* [62] Tomohiro Ogawa and Hiroshi Nagaoka.  _Strong Converse and Stein’s Lemma in Quantum Hypothesis Testing_ , pages 28–42.  doi: 10.1142/9789812563071_0003.  URL https://www.worldscientific.com/doi/abs/10.1142/9789812563071_0003.
  * Lostaglio et al. [2015b] Matteo Lostaglio, David Jennings, and Terry Rudolph.  Description of quantum coherence in thermodynamic processes requires constraints beyond free energy.  _Nature Communications_ , 6(1):6383, 2015b.  doi: 10.1038/ncomms7383.  URL https://doi.org/10.1038/ncomms7383.
  * Ahmadi et al. [2013] Mehdi Ahmadi, David Jennings, and Terry Rudolph.  The wigner–araki–yanase theorem and the quantum resource theory of asymmetry.  _New Journal of Physics_ , 15(1):013057, jan 2013\.  doi: 10.1088/1367-2630/15/1/013057.  URL https://doi.org/10.1088%2F1367-2630%2F15%2F1%2F013057.
  * Watrous [2018] John Watrous.  _The Theory of Quantum Information_.  Cambridge University Press, 2018.  doi: 10.1017/9781316848142.
  * Gühne et al. [2005] Otfried Gühne, Géza Tóth, and Hans J Briegel.  Multipartite entanglement in spin chains.  _New Journal of Physics_ , 7:229–229, nov 2005.  doi: 10.1088/1367-2630/7/1/229.  URL https://doi.org/10.1088%2F1367-2630%2F7%2F1%2F229.
  * Tóth [2012] Géza Tóth.  Multipartite entanglement and high-precision metrology.  _Phys. Rev. A_ , 85:022322, Feb 2012.  doi: 10.1103/PhysRevA.85.022322.  URL https://link.aps.org/doi/10.1103/PhysRevA.85.022322.
  * Chen [2005] Zeqian Chen.  Wigner-Yanase skew information as tests for quantum entanglement.  _Phys. Rev. A_ , 71:052302, May 2005.  doi: 10.1103/PhysRevA.71.052302.  URL https://link.aps.org/doi/10.1103/PhysRevA.71.052302.
  * Goold et al. [2018] John Goold, Francesco Plastina, Andrea Gambassi, and Alessandro Silva.  _The Role of Quantum Work Statistics in Many-Body Physics_ , pages 317–336.  Springer International Publishing, Cham, 2018.  ISBN 978-3-319-99046-0.  doi: 10.1007/978-3-319-99046-0_13.  URL https://doi.org/10.1007/978-3-319-99046-0_13.
  * Said [1958] A. S. Said.  Some properties of the poisson distribution.  _AIChE Journal_ , 4(3):290–292, 1958.  doi: 10.1002/aic.690040311.  URL https://aiche.onlinelibrary.wiley.com/doi/abs/10.1002/aic.690040311.
  * Prasad et al. [1987] Sudhakar Prasad, Marlan O. Scully, and Werner Martienssen.  A quantum description of the beam splitter.  _Optics Communications_ , 62(3):139 – 145, 1987\.  ISSN 0030-4018.  doi: https://doi.org/10.1016/0030-4018(87)90015-0.  URL http://www.sciencedirect.com/science/article/pii/0030401887900150.
  * Petz [1994] Dénes Petz.  A survey of certain trace inequalities.  _Banach Center Publications_ , 30(1):287–298, 1994.  doi: 10.4064/-30-1-287-298.  URL http://eudml.org/doc/262566.
  * Massart [2007] Pascal Massart.  _Concentration Inequalities and Model Selection: Ecole d’Eté de Probabilités de Saint-Flour XXXIII - 2003_.  Lecture Notes in Mathematics. Springer, 2007.
  * Billingsley [1995] P. Billingsley.  _Probability and Measure_.  Wiley Series in Probability and Statistics. Wiley, 1995.  ISBN 9780471007104.

##  Appendix A Coherent work processes

Here we gather the proofs of the statements for coherent work processes. First
we show that for auxiliary ‘weight’ systems with non-degenerate spectra, the
coherent work output is effectively unique.

######  Proposition A.1.

Let $|\omega\rangle_{A}=\sum_{k}e^{i\theta_{k}}\sqrt{p_{k}}|k\rangle_{A}$ be
the coherent work output from a coherent work process
$|\psi\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\phi\rangle_{S}$
where $\\{|k\rangle_{A}\\}$ is the eigenbasis of the non-degenerate
Hamiltonian $H_{A}$. Then the distribution $\\{p_{k}\\}$ is uniquely
determined, while the phases $\\{\theta_{k}\\}$ are arbitrary.

###### Proof.

Assume $|\omega\rangle_{A}$ is the coherent work output of the process
$|\psi_{S}\rangle\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\phi_{S}\rangle$,
enacted by the unitary $U$. Define the characteristic function
$\varphi_{X,\eta}(t):=\mbox{tr}(e^{iH_{X}t}|\eta\rangle\langle\eta|_{X})$.
Then under a coherent work process,

$\displaystyle\varphi_{S,\psi}(t)\varphi_{A,0}(t)$
$\displaystyle=\mbox{tr}(e^{i(H_{S}+H_{A})t}|\psi\rangle\langle\psi|_{S}\otimes|0\rangle\langle
0|_{A})$ (73)
$\displaystyle=\mbox{tr}(e^{i(H_{S}+H_{A})t}U^{\dagger}U|\psi\rangle\langle\psi|_{S}\otimes|0\rangle\langle
0|_{A})$ (74) $\displaystyle=\varphi_{S,\phi}(t)\varphi_{A,\omega}(t),$ (75)

where $U$ is a unitary and $[U,H_{S}+H_{A}]=0$ by the construction of coherent
work processes. Due to the uniqueness property of characteristic functions
[13], $\varphi_{A,\omega}(t)=\varphi_{A,\omega^{\prime}}(t)$ for all
$t\in\mathbb{R}$ if and only if the distribution for $\omega$ and
$\omega^{\prime}$ are equal. Therefore with $\psi$ and $\phi$ pre-determined,
the distribution of $\omega$ is uniquely determined.

Since the only constraint on $U$ is that it is energy conserving, then for any
Hermitian operator $L_{A}$ such that $[L_{A},H_{A}]=0$, we can define a
unitary $V=(\mathds{1}_{S}\otimes e^{-iL_{A}\theta})U$ that also conserves
energy, with $[V,H_{S}+H_{A}]=0$. Furthermore, with appropriate choice of
$L_{A}$, any set of phases $\\{\theta_{k}\\}$ can be selected. ∎

It is necessary to constrain the spectrum of $A$ to be non-degenerate, as
given a degenerate Hamiltonian $H_{A^{\prime}}$ for a system $A^{\prime}$, one
could enact a unitary $T$ that locally conserves energy
($[H_{A^{\prime}},T]=0$) and can arbitrarily change the state
$|\omega\rangle_{A}$ in the degenerate subspace.

Though we describe the coherent work output as effectively unique, one is
always free to enact evolution locally on the auxiliary system that will
induce a set of relative phases. In this way, any phases gained from the joint
evolution between $S$ and $A$ can be arbitrarily changed with local evolution.
Since relative phases are a signature of time-evolution, the result
distinguishes between global and local effects of evolution on states.

We can also prove that all reversible transformations involve trivial coherent
work processes.

######  Proposition A.2.

A coherent work process is trivial if and only if it is reversible.

###### Proof.

Let
$V|\psi_{0}\rangle_{S}\otimes|E_{0}\rangle_{A}=|\psi_{1}\rangle_{S^{\prime}}\otimes|\omega\rangle_{A}$
be a coherent work process. In terms of random variables with respect to
Hamiltonians $H_{S}$ and $H_{A}$, an equivalent condition for the existence of
this process is $\hat{S}=\hat{S}^{\prime}+\hat{A}$, where
$\hat{S},\,\hat{S^{\prime}},\,\hat{A}$ are the random variables for
$|\psi_{0}\rangle_{S},\,|\psi_{1}\rangle_{S^{\prime}},\,|\omega\rangle_{A}$
respectively. If it is a trivial coherent work process, then at least one of
$\hat{S}^{\prime}$ or $\hat{A}$ is a constant random variable. Without loss of
generality, we assume $\hat{A}=a$ is constant and so we must have that
$\hat{S}^{\prime}=\hat{S}-a$ and $\hat{S}=\hat{S^{\prime}}+a$. This gives
$|\psi_{0}\rangle_{S}\stackrel{{\scriptstyle
a}}{{\longrightarrow}}|\psi_{1}\rangle_{S^{\prime}}$ and
$|\psi_{1}\rangle_{S^{\prime}}\stackrel{{\scriptstyle-a}}{{\longrightarrow}}|\psi_{0}\rangle_{S}$.

Conversely, assume we have a reversible coherent work process

$\displaystyle V|\psi_{0}\rangle_{S}\otimes|E_{0}\rangle_{A_{1}}$
$\displaystyle=|\psi_{1}\rangle_{S^{\prime}}\otimes|\omega_{1}\rangle_{A_{1}},$
(76) $\displaystyle
U|\psi_{1}\rangle_{S^{\prime}}\otimes|E_{0}^{\prime}\rangle_{A_{2}}$
$\displaystyle=|\psi_{0}\rangle_{S}\otimes|\omega_{2}\rangle_{A_{2}},$ (77)

where we introduce a different auxiliary system for each process. Then
$\hat{S}=\hat{S}^{\prime}+\hat{A}_{1}$ and
$\hat{S}^{\prime}=\hat{S}+\hat{A}_{2}$. As the random variables are all
independent, we have both
$\text{Var}(\hat{S})=\text{Var}(\hat{S}^{\prime})+\text{Var}(\hat{A}_{1})$ and
$\text{Var}(\hat{S}^{\prime})=\text{Var}(\hat{S})+\text{Var}(\hat{A}_{2})$.
Therefore

$\mbox{Var}(\hat{A}_{1})+\mbox{Var}(\hat{A}_{1})=0.$ (78)

Due to the non-negativity of the variance for real-valued random variables
this implies that $\text{Var}(\hat{A}_{1})=\text{Var}(\hat{A}_{2})=0$. This
implies that both $\hat{A}_{1}$ $\hat{A}_{2}$ are constant random variables
and so both processes are trivial as required. ∎

The consequence of this result is that any reversible coherent work process
can simply be labeled by the constant value of $\hat{A}_{1}$. For example, if
we have
$|\psi_{0}\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\psi_{1}\rangle_{S^{\prime}}$
and
$|\psi_{1}\rangle_{S^{\prime}}\stackrel{{\scriptstyle\omega^{\prime}}}{{\longrightarrow}}|\psi_{0}\rangle_{S}$
then this implies $|\omega\rangle=|E_{k}\rangle$ and
$|\omega^{\prime}\rangle=|-E_{k}\rangle$ for some energy $E_{k}$ (if
$E_{0}=0$), then we label the reversible process simply as
$|\psi_{0}\rangle_{S}\stackrel{{\scriptstyle
E_{k}}}{{\longleftrightarrow}}|\psi_{1}\rangle_{S^{\prime}}$.

###  A.1 Decomposition of Poisson distributions

The main theorem used in this section is Theorem A.3 due to Raikov and later
generalized by others, but we first re-cap on some background details [13].

The cumulative distribution function (CDF) of a random variable $\hat{X}$ on
the real line is defined as $F(x):=Prob[\hat{X}\leq x]$. If $\hat{X}$ has CDF
$F_{1}$ and $\hat{Y}$ has CDF $F_{2}$, and $\hat{X}$ and $\hat{Y}$ are
independent, then the CDF of the random variable $\hat{Z}:=\hat{X}+\hat{Y}$
has CDF given by the convolution

$\displaystyle F_{\hat{Z}}(x)$
$\displaystyle=\int_{-\infty}^{\infty}F_{1}(x-y)dF_{2}(y)$
$\displaystyle=\int_{-\infty}^{\infty}F_{1}(x-y)\frac{dF_{2}}{dy}(y)dy$
$\displaystyle=\int_{-\infty}^{\infty}F_{1}(x-y)f_{2}(y)dy,$ (79)

where $f_{2}(y)$ is the _probability density_ for $\hat{Y}$. For completeness
of notation, we also define the step function $\epsilon(x)=1$ if $x\geq 0$ and
zero otherwise.

With this notation in place, we can now state the theorem by Raikov.

######  Theorem A.3 (Raikov, 1938).

The Poisson law has cumulative distribution function

$F\left(\frac{x-\alpha}{\sigma},\lambda\right),\sigma>0,\alpha\in\mathbb{R},\lambda\geq
0,$ (80)

where

$F(x;\lambda)=e^{-\lambda}\sum_{k\leq x}\frac{\lambda^{k}}{k!}.$ (81)

(for $x<0$ the sum is empty and $F(x;\lambda)=0$). $F(x;\lambda)$ corresponds
to the Poisson cumulative distribution for a random variable that takes only
integer values at non-negative values $k$ with probability
$\frac{\lambda^{k}}{k!}e^{-\lambda}$. When $\lambda=0$ we obtain the singular
distribution law $\epsilon(x-\alpha)$.

