Clock-driven quantum thermal engines

We now constructively show that it is possible within our framework to exactly implement any energy-conserving unitary U_e on the engine via interactions with the clock. This succeeds for any initial state satisfying equation (2), and the clock and engine are always unentangled at the end.

We start by choosing an interaction Hamiltonian of the form

$$\begin{eqnarray}&&{{H}^{\operatorname{int}}}={{\int }_{\mathbb{R}}}H_{e}^{\operatorname{int}}(x)\;\otimes \;|x{{\rangle }_{c}}\langle x|{\rm d}x,\end{eqnarray} \tag{ 3 }$$

where $|x{{\rangle }_{c}}$ are the clock's position eigenstates. Since the clock's position increases linearly with time, this means the clock is driving the engine by applying on it an effective time-dependent Hamiltonian.

We then choose $H_{e}^{\operatorname{int}}(x)$ such that

$$\begin{eqnarray}&&[{{H}^{\operatorname{int}}},{{H}_{e}}]=0,\end{eqnarray} \tag{ 4 }$$

thus guaranteeing that the interaction will never transfer energy between the clock and the engine. This also prevents the clock and engine from becoming entangled.

Given this commutation relation, ${{H}^{\operatorname{int}}}$ described in the interaction picture takes the form

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} {{\tilde{H}}^{\operatorname{int}}}(t) & = & {{{\rm e}}^{\frac{{\rm i}}{\hbar }({{H}_{e}}+{{H}_{c}})t}}{{H}^{\operatorname{int}}}{{{\rm e}}^{-\frac{{\rm i}}{\hbar }({{H}_{e}}+{{H}_{c}})t}} \\ {} & = & {{\int }_{\mathbb{R}}}H_{e}^{\operatorname{int}}(x)\;\otimes \;|x-t{{\rangle }_{c}}\langle x-t|{\rm d}x. \\ \end{array}\end{eqnarray} \tag{ 5 }$$

Thus, shifting $x\to x+t$, one can see that the total evolution operator between times 0 and t is given by

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} \mathcal{U}(t) & = & {{{\rm e}}^{-\frac{{\rm i}}{\hbar }({{H}_{e}}+{{H}_{c}})t}}\mathcal{T}\left[ {{{\rm e}}^{-\frac{{\rm i}}{\hbar }\int _{0}^{t}{{\tilde{H}}^{\operatorname{int}}}(t^{\prime} ){\rm d}t^{\prime} }} \right] \\ {} & = & {{{\rm e}}^{-\frac{{\rm i}}{\hbar }({{H}_{e}}+{{H}_{c}})t}}{{\int }_{\mathbb{R}}}\mathcal{T}\left[ {{{\rm e}}^{-\frac{{\rm i}}{\hbar }\int _{0}^{t}H_{e}^{\operatorname{int}}(x+t^{\prime} ){\rm d}t^{\prime} }} \right]\;\otimes \;|x{{\rangle }_{c}}\langle x|{\rm d}x \\ {} & = & {{{\rm e}}^{-\frac{{\rm i}}{\hbar }{{H}_{e}}t}}{{\int }_{\mathbb{R}}}\mathcal{T}\left[ {{{\rm e}}^{-\frac{{\rm i}}{\hbar }\int _{0}^{t}H_{e}^{\operatorname{int}}(x+t^{\prime} ){\rm d}t^{\prime} }} \right]\;\otimes \;|x+t{{\rangle }_{c}}\langle x|{\rm d}x, \\ \end{array}\end{eqnarray} \tag{ 6 }$$

where $\mathcal{T}[\cdot ]$ is the time ordering operation.

Note that the clock moves forward at constant speed, despite the interaction. We now choose $H_{e}^{\operatorname{int}}$ to have support inside an interval of size L immediately to the right of the clock state's support (the interaction region), define $\tau =K+L$ and choose $t\gt \tau$ so the clock has time to completely 'cross-over' the support of $H_{e}^{\operatorname{int}}$ (see figure 1). This causes the time integral in equation (6) to go over the entire support of $H_{e}^{\operatorname{int}}$, becoming independent of x.

