta_{1}S}e^{i\theta_{2}\frac{\sigma_{z}\otimes\sigma_{z}}{2}}(\tilde{U}^{\dagger}\otimes\mathbb{I}),$
(7)

where $\tilde{U}$ is a one-qubit unitary such that
$\tilde{U}\sigma_{z}\tilde{U}^{\dagger}=\vec{n}\cdot\vec{\sigma}$ and the swap
operator $S(\ket{a}\ket{b})=\ket{b}\ket{a}$. The associated Kraus operators on
$M$ do not commute (see Appendix A3). Since $U(\theta):=e^{i\theta
S}=\cos(\theta)\mathbb{I}+i\sin(\theta)S$, we see $\sin(\theta):=s\in[0,1]$
quantifies the thermalisation strength (and denote $\cos(\theta):=c)$.

The overall interaction, as depicted in Figure 1, is then the switched version
of the unitaries $U_{ME_{1}}(\theta)$ and $U_{ME_{2}}(\theta)$:

$\displaystyle L\\!=\\!\ket{\psi}_{c}\bra{\psi}\otimes
U_{ME_{2}}U_{ME_{1}}\\!+\\!\ket{\psi^{\perp}}_{c}\bra{\psi^{\perp}}\otimes
U_{ME_{1}}U_{ME_{2}},$ (8)

where $\ket{\psi},\ket{\psi^{\perp}}$ are eigenvectors of $H_{C}$. $L$ is
unitary and energy-preserving (see Appendix A4).

Bound on information capacity from thermodynamics— We now derive an upper
bound on the mutual information of channels that are Gibb’s-preserving and
energy-preserving. For classical channels the mutual information $I(A:B)$
between an input-record A and output B (optimized over inputs $A$) can be
interpreted as the optimal communication rate, termed the channel’s
information capacity [20]. The quantum mutual information of the input-record
and the output of a quantum channel, $I(A:B):=S(A)+S(B)-S(AB)$, can be
interpreted similarly [7]. Quantum channels followed by measurements induce
classical channels whose optimized mutual information (the so-called
accessible information of the quantum channel [7]) is upper bounded by the
optimized pre-measurement $I(A:B)$. Thus our bound on $I(A:B)$ can be
interpreted as a bound on the classical information capacity.

######  Theorem 1 (Bound on information capacity from thermodynamics).

Consider the mutual information between the message system $M$ and a record of
the message $A$. Initially it is $I(A:M^{\textrm{in}})$ and finally, after the
quantum switched energy-preserving and Gibbs-preserving channels, it is
$I(A:CM^{\textrm{fin}})$, where $C$ is the control system. Then

$\displaystyle I(A:CM^{\textrm{fin}})\leq$
$\displaystyle(c^{4}+\lambda\left(q^{2}-q+\frac{1}{2}\right)s^{4})I(A:M^{\textrm{in}})$
$\displaystyle+\lambda G_{\geq 0},$ (9)

where $\lambda\in[0,1]$ represents how much the switch is ON, $s$ is the
thermalisation strength ($s^{2}+c^{2}=1$) and $q$ is the ground state
probability of the environment thermal state. $G_{\geq 0}$ is a non-negative
function of $s$ and $q$ given in Eq. (A5) of Appendix A5.

Before describing the proof, we note that the bound captures that a certain
amount of increase in mutual information is possible from from the switch
being ON: the quantities multiplied by $\lambda$. Moreover only a limited
amount of increase is allowed from the switch. Two extreme temperature cases
illustrate this: (i) if the temperature of the heat bath is infinite
($q=1/2$),

$I(A:CM^{\text{fin}})\leq(c^{4}+\frac{\lambda}{4}s^{4})I(A:M^{\text{in }}),$
(10)

and (ii) if the temperature of the heat bath is zero ($q=1$),

$I(A:CM^{\text{fin}})\leq(c^{4}+\frac{\lambda}{2}s^{4})I(A:M^{\text{in}})+\lambda
G_{\geq 0}.$ (11)

It follows, for example, that for for infinite temperature, energy conserving
Gibbs-preserving channels can at most have an increase
$\frac{\lambda}{4}s^{4}I(A:M^{\text{in }})\leq\frac{1}{4}I(A:M^{\text{in }})$.
We moreover see that the low temperature allows for more information capacity
as well as a greater difference between the cases where the switch is ON and
OFF. This is consistent with how for the infinite temperature case only,
$I(A:C)=0$ (as per Eq. (A5) in the Appendix). The bound is tight for $s=0$
(see Figure 2). An example of a pair of channels–one of which is not energy-
preserving–violating the bound is given in Appendix A6, demonstrating that the
bound adds a significant restriction.

Figure 2: We derived, from the first and second laws, a general bound on the
final mutual information between a copy of the input message and the output
after the switch $I(A:CM^{\textrm{fin}})$. Here we show the bound together
with a curve that is realizable for two different temperatures. $s\in[0,1]$ is
the thermalization strength and $\lambda\in[0,1]$ the extent to which the
switch is ON. The red circle proves that low temperature can outperform high
temperature in terms of mutual information and the blue circle illustrates the
increased mutual info when the switch is ON.

To illustrate the approach to proving Theorem 1 (see Appendix A5 for the full
proof), consider the case given in [8] corresponding to an infinite
temperature heat bath with the quantum switch fully ON ($q=1/2,\lambda=1$).
Firstly apply the dynamics given in Eq. (8) to the initial state
$\sigma_{C}\otimes\rho^{\textrm{in}}_{AM}$. The final state
$\rho^{\textrm{fin}}_{CAM}$ then turns out to obey

$\displaystyle\rho^{\textrm{fin}}_{CAM}=$
$\displaystyle\ket{-}_{C}\bra{-}\otimes\left(k\left(\sigma_{x}^{M}\rho^{\textrm{in}}_{AM}\sigma_{x}^{M}\right)+l\left(\rho^{\textrm{in}}_{A}\otimes\frac{\mathbb{I}_{M}}{2}\right)\right)$
$\displaystyle+$ $\displaystyle\ket{+}_{C}\bra{+}\otimes\left(m(\rho^{\text{in
}}_{AM})+n(\rho^{\textrm{in}}_{A}\otimes\tau_{M})\right),$ (12)

where $\rho_{A}^{\textrm{in}}:=\textrm{Tr}_{M}\rho^{\textrm{in}}_{AM}$ and
$k$,$l$,$m$ and $n$ are all non-negative real numbers. This makes the state
amenable to conditioning on $C$, such that one can employ the mutual
information (chain) rule

$\displaystyle I(A:CM^{\textrm{fin}})=I(A:C)+I(A:M^{\textrm{fin}}|C).$ (13)

This breaks the task into calculating $I(A:C)$ as well as
$I(A:M^{\textrm{fin}}|C)=p(\ket{+}_{C})I(A:M^{\textrm{fin}}|\ket{+}_{C})+p(\ket{-}_{C})I(A:M^{\textrm{fin}}|\ket{-}_{C})$.
The reduced states in question turn out to be mixtures of product states with
another state with the same reduced state on A, i.e.
$\rho^{\text{fin}}_{AB}=p\rho^{\text{in}}_{AB}+(1-p)\rho^{\text{in}}_{A}\otimes\sigma_{B}$.
We show (via the data processing inequality) that for such states

$\displaystyle I(A:B^{\text{fin}})\leq pI(A:B^{\text{in}}).$ (14)

We use Eq. (14) to upper bound the two terms in the RHS of Eq. (13). (The
argument details, including the extension to arbitrary temperature of the heat
bath and $\lambda\in[0,1]$, are given in Appendix A5).

Information capacity increase consumes a thermodynamical resource—We now argue
that $\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})(\cdot)$, where
$\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are both Gibbs-preserving channels
leaving the same thermal state invariant, should not be regarded as
thermodynamically free, except for the case of $\sigma_{C}=\tau_{C}$, the
thermal state of $C$ for the ambient temperature. To formalise this argument
we firstly consider the impact of $\mathcal{S}_{\sigma_{C}}$ on $M$, before
including the impact on $C$.

######  Theorem 2 (Switched Gibbs-preserving channels is a Gibbs-preserving
channel on $M$).

If $\mathcal{C}_{1}(\tau_{M})=\mathcal{C}_{2}(\tau_{M})=\tau_{M}$, where
$\tau_{M}$ is the thermal state on $M$ then, for any state $\sigma_{C}$,
$Tr_{C}\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})(\tau_{M})=\tau_{M}$.

The proof of theorem 2 can be found in Appendix A7. A similar claim was made
independently in Ref. [27].

Theorem 2 leaves it open for $\mathcal{S}_{\sigma_{C}}$ to be viewed as a
thermodynamically free operation. However, it is important to also take the
presence of $C$ into account.

######  Theorem 3 (Including the control system).

(i) For the control system initially in a thermal state $\tau_{C}$, and
$\mathcal{C}_{1},\mathcal{C}_{2}$ being Gibbs-preserving channels,
$\mathcal{S}_{\tau_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})$ leaves
$\tau_{C}\otimes\tau_{M}$ invariant. (ii) For the control system initially in
$\ket{+}\bra{+}$, and $\mathcal{C}_{1},\mathcal{C}_{2}$ being Gibbs-preserving
channels for infinite temperature,
$\mathcal{S}_{\ket{+}}(\mathcal{C}_{1},\mathcal{C}_{2})$ followed by a
particular free operation does not leave the thermal state on $M$, $\tau_{M}$
invariant.