The integral equation

$\int_{-\infty}^{\infty}F_{1}(x-y)dF_{2}(y)=F\left(\frac{x-\alpha}{\sigma},\lambda\right),$
(82)

where $F_{1}(x)$ and $F_{2}(x)$ are cumulative distribution functions only has
solutions

$\displaystyle F_{1}(x)$
$\displaystyle=F\left(\frac{x-\alpha_{1}}{\sigma},\mu\right)$ (83)
$\displaystyle F_{2}(x)$
$\displaystyle=F\left(\frac{x-\alpha_{2}}{\sigma},\nu\right),$ (84)

where $\mu\geq 0$, $\nu\geq 0$, $\mu+\nu=\lambda$ and
$\alpha_{1}+\alpha_{2}=\alpha$, with $\alpha_{1},\alpha_{2}\in\mathbb{R}$.

We refer the reader to [15] for the proof.

We note that the original statement of Raikov’s theorem found in [15] has a
typographical error to do with the shifting parameters $\alpha$ and $\beta$.
Namely, $F_{1}(x)=F(\frac{x-\alpha-\beta}{\sigma},\mu)$ and
$F_{2}(x)=F(\frac{x-\alpha+\beta}{\sigma},\nu)$, with $\beta\in\mathbb{R}$.
This error can be seen in the following manner. Suppose we have a Poisson
distributed random variable $\hat{Y}$ shifted by $a\in\mathbb{R}$, such that
$\hat{Y}^{\prime}=(\hat{Y}-a)\sim\text{Pois}(\mu)$, where $\text{Pois}(\mu)$
denotes a Poisson distribution on the non-negative integers. Similarly, let us
define another Poisson distributed random variable $\hat{Z}$ shifted by
$b\in\mathbb{R}$, such that
$\hat{Z}^{\prime}=(\hat{Z}-b)\sim\text{Pois}(\nu)$. If we assume $\hat{Y}$ and
$\hat{Z}$ are independent, then the sum of the random variables is

$\displaystyle\hat{X}$ $\displaystyle=\hat{Y}+\hat{Z}$ (85)
$\displaystyle=(\hat{Y}-a)+(\hat{Z}-b)+(a+b).$ (86)

Since $a$ and $b$ are constants, we are able to re-arrange this to find

$\hat{X}-(a+b)=(\hat{Y}-a)+(\hat{Z}-b).$ (87)

However, the right hand side of this equation corresponds to the sum of two
unshifted Poisson distributed random variables and thus
$\hat{X}-(a+b)=\hat{Y}^{\prime}+\hat{Z}^{\prime}\sim\text{Pois}(\mu+\nu)$,
where we deduce this from Raikov’s theorem. From this we see that $\hat{X}$ is
a Poisson distributed random variable shifted by an amount $(a+b)$. Now let us
choose $a=\alpha+\beta$ and $b=\alpha-\beta$. Thus $a+b=2\alpha$ and we expect
the sum of the shifted random variables to have a shifting of $2\alpha$,
rather than the $\alpha$ that is claimed.

Another consistency check is to consider the limit of a Poisson cumulative
distribution function. We have that

$\displaystyle\lim_{\lambda\to
0}F\left(\frac{x-\alpha}{\sigma};\lambda\right)$
$\displaystyle=\lim_{\lambda\to
0}\sum_{k\leq\frac{x-\alpha}{\sigma}}e^{-\lambda}\frac{\lambda^{k}}{k!}$ (88)
$\displaystyle=\begin{cases}1,&\text{if }x\geq\alpha\\\
0,&\text{otherwise}\end{cases}$ (89)

where we make use of the limit of a Poisson probability mass function [70].
Now suppose $\hat{Y}^{\prime}\sim\text{Pois}(\mu)$ and
$\hat{Y}=\hat{Y}^{\prime}+a$. Then $\lim_{\mu\to 0}\hat{Y}=\lim_{\mu\to
0}\hat{Y}^{\prime}+a=a$. If we similarly define
$\hat{Z}^{\prime}\sim\text{Pois}(\nu)$, with $\hat{Z}=\hat{Z}^{\prime}+b$,
then $\lim_{\nu\to 0}\hat{Z}=b$. Thus when we sum the two shifted random
variables, $\lim_{\mu,\nu\to 0}\hat{Y}+\hat{Z}=a+b$, as expected.

###  A.2 Proof for decomposability theorem

######  Theorem A.4.

Given a quantum system $X$ with Hamiltonian $H_{X}$ with a discrete spectrum,
a state $|\psi\rangle_{X}$ admits a non-trivial coherent work process if and
only if the associated classical random variable $\hat{X}$ is a decomposable
random variable. Furthermore, for a coherent work process

$|\psi\rangle_{X}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\phi\rangle_{Z},$
(90)

the associated classical random variables are given by
$\hat{X}=\hat{Z}+\hat{W}$, where $\hat{Z}$ and $\hat{W}$ correspond to the
measurements of $H_{Z}$ and $H_{W}$ in $|\phi\rangle_{Z}$ and
$|\omega\rangle_{W}$ respectively.

###### Proof.

Given quantum systems $X,W,Y,Z$ with corresponding Hamiltonians, suppose first
the state $|\psi\rangle_{X}$ has decomposable coherence. This implies that
there exists an auxiliary system $Y$ with zero energy eigenstate
$|0\rangle_{Y}$ and an isometry $V$ from
${\mathcal{H}}_{X}\otimes{\mathcal{H}}_{Y}$ into
${\mathcal{H}}_{Z}\otimes{\mathcal{H}}_{W}$ with

$V(H_{X}\otimes\mathbbm{1}_{Y}+\mathbbm{1}_{X}\otimes
H_{Y})=(H_{Z}\otimes\mathbbm{1}_{W}+\mathbbm{1}_{Z}\otimes H_{W})V^{\dagger},$
(91)

such that

$V[|\psi\rangle_{X}\otimes|0\rangle_{Y}]=|\phi\rangle_{Z}\otimes|\omega\rangle_{W},$
(92)

and where neither $|\phi\rangle_{Z}$ nor $|\omega\rangle_{W}$ are energy
eigenstates of their respective Hamiltonians. The associated classical random
variables for the input systems are $\hat{X}$ and $\hat{Y}$, while the
associated classical random variables on the output systems are $\hat{Z}$ and
$\hat{W}$.

Now since $\hat{Y}=0$ the distribution for $\hat{X}$ coincides with the
distribution $\hat{X}+\hat{Y}$, which is simply the total energy of the input
systems. However equation (91) implies that $V$ is block diagonal in the total
energy and thus preserves the measurement statistics for the total energy.
Therefore $\hat{X}=\hat{Z}+\hat{W}$. Since $|\phi\rangle_{Z}$ and
$|\omega\rangle_{W}$ are not energy eigenstates this means that both $\hat{Z}$
and $\hat{W}$ have non-trivial distributions and are independent since the
composite state on the output is a product state, and thus $\hat{X}$ is a
decomposable random variable.

Conversely, given an initial state $|\psi\rangle_{X}$ with associated
classical random variable $\hat{X}$, now suppose that $\hat{X}$ is
decomposable. This means there exist classical random variables $\hat{Z}$ and
$\hat{W}$, both with independent, non-trivial distributions, such that
$\hat{X}=\hat{Z}+\hat{W}$. Since $H_{X}$ is assumed to have a discrete
spectrum the classical random variable $\hat{X}$ only has support on a
discrete set of values, and thus $\hat{Z}$ and $\hat{W}$ only take on discrete
values too. Let $E_{1},E_{2},\dots$ denote the values that the random
variables $\hat{X},\hat{Z},\hat{W}$ take on, and define the set
$\Omega=\\{E_{1},\dots,E_{d},\dots\\}$. Denote by $(r_{k})$ and $(s_{k})$
respectively the distributions of $\hat{Z}$ and $\hat{W}$ over $\Omega$.

Given these distributions, we first introduce two quantum systems $Z,W$, each
of countable dimension $|\Omega|$ with Hamiltonians $H_{Z},H_{W}$ and each
with an energy spectrum lying in $\Omega$. Define the states

$\displaystyle|\phi\rangle_{Z}$
$\displaystyle:=\sum_{k=1}^{|\Omega|}\sqrt{r_{k}}|E_{k}\rangle_{Z}$ (93)
$\displaystyle|\omega\rangle_{W}$
$\displaystyle:=\sum_{k=1}^{|\Omega|}\sqrt{s_{k}}|E_{k}\rangle_{W},$ (94)

on quantum systems $Z$ and $W$ where we label the eigenstates of each
Hamiltonian with their corresponding eigenvalues. We also introduce a third
system auxiliary $Y$ of arbitrary dimension, but with a zero energy eigenstate
$|0\rangle_{Y}$ for some Hamiltonian $H_{Y}$. We now construct an isometry
$V_{\mbox{\tiny tot}}$ from ${\mathcal{H}}_{X}\otimes{\mathcal{H}}_{Y}$ to
${\mathcal{H}}_{Z}\otimes{\mathcal{H}}_{W}$ that sends
$|\psi\rangle_{X}\otimes|0\rangle_{Y}$ to
$|\phi\rangle_{Z}\otimes|\omega\rangle_{W}$.

First observe that

$|\phi\rangle_{Z}\otimes|\omega\rangle_{W}=\sum_{z,w=1}^{|\Omega|}\sqrt{r_{z}s_{w}}|E_{z}\rangle_{Z}\otimes|E_{w}\rangle_{W}.$
(95)

However $\hat{X}=\hat{Z}+\hat{W}$, with $\hat{Z}$ and $\hat{W}$ independent
random variables. This implies that the distribution $(p_{x})$ of $\hat{X}$
over $\Omega$ is given by the convolution

$p_{x}=\sum_{\begin{subarray}{c}z,w:\\\
E_{x}=E_{z}+E_{w}\end{subarray}}q_{z}r_{w}.$ (96)

Therefore we instead write the above expression for the output states as

$\displaystyle|\phi\rangle_{Z}\otimes|\omega\rangle_{W}$
$\displaystyle=\sum_{z,w=1}^{|\Omega|}\sqrt{r_{z}s_{w}}|E_{z}\rangle_{Z}\otimes|E_{w}\rangle_{W}$
$\displaystyle=\sum_{E_{x}}\sum_{\begin{subarray}{c}z,w:\\\
E_{x}=E_{z}+E_{w}\end{subarray}}\sqrt{r_{z}s_{w}}|E_{z}\rangle_{Z}\otimes|E_{w}\rangle_{W}.$
(97)

We let $\Pi_{x}$ denote the projector onto the $E_{x}$ eigenspace of
${\mathcal{H}}_{X}$, and so for those $x$ for which $p_{x}\neq 0$ we have

$\Pi_{x}|\psi\rangle_{X}=\sqrt{p_{x}}|\psi_{x}\rangle_{X},$ (98)

which defines a normalised pure state $|\psi_{x}\rangle_{X}$ within the
$E_{x}$ eigenspace of ${\mathcal{H}}_{X}$. Thus

$|\psi\rangle_{X}=\sum_{x=1}^{|\Omega|}\sqrt{p_{x}}|\psi_{x}\rangle_{X}.$ (99)

We next define a linear mapping $V$ from ${\mathcal{H}}_{\mbox{\tiny
sup}}:=\mbox{span}\\{|\psi_{x}\rangle_{X}\otimes|0\rangle_{Y}\\}_{x}$ into
${\mathcal{H}}_{Z}\otimes{\mathcal{H}}_{W}$ for which

$V\left[|\psi_{x}\rangle_{X}\otimes|0\rangle_{Y}\right]=\frac{1}{\sqrt{p_{x}}}\sum_{\begin{subarray}{c}z,w:\\\
E_{x}=E_{z}+E_{w}\end{subarray}}\sqrt{r_{z}s_{w}}|E_{z}\rangle_{Z}\otimes|E_{w}\rangle_{W},$
(100)

for all $x=1,\dots|\Omega|$ for which $p_{x}\neq 0$. If $p_{x}\neq 0$ for some
$x$ then equation (96) implies that there is at least one pair $(z,w)$ such
that $q_{z}$ and $r_{w}$ are non-zero and so for such an $x$ the image under
$V$ is a non-zero vector.

_V is an isometry_ : It is readily checked, using the orthogonality of the
eigenstates of the output systems and equation (96) that

$\langle\psi_{x}|_{X}\otimes\langle
0|_{Y}V^{\dagger}V|\psi_{x}\rangle_{X}\otimes|0\rangle_{Y}=1$ (101)

for all $x$ for which $p_{x}\neq 0$. Note that since different energy
eigenspaces are orthogonal, we have both that the input states
$\\{|\psi_{x}\rangle_{X}\otimes|0\rangle_{Y}\\}_{x}$ are an orthonormal basis
for ${\mathcal{H}}_{\mbox{\tiny sup}}$, and also the output states are
orthonormal

$\langle\psi_{x}|_{X}\otimes\langle
0|_{Y}V^{\dagger}V|\psi_{y}\rangle_{X}\otimes|0\rangle_{Y}=0,$ (102)

for any $x\neq y$. The above provide the matrix components for $V$ and imply
that $V$ is an isometry from ${\mathcal{H}}_{\mbox{\tiny sup}}$ into
${\mathcal{H}}_{Z}\otimes{\mathcal{H}}_{W}$.