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In other words, denoting as ${\rm supp}[f(x)]$ the support of f on x, we have ${\rm supp}[H_{e}^{\operatorname{int}}(x)]\subset [x^{\prime} ,x^{\prime} +\tau ]$ for all $x^{\prime} \in {\rm supp}[\langle x|{{\rho }_{c}}(0)|x\rangle ]$. And so, for all $t\gt \tau$, the time evolution acts separately on the clock and on the engine

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} \rho (t) & = & {{\mathcal{U}}_{e}}(t)\;{{\rho }_{e}}(0)\ \mathcal{U}_{e}^{\dagger }(t)\;\otimes \;{{{\rm e}}^{-\frac{{\rm i}}{\hbar }{{H}_{c}}t}}{{\rho }_{c}}(0){{{\rm e}}^{\frac{{\rm i}}{\hbar }{{H}_{c}}t}} \\ {{\mathcal{U}}_{e}}(t) & = & {{{\rm e}}^{-\frac{{\rm i}}{\hbar }{{H}_{e}}t}}{{U}_{e}} \\ {{U}_{e}} & = & \mathcal{T}\left[ {{{\rm e}}^{-\frac{{\rm i}}{\hbar }{{\int }_{{\rm supp}(H_{e}^{\operatorname{int}})}}H_{e}^{\operatorname{int}}(t^{\prime} ){\rm d}t^{\prime} }} \right], \\ \end{array}\end{eqnarray} \tag{ 7 }$$

which proves that the engine is acted on by U_e and then undergoes free evolution indefinitely.

Finally, for any unitary V, there is always a Hermitian operator acting on the space, G_V, such that $V={{{\rm e}}^{-{\rm i}{{G}_{V}}}}$. Thus, one needs simply to choose $H_{e}^{\operatorname{int}}$ such that ${{\int }_{{\rm supp}(H_{e}^{\operatorname{int}})}}H_{e}^{\operatorname{int}}(t^{\prime} ){\rm d}t^{\prime} ={{G}_{{{U}_{e}}}}$ in order to implement any desired energy-conserving U_e. One possibility is a fixed Hamiltonian which switches on and off, i.e. $H_{e}^{\operatorname{int}}(t)={\rm i}f(t)\;{\rm ln} {{U}_{e}}$, where f is a normalized function with support inside the interaction region. However, note that our approach applies to any form of time dependence, which incorporates a broader range of experimental procedures.

Of course, the clock and engine can become entangled during the procedure, but they will always be in a product state at the end ($t\gt \tau$). In fact, the state of the clock doesn't change other than being translated. This means the operation never degrades the clock.

Here, we constructively prove that for any unitary V_s on s there is a choice of ${{H}^{\operatorname{int}}}$ which approximately implements it. First, we choose U_e to have the form

$$\begin{eqnarray}&&{{U}_{e}}=\mathop{\sum }\limits_{n,j}\langle E_{n}^{s}|{{V}_{s}}|E_{j}^{s}\rangle |E_{n}^{s}\rangle \langle E_{j}^{s}|\;\otimes \;{{{\rm e}}^{-\frac{{\rm i}}{\hbar }({{P}_{w}}-{{p}_{0}})(E_{j}^{s}-E_{n}^{s})}},\end{eqnarray} \tag{ 8 }$$

where $E_{j}^{s}$ are the energy levels of ${{H}_{s}}$, and $|E_{j}^{s}\rangle$ are their respective eigenstates. Note that this unitary satisfies energy-conservation. Furthermore, it is translation-invariant on w, i.e.

$$\begin{eqnarray}&&[{{U}_{e}},{{P}_{w}}]=0,\end{eqnarray} \tag{ 9 }$$

where P_w is the weight's momentum operator. This serves two purposes: (i) it guarantees the protocol works regardless of the initial height of the weight (or how much energy is stored in it), (ii) it enables the catalytic use of the coherences in the weight to transform the system [3]. By the proof above, in order to implement this U_e, we may choose an ${{H}^{\operatorname{int}}}$ which also satisfies $[{{H}^{\operatorname{int}}},{{P}_{w}}]=0$. In particular, one possible example is the aforementioned $H_{e}^{\operatorname{int}}(t)={\rm i}f(t){\rm ln} {{U}_{e}}$.

Alternatively, equation (8) can be written as

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} {{U}_{e}} & = & {{\int }_{\mathbb{R}}}{{U}_{s}}(p)\;\otimes \;|p{{\rangle }_{w}}\langle p|{\rm d}p \\ {{U}_{s}}(p) & = & \mathop{\sum }\limits_{n,j}{{{\rm e}}^{-\frac{{\rm i}}{\hbar }(p-{{p}_{0}})(E_{j}^{s}-E_{n}^{s})}}\langle E_{n}^{s}|{{V}_{s}}|E_{j}^{s}\rangle |E_{n}^{s}\rangle \left\langle E_{j}^{s} \right|, \\ \end{array}\end{eqnarray} \tag{ 10 }$$

where $|p{{\rangle }_{w}}$ are the momentum eigenstates of the weight. In this form, one sees that, upon applying U_e on the engine state, the reduced state of the system becomes

$$\begin{eqnarray}&&{\rm T}{{{\rm r}}_{w}}\left[ {{U}_{e}}{{\rho }_{e}}(0)U_{e}^{\dagger } \right]={{\int }_{\mathbb{R}}}{{\mu }_{w}}(p)\;{{U}_{s}}(p){{\rho }_{s}}(0)U_{s}^{\dagger }(p){\rm d}p,\end{eqnarray} \tag{ 11 }$$

where ${{\mu }_{w}}(p)=\langle p|{{\rho }_{w}}(0)|p\rangle$ is the initial momentum distribution of the weight.