Statement (i) in Theorem 3 follows because $\tau_{C}\otimes\tau_{M}$ can be
interpreted as a mixture of two cases $\sigma_{C}=\ket{0/1}\bra{0/1}$. By
inspection, from the definition of the switch $\mathcal{S}$, in each case
$\sigma_{C}$ is unchanged by the switch. Moreover the thermal state on $M$ is
unchanged since only Gibbs-preserving channels are applied to $M$. Hence it
appears fair to call
$\mathcal{S}_{\tau_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})(\cdot)$ a
thermodynamically free operation, at least in so far as Eq.(1) is demanded.

If the control system is instead in $\ket{+}$ as in statement (ii), things are
subtly but importantly different. Consider for simplicity the case of infinite
temperature (or null Hamiltonians) and full swap ($s=1$). Whilst Theorem 2
implies $\tau_{M}$ is invariant it is a priori not sensible to ask whether
$\tau_{C}\otimes\tau_{M}$ is invariant in this case, since
$\sigma_{C}\neq\tau_{C}$. However we can answer the question of whether it is
a free operation in another way: acting on the final $CM$ with a certain free
operation changes $M$ into something that is different to $\tau_{M}$. The swap
unitary $S_{CM}$ should be treated as free here, as discussed in the model
section. Now,

$\displaystyle\textrm{Tr}_{C}(S_{CM}\circ\mathcal{S}_{\ket{+}}(\mathcal{C}_{1},\mathcal{C}_{2})(\tau_{M}))$
$\displaystyle=\frac{3}{8}\ket{+}_{M}\bra{+}+\frac{5}{8}\ket{-}_{M}\bra{-}\neq\frac{1}{2}\mathbbm{1}=\tau_{M},$
(15)

as shown in Eq. (A5). Thus the switched Gibbs-preserving channels under such
an initial state on the control system should not be termed a free thermal
operation (else one would need to declare a violation of the Second Law of Eq.
(1)).

The above argument is analogous to Bennett’s famous exorcism of the Szilard-
engine Maxwell’s demon [32]. Bennett’s argument contains two (qu)bits
(sometimes one of them is a trit but for simplicity we take two qubits): the
demon’s memory $A$ and the working medium system $B$. Initially
$\rho_{AB}=\ket{0}_{A}\bra{0}\otimes\mathbbm{1}_{B}/2$. The demon performs its
famous measurement modelled as a
$\mathrm{CNOT}=\mathbbm{1}_{A}\otimes\ket{0}_{B}\bra{0}+(\sigma_{x})_{A}\otimes\ket{1}_{B}\bra{1}$.
Bennett notes, crucially[32], that the CNOT operation may in principle be
thermodynamically free, considering that the initial and final Hamiltonian may
be null ($H=0$), and noting that CNOT leaves the corresponding thermal state
($\frac{1}{4}\mathbbm{1}_{AB}$) invariant. There can now, analogously to the
swap unitary above, be a second free operation, namely CNOT in the opposite
direction, setting $B$ to $\ket{0}$ regardless of its initial state. Thus, the
end effect is a map on $B$ respecting
$\varepsilon(\rho_{B}=\frac{1}{2}\mathbbm{1}=\tau_{B})=\ket{0}_{B}\bra{0}\neq\tau_{B}$.
Such a map creates free energy, e.g. taking a thermal state to a non-thermal
state, and should be treated as a resource.

The switch can be said to harness only one type of free energy from
$\sigma_{C}$, known as the free energy of coherence. If $\sigma_{C}$ is
diagonal in the computational basis, then, as discussed above, the action of
the switch is trivial. Instead there must be coherence. The free energy of
coherence can be defined as $F_{coh}=F(\sigma)-F(\mathcal{D}_{H}(\sigma))$,
where $\mathcal{D}_{H}(\cdot)$ diagonalises the state in the Hamiltonian
eigenbasis [33]. From the definition, $F=F_{coh}+F(\mathcal{D}_{H}(\sigma))$.
$F_{coh}(\sigma_{C})$ is a natural quantifier of the thermodynamic resource
value of the switch. We can explicitly mathematically capture this by noting
that $F_{coh}(\sigma,H)\geq kT\frac{\lambda^{2}}{\ln 16}$ (see Appendix A8)
such that the information capacity activation in Theorem 1 is directly
governed by $F_{coh}$.

Relation to existing work–The conclusion that the switch is a thermodynamical
resource does not contradict the resource theory created in Ref. [15], in
which the switch is taken as free, but rather shows that these are separate
resource theories. The result that the switch is not thermodynamically free is
consistent with the conclusion from Ref. [21] that it could be possible to use
the switch to perform refrigeration. The results are also consistent with Ref.
[9] which show how the switch gives an apparent violation of the bottleneck
inequality, which holds for all channels. The bottleneck inequality (which is
equivalent to the data processing inequality) for two maps
$\varepsilon_{1}(\cdot)$ and $\varepsilon_{2}(\cdot)$ says that
$I(A:\varepsilon_{2}\circ\varepsilon_{1}(B))\leq I(A:\varepsilon_{1}(B))$. The
information capacity activation phenomenon can be expressed as
$I(A:\mathcal{S}_{\ket{+}}(\mathcal{C}_{1},\mathcal{C}_{2})(B))>I(A:\mathcal{C}_{1}(B))$,
indicating a possible violation of the bottle-neck inequality. As a vivid
example, let $\mathcal{C}_{2}=\mathcal{C}_{1}:=\mathcal{C}$ correspond to the
full swap case (s=1) such that $\mathcal{C}(\tau_{M})=\tau_{M}$. If we
denoted, carelessly, the action on $M$ of the switch channels followed by the
conditioning on $C$, as $\mathcal{C}\circ\mathcal{C}$, there would indeed be a
violation of Eq. (1) in the sense that in the second application of
$\mathcal{C}$ the state goes from being thermal to becoming athermal. This
again highlights how the switch should be treated as a thermal resource.

Summary and Outlook– We derived an upper bound on the information capacity of
energy-preserving Gibbs-preserving channels with or without indefinite causal
order. The bound captures the capacity activation associated with indefinite
causal order as well as revealing that this is stronger for low temperatures.
We argued that there is a thermodynamic cost to this activation (or else the
Second Law is violated).

It remains to be investigated whether the arguments here directly generalise
to single-shot thermodynamics where statements concern other quantities than
average work, e.g. the worst-case work [34, 35, 3, 36]. It is intriguing that
only a limited activation capacity from the switch is allowed for Gibbs-
preserving channels, deserving further investigation. Our approach may help
clarify what the difference is between the switch and other coherent
combinations of channels [37, 38, 39, 26, 40, 24, 25, 41, 42]. Finally, it
would be intriguing to extend our upper bound to the case of an environment
with memory, e.g. via the approach of Ref. [43].

Acknowledgements– We acknowledge discussions with Qian Chen, Giulio
Chiribella, Yutong Luo, Yi-Zheng Zhen and early discussions with David Felce,
Sina Salek and Vlatko Vedral. We acknowledge support from the National Natural
Science Foundation of China (Grants No. 12050410246, No. 1200509, No.
12050410245).

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##  Appendix A1 Thermal state preserving condition and the second law

A thermal channel $\mathcal{E}$ is a completely positive, trace-preserving
(CPTP) map that leaves the thermal state in question invariant.

Proposition:  The thermal state preserving condition, i.e.,
$\mathcal{E}(\tau)=\tau$, is equivalent to the condition that the relative
entropy between the state $\rho$ and the thermal state $\tau$ of the same
system cannot increase, i.e., $\Delta
S(\rho||\tau):=S(\mathcal{E}(\rho)||\tau)-S(\rho||\tau)\leq 0$ .

###### Proof.

$\Rightarrow$. According to the property of monotonicity of quantum relative
entropy under CPTP maps, we have

$\displaystyle S(\rho||\tau)\geq
S(\mathcal{E}(\rho)||\mathcal{E}(\tau))=S(\mathcal{E}(\rho)||\tau),$ (16)

where the thermal state preserving condition $\mathcal{E}(\tau)=\tau$ was used
in the equality. This leads to

$\Delta S(\rho||\tau)\leq 0.$

$\l\Leftarrow$. Consider initially $\rho=\tau$. Then, the change of the
relative entropy is

$\displaystyle\Delta S(\tau||\tau)$
$\displaystyle=S(\mathcal{E}(\tau)||\tau)-S(\tau||\tau)$
$\displaystyle=S(\mathcal{E}(\tau)||\tau)\leq 0.$ (17)

Combining the above expression with the fact that relative entropy is non-
negative gives us $S(\mathcal{E}(\tau)||\tau)=0.$ Finally, we arrive at
$\mathcal{E}(\tau)=\tau$.