_V is energy conserving:_ Let
$H_{XY}:=H_{X}\otimes\mathbbm{1}_{Y}+\mathbbm{1}_{X}\otimes H_{Y}$ be the
initial Hamiltonian. Now since
${\mathcal{H}}_{X}\otimes{\mathcal{H}}_{Y}={\mathcal{H}}_{\mbox{\tiny
sup}}\oplus{\mathcal{H}}_{\mbox{\tiny sup}}^{\perp}$ and since
$|\psi_{x}\rangle_{X}\otimes|0\rangle_{Y}$ are eigenstates of $H_{XY}$ we can
write $H_{XY}=H_{XY,s}\oplus H_{XY,p}$, where $H_{XY,s}$ is simply the
restriction of the total Hamiltonian to the subspace
${\mathcal{H}}_{\mbox{\tiny sup}}$ and $H_{XY,p}$ is the component on
${\mathcal{H}}_{\mbox{\tiny sup}}^{\perp}$. Now each pair $(z,w)$ in equation
(100) contributes a vector
$\sqrt{r_{z}}|E_{z}\rangle_{Z}\otimes\sqrt{s_{w}}|E_{w}\rangle_{W}$ that is an
$E_{x}$ energy eigenvector of the total Hamiltonian
$H_{ZW}:=H_{Z}\otimes\mathbbm{1}_{W}+\mathbbm{1}_{Z}\otimes H_{W}$, and so
$V|\psi_{x}\rangle_{X}\otimes|0\rangle_{Y}$ is an $E_{x}$ energy eigenstate of
$H_{ZW}$ for any $x$. Thus $V$ sends $E_{x}$ eigenstates of $H_{XY,s}$ to
$E_{x}$ eigenstates of $H_{ZW}$ as required by energy conservation.

Finally, by rearranging equation (100) and summing over $x$ we see that
$V[|\psi\rangle_{X}\otimes|0\rangle_{Y}]=|\phi\rangle_{Z}\otimes|\omega\rangle_{W}$.
To get an isometry on the full space, we again write
${\mathcal{H}}_{X}\otimes{\mathcal{H}}_{Y}={\mathcal{H}}_{\mbox{\tiny
sup}}\oplus{\mathcal{H}}_{\mbox{\tiny sup}}^{\perp}$, and extend the systems
$Z$ and $W$ and their Hamiltonians so that now

${\mathcal{H}}_{Z}\otimes{\mathcal{H}}_{W}=\mbox{span}\\{|E_{z}\rangle_{Z}\otimes|E_{w}\rangle_{W}\\}_{z,w}\oplus{\mathcal{H}}_{\mbox{\tiny
other}}$ (103)

with $\mbox{dim}({\mathcal{H}}_{\mbox{\tiny
other}})\geq\mbox{dim}({\mathcal{H}}_{\mbox{\tiny sup}}^{\perp})$. The final
isometry on the input system is then $V_{\mbox{\tiny tot}}=V\oplus
V_{\mbox{\tiny other}}$, where $V_{\mbox{\tiny other}}$ is an arbitrary energy
conserving isometry from ${\mathcal{H}}_{\mbox{\tiny sup}}^{\perp}$ into
${\mathcal{H}}_{\mbox{\tiny other}}$ . This constructs a globally energy
conserving $V_{\mbox{\tiny tot}}$ such that

$V_{\mbox{\tiny
tot}}|\psi\rangle_{X}\otimes|0\rangle_{Y}=|\phi\rangle_{Z}\otimes|\omega\rangle_{W},$
(104)

and since both of $\hat{Z}$ and $\hat{W}$ are non-trivial random variables we
have that neither of the output states $|\omega\rangle_{W},|\phi\rangle_{Z}$
is an energy eigenstate of its respective Hamiltonian. Thus the state
$|\psi\rangle_{X}$ has decomposable coherence as required.

∎

###  A.3 Proof of semi-classical coherent work processes theorem

The proof of the semi-classical theorem relies on Raikov’s theorem, which is a
result for classical probability distributions. The core idea behind the proof
is outlined in Fig. 7. For the non-trivial direction, one starts with a tuple
$(S,|\psi\rangle_{S},H_{S})$ with $S$ a quantum system with Hamiltonian
$H_{S}$, and $|\psi\rangle_{S}$ a state of the system that admits a non-
trivial coherent work process. Any state $|\psi\rangle_{S}$ uniquely
determines a classical random variable $\hat{S}$, although the converse is not
true since we have freedom in, for example, choosing phases or the dimension
of $S$. If $\hat{S}$ is decomposable, then Raikov’s theorem tells us the
relationship between the components of the decomposition
$\hat{S}=\hat{S}^{\prime}+\hat{A}$. From this we can construct (non-unique)
tuples $(S^{\prime},|\phi\rangle_{S^{\prime}},H_{S^{\prime}})$ and
$(A,|\omega\rangle_{A},H_{A})$ associated to the random variables
$\hat{S}^{\prime}$ and $\hat{A}$ respectively. On these quantum systems we
then show that a coherent work process can be constructed. The precise
statement and proof are as follows.

Figure 7: Raikov’s theorem and coherent work processes. Given any quantum
system $S$ with Hamiltonian $H_{S}$ in a state $|\psi\rangle_{X}$ there is a
unique classical random variable $\hat{X}_{S}$ corresponding to the energy
measurement on $S$. Conversely, given a random variable $\hat{X}_{S}$ we may
construct a (non-unique) tuple $(S,|\psi\rangle_{S},H_{S})$ associated to this
random variable.

######  Theorem A.5.

Let $S$ be a harmonic oscillator system, with Hamiltonian $H_{S}=h\nu
a^{\dagger}a$. Let $\mathcal{C}$ be the set of quantum states for $S$ as
defined above. Then:

  1. 1.

The set of quantum states $\mathcal{C}$ is closed under all possible coherent
work processes from $S$ to $S$ with fixed Hamiltonian $H_{S}$ for both the
input and output.

  2. 2.

Given any quantum state $|\psi\rangle_{S}\in\mathcal{C}$ there is a unique,
canonical state $|\alpha,k\rangle_{S}\in\mathcal{C}$ such that we have a
reversible transformation

$|\psi\rangle_{S}\stackrel{{\scriptstyle\omega_{c}}}{{\longleftrightarrow}}|\alpha,k\rangle_{S},$
(105)

with $|\omega_{c}\rangle_{A}=|0\rangle_{A}$ and $\alpha=|\alpha|$. Moreover,
the only coherent work processes possible between canonical states are

$|\alpha,k\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\alpha^{\prime},k\rangle_{S},$
(106)

such that $|\alpha^{\prime}|\leq|\alpha|$. Modulo phases, the coherent work
output is of the form $|\omega\rangle_{A}=|\lambda,n\rangle_{A}$, where
$\lambda=\sqrt{|\alpha|^{2}-|\alpha^{\prime}|^{2}}$ and $n,k^{\prime}$ are any
integers that obey $n+k^{\prime}=k$.

  3. 3.

In the classical limit of large displacements $|\alpha|\gg 1$, from coherent
processes on $\mathcal{C}$ we recover all classical work processes on the
system $S$ under a conservative force.

###### Proof.

(Proof of 1.) Let $|\psi\rangle_{S}=e^{iL_{S}\theta}|\lambda,m\rangle_{S}$ be
a state in $\mathcal{C}$ defined with respect to a Hamiltonian $H_{S}=h\nu
a^{\dagger}a$, with $[L_{S},H_{S}]=0$, $L_{S}=L_{S}^{\dagger}$ and
$\theta\in\mathbb{R}$. Denote by $\hat{S}$ the classical random variable
obtained from an energy measurement of $S$, which takes values in
$\\{s_{n}=nh\nu:n\in\mathbb{Z}\\}$, and with distribution
$\bm{p}=(p_{n}=|\langle\psi|n\rangle|^{2})$. We can expand the state
$|\psi\rangle_{S}$ in the energy basis as

$|\psi\rangle=e^{-\lambda/2}\sum_{n=0}^{\infty}e^{i\theta_{n}}\sqrt{\frac{\lambda^{n}}{n!}}|n+m\rangle,$
(107)

for some arbitrary phases $\\{\theta_{n}\\}$ and some $m\in\mathbb{N}$. Denote
by $\sigma=h\nu$ the energy spacing of $H_{S}$. Then the energy measurement
distribution is Poissonian and given by

$p_{n}=P\left(\hat{S}=(n+m)h\nu\right)=e^{-\lambda}\frac{\lambda^{n}}{n!},$
(108)

for $n\geq 0$ and $p_{n}=0$ for $n<0$. The associated cumulative distribution
function is given by

$F_{\hat{S}}(x)=F\left(\frac{x-mh\nu}{\sigma},\lambda\right).$ (109)

The set $\mathcal{C}$ is clearly closed under trivial coherent work processes.
Now suppose that $|\psi\rangle_{S}$ admits some non-trivial coherent work
process given by

$|\psi\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\phi\rangle_{S^{\prime}}.$
(110)

By Theorem 2.5 this implies that $\hat{S}=\hat{S}^{\prime}+\hat{A}$, where
$\hat{S}^{\prime}$ and $\hat{A}$ are non-trivial, independent classical random
variables. From these we can construct tuples
$(S^{\prime},|\phi\rangle_{S^{\prime}}.H_{S^{\prime}})$ and
$(A,|\omega\rangle_{A},H_{A})$ that realise the associated coherent work
process, as detailed in the proof of Theorem 2.5. For any
$|\psi\rangle_{S}\in\mathcal{C}$ we have that
$\hat{S}=\hat{S}^{\prime}+\hat{A}$ is a decomposable Poisson random variable,
and thus by Raikov’s theorem the cumulative distribution functions for
$\hat{S}^{\prime}$ and $\hat{A}$ are also Poissonian, with

$\displaystyle F_{\hat{S}^{\prime}}(x)$
$\displaystyle=F\left(\frac{x-\alpha_{1}}{\sigma},\mu\right),$ (111)
$\displaystyle F_{\hat{A}}(x)$
$\displaystyle=F\left(\frac{x-\alpha_{2}}{\sigma},\kappa\right),$ (112)

where $\nu,\kappa\geq 0$ and constrained as $\mu+\kappa=\lambda$,
$\alpha_{1}+\alpha_{2}=mh\nu$ and $\alpha_{1},\alpha_{2}\in\mathbb{R}$.

As $\sigma$ appears identically in all cumulative distribution functions this
implies that both $H_{S^{\prime}}$ and $H_{A}$ must have within their spectral
decompositions, countably infinite discrete energy eigenstates with equal
spacings $h\nu$, corresponding to the outcomes of the energy measurements for
the constructed states $|\phi\rangle_{S^{\prime}}$ and $|\omega\rangle_{A}$.
We write the corresponding distributions for $\hat{S}^{\prime}$ and $\hat{A}$
as

$\displaystyle q_{n}:=P[\hat{S}^{\prime}=nh\nu+c_{S^{\prime}}]$
$\displaystyle=e^{-\mu}\frac{\mu^{n}}{n!}$ $\displaystyle
r_{a}:=P[\hat{A}=ah\nu+c_{A}]$
$\displaystyle=e^{-\kappa}\frac{\kappa^{a}}{a!},$ (113)

for integers $n,a\geq 0$, and where $c_{S^{\prime}}$ and $c_{A}$ are any
constants that obey

$c_{S^{\prime}}+c_{A}=mh\nu.$ (114)

By suitable constant shifts of the Hamiltonians $H_{S^{\prime}}$ and $H_{A}$,
that shift the zero energy level, we can ensure that
$c_{S^{\prime}}=m^{\prime}h\nu$ for some integer $m^{\prime}$, and thus
$c_{A}=(m-m^{\prime})h\nu$. This implies that both $(q_{n})$ and $(r_{a})$ are
both Poissonian distributions on a discrete set of energy eigenvalues with
constant separation $h\nu$.

Let $\Pi_{n}^{S^{\prime}}$ and $\Pi^{A}_{a}$ denote the projectors for the
discrete energy eigenvalues in the energy measurements of $S^{\prime}$ and $A$
with probabilities $q_{n}$ and $r_{a}$ on $|\phi\rangle_{S^{\prime}}$ and
$|\omega\rangle_{A}$ respectively. We now define the states
$|n+m^{\prime}\rangle_{S^{\prime}}$ and $|a+m-m^{\prime}\rangle_{A}$ via

$\displaystyle\sqrt{q_{n}}|n+m^{\prime}\rangle_{S^{\prime}}$
$\displaystyle:=\Pi_{n}^{S^{\prime}}|\phi\rangle_{S^{\prime}}$ (115)
$\displaystyle\sqrt{r_{a}}|a+m-m^{\prime}\rangle_{A}$
$\displaystyle:=\Pi_{a}^{A}|\omega\rangle_{A},$ (116)

with $n,a\geq 0$ and where we have ensured trivial phases via the definition
of the states. Therefore we can write the two output states as

$\displaystyle|\phi\rangle_{S^{\prime}}$
$\displaystyle=e^{-\mu/2}\sum_{n=0}^{\infty}\sqrt{\frac{\mu^{n}}{n!}}|n+m^{\prime}\rangle_{S^{\prime}}$
$\displaystyle|\omega\rangle_{A}$
$\displaystyle=e^{-\kappa/2}\sum_{a=0}^{\infty}\sqrt{\frac{\kappa^{a}}{a!}}|a+m-m^{\prime}\rangle_{A}.$
(117)

Recall that the translated coherent states are given by

$\displaystyle D(\alpha)|0\rangle=|\alpha\rangle$
$\displaystyle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}}{n!}|n\rangle$
(118) $\displaystyle|\alpha,m\rangle$
$\displaystyle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}}{n!}|n+m\rangle.$
(119)

and so we can define real constants $\beta_{1}=\sqrt{\mu}$ and
$\beta_{2}=\sqrt{\kappa}$ that allow us to write
$|\phi\rangle_{S^{\prime}}=|\beta_{1},m^{\prime}\rangle_{S^{\prime}}$ and
$|\omega\rangle_{A}=|\beta_{2},m-m^{\prime}\rangle_{A}$, in terms of the
notation for $\mathcal{C}$ states. Therefore
$|\phi\rangle_{S^{\prime}},|\omega\rangle_{A}\in\mathcal{C}$ and so
$\mathcal{C}$ is closed under coherent work processes.

