A measure of majorization emerging from single-shot statistical mechanics

There are two central pieces of information about the system, the energy eigenvalues, and their occupation probabilities. We shall find it very powerful to follow Ruch (1975), Ruch and Mead (1976), Mead (1977) and combine them into one object, the Gibbs-rescaled distribution.

Consider states with discrete spectra $\{{\lambda }_{i}\}$. We firstly transform the spectrum into the associated step-function. Then we take each block, rescale its height as ${\lambda }_{i}\mapsto {\lambda }_{i}/\mathrm{exp}\left(-\frac{{E}_{i}}{{kT}}\right)$, and its width $l=1\mapsto \mathrm{exp}\left(-\frac{{E}_{i}}{{kT}}\right)$ such that the area of the new block is ${\lambda }_{i}$ as before. We write this operation applied to a density matrix ρ as ${G}^{T}(\rho )$. It is depicted in figure A1 . Gibbs rescaling can, as will prove useful in later proofs, be written out in the language of continuous functions in the following manner:

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Definition 2 (Gibbs rescaling). Consider a density matrix $\rho ={\displaystyle \sum }_{i=1}^{n}{\lambda }_{i}| {e}_{i}\rangle \langle {e}_{i}|$ with eigenvalues ${\{{\lambda }_{i}\}}_{i=1}^{n}$ and take the energy eigenstates of the system to be $\{| {e}_{i}\rangle \}{}_{i=1}^{n}$ with energies ${\{{E}_{i}\}}_{i=1}^{n}$ respectively. There is an associated step function for the spectrum, $\lambda ({xn})={\lambda }_{\lceil {xn}\rceil }$ where $x\in (0,1]$. Similarly there is an energy step function $E({xn})={E}_{\lceil {xn}\rceil }$ where $x\in (0,1]$. The Gibbs rescaling associated with temperature T combines $\lambda ({xn})$ and $E({xn})$ to a new function ${G}^{T}(y)$ implicitly defined by

$$\begin{eqnarray*}&&{G}^{T}\left({\displaystyle \int }_{0}^{x}{{\rm{e}}}^{-\displaystyle \frac{{E}_{\lceil {zn}\rceil }}{{kT}}}{\rm{d}}z\right)=\displaystyle \frac{{\lambda }_{\lceil {xn}\rceil }}{{{\rm{e}}}^{-\displaystyle \frac{{E}_{\lceil {xn}\rceil }}{{kT}}}}.\end{eqnarray*}$$

It follows that ${G}^{T}(y)$ is defined on $(0,Z]$, with $Z={\displaystyle \sum }_{j=1}^{n}\mathrm{exp}\left(-\displaystyle \frac{{E}_{j}}{{kT}}\right)$ the partition function. Moreover ${G}^{T}(y)$ is a probability distribution satisfying ${\displaystyle \int }_{0}^{Z}{G}^{T}(y){\rm{d}}y=1$.

We now turn to how interactions with the heat bath, thermalizations, act on the state of the system. Roughly speaking these take the density matrix closer to the associated Gibbs state, similar statements can be found in Ruch (1975), Ruch and Mead (1976), Mead (1977) (especially see section 4 of Ruch and Mead 1976, where also a different argument is given for the result below concerning thermalizations). As already mentioned the thermalization is taken to only change occupation probabilities and not energy eigenvalues. We take the thermalization to act as a stochastic process on the energy eigenstates, in that the probability of occupying a given energy state, $P(i)$, becomes $P\prime (i)={\displaystyle \sum }_{j}P(j\to i)P(j)$ where the summation is over all eigenstates, $P(j\to i)$ is a transition probability, and $P(j)$ an occupation probability (before the interaction with the heat bath). This can equivalently be written as $\vec{P\prime }=B\vec{P}$ where B is a stochastic matrix (entries are probabilities and columns sum to 1).

Not every stochastic matrix B is allowed however. The Gibbs state (associated with temperature T) is taken to be invariant under a thermalization. Consider the implications firstly for the fully degenerate case of all energies being the same. In this case the Gibbs state is the uniform distribution. The only stochastic matrices that leave the uniform distribution invariant are bistochastic ones (rows also sum to 1). Thus in the fully degenerate case B must be bistochastic. We see no reason to impose further restrictions, so any such B is allowed.

Consider secondly the non-degenerate case. Here it is again convenient to use the Gibbs rescaled distribution. Note that the Gibbs state becomes uniform after the Gibbs rescaling. Thus one may hope that a thermalization, i.e.a Gibbs state preserving stochastic matrix on the occupation probabilities, acts as a bi-stochastic matrix on the Gibbs-rescaled distribution, and we now show that is indeed the case.

Before considering the general case, we look at a simple example of a two-level system.

Let B be the stochastic matrix⁶ defined by the transition-probabilities, i.e:

$$\begin{eqnarray*}&&\left(\begin{array}{c}P(1)\\ P(2)\end{array}\right)\to \left(\begin{array}{c}P\prime (1)\\ P\prime (2)\end{array}\right)=\left(\begin{array}{cc}{p}_{(1\to 1)} & {p}_{(2\to 1)}\\ {p}_{(1\to 2)} & {p}_{(2\to 2)}\end{array}\right)\left(\begin{array}{c}P(1)\\ P(2)\\ \end{array}\right)=B\left(\begin{array}{c}P(1)\\ P(2)\end{array}\right).\end{eqnarray*}$$

The stochastic matrix should leave the thermal state invariant:

$$\begin{eqnarray*}&&\left(\begin{array}{c}e(1)/Z\\ e(2)/Z\end{array}\right)\to \left(\begin{array}{c}e\prime (1)/Z\\ e\prime (2)/Z\end{array}\right)=B\left(\begin{array}{c}e(1)/Z\\ e(2)/Z\end{array}\right)=\left(\begin{array}{c}e(1)/Z\\ e(2)/Z\end{array}\right),\end{eqnarray*}$$

where ${e}^{(\prime )}(i)/Z=\mathrm{exp}(-{E}^{(\prime )}(i)/({kT}))/Z$ is the Gibbs state (which should be invariant as the energy does not change) and $Z=e(1)+e(2)$.

Look at what happens with $e(1)=2$ and $e(2)=1$. For the Gibbs rescaling this means that $P(1)\to P(1)/2$ on the length 2 and $P(2)\to P(2)$ on the length 1. We can split the first level into two parts (in our mind) and consider new levels $(P({1}_{1}),P({1}_{2}),P({2}_{1}))=P(1)/2,P(1)/2,P(2)/1)$ all having the same length after Gibbs rescaling. For the thermal state this means:

$$\begin{eqnarray*}&&\left(\begin{array}{c}2/3\\ 1/3\\ \end{array}\right)\to \left(\begin{array}{c}1/3\\ 1/3\\ 1/3\\ \end{array}\right).\end{eqnarray*}$$

The transition matrix becomes:

$$\begin{eqnarray*}\left(\begin{array}{cc}{p}_{(1\to 1)} & {p}_{(2\to 1)}\\ {p}_{(1\to 2)} & {p}_{(2\to 2)}\end{array}\right)\to \left(\begin{array}{ccc}{p}_{(1\to 1)}/2 & {p}_{(1\to 1)}/2 & {p}_{(2\to 1)}/2\\ {p}_{(1\to 1)}/2 & {p}_{(1\to 1)}/2 & {p}_{(2\to 1)}/2\\ {p}_{(1\to 2)} & {p}_{(1\to 2)} & {p}_{(2\to 2)}\end{array}\right)\end{eqnarray*}$$

which is still stochastic, because the initial matrix was. Since the thermal state has to be invariant under the action of this matrix and the thermal state in this case is proportional to the identity, it is straightforward to check that the matrix has to be bistochastic (rows and columns sum to 1).

For the general case consider dividing the Gibbs-rescaled distribution into fine blocks such that all fine blocks have the same width w. Let N be the number of fine blocks. (As the maximum support is given by the partition function Z we have $w=Z/N$). Let N_k be the number of fine grained blocks associated with level k, such that ${\displaystyle \sum }_{k=1}^{n}{N}_{k}=N$. Each energy level is associated with one block only labelled by k. Each lth fine block is associated with a level k_l.

Fine blocks associated with the same energy level k must all have the same height, given by $P({k}_{l})/e({k}_{l})$ (where $e({k}_{l})=\mathrm{exp}(-E({k}_{l})/{kT})={{wN}}_{{k}_{l}}={{ZN}}_{{k}_{l}}/N$ is the total width of the level k_l after Gibbs rescaling. See the comment after the definition of Gibbs rescaling 2). Let $\vec{f}$ contain the N heights of the fine blocks, with $P({k}_{l})/e({k}_{l})=P({k}_{l})N/({{ZN}}_{{k}_{l}})$ as its lth entry. Now when the occupation probabilities transform under B, $\vec{f}$ undergoes an associated transform. We will argue it is given by a matrix F whose entry in the l-th row and m-th column is given by

$$\begin{eqnarray}&&{F}_{{lm}}:= \displaystyle \frac{{B}_{{k}_{l}{k}_{m}}}{{N}_{{k}_{l}}}.\end{eqnarray} \tag{ A.1 }$$

To see this note firstly that ${P}_{i}^{\prime }={\displaystyle \sum }_{j}{B}_{{ij}}{P}_{j}={\displaystyle \sum }_{j}\frac{{N}_{j}}{{N}_{j}}{B}_{{ij}}{P}_{j}$, and recall that ${f}_{l}^{\prime }={P}_{{k}_{l}}^{\prime }N/({{ZN}}_{{k}_{l}})$. Thus

$$\begin{eqnarray}&&{f}_{l}^{\prime }=\displaystyle \sum _{j=1}^{n}\displaystyle \frac{{N}_{j}}{{N}_{j}}{B}_{{k}_{l}j}{P}_{j}\displaystyle \frac{N}{{{ZN}}_{{k}_{l}}}\end{eqnarray} \tag{ A.2 }$$

$$\begin{eqnarray}&&=\displaystyle \sum _{j=1}^{n}\displaystyle \frac{{N}_{j}{B}_{{k}_{l}j}}{{N}_{{k}_{l}}}\displaystyle \frac{{P}_{j}N}{{{ZN}}_{j}}\end{eqnarray} \tag{ A.3 }$$

$$\begin{eqnarray}&&=\displaystyle \sum _{j=1}^{n}\displaystyle \sum _{m| {k}_{m}=j}\displaystyle \frac{{B}_{{k}_{l}j}}{{N}_{{k}_{l}}}{f}_{m}\end{eqnarray} \tag{ A.4 }$$

$$\begin{eqnarray}&&=\displaystyle \sum _{m=1}^{N}\displaystyle \frac{{B}_{{k}_{l}{k}_{m}}}{{N}_{{k}_{l}}}{f}_{m}.\end{eqnarray} \tag{ A.5 }$$

As B_ij and N are non-negative real numbers F has non-negative real entries only. To see that the columns sum to 1 so that F is a stochastic matrix, note that the column sums are the same as for B which is stochastic. Moreover as B must leave the Gibbs state invariant, and this is a uniform distribution after the Gibbs rescaling, F must leave the uniform distribution (or anything proportional to it) invariant. Then for any row i: ${\displaystyle \sum }_{j}{F}_{{ij}}(1/N)=1/N$ so each row of F must sum to 1. Therefore F is a bistochastic matrix. Note that F is additionally restricted, through being defined via B, to keep the heights of fine blocks the same whenever these are associated with the same level.

Accordingly we define interactions with the heat-baths, thermalizations, to act in the following way on the system.

Definition 3 (Thermalization). A thermalization leaves the energy eigenvalues invariant. It acts on the occupation probabilities, i.e. the eigenvalues of the density matrix, as a stochastic matrix. This stochastic matrix leaves the Gibbs state $\mathrm{exp}(\beta H)/Z$ invariant. It follows from this definition and the definition of the Gibbs-rescaled distribution that a thermalization acts on the Gibbs-rescaled distribution as a bistochastic matrix.