∎

##  Appendix A2 Energy conservation gives an explicit parametrised two-qubit
unitary collision model

Consider a qubit collision model. Without loss of generality, we assume that
$H_{E_{i}}=-\sigma_{z}$ and $H_{s}=-\vec{n}\cdot\vec{\sigma}$, and that there
exists a one-qubit unitary transformation $\tilde{U}$, such that

$\displaystyle\tilde{U}:\,\,$
$\displaystyle\\{\ket{0},\ket{1}\\}\rightarrow\\{\ket{\phi}:=\tilde{U}\ket{0},\ket{\phi^{\perp}}:=\tilde{U}\ket{1}\\},$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\sigma_{z}\,\,\,\,\,\,\,\rightarrow\tilde{U}\sigma_{z}\tilde{U}^{\dagger}=\vec{n}\cdot\vec{\sigma}.$
(18)

That is to say that $\ket{\phi},\ket{\phi^{\perp}}$ are the eigenstates of
$H_{s}$ with eigenvalues -1 and 1, respectively.

######  Lemma 1.

For unitary collisions $U$ between qubits $M$ and $E_{j}$ obeying the energy
conservation condition

$\displaystyle[U_{ME_{j}},H_{M}+H_{E_{j}}]=0,$ (19)

it must be that $U_{ME_{j}}$ has the following parametrised form:

$\displaystyle
U_{ME_{j}}(\theta_{1},\theta_{2})=(\tilde{U}\otimes\mathbb{I})e^{i\theta_{1}S}e^{i\theta_{2}\frac{\sigma_{z}\otimes\sigma_{z}}{2}}(\tilde{U}^{\dagger}\otimes\mathbb{I}),$
(20)

where $\tilde{U}$ is a one-qubit unitary such that
$\tilde{U}\sigma_{z}\tilde{U}^{\dagger}=\vec{n}\cdot\vec{\sigma}$ and the swap
operator
$S(\ket{\phi_{1}}\otimes\ket{\phi_{2}})=\ket{\phi_{2}}\otimes\ket{\phi_{1}}\,\forall\ket{\phi_{1}},\ket{\phi_{2}}\in\mathbb{C}^{2}$.

###### Proof.

First, we prove that, for qubit collisions, the energy conservation condition
is equivalent to the global Gibbs state preserving condition, $i.e.,$

$U(\tau_{s}\otimes\gamma_{E_{i}})U^{\dagger}=\tau_{s}\otimes\gamma_{E_{i}}.$

Denote $a:=e^{{\beta}}/(e^{{\beta}}+e^{{-\beta}})$.

$\Rightarrow$. The proof is trivial for this direction.

$\Leftarrow$. The global Gibbs state preserving condition gives

$\displaystyle U(\gamma_{A}\otimes\tau_{E_{i}})U^{\dagger}$ $\displaystyle=$
$\displaystyle
U\Bigl{(}(a\ket{\phi}\bra{\phi}+(1-a)\ket{\phi^{\perp}}\bra{\phi^{\perp}})\otimes$
$\displaystyle(a\ket{0}\bra{0}+(1-a)\ket{1}\bra{1})\Bigl{)}U^{\dagger}$
$\displaystyle=$ $\displaystyle\gamma_{A}\otimes\tau_{E_{i}}.$ (21)

Demanding that Eq. (A2) holds for all temperatures implies that the unitary
interactions $U$ must preserve three subspaces:
$\ket{\phi,0}\bra{\phi,0},\ket{\phi^{\perp},1}\bra{\phi^{\perp},1}$ and
$\ket{\phi,1}\bra{\phi^{\perp},0}+\ket{\phi^{\perp},0}\bra{\phi,1}$. One sees
that each subspace consists of the states with the same energy, which means
that $U$ preserves energy, $i.e,[U,H_{s}+H_{E_{i}}]=0$ holds.

Next, we use the result from Ref [31], which gives an explicit parametrised
form of two-qubit unitaries obeying that global Gibbs state preserving
condition.

Denote $S$ as the swap operator, $i.e.,$

$\displaystyle S=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&0&1&0\\\ 0&1&0&0\\\
0&0&0&1\\\ \end{array}\right).$ (26)

The general form of Gibbs state preserving unitaries [31], up to a global
phase factor, is
$U(\theta_{1},\theta_{2}):=U_{partialsw}(\theta_{1})U_{phasefluct}(\theta_{2})$
where

$\displaystyle
U_{partialsw}(\theta_{1})=(\tilde{U}\otimes\mathbb{I})e^{i\theta_{1}S}(\tilde{U}^{\dagger}\otimes\mathbb{I})$
$\displaystyle=$ $\displaystyle
c\mathbb{I}+is\Bigl{(}\ket{\phi,0}\bra{\phi,0}+\ket{\phi^{\perp},1}\bra{\phi^{\perp},1}+\ket{\phi,1}\bra{\phi^{\perp},0}$
$\displaystyle+\ket{\phi^{\perp},0}\bra{\phi,1}\Bigl{)},$ $\displaystyle
U_{phasefluct}(\theta_{2})=(\tilde{U}\otimes\mathbb{I})e^{i\theta_{2}\frac{\sigma_{z}\otimes\sigma_{z}}{2}}(\tilde{U}^{\dagger}\otimes\mathbb{I})$
$\displaystyle=$ $\displaystyle
e^{i\theta_{2}}\ket{\phi,0}\bra{\phi,0}+e^{-i\theta_{2}}\ket{\phi,1}\bra{\phi,1}$
$\displaystyle+e^{-i\theta_{2}}\ket{\phi^{\perp},0}\bra{\phi^{\perp},0}+e^{i\theta_{2}}\ket{\phi^{\perp},1}\bra{\phi^{\perp},1}.$
(27)

It is straightforward to verify that
$[U_{partialsw}(\theta_{1}),U_{phasefluct}(\theta_{2})]=0$. ∎

##  Appendix A3 Kraus decomposition of the quantum partial swap unitary

Consider a unitary interaction $U_{ME}(\theta):=e^{i\theta
S}=(c\mathbb{I}+isS)$ where $S$ is the swap operator. If the heat bath is in
the state $P_{0}=\ket{0}_{E}\bra{0}$, one has

$\displaystyle\tilde{E}_{0}=_{E}\bra{0}U\ket{0}_{E}=(c\mathbbm{1}+is\ket{0}\bra{0}),$
$\displaystyle\tilde{E}_{1}=_{E}\bra{1}U\ket{0}_{E}=is\ket{0}\bra{1}.$

If the heat bath in the state $P_{1}=\ket{1}_{E}\bra{1}$, one has

$\displaystyle\tilde{E}^{\prime}_{0}=_{E}\bra{0}U\ket{1}_{E}=is\ket{1}\bra{0}),$
$\displaystyle\tilde{E}^{\prime}_{1}=_{E}\bra{1}U\ket{1}_{E}=(c\mathbbm{1}+is\ket{1}\bra{1}).$

Hence, if the heat bath is in the state of
$\tau=q\ket{0}\bra{0}+(1-q)\ket{1}\bra{1}$, we have the set of Kraus operators

$\displaystyle K_{1}\equiv\sqrt{q}\tilde{E}_{0}=$
$\displaystyle\sqrt{q}(c\mathbbm{1}+is\ket{0}\bra{0}),$ $\displaystyle
K_{2}\equiv\sqrt{q}\tilde{E}_{1}=$ $\displaystyle\sqrt{q}is\ket{0}\bra{1},$
$\displaystyle K_{3}\equiv\sqrt{1-q}\tilde{E}^{\prime}_{0}=$
$\displaystyle\sqrt{1-q}is\ket{1}\bra{0},$ $\displaystyle
K_{4}\equiv\sqrt{1-q}\tilde{E}^{\prime}_{1}=$
$\displaystyle\sqrt{1-q}(c\mathbbm{1}+is\ket{1}\bra{1}).$ (28)

##  Appendix A4  Unitary dilation of control-system interaction is energy
preserving

Consider two thermal operations $\mathcal{E}_{1},\mathcal{E}_{2}$ acting on
the system $S$. $\mathcal{E}_{i},i=1,2$ are defined by $(U_{i},\tau_{E_{i}})$
where $U_{i}$ is a unitary dilation of $\mathcal{E}_{i}$ satisfying

$\displaystyle[U_{i},H_{S}+H_{E_{i}}]=0,$ (29)

and $\tau_{E_{i}}:=e^{-\beta H_{E_{i}}}/Z_{i}$ is the corresponding thermal
environment state. Further, the energy conservation condition ensures global
Gibbs state conservation, i.e.,

$\displaystyle U_{i}(e^{-\beta H_{S}}\otimes e^{-\beta
H_{E_{i}}})U_{i}^{\dagger}=e^{-\beta H_{S}}\otimes e^{-\beta H_{E_{i}}}.$ (30)

Then, tracing out $E_{i}$, one arrives at $\mathcal{E}_{i}(e^{-\beta
H_{S}})=e^{-\beta H_{S}}.$

Let $C$ be the system controlling the orders of operations. Denote by
$\ket{\psi},\ket{\psi^{\perp}}$ the two different eigenvectors of the
Hamiltonian $H_{C}$ of the control system $C$. We then construct a unitary