(Proof of 2.) Any state in $\mathcal{C}$ can be expressed in the form

$|\psi\rangle_{S}=e^{-\lambda/2}\sum_{n=0}^{\infty}e^{i\theta_{n}}\sqrt{\frac{\lambda^{n}}{n!}}|n+m\rangle,$
(120)

where $m$ is some non-negative integer, $\lambda\geq 0$ and $\\{\theta_{n}\\}$
are phases. We now define a unitary of the form $V=V_{S}\otimes\mathds{1}_{A}$
where

$V_{S}=\sum_{n=0}^{\infty}e^{-i\theta_{n}}|n+m\rangle\langle
n+m|+\sum_{k=0}^{m-1}|k\rangle\langle k|,$ (121)

which realises the transformation
$V[|\psi\rangle_{S}\otimes|0\rangle_{A}]=|\sqrt{\lambda},m\rangle_{S}\otimes|0\rangle_{A}$
with $\lambda\geq 0$. It is clear that both $V$ and its inverse $V^{\dagger}$
are energy conserving and so we have a reversible coherent work process with
$|\omega_{c}\rangle_{A}=|0\rangle_{A}$ as claimed.

As constructed in the proof of part 1, we saw that modulo arbitrary phases
between the energy eigenstates, any coherent work process on a state
$|\alpha,m\rangle_{S}$ takes the form

$|\alpha,m\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\beta_{1},m^{\prime}\rangle_{S^{\prime}}$
(122)

and with $|\omega\rangle_{A}=|\beta_{2},m-m^{\prime}\rangle_{A}$. With a
freedom to shift the energies of $S^{\prime}$ and $A$ by a constant factor
$\pm c$ respectively. Moreover, since the Poisson parameters are given, as
above, by $\lambda=|\alpha|^{2}$, $\mu=|\beta_{1}|^{2}$ and
$\kappa=|\beta_{2}|^{2}$, Raikov’s theorem tells us that we must have
$|\alpha|^{2}=|\beta_{1}|^{2}+|\beta_{2}|^{2}$, and therefore
$|\beta_{2}|=\sqrt{|\alpha|^{2}-|\beta_{1}|^{2}}$ as claimed.

(Proof of 3.) For a classical work process under a conservative force, we need
two conditions to hold: (i) the position, momentum and energy are sharp and
(ii) the process depends only on the initial and final energies, not the path.
For the first, consider the position and momentum operators
$q=\sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a)$ and $p=i\sqrt{\frac{\hbar
m\omega}{2}}(a^{\dagger}-a)$ with the canonical commutation relation
$[q,p]=i\hbar$, and $\hbar\omega=h\nu$. It is straightforward to verify that
the variances satisfy $\sigma_{q}^{2}=\frac{\hbar}{2m\omega}$ and
$\sigma_{p}^{2}=\frac{\hbar m\omega}{2}$ for any coherent state. Then coherent
states are minimal uncertainty states with Heisenberg uncertainty relation
$\sigma_{q}^{2}\sigma_{p}^{2}=\frac{\hbar^{2}}{4}$. For a coherent state
$|\alpha\rangle$, the expectation value for the position and momentum are
$\langle q\rangle_{\alpha}=\sqrt{\frac{\hbar}{2m\omega}}(\alpha^{*}+\alpha)$
and $\langle p\rangle_{\alpha}=i\sqrt{\frac{\hbar
m\omega}{2}}(\alpha^{*}-\alpha)$. Therefore

$\displaystyle\frac{\sigma_{q}}{\langle q\rangle_{\alpha}}$
$\displaystyle=\frac{1}{(\alpha^{*}+\alpha)},$ (123)
$\displaystyle\frac{\sigma_{p}}{\langle p\rangle_{\alpha}}$
$\displaystyle=\frac{i}{(\alpha-\alpha^{*})}.$ (124)

In the limit of large displacement $|\alpha|\gg 1$, the values for $\langle
q\rangle_{\alpha}$ and $\langle p\rangle_{\alpha}$ become arbitrarily sharp
with the ratio of standard deviation to value tending to zero. Similarly, for
a Harmonic oscillator Hamiltonian $H_{S}=\hbar\omega a^{\dagger}a$, the
variance of energy is $\sigma_{H}^{2}=\hbar^{2}\omega^{2}|\alpha|^{2}$ leading
to the ratio of the standard deviation to the mean

$\frac{\sigma_{H}}{\langle H_{S}\rangle_{\alpha}}=\frac{1}{|\alpha|},$ (125)

which again becomes sharp in the limit in the same manner. Thus, all energy
transfers, positions and momenta become sharp in the limit of large
displacements and the classical mechanics of a harmonic oscillator is
recovered.

For (ii), the energy transfer depends only on the initial and final states of
the system $|\alpha\rangle_{S}$ and $|\alpha^{\prime}\rangle_{S}$. As all
coherent states are included in $\mathcal{C}$, then coherent work processes on
$\mathcal{C}$ states with large displacement are classical work processes
under a conservative force. Furthermore, for the transition
$|\alpha\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\alpha^{\prime}\rangle_{S}$,
we have that $|\alpha|\geq|\alpha^{\prime}|$ and therefore $\langle
H_{S}\rangle_{\alpha}\geq\langle H_{S}\rangle_{\alpha^{\prime}}$ – meaning the
process on $S$ is work extracting only for the coherence component of
$|\alpha,k\rangle$. However variations in the $k$ label are always sharp and
can involve either positive or negative work and it is clear that for large
displacements the fractional energy separation of levels tends to zero and one
has an effective continuum of energy values that recovers the classical case.

∎

###  A.4 Proof of Theorem II.6 (coherent work processes on a ladder system)

######  Theorem A.6.

Let $S$ be a quantum system, with energy eigenbasis
$\\{|n\rangle:n\in\mathbb{Z}\\}$ with fixed Hamiltonian
$H_{S}=\sum_{n\in\mathbb{Z}}n|n\rangle\langle n|$. Given an initial quantum
state $|\psi\rangle_{S}=\sum_{i}\sqrt{p_{i}}|i\rangle_{S}$, a coherent work
process

$|\psi\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\phi\rangle_{S},$
(126)

is possible with

$\displaystyle|\phi\rangle_{S}$
$\displaystyle=\sum_{j}e^{i\theta_{j}}\sqrt{q_{j}}|j\rangle_{S}$ (127)
$\displaystyle|\omega\rangle_{A}$
$\displaystyle=\sum_{k}e^{i\varphi_{k}}\sqrt{r_{k}}|k\rangle_{A}$ (128)

for arbitrary phases $\\{\theta_{j}\\}$ and $\\{\varphi_{k}\\}$, with output
Hamiltonian $H_{S}$ as above and
$H_{A}=\sum_{n\in\mathbb{Z}}n|n\rangle_{A}\langle n|$, if and only the
distribution $(p_{n})$ over $\mathbb{Z}$ can be written as

$\displaystyle p_{n}$
$\displaystyle=\sum_{j\in\mathbb{Z}}r_{j}(\Delta^{j}\mathbf{q})_{n}.$ (129)

for distributions $(q_{m})$ and $(r_{j})$ over $\mathbb{Z}$.

###### Proof.

Assume
$|\psi\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\phi\rangle_{S}$.
Then there exists a unitary $V$ such that
$V|\psi\rangle_{S}\otimes|0\rangle_{A}=|\phi\rangle_{S}\otimes|\omega\rangle_{A}$.
Furthermore, the unitary satisfies the global energy constraint
$[V,H_{S}+H_{A}]=0$, where $H_{S}$ and $H_{A}$ are the Hamiltonians of $S$ and
$A$ respectively.

We expand $|\phi\rangle_{S}$ and $|\omega\rangle_{A}$ in their energy
eigenbases as

$\displaystyle|\phi\rangle_{S}$
$\displaystyle=\sum_{j}e^{i\theta_{j}}\sqrt{q_{j}}|j\rangle_{S}$
$\displaystyle|\omega\rangle_{A}$
$\displaystyle=\sum_{k}e^{i\varphi_{k}}\sqrt{r_{k}}|k\rangle_{A}.$

If $V$ provides a valid coherent work process then so too does the unitary
$UV$ where $U=e^{iB_{S}\theta}\otimes e^{iC_{A}\varphi}$ for any Hermitian
$B_{S}$ and $C_{A}$ such that $[B_{S},H_{S}]=[C_{A},H_{A}]=0$. By choosing the
operators $B_{S}$ and $C_{A}$ appropriately one can freely vary the phases
$\theta_{j}$ and $\varphi_{k}$. We have that $A$ has the same Hamiltonian
ladder structure as $S$. Energy conservation implies that for any
$n\in\mathbb{Z}$, the action of $V$ on the state
$|n\rangle_{S}\otimes|0\rangle_{A}$ gives

$V|n\rangle_{S}\otimes|0\rangle_{A}=\sum_{j\in\mathbb{Z}}e^{i\alpha_{n,j}}\sqrt{m_{j|n}}|n-j\rangle_{S}\otimes|j\rangle_{A}.$
(130)

for some distribution $(m_{j|n})_{j}$ over $j$, and phases $(\alpha_{n,j})$.
This implies that

$\displaystyle|\phi\rangle_{S}\otimes|\omega\rangle_{A}$
$\displaystyle=\sum_{n,j\in\mathbb{Z}}e^{i\alpha_{n,j}}\sqrt{m_{j|n}p_{n}}|n-j\rangle_{S}\otimes|j\rangle_{A},$
(131)
$\displaystyle=\sum_{k,l\in\mathbb{Z}}e^{i(\theta_{k}+\varphi_{l})}\sqrt{q_{k}r_{l}}|k\rangle_{S}\otimes|l\rangle_{A}.$
(132)

Let $\Pi_{n}$ be the projector onto the energy eigenspace of total energy
$n\in\mathbb{Z}$ for the joint system $SA$. We thus have that

$\displaystyle\langle\phi|_{S}\otimes\langle\omega|_{A}\Pi_{n}|\phi\rangle_{S}\otimes|\omega\rangle_{A}$
$\displaystyle=\sum_{j\in\mathbb{Z}}m_{j|n}p_{n},$ (133)
$\displaystyle=\sum_{j\in\mathbb{Z}}q_{n-j}r_{j},$ (134)

However $\sum_{j}m_{j|n}=1$ and so we have that the distribution $(p_{n})$
decomposes as

$p_{n}=\sum_{j\in\mathbb{Z}}r_{j}q_{n-j}=\sum_{j\in\mathbb{Z}}r_{j}(\Delta^{j}\mathbf{q})_{n},$
(135)

as required. Conversely, suppose we have a distribution $(p_{n})$ over
$\mathbb{Z}$ which can be decomposed as

$p_{n}=\sum_{j\in\mathbb{Z}}r_{j}q_{n-j}=\sum_{j\in\mathbb{Z}}r_{j}(\Delta^{j}\mathbf{q})_{n},$
(136)

for two distributions $(r_{j})$ and $(q_{k})$. From these we define for ladder
systems $S$ and $A$ the states

$|\psi\rangle_{S}=\sum_{i}\sqrt{p_{i}}|i\rangle_{S},$ (137)

and

$\displaystyle|\phi\rangle_{S}$
$\displaystyle=\sum_{j}e^{i\theta_{j}}\sqrt{q_{j}}|j\rangle_{S}$ (138)
$\displaystyle|\omega\rangle_{A}$
$\displaystyle=\sum_{k}e^{i\varphi_{k}}\sqrt{r_{k}}|k\rangle_{A}$ (139)

where $\theta_{j}$ and $\varphi_{k}$ are arbitrary phases. The joint system
$SA$ therefore has the two states

$\displaystyle|\psi\rangle_{S}\otimes|0\rangle_{A}$
$\displaystyle=\sum_{i}\sqrt{p_{i}}|i\rangle_{S}\otimes|0\rangle_{A}$
$\displaystyle|\phi\rangle_{S}\otimes|\omega\rangle_{A}$
$\displaystyle=\sum_{k,l\in\mathbb{Z}}e^{i(\theta_{k}+\varphi_{l})}\sqrt{q_{k}r_{l}}|k\rangle_{S}\otimes|l\rangle_{A}.$
$\displaystyle=U^{\dagger}\sum_{k,l\in\mathbb{Z}}\sqrt{q_{k}r_{l}}|k\rangle_{S}\otimes|l\rangle_{A},$
(140)

where $U^{\dagger}$ is a phase unitary that generates the phases $\theta_{k}$,
$\varphi_{j}$ and which is diagonal in the energy eigenbasis. Again, we let
$\Pi_{n}$ denote the projector onto the $n$–eigenspace of the system $SA$ and
so