The second elementary process is changing the Hamiltonian of the system through shifting a set of energy levels by some predetermined amount $\Delta E(j)$, where j labels the jth work extraction. This may involve a work gain/cost, because if the system occupies one of the energy eigenstates that get shifted by $\Delta E(j)$ this counts as work done on the system and we write ${W}_{j}=\Delta E(j)$. It is assumed that this entails an energy transfer of $\Delta E(j)$ to the work reservoir system, so that energy is conserved. If the system does not occupy the eigenstate that gets shifted there is no work cost, ${W}_{j}=0$. To reduce the notation later on we will also find it convenient to define the 'logarithmic' work w^j s.t. ${W}_{j}:= {kT}\mathrm{ln}{w}^{j}$ (or equivalently ${w}^{j}:= \mathrm{exp}({W}_{j}/{kT})$).

There is thus for each elementary work extraction a probability distribution over work transfer, with two elements, [$p({W}_{j}=0)$, $p({W}_{j}=\Delta E(j))$]. A sequence of work extractions generates a randomly picked sequence of energy transfers to the work reservoir by, e.g. $\{0,0,\Delta E(3),0,\Delta E(5)...\}$. There is an associated vector of 0's and 1's where a 1 as the j-th entry indicates that there was indeed a work transfer of $\Delta E(j)$ in the j-th step. We call this latter vector $\vec{s}$, and the jth entry thereof s_j. ${s}_{j}=0$ means that the levels shifted in work extraction step j were not occupied, and ${s}_{j}=1$ means that they were.

From the perspective of someone who learns s_j, the occupation probabilities $\{{\lambda }_{i}\}$ change. If ${s}_{j}=1$ one projects the state ρ with projector ${\Pi }_{\mathrm{shifted}}$ onto the set of levels shifted so that the new state is

$$\begin{eqnarray*}&&\displaystyle \frac{{\Pi }_{\mathrm{shifted}}{\displaystyle \sum }_{i}{\lambda }_{i}| i\rangle \langle i| {\Pi }_{\mathrm{shifted}}}{\mathrm{tr}(\rho {\Pi }_{\mathrm{shifted}})}.\end{eqnarray*}$$

If instead ${s}_{j}=0$ one replaces the projector with one onto the levels that were not shifted.

We accordingly represent a work extraction in the following manner:

Definition 4 (Work extraction). We define a work extraction on the first l levels, which are all to get shifted in energy by $\Delta E(j)=-{kT}\mathrm{ln}({w}_{\vec{s}| {s}_{j}=1}^{j})$, while the remaining levels are untouched as follows. Letting ${\Theta }_{U}(y)$ denote the function that is 1 if $y\in U$ and else 0, the new occupation probabilities and energies are given by:

• In the case when ${s}_{j}=1$ (state of the system is found to be in the levels $(1,\ldots ,l)$):
  $$\begin{eqnarray*}&&{\lambda }_{\vec{s}| {s}_{j}=1}^{j}(k)={\Theta }_{\{1,\ldots ,l\}}(k)\displaystyle \frac{{\lambda }_{\vec{s}}^{j-1}(k)}{{\eta }_{\vec{s}| {s}_{j}=1}^{j}},\end{eqnarray*}$$
  where ${\eta }_{\vec{s}| {s}_{j}=1}^{j}={\displaystyle \sum }_{i=1}^{l}{\lambda }_{\vec{s}}^{j-1}(i)$. In this case there is an energy transfer to the reservoir given, in terms of the logarithmic work, by ${w}_{\vec{s}| {s}_{j}=1}^{j}=\mathrm{exp}(\Delta E(j)/{kT})$.
• In the case when ${s}_{j}=0$ (state of the system is not found to be in the levels $(1,\ldots ,l)$):
  $$\begin{eqnarray*}&&{\lambda }_{\vec{s}| {s}_{j}=0}^{j}(k)={\Theta }_{\{l+1,\ldots ,n\}}(k)\displaystyle \frac{{\lambda }_{\vec{s}}^{j-1}(k)}{{\eta }_{\vec{s}| {s}_{j}=0}^{j}},\end{eqnarray*}$$
  where ${\eta }_{\vec{s}| {s}_{j}=0}^{j}={\displaystyle \sum }_{i=l+1}^{n}{\lambda }_{\vec{s}}^{j-1}(i)$. In this case there is no energy transfer to the work reservoir, i.e. ${w}_{\vec{s}| {s}_{j}=0}^{j}=1$.

This next lemma considers how the work extraction in the preceding definition acts on the Gibbs rescaled distribution. This is also depicted in figure A2.

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Lemma 2. Let the levels $\{1,\ldots ,l\}$ be used for work extraction as in the above definition. Let $a\in {\mathbb{R}}$ be the combined width of the blocks of the Gibbs-rescaled distribution corresponding to the levels $\{1,\ldots ,l\}$, i.e. $a={\displaystyle \sum }_{i=1}^{l}{e}^{\displaystyle \frac{-{E}_{i}^{j-1}}{{kT}}}$. Let $x\in (0,{Z}_{j}]$ (with Z_j the partition function after step j).

Then following a work extraction in step j, the resulting Gibbs rescaled probability distribution, conditioned on the previous steps on path $\vec{s}$, is given by the following. In the case where ${s}_{j}=1$:

$$\begin{eqnarray*}&&{p}_{\vec{s}| {s}_{j}=1}^{j}(x)={\Theta }_{(0,{{aw}}_{\vec{s}| {s}_{j}=1}^{j}]}(x)\displaystyle \frac{{p}_{\vec{s}}^{j-1}(\displaystyle \frac{x}{{w}_{\vec{s}| {s}_{j}=1}^{j}})}{{w}_{\vec{s}| {s}_{j}=1}^{j}{\eta }_{\vec{s}| {s}_{j}=1}^{j}}.\end{eqnarray*}$$

In the case where ${s}_{j}=0$:

$$\begin{eqnarray*}&&{p}_{\vec{s}| {s}_{j}=0}^{j}(x)={\Theta }_{({{aw}}_{\vec{s}| {s}_{j}=1}^{j},{Z}_{j}]}(x)\displaystyle \frac{{p}_{\vec{s}}^{j-1}(x-{{aw}}_{\vec{s}| {s}_{j}=1}^{j}+a)}{{\eta }_{\vec{s}| {s}_{j}=0}^{j}}.\end{eqnarray*}$$

Proof. Case ${s}_{j}=1$:

Let the logarithmical work in step j be denoted by $w={w}_{\vec{s}}^{j}$,

let the Gibbs rescaled probability distribution after step j be ${p}^{j}={p}_{\vec{s}}^{j}$ and the one before the step j: ${p}^{j-1}={p}_{\vec{s}}^{j-1}$,

let the occupation probabilities be ${\lambda }^{j}={\lambda }_{\vec{s}}^{j}$ and the sum of the relevant occupation probabilities (as in definition 4): ${\eta }^{j}={\eta }_{\vec{s}}^{j}$

Let $x\in (0,{aw}]\bigcap (0,{Z}_{j}]$ and $b\in (0,\infty )$ such that ${\displaystyle \int }_{0}^{b}\mathrm{exp}(-\displaystyle \frac{{E}_{\lceil \displaystyle \frac{{yn}}{w}\rceil }^{j-1}}{{kT}}){\rm{d}}y=x$.

$$\begin{eqnarray*}{p}^{j}(x) & = & {p}^{j}\left({\displaystyle \int }_{0}^{b}\mathrm{exp}\left(-\displaystyle \frac{{E}_{\lceil \displaystyle \frac{{yn}}{w}\rceil }^{j-1}}{{kT}}\right){\rm{d}}y\right)\\ & = & {p}^{j}\left({\displaystyle \int }_{0}^{b}\mathrm{exp}\left(-\left(\displaystyle \frac{{E}_{\lceil \displaystyle \frac{{yn}}{w}\rceil }^{j-1}-{kT}\mathrm{ln}(w)}{{kT}}\right)\right)\displaystyle \frac{1}{w}{\rm{d}}y\right)\\ & = & {p}^{j}\left({\displaystyle \int }_{0}^{b/w}\mathrm{exp}\left(-\left(\underset{\stackrel{(*)}{=}{E}_{\lceil {zn}\rceil }^{j}/({kT})}{\underbrace{\displaystyle \frac{{E}_{\lceil {zn}\rceil }^{j-1}-{kT}\mathrm{ln}(w)}{{kT}}}}\right)\right){\rm{d}}z\right)\\ & \stackrel{(**)}{=} & \displaystyle \frac{{\lambda }_{\lceil \displaystyle \frac{{bn}}{w}\rceil }^{j}}{\mathrm{exp}\left(-\left(\displaystyle \frac{{E}_{\lceil \displaystyle \frac{{bn}}{w}\rceil }^{j-1}-{kT}\mathrm{ln}(w)}{{kT}}\right)\right)}=\displaystyle \frac{{\lambda }_{\lceil \displaystyle \frac{{bn}}{w}\rceil }^{j-1}}{\mathrm{exp}\left(-\left(\displaystyle \frac{{E}_{\lceil \displaystyle \frac{{bn}}{w}\rceil }^{j-1}}{{kT}}\right)\right)w{\eta }^{j}}\\ & = & \displaystyle \frac{1}{w{\eta }^{j}}{p}^{j-1}\left({\displaystyle \int }_{0}^{b/w}\mathrm{exp}\left(-\left(\displaystyle \frac{{E}_{\lceil {zn}\rceil }^{j-1}}{{kT}}\right)\right){\rm{d}}z\right)\\ & = & \displaystyle \frac{1}{w{\eta }^{j}}{p}^{j-1}\left({\displaystyle \int }_{0}^{b}\displaystyle \frac{\mathrm{exp}\left(-\left(\displaystyle \frac{{E}_{\lceil \displaystyle \frac{{yn}}{w}\rceil }^{j-1}}{{kT}}\right)\right)}{w}{\rm{d}}y\right)=\displaystyle \frac{1}{w{\eta }^{j}}{p}^{j-1}\left(\displaystyle \frac{x}{w}\right),\end{eqnarray*}$$

where the equation $(*)$ follows by definition 4 and the equation $(**)$ follows by definition 2.

One easily sees that ${p}^{j}(x)=0$ for $x\geqslant {aw}$, since then ${\Theta }_{\{0,\ldots ,l\}}(x)=0$ in definition 4.

The proof for the case ${s}_{j}=0$ is analogous. □

This next lemma shows how the partition function changes during a work extraction, as a function of how much the chosen levels are stretched (encoded in w) and how many levels are shifted (encoded in a as described above).

Lemma 3. The partition function Z_j immediately after step j is given by:

$$\begin{eqnarray*}&&{Z}_{j}={Z}_{j-1}+a({w}_{1}-1),\end{eqnarray*}$$

where $(0,a]$ is the interval on which the Gibbs-rescaled distribution is associated with the stretched levels, and w₁ is the logarithmic work extracted if the extraction is successful.

Proof. Let the $(0,a]$ interval be associated with blocks corresponding to the levels $\{1,\ldots ,l\}$ and split the interval $(a,{Z}_{j}]$ into $n-l$ blocks for some n.

$$\begin{eqnarray*}{Z}_{j} & = & \displaystyle \sum _{k}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j}}{{kT}}}=\displaystyle \sum _{k=1}^{l}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j}}{{kT}}}+\displaystyle \sum _{k=l+1}^{n}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j}}{{kT}}}\\ & = & \displaystyle \sum _{k=1}^{l}{{\rm{e}}}^{\displaystyle \frac{-\left({E}_{k}^{j-1}-{kT}\mathrm{ln}\left({w}_{1}\right)\right)}{{kT}}}+\displaystyle \sum _{k=l+1}^{n}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j}}{{kT}}}={w}_{1}\underset{a}{\underbrace{\displaystyle \sum _{k=1}^{l}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j-1}}{{kT}}}}}+\displaystyle \sum _{k=l+1}^{n}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j-1}}{{kT}}}\\ & = & {w}_{1}a-a+\displaystyle \sum _{k=1}^{l}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j-1}}{{kT}}}+\displaystyle \sum _{k=l+1}^{n}{{\rm{e}}}^{\displaystyle \frac{-{E}_{k}^{j-1}}{{kT}}}={Z}_{j-1}+{{aw}}_{1}-a\end{eqnarray*}$$

out of which the lemma follows. □

We consider scenarios where there is an external agent who wants to use thermalizations and work extractions to transform a system with an initial Hamiltonian H_i and density matrix ρ, to a given final Hamiltonian H_f and density matrix σ. In the process the agent will want to keep the energy of the work reservoir as high as possible, in a way that will be made more precise below.