As shown in the appendix, this can be made arbitrarily close to ${{V}_{s}}{{\rho }_{s}}(0)V_{s}^{\dagger }$ in trace distance, by making ${{\rho }_{w}}(0)$ narrow enough in momentum space. Intuitively speaking, the closer ${{\mu }_{w}}(p)$ is to a delta function, the closer this operation is to the exact V_s (which is ${{U}_{s}}({{p}_{0}})$). Therefore, for any $\epsilon \gt 0$, there is always a good enough ${{\rho }_{w}}(0)$ such that

$$\begin{eqnarray}&&\left \|{\rm T}{{{\rm r}}_{w}}\left[ {{U}_{e}}{{\rho }_{e}}(0)U_{e}^{\dagger } \right]-{{V}_{s}}{{\rho }_{s}}(0)V_{s}^{\dagger }\right\|{_{1}}\leqslant \epsilon .\end{eqnarray} \tag{ 12 }$$

Moreover, since $[{{U}_{e}},{{P}_{w}}]=0$, ${{\mu }_{w}}(p)$ is conserved by the operations. Also, note that the full evolution includes both the operations and the free evolution, but the weight's free evolution only shifts p₀ without affecting the shape of ${{\mu }_{w}}(p)$.

As the error in implementing V_s depends only on how narrow ${{\mu }_{w}}(p)$ is, this implies that the usefulness of the weight is not degraded.

An interesting special case arises if V_s only permutes energy levels, ${{V}_{s}}|E_{j}^{s}\rangle =|E_{\pi (j)}^{s}\rangle$ for some permutation π, and if the initial state is diagonal in energy, ${{\rho }_{s}}(0)={{\sum }_{n}}p_{n}^{s}|E_{n}^{s}\rangle \langle E_{n}^{s}|$. Joining these two, equation (11) simplifies to

$$\begin{eqnarray}&&\begin{array}{ccccccccccccccc} {\rm T}{{{\rm r}}_{w}}\left[ {{U}_{e}}{{\rho }_{e}}(0)U_{e}^{\dagger } \right] \\ \quad =\mathop{\sum }\limits_{n,m}|E_{\pi (m)}^{s}\rangle \left\langle E_{m}^{s} \right|{{\rho }_{s}}(0)|E_{n}^{s}\rangle \langle E_{\pi (n)}^{s}| \\ \quad \quad {{\int }_{\mathbb{R}}}{{{\rm e}}^{-\frac{{\rm i}}{\hbar }(p-{{p}_{0}})(E_{n}^{s}-E_{\pi (n)}^{s})}}{{\mu }_{w}}(p){{{\rm e}}^{\frac{{\rm i}}{\hbar }(p-{{p}_{0}})(E_{m}^{s}-E_{\pi (m)}^{s})}}{\rm d}p. \\ \quad =\mathop{\sum }\limits_{n}p_{n}^{s}|E_{\pi (n)}^{s}\rangle \langle E_{\pi (n)}^{s}| \\ \quad ={{V}_{s}}{{\rho }_{s}}(0)V_{s}^{\dagger }, \\ \end{array}\end{eqnarray} \tag{ 13 }$$

which means the transformation can be performed exactly regardless of the state of the weight.

Consider a quantum machine whose Hilbert space and Hamiltonian are as described above, and take as assumptions the following identities:

$$\begin{eqnarray}&&[{{H}^{\operatorname{int}}},{{H}_{e}}]=0,\end{eqnarray} \tag{ 15a }$$

$$\begin{eqnarray}&&[{{H}^{\operatorname{int}}},{{X}_{c}}]=[{{H}^{\operatorname{int}}},{{P}_{w}}]=0,\end{eqnarray} \tag{ 15b }$$

and $\rho (0)={{\rho }_{u}}(0)\;\otimes \;{{\rho }_{b}}(0)\;\otimes \;{{\rho }_{w}}(0)\;\otimes \;{{\rho }_{c}}(0)$. Note that these are all satisfied by the constructions thus far.