$\displaystyle\Pi_{n}|\psi\rangle_{S}$
$\displaystyle\otimes|0\rangle_{A}=\sqrt{p_{n}}|i\rangle_{S}\otimes|0\rangle_{A}$
$\displaystyle\\!\\!\\!\Pi_{n}U|\phi\rangle_{S}\otimes|\omega\rangle_{A}$
$\displaystyle=\\!\\!\\!\sum_{m\in\mathbb{Z}}\sqrt{q_{n-m}r_{m}}|n-m\rangle_{S}|m\rangle_{A}.$
(141)

Thus we have that

$\langle\phi|_{S}\otimes\langle\omega|_{A}U^{\dagger}\Pi_{n}U|\phi\rangle_{S}\otimes|\omega\rangle_{A}=\sum_{m}q_{n-m}r_{m}=p_{n},$
(142)

and thus the vectors $\Pi_{n}|\psi\rangle_{S}\otimes|0\rangle_{A}$ and
$\Pi_{n}U|\phi\rangle_{S}\otimes|\omega\rangle_{A}$ both lie in the
$n$–eigenspace of the total hamiltonian and have equal norms. Therefore, we
may define a unitary $V_{n}$ within each energy eigenspace such that

$V_{n}\Pi_{n}|\psi\rangle_{S}\otimes|0\rangle_{A}=\Pi_{n}U|\phi\rangle_{S}\otimes|\omega\rangle_{A}.$
(143)

For those $n$–eigenspaces outside the support of
$|\psi\rangle_{S}\otimes|0\rangle_{A}$ we simply let $V_{n}$ be the identity
unitary on that subspace. We now write $V:=\oplus_{n}V_{n}$ to obtain an
energy conserving unitary on the full space, and using the fact that $U$
commutes with the energy projectors we have that

$U^{\dagger}V|\psi\rangle_{S}\otimes|0\rangle_{A}=|\phi\rangle_{S}\otimes|\omega\rangle_{A},$
(144)

where $U^{\dagger}V$ is an energy conserving unitary on the joint system.
Therefore the required coherent work process exists.

∎

###  A.5 Physical example of decomposable coherence

In the main text, an example of infinitely divisible coherence was provided in
the form of a coherent state. Here we provide a physical system in which one
might encounter such examples of coherent state manipulations. Consider a beam
splitter set up, with a coherent state of light injected as the input on one
mode and the vacuum on the other mode, such that the initial state is
$|\alpha\rangle_{S}\otimes|0\rangle_{A}$. We use the beam splitter unitary

$B(\theta)=e^{\theta(ab^{\dagger}-a^{\dagger}b)}$ (145)

where $a$ and $b$ are the annihilation operators for $S$ and $A$ respectively
[71]. Under the action of a beam splitting, which conserves photon number, we
want to achieve

$\displaystyle
B(\theta)|\alpha\rangle\otimes|0\rangle=|\beta\rangle\otimes|\omega\rangle,$
(146)

for some tunable parameter $\theta$. It is well known that the two mode
annihilation operators satisfy the algebraic property:

$\displaystyle B^{\dagger}(\theta)aB(\theta)$
$\displaystyle=a\cos\theta+b\sin\theta$ (147) $\displaystyle
B^{\dagger}(\theta)bB(\theta)$ $\displaystyle=b\cos\theta-a\sin\theta.$ (148)

Since coherent states are displaced vacuum states, we have that

$\displaystyle B(\theta)|\alpha\rangle\otimes|0\rangle$
$\displaystyle=B(\theta)e^{-\frac{1}{2}|\alpha|^{2}}e^{\alpha
a^{\dagger}}e^{-\alpha^{*}a}|0\rangle\otimes|0\rangle$ (149)
$\displaystyle=e^{-\frac{1}{2}|\alpha|^{2}}B(\theta)e^{\alpha
a^{\dagger}}B^{\dagger}(\theta)|0\rangle\otimes|0\rangle$ (150)
$\displaystyle=e^{-\frac{1}{2}|\alpha|^{2}}e^{\alpha
B(\theta)a^{\dagger}B^{\dagger}(\theta)}|0\rangle\otimes|0\rangle$ (151)
$\displaystyle=e^{-\frac{1}{2}|\alpha|^{2}}e^{\alpha(a^{\dagger}\cos\theta-b^{\dagger}\sin\theta)}|0\rangle\otimes|0\rangle$
(152) $\displaystyle=|\alpha\cos\theta\rangle\otimes|-\alpha\sin\theta\rangle$
(153)

where we used the fact that
$B(\theta)|0\rangle\otimes|0\rangle=|0\rangle\otimes|0\rangle$ and $B(\theta)$
is unitary. Here, $r=\cos\theta$ and $t=\sin\theta$ are the transmission and
reflection coefficients. The consequence is that we have achieved the desired
coherent state splitting with a photon number conserving unitary. The second
mode that began in the vacuum therefore carries information of the coherent
energy transfer. As expected, the condition
$|\alpha|^{2}=|\alpha\cos\theta|^{2}+|\alpha\sin\theta|^{2}$ is satisfied.
If one were to perform measurements on the ancilla system, they would extract
information on the energetics and coherences. However, by performing the
measurement, the state would no longer be available for manipulations that
involve interference effects. For this reason, the coherent work output is the
state $|\alpha\sin\theta\rangle$ rather than a post-processed state.

##  Appendix B Fluctuation relations and the effective potential

Here we gather together the proofs for the fluctuation relation analysis.

###  B.1 Basic properties of the effective potential.

It is worth describing some properties of the function $\Lambda(\beta,\rho)$
to shed light on the fluctuation relations.

We first note that the effective potential has an invariance under the group
generated by the observable $H$. More precisely, let $U(t)$ be the unitary
representation of time-translations generated by $H$. Then
$\Lambda(\beta,U(t)\rho U(t)^{\dagger})=\Lambda(\beta,\rho)$ for any
$t\in\mathbb{R}$, and so we have that
$\Lambda(\beta,\rho(t_{1}))=\Lambda(\beta,\rho(t_{2}))$ for any two times
$t_{1}$, and $t_{2}$. Thus, $\Lambda$ is a constant of motion under the free
Hamiltonian evolution. For a pure quantum state
$|\psi\rangle=\sum_{k}e^{i\theta_{k}}\sqrt{p_{k}}|E_{k}\rangle$, $\Lambda$ is
purely a function of the classical distribution $p_{k}$ over the energy
eigenstates. For a projective measurement of energy in this eigenbasis we can
denote by $\hat{X}$ the classical random variable for the energy obtained in
the measurement, then it is readily seen that $\Lambda(\beta,\psi)$ is
essentially the negative cumulant generating function for $\hat{X}$ in the
inverse temperature $\beta$,

$\Lambda(\beta,\psi)=-\sum_{n\geq
1}\frac{(-1)^{n}\beta^{n}}{n!}\kappa_{n}(\hat{X}).$ (154)

We shall make use of the following theorem for our analysis.

######  Theorem B.1.

If $A$ is Hermitian and $Y$ is strictly positive, then

$\displaystyle\ln\mbox{tr}(e^{A+\ln Y})=\max\\{$
$\displaystyle\mbox{tr}(XA)-S(X||Y)\ :\ X\ \text{is positive }$
$\displaystyle\&\ \mbox{tr}X=1\\}.$

Alternatively, if $X$ is positive with $\operatorname{Tr}(X)=1$ and $B$ is
Hermitian, then

$S(X||e^{B})=\max\\{\mbox{tr}(XA)-\log\mbox{tr}(e^{A+B})\ :\
A=A^{\dagger}\\}.$

The proof of this can be found in [72].

We now establish basic properties of the effective potential
$\Lambda(\beta,\rho)$. As a function of the quantum state $\rho$, the
effective potential has the following properties.

######  Lemma B.2.

Given a quantum system $S$ with Hamiltonian $H$ , and inverse temperature
$\beta>0$. We have that

  1. 1.

(Invariance) Let $\rho(t)=e^{-iHt}\rho e^{+iHt}$ for all $t\in\mathbb{R}$. For
any state $\rho$ and any $t\in\mathbb{R}$ we have
$\Lambda(\beta,\rho(t))=\Lambda(\beta,\rho(0))$. Moreover, we have that
$\Lambda(\beta,\rho)=\Lambda(\beta,{\mathcal{D}}(\rho))$ where
${\mathcal{D}}(\rho)=\sum_{k}\Pi_{k}\rho\Pi_{k}$ is the dephasing of $\rho$ in
the energy basis, with $\Pi_{k}$ being the projector onto the $k$’th energy
eigenspace of $H$.

  2. 2.

For any state $\rho$, we have that

$\beta\langle H\rangle_{\rho}-\frac{\beta^{2}}{8}(E^{\rm max}-E^{\rm
min})^{2}\leq\Lambda(\beta,\rho)\leq\beta\langle H\rangle_{\rho}.$ (155)

  3. 3.

Let $E^{\rm min}$ and $E^{\rm max}$ be the smallest and largest eigenvalues of
$H$ respectively in the support of $\rho$. Then

$\beta E^{\rm min}\leq\Lambda(\beta,\rho)\leq\beta E^{\rm max}.$ (156)

  4. 4.

(Variational form) For states $\rho$ of full rank and $[\rho,H]=0$, we have
that

$\Lambda(\beta,\rho)=\min_{\sigma}\left[\beta\langle
H\rangle_{\sigma}+S(\sigma||\rho)\right],$ (157)

where the minimization is taken over all quantum states $\sigma$ of the system
$S$.

  5. 5.

(Additivity) For a bipartite system $S_{1}S_{2}$ with total Hamiltonian
$H_{12}=H_{1}\otimes\mathbbm{1}_{2}+\mathbbm{1}_{1}\otimes H_{2}$ and a
bipartite product state $\rho_{12}=\rho_{1}\otimes\rho_{2}$ we have
$\Lambda(\beta,\rho_{1}\otimes\rho_{2})=\Lambda_{1}(\beta,\rho_{1})+\Lambda_{2}(\beta,\rho_{2})$,
where the effective potentials for subsystems are defined solely with respect
to the corresponding Hamiltonian on that system.

###### Proof.

1\. For any $t\in\mathbb{R}$ we have
$\Lambda(\beta,\rho(t))=-\ln\mbox{tr}(e^{-\beta H}e^{-iHt}\rho
e^{+iHt})=\Lambda(\beta,\rho)$, by cyclicity of the trace. Moreover, we can
evaluate the trace in an energy eigenbasis of $H$, and so denoting by
$\Pi_{k}$ the projector onto the $k$’th eigenspace of $H$ we have that

$\displaystyle-\ln\mbox{tr}(e^{-\beta H}\rho)$
$\displaystyle=-\ln\mbox{tr}(\sum_{k}\Pi_{k}\Pi_{k}(e^{-\beta H}\rho))$
$\displaystyle=-\ln\mbox{tr}(\sum_{k}\Pi_{k}(e^{-\beta H}\rho)\Pi_{k})$
$\displaystyle=-\ln\mbox{tr}(\sum_{k}e^{-\beta H}\Pi_{k}\rho\Pi_{k})$
$\displaystyle=-\ln\mbox{tr}(e^{-\beta
H}{\mathcal{D}}(\rho))=\Lambda(\beta,{\mathcal{D}}(\rho)).$ (158)

2\. The proof of the upper bound is a direct result of Jensen’s inequality,
$\langle e^{-\beta H}\rangle_{\rho}\geq e^{-\beta\langle H\rangle_{\rho}}$,
which implies

$\Lambda(\beta,\rho)\leq-\ln e^{-\beta\langle H\rangle_{\rho}}=\beta\langle
H\rangle_{\rho},$ (159)

achieving the desired result.

The lower bound is obtained as follows. Given a state $\rho$, let us define
the zero mean observable $H_{0}=H-\langle H\rangle_{\rho}\mathds{1}$. It is
readily seen that the effective potential satisfies
$\Lambda(\beta,\rho)=\Lambda_{0}(\beta,\rho)+\beta\langle H\rangle_{\rho}$,
where $\Lambda_{0}(\beta,\rho)$ is evaluated for the Hamiltonian $H_{0}$. Let
us denote the random variable $\hat{X}_{0}$ with sample space
$\text{eigs}(H_{0})$, distributed by $\rho$. Then $\hat{X}_{0}$ has zero mean
and $E^{\rm min}\leq\hat{X}_{0}\leq E^{\rm max}$ almost surely. Thus we can
apply Hoeffding’s lemma [73], which states

$\mathbb{E}[e^{\lambda\hat{X}_{0}}]\leq e^{\frac{\lambda^{2}}{8}(E^{\rm
max}-E^{\rm min})^{2}}$ (160)

for any $\lambda\in\mathbb{R}$. Taking the logarithm and re-ordering the
inequality, $\Lambda_{0}(\beta,\rho)\geq-\frac{\beta^{2}}{8}(E^{\rm
max}-E^{\rm min})^{2}$, which gives the lower bound.