Definition 5 (The work extraction game). There are three systems and a work-extraction agent. One system is the working medium, another is a heat bath of temperature T, and the last is the work reservoir.

The initial energy spectrum $\{E\}$ of the working medium is arbitrary but given. The initial density matrix ρ of the same is diagonal in the energy basis. The final energy spectrum $\{F\}$ and diagonal density matrix σ are also arbitrary but given.

The agent can combine thermalization (defined above) and work extraction (also defined above) in any sequence. This sequence, together with the specifications for each step is called the agent's strategy.

In a single-shot implementation of the strategy there will be a transfer of some energy ν to the work extraction reservoir. Before the extraction the agent must specify W. If $\nu \geqslant W$ and the final state conditioned on $\nu \geqslant W$ is σ, the work extraction is termed successful (or else a failure). The probability of success is called $1-\varepsilon$.

A crucial quantity we will be interested in calculating is the optimal work that the agent can be guaranteed to extract or need to insert. Before defining this quantity mathematically we recall a motivation for being interested in it: consider a scenario where some process is activated only if the the work reservoir energy goes above a certain threshold. One is then interested in whether this threshold is guaranteed to be exceeded. This is as opposed to the standard paradigm of focussing on the average energy increase in the reservoir. This is a key difference between the single-shot paradigm and average paradigm.

Definition 6 (Guaranteed work). For a given strategy S, and a given initial state there is a probability distribution of work transferred to the reservoir, ${p}_{S}({\mathcal{W}})$. We denote the work guaranteed up to a probability of failure associated with that strategy as ${W}_{S}^{\varepsilon }$, and define it through the equation

$$\begin{eqnarray*}&&{W}_{S}^{\varepsilon }=\mathrm{max}y\;:{\displaystyle \int }_{0}^{y}{p}_{S}({\mathcal{W}}){\rm{d}}{\mathcal{W}}\leqslant \varepsilon .\end{eqnarray*}$$

For an initial Hamiltonian H_i, density matrix ρ and tolerated probability of failure , there is a set ${\mathbb{S}}$ of allowed strategies which succeed with probability greater than or equal to $1-\varepsilon$. We denote the optimal work guaranteed (up to failure probability ) for the given initial and final conditions by ${W}^{\varepsilon }(\rho ,{H}_{i}\to \sigma ,{H}_{f})$ and define it as the optimal work over all the allowed strategies in the set:

$$\begin{eqnarray*}&&{W}^{\varepsilon }(\rho ,{H}_{i}\to \sigma ,{H}_{f}):= \underset{S\in {\mathbb{S}}}{\mathrm{sup}}{W}_{S}^{\varepsilon }(\rho ,{H}_{i}\to \sigma ,{H}_{f}).\end{eqnarray*}$$

(Note that this quantity may be negative in the case where work is required to effect the given change in state and Hamiltonian).

To assist the reading of the proofs below we collect key notation in the following:

Definition 7 (Notation). We shall use the following notation:

• a vector with one entry for each of m work extractions (subsequently called 'steps'): ${s}_{j}\;=\;1$: system is in one of the energy levels chosen for work extraction.
• system is not in one of the states chosen for work extraction $\vec{s}$ is called a path. ${\hat{s}}_{j}$ is the complement of s_j: ${s}_{j}=1\;\iff \;{\hat{s}}_{j}=0$ and ${s}_{j}=0\;\iff \;{\hat{s}}_{j}=1$.
• logarithmical work (${kT}\mathrm{ln}({w}_{\vec{s}}^{j})={W}_{\vec{s}}^{j}$) extracted in step j on path $\vec{s}$.
• The logarithmical work one extracts in step j if the specified level is occupied.
• work demanded in order to call the total extraction successful (see definition 5).
• total logarithmical work demanded in order to call the total extraction successful.

G is the set of successful paths, i.e. those yielding as much work as demanded:

$$\begin{eqnarray*}&&G=\left\{\vec{s}\left|\displaystyle \prod _{j=1}^{m}{w}_{\vec{s}}^{j}\geqslant w\right.\right\}.\end{eqnarray*}$$

• probability of picking step j on the path $\vec{s}$. I.e. as in definition 4: ${\eta }_{\vec{s}| {s}_{j}=1}^{j}={\displaystyle \sum }_{i=1}^{l}{\lambda }_{\vec{s}}^{j-1}(i)$, if the chosen energy levels for work-extraction in step j are $\{1,\ldots ,l\}$ and ${\lambda }_{\vec{s}}^{j-1}$ as defined below.
• total probability of success:${P}_{S}={\displaystyle \sum }_{\vec{s}\in G}{\prod }_{j}{\eta }_{\vec{s}}^{j}$.
• occupation probabilities after step j if the previous evolution of the system is given by the path $\vec{s}$.
• Gibbs rescaled probability distribution after step j (before thermalizing) conditioned on the previous steps on path $\vec{s}$.
• Gibbs rescaled probability distribution after step j (after thermalizing) conditioned on the previous steps on path $\vec{s}$.
• for $a\lt b$ the interval $(a,b]$ is said to be a block corresponding to a level k, if ${p}_{\vec{s}}^{j}$ is constant on this interval $\forall \vec{s}$.
• final Gibbs rescaled probability distribution, conditioned on successful work extraction:
  $$\begin{eqnarray*}&&q=\sum _{\vec{s}\in G}{p}_{\vec{s},t}^{m}\displaystyle \frac{\displaystyle \prod _{j}{\eta }_{\vec{s}}^{j}}{{P}_{S}}.\end{eqnarray*}$$
• Bistochastic matrix one chooses after step j by thermalizing the system (this has to be the same for all paths).
• Energy of the level labelled by x after step j.
• Step function associated with an interval U:
  $$\begin{eqnarray*}{\Theta }_{U}(x)=\left\{\begin{array}{ll}1 & {\rm{:}}\;{\rm{for}}\;{\rm{}}x\in U\\ 0 & {\rm{:}}\;{\rm{else}}\end{array}\right..\end{eqnarray*}$$

Consider zero-risk work extraction such that all levels with non-zero occupation probability are shifted. Note firstly that after a work extraction by $W={kT}\mathrm{ln}(w)$ the height of the Gibbs-rescaled probability distribution is given by ${\lambda }_{i}/\mathrm{exp}\left(-\left(\frac{({E}_{i}-W)}{{kT}}\right)\right)={\lambda }_{i}/\left(\mathrm{exp}\left(-\frac{{E}_{i}}{{kT}}\right)w\right)$, while the width gets stretched by a factor w. So the new Gibbs-rescaled probability distribution is given in terms of the old one as follows: ${p}_{\mathrm{new}}(x)=\frac{{P}_{\mathrm{old}}(x/w)}{w}$ (see lemma 2 for more details).

Thermalization acts as a bistochastic matrix on the Gibbs-rescaled probability distribution and therefore (see Hardy et al 1952) ${\displaystyle \int }_{0}^{l}p(x){\rm{d}}{\text{}}x\geqslant {\displaystyle \int }_{0}^{l}{p}_{\mathrm{thermalized}}(x){\rm{d}}{\text{}}x$, if both distributions are monotonically falling, which we will now assume w.l.o.g. Thus after a thermalization and a work extraction the following holds:

$$\begin{eqnarray*}{\displaystyle \int }_{0}^{l}{p}_{\mathrm{new},\mathrm{thermalized}}(x){\rm{d}}{\text{}}x & \leqslant & {\displaystyle \int }_{0}^{l}{p}_{\mathrm{new}}(x){\rm{d}}{\text{}}x\\ & = & {\displaystyle \int }_{0}^{l}\displaystyle \frac{{p}_{\mathrm{old}}(x/w)}{w}{\rm{d}}{\text{}}x={\displaystyle \int }_{0}^{{wl}}{\text{}}{p}_{\mathrm{old}}(x){\rm{d}}{\text{}}x.\end{eqnarray*}$$

Inductively, after any number of work extractions and thermalizations and total work ${kT}\mathrm{ln}(w)$:

$$\begin{eqnarray*}&&{\displaystyle \int }_{0}^{{wl}}{p}_{\mathrm{initial}}(x){\rm{d}}{\text{}}x\geqslant {\displaystyle \int }_{0}^{l}{p}_{\mathrm{final}}({\text{}}x){\rm{d}}{\text{}}x.\end{eqnarray*}$$

It follows that the maximal logarithmical work given the initial and final Gibbs-rescaled distributions is given by

$$\begin{eqnarray*}&&\mathrm{max}w{\rm{}}\;{\rm{s.t.}}\;{\rm{}}\;{\displaystyle \int }_{0}^{{wl}}{p}_{\mathrm{initial}}(x){\rm{d}}{\text{}}x\geqslant {\displaystyle \int }_{0}^{l}{p}_{\mathrm{final}}(x){\rm{d}}{\text{}}x,\end{eqnarray*}$$

or equivalently in terms of the cumulative distribution functions ${\mathcal{F}}$,

$$\begin{eqnarray*}&&\mathrm{max}w{\rm{}}\;{\rm{s.t.}}\;{\rm{}}\;{{\mathcal{F}}}_{{\rm{p(initial)}}}\left(\displaystyle \frac{x}{w}\right)\geqslant {{\mathcal{F}}}_{{\rm{p(final)}}}(x)\;\;\forall x.\end{eqnarray*}$$

This is precisely the relative mixedness defined in the main section. In Horodecki and Oppenheim (2011/2013) they also arrive at the same result for the zero-risk case (starting from an a priori different model and using different arguments).

We now turn to the general case. We combine the two previous lemmas to gain another relation between the Gibbs rescaled distribution at steps j and $j-1$. We shall use this later in an iterative manner to relate the very first and final Gibbs rescaled distributions.

Lemma 4. The Gibbs rescaled probability distributions at steps j and $j-1$ respectively satisfy the relation

$$\begin{eqnarray*}&&{p}_{\vec{s},t}^{j-1}(x)=\displaystyle \sum _{k=0,1}{w}_{\vec{s}| {s}_{j}=k}^{j}{\eta }_{\vec{s}| {s}_{j}=k}^{j}{p}_{\vec{s}| {s}_{j}=k}^{j}\left({{xw}}_{\vec{s}| {s}_{j}=k}^{j}+{c}_{\vec{s}| {s}_{j}=k}^{j}\right)\end{eqnarray*}$$

with constants ${c}_{\vec{s}| {s}_{j}=1}^{j}=0$ and ${c}_{\vec{s}| {s}_{j}=0}^{j}={{aw}}^{j}-a$.

Proof. Let ${w}_{k}={w}_{\vec{s}| {s}_{j}=k}^{j}$, ${p}_{k}^{j}={p}_{\vec{s}| {s}_{j}=k}^{j}$, ${p}^{j-1}={p}_{\vec{s},t}^{j-1}$, ${\eta }_{k}={\eta }_{\vec{s}| {s}_{j}=k}^{j}$. Let ${c}_{0}={{aw}}_{1}-a$ and ${c}_{1}=0$. Then:

$$\begin{eqnarray*}&&{\eta }_{0}{w}_{0}{p}_{0}^{j}({{xw}}_{0}+{c}_{0})+{\eta }_{1}{w}_{1}{p}_{1}^{j}({{xw}}_{1}+{c}_{1})\\ &&\quad ={\eta }_{0}{p}_{0}^{j}(x+{{aw}}_{1}-a)+{\eta }_{1}{w}_{1}{p}_{1}^{j}({{xw}}_{1})\\ &&\quad ={\Theta }_{({{aw}}_{1},{Z}_{j}]}(x+{{aw}}_{1}-a){p}^{j-1}(x)+{\Theta }_{(0,{{aw}}_{1}]}({{xw}}_{1}){p}^{j-1}(x)\\ &&\quad ={\Theta }_{(a,{Z}_{j}-{{aw}}_{1}+a]}(x){p}^{j-1}(x)+{\Theta }_{(0,a]}(x){p}^{j-1}(x)\\ &&\quad ={\Theta }_{(0,{Z}_{j-1}]}(x){p}^{j-1}(x).\end{eqnarray*}$$

□

We now use the above to make a statement about the relation between the integrals of the Gibbs rescaled distribution at steps j and $j-1$. We show that the distribution before step j majorizes the distribution after the step, even after the latter has been stretched by the logarithmical work done (w in the case ${s}_{j}=1$, 1 else). This can be seen as a generalization of the inequality: ${\displaystyle \int }_{0}^{l}\frac{{p}_{\mathrm{old}}(x/w)}{w}{\rm{d}}{\text{}}x\geqslant {\displaystyle \int }_{0}^{l}{p}_{\mathrm{new},\mathrm{thermalized}}({\text{}}x){\rm{d}}{\text{}}x$ from the above special instructive case to the case where ${s}_{j}=0$ is also possible.

