Fluctuating Work in Coherent Quantum Systems: Proposals and Limitations

Abstract

One of the most important goals in quantum thermodynamics is to demonstrate advantages of themodynamic protocols over their classical counterparts. For that, it is necessary to (i) develop theoretical tools and experimental set-ups to deal with quantum coherence in thermodynamic contexts, and to (ii) elucidate which properties are genuinely quantum in a thermodynamic process. In this short review, we discuss proposals to define and measure work fluctuations that allow to capture quantum interference phenomena. We also discuss fundamental limitations arising due to measurement back-action, as well as connections between work distributions and quantum contextuality. We hope the different results summarised here motivate further research on the role of quantum phenomena in thermodynamics.

Appendix: Loss in Maximal Amount of Average Extractable Work Due to Energy Measurement

The maximal average amount of work that can be extracted from a quantum system in state $\rho$ with internal Hamiltonian $H_S$ and unitary operations on systems and a thermal bath at temperature T is

$$\begin{aligned} \langle W \rangle _{max}(\rho ) = F(\rho ,H_S) - F(\tau ,H_S) \equiv kT S(\rho ||\tau ), \end{aligned}$$

(11.15)

where $\tau = \frac{e^{-\beta H_S}}{Z}$, $Z = \mathrm {Tr}{}{[e^{-\beta H_S}]}$, $\beta = 1/(kT)$ with k Boltzmann constant, $F(X,H) = \mathrm {Tr}{}{[H X]} - kT S(X)$ with $S(X):= - \mathrm {Tr}{}{[X \log X]}$ the von Neumann entropy, $S(X||Y):= \mathrm {Tr}{}{[X (\log X - \log Y)]}$ the quantum relative entropy.

The result (11.15) can be inferred from, e.g., the derivations in [60, 109,110,111]. However, for clarity of the exposition, we present here a simple proof. Consider a bipartite system SB with an initial non-interacting Hamiltonian

$$\begin{aligned} H=H_\mathrm{S}+H_\mathrm{B}, \end{aligned}$$

(11.16)

prepared in a product state

$$\begin{aligned} \rho _{SB}=\rho _\mathrm{S}\otimes \rho _\mathrm{B}, \end{aligned}$$

(11.17)

with B a Gibbs state at inverse temperature $\beta = (kT)^{-1}$:

$$\begin{aligned} \rho _\mathrm{B}=\tau _{B}=\frac{e^{-\beta H_\mathrm{B}}}{\mathcal {Z}}. \end{aligned}$$

(11.18)

Hence we can think of B as an auxiliary thermal state. Note that B is not necessarily a thermal bath, in the sense that it may not be macroscopic. We now consider a generic closed evolution in SB, which can always be described as a unitary operation U:

$$\begin{aligned} \rho '_{SB}=U \rho _S \otimes \rho _B U^{\dagger }, \end{aligned}$$

(11.19)

The average extracted work is given by the energy change on SB,

$$\begin{aligned} W=\mathrm {Tr}[H \rho _{SB}]-\mathrm {Tr}[H \rho '_{SB}] = F(\rho _{SB},H) - F(\rho '_{SB},H), \end{aligned}$$

(11.20)

where we assumed that at the end of the interaction the final Hamiltonian is again H, and for the second equality we used $S(\rho '_{SB}) = S(\rho _{SB})$.

For any bipartite system $X_{SB}$ with Hamiltonian H, denoting by $X_{S/B} = \mathrm {Tr}_{B/S}X_{SB}$, one has that the non-equilibrium free energy decomposes into local parts plus correlations:

$$\begin{aligned} F(X_{SB}, H) = F(X_S,H_S) + F(X_B,H_B) + kT I(X_{SB}), \end{aligned}$$

(11.21)

where $I(X_{SB}) = S(X_S) + S(X_B) - S(X_{SB})$ is the mutual information (this can be verified by summing and subtracting the local entropies to the expression for $F(X_{SB}, H)$).