$\displaystyle L=\ket{\psi}_{C}\bra{\psi}\otimes
U_{2}U_{1}+\ket{\psi^{\perp}}_{C}\bra{\psi^{\perp}}\otimes U_{1}U_{2}.$ (31)

One can verify that,

$\displaystyle[L,H_{C}+H_{S}+H_{E_{1}}+H_{E_{2}}]$ $\displaystyle=$
$\displaystyle\ket{\psi}_{C}\bra{\psi}\otimes[U_{2}U_{1},H_{S}+H_{E_{1}}+H_{E_{2}}]$
$\displaystyle+\ket{\psi^{\perp}}_{C}\bra{\psi^{\perp}}\otimes[U_{1}U_{2},H_{S}+H_{E_{1}}+H_{E_{2}}]$
$\displaystyle=$
$\displaystyle\ket{\psi}_{C}\bra{\psi}\otimes\Bigl{(}U_{2}[U_{1},H_{S}+H_{E_{1}}]+[U_{2},H_{S}$
$\displaystyle+H_{E_{2}}]U_{1}\Bigl{)}+\ket{\psi^{\perp}}_{C}\bra{\psi^{\perp}}\otimes\Bigl{(}U_{1}[U_{2},H_{S}$
$\displaystyle+H_{E_{2}}]+[U_{1},H_{S}+H_{E_{1}}]U_{2}\Bigl{)}$
$\displaystyle=$ $\displaystyle 0,$ (32)

where energy conservation conditions are used in last equality. Thus, $L$ is
an energy preserving unitary on the compound system $CSE_{1}E_{2}$.

We note for completeness that it follows that

$\displaystyle L(e^{-\beta H_{C}}\otimes e^{-\beta H_{S}}\otimes e^{-\beta
H_{E_{1}}}\otimes e^{-\beta H_{E_{2}}})L^{\dagger}$ $\displaystyle=$
$\displaystyle e^{-\beta H_{C}}\otimes e^{-\beta H_{S}}\otimes e^{-\beta
H_{E_{1}}}\otimes e^{-\beta H_{E_{2}}}.$ (33)

##  Appendix A5 Bound on information capacity from thermodynamics

Theorem 1. For our model of two identical energy preserving thermal channels
$\\{\mathcal{C}\\}$ undergoing a quantum switch, with $\lambda\in[0,1]$
representing the switch strength, $s$ the thermalisation strength, $q$ the
probability of the ground state of the environment qubit thermal states, the
final (fin) and initial (in) mutual informations between the untouched quantum
system $A$ on the one hand and the control system $C$ plus the memory $M$
undergoing the channel on the other, obey

$\displaystyle I(A:CM^{\textrm{fin}})\leq$
$\displaystyle\Bigl{(}c^{4}+\lambda(q^{2}-q+\frac{1}{2})s^{4}\Bigl{)}I(A:M^{\textrm{in}})$
$\displaystyle+\lambda G_{\geq 0},$ (34)

where $G_{\geq
0}:=fI(A:D_{0})+(1-f)I(A:D_{1})$,$f=1-\frac{(1-q^{2})+(2q-1)(1-p)}{2}s^{4}$
and the non-negative terms

$\displaystyle I(A:D_{0})=$ $\displaystyle
S(p_{0}\rho_{A}+(1-p_{0})\rho^{(0)}_{A})-p_{0}S(\rho_{A})$
$\displaystyle-(1-p_{0})S(\rho^{(0)}_{A}),$ $\displaystyle I(A:D_{1})=$
$\displaystyle S(p_{1}\rho_{A}+(1-p_{1})\rho^{(1)}_{A})-p_{1}S(\rho_{A})$
$\displaystyle-(1-p_{1})S(\rho^{(1)}_{A}),$ (35)

with
$p_{0}:=(1+\frac{(q^{2}-2q)s^{4}}{2})/f,p_{1}=\frac{(1-q^{2})s^{4}}{2}/(1-f)$,
which involves virtual ancillary qubits $D_{0}$ and $D_{1}$ that are fully
determined by the initial conditions.

###### Proof.

In our model, the heat bath’s Hamiltonian $H_{E_{j}}=-\sigma_{z}\,\forall j$,
and memory system $M$’s Hamiltonian $H_{M}=-\vec{n}\cdot\vec{\sigma}$, where
$\vec{n}$ is a three-dimensional unit vector and $\sigma$ indicates Pauli-
matrices. Without loss of generality, we assume the Hamiltonian of the two-
dimensional control system $H_{c}=-\sigma_{z}$. The energy eigenstates of the
control system are $\\{\ket{0},\ket{1}\\}$. The initial state of the joint
system $CAME_{1}E_{2}$ is

$\displaystyle\tilde{U}_{M}^{\dagger}\rho^{\textrm{in}}\tilde{U}_{M}$
$\displaystyle=$
$\displaystyle\sigma_{C}\otimes\tilde{U}_{M}^{\dagger}\rho^{\text{in}}_{AM}\tilde{U}_{M}\otimes\tau_{E_{1}}\otimes\tau_{E_{2}}$
$\displaystyle=$
$\displaystyle\Bigl{(}\lambda\ket{+}_{c}\bra{+}+(1-\lambda)\ket{0}_{c}\bra{0}\Bigl{)}\otimes\Bigl{(}p\rho^{(0)}_{A}\otimes\ket{0}_{M}\bra{0}$
$\displaystyle+(1-p)\rho^{(1)}_{A}\otimes\ket{1}_{M}\bra{1}\Bigl{)}\otimes\tau_{E_{1}}\otimes\tau_{E_{2}},$
(36)

where $\tilde{U}\sigma_{z}\tilde{U}^{\dagger}=\vec{n}\cdot\vec{\sigma}$ .

According to lemma 1, the energy conservation condition of thermal channels
$\\{\mathcal{C}\\}$ gives a parametrized form of the two-qubit unitary
dilation:

$\displaystyle U_{ME_{i}}(\alpha,\beta)$
$\displaystyle=(\tilde{U}\otimes\mathbb{I})e^{i\alpha
S}e^{i\beta\frac{\sigma_{z}\otimes\sigma_{z}}{2}}(\tilde{U}^{\dagger}\otimes\mathbb{I})$
$\displaystyle:=(\tilde{U}\otimes\mathbb{I})U_{i}(\tilde{U}^{\dagger}\otimes\mathbb{I}),$

where $S$ is the swap operator, i.e.,
$S\ket{\phi_{1}}\otimes\ket{\phi_{2}}=\ket{\phi_{2}}\otimes\ket{\phi_{1}}\,\forall\ket{\phi_{1}},\ket{\phi_{2}}\in\mathbb{C}^{2}$.
For the joint system $CAME_{1}E_{2}$, the corresponding Kraus operator of
$\mathcal{S}(U_{ME_{1}},U_{ME_{2}})$ is

$\displaystyle L=$ $\displaystyle\ket{0}_{c}\bra{0}\otimes
U_{ME_{2}}U_{ME_{1}}+\ket{1}_{c}\bra{1}\otimes U_{ME_{1}}U_{ME_{2}}$
$\displaystyle=$ $\displaystyle\tilde{U}(\ket{0}_{c}\bra{0}\otimes
U_{2}U_{1}+\ket{1}_{c}\bra{1}\otimes U_{1}U_{2})\tilde{U}^{\dagger}$
$\displaystyle:=$ $\displaystyle\tilde{U}\tilde{L}\tilde{U}^{\dagger}.$ (37)

Then, the final state of $CAME_{1}E_{2}$ is

$\displaystyle\rho^{\text{fin}}$
$\displaystyle=L\rho^{\text{{in}}}L^{\dagger}=\tilde{U}\tilde{L}\tilde{U}^{\dagger}\rho^{\textrm{in}}\tilde{U}\tilde{L}\tilde{U}^{\dagger}.$
(38)

Note that $\tilde{L}\tilde{U}^{\dagger}\rho^{\textrm{in}}\tilde{U}\tilde{L}$
is equivalent to the final state of $CAME_{1}E_{2}$ when system $M$’s
Hamiltonian $H_{M}=-\sigma_{z}$. Since we are interested in the mutual
information between $A$ and $CM$ and local unitaries do not change mutual
information, we restrict ourselves to the case of $H_{M}=-\sigma_{z}$ in the
following. Moreover, $[e^{i\alpha
S},e^{i\beta\frac{\sigma_{z}\otimes\sigma_{z}}{2}}]=0$, $M$ and $E_{i}$ are
initially diagonal in the computational basis. This implies that
$e^{i\beta\frac{\sigma_{z}\otimes\sigma_{z}}{2}}$ does not impact the final
states. Thus, we only need to consider $U_{i}=e^{i\theta S_{ME_{i}}},i=1,2,$
in the following.