Lemma 5. Let $j\in \{1,\ldots ,m\}$. Let $l\in (0,{Z}_{j}]$. Let $\vec{s}\prime \in \{0,1\}{}^{m-j-1}$. Define ${\vec{s}}_{1}=({s}_{1},\ldots ,{s}_{j},1,{s}_{1}^{\prime },\ldots ,{s}_{m-j-1}^{\prime })$ and ${\vec{s}}_{0}=({s}_{1},\ldots ,{s}_{j},0,{s}_{1}^{\prime },\ldots ,{s}_{m-j-1}^{\prime })$. Then:

$$\begin{eqnarray*}&&\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}{\tau }_{t}^{j}\;\circ \;{p}_{{\vec{s}}_{0},t}^{j}(x){\rm{d}}{\text{}}x\\ &&\quad \geqslant \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}\left({w}^{j+1}{\eta }_{{\vec{s}}_{1}}^{j+1}{\tau }_{t}^{j+1}\;\circ \;{p}_{{\vec{s}}_{1},t}^{j+1}({{xw}}^{j+1})\right.{\rm{d}}{\text{}}x\\ &&\quad +\left.{\eta }_{{\vec{s}}_{0}}^{j+1}{\tau }_{t}^{j+1}\;\circ \;{p}_{{\vec{s}}_{0},t}^{j+1}(x)\right){\rm{d}}{\text{}}x,\end{eqnarray*}$$

where ${\tau }_{t}^{j}$ is the permutation of any blocks, which maximizes the left hand side, while ${\tau }_{t}^{j+1}$ is the one which maximizes the right hand side.

Proof. Let ${p}_{1}={p}_{{\vec{s}}_{1},t}^{j+1}$, ${p}_{0}={p}_{{\vec{s}}_{0},t}^{j+1}$, ${\eta }_{1}={\eta }_{{\vec{s}}_{1}}^{j+1}$, ${\eta }_{0}={\eta }_{{\vec{s}}_{0}}^{j+1}$, $w={w}^{j+1}$.

$$\begin{eqnarray*}&&\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}{\tau }_{t}^{j}\;\circ \;{p}_{{\vec{s}}_{0},t}^{j}(x){\rm{d}}{\text{}}x\\ &&\quad =\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}\left({\eta }_{1}w{\tau }_{t}^{j}\;\circ \;{p}_{1}({xw})+{\eta }_{0}{\tau }_{t}^{j}\;\circ \;{p}_{0}(x+{aw}-a)\right){\rm{d}}{\text{}}x\\ &&\quad =\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{{l}_{1}}{\eta }_{1}w\tilde{\tau }\;\circ \;{p}_{1}({xw})+\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{a}^{l+a-{l}_{1}}{\eta }_{0}\tilde{\tau }\;\circ \;{p}_{0}(x+{aw}-a){\rm{d}}{\text{}}x,\end{eqnarray*}$$

where the first equality is exactly lemma 4. In the second equality ${l}_{1}\in (0,\mathrm{min}(a,l)]$ is a value which maximizes the right hand side of the last line and $\tilde{\tau }$ reorders ${\displaystyle \sum }_{\vec{s}\in \{\mathrm{0,1}\}{}^{j}}{p}_{1}$ in descending order in $(0,{aw}]$ and ${\displaystyle \sum }_{\vec{s}\in \{\mathrm{0,1}\}{}^{j}}{p}_{0}$ in $({aw},{Z}_{j}]$. This is possible since p₁ and p₀ have disjoint support, also for different $\vec{s}$, since a in definition 4 has to be chosen independently of the path. (See lemma 2). This reordering maximizes the last line, thus it is equal to the line above.

After changing variables in the second integral we can translate its bounds by $-{aw}+{l}_{1}$, if we translate the integrand in the opposite direction applying a second permutation. Thus:

$$\begin{eqnarray*}\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}{\tau }_{t}^{j}\;\circ \;{p}_{{\vec{s}}_{0},t}^{j}(x){\rm{d}}{\text{}}x & = & \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{{l}_{1}}{\eta }_{1}w\tilde{\tau }\;\circ \;{p}_{1}({xw})\\ & & +\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{{aw}}^{l+{aw}-{l}_{1}}{\eta }_{0}\tilde{\tau }\;\circ \;{p}_{0}(x){\rm{d}}{\text{}}x\\ & = & \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}\left({\eta }_{1}w{\tau }^{j+1}\;\circ \;{p}_{1}({xw})+{\eta }_{0}{\tau }^{j+1}\;\circ \;{p}_{0}(x)\right){\rm{d}}{\text{}}x.\end{eqnarray*}$$

Applying any bistochastic matrix $\tilde{B}$ on the probabilities p₀ and p₁ and reordering in descending order with ${\tau }_{t}^{j+1}$ afterwards, we get (we write $\tilde{B}=B\;\circ \;{({\tau }^{j+1})}^{-1}$ for convenience, then B is again bistochastic):

$$\begin{eqnarray*}\displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}{\tau }_{t}^{j}\;\circ \;{p}_{{\vec{s}}_{0},t}^{j}(x){\rm{d}}{\text{}}x & \geqslant & {\displaystyle \int }_{0}^{l}{\tau }_{t}^{j+1}\;\circ \;B\;\circ \;{({\tau }^{j+1})}^{-1}\;\circ \;\\ & & \times \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}\left({\eta }_{1}w{\tau }^{j+1}\;\circ \;{p}_{1}({xw})+{\eta }_{0}{\tau }^{j+1}\;\circ \;{p}_{0}(x)\right){\rm{d}}{\text{}}x\\ & = & \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}\left(w{\eta }_{1}{\tau }_{t}^{j+1}\;\circ \;B\;\circ \;{p}_{1}({xw})+{\eta }_{0}{\tau }_{t}^{j+1}\;\circ \;B\;\circ \;{p}_{0}(x)\right){\rm{d}}{\text{}}x\\ & = & \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{j}}{\displaystyle \int }_{0}^{l}\left(w{\eta }_{1}{\tau }_{t}^{j+1}\;\circ \;{p}_{{\vec{s}}_{1},t}^{j+1}({xw})+{\eta }_{0}{\tau }_{t}^{j+1}\;\circ \;{p}_{{\vec{s}}_{0},t}^{j+1}(x)\right){\rm{d}}{\text{}}x,\end{eqnarray*}$$

where the inequality follows out of the inequality ${Bp}\;\succ \;p$ for any bistochastic matrix B and vector p, which is proved in Hardy et al (1952). □

The above lemma is the main ingredient for the first part of the main theorem and the rest of the proof is straightforward.

Theorem (First part of theorem 1 in main body, giving the bound). In the work extraction game defined above, if one is given an initial density matrix $\rho ={\displaystyle \sum }_{i}{\lambda }_{i}| {e}_{i}\rangle \langle {e}_{i}|$ and final density matrix $\sigma ={\displaystyle \sum }_{j}{\nu }_{j}| {f}_{j}\rangle \langle {f}_{j}|$ with $\{| {e}_{i}\rangle \}$, $\{| {f}_{j}\rangle \}$ the respective energy eigenstates and both ρ and σ having finite rank, then the work ${W}^{\varepsilon }$ one can extract with certainty except with probability respects

$$\begin{eqnarray*}&&{W}_{S}^{\varepsilon }\leqslant {kT}\mathrm{ln}\left(M\left(\displaystyle \frac{{G}^{(T,{H}_{i})}(\rho )}{1-\varepsilon }\parallel {G}^{(T,{H}_{f})}(\sigma )\right)\right).\end{eqnarray*}$$

Proof. Define ${p}_{{\vec{s}}^{\prime }}^{0}=p$. W.l.o.g. ${\vec{s}}^{\prime }=\{0,\ldots ,0\}$ (the first probability distribution is independent of the path afterwards). Inductively using lemma 5 one gets:

$$\begin{eqnarray*}{\displaystyle \int }_{0}^{l}p(x){\rm{d}}{\text{}}x & = & {\displaystyle \int }_{0}^{l}{p}_{{\vec{s}}^{\prime }}^{0}(x){\rm{d}}{\text{}}x\\ & \geqslant & \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{m}}{\displaystyle \int }_{0}^{l}\left(\displaystyle \prod _{j=1}^{m}{\eta }_{\vec{s}}^{j}\right)\left(\displaystyle \prod _{j=1}^{m}{w}_{\vec{s}}^{j}\right){\tau }_{t}^{m}\;\circ \;{p}_{\vec{s}}\left(x\displaystyle \prod _{j=1}^{m}{w}_{\vec{s}}^{j}\right){\rm{d}}{\text{}}x\\ & = & \displaystyle \sum _{\vec{s}\in \{0,1\}{}^{m}}{\displaystyle \int }_{0}^{l\left(\displaystyle \prod _{j=1}^{m}{w}_{\vec{s}}^{j}\right)}\left(\displaystyle \prod _{j=1}^{m}{\eta }_{\vec{s}}^{j}\right){\tau }_{t}^{m}\;\circ \;{p}_{\vec{s}}(x){\rm{d}}{\text{}}x\\ & \geqslant & \displaystyle \sum _{\vec{s}\in G}{\displaystyle \int }_{0}^{l\left(\displaystyle \prod _{j=1}^{m}{w}_{\vec{s}}^{j}\right)}\left(\displaystyle \prod _{j=1}^{m}{\eta }_{\vec{s}}^{j}\right){\tau }_{t}^{m}\;\circ \;{p}_{\vec{s}}(x){\rm{d}}{\text{}}x\\ & \geqslant & \displaystyle \sum _{\vec{s}\in G}{\displaystyle \int }_{0}^{{lw}}\left(\displaystyle \prod _{j=1}^{m}{\eta }_{\vec{s}}^{j}\right){\tau }_{t}^{m}\;\circ \;{p}_{\vec{s}}(x){\rm{d}}{\text{}}x={P}_{S}{\displaystyle \int }_{0}^{{lw}}q(x){\rm{d}}{\text{}}x,\end{eqnarray*}$$

where ${\tau }_{t}^{m}$ is the permutation which maximizes the expression of the right hand side of the first inequality (t stands for 'after thermalizing', while m stands for the mth time one applies lemma 5). Therefore (with ${P}_{S}=1-\varepsilon$):

$$\begin{eqnarray*}&&{W}^{\varepsilon }={kT}\mathrm{ln}(w)\\ &&\leqslant {kT}\mathrm{ln}\left(\mathrm{max}\left\{m\left|{\displaystyle \int }_{0}^{l}p({x}_{1}){\rm{d}}{x}_{1}\geqslant {\displaystyle \int }_{0}^{{lm}}(1-\varepsilon )q({x}_{2}){\rm{d}}{x}_{2}\;\forall l\right\}\right.\right)\\ &&={kT}\mathrm{ln}\left(M\left(\displaystyle \frac{{G}^{(T,{H}_{i})}(\rho )}{1-\varepsilon }\parallel {G}^{(T,{H}_{f})}(\sigma )\right)\right).\end{eqnarray*}$$

This proves the first part of the main theorem. □

Before giving the general protocol it is instructive to consider an example. We begin with a density matrix ϕ with energy eigenvalues ${E}_{i}(j)$, occupation probabilities ${\lambda }_{i}(j)$ and A_i defined by ${A}_{i}(j)=\mathrm{exp}\left(\frac{-{E}_{i}(j)}{{kT}}\right)$. These are given by:

$$\begin{eqnarray*}&&{\lambda }_{i}=\left(\displaystyle \frac{2}{3},\displaystyle \frac{1}{3},0\right){A}_{i}=\left(\displaystyle \frac{1}{3},\displaystyle \frac{1}{3},\displaystyle \frac{1}{3}\right)\end{eqnarray*}$$

and therefore:

$$\begin{eqnarray}{p}_{i}(x)=\left\{\begin{array}{ll}2,\ & x\in \left(0,\displaystyle \frac{1}{3}\right]\\ 1,\ & x\in \left(\displaystyle \frac{1}{3},\displaystyle \frac{2}{3}\right]\\ 0,\ & x\in \left(\displaystyle \frac{2}{3},1\right]\end{array}\right..\end{eqnarray} \tag{ C.1 }$$

The final state we want to reach is defined through:

$$\begin{eqnarray*}&&{\lambda }_{f}=\left(\displaystyle \frac{1}{2},\displaystyle \frac{1}{2},0\right){A}_{f}=\left(\displaystyle \frac{1}{6},\displaystyle \frac{1}{3},0\right),\end{eqnarray*}$$

and therefore:

$$\begin{eqnarray}{p}_{f}(x)=\left\{\begin{array}{ll}3,\ & x\in \left(0,\displaystyle \frac{1}{6}\right]\\ \displaystyle \frac{3}{2},\ & x\in \left(\displaystyle \frac{1}{6},\displaystyle \frac{1}{2}\right]\end{array}\right..\end{eqnarray} \tag{ C.2 }$$

With a risk $\varepsilon =\frac{1}{2}$ the work for this game is limited by $W={kT}\mathrm{ln}\left(M\left(\frac{{p}_{i}}{1-\varepsilon }| | {p}_{f}\right)\right)={kT}\mathrm{ln}\left(\frac{4}{3}\right)$. In this example we show how this amount of work can be extracted.