3\. From 2, we know that $\Lambda(\beta,\rho)\leq\beta\langle
H\rangle_{\rho}\leq\beta E^{\rm max}$ as the expectation value cannot exceed
the greatest eigenvalue. To prove the lower bound, consider a state $\rho$
with $\text{diag}(\rho)=\bm{p}$ and assume $E_{k}=E^{\rm min}\implies
p_{k}\neq 0$ without loss of generality. Then

$\displaystyle\Lambda(\beta,\rho)$ $\displaystyle=-\ln\big{(}e^{-\beta E^{\rm
min}}[p_{k}+\sum_{j\neq k}p_{j}e^{-\beta(E_{j}-E^{\rm min})}]\big{)}$ (161)
$\displaystyle=\beta E^{\rm min}-\ln\big{(}p_{k}+\sum_{j\neq
k}p_{j}e^{-\beta(E_{j}-E^{\rm min})}\big{)}$ (162)

By assumption, $E^{\rm min}$ is the smallest eigenvalue of $H$ in the support
of $\rho$ and therefore $e^{-\beta(E_{n}-E^{\rm min})}\leq 1$ for all $n$.
From the property $\sum_{i}p_{i}=1$, the argument of the logarithm is smaller
than one and therefore the latter term is positive. It therefore follows that
$\beta E^{\rm min}\leq\Lambda(\beta,\rho)\leq\beta E^{\rm max}$.

4\. With an appropriate choice of operators $A=-\beta H$ and $Y=\rho$, Theorem
B.1 implies $\Lambda(\beta,\rho)=-\max_{\sigma}\\{-\beta\langle
H\rangle_{\sigma}-S(\sigma||\rho)\\}=\min_{\sigma}\\{\beta\langle
H\rangle_{\sigma}+S(\sigma||\rho)\\}$ for states $\sigma$. However as
$\ln\rho$ will diverge for non-full rank states, and so the assumption that
$\rho$ must be full rank is required.
5\. Since $H_{12}=H_{1}\otimes\mathbbm{1}_{2}+\mathbbm{1}_{1}\otimes H_{2}$ we
obtain

$\displaystyle\Lambda(\beta,\rho_{1}\otimes\rho_{2})$
$\displaystyle=-\ln\mbox{tr}(e^{-\beta(H_{1}\otimes\mathds{1}+\mathds{1}\otimes
H_{2})}\rho_{1}\otimes\rho_{2})$ (163)
$\displaystyle=-\ln\operatorname{Tr}(e^{-\beta
H_{1}}\rho_{1})\mbox{tr}(e^{-\beta H_{2}}\rho_{2})$ (164)
$\displaystyle=\Lambda_{1}(\beta,\rho_{1})+\Lambda_{2}(\beta,\rho_{2}).$ (165)

∎

The effective potential has the following properties in terms of the parameter
$\beta$.

######  Lemma B.3.

Given a quantum system $S$ with Hamiltonian $H$. For any state $\rho$ of $S$,
the effective potential $\Lambda(\beta,\rho)$ obeys

  1. 1.

$\Lambda(\beta,\rho)$ is concave in $\beta$ for all $\beta\geq 0$.

  2. 2.

$\Lambda(\beta,\rho)$ is monotone increasing in the variable $\beta$, for all
states $\rho$, if and only if $H\geq 0$.

  3. 3.

(High temperature regime) $\lim_{s\rightarrow
0}s^{-1}\Lambda(s\beta,\rho)=\beta\langle H\rangle_{\rho}$.

  4. 4.

(Low temperature regime)
$\lim_{s\rightarrow\infty}s^{-1}\Lambda(s\beta,\rho)=\beta E^{\rm min}$, where
$E^{\rm\tiny min}$ is the smallest eigenvalue of $H$ in the support of $\rho$.

###### Proof.

1\. Let $a,b>0$ satisfy $a+b=1$. For two different inverse temperatures
$\beta_{1}$ and $\beta_{2}$ and Hamiltonian $H$, we have:

$\displaystyle\Lambda(a\beta_{1}+b\beta_{2},\rho)$
$\displaystyle=-\ln\mbox{tr}(\rho e^{-a\beta_{1}H}e^{-b\beta_{2}H}).$ (166)

Let us define the operators $X=e^{-a\beta_{1}H}$ and $Y=e^{-b\beta_{2}H}$,
where $X,Y\geq 0$. Then Hölder’s inequality tells us expectation values
satisfy
$\mathbb{E}[|XY|]\leq(\mathbb{E}[|X|^{p}])^{1/p}(\mathbb{E}[|Y|^{q}])^{1/q}$,
where $\frac{1}{p}+\frac{1}{q}=1$. Choosing $p=\frac{1}{a}$ and
$q=\frac{1}{b}$, we get:

$\displaystyle\Lambda(a\beta_{1}+b\beta_{2},\rho)\geq
a\Lambda(\beta_{1},\rho)+b\Lambda(\beta_{2},\rho).$ (167)

Therefore $\Lambda$ is concave in the inverse temperature.
2\. First suppose that $H\geq 0$, then

$\frac{\partial\Lambda(\beta,\rho)}{\partial\beta}=\frac{\mbox{tr}(He^{-\beta
H}\rho)}{\mbox{tr}(e^{-\beta H}\rho)}$ (168)

For $\beta\in\mathbb{R}$, the denominator is always strictly positive. The
nominator can be expanded using the spectral decomposition
$H=\sum_{k}E_{k}\Pi_{k}$ and so

$\mbox{tr}(He^{-\beta H}\rho)=\sum_{k}E_{k}e^{-\beta
E_{k}}\mbox{tr}(\Pi_{k}\rho\Pi_{k}).$ (169)

Since $H\geq 0$ all terms in this expression are non-negative for all $\beta$
and thus $\partial_{\beta}(\Lambda(\beta,\rho)\geq 0$ which implies the
function is monotonely increasing in $\beta$.

Conversely, assume $\partial_{\beta}\Lambda(\beta,\rho)\geq 0$ for all states
$\rho$. This implies that $\mbox{tr}(He^{-\beta H}\rho)\geq 0$ for all states
$\rho$. Again, using the spectral decomposition of $H$ and the fact that we
have that

$\sum_{k}E_{k}e^{-\beta E_{k}}p_{k}\geq 0$ (170)

for all distributions $p_{k}$ over the eigenvalues $E_{k}$ of $H$. For
$\bm{p}=(p_{k})$ sharp on an eigenvalue $E_{m}$ the above equation implies
that $E_{m}\geq 0$, and since this holds for all $m$ this in turn implies that
$H\geq 0$ as required.
3\. To prove this, we Taylor expand in orders of $s$. We find:

$\displaystyle\lim_{s\to 0}s^{-1}\Lambda(s\beta,\rho)$
$\displaystyle=\lim_{s\to 0}(-s^{-1})\ln\langle e^{-s\beta H}\rangle_{\rho}$
(171) $\displaystyle=\lim_{s\to 0}(-s^{-1})\ln[1-s\beta\langle
H\rangle_{\rho}+{\mathcal{O}}(s^{2})]$ (172) $\displaystyle=\lim_{s\to
0}(-s^{-1})[-s\beta\langle H\rangle_{\rho}+\mathcal{O}(s^{2})]$ (173)
$\displaystyle=\beta\langle H\rangle_{\rho}$ (174)

as claimed.
4\. Let $E^{\rm min}=E_{k}$ be the smallest energy eigenvalue in the support
of $\rho$, and let $\rho_{i}:=\mbox{tr}(\Pi_{i}\rho)$ for any $i$. Then:

$\displaystyle\lim_{s\to\infty}s^{-1}\Lambda(s\beta,\rho)=\lim_{s\to\infty}(-s^{-1})\ln\left(\sum_{i}e^{-s\beta
E_{i}}\rho_{i}\right)$ (175)
$\displaystyle=\lim_{s\to\infty}(-s^{-1})\ln\big{(}e^{-s\beta E^{\rm
min}}[\rho_{k}+\sum_{i\neq k}e^{-s\beta(E_{i}-E^{\rm min})}\rho_{i}]\big{)}$
$\displaystyle=\lim_{s\to\infty}(-s^{-1})[-s\beta E^{\rm
min}+\ln(\rho_{k}+\sum_{i\neq k}e^{-s\beta(E_{i}-E^{\rm min})}\rho_{i})]$
$\displaystyle=\beta E^{\rm
min}+\lim_{s\to\infty}(-s^{-1})\ln(\rho_{k}+\sum_{i\neq
k}e^{-s\beta(E_{i}-E^{\rm min})}\rho_{i}).$ (176)

Now since $E_{i}-E_{k}>0$ we have that $e^{-s\beta(E_{i}-E_{\rm min})}<1$, for
all $i\neq k$ and for all $s>0,\beta>0$ and this term decreases to zero as
$s\rightarrow\infty$. Moreover $\rho_{k}\neq 0$ by assumption, and therefore

$\displaystyle\lim_{s\to\infty}(-s^{-1})\ln(\rho_{k}+\sum_{i\neq
k}e^{-s\beta(E_{i}-E^{\rm min})}\rho_{i})$
$\displaystyle=\lim_{s\to\infty}(-s^{-1})\ln(\rho_{k})=0.$ (177)

which implies that $\lim_{s\to\infty}s^{-1}\Lambda(s\beta,\rho)=\beta E^{\rm
min}$ as required.

∎

###  B.2 Mean coherence representation

As in the main text, we have that $\chi_{m}$, the _mean coherence at inverse
temperature_ $\beta$, is given by the equation

$\Lambda(\beta,\rho)=\beta\langle
H\rangle_{\rho}-\frac{1}{8\pi}\beta^{2}\chi_{m}(\tilde{\rho}),$ (178)

where $\chi_{m}$ is evaluated at the inverse temperature $\beta_{m}\leq\beta$
determined from the Mean Value Theorem.

######  Theorem B.4.

Given a quantum system $S$ with Hamiltonian $H_{S}$ with spectral
decomposition $H_{S}=\sum_{k}E_{k}\Pi_{k}$, the mean coherence at inverse
temperature $\beta$ of a quantum state $|\psi\rangle$ is given by

$\frac{\beta^{2}}{8\pi}\chi_{m}(\tilde{\psi})=\beta(\langle
H_{S}\rangle_{\bm{p}}-\langle
H_{S}\rangle_{\tilde{\bm{p}}})-S(\tilde{\bm{p}}||\bm{p}).$ (179)

where $\bm{p}$ is the distribution $p_{k}:=\mbox{tr}[\Pi_{k}\psi]$ over energy
of $\psi$ and $\tilde{\bm{p}}$ is given by
$\tilde{p}_{k}:=\mbox{tr}[\Pi_{k}\tilde{\psi}]$, the distribution over energy
for
$|\tilde{\psi}\rangle\langle\tilde{\psi}|=\Gamma(|\psi\rangle\langle\psi|)$.

###### Proof.

From Lemma B.2 we have that
$\Lambda(\beta,\rho)=\Lambda(\beta,{\mathcal{D}}(\rho))$ where
${\mathcal{D}}(\rho)$ is the dephased state across the energy eigenspaces, and
so because of this it will suffice to restrict to this dephased state, as we
now show.

Let $\sigma={\mathcal{D}}(\rho)$ for compactness. Noting that $[\sigma,H]=0$
we now expand $S(\tilde{\sigma})$ as

$\displaystyle S(\tilde{\sigma})$
$\displaystyle=-\mbox{tr}[\tilde{\sigma}\ln\tilde{\sigma}]$
$\displaystyle=-\mbox{tr}\left[\tilde{\sigma}\ln\left\\{e^{-\beta H/2}\sigma
e^{-\beta H/2}\frac{1}{\mbox{tr}(e^{-\beta H}\sigma)}\right\\}\right]$
$\displaystyle=-\mbox{tr}\left[\tilde{\sigma}\left\\{\ln(e^{-\beta H/2}\sigma
e^{-\beta H/2})+\Lambda(\beta,\sigma)\right\\}\right]$
$\displaystyle=-\mbox{tr}\left[\tilde{\sigma}\ln(e^{-\beta H/2}\sigma
e^{-\beta H/2})\right]-\Lambda(\beta,\sigma)$
$\displaystyle=-\mbox{tr}\left[\tilde{\sigma}(-\beta
H+\ln\sigma)\right]-\Lambda(\beta,\sigma)$ $\displaystyle=\beta\langle
H\rangle_{\tilde{\sigma}}-\mbox{tr}[\tilde{\sigma}\ln\sigma]-\Lambda(\beta,\sigma)$
$\displaystyle=\beta\langle
H\rangle_{\tilde{\sigma}}+S(\tilde{\sigma})+S(\tilde{\sigma}||\sigma)-\Lambda(\beta,\sigma),$
(180)

which implies that

$\Lambda(\beta,\sigma)=\beta\langle
H\rangle_{\tilde{\sigma}}+S(\tilde{\sigma}||\sigma).$ (181)

Now since $\sigma={\mathcal{D}}(\rho)$ we have that
$\mbox{tr}[\Pi_{k}\sigma]=\mbox{tr}[\Pi_{k}\rho]$ and likewise it can be
checked that
$\mbox{tr}[\Pi_{k}\tilde{\sigma}]=\mbox{tr}[\Pi_{k}\tilde{\rho}]$, where we
use that ${\mathcal{D}}(\Gamma(\rho))=\Gamma({\mathcal{D}}(\rho))$ for any
$\rho$. Thus the preceding equation can be written

$\Lambda(\beta,{\mathcal{D}}(\rho))=\Lambda(\beta,\rho)=\beta\langle
H\rangle_{\tilde{\rho}}+S(\tilde{\sigma}||\sigma).$ (182)

Restricting to $\rho=\psi$, a pure state, and using
$p_{k}=\mbox{tr}[\Pi_{k}\psi]$ and
$\tilde{p}_{k}=\mbox{tr}[\Pi_{k}\tilde{\psi}]$ we have that

$\Lambda(\beta,\psi)=\beta\langle
H\rangle_{\tilde{\bm{p}}}+S(\tilde{\bm{p}}||\bm{p}),$ (183)

where we use that $\tilde{\sigma}$ commutes with $\sigma$ and so
$S(\tilde{\sigma}||\sigma)$ can be replaced with the classical divergence
between the respective probability distributions over energy
$S(\tilde{\bm{p}}||\bm{p})$. Finally, using equation (178) we have that

$\frac{1}{8\pi}\beta^{2}\chi_{m}(\tilde{\psi})=\beta(\langle
H_{S}\rangle_{\bm{p}}-\langle
H_{S}\rangle_{\tilde{\bm{p}}})-S(\tilde{\bm{p}}||\bm{p}),$ (184)

as required. ∎

From this the fluctuation relation in Theorem 3.1 can be stated in the
following way.