The Kraus operators of energy-preserving thermal channels $\mathcal{C}$ with a
unitary dilation $U=e^{i\theta S}$ and thermal state $\tau$ are given in
Appendix A3. The corresponding Kraus Operators for the resulting switched
channels $\mathcal{S}_{\sigma_{C}}(\mathcal{C},\mathcal{C})$ are

$\displaystyle S_{ij}=$
$\displaystyle\ket{0}_{C}\bra{0}\otimes(\mathbb{I}^{A}\otimes
K_{i}^{M})(\mathbb{I}^{A}\otimes K_{j}^{M})$
$\displaystyle+\ket{1}_{C}\bra{1}\otimes(\mathbb{I}^{A}\otimes
K_{j}^{M})(\mathbb{I}^{A}\otimes K_{i}^{M}).$ (39)

The final state of $CAM$ can be obtained by

$\displaystyle\rho_{CMA}^{\textrm{fin}}=\textrm{Tr}_{E_{1}E_{2}}\rho^{\textrm{fin}}=\mathcal{S}_{\sigma_{C}}(\mathcal{E},\mathcal{E})(\rho_{CAM}^{\textrm{in}}).$
(40)

Consider $\lambda=1$. Based on Eq. (70), Eq. (A3) and Eq. (40), we further
find that the final state of $CAM$ has the structure of

$\displaystyle\rho^{\text{ {fin}}}_{CMA}$ $\displaystyle=$
$\displaystyle\ket{+0}_{CM}\bra{+0}\otimes\rho^{(1^{\prime})}_{A}+\ket{+1}_{CM}\bra{+1}\otimes\rho^{(2^{\prime})}_{A}$
$\displaystyle+\ket{-0}_{CM}\bra{-0}\otimes\rho^{(3^{\prime})}_{A}+\ket{-1}_{CM}\bra{-1}\otimes\rho^{(4^{\prime})}_{A}.$
(41)

where

$\displaystyle\rho^{(1^{\prime})}_{A}=\sum_{ij}\textrm{Tr}_{CM}\ket{+0}_{CM}\bra{+0}\Bigl{[}S_{ij}(\sigma_{C}\otimes\rho_{AM}^{\textrm{in}})S_{ij}^{\dagger}\Bigl{]}$
$\displaystyle=$
$\displaystyle\sum_{ij}\textrm{Tr}_{CM}\ket{+0}_{CM}\bra{+0}S_{ij}\Bigl{(}\ket{+}_{C}\bra{+}\otimes(p\ket{0}_{M}\bra{0}\otimes\rho^{(0)}_{A}$
$\displaystyle+(1-p)\ket{1}_{M}\bra{1}\otimes\rho^{(1)}_{A})\Bigl{)}S_{ij}^{\dagger}$
$\displaystyle=$ $\displaystyle
p_{0}^{1^{\prime}}\rho^{(0)}_{A}+p_{1}^{1^{\prime}}\rho^{(1)}_{A}.$ (42)

Similarly, we have

$\displaystyle\rho^{(2^{\prime})}=$ $\displaystyle
p_{0}^{2^{\prime}}\rho^{(0)}_{A}+p_{1}^{2^{\prime}}\rho^{(1)}_{A},$
$\displaystyle\rho^{(3^{\prime})}=$ $\displaystyle
p_{0}^{3^{\prime}}\rho^{(0)}_{A}+p_{1}^{3^{\prime}}\rho^{(1)}_{A},$
$\displaystyle\rho^{(4^{\prime})}=$ $\displaystyle
p_{0}^{4^{\prime}}\rho^{(0)}_{A}+p_{1}^{4^{\prime}}\rho^{(1)}_{A}.$ (43)

The coefficients $p_{0}^{1^{\prime}},p_{1}^{1^{\prime}}$ are obtained by

$\displaystyle p_{0}^{1^{\prime}}=$
$\displaystyle\sum_{ij}\textrm{Tr}_{CM}\ket{+0}_{CM}\bra{+0}$ $\displaystyle
S_{ij}\Bigl{(}\ket{+}_{C}\bra{+}\otimes(p\ket{0}_{M}\bra{0}\Bigl{)}S^{{\dagger}}_{ji}$
$\displaystyle=$ $\displaystyle p(c^{4}+2c^{2}s^{2}q+\frac{1}{2}s^{4}q(1+q)),$
$\displaystyle p_{1}^{1^{\prime}}=$
$\displaystyle\sum_{ij}\textrm{Tr}_{CM}\ket{+0}_{CM}\bra{+0}$ $\displaystyle
S_{ij}\Bigl{(}\ket{+}_{C}\bra{+}\otimes(1-p)\ket{1}_{M}\bra{1}\Bigl{)}S^{{\dagger}}_{ji}$
$\displaystyle=$ $\displaystyle\frac{1-p}{2}s^{2}(4c^{2}+s^{2})q.$ (44)

Similarly, one obtains the coefficients :

$\displaystyle p_{0}^{2^{\prime}}=$
$\displaystyle\frac{p}{2}s^{2}(4c^{2}+s^{2})(1-q),$ $\displaystyle
p_{1}^{2^{\prime}}=$
$\displaystyle(1-p)\bigl{(}c^{4}+2c^{2}s^{2}(1-q)+\frac{1}{2}s^{4}(2-q)(1-q)\bigl{)},$
$\displaystyle p_{0}^{3^{\prime}}=$ $\displaystyle\frac{p}{2}s^{4}q(1-q),\ \
p_{1}^{3^{\prime}}=\frac{1-p}{2}s^{4}q,$ $\displaystyle p_{0}^{4^{\prime}}=$
$\displaystyle\frac{p}{2}s^{4}(1-q),\ \
p_{1}^{4^{\prime}}=\frac{1-p}{2}s^{4}q(1-q).$ (45)

Substituting Eqs.(A5)-(A5) in Eq.(A5), one can write down the explicit form of
the final state of $CMA$

$\displaystyle\rho^{\lambda=1}_{CMA}$ $\displaystyle=$
$\displaystyle\ket{-}_{C}\bra{-}\otimes
s^{4}\Bigl{(}\frac{(1-q)^{2}}{2}(\sigma_{x})_{M}\rho_{MA}(\sigma_{x})_{M}+q(1-q)$
$\displaystyle\frac{\mathbb{I}_{M}}{2}\otimes\rho_{A}+\frac{2q-1}{2}(1-p)\ket{0}_{M}\bra{0}\otimes\rho^{(1)}_{A}\Bigl{)}$
$\displaystyle+\ket{+}_{C}\bra{+}\otimes\Bigl{(}(c^{4}+\frac{(1-q)^{2}}{2}s^{4})\rho^{\text{in
}}_{MA}+(2c^{2}s^{2}+\frac{s^{4}}{2})$
$\displaystyle\tau_{M}\otimes\rho_{A}+s^{4}\frac{2q-1}{2}p\ket{0}_{M}\bra{0}\otimes\rho^{(0)}_{A}\Bigl{)}.$
(46)

Consider $\lambda\in[0,1]$. It is straightforward to write down the final
state of $CMA$

$\displaystyle\rho^{\textrm{fin}}_{CMA}=$
$\displaystyle\lambda\rho^{\lambda=1}_{CMA}+(1-\lambda)\ket{0}_{c}\bra{0}\otimes(c^{4}\rho_{MA}^{\textrm{in}}$
$\displaystyle+(1-c^{4})\tau_{M}\otimes\rho_{A}).$ (47)

Adding an ancilla system $B$, we have

$\displaystyle\rho^{\textrm{fin}}_{BCMA}$ $\displaystyle=$
$\displaystyle\lambda\ket{0}_{B}\bra{0}\otimes\ket{-}_{C}\bra{-}\otimes
s^{4}\Bigl{(}\frac{(1-q)^{2}}{2}(\sigma_{x})_{M}\rho_{MA}(\sigma_{x})_{M}$
$\displaystyle+q(1-q)\frac{\mathbb{I}_{M}}{2}\otimes\rho_{A}+\frac{2q-1}{2}(1-p)\ket{0}_{M}\bra{0}\otimes\rho^{(1)}_{A}\Bigl{)}$
$\displaystyle+\lambda\ket{1}_{B}\bra{1}\otimes\ket{+}_{C}\bra{+}\otimes\Bigl{(}(c^{4}+\frac{(1-q)^{2}}{2}s^{4})\rho^{\text{in
}}_{MA}$
$\displaystyle+(2c^{2}s^{2}+\frac{s^{4}}{2})\tau_{M}\otimes\rho_{A}+s^{4}\frac{2q-1}{2}p\ket{0}_{M}\bra{0}\otimes\rho^{(0)}_{A}\Bigl{)}$
$\displaystyle+(1-\lambda)\ket{2}_{B}\bra{2}\otimes\ket{0}_{C}\bra{0}\otimes(c^{4}\rho_{MA}^{\textrm{in}}+(1-c^{4})\tau_{M}\otimes\rho_{A}).$
(48)

To bound the final mutual information between $A$ and $CM^{\text{fin}}$, we
apply the quantum data-processing inequality and the chain rule

$\displaystyle I(A:CM^{\text{fin}})\leq I(A:BCM^{\text{fin}})$
$\displaystyle=I(A:B)+I(A:CM^{\text{fin}}|B).$ (49)

First, we bound the term $I(A:B)$. Tracing out $CM$ from Eq.(A5), we obtain
the state of $BA$ as follows