The first law of thermodynamics is trivially satisfied by equation (15a ), which implies that $\langle {{H}_{e}}\rangle$ is a conserved quantity⁴ , and so

$$\begin{eqnarray}&&\Delta \langle {{H}_{u}}\rangle +\Delta \langle {{H}_{w}}\rangle +\Delta \langle {{H}_{b}}\rangle =\Delta \langle {{H}_{e}}\rangle =0\Rightarrow \quad \quad \Delta U+W+Q=0.\end{eqnarray} \tag{ 16 }$$

To prove the second law, we construct a Kelvin–Planck statement [19, 20], showing that it is impossible to extract positive work in a cyclic process. From equation (15b ), we know ${{H}^{\operatorname{int}}}$ must satisfy

$$\begin{eqnarray}&&{{H}^{\operatorname{int}}}=\int {{\int }_{{{\mathbb{R}}^{2}}}}H_{ub}^{\operatorname{int}}(x,p)\;\otimes \;|p{{\rangle }_{w}}{{\langle p|\;\otimes \;|x\rangle }_{c}}\langle x|{\rm d}p{\rm d}x.\end{eqnarray} \tag{ 17 }$$

As detailed in the appendix, this means the reduced state on $u\;\otimes \;b$ is a mixture of unitaries $V(x,p)$ applied on the initial state,

$$\begin{eqnarray}&&{\rm T}{{{\rm r}}_{wc}}[\rho (t)]=\int {{\int }_{{{\mathbb{R}}^{2}}}}\mu (x,p)V(x,p,t){{\rho }_{ub}}(0){{V}^{\dagger }}(x,p,t){\rm d}p{\rm d}x,\end{eqnarray} \tag{ 18 }$$

where $\mu (x,p)=\langle x,p|{{\rho }_{cw}}(0)|x,p\rangle$. This means the entropy of $u\;\otimes \;b$ never decreases

$$\begin{eqnarray}&&0\leqslant \Delta S({{\rho }_{ub}})\quad {\rm or}\quad -\Delta S({{\rho }_{b}})\leqslant \Delta S({{\rho }_{u}}),\end{eqnarray} \tag{ 19 }$$

where Δ denotes a difference between times 0 and t, and we have used the subadditivity of the entropy and the assumption that the initial state is a product state.

Since the initial bath state has minimum free energy for a given temperature, one finds

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} 0\leqslant \Delta F({{\rho }_{b}}) & = & \Delta E({{\rho }_{b}})-T\Delta S({{\rho }_{b}}) \\ {} & \Longrightarrow & T\Delta S({{\rho }_{b}})\leqslant \Delta E({{\rho }_{b}})=-\Delta E({{\rho }_{w}})-\Delta E({{\rho }_{u}}) \\ {} & \Longrightarrow & W=\Delta E({{\rho }_{w}})\leqslant -T\Delta S({{\rho }_{b}})-\Delta E({{\rho }_{u}})\leqslant -\Delta F({{\rho }_{u}}). \\ \end{array}\end{eqnarray} \tag{ 20 }$$

This means one cannot extract more energy than the reduction in free energy of the subsystem. Clearly, if the thermodynamic properties of the subsystem are the same in its initial and final state, $\Delta F({{\rho }_{u}})=0$ and hence positive work cannot be extracted from the bath. That is, 'one cannot turn heat purely into work'.

Note that this result is a direct consequence of the assumptions in equation (15), and not of any implementation details. So the first and second laws hold given an arbitrary protocol which follows these assumptions, even one which is far from optimal, or which acts on a different initial state from the one it was designed for.

We now show that this framework allows thermodynamically optimal state transformation protocols. In particular, a protocol exists such that the work extracted is arbitrarily close to the reduction in free energy of the subsystem, and the final state of the subsystem is arbitrarily close to the desired final state. Here, a protocol represents a unitary operation on $u\;\otimes \;b\;\otimes \;w$ or, equivalently, an ${{H}^{\operatorname{int}}}$ on $u\;\otimes \;b\;\otimes \;w\;\otimes \;c$.

This section is based on the protocols in [2, 4], but our framework is slightly different and involves a simpler proof strategy. Given the above results, we need only find a unitary V_s on $s=u\;\otimes \;b$ with the desired effect. Then, the existence of an interaction Hamiltonian which implements this unitary relies only on the weight state being good enough.

Let us write the initial state of the subsystem as

$$\begin{eqnarray}&&{{\rho }_{u}}(0)=\mathop{\sum }\limits_{n}{{p}_{n}}|{{\psi }_{n}}\rangle \left\langle {{\psi }_{n}} \right|,\end{eqnarray} \tag{ 21 }$$

and the desired target state of the subsystem by

$$\begin{eqnarray}&&{{\sigma }_{u}}=\mathop{\sum }\limits_{m}{{q}_{m}}|{{\phi }_{m}}\rangle \left\langle {{\phi }_{m}} \right|,\end{eqnarray} \tag{ 22 }$$

where $\{|{{\psi }_{n}}\rangle \}$ and $\{|{{\phi }_{n}}\rangle \}$ are respective eigenbases, labeled so that ${{p}_{n}}\gt {{p}_{n+1}}$ and ${{q}_{n}}\gt {{q}_{n+1}}$. Thus, we desire that ${{\rho }_{u}}(t)={{{\rm e}}^{-\frac{{\rm i}}{\hbar }{{H}_{u}}t}}{{\sigma }_{u}}{{{\rm e}}^{\frac{{\rm i}}{\hbar }{{H}_{u}}t}}$, $\forall t\gt \tau$, so the interacting initial state becomes the freely evolving target state.