We first want to raise as many energy levels as we can to infinite energy, such that if we succeed (i.e. if these levels are empty and the action therefore costs 0 work) we start with a more known state. Unfortunately the sum of the occupation probabilities of the lowest levels will never yield exactly , so we need to change this first.

We start by raising the empty energy level to infinite energy, such that even if one mixes it completely with any other energy level it will stay empty. Then we lower the energy of the empty level, while constantly mixing this level with the first one. At the same time we enhance the energy of the first level, such that in total the energy of the work reservoir is unchanged with probability 1 (the details of this action can be found below in definition 9 and the following lemma). We then have:

$$\begin{eqnarray*}&&{\lambda }_{1}=\left(\displaystyle \frac{1}{2},\displaystyle \frac{1}{3},\displaystyle \frac{1}{6}\right){A}_{1}=\left(\displaystyle \frac{1}{4},\displaystyle \frac{1}{3},\displaystyle \frac{1}{3}\right)\end{eqnarray*}$$

$$\begin{eqnarray*}{p}_{1}(x)=\left\{\begin{array}{ll}2,\ & x\in \left(0,\displaystyle \frac{1}{3}\right]\\ 1,\ & x\in \left(\displaystyle \frac{1}{3},\displaystyle \frac{2}{3}\right].\end{array}\right.\end{eqnarray*}$$

The lowest two occupation probabilities now sum up to . We enhance the energy of these two levels by doing a work extraction changing the energy of their states by $\infty$. With probability $1-\varepsilon =\frac{1}{2}$ we get the work 0 and the state:

$$\begin{eqnarray*}&&{\lambda }_{2}=(1,0,0){A}_{2}=\left(\frac{1}{4},0,0\right)\end{eqnarray*}$$

$$\begin{eqnarray*}{p}_{2}(x)=4,\ & x\in \left(0,\displaystyle \frac{1}{4}\right]\end{eqnarray*}$$

which in this case is a pure state (the state would not have been pure if we had chosen to be smaller than $\frac{1}{3}$). With probability $\frac{1}{2}$ we get the work $-\infty$, in which case the work extraction cannot be successful in total. So in the case where the work extraction is successful the above state is the only one we need to consider.

Now we extract the work $W={kT}\mathrm{ln}\left(\frac{4}{3}\right)$ on all the levels. This succeeds with probability 1. The state afterwards is given by:

$$\begin{eqnarray*}&&{\lambda }_{3}=(1,0,0)\end{eqnarray*}$$

$$\begin{eqnarray*}&&{A}_{3}=\left(\displaystyle \frac{1}{3},0,0\right)\end{eqnarray*}$$

$$\begin{eqnarray*}{p}_{3}(x)=3,\ & x\in \left(0,\displaystyle \frac{1}{3}\right].\end{eqnarray*}$$

Again we need two levels where we only have one. Acting again as defined in definition 9 on the first two levels we can get:

$$\begin{eqnarray*}&&{\lambda }_{4}=\left(\displaystyle \frac{1}{2},\displaystyle \frac{1}{2},0\right)\end{eqnarray*}$$

$$\begin{eqnarray*}&&{A}_{4}=\left(\displaystyle \frac{1}{6},\displaystyle \frac{1}{6},0\right)\end{eqnarray*}$$

$$\begin{eqnarray*}{p}_{4}(x)=3,\ & x\in \left(0,\displaystyle \frac{1}{3}\right].\end{eqnarray*}$$

The energy of the second level is now too high and we need to lower it by ${kT}\mathrm{ln}(2)$:

$$\begin{eqnarray*}{\lambda }_{5} & = & \left\{\begin{array}{ll}(1,0,0),\ & \text{with probability}\;\displaystyle \frac{1}{2}\\ (0,1,0),\ & \text{with probability}\;\displaystyle \frac{1}{2}\end{array}\right.\\ {A}_{5} & = & \left(\displaystyle \frac{1}{6},\displaystyle \frac{1}{3},0\right).\end{eqnarray*}$$

The work extracted in this step is in both cases at least 0. So by measuring whether the energy in the work-reservoir has been enhanced by at least $W={kT}\mathrm{ln}\left(\frac{4}{3}\right)$, we get a 'yes' and the wanted final state with probability $\frac{1}{2}$.

To make the idea clearer we start giving the general algorithm and will then give the proof of the second part of the main theorem, which builds on lemmas proved later on. We assume here that we have at least $n/2$ energy levels with 0 occupation probability, but make sure that in the end these levels have again 0 occupation probability (note, that this does not change the upper bound for the work). We assume that the levels are ordered in descending order of their Gibbs rescaled probability.

Definition 8 (Work extraction algorithm (see figure C1)). Let p and p_f be Gibbs rescaled probability distributions of two states ρ and σ, with the same number of levels n.

Let ρ, σ have at least $n/2$ levels with occupation probabilities ${\lambda }_{e}=0$.

Define $W={kT}\mathrm{ln}({M}^{\varepsilon }(p,q))$.

• (i)  
  Do a work extraction on the levels $k+1,\ldots ,n$ by $-\infty$ (such that their width becomes 0).If there is no k for which $1-\varepsilon ={\displaystyle \sum }_{i=1}^{k}\lambda (i)$:Split the level k for which ${\displaystyle \sum }_{i=1}^{k-1}\lambda (i)\lt 1-\varepsilon \lt {\displaystyle \sum }_{i=1}^{k}\lambda (i)$ (see the corollary to lemma 7, below).
• (ii)  
  Make a work extraction on all levels by W (i.e. stretch their Gibbs rescaled probability distributions such that it just majorizes the final one).
• (iii)  
  Thermalize the obtained state to get the final state (up to permutation).
• (iv)  
  Permute the levels of the obtained state such, that one gets the final state.

Theorem 6 (Bound can be achieved (second part of main theorem)). Let p and p_f be Gibbs rescaled probability distributions of two states ρ and σ, with the same number of levels n.

• Let ρ, σ have at least $n/2$ levels with occupation probabilities ${\lambda }_{e}=0$.
• Define $W={kT}\mathrm{ln}({M}^{\varepsilon }(p,q))$.

The work extraction algorithm on ρ yields the work W with probability $1-\varepsilon$. If the work extraction is successful, the final state is given by σ with probability 1.

Proof. The work extraction in step 1. succeeds with probability and if it does not succeed it yields 0 work (else $-\infty$).

• After step 1. The occupation probabilities are given by ${\lambda }_{1}(i)=\frac{\lambda (i)}{1-\varepsilon }$ for $i=1,\ldots ,k$ (post-selecting on the case, in which the state was not one of the less likelier) and ${\lambda }_{1}(i)=0$ else (if the work extraction 'succeeds' and our algorithm fails). See the corollary to lemma 7, below.
• After step 2. By the definition of W we have that ${p}_{2}(i)\;\succ \;{p}_{f}(i)$, the extracted work is W. Therefore one can thermalize the obtained state to get the final state ρ (up to permutation) with probability 1 (see lemma 8, below). After the permutation (if the levels have some special physical meaning) we get the final state ρ with probability 1.
• In total we get the final state ρ with probability 1, if the work extraction succeeds and the extracted work is W with probability $1-\varepsilon$.

□

To start, we need some algorithm which allows us to shift some probability from one level to the other, if they are in thermal equilibrium. We only want to change these two levels (say j, k), so the sum of their occupation probabilities remains constant (${\lambda }_{j}+{\lambda }_{k}=\mathrm{const}$). Also we hope to be able to do this without needing to do any work, so we keep our total knowledge of these levels constant. To achieve this it seems a good idea to have ${p}_{j}+{p}_{k}=\mathrm{const}$ and constantly thermal equilibrium. This is the guiding idea for the following algorithm. Instead of doing this (rather complicated) proof one also could have assumed that one can split levels in a physical fashion (see the corollary to the next lemma for details). Then one would have got the 'isothermal shift' for free, by simply splitting the level k in two parts and afterwards removing the level j. But this would have been a further assumption. So the following definition and subsequent lemma can also be seen to show it possible (in principle) to achieve a splitting of a level by just having one further empty level a heat bath and a work reservoir (which remains untouched with probability 1).

Definition 9 (Isothermal shift of boundary (see figure C2)). Let $A(j)=\mathrm{exp}\left(\frac{-{E}_{j}}{{kT}}\right)$, where E_j is the energy eigenvalue of the jth level.

Let the levels j, $k=j+1$ have the same Gibbs rescaled probability. We call the limit $n\to \infty$ of the following process an isothermal shift of the boundary between j and k by $w\in \left(-\frac{A(j)}{A(j)+A(k)},\frac{A(k)}{A(j)+A(k)}\right)$ in direction k:

• (i)  
  Do a permutation, which brings the level j in front and level k as second.
• (ii)  
  Do a work extraction on level j by:

  $$\begin{eqnarray*}&&{w}_{1}=1+\displaystyle \frac{w}{n}\displaystyle \frac{A(j)+A(k)}{A(j)}.\end{eqnarray*}$$
• (iii)  
  Do a permutation, which brings the level k in front and level j second.
• (iv)  
  Do a work extraction on level k by:

  $$\begin{eqnarray*}&&{w}_{2}=1-\displaystyle \frac{w}{n}\displaystyle \frac{A(j)+A(k)}{A(k)}.\end{eqnarray*}$$
• (v)  
  Do a thermalization totally mixing the two levels j and k and letting all others untouched (i.e. the matrix with entries $1/2$ in $(1,1)$, $(1,2)$, $(2,1)$ and $(2,2)$ and ${\delta }_{m,l}$ everywhere else, such that the first entry of the vector it is applied on, is the probability of the level j after work extraction and the second is the probability of the level k).
• (vi)  
  Restart with 1. n times in total, redefining $A(j)$ and $A(k)$ as above for the probabilities after this process.
• (viii)  
  Do a permutation, which brings back the levels j and $k=j+1$ at their position at the beginning (we show below, that this is possible).

Instead of the first four actions, we could have simply said we do extract the work w₁ on the level j and the work w₂ on the level k. Then we would have had to continue with doing the total mixing also between these levels (instead of at the first and second position of the matrix) and so on. What we mean here with doing a work extraction on the level j is the action: do a permutation bringing the level j in front, extract work, permute the level back.

In later definitions we will make use of this. Here we do not, since the algebra would get slightly more complicated.

The following lemma shows that the above process costs no work with probability 1 and that it can indeed be seen as a shift of the separation between the levels.

Lemma 7 (Action of the isothermal shift of boundary). Let $A(j)=\mathrm{exp}\left(\frac{-E(j)}{{kT}}\right)$, where $E(j)$ is the energy eigenvalue of the jth level.