######  Theorem B.5.

Given the assumptions to Theorem 3.1, let $\bm{p}_{k}$ and
$\tilde{\bm{p}}_{k}$ denote the probability distribution for an energy
measurement of the state $|\psi_{k}\rangle$ and $|\tilde{\psi}_{k}\rangle$
respectively. Then

$\frac{P[\psi_{1}|\tilde{\psi}_{0}]}{P[\psi_{0}^{*}|\tilde{\psi}_{1}^{*}]}=e^{-\beta\Delta
F-\beta\tilde{W}_{S}}e^{S(\mathbf{\tilde{p}}_{0}||\mathbf{p}_{0})-S(\mathbf{\tilde{p}}_{1}||\mathbf{p}_{1})},$
(185)

where $\tilde{W}_{S}:=\langle H_{S}\rangle_{\tilde{\psi}_{1}}-\langle
H_{S}\rangle_{\tilde{\psi}_{0}}$ and $S(\mathbf{p}||\mathbf{q})$ is the
Kullback-Leibler divergence.

###  B.3 Variational expression for the effective potential

The following is the same as Lemma 3.3 in the main text.

######  Lemma B.6.

Given a quantum system $S$ with Hamiltonian $H_{S}$ and $\\{\Pi_{k}\\}$ the
projectors onto the energy eigenspaces of $H_{S}$. Let us denote the de-phased
form of state in the energy basis by
${\mathcal{D}}(\rho)=\sum_{k}\Pi_{k}\rho\Pi_{k}$. Then, for any full rank
quantum state $\rho$, the effective potential $\Lambda(\beta,\rho)$ is given
by

$\Lambda(\beta,\rho)=\min_{\tau}\\{\beta\langle
H_{S}\rangle_{\tau}+S(\tau||{\mathcal{D}}(\rho))\\},$ (186)

where the minimization is over all quantum states $\tau$ of $S$, and
$S(\tau||\rho)=\mbox{tr}[\tau\log\tau-\tau\log\rho]$ is the relative entropy
function. Moreover, the minimization is attained for the state
$\tau=\Gamma({\mathcal{D}}(\rho))$.

###### Proof.

From Lemma B.2, we see that for any full rank $\rho$ we have

$\displaystyle\Lambda(\beta,\rho)$
$\displaystyle=\Lambda(\beta,{\mathcal{D}}(\rho))$ (187)
$\displaystyle=\min_{\tau}[\beta\langle
H_{S}\rangle_{\tau}+S(\tau||{\mathcal{D}}(\rho))].$

However equation (181) says

$\Lambda(\beta,\rho)=\Lambda(\beta,\sigma)=\beta\langle
H_{S}\rangle_{\tilde{\sigma}}+S(\tilde{\sigma}||\sigma),$ (188)

where $\sigma={\mathcal{D}}(\rho)$. Comparing the two equations we see that
the minimization is realised for
$\tau=\tilde{\sigma}=\Gamma(\sigma)=\Gamma({\mathcal{D}}(\rho))$ as claimed.

∎

###  B.4 Coherent work and the quantum fluctuation relation.

We now prove the coherent work form of the coherent fluctuation theorem and
provide an account of how it naturally connects with the resource theory of
asymmetry, inducing a natural measure for coherent work. However, first we
require the conditions in which the moments of a probability distribution
uniquely determine the distribution.

######  Theorem B.7 ([74]).

Let $\mu$ be a probability measure on the line having finite moments
$m_{k}=\int_{-\infty}^{\infty}x^{k}\mu(dx)$ of all orders. If the power series
$\sum_{k=1}^{\infty}\frac{r^{k}}{k!}m_{k}$ has positive radius of convergence,
then $\mu$ is the unique probability measure with the moments
$m_{1},m_{2},\dots$.

When the full set of moments uniquely define the probability distribution,
then so too do the cumulants. This follows since in such cases the moments can
be expressed in terms of the cumulants. Furthermore, finite moments correspond
to finite cumulants since once can obtain the cumulants from a recursive
formula that is polynomial in the moments. Thus the full set of cumulants
uniquely determines a distribution when the moment generating function of the
distribution has a positive radius of convergence. When the moment generating
function is finite and non-zero, then the cumulant generating function is
finite and non-zero as it is obtained by taking the logarithm.

In the result that follows, we assume the moment generating function and by
extension the effective potential, have infinite radius of convergence and
therefore always exist and uniquely specify the distribution. This choice is
for simplicity in the result, though one could generalise it to hold for any
positive radius of convergence. Requiring an infinite radius of convergence is
a restriction on the states permitted. For example, the negative binomial
distribution $\text{NB}(r,p)$ has finite radius of convergence and therefore
any state with such a distribution would not be included in the statement.
This can be seen from the moment generating function $M(t)$ for a random
variable $\hat{X}\sim\text{NB}(r,p)$, with [13]

$M(t)=\left(\frac{p}{1-(1-p)e^{t}}\right)^{r},$ (189)

which is valid for $t<-\ln(1-p)$.

######  Theorem B.8.

Let $S$ be a quantum system with Hamiltonian $H_{S}$ and let
$|\psi_{0}\rangle_{S}$ and $|\psi_{1}\rangle_{S}$ be two pure states of $S$
that have energy statistics with finite cumulants of all orders. Then the
following are equivalent:

  1. 1.

The pure states $|\psi_{0}\rangle_{S}$ and $|\psi_{1}\rangle_{S}$ are
coherently connected with coherent work output/input $\omega$ on an auxiliary
system $A$.

  2. 2.

Within the fluctuation relation context with a thermal system $B$ and quantum
system $S$ we have that

$\frac{P[\psi_{1}|\tilde{\psi}_{0}]}{P[\psi^{*}_{0}|\tilde{\psi}^{*}_{1}]}=e^{-\beta(\Delta
F\pm kT\Lambda(\beta,\omega))},$ (190)

for all inverse temperatures $\beta\geq 0$, for some state
$|\omega\rangle_{A}$ with finite cumulants on an auxiliary system $A$ and some
choice of sign before $\Lambda(\beta,\omega)$.

###### Proof.

First assume that $|\psi_{0}\rangle_{S}$ and $|\psi_{1}\rangle_{S}$ are
coherently connected states. Suppose there is a coherent work process
transforming
$|\psi_{0}\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\psi_{1}\rangle_{S}$.
Without loss of generality we can choose the initial state $|0\rangle_{A}$ of
the auxiliary system $A$ to have zero energy.

By unitary invariance of the effective potential we have that

$\displaystyle\Lambda(\beta,|\psi_{0}\rangle_{S}\otimes|0\rangle_{A})$
$\displaystyle=\Lambda(\beta,V(|\psi_{0}\rangle_{A}\otimes|0\rangle_{A}))$
$\displaystyle=\Lambda(\beta,|\psi_{1}\rangle_{A}\otimes|\omega\rangle_{A}),$
(191)

where $V$ is a unitary that realises the process. This conserves energy and so
commutes with $e^{-\beta(H_{S}\otimes\mathds{1}_{A}+\mathds{1}_{S}\otimes
H_{A})}$. Since the effective potential is additive for independent
Hamiltonians and product states, we have that

$\Lambda(\beta,\psi_{0})=\Lambda(\beta,\psi_{1})+\Lambda(\beta,\omega).$ (192)

Therefore the effective potential change in the fluctuation theorem can be
expressed in terms of the coherent work output $|\omega\rangle$ as

$e^{\Lambda(\beta,\psi_{0})-\Lambda(\beta,\psi_{1})}=e^{\Lambda(\beta,\omega)}.$
(193)

and so the fluctuation theorem can be expressed as

$\frac{P[\psi_{1}|\tilde{\psi}_{0}]}{P[\psi_{0}^{*}|\tilde{\psi}_{1}^{*}]}=e^{\Lambda(\beta,\omega)-\beta\Delta
F}=\frac{e^{-\beta\Delta F}}{\langle\omega|e^{-\beta H_{A}}|\omega\rangle}.$
(194)

If instead
$|\psi_{1}\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\psi_{0}\rangle_{S}$,
then the same argument applies with the only difference being the sign in
front of $\Lambda(\beta,\omega)$.

Conversely, suppose the above statement 2 holds for all inverse temperatures
$\beta\geq 0$, for some pure state $|\omega\rangle_{A}$ on a system $A$, and
choice of sign in the exponent. The exponent in Theorem 3.1 implies that we
have

$\Lambda(\beta,\psi_{0})=\Lambda(\beta,\psi_{1})\mp\Lambda(\beta,\omega),$
(195)

for all inverse temperatures $\beta\in[0,\infty)$. In the event of the minus
sign, we write this as

$\displaystyle\Lambda(\beta,\psi_{0})+\Lambda(\beta,\omega)$
$\displaystyle=\Lambda(\beta,|\psi_{0}\rangle_{A}\otimes|\omega\rangle_{A})$
$\displaystyle=\Lambda(\beta,|\psi_{1}\rangle_{S}\otimes|0\rangle_{A}),$ (196)

where we can assume that $A$ has a zero energy eigenstate $|0\rangle_{A}$.
However, since all the above cumulant generating functions have infinite
radius of convergence and are finite, the energy distributions of measuring
$|\psi_{0}\rangle_{S}$, $|\psi_{1}\rangle_{S}$ and $|\omega\rangle_{A}$ are
uniquely determined. Moreover the distributions of a joint energy measurement
of composite system $SA$ in the state
$|\psi_{0}\rangle_{S}\otimes|\omega\rangle_{A}$ and
$|\psi_{1}\rangle_{S}\otimes|0\rangle_{A}$ are identical. This implies that

$\hat{S}_{0}+\hat{A}=\hat{S}_{1},$ (197)

where $\hat{S}_{0},\hat{S}_{1},\hat{A}$ are the associated random variables
for measuring $|\psi_{0}\rangle_{S},|\psi_{1}\rangle_{S},|\omega\rangle_{A}$
respectively in their energy eigenbases. By Theorem 2.5 this implies there
exists a non-trivial coherent work process

$|\psi_{1}\rangle\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\psi_{0}\rangle,$
(198)

and so the states are coherently connected. In the event of the plus sign in
the exponent we instead write

$\Lambda(\beta,\psi_{0})=\Lambda(\beta,\psi_{1})+\Lambda(\beta,\omega),$ (199)

and then express the argument of the left-hand side using
$|\psi_{0}\rangle_{S}\otimes|0\rangle_{A}$ as before, using the additivity of
the effective potential for the left-hand side. The same argument as before
then implies the existence of a coherent work process

$|\psi_{0}\rangle\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\psi_{1}\rangle,$
(200)

and again the states are coherently connected, which completes the proof.

∎

###  B.5 Proof of semi-classical fluctuation theorem

######  Theorem B.9 (Semi-classical relation [25]).

Let the assumptions of Theorem 3.1 hold, and let $S$ be a harmonic oscillator
with Hamiltonian $H_{S}=h\nu(a^{\dagger}a+\frac{1}{2})$. Then for two coherent
states of the system $|\alpha_{0}\rangle$ and $|\alpha_{1}\rangle$, the
following holds

$\displaystyle\frac{P[\alpha_{1}|\tilde{\alpha}_{0}]}{P[\alpha_{0}^{*}|\tilde{\alpha}_{1}^{*}]}$
$\displaystyle=\exp\left[-\frac{\Delta
F}{kT}+\frac{\bar{W}_{B}}{h\nu_{\rm\tiny{th}}}\right],$ (201)

where $h\nu_{th}=\langle H_{S}\rangle_{\gamma}$ is the average energy of a
Gibbs state $\gamma$ of a quantum harmonic oscillator, related to the thermal
de Broglie wavelength $\lambda_{\rm\tiny{dB}}(T)$ via

$h\nu_{\rm\tiny{th}}=\frac{h^{2}}{m\lambda_{\rm\tiny{dB}}(T)^{2}}+\frac{1}{2}h\nu,$
(202)

and $\bar{W}_{B}:=-\frac{1}{2}(W_{S}+\tilde{W}_{S})$, $W_{S}=\langle
H_{S}\rangle_{\alpha_{1}}-\langle H_{S}\rangle_{\alpha_{0}}$,
$\tilde{W}_{S}=\langle H_{S}\rangle_{\tilde{\alpha}_{1}}-\langle
H_{S}\rangle_{\tilde{\alpha}_{0}}$.