For simplicity we assume that ${{\sigma }_{u}}$ is full rank. If the final state has lower rank, we can instead transform the subsystem to a state $\sigma _{s}^{\prime }$ with full rank and close to ${{\sigma }_{u}}$ in trace distance.

The desired subsystem-bath unitary, ${{V}_{ub}}={{V}_{s}}$, is composed of three stages. The first stage acts only on u, rotating ${{\rho }_{u}}$ to be diagonal in its energy basis. That is

$$\begin{eqnarray}&&V_{ub}^{(1)}=\mathop{\sum }\limits_{n}|E_{n}^{u}\rangle \left\langle {{\psi }_{n}} \right|\;\otimes \;{\mathbb{1}_{b}},\end{eqnarray} \tag{ 23 }$$

where $|E_{n}^{u}\rangle$ are the eigenstates of ${{H}_{u}}$. After this step the state of the subsystem will be

$$\begin{eqnarray}&&{{\sum }_{n}}{{p}_{n}}|E_{n}^{u}\rangle \left\langle E_{n}^{u} \right|.\end{eqnarray} \tag{ 24 }$$

Note that entropy is conserved in this stage, so the total energy change is equal to the change in free energy of ${{\rho }_{u}}$.

The second stage, $V_{ub}^{(2)}$, takes this diagonal state into another diagonal state, and has been studied before [1–4]. We present here a proof of its energy efficiency which is simpler than previous ones. The unitary is divided into many small steps, acting on $u\;\otimes \;b$ and gradually changing the probabilities of the subsystem's energy levels from p_n to q_n, by performing the swap operation on the subsystem state and a similar state from the bath. The first step uses a thermal state in the bath of dimension d_u, whose probabilities $p_{n}^{\prime }$ lie between p_n and q_n and are very close to the former, i.e., $|{{p}_{n}}-p_{n}^{\prime }|\lt \delta p$ for all n.

Since this thermal state has minimum free energy, $\Delta {{F}_{b}}$ must be of order $\delta {{p}^{2}}$, so $\Delta {{E}_{b}}=T\Delta {{S}_{b}}+\mathcal{O}\left( \delta {{p}^{2}} \right)$. Furthermore, due to the swap operation, $\Delta {{S}_{b}}=-\Delta {{S}_{u}}$, and so the energy variation of this step is

$$\begin{eqnarray}&&\Delta E_{ub}^{1}=\Delta E_{b}^{1}+\Delta E_{u}^{1}=\Delta F_{u}^{1}+\mathcal{O}\left( \delta {{p}^{2}} \right).\end{eqnarray} \tag{ 25 }$$

Since $|{{p}_{n}}-{{q}_{n}}|\lt 1$ for all n, we can choose a unitary $V_{ub}^{(2)}$ such that only $\delta {{p}^{-1}}$ steps are necessary, and so the total energy variation must be

$$\begin{eqnarray}&&\Delta {{E}_{ub}}=\mathop{\sum }\limits_{j=1}^{\delta {{p}^{-1}}}\Delta F_{u}^{j}+\mathcal{O}\left( \delta {{p}^{2}} \right)=\Delta {{F}_{u}}+\mathcal{O}(\delta p).\end{eqnarray} \tag{ 26 }$$

After this stage the state of the subsystem will be exactly

$$\begin{eqnarray}&&\mathop{\sum }\limits_{n}{{q}_{n}}|E_{n}^{u}\rangle \left\langle E_{n}^{u} \right|.\end{eqnarray} \tag{ 27 }$$

Note that this step only consists of permutations on the ${{H}_{u}}+{{H}_{b}}$ energy basis, which means that if both ${{\rho }_{u}}(0)$ and ${{\sigma }_{u}}$ are diagonal then the transformation can be performed perfectly regardless of the state of the weight.

In the final stage, we again act only on the subsystem, rotating it into the eigenstates of ${{\sigma }_{u}}$ via the unitary

$$\begin{eqnarray}&&V_{ub}^{(3)}=\mathop{\sum }\limits_{n}|{{\phi }_{n}}\rangle \left\langle E_{n}^{u} \right|\;\otimes \;{\mathbb{1}_{b}},\end{eqnarray} \tag{ 28 }$$

which takes the final state of the subsystem into ${{\sigma }_{u}}$

$$\begin{eqnarray}&&{\rm T}{{{\rm r}}_{b}}\left[ {{V}_{ub}}{{\rho }_{ub}}(0)V_{ub}^{\dagger } \right]={{\sigma }_{u}},\end{eqnarray} \tag{ 29 }$$