$\displaystyle\rho^{\textrm{fin}}_{BA}$ $\displaystyle=$
$\displaystyle\lambda\Bigl{(}s^{4}\frac{1-q^{2}}{2}\ket{0}_{B}\bra{0}+(c^{4}+\frac{1+(1-q)^{2}}{2}s^{4}+2c^{2}s^{2})$
$\displaystyle\ket{1}_{B}\bra{1}\Bigl{)}\otimes\rho_{A}+\lambda\frac{2q-1}{2}s^{4}\Bigl{(}(1-p)\ket{0}_{B}\bra{0}\otimes\rho^{(1)}_{A}$
$\displaystyle+p\ket{1}_{B}\bra{1}\otimes\rho^{(0)}_{A}\Bigl{)}+(1-\lambda)\ket{2}_{B}\bra{2}\otimes\rho_{A}.$
(50)

Adding an ancilla system $D$ to $\rho^{\text{fin}}_{BA}$, we have

$\displaystyle\rho^{\text{fin}}_{DBA}$ $\displaystyle=$
$\displaystyle\ket{0}_{D}\bra{0}\otimes\Bigl{(}\lambda
s^{4}\frac{1-q^{2}}{2}\ket{0}_{B}\bra{0}+\lambda(c^{4}+\frac{1+(1-q)^{2}}{2}s^{4}$
$\displaystyle+2c^{2}s^{2})\ket{1}_{B}\bra{1}+(1-\lambda)\ket{2}_{B}\bra{2}\Bigl{)}\otimes\rho_{A}+\lambda\ket{1}_{D}\bra{1}$
$\displaystyle\otimes(q-\frac{1}{2})s^{4}\Bigl{(}(1-p)\ket{0}_{B}\bra{0}\otimes\rho^{(1)}_{A}+p\ket{1}_{B}\bra{1}\otimes\rho^{(0)}_{A}\Bigl{)}.$
(51)

It is straightforward to prove that $I(A:D)_{\rho^{\textrm{fin}}_{DBA}}=0$.
Then, one can calculate

$\displaystyle I(A:B)_{\rho^{\textrm{fin}}_{BA}}\leq
I(A:DB)_{\rho^{\textrm{fin}}_{DBA}}$ $\displaystyle=$ $\displaystyle
I(A:D)_{\rho^{\text{fin}}_{DBA}}+I(A:B|D)_{\rho^{\text{fin}}_{DBA}}$
$\displaystyle=$ $\displaystyle\lambda
s^{4}(q-\frac{1}{2})I(A:M^{\textrm{in}}),$ (52)

where the quantum data-processing inequality is used in the first inequality.

Next, we proceed to bound the second term $I(A:CM^{\text{fin}}|B)$. According
to the definition of conditional mutual information, one can write down

$\displaystyle I(A:CM^{\text{fin}}|B)=P_{0}I(A:CM^{\text{fin}}|\ket{0}_{B})$
$\displaystyle+P_{1}I(A:CM^{\text{fin}}|\ket{1}_{B})+P_{2}I(A:CM^{\text{fin}}|\ket{2}_{B}),$
(53)

where

$\displaystyle P_{0}$
$\displaystyle=\lambda(1-\frac{(1-q^{2})+(2q-1)(1-p)}{2}s^{4}),$
$\displaystyle P_{1}$ $\displaystyle=\lambda-P_{0},P_{2}=1-\lambda.$ (54)

We then introduce an ancilla system $D_{0}$ for $\rho_{ACM|\ket{0}_{B}}$ such
that

$\displaystyle\rho_{D_{0}ACM|\ket{0}_{B}}$ $\displaystyle=$
$\displaystyle\frac{\lambda}{P(\ket{+}_{C})}\ket{+}_{c}\bra{+}\otimes\Bigl{(}(c^{4}+\frac{(1-q)^{2}}{2}s^{4})\ket{0}_{D_{0}}\bra{0}\otimes\rho^{\text{in
}}_{MA}$
$\displaystyle+(2c^{2}s^{2}+\frac{s^{4}}{2})\ket{1}_{D_{0}}\bra{1}\otimes\tau_{M}\otimes\rho_{A}+\frac{(2q-1)ps^{4}}{2}$
$\displaystyle\ket{2}_{D_{0}}\bra{2}\otimes\ket{0}_{M}\bra{0}\otimes\rho^{(0)}_{A}\Bigl{)}.$
(55)

Tracing out $CM$, one obtains

$\displaystyle\rho_{D_{0}A|\ket{0}_{B}}$ $\displaystyle=$
$\displaystyle\frac{\lambda}{P_{0}}\Bigl{(}(c^{4}+\frac{(1-q)^{2}}{2}s^{4})\ket{0}_{D_{0}}\bra{0}\otimes\rho_{A}+(2c^{2}s^{2}+\frac{s^{4}}{2})$
$\displaystyle\ket{1}_{D_{0}}\bra{1}\otimes\rho_{A}+\frac{(2q-1)ps^{4}}{2}\ket{2}_{D_{0}}\bra{2}\otimes\rho^{(0)}_{A}\Bigl{)}.$

The mutual information between $A$ and $D_{0}$ is

$\displaystyle I(A:D_{0})=$ $\displaystyle
S(p_{0}\rho_{A}+(1-p_{0})\rho^{(0)}_{A})$ $\displaystyle-
p_{0}S(\rho_{A})-(1-p_{0})S(\rho^{(0)}_{A}),$ (56)

where $p_{0}:=\lambda(1+\frac{(q^{2}-2q)s^{4}}{2})/P_{0})$. Then,

$\displaystyle
I(A:CM)_{\rho_{ACM|\ket{0}_{B}}}=I(A:M)_{\rho_{ACM|\ket{0}_{B}}}$
$\displaystyle\leq$ $\displaystyle
I(A:D_{0}M)_{\rho_{D_{0}ACM|\ket{0}_{B}}}=I(A:D_{0})+I(A:M|D_{0})$
$\displaystyle=$
$\displaystyle\frac{\lambda}{P_{0}}(c^{4}+\frac{(1-q)^{2}}{2}s^{4})I(A:M^{\text{in}})+I(A:D_{0}),$
(57)

where the quantum data-processing inequality is used in the first inequality.

Again, we introduce an ancilla system $D_{1}$ for $\rho_{CAM|\ket{1}_{B}}$
such that

$\displaystyle\rho_{D_{1}CMA|\ket{1}_{B}}$ $\displaystyle=$
$\displaystyle\frac{\lambda
s^{4}}{P_{1}}\Bigl{(}\frac{(1-q)^{2}}{2}\ket{0}_{D_{1}}\bra{0}\otimes(\sigma_{x})_{M}\rho_{MA}(\sigma_{x})_{M}+q(1-q)$
$\displaystyle\ket{1}_{D_{1}}\bra{1}\otimes\frac{\mathbb{I}_{M}}{2}\otimes\rho_{A}+\frac{2q-1}{2}(1-p)\ket{2}_{D_{1}}\bra{2}\otimes$
$\displaystyle\ket{0}_{M}\bra{0}\otimes\rho^{(1)}_{A}\Bigl{)}\otimes\ket{-}_{c}\bra{-}.$
(58)

Tracing out $CM$, one obtains

$\displaystyle\rho_{D_{1}A|\ket{1}_{B}}$ $\displaystyle=$
$\displaystyle\frac{\lambda
s^{4}}{P_{1}}\Bigl{(}\frac{(1-q)^{2}}{2}\ket{0}_{D_{1}}\bra{0}\otimes\rho_{A}+q(1-q)$
$\displaystyle\ket{1}_{D_{1}}\bra{1}\otimes\rho_{A}+\frac{2q-1}{2}(1-p)\ket{2}_{D_{1}}\bra{2}\otimes\rho^{(1)}_{A}\Bigl{)}.$

The mutual information between $A$ and $D_{1}$ is

$\displaystyle I(A:D_{1})=$ $\displaystyle
S(p_{1}\rho_{A}+(1-p_{1})\rho^{(1)}_{A})$ $\displaystyle-
p_{1}S(\rho_{A})-(1-p_{1})S(\rho^{(1)}_{A}),$ (59)

where $p_{1}:=\frac{\lambda(1-q^{2})s^{4}}{2}/P_{1}$. Similarly, we have

$\displaystyle
I(A:CM)_{\rho_{CMA|\ket{1}_{B}}}=I(A:M)_{\rho_{CMA|\ket{1}_{B}}}$
$\displaystyle\leq$ $\displaystyle
I(A:D_{1}CM)_{\rho_{D_{1}CAM|\ket{1}_{B}}}=I(A:D_{1})+I(A:M|D_{1})$
$\displaystyle=$
$\displaystyle\frac{\lambda(1-q)^{2}s^{4}}{2P_{1}}I(A:M^{\text{in}})+I(A:D_{1}).$
(60)

Similarly, we introduce an ancilla system $D_{2}$ for $\rho_{CAM|\ket{2}_{B}}$
such that

$\displaystyle\rho_{CD_{2}MA|\ket{2}_{B}}=$
$\displaystyle\ket{0}_{C}\bra{0}\otimes\Bigl{(}c^{4}\ket{0}_{D_{2}}\bra{0}\otimes\rho_{MA}^{\textrm{in}}$
$\displaystyle+(1-c^{4})\ket{1}_{D_{2}}\bra{1}\otimes\tau_{M}\otimes\rho_{A}\Bigl{)}.$
(61)