###### Proof.

Consider a system with Hamiltonian $H_{S}=h\nu(a^{\dagger}a+\frac{1}{2})$. For
simplicity, we take an arbitrary state $|\alpha,k\rangle\in\mathcal{C}$ and
assume $k=0$, as the extension is simple but provides little benefit. The
effective potential for a coherent state $|\alpha\rangle$:

$\displaystyle\Lambda(\beta,\alpha)$ $\displaystyle=\frac{\beta
h\nu}{2}+|\alpha|^{2}(1-e^{-\beta h\nu}).$ (203)

Now consider the average energy of a thermally distributed quantum harmonic
oscillator

$\displaystyle\langle H_{S}\rangle_{\gamma}$ $\displaystyle=\frac{\sum_{n\geq
0}h\nu(n+\frac{1}{2})e^{-\beta h\nu(n+\frac{1}{2})}}{\sum_{m\geq 0}e^{-\beta
h\nu(m+\frac{1}{2})}}$ (204)
$\displaystyle=\frac{h\nu}{2}\left(\frac{1+e^{-\beta h\nu}}{1-e^{-\beta
h\nu}}\right).$ (205)

Let us define a thermal frequency $h\nu_{th}:=\langle H_{S}\rangle_{\gamma}$.
Then the effective potential change can be expressed as

$\displaystyle\Lambda(\beta,\alpha_{1})-\Lambda(\beta,\alpha_{0})$
$\displaystyle=(|\alpha_{1}|^{2}-|\alpha_{0}|^{2})(1-e^{-\beta h\nu})$ (206)
$\displaystyle=\frac{\nu}{2\nu_{th}}(|\alpha_{1}|^{2}-|\alpha_{0}|^{2})(1+e^{-\beta
h\nu}).$ (207)

Since the Gibbs rescaled coherent state $|\tilde{\alpha}\rangle=|\alpha
e^{-\beta h\nu/2}\rangle$, the change in effective potential can be expressed
in terms of the unscaled and rescaled coherent state energy transfers,

$\displaystyle W_{S}$ $\displaystyle=\langle H_{S}\rangle_{\alpha_{1}}-\langle
H_{S}\rangle_{\alpha_{0}}$ (208) $\displaystyle\tilde{W}_{S}$
$\displaystyle=\langle H_{S}\rangle_{\tilde{\alpha}_{1}}-\langle
H_{S}\rangle_{\tilde{\alpha}_{0}},$ (209)

as

$\Lambda(\beta,\alpha_{1})-\Lambda(\beta,\alpha_{0})=\frac{1}{h\nu_{th}}\left(\frac{W_{S}+\tilde{W}_{S}}{2}\right).$
(210)

which gives the claimed result upon insertion into Theorem 3.1. For the
breakdown of $h\nu_{th}$ in terms of $\lambda_{dB}(T)$, we refer to [25]. ∎

For completeness, we can also show the form of the fluctuation theorem using
the coherent work representation. Consider an auxillary weight system $A$ with
Hamiltonian $H_{A}=h\nu a^{\dagger}a$. Once again, this system exists as a
hypothetical process in which
$|\psi_{0}\rangle\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\psi_{1}\rangle$
under a coherent work process. We note that

$e^{\Lambda(\beta,\omega)}=\frac{\langle\alpha_{1}|e^{-\beta
H_{S}}|\alpha_{1}\rangle}{\langle\alpha_{0}|e^{-\beta
H_{S}}|\alpha_{0}\rangle}.$ (211)

From the semi-classical result Theorem 2.4, we know that
$|\alpha_{0}\rangle_{S}\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}|\alpha_{1}\rangle_{S}$
under a coherent work process if and only if $|\alpha_{1}|\leq|\alpha_{0}|$.
Assuming this condition to be true, then we can simply write down:

$\frac{P[\alpha_{1}|\tilde{\alpha}_{0}]}{P[\alpha_{0}^{*}|\tilde{\alpha}_{1}^{*}]}=e^{|\alpha|^{2}(1-e^{-\beta
h\nu})-\beta\Delta F}$ (212)

where $|\alpha|^{2}=|\alpha_{0}|^{2}-|\alpha_{1}|^{2}$ and $||\alpha|\rangle$
is the coherent work output of the hypothetical process. Owing to the simple
form of the effective potential for coherent states, this is fairly trivial to
write down. However generally, the effective potential fails to obtain such a
simplified form, and simplifying the change in effective potential
$\Lambda(\beta,\psi_{1})-\Lambda(\beta,\psi_{0})$ may be harder than finding
the analytic form for $\Lambda(\beta,\omega)$.

###  B.6 Macrosopic limit of the quantum fluctuation relation

For simplicity, we choose to make the random variable dimensionless. We have
that $\langle\psi_{i}|H_{i}|\psi_{i}\rangle=\mu$, which is independent of $i$.
We define the operator $X_{i}:=\frac{1}{\mu}H_{i}-\mathbbm{1}$, which has mean
zero in the state $|\psi_{i}\rangle$. Note that the variance of $X_{i}$ is
also dimensionless and given by

$\displaystyle\langle X_{i}^{2}\rangle-\langle X_{i}\rangle^{2}$
$\displaystyle=\langle X_{i}^{2}\rangle$ (213)
$\displaystyle=\langle\frac{1}{\mu^{2}}H_{i}^{2}-\frac{2}{\mu}H_{i}+\mathbbm{1}\rangle$
(214) $\displaystyle=\frac{1}{\mu^{2}}\langle H_{i}^{2}\rangle-2+1$ (215)
$\displaystyle=\frac{1}{\mu^{2}}(\langle H_{i}^{2}\rangle-\mu^{2})$ (216)
$\displaystyle\text{Var}(X_{i})$
$\displaystyle=\frac{\sigma^{2}}{\mu^{2}}\equiv\tilde{\sigma}^{2}.$ (217)

The cumulant generating function of the total state is

$\displaystyle e^{-\Lambda(\beta,\psi_{n})}$
$\displaystyle=\mbox{tr}[e^{-\beta\sum_{i}H_{i}}\psi^{\otimes n}]$ (218)
$\displaystyle=\mbox{tr}[e^{-\beta\mu\sum_{i}(X_{i}+\mathbbm{1})}\psi^{\otimes
n}]$ (219) $\displaystyle=e^{-\beta
n\mu}\mbox{tr}[e^{-\beta\mu\sum_{i}X_{i}}\psi^{\otimes n}]$ (220)
$\displaystyle=e^{-\beta
n\mu}\mbox{tr}[e^{-\beta\sqrt{n}\mu\sum_{i}\frac{1}{\sqrt{n}}X_{i}}\psi^{\otimes
n}]$ (221) $\displaystyle=e^{-\beta
n\mu}e^{-\tilde{\Lambda}(\beta\sqrt{n}\mu,\psi_{n})}.$ (222)

Here $\tilde{\Lambda}$ is the CGF corresponding to the dimensionless mean zero
observable $\frac{1}{\sqrt{n}}(X_{1}+\dots+X_{n})$. Thus we have that

$\Lambda(\beta,\psi_{n})=\beta
n\mu+\tilde{\Lambda}(\beta\sqrt{n}\mu,\psi_{n}),$ (223)

for all $\beta,n$.
A subtlety comes in here – if one wants to take a limit and obtain a Central
Limit Theorem (CLT) statement we need to scale $\beta$ also in order to get a
non-trivial statement. This is a mathematical re-formulation so as to maintain
the correct scaling to obtain a non-trivial statement the mean value of the
sequence (in $n$) of random variables.
Since $\beta$ has units of reciprocal energy we can set
$\beta=\frac{\beta_{0}}{\sqrt{n}\lambda}$, for some arbitrary energy scale
$\lambda$, where $\beta_{0}$ is dimensionless and we count in energy units of
$\sqrt{n}\lambda$ when we have $n$ systems. It might look like the inverse
temperature is decreasing as we increase the number of systems, but this is
not what is intended by this – in the description of the sequence of systems
one scales the units of energy as we increase $n$ when we specify the
temperature. Note also that the first and second cumulants scale differently
with energy, and it is this that makes the scaling trick needed to obtain a
statement containing both in the large $n$ limit.
Expressed in these units we have

$\Lambda(\frac{\beta_{0}}{\sqrt{n}\lambda},\psi_{n})=\beta_{0}\mu_{0}\sqrt{n}+\tilde{\Lambda}(\beta_{0}\mu_{0},\psi_{n}),$
(224)

where $\mu_{0}:=\frac{\mu}{\lambda}$ is dimensionless, and $\beta_{0}$ does
not vary in $n$.

This holds for all $n$, and in the $n\rightarrow\infty$ limit we have that

$\lim_{n\rightarrow\infty}\left(\Lambda(\frac{\beta_{0}}{\sqrt{n}\lambda},\psi_{n})-\beta_{0}\mu_{0}(\psi)\sqrt{n}\right)=-\frac{1}{2}\tilde{\sigma}_{\psi}^{2}\beta_{0}^{2}\mu_{0}(\psi)^{2},$
(225)

and also

$\lim_{n\rightarrow\infty}\left(\Lambda(\frac{\beta_{0}}{\sqrt{n}\lambda},\phi_{n})-\beta_{0}\mu_{0}(\phi)\sqrt{n}\right)=-\frac{1}{2}\tilde{\sigma}_{\phi}^{2}\beta_{0}^{2}\mu_{0}(\phi)^{2},$
(226)

where we have included the state variables to distinguish the energies of the
two states.

This implies that for any $\epsilon>0$ there is an integer $M$ such that for
all $n>M$ we have the following two conditions hold

$\displaystyle\left|\Lambda(\frac{\beta_{0}}{\sqrt{n}\lambda},\psi_{n})-[\beta_{0}\mu_{0}(\psi)\sqrt{n}-\frac{1}{2}\sigma_{\psi}^{2}\beta_{0}^{2}]\right|$
$\displaystyle\leq\epsilon$ (227)
$\displaystyle\left|\Lambda(\frac{\beta_{0}}{\sqrt{n}\lambda},\phi_{n})-[\beta_{0}\mu_{0}(\phi)\sqrt{n}-\frac{1}{2}\sigma_{\phi}^{2}\beta_{0}^{2}]\right|$
$\displaystyle\leq\epsilon,$ (228)

where we also used the definition of $\tilde{\sigma}^{2}$. We therefore have
that for any $\epsilon>0$ there is an $M>0$ such that for all $n>M$ we have

$|\Lambda(\frac{\beta_{0}}{\sqrt{n}\lambda},\psi_{n})-\Lambda(\frac{\beta_{0}}{\sqrt{n}\lambda},\phi_{n})-[\beta_{0}\sqrt{n}\Delta\mu_{0}-\frac{1}{2}\beta_{0}^{2}\Delta\sigma_{0}^{2}]|\leq
2\epsilon.$ (229)

Thus, for any $\epsilon>0$ there is an $M$ such that for all $n>M$ we have

$\displaystyle\left|\frac{P[\psi_{n}|\phi_{n}]}{P[\phi^{*}_{n}|\psi^{*}_{n}]}-e^{-\sqrt{n}(\beta_{0}\Delta
f_{0}+\beta_{0}\Delta\mu_{0})+\frac{1}{2}\beta_{0}^{2}\Delta\sigma_{0}^{2}}\right|\leq\epsilon$
(230)

where $\Delta f_{0}:=\frac{\Delta F}{n\lambda}$ is a dimensionless number. We
could put back in $\beta$, but only with the proviso that it is assumed to
scale with $n$ in the way stated. This gives

$\displaystyle\left|\frac{P[\psi_{n}|\tilde{\phi}_{n}]}{P[\phi^{*}_{n}|\tilde{\psi}^{*}_{n}]}-e^{-n(\beta\Delta
f+\beta\Delta\mu-\frac{1}{2}\beta^{2}\Delta\sigma^{2})}\right|\leq\epsilon$
(231)

with the proviso that the scaling $\beta=\frac{\beta_{0}}{\sqrt{n}\lambda}$ is
adopted. This captures the informal statement that for the IID case the states
“become more like Gaussians” and thus the fluctuation theorem behaves as

$\frac{P[\psi_{n}|\tilde{\phi}_{n}]}{P[\phi^{*}_{n}|\tilde{\psi}^{*}_{n}]}\stackrel{{\scriptstyle
n\rightarrow\infty}}{{\sim}}e^{-n(\beta\Delta
f+\beta\Delta\mu-\frac{1}{2}\beta^{2}\Delta\sigma^{2})}.$ (232)