with ${{V}_{ub}}=V_{ub}^{(3)}V_{ub}^{(2)}V_{ub}^{(1)}$. Note that the protocol also involves free evolution, see equation (7), so the subsystem's state becomes the freely evolving ${{\sigma }_{u}}$. That is, ${{\rho }_{u}}(t)={{{\rm e}}^{-\frac{{\rm i}}{\hbar }{{H}_{u}}t}}{{\sigma }_{u}}{{{\rm e}}^{\frac{{\rm i}}{\hbar }{{H}_{u}}t}}$ for all $t\gt \tau$. If one desires to have exactly ${{\sigma }_{u}}$ at a specific time t', one simply needs to add a fourth step to the unitary, ${{V}_{ub}}={{{\rm e}}^{{\rm i}{{H}_{u}}(t-t^{\prime} )}}V_{ub}^{(3)}V_{ub}^{(2)}V_{ub}^{(1)}$.

Looking at the energy balance, we find that the energy change of the subsystem and bath is as close as desired to the free energy change of the subsystem, making it a thermodynamically optimal transformation. That is

$$\begin{eqnarray}&&\Delta {{E}_{ub}}=\Delta {{F}_{u}}+\epsilon ,\end{eqnarray} \tag{ 30 }$$

where $\epsilon \gt 0$ can be made as small as desired.

By the results above, if the state of the weight is narrow enough in momentum, there exists an ${{H}^{\operatorname{int}}}$ which implements V_ub after a time τ, so that $\left \|{\rm T}{{{\rm r}}_{bwc}}\rho (t)-{{{\rm e}}^{-\frac{{\rm i}}{\hbar }{{H}_{u}}t}}{{\sigma }_{u}}{{{\rm e}}^{\frac{{\rm i}}{\hbar }{{H}_{u}}t}}\right\|_{1}$ is small for all $t\gt \tau$. This means equation (30) still holds up to a small error and, by energy conservation, this energy difference must have been transferred to the weight.

Thus for any upper bound $\epsilon ^{\prime} \gt 0$ we wish to impose on the error, there is always a good enough weight state (small trace distance) and a gradual enough protocol (small $\delta p$) such that, by our definition of work

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} W & = & {\rm Tr}\left[ {{H}_{w}}(\rho (\tau )-\rho (0)) \right] \\ {} & \geqslant & -\Delta {{F}_{u}}-\epsilon ^{\prime} . \\ \end{array}\end{eqnarray} \tag{ 31 }$$

To say that a statement is true for ${{\mu }_{w}}(p)$ narrow enough and centered around p₀ is equivalent to saying it is true for $\frac{1}{\delta }{{\mu }_{w}}(\frac{p-{{p}_{0}}}{\delta }+{{p}_{0}})$ with a small enough $\delta \gt 0$.

Theorem 1. For any finite dimensional Hilbert space $\mathcal{H}$, let $\mathcal{D}(\mathcal{H})$ be its set of density matrices. Given any $\epsilon \gt 0$, any probability distribution ${{\mu }_{w}}:\mathbb{R}\to [0,\infty )$ with a well defined first moment, and any continuous function $\nu \;:\mathbb{R}\to \mathcal{D}(\mathcal{H})$, there is always a $\delta \gt 0$ such that

$$\begin{eqnarray}&&\left\|\frac{1}{\delta }{{\int }_{\mathbb{R}}}{{\mu }_{w}}\left( \frac{p-{{p}_{0}}}{\delta }+{{p}_{0}} \right)\nu (p){\rm d}p-\nu ({{p}_{0}})\right\|{_{1}}\leqslant \epsilon ,\end{eqnarray} \tag{ 32 }$$

where ${{p}_{0}}=\int p\;{{\mu }_{w}}(p){\rm d}p$.

Proof. By virtue of the continuity of ν, there is always a $\delta ^{\prime}$ such that

$$\begin{eqnarray}&&{{{\rm max} }_{p\in \Delta }}\left\|\nu (p)-\nu ({{p}_{0}})\right\|{_{1}}\leqslant \frac{\epsilon }{2},\end{eqnarray} \tag{ 33 }$$

where $\Delta =({{p}_{0}}-\delta ^{\prime} ,{{p}_{0}}+\delta ^{\prime} )$. Furthermore, since ${{\mu }_{w}}$ is a normalized probability distribution on $\mathbb{R}$, for any $\delta ^{\prime} \gt 0$ there is always a δ such that

$$\begin{eqnarray}&&\int _{{{p}_{0}}-\frac{\delta ^{\prime} }{\delta }}^{{{p}_{0}}+\frac{\delta ^{\prime} }{\delta }}{{\mu }_{w}}(p){\rm d}p\gt 1-\frac{\epsilon }{4}.\end{eqnarray} \tag{ 34 }$$