Then, the mutual information $I(A:CM^{\text{fin}}|\ket{2}_{B})$ is upper
bounded by

$\displaystyle I(A:CM^{\textrm{fin}})_{\rho_{CMA|\ket{2}_{B}}}\leq
I(A:CD_{2}M)_{\rho_{CD_{2}AM|\ket{1}_{B}}}$ $\displaystyle=$ $\displaystyle
I(A:D_{2})+I(A:M|D_{2})=c^{4}I(A:M^{\text{in}}).$ (62)

Finally, combining Eq.(A5), Eq.(A5), Eq.(A5), Eq.(A5) with Eq. (A5), we obtain

$\displaystyle I(A:CM^{\text{fin}})\leq I(A:BCM^{\text{fin}})$
$\displaystyle=$ $\displaystyle I(A:B)+I(A:CM^{\text{fin}}|B)$
$\displaystyle\leq$
$\displaystyle\lambda(q-\frac{1}{2})s^{4}I(A:M^{\text{in}})+P_{0}I(A:M^{\text{fin}}|\ket{0}_{B})$
$\displaystyle+P_{1}I(A:M^{\text{fin}}|\ket{1}_{B})+P_{2}I(A:M^{\text{fin}}|\ket{2}_{B})$
$\displaystyle\leq$
$\displaystyle(c^{4}+\lambda(q^{2}-q+\frac{1}{2})s^{4})I(A:M^{\text{in}})+P_{0}I(A:D_{0})$
$\displaystyle+P_{1}I(A:D_{1}),$ (63)

where $P_{0}:=\lambda(1-\frac{(1-q^{2})+(2q-1)(1-p)}{2}s^{4}),P_{1}:=\lambda-
P_{0}$ are the probabilities conditioning on $B$ , and

$\displaystyle I(A:D_{0/1}):=S(p_{0/1}\rho_{A}+(1-p_{0/1})\rho^{(0/1)}_{A})$
$\displaystyle-p_{0/1}S(\rho_{A})-(1-p_{0/1})S(\rho^{(0/1)}_{A})$ (64)

with
$p_{0}:=\lambda(1+\frac{(q^{2}-2q)s^{4}}{2})/P_{0},p_{1}=\lambda\frac{(1-q^{2})s^{4}}{2}/P_{1}$.

∎

Infinite temperature case  In this case, the thermal state is
$\tau=\frac{\mathbb{I}}{2}$, $i.e.,q=1/2$. Substituting $q=1/2$ into the
bound, one has

$\displaystyle P_{0}=\lambda(1-\frac{3}{8}s^{4}),P_{1}=\frac{3}{8}\lambda
s^{4},p_{0}=p_{1}=1.$ (65)

Thus, according to Theorem 1, the mutual information $I(A:CM^{\textrm{fin}})$
is bounded by

$\displaystyle
I(A:CM^{\text{fin}})\leq(c^{4}+\frac{\lambda}{4}s^{4})I(A:M^{\text{in}}).$
(66)

Zero temperature case:  In this case, the thermal state is
$\tau=\ket{0}\bra{0}$, $i.e.,q=1$. According to Theorem 1, the mutual
information $I(A:CM^{\textrm{fin}})$ is bounded by

$\displaystyle I(A:CM^{\text{fin}})$ $\displaystyle\leq$
$\displaystyle(c^{4}+\frac{\lambda}{2}s^{4})I(A:M^{\text{in}})+P_{0}I(A:D_{0}),$
(67)

where $P_{0}=\lambda(1-\frac{1-p}{2}s^{4})$ and

$\displaystyle I(A:D_{0}):=$ $\displaystyle
S(p_{0}\rho_{A}+(1-p_{0})\rho^{(0)}_{A})$ $\displaystyle-
p_{0}S(\rho_{A})-(1-p_{0})S(\rho^{(0)}_{A})$

with $p_{0}=\lambda(1-\frac{s^{4}}{2})/P_{0}$.

##  Appendix A6 Violation of the first law of thermodynamics

Figure A 1: Demonstration of violation of the bound by the use of an energy-
altering channel. The blue line is the bound and the orange line is the mutual
information achieved by two switched channels, one being associated with a
partial swap unitary and the other a ‘partial CNOT’.

We take the Hamiltonians of the system and the environment to be
$H_{M}=H_{E}=\sigma_{z}$. When the temperature is infinity, the thermal state
is then $\tau:=\frac{\mathbb{I}}{2}$. Consider the CNOT gate

$V:=\ket{0}\bra{0}\otimes\mathbb{I}+\ket{1}\bra{1}\otimes\sigma_{x},$

Then, $e^{i\theta V}=\cos\theta\mathbb{I}+i\sin\theta V:=c\mathbb{I}+isV$. One
can check that

$\displaystyle e^{i\theta
V}\left(\frac{\mathbb{I}_{M}}{2}\otimes\frac{\mathbb{I}_{E}}{2}\right)e^{-i\theta
V}=\frac{\mathbb{I}_{M}}{2}\otimes\frac{\mathbb{I}_{E}}{2}.$ (68)

Tracing out $E$ in the above equation results in
$\mathcal{N}(\frac{\mathbb{I}}{2})=\frac{\mathbb{I}}{2}.$ Thus, $\mathcal{N}$
is a thermal channel and the second law is obeyed. The set of Kraus operators
for $\mathcal{N}$ is

$\displaystyle\\{(c\mathbb{I}+is\ket{0}\bra{0}),is\ket{1}\bra{1}\\}.$ (69)

However, one sees that

$[e^{i\theta V},H_{E}+H_{M}]\neq 0,$

i.e., the first law of thermodynamics is violated. Recall that the Kraus
operators for the quantum partial swap unitary given in the Appendix A3. We
see that some elements of the two sets of Kraus operators do not commute,
enabling information capacity increase.

As shown in Figure A 1, we plotted the mutual information
$I(A:CM^{\textrm{fin}})$ achieved by two cases of switched channels, one being
associated with a partial swap unitary and the other a ‘partial CNOT’. The
figure shows that our bound is violated in some region by the ‘partial CNOT’
case, consistent with the energy conservation violation of that channel.

##  Appendix A7 Proof of Theorem 2

Recall that the quantum switch $\mathcal{S}_{\sigma_{C}}$ is a supermap [30],
which takes two quantum channels $\mathcal{C}_{1},\mathcal{C}_{2}$ as input
and output another quantum channel
$\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})$. In the language
of Kraus operators [28], if we denote the Kraus operators of the channel
$\mathcal{C}_{1}$ as $\\{K^{(1)}_{i}\\}$ and $\mathcal{C}_{2}$ as
$\\{K^{(2)}_{j}\\}$, the action of the channel
$\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})$ is expressed by

$\displaystyle\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})(\rho)=\sum_{ij}S_{ij}(\sigma_{C}\otimes\rho)S_{ij}^{\dagger}$
(70)

where $\sigma_{C}$ is the state of the control system, and $S_{ij}$ denotes
the Kraus operators for
$\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})$, namely,

$\displaystyle S_{ij}=|0\rangle_{C}\langle 0|\otimes
K^{(2)}_{j}K^{(1)}_{i}+|1\rangle_{C}\langle 1|\otimes K^{(1)}_{i}K^{(2)}_{j}\
.$ (71)

Theorem 2:  If $\mathcal{C}_{1}(\tau_{M})=\mathcal{C}_{2}(\tau_{M})=\tau_{M}$,
where $\tau_{M}$ is the thermal state on $M$ then, for any state $\sigma_{C}$,
$\textrm{Tr}_{C}\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})(\tau_{M})=\tau_{M}$.

###### Proof.

Denote by $\sigma_{C}^{ij}$ the matrix element of $i$th row and $j$th column
of $\sigma_{C}$. The action of the channel
$\textrm{Tr}_{C}\mathcal{S}_{\sigma_{C}}$ on the thermal state $\tau_{M}$ is
given by

$\displaystyle\textrm{Tr}_{C}\mathcal{S}_{\sigma_{C}}(\mathcal{C}_{1},\mathcal{C}_{2})(\tau_{M})$
$\displaystyle=$
$\displaystyle\textrm{Tr}_{C}\sum_{i,j}S_{ij}(\sigma_{C}\otimes\tau_{M})S_{ij}^{\dagger}$
$\displaystyle=$
$\displaystyle\sigma_{C}^{00}\sum_{i,j}K^{(2)}_{j}K^{(1)}_{i}\tau_{M}K^{(1){\dagger}}_{i}K^{(2){\dagger}}_{j}$
$\displaystyle+\sigma_{C}^{11}\sum_{i,j}K^{(1)}_{i}K^{(2)}_{j}\tau_{M}K^{(2){\dagger}}_{j}K^{(1){\dagger}}_{i}$
$\displaystyle=$
$\displaystyle\sigma_{C}^{00}\mathcal{C}_{2}\circ\mathcal{C}_{1}(\tau_{M})+\sigma_{C}^{11}\mathcal{C}_{1}\circ\mathcal{C}_{2}(\tau_{M})$
$\displaystyle=$ $\displaystyle\tau_{M},$ (72)

where $\mathcal{C}_{1}(\tau_{M})=\mathcal{C}_{2}(\tau_{M})=\tau_{M}$ is used
in the last equality. ∎