Let the levels j, $k=j+1$ have the same Gibbs rescaled probability.

After an isothermal shift of the boundary between j and k by $w\in \left(-\frac{A(j)}{A(j)+A(k)},\frac{A(k)}{A(j)+A(k)}\right)$ in direction k:

• (i)  
  • (a)  
    The energy eigenvalues of all levels but j and k remain constant.
  • (b)  
    At the end ${A}_{f}(j)=\mathrm{exp}\left(\frac{-{E}_{f}(j)}{{kT}}\right)$ is given by ${A}_{f}(j)=A(j)+w(A(j)+A(k))$ and for the level k: ${A}_{f}(k)=A(k)-w(A(j)+A(k))$ (${E}_{f}(j)$ is the energy of the eigenvalue j after the shift).
• (ii)  
  With probability $1-(\lambda (j)+\lambda (k))$, the occupation probabilities of the final state are given by $\frac{\lambda (l)}{1-(\lambda (j)+\lambda (k))}$ for $l\ne j,k$ and 0 for $l=j,k$.
• (iii)  
  With probability $\lambda (j)+\lambda (k)$, the occupation probabilities of the final state are given by $\frac{{A}_{f}(l)}{A(j)+A(k)}$ for $l=j,k$ and 0 else.
• (iv)  
  With probability 1 the energy in the work reservoir is changed by W = 0.

Proof. 1(a) Just follows out of the algorithm, since we did not do any work extraction on any levels and this is the only way we can change energies in our game. For 1.(b) we need to look at how the energy eigenvalues of the jth and kth level change each of the n times one goes through the algorithm in definition 9. directly from the algorithm we get, that in the first time one goes through it $A(j)$ changes to ${A}_{1}(j)=\mathrm{exp}\left(\frac{-E(j)+{kT}\mathrm{ln}({w}_{1})}{{kT}}\right)$ and we get ${A}_{1}(j)={w}_{1}A(j)=A(j)+\frac{w}{n}(A(j)+A(k))$ and by the same argument ${A}_{1}(k)={w}_{2}A(k)=A(k)-$ $\frac{w}{n}(A(j)+A(k))$. Since ${A}_{1}(j)+{A}_{1}(k)=A(j)+A(k)$ we see, that after l times one goes through the algorithm, one ends up with: ${A}_{l}(j)=A(j)+(l-1)\frac{w}{n}(A(j)+A(k))+\frac{w}{n}(A(j)+A(k))=A(j)+l\frac{w}{n}(A(j)+A(k))$ and ${A}_{l}(k)=A(k)-l\frac{w}{n}(A(j)+A(k))$. With l = n we get what is stated in 1 (b).

In order to derive 2 and 3 we need to have a closer look at how the occupation probabilities change each of the n times we go through the algorithm. The occupation probabilities are given by the Gibbs rescaled probabilities multiplied with the corresponding $A(l)$.

Let q be the Gibbs rescaled probability distribution after step 1. of the ith time one goes through the algorithm in definition 9. After step 2 we have:

$$\begin{eqnarray*}q(x)\Rightarrow \left\{\begin{array}{ll}\displaystyle \frac{q\left(\displaystyle \frac{x}{{w}_{1}}\right){\Theta }_{(0,{A}_{j}]}(x)}{{w}_{1}\eta ({q}_{j})}, & {\rm{with}}\;{\rm{prob.}}\;{\rm{}}\eta ({q}_{j})\\ \displaystyle \frac{q(x-{A}_{j}{w}_{1}+{A}_{j}){\Theta }_{({A}_{j},Z(q)]}(x)}{1-\eta ({q}_{j})}, & \text{with prob.}1-\eta ({q}_{j}),\end{array}\right.\end{eqnarray*}$$

where $\eta ({q}_{j})={\displaystyle \int }_{0}^{{A}_{j}}q(x){\rm{d}}{\text{}}x$ and $Z(q)$ is the partition function of q.

After step 4 we thus have:

$$\begin{eqnarray*}\left\{\begin{array}{ll}\displaystyle \frac{q\left(\displaystyle \frac{x}{{w}_{2}}\right){\Theta }_{(0,{A}_{k}]}(x)}{{w}_{2}\eta ({q}_{k})}, & {\rm{w.}}\;{\rm{prob.}}\;{\rm{}}\eta ({q}_{k})\\ \displaystyle \frac{q\left(\displaystyle \frac{x-{A}_{j}}{{w}_{1}}+{A}_{j}-{A}_{k}{w}_{2}+{A}_{k}\right){\Theta }_{({A}_{k},{A}_{k}+{A}_{j}]}(x)}{{w}_{1}\eta ({q}_{j})}, & {\rm{w.}}\;{\rm{prob.}}\;{\rm{}}\eta ({q}_{j})\\ \displaystyle \frac{q\left(x-{A}_{j}{w}_{1}-{A}_{k}{w}_{2}+{A}_{k}+{A}_{j}\right){\Theta }_{({A}_{j}+{A}_{k},Z]}(x)}{1-\eta ({q}_{j})-\eta ({q}_{k})} & ,\ 1-\eta ({q}_{j})-\eta ({q}_{k}).\end{array}\right.\end{eqnarray*}$$

Noting that $q(x)=q(x/{w}_{2})$ for $x\in (0,{A}_{k}]$ and similarly for $x\in ({A}_{j},{A}_{k}+{A}_{j}]$ and $x-{A}_{j}{w}_{1}-{A}_{k}{w}_{2}+{A}_{k}+{A}_{j}=x$, we can rewrite this as:

$$\begin{eqnarray*}\left\{\begin{array}{ll}\displaystyle \frac{q(x){\Theta }_{(0,{A}_{k}{w}_{2}]}(x)}{{w}_{2}\eta ({q}_{k})}, & {\rm{w.}}\;{\rm{prob.}}\;{\rm{}}\eta ({q}_{k})\\ \displaystyle \frac{q(x){\Theta }_{({A}_{k}{w}_{2},{A}_{k}+{A}_{j}]}(x)}{{w}_{1}\eta ({q}_{j})}, & {\rm{w.}}\;{\rm{prob.}}\;{\rm{}}\eta ({q}_{j})\\ \displaystyle \frac{q(x){\Theta }_{({A}_{j}+{A}_{k},Z]}(x)}{1-\eta ({q}_{j})-\eta ({q}_{k})}, & \ 1-\eta ({q}_{j})-\eta ({q}_{k}).\end{array}\right.\end{eqnarray*}$$

Which means that after step 5 we get:

$$\begin{eqnarray*}\left\{\begin{array}{ll}\displaystyle \frac{q(x){\Theta }_{(0,{A}_{k}+{A}_{j}]}(x)}{{w}_{2}\eta ({q}_{k})}\displaystyle \frac{{w}_{2}\eta ({q}_{k})}{\eta ({q}_{j})+\eta ({q}_{k})}, & \ \eta ({q}_{k})\\ \displaystyle \frac{q(x){\Theta }_{(0,{A}_{k}+{A}_{j}]}(x)}{{w}_{1}\eta ({q}_{j})}\displaystyle \frac{{w}_{1}\eta ({q}_{j})}{\eta ({q}_{j})+\eta ({q}_{k})}, & \ \eta ({q}_{j})\\ \displaystyle \frac{q(x){\Theta }_{({A}_{j}+{A}_{k},Z]}(x)}{1-\eta ({q}_{j})-\eta ({q}_{k})}, & \ 1-\eta ({q}_{j})-\eta ({q}_{k}).\end{array}\right.\end{eqnarray*}$$

For 2 note that with probability $1-(\lambda (j)+\lambda (k))$ we get after the first time one goes through the algorithm: ${q}_{j}={q}_{k}=0$ (which just means, that the state is measured to be orthogonal to j and k). And therefore in the subsequent steps we have $\eta ({q}_{j})=\eta ({q}_{k})=0$. So we get with probability $1-(\lambda (j)+\lambda (k))$, the final probability distribution:

$$\begin{eqnarray*}&&\displaystyle \frac{p(x){\Theta }_{({A}_{j}+{A}_{k},Z]}(x)}{1-(\lambda (j)+\lambda (k))}.\end{eqnarray*}$$

Since the energy eigenvalues of these levels are unchanged, we get $\frac{\lambda (l)}{1-(\lambda (j)+\lambda (k))}$ for $l\ne j,k$ and 0 for $l=j,k$ for the occupation probabilities, which proves 2.

The final Gibbs rescaled probabilities of the levels j and k have the same value (since we completely mix them in step 5). Their integral (${\displaystyle \int }_{0}^{{A}_{j}+{A}_{k}}q(x){\rm{d}}{\text{}}x$), after the first time one goes through the algorithm keeps 1 (with probability $\lambda (j)+\lambda (k)$). As noticed before, ${A}_{f}(j)+{A}_{f}(k)=A(j)+A(k)$. Thus we get that with probability $\lambda (j)+\lambda (k)$ the occupation probabilities of the levels are given by: $\frac{{A}_{f}(l)}{A(j)+A(k)}$ for $l=j,k$ and 0 else. Which proves 3.

Suppose in the first time one goes through the algorithm the state is orthogonal to the levels $j,k$: then the energy in the work reservoir is unchanged throughout the whole n times one goes through the algorithm and for this case, 4 follows trivially.

We now look at the other case (the case where the state is projected onto the levels $j,k$ the first time one goes through the algorithm).

Let $\vec{s}\in \{1,2\}{}^{n}$. Define $\sigma (2)=1$ and $\sigma (1)=-1$. Define ${\alpha }_{1}=\displaystyle \frac{A(j)}{A(j)+A(k)}$ and ${\alpha }_{2}=1-{\alpha }_{1}$. In the lth time one goes through the algorithm one either gets the logarithmical work

$$\begin{eqnarray*}{w}_{l}(1) & = & 1+\displaystyle \frac{w}{n}\displaystyle \frac{{A}_{l}(j)+{A}_{l}(k)}{{A}_{l}(j)}\\ & = & 1+\displaystyle \frac{w}{n}\displaystyle \frac{A(j)+A(k)}{A(j)+(l-1)\displaystyle \frac{w}{n}(A(j)+A(k))}=\displaystyle \frac{{\alpha }_{1}+l\displaystyle \frac{w}{n}}{{\alpha }_{1}+(l-1)\displaystyle \frac{w}{n}}\end{eqnarray*}$$

or the similarly derivable value for ${w}_{l}(2)$ (A_l is defined in the proof of 1(b)). Thus we can write:

$$\begin{eqnarray*}&&{w}_{l}({s}_{l})=\displaystyle \frac{{\alpha }_{{s}_{l}}+\sigma ({s}_{l})l\displaystyle \frac{w}{n}}{{\alpha }_{{s}_{l}}+\sigma ({s}_{l})(l-1)\displaystyle \frac{w}{n}}.\end{eqnarray*}$$

In total we get the logarithmical work:

$$\begin{eqnarray*}&&{w}_{\mathrm{tot}}=\displaystyle \prod _{l=1}^{n}\displaystyle \frac{{\alpha }_{{s}_{l}}+\sigma ({s}_{l})l\displaystyle \frac{w}{n}}{{\alpha }_{{s}_{l}}+\sigma ({s}_{l})(l-1)\displaystyle \frac{w}{n}}\end{eqnarray*}$$

with probability (given, that we have the case where the state is projected onto the levels $j,k$ the first time one goes through the algorithm):

$$\begin{eqnarray*}&&P(\vec{s}| j\vee k)=\displaystyle \prod _{l=1}^{n}\left({\alpha }_{{s}_{l}}+\sigma ({s}_{l})(l-1)\displaystyle \frac{w}{n}\right).\end{eqnarray*}$$

The expectation value of w_tot can be computed as follows (for $n\lt \infty$):

$$\begin{eqnarray*}E({w}_{\mathrm{tot}}) & = & \displaystyle \sum _{\vec{s}}P(\vec{s}| j\vee k){w}_{\mathrm{tot}}(\vec{s})=\displaystyle \sum _{\vec{s}}\displaystyle \prod _{l=1}^{n}\left({\alpha }_{{s}_{l}}+\sigma ({s}_{l})l\displaystyle \frac{w}{n}\right)\\ & = & \displaystyle \prod _{l=1}^{n}\left(\displaystyle \sum _{\vec{s}}{\alpha }_{{s}_{l}}+\sigma ({s}_{l})l\displaystyle \frac{w}{n}\right)\\ & = & \displaystyle \prod _{l=1}^{n}\left(\underset{=1}{\underbrace{{\alpha }_{1}+{\alpha }_{2}}}+\underset{=0}{\underbrace{(\sigma (l)+\sigma (2))}}l\displaystyle \frac{w}{n}\right)=1.\end{eqnarray*}$$