Given that

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} \int _{{{p}_{0}}-\frac{\delta ^{\prime} }{\delta }}^{{{p}_{0}}+\frac{\delta ^{\prime} }{\delta }}{{\mu }_{w}}(p){\rm d}p & = & \int _{-\delta ^{\prime} }^{\delta ^{\prime} }{{\mu }_{w}}\left( \frac{p}{\delta }+{{p}_{0}} \right)\frac{{\rm d}p}{\delta } \\ {} & = & {{\int }_{\Delta }}{{\mu }_{w}}\left( \frac{p-{{p}_{0}}}{\delta }+{{p}_{0}} \right)\frac{{\rm d}p}{\delta } \\ \end{array}\end{eqnarray} \tag{ 35 }$$

this implies

$$\begin{eqnarray}&&{{\int }_{\Delta }}{{\mu }_{w}}\left( \frac{p-{{p}_{0}}}{\delta }+{{p}_{0}} \right)\frac{{\rm d}p}{\delta }\gt 1-\frac{\epsilon }{4}.\end{eqnarray} \tag{ 36 }$$

To simplify the following equations, let us define $f(p)={{\mu }_{w}}(\frac{p-{{p}_{0}}}{\delta }+{{p}_{0}})$. Therefore, for any we have

$$\begin{eqnarray}&&\begin{array}{ccccccccccccccc} \left\|\frac{1}{\delta }{{\int }_{\mathbb{R}}}f(p)\nu (p){\rm d}p-\nu ({{p}_{0}})\right\|{_{1}} \\ \quad =\left\|{{\int }_{\mathbb{R}}}f(p)\left[ \nu (p)-\nu ({{p}_{0}}) \right]\frac{{\rm d}p}{\delta }\right\|{_{1}} \\ \quad \leqslant {{\int }_{\mathbb{R}}}f(p)\left\|\nu (p)-\nu ({{p}_{0}})\right\|{_{1}}\frac{{\rm d}p}{\delta } \\ \quad \leqslant {{\int }_{\Delta }}f(p)\left\|\nu (p)-\nu ({{p}_{0}})\right\|{_{1}}\frac{{\rm d}p}{\delta }+2{{\int }_{\mathbb{R}/\Delta }}f(p)\frac{{\rm d}p}{\delta } \\ \quad \leqslant {{\int }_{\Delta }}f(p)\frac{{\rm d}p}{\delta }{{{\rm max} }_{p^{\prime} \in \Delta }}\left\|\nu (p^{\prime} )-\nu ({{p}_{0}})\right\|{_{1}}+\frac{\epsilon }{2} \\ \quad \leqslant \epsilon , \\ \end{array}\end{eqnarray} \tag{ 37 }$$

where the third inequality is due to $\left\|\nu (p)-\nu ({{p}_{0}})\right\|{_{1}}\leqslant \parallel \nu (p){{\parallel }_{1}}+\left\|\nu ({{p}_{0}})\right\|{_{1}}=2$.□

Finally, combining equation (11) with theorem 1 while setting $\nu (p)={\rm T}{{{\rm r}}_{w}}\left[ {{U}_{e}}(p){{\rho }_{e}}(0)U_{e}^{\dagger }(p) \right]$ leads to the statement in equation (12).

Given an initial state of the form $\rho (0)={{\rho }_{ub}}(0)\;\otimes \;{{\rho }_{w}}(0)\;\otimes \;{{\rho }_{c}}(0)$, we calculate the reduced subsystem-bath state evolving under the Hamiltonian

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} H & = & {{H}_{u}}+{{H}_{b}}+{{H}_{w}}+{{H}_{c}}+{{H}^{\operatorname{int}}} \\ {} & = & {{H}_{0}}+{{H}^{\operatorname{int}}} \\ \end{array}\end{eqnarray} \tag{ 38 }$$

$$\begin{eqnarray}&&{{H}^{\operatorname{int}}}=\int {{\int }_{{{\mathbb{R}}^{2}}}}H_{ub}^{\operatorname{int}}(x,p)\;\otimes \;|p{{\rangle }_{w}}{{\langle p|\;\otimes \;|x\rangle }_{c}}\langle x|{\rm d}p{\rm d}x.\end{eqnarray} \tag{ 39 }$$