##  Appendix A8 Free energy of coherence of the control

Denote by $H$ the Hamiltonian of the system. The free energy of coherence of
the system is defined as [33]

$\displaystyle
F_{coh}(\sigma,H):=kT\Bigl{(}S(\mathcal{D}_{H}(\sigma))-S(\sigma)\Bigl{)},$
(73)

where $\mathcal{D}_{H}$ is the map that kills all the off-diagonal elements in
the energy eigenbasis. The relation between the free energy $F(\sigma,H)$ and
the free energy of coherence is

$\displaystyle F(\sigma,H)$ $\displaystyle:=\textrm{Tr}(\sigma H)-kTS(\sigma)$
$\displaystyle=F(\mathcal{D}_{H}(\sigma),H)+F_{coh}(\sigma,H).$ (74)

Let the Hamiltonian of the system $H=\sigma_{z}$, and the initial state of the
system
$\sigma=\lambda\ket{+}\bra{+}+(1-\lambda)\ket{0}\bra{0},\lambda\in[0,1]$.
Then,
$\mathcal{D}_{H}(\sigma)=\lambda\frac{\mathbb{I}}{2}+(1-\lambda)\ket{0}\bra{0}$.
The set of eigenvalues of $\mathcal{D}_{H}(\sigma)$ is
$\\{1-\frac{\lambda}{2},\frac{\lambda}{2}\\}$. The set of eigenvalues of
$\sigma$ is $\\{\frac{1}{2}(1\pm\sqrt{1-2\lambda+\lambda^{2}})\\}$. Then,
applying a Taylor expansion and some manipulations, we arrive at

$\displaystyle F_{coh}(\sigma,H)\geq kT\frac{\lambda^{2}}{\ln 16}.$ (75)

##  Appendix A9 Interpolating between switch being ON and OFF

The main text utilized a control parameter $\lambda\in[0,1]$ to denote the
level of mixture of the state
$\sigma_{c}=\lambda\ket{+}\bra{+}+(1-\lambda)\ket{0}\bra{0}$ of the control.
Naturally, when $\lambda$ is closer to one, the control system has more
coherence – regulating the processing orders of the two energy preserving
thermal channels. In this Section, we explore whether a second method for
gradually turning a switch ON and OFF, used in Ref. [8], yields the same
conclusions. In this scenario, the control state is pure and reads
$\ket{\Psi}_{c}=\sqrt{\alpha}\ket{0}+\sqrt{1-\alpha}\ket{1},\alpha\in[0,1]$.
Here, the switch is maximally turned on for $\alpha=1/2$, such that
$\ket{\Psi}_{c}=\ket{+}$.

Following the same procedure as in the Appendix A5 but replacing the control
state with $\ket{\Psi}_{c}$, we calculate the final state of the control
system $C$, the record of the message $A$ and the system undergoing the
channel $M$ as

$\displaystyle\rho^{\text{ {fin}}}_{CMA}=$
$\displaystyle\mathcal{S}_{\ket{\Psi}_{c}}(\mathcal{C},\mathcal{C})(\rho_{AM})=\sum_{i,j}S_{ij}\rho_{AM}S_{ij}^{\dagger}$
$\displaystyle=$ $\displaystyle\alpha|0\rangle_{c}\langle
0|\otimes\sum_{i,j}K^{M}_{j}K^{M}_{i}\rho_{AM}K^{M{\dagger}}_{i}K^{M{\dagger}}_{j}$
$\displaystyle+$ $\displaystyle(1-\alpha)|1\rangle_{c}\langle
1|\otimes\sum_{i,j}K^{M}_{i}K^{M}_{j}\rho_{AM}K^{M{\dagger}}_{j}K^{M{\dagger}}_{i}$
$\displaystyle+$ $\displaystyle\sqrt{\alpha(1-\alpha)}|0\rangle_{c}\langle
1|\otimes\sum_{i,j}K^{M}_{j}K^{M}_{i}\rho_{AM}K^{M{\dagger}}_{j}K^{M{\dagger}}_{i}$
$\displaystyle+$ $\displaystyle\sqrt{\alpha(1-\alpha)}|1\rangle_{c}\langle
0|\otimes\sum_{i,j}K^{M}_{i}K^{M}_{j}\rho_{AM}K^{M{\dagger}}_{i}K^{M{\dagger}}_{j}.$
(76)

Note that here we have used the shorthand
$K^{M}_{i}:=\mathbb{I}_{A}\otimes(K_{i})_{M}$. We then calculate the two
summations appearing in the above expression:

$\displaystyle\sum_{i,j}K^{M}_{j}K^{M}_{i}\rho_{AM}K^{M{\dagger}}_{i}K^{M{\dagger}}_{j}=c^{4}\rho^{\textrm{in}}_{AM}+(1-c^{4})\rho_{A}\otimes\tau_{M},$
$\displaystyle\sum_{i,j}K^{M}_{i}K^{M}_{j}\rho_{AM}K^{M{\dagger}}_{i}K^{M{\dagger}}_{j}=(c^{2}+s^{2}q)^{2}\rho^{(0)}_{A}\otimes
p\ket{0}_{M}\bra{0}$
$\displaystyle+(c^{2}+s^{2}(1-q))^{2}\rho^{(1)}_{A}\otimes(1-p)\ket{1}_{M}\bra{1}+2c^{2}s^{2}(1-q)$
$\displaystyle\rho^{(0)}_{A}\otimes
p\ket{1}_{M}\bra{1}+2c^{2}s^{2}q\rho^{(1)}_{A}\otimes(1-p)\ket{0}_{M}\bra{0}$
$\displaystyle=(c^{2}+s^{2}(1-q))^{2}\rho^{\textrm{in}}_{AM}+2c^{2}s^{2}(1-q)(\sigma_{x})_{M}\rho^{\textrm{in}}_{AM}(\sigma_{x})_{M}$
$\displaystyle+(ps^{4}+2s^{2}c^{2})(2q-1)\rho^{(0)}_{A}\otimes\ket{0}_{M}\bra{0}.$
(77)

Substituting the two summations into Eq. (A91), we obtain

$\displaystyle\rho^{\text{ {fin}}}_{CMA}$ $\displaystyle=$
$\displaystyle(\alpha|0\rangle_{c}\langle 0|+(1-\alpha)|1\rangle_{c}\langle
1|)\otimes\Bigl{(}c^{4}\rho^{\textrm{in}}_{AM}+(1-c^{4})\rho_{A}\otimes\tau_{M}\Bigl{)}$
$\displaystyle+\sqrt{\alpha(1-\alpha)}(|0\rangle_{c}\langle
1|+|1\rangle_{c}\langle
0|)\otimes\Bigl{(}(c^{2}+s^{2}(1-q))^{2}\rho^{\textrm{in}}_{AM}$
$\displaystyle+2c^{2}s^{2}(1-q)(\sigma_{x})_{M}\rho^{\textrm{in}}_{AM}(\sigma_{x})_{M}+(ps^{4}+2s^{2}c^{2})(2q-1)$
$\displaystyle\rho^{(0)}_{A}\otimes\ket{0}_{M}\bra{0}\Bigl{)}.$ (78)

We can break the general control system state into three cases:

  1. i)

Switch is fully ON: $\alpha=\frac{1}{2}$, i.e, $\ket{\Psi}_{c}=\ket{+}$. By
inspection Eq.(A93) and Eq.(A514) denote the same state, that is, this
corresponds to $\lambda=1,\sigma_{c}=\ket{+}\bra{+}$. Thus our bound (1) is
directly applicable to this case.

  2. ii)

Switch is fully OFF: $\alpha=0/1$, i.e, $\ket{\Psi}_{c}=\ket{0}/\ket{1}$.
Here, there is no superposition of orders. This corresponds to the switch
being off, i.e., $\lambda=0,\sigma_{c}=\ket{0}\bra{0}$. By inspection Eq.
(A93) and Eq. (A514) denote the same state in this case. Thus our bound is
directly applicable to this case as well.

  3. iii)

Other values of $\alpha$. The final state of the compound system $CMA$
expressed by Eq. (A93) is not identical to that of Eq. (A514). Nevertheless,
the mutual information after the interaction is very similar whether the
control system is taken as pure or mixed, as shown in Figure A2.

We conclude that the two possible choices of the general state of the control
system lead to qualitatively and quantitatively very similar behaviour and
focus on the case of the mixed control as it gives neater mathematical
expressions.

Figure A 2: The mutual information between memory and post-interaction system
plus control is very similar whether the control is in the mixed state
$\sigma_{c}=\lambda\ket{+}\bra{+}+(1-\lambda)\ket{0}\bra{0},\lambda\in[0,1]$
vs. the pure state
$\ket{\Psi}_{c}=\sqrt{\alpha}\ket{0}+\sqrt{1-\alpha}\ket{1},\alpha\in[0,1/2]$.
The plot is for the case of the temperature $T=0$ (we observe similar
behaviour for all temperatures). $s$ is a thermalisation strength parameter.