We now look at how much the work $W=\mathrm{ln}({w}_{\mathrm{tot}})$ changes, if in step l one replaces s_l by ${\hat{s}}_{l}$ (remember that ${s}_{j}=0\;\iff \;{\hat{s}}_{j}=1$ and vice versa):

$$\begin{eqnarray*}&&W({s}_{1},\ldots ,{s}_{n})-W({s}_{1},\ldots ,{\hat{s}}_{l},\ldots ,{s}_{n})\\ &&\quad =\mathrm{ln}\left(\displaystyle \frac{{\alpha }_{{s}_{l}}+\sigma ({s}_{l})l\displaystyle \frac{w}{n}}{{\alpha }_{{s}_{l}}+\sigma ({s}_{l})(l-1)\displaystyle \frac{w}{n}}\right)-\mathrm{ln}\left(\displaystyle \frac{{\alpha }_{{\hat{s}}_{l}}+\sigma ({\hat{s}}_{l})l\displaystyle \frac{w}{n}}{{\alpha }_{{\hat{s}}_{l}}+\sigma ({\hat{s}}_{l})(l-1)\displaystyle \frac{w}{n}}\right)\end{eqnarray*}$$

with $c=w\sigma ({s}_{l})$ (and therefore $w\sigma ({\hat{s}}_{l})=-c$), $a={\alpha }_{{s}_{l}}$ (and ${\alpha }_{{\hat{s}}_{l}}=1-a$), $x=a+c\frac{l}{n}$ and $y=1-a-c\frac{l}{n}$ we get:

$$\begin{eqnarray*}\left|W({s}_{1},\ldots ,{s}_{n})-W({s}_{1},\ldots ,{\hat{s}}_{l},\ldots ,{s}_{n})\right| & = & \left|\mathrm{ln}\left(\displaystyle \frac{x\left(y+\displaystyle \frac{c}{n}\right)}{\left(x-\displaystyle \frac{c}{n}\right)y}\right)\right|\\ & = & \left|\mathrm{ln}\left(1+\underset{z}{\underbrace{\displaystyle \frac{1}{n}\displaystyle \frac{{cy}+{cx}}{{xy}\left(1-\displaystyle \frac{c}{{xn}}\right)}}}\right)\right|\leqslant \displaystyle \frac{1}{n}| z| \;=:\;{q}_{l}.\end{eqnarray*}$$

Using the McDiarmid inequality (McDiarmid 1989) we get that the probability that W differs from its expectation value is bounded by:

$$\begin{eqnarray*}&&P\left(| W(\vec{s})-E(W)| \geqslant \delta \right)\leqslant 2\mathrm{exp}\left(\displaystyle \frac{-2{\delta }^{2}}{\displaystyle \sum _{l}{q}_{l}^{2}}\right)\leqslant 2\mathrm{exp}\left(\displaystyle \frac{-2{\delta }^{2}}{\displaystyle \frac{1}{n}{| z| }^{2}}\right)\end{eqnarray*}$$

which tends to 0 for any $\delta \gt 0$. Therefore we get that the work in this process is given by 0 with probability 1, which proves 4. □

Corollary. Using the above lemma one can split up any level k into two parts by using an empty level e:

• (i)  
  Permuting the levels such, that the empty level e comes before the level k.
• (ii)  
  Doing a work extraction by $\infty$ on the level e (such that its energy is $\infty$, while its width is 0, this costs no work, since the level is empty).
• (iii)  
  Do an isothermal shift of the level e in direction k by $w\in (0,1)$.

Then by the previous lemma the final overall distribution is the same as the initial, apart from the two levels e and k, which have now occupation probabilities:

$$\begin{eqnarray*}&&{\lambda }_{f}(e)=w\lambda (k),\end{eqnarray*}$$

$$\begin{eqnarray*}&&{\lambda }_{f}(k)=(1-w)\lambda (k)\end{eqnarray*}$$

and have energies E with $\mathrm{exp}(-{E}_{k}/{kT})=A$:

$$\begin{eqnarray*}&&{A}_{f}(e)={wA}(k),\end{eqnarray*}$$

$$\begin{eqnarray*}&&{A}_{f}(k)=(1-w)A(k).\end{eqnarray*}$$

The corollary directly follows from the lemma. Next we need an algorithm which makes it possible to get the end state σ out of the initial state ρ, if $p\;\succ \;{p}_{f}$ (the generalization of the step $4\to 5$ in the example).

The idea for the algorithm is that we first take the biggest eigenvalues of ρ, such that their area (i.e. the sum of their occupation probabilities) is equal to the biggest occupation probability (${\lambda }_{f}(1)$ of σ). Then we mix them and make a work extraction, such that their total width (i.e. the sum of $\mathrm{exp}(-E(j)/{kT})$) is the same as that of the final energy level 1. then we continue with the second and so forth.

To write down the algorithm, we first need two definitions simplifying the notation:

Definition 10 (Generalized sum). If $c\in {\mathbb{R}}$, $c\geqslant 1$, we define $\displaystyle \sum _{i=1}^{c}{d}_{i}:= \displaystyle \sum _{i=1}^{\lfloor c\rfloor }{d}_{i}+(c-\lfloor c\rfloor ){d}_{\lceil c\rceil }$. If $c\in {\mathbb{R}}$, $0\leqslant c\lt 1$, we define $\displaystyle \sum _{i=1}^{c}{d}_{i}:= c\cdot {d}_{1}$.

(Note that the above definition reduces to the usual sum if $c\in {\mathbb{N}}$).

Definition 11 (Gibbs-equivalent and Gibbs-expanding (see figure C3)). We say two tuples of $(\rho ,{H}_{i})$, $(\sigma ,{H}_{f})$ are Gibbs-equivalent (for a given temperature) if they give rise to the same Gibbs-rescaled distribution (where both are defined, 0 else). A transform is similarly said to be Gibbs-equivalent if it changes a tuple to a Gibbs-equivalent one. Finally a transform is said to be Gibbs-expanding if it changes a tuple $(\rho ,{H}_{i})$ to another one $(\sigma ,{H}_{f})$ with ${G}^{T}(\rho )\;\succ \;{G}^{T}(\sigma )$.

Lemma 8 (Optimal Gibbs-expanding transforms). Let ρ, σ be two states, diagonal in their energy-basis of dimension n.Let ρ and σ have at least $n/2$ empty levels. Let ${G}^{T}(\rho )\;\succ \;{G}^{T}(\sigma )$.

• Then one can transform ρ into σ with 0 work with probability 1.
• In other words: optimal Gibbs-expanding transforms exist and yield at least 0 work.

Proof. W.l.o.g. let the levels of ρ and σ be ordered in descending order.

• Let ${\lambda }_{i(f)}(j)$ denote the jth level of the initial (final) state.
• Define ${a}_{1}\in {\mathbb{R}}$ as the number of needed levels of ρ s.t. the total area is equal to the area at the end:

  $$\begin{eqnarray*}&&{\displaystyle \sum }_{j=1}^{{a}_{1}}{\lambda }_{i}(j)={\lambda }_{f}(1)\end{eqnarray*}$$

  (if ${a}_{1}\notin {\mathbb{N}}$ one needs to split the level $\lceil {a}_{1}\rceil$ as in the above corollary).

Define c as the width of the final first level:

$$\begin{eqnarray*}&&{\displaystyle \sum }_{j=1}^{c}{A}_{i}(j)={A}_{f}(1),\end{eqnarray*}$$

where ${A}_{i(f)}(j)=\mathrm{exp}(-{E}_{i(f)}(j)/{kT})$.

Now we get because of ${G}^{T}(\rho )\;\succ \;{G}^{T}(\sigma )$:

$$\begin{eqnarray*}&&{\displaystyle \int }_{0}^{{A}_{f}(1)}{G}^{T}(\rho ){\rm{d}}{\text{}}x\geqslant {\displaystyle \int }_{0}^{{A}_{f}(1)}{G}^{T}(\sigma ){\rm{d}}{\text{}}x\end{eqnarray*}$$

which by ${A}_{f}(1)={\displaystyle \sum }_{j=1}^{c}{A}_{i}(j)$ can be stated as:

$$\begin{eqnarray*}&&\displaystyle \sum _{j=1}^{c}{A}_{i}(j)\cdot \left(\displaystyle \frac{{\lambda }_{i}(j)}{{A}_{i}(j)}\right)\geqslant {A}_{f}(1)\cdot \left(\displaystyle \frac{{\lambda }_{f}(1)}{{A}_{f}(1)}\right)={\lambda }_{f}(1)=\displaystyle \sum _{j=1}^{{a}_{1}}{\lambda }_{i}(j)\end{eqnarray*}$$

therefore: $c\geqslant {a}_{1}$ and finally:

$$\begin{eqnarray*}&&\displaystyle \frac{{\displaystyle \sum }_{j=1}^{{a}_{1}}{\lambda }_{i}(j)}{{\displaystyle \sum }_{j=1}^{{a}_{1}}{A}_{i}(j)}\geqslant \displaystyle \frac{{\displaystyle \sum }_{j=1}^{{a}_{1}}{\lambda }_{i}(j)}{{\displaystyle \sum }_{j=1}^{c}{A}_{i}(j)}=\displaystyle \frac{{\lambda }_{f}(1)}{{A}_{f}(1)}\end{eqnarray*}$$

which means that one can change the energy of the first a₁ such that it is equal to the energy of the level 1 at the end, with 0 risk at no cost, since either successful or not, the energy gained will be at least 0. The occupation probabilities λ will obviously not be changed by this (apart the total mixing of the first a₁ levels). Now we could go on and prove the same for the second level and so forth, but there is an easier way:

The only ingredient we needed for the above reasoning to work was ${G}^{T}(\rho )\;\succ \;{G}^{T}(\sigma )$. But this is equivalent to ${G}^{T}(\rho )-K\;\succ \;{G}^{T}(\sigma )-K$ for any constant K, especially for $K={\lambda }_{f}(1)$. Explicitly:

$$\begin{eqnarray*}&&{\displaystyle \int }_{0}^{l}{G}^{T}(\rho ){\rm{d}}{\text{}}x-{\lambda }_{f}(1)\geqslant {\displaystyle \int }_{0}^{l}{G}^{T}(\sigma ){\rm{d}}{\text{}}x-{\lambda }_{f}(1)\ \forall {\text{}}l.\end{eqnarray*}$$

Remembering ${\lambda }_{f}(1)={\displaystyle \sum }_{j=1}^{{a}_{1}}{\lambda }_{i}(j)={\displaystyle \sum }_{j=1}^{{a}_{1}}{A}_{i}(j)\cdot \left(\displaystyle \frac{{\lambda }_{i}(j)}{{A}_{i}(j)}\right)$ the above can be rewritten as:

$$\begin{eqnarray*}&&{\displaystyle \int }_{{\displaystyle \sum }_{j=1}^{{a}_{1}}{A}_{i}(j)}^{l}{G}^{T}(\rho ){\rm{d}}{\text{}}x\geqslant {\displaystyle \int }_{{A}_{f}(1)}^{l}{G}^{T}(\sigma ){\rm{d}}{\text{}}x\ \forall {\text{}}l\end{eqnarray*}$$

i.e. we get the same requirement for the remaining levels. Which means, that we can inductively apply our argument. Since the number of non-empty levels of σ is at most $n/2$ it follows that we need at most $n/2$ empty levels to be able to split all the levels at the right place. □

With this lemma we can now classify the operations which cost 0 work (with risk 0) and their reverse also costs 0 work: these are exactly those which do not change the Gibbs-rescaled probability distribution and are optimal:

From the above lemma it follows that any optimal Gibbs-equivalent transform costs no work. Secondly, if the initial and the final state are Gibbs-equivalent such a transform exists (again by the above lemma), so it is reversible. On the other hand if a transform is not Gibbs-equivalent either it or its reverse cost more than 0 work (by the first part of theorem 1).