In the interaction picture ${{H}^{\operatorname{int}}}$ takes the form

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} {{\tilde{H}}^{\operatorname{int}}}(t) & = & {{{\rm e}}^{{\rm i}{{H}_{0}}t}}{{H}^{\operatorname{int}}}{{{\rm e}}^{-{\rm i}{{H}_{0}}t}} \\ {} & = & {{{\rm e}}^{{\rm i}{{H}_{0}}t}}\int {{\int }_{{{\mathbb{R}}^{2}}}}H_{ub}^{\operatorname{int}}(x,p)\;\otimes \;|p{{\rangle }_{w}}{{\langle p|\;\otimes \;|x\rangle }_{c}}\langle x|{\rm d}p{\rm d}x{{{\rm e}}^{-{\rm i}{{H}_{0}}t}} \\ {} & = & \int {{\int }_{{{\mathbb{R}}^{2}}}}\tilde{H}_{ub}^{\operatorname{int}}(x,p,t)\;\otimes \;|p-t{{\rangle }_{w}}{{\langle p-t|\;\otimes \;|x-t\rangle }_{c}}\langle x-t|{\rm d}p{\rm d}x \\ {} & = & \int {{\int }_{{{\mathbb{R}}^{2}}}}\tilde{H}_{ub}^{\operatorname{int}}(x+t,p+t,t)\;\otimes \;|p{{\rangle }_{w}}{{\langle p|\;\otimes \;|x\rangle }_{c}}\langle x|{\rm d}p{\rm d}x, \\ \end{array}\end{eqnarray} \tag{ 40 }$$

where

$$\begin{eqnarray}&&\tilde{H}_{ub}^{\operatorname{int}}(x,p,t)={{{\rm e}}^{{\rm i}({{H}_{u}}+{{H}_{b}})t}}H_{ub}^{\operatorname{int}}(x,p){{{\rm e}}^{-{\rm i}({{H}_{u}}+{{H}_{b}})t}}.\end{eqnarray} \tag{ 41 }$$

With this, the time evolution operator between times 0 and t can be written as

$$\begin{eqnarray}&&\mathcal{U}(t)={{{\rm e}}^{-{\rm i}{{H}_{0}}t}}\mathcal{T}\left[ {{{\rm e}}^{-{\rm i}\int _{0}^{t}{{\tilde{H}}^{\operatorname{int}}}(t^{\prime} ){\rm d}t^{\prime} }} \right]\end{eqnarray} \tag{ 42 }$$

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} {} & = & {{{\rm e}}^{-{\rm i}{{H}_{w}}t}}{{{\rm e}}^{-{\rm i}{{H}_{c}}t}}{{{\rm e}}^{-{\rm i}{{H}_{u}}t}}{{{\rm e}}^{-{\rm i}{{H}_{b}}t}} \\ {} & {} & \times \int {{\int }_{{{\mathbb{R}}^{2}}}}\mathcal{T}\left[ {{{\rm e}}^{-{\rm i}\int _{0}^{t}\tilde{H}_{ub}^{\operatorname{int}}(x+t^{\prime} ,p+t^{\prime} ,t^{\prime} ){\rm d}t^{\prime} }} \right] \\ {} & {} & \times |p{{\rangle }_{w}}{{\langle p|\;\otimes \;|x\rangle }_{c}}\langle x|{\rm d}p{\rm d}x. \\ \end{array}\end{eqnarray} \tag{ 43 }$$

When calculating the reduced time-evolved subsystem-bath state, the ${{H}_{w}}$ and ${{H}_{c}}$ exponentials vanish by ciclicity of the trace, and one is left with a mixture of unitaries. That is

$$\begin{eqnarray}\begin{array}{ccccccccccccccc} {\rm T}{{{\rm r}}_{wc}}[\rho (t)] & = & {\rm T}{{{\rm r}}_{wc}}[\mathcal{U}(t)\rho (0){{\mathcal{U}}^{\dagger }}(t)] \\ {} & = & \int {{\int }_{{{\mathbb{R}}^{2}}}}{\rm T}{{{\rm r}}_{wc}}\left[ ({{\rho }_{w}}(0)\;\otimes \;{{\rho }_{c}}(0)){{(|p\rangle }_{w}}{{\langle p|\;\otimes \;|x\rangle }_{c}}\langle x|) \right] \\ {} & {} & \times V(x,p){{\rho }_{ub}}(0){{V}^{\dagger }}(x,p){\rm d}p{\rm d}x, \\ {} & = & \int {{\int }_{{{\mathbb{R}}^{2}}}}\mu (x,p)V(x,p){{\rho }_{ub}}(0){{V}^{\dagger }}(x,p){\rm d}p{\rm d}x, \\ \end{array}\end{eqnarray} \tag{ 44 }$$

where $\mu (x,p)=\langle x,p|{{\rho }_{cw}}(0)|x,p\rangle$ and

$$\begin{eqnarray}&&V(x,p)={{{\rm e}}^{-{\rm i}{{H}_{u}}t}}{{{\rm e}}^{-{\rm i}{{H}_{b}}t}}\mathcal{T}\left[ {{{\rm e}}^{-{\rm i}\int _{0}^{t}\tilde{H}_{ub}^{\operatorname{int}}(x+t^{\prime} ,p+t^{\prime} ,t^{\prime} ){\rm d}t^{\prime} }} \right].\end{eqnarray} \tag{ 45 }$$