As an aside: this, together with the triangle inequality, proves that the symmetrized version of the mixing distance $D(a,b)={\mathsf{M}}(a\parallel b)+{\mathsf{M}}(b\parallel a)\geqslant 0$ is a metric on the set of probability distributions on the positive reals ordered in descending order.

Lemma 9. In the model for thermalization used here equation D.1 is always respected.

Proof. We firstly recall the model and define certain notation.

Recall that the thermalization model states that when two levels, 1 and 2, are coupled to the heat bath, their ratio ${\lambda }_{1}/{\lambda }_{2}$ gets closer to $\mathrm{exp}(-\beta ({E}_{1}-{E}_{2}))$, and the other λ's are untouched. In our model one may concatenate several such interactions to implement any allowed multi-level interaction with the bath. It will therefore suffice to show that equation (D.1) holds for a single two-level interaction with the heat bath.

For notational convenience let the probability of being in level 1 or 2 be called ${\lambda }_{12}:= {\lambda }_{1}+{\lambda }_{2}$. This is then constant for the given two-level interaction with the bath. In the extreme case of the two levels interacting with the bath for an arbitrary amount of time we have ${\lambda }_{1}:= {\lambda }_{1}^{T}$ and ${\lambda }_{2}:= {\lambda }_{2}^{T}$ (T reminds us of the temperature dependence). These values must then obey the relation

$$\begin{eqnarray}&&{\lambda }_{1}^{T}/{\lambda }_{2}^{T}=\mathrm{exp}(-\beta ({E}_{1}-{E}_{2})).\end{eqnarray} \tag{ D.3 }$$

We also assume without loss of generality that ${E}_{2}\leqslant {E}_{1}$. This implies that ${\lambda }_{1}^{T}\leqslant 0.5{\lambda }_{12}$.

Now we begin to prove the statement. Firstly we simplify $\Delta S$ by noting that only two levels change their probabilities. We write

$$\begin{eqnarray*}S & = & -\displaystyle \sum _{i}{\lambda }_{i}\mathrm{log}{\lambda }_{i}\\ & = & -{\lambda }_{1}\mathrm{log}{\lambda }_{1}-({\lambda }_{12}-{\lambda }_{1})\mathrm{log}({\lambda }_{12}-{\lambda }_{1})-\displaystyle \sum _{i=3}^{{i}_{\mathrm{max}}}{\lambda }_{i}\mathrm{log}{\lambda }_{i}\\ & \equiv & {S}_{12}-\displaystyle \sum _{i=3}^{{i}_{\mathrm{max}}}{\lambda }_{i}\mathrm{log}{\lambda }_{i}.\end{eqnarray*}$$

We see that in any two-level interaction

$$\begin{eqnarray}&&\Delta S=\Delta {S}_{12}.\end{eqnarray} \tag{ D.4 }$$

It is helpful to re-express ${S}_{12}$ in terms of an actual entropy $\bar{{S}_{12}}$, so that we can use known properties of entropies to make statements about ${S}_{12}$. We let $\bar{{\lambda }_{1}}:= {\lambda }_{1}/{\lambda }_{12}$ and $\bar{{\lambda }_{2}}:= {\lambda }_{2}/{\lambda }_{12}$ such that $\bar{{\lambda }_{1}}+\bar{{\lambda }_{2}}=1$. We define

$$\begin{eqnarray*}&&\bar{{S}_{12}}:= -\bar{{\lambda }_{1}}\mathrm{log}\bar{{\lambda }_{1}}-\bar{{\lambda }_{2}}\mathrm{log}\bar{{\lambda }_{2}}.\end{eqnarray*}$$

One can then see in a few lines of algebra that

$$\begin{eqnarray*}&&{S}_{12}={\lambda }_{12}\bar{{S}_{12}}-{\lambda }_{12}\mathrm{log}{\lambda }_{12}.\end{eqnarray*}$$

It follows that

$$\begin{eqnarray}&&\Delta {S}_{12}={\lambda }_{12}\Delta \bar{{S}_{12}}.\end{eqnarray} \tag{ D.5 }$$

We accordingly now want to show that ${\lambda }_{12}\Delta \bar{{S}_{12}}\geqslant \beta \Delta \langle E\rangle .$

We can now use a well known property of the Shannon/von Neumann entropy: $\bar{{S}_{12}}$ is concave in $\bar{{\lambda }_{1}}={\lambda }_{1}/{\lambda }_{12}$. The function is accordingly upper bounded by any tangential line, as in figure D1 . Consider the tangential line at ${\lambda }_{1}={\lambda }_{1}^{T}$. At that point it follows from a few lines that

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}}{{\rm{d}}{\lambda }_{1}}{S}_{12}{| }_{{\lambda }_{1}={\lambda }_{1}^{T}}=\displaystyle \frac{{\rm{d}}}{{\rm{d}}\bar{{\lambda }_{1}}}\bar{{S}_{12}}=\beta ({E}_{1}-{E}_{2}).\end{eqnarray} \tag{ D.6 }$$

Note now that $\langle E\rangle$ may similarly to the entropy be written as

$$\begin{eqnarray*}&&\langle E\rangle =-\displaystyle \sum _{i}{\lambda }_{i}{E}_{i}\equiv {\langle E\rangle }_{12}+{\langle E\rangle }_{\mathrm{rest}},\end{eqnarray*}$$

such that $\Delta \langle E\rangle =\Delta \langle E{\rangle }_{12}=(\Delta {\lambda }_{1})({E}_{1}-{E}_{2})$, with $\Delta {\lambda }_{1}={\lambda }_{1}^{\prime }-{\lambda }_{1}$ the change in ${\lambda }_{1}$. So $\langle E\rangle ({\lambda }_{1})$ is a line with gradient given by

$$\begin{eqnarray*}&&\displaystyle \frac{\Delta \langle E\rangle }{\Delta {\lambda }_{1}}={E}_{1}-{E}_{2}.\end{eqnarray*}$$

Similarly

$$\begin{eqnarray*}&&\displaystyle \frac{\Delta \langle E\rangle }{\Delta \bar{{\lambda }_{1}}}=\displaystyle \frac{1}{{\lambda }_{12}}({E}_{1}-{E}_{2}).\end{eqnarray*}$$

Comparing this with the gradient of the tangential line to $\bar{{S}_{12}}$ in equation (D.6), we see that $\displaystyle \frac{1}{{\lambda }_{12}}\beta \langle E{\rangle }_{12}$ has the same gradient as the tangential line. We therefore only need to show that the change in the tangential line is upper bounded by the change in the entropy curve, as it is equivalent to showing that $\Delta \bar{{S}_{12}}\geqslant \frac{1}{{\lambda }_{12}}\beta \langle E{\rangle }_{12}$. This must hold for all possible initial and final values of $\bar{{\lambda }_{1}}$ and all possible values of ${\bar{{\lambda }_{1}}}^{T}$(recall that we assumed without loss of generality that ${\bar{{\lambda }_{1}}}^{T}\geqslant 0.5$ ). These can be grouped into three cases.

• (i)  
  $\bar{{\lambda }_{1}}\leqslant {\bar{{\lambda }_{1}}}^{T}$. Here the tangential bound above implies that $\Delta \bar{{S}_{12}}\geqslant \frac{\beta }{{\lambda }_{12}}\langle E{\rangle }_{12}\geqslant 0$.
• (ii)  
  ${{\lambda }_{1}}^{T}\leqslant \bar{{\lambda }_{1}}\leqslant 0.5$. Here the tangential bound implies that $0\geqslant \Delta \bar{{S}_{12}}\geqslant \frac{\beta }{{\lambda }_{12}}\langle E{\rangle }_{12}$.
• (iii)  
  $\bar{{\lambda }_{1}}\geqslant 0.5$, also after the interaction. Here the tangential bound implies that $\bar{\Delta {S}_{12}}\geqslant 0\geqslant \frac{\beta }{{\lambda }_{12}}\langle E{\rangle }_{12}$.

This implies the lemma. □

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Recall that our condition on thermalizing evolutions was stronger than equation (D.1). There are, as mentioned in the main body, examples of evolutions that respect equation (D.1) but violate our condition: equation (D.2). In this section we consider whether these evolutions may violate Kelvin's second law: no process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

We use standard results concerning majorization, as well as our main theorem. We will consider degenerate energy levels for simplicity so that equation (D.1) reduces to $\Delta S\geqslant 0$. We now only assume that the evolution is represented by a stochastic matrix (which it is if the map is Markovian). We do not assume it is the type of thermalization used hitherto as that would automatically respect equation (D.2).

Lemma 10. Any stochastic matrix A which for some state violates equation (D.2) but respects the entropy condition $\Delta S\geqslant 0$ will for some input state, namely the uniform distribution, violate $\Delta S\geqslant 0$.

Proof. 

• (i)  
  Equation (D.2) is respected iff the matrix is bistochastic. Thus A is NOT bistochastic.
• (ii)  
  The uniform distribution is invariant under a stochastic matrix iff it is bistochastic.

Thus A does NOT preserve the uniform distribution. Now the uniform distribution is unique in having maximal von Neumann entropy. Thus $\Delta S\geqslant 0$ is violated if the input state is the uniform distribution.

Lemma 11. Consider a state changing to another one. Suppose: (i) the von Neumann entropy is increased, (ii) equation (D.2) is violated , and (iii) the evolution is a stochastic matrix. Then this evolution–applied to the thermal state–would allow for the violation of Kelvin's second law within our game: deterministic work extraction would be possible from a cycle where the system is in the thermal state both initially and finally.

Proof. Recall that we are for simplicity considering degenerate energy levels in this section. The thermal state is then the uniform distribution. Apply A to this (at no work cost as it represents an interaction with the heat bath). Now we have a state σ other than the uniform distribution, so it must majorize the uniform distribution.

To see that this implies deterministic work extraction we firstly show that ${W}^{0}\gt 0$ for some process using A and allowed operations within the game. Consider taking n copies of σ and going to the von Neumann limit by taking n to infinity as well as taking the risk of failure to 0. To evaluate ${W}^{\varepsilon }$ in this limit it is convenient to use theorem 12 which re-expresses ${W}^{\varepsilon }$. Recall that in the von Neumann limit the smooth max entropy reduces to the von Neumann entropy S. We therefore have, for the case of degenerate levels:

$$\begin{eqnarray*}&&\underset{n\to \infty ,\varepsilon \to 0}{\mathrm{lim}}\displaystyle \frac{{W}^{\varepsilon }({\sigma }^{\otimes n}\to {\tau }^{\otimes n})}{n}=({H}_{\mathrm{max}}(\tau )-S(\sigma )){kT}\mathrm{ln}2,\end{eqnarray*}$$

where we have also used the well-known additivity of both entropies: ${H}_{\mathrm{max}}({\rho }^{\otimes n})={{nH}}_{\mathrm{max}}(\rho )$ and $S({\rho }^{\otimes n})={nS}(\rho )$ In this case $\tau ={\mathbb{1}}/d$, i.e. the maximally mixed state associated with a d-dimensional Hilbert space. Moreover ${H}_{\mathrm{max}}({\mathbb{1}}/d)-S(\sigma )\gt 0$ since the uniform distribution is unique in having maximal von Neumann entropy and ${H}_{\mathrm{max}}\geqslant S$. Thus ${W}^{0}\gt 0$ for that process.

Recall secondly the subtlety that we proved that ${W}^{\varepsilon }(\sigma \to {\sigma }^{\prime })$ is achievable within the game when there is access to a catalyst system. Consider extracting work from n copies of $\sigma \otimes | \xi \rangle \langle \xi |$ which will be set to n copies of ${\mathbb{1}}/d\otimes | \xi \rangle \langle \xi |$ at the end. Now ${H}_{\mathrm{max}}({\mathbb{1}}/d\otimes | \xi \rangle \langle \xi | )-S(\sigma \otimes | \xi \rangle \langle \xi | )\gt 0$ as neither entropy of a state is changed by adding a pure system in this way. Thus including the catalyst system does not change the statement that ${W}^{0}\gt 0$ for the above procedure in the von Neumann limit. Accordingly this process violates Kelvin's law. □