Denoting now $\rho '_{S/B}=\mathrm {Tr}_{B/S}[\rho '_{SB}]$ and using the formula above, together with the fact that $I(\rho _{SB}) = 0$, we obtain

$$\begin{aligned} W&=F(\rho _S,H_S)-F(\rho '_S,H_S) +F(\rho _B,H_B)-F(\rho '_B,H_B) -kT I(\rho '_{SB}) \nonumber \\&=F(\rho _S,H_S)-F(\rho '_S,H_S) -kT (S(\rho '_{B}||\tau _B) +I(\rho '_{SB})), \end{aligned}$$

(11.22)

where we used $F(\rho '_B,H_B)-F(\tau _B,H_B)=kT S(\gamma _B||\tau _{B})$. As both $I(\cdot )$ and $S(\cdot ||\cdot )$ are non-negative quantities, we immediately obtain

$$\begin{aligned} W\le F(\rho _S,H_S)-F(\rho '_S,H_S). \end{aligned}$$

(11.23)

Note that this expression only depends on the (initial and final) state of S and the temperature of B. We can obtain a bound that is independent of the final state by adding and subtracting $F(\tau _S,H_S)$ in Eq. (11.22) (where $\tau _\mathrm{S}=e^{-\beta H_S}/\mathrm {Tr}(e^{-\beta H_S})$), which gives

$$\begin{aligned} W= F(\rho _\mathrm{S},H_S)-F(\tau _\mathrm{S},H_S)-T\left[ S(\rho '_S||\tau _\mathrm{S})+I(\rho '_{SB})+S(\rho '_B||\tau _\mathrm{B}) \right] , \end{aligned}$$

(11.24)

Note again that all terms in the square parenthesis are non-negative, and each of them has an intuitive physical meaning: in order of appearance, the athermality of the final state of S (when $\rho '_S\ne \tau _S$), the correlations created between S and B, and the athermality of final state of B (when $\rho '_B \ne \tau _B$). With the above equality we finally obtain

$$\begin{aligned} W\le F(\rho _{S},H_S)-F(\tau _{S},H_S) = \langle W \rangle _{max}(\rho ). \end{aligned}$$

(11.25)

The remaining interesting question is whether these bounds can be saturated: protocols that achieve (11.15) are constructed in [109,110,111] which, interestingly, require B to be of macroscopic size.

We now move to the question of how energy measurements change Eq. (11.15), by making it unattainable. For simplicity of exposition, we will assume that $H_S$ is not degenerate. If one performs an energy measurement, one obtains a pure energy state $E_i$ with probability $p_i = \langle E_i| \rho |E_i\rangle$. Using Eq. (11.15), one can see that from state $|E_i\rangle$ one can extract a maximum amount of work equal to $F(|E_i\rangle ) - F(\tau ) = E_i + kT \log Z$ (this can be understood as the conjunction of the unitary process that maps $|E_i\rangle$ into the ground state, extracting energy $E_i$, followed by a protocol that extracts work $kT \log Z$ from the purity of the ground state). To complete the process one needs to reset the memory, implicitly used in the measurement, to its “blank state”; in the presence of a bath at temperature T, this requires an investment of $kT H(\mathbf {p})$, where $\mathbf {p}$ is the distribution $p_i$ and $H(\mathbf {p}) = - \sum _i p_i \log p_i$ is the Shannon entropy (this is known as Landauer erasure). Overall the protocol that extracts work after an energy measurement optimally achieves the average $\langle W \rangle _{meas}(\rho ) = \sum _i p_i E_i - kT H(\mathbf {p}) + kT \log Z$. A direct calculation shows $\langle W \rangle _{meas} = F(\mathcal {D}(\rho )) - F(\tau )$, with $\mathcal {D}$ the dephasing operation $\mathcal {D}(\rho ) = \sum _i p_i |E_i\rangle \langle E_i|$. This implies

$$\begin{aligned} \langle W \rangle _{meas}(\rho ) = \langle W \rangle _{max}(\mathcal {D}(\rho )), \end{aligned}$$

(11.26)

i.e. the energy measurement protocol optimally extracts an average amount of work equal to the maximum that can be extracted by first dephasing the state and then performing work extraction. This implies a loss

$$\begin{aligned} \langle W \rangle _{max}(\rho ) - \langle W \rangle _{max}(\mathcal {D}(\rho )) = kT S(\mathcal {D}(\rho )) - kT S(\rho ) \equiv kT S(\rho ||\mathcal {D}(\rho )):= kT A(\rho ), \end{aligned}$$

(11.27)

proportional to a quantity $A(\rho )$ called asymmetry or relative entropy of coherence, which is a measure of quantum coherence in the eigenbasis of $H_S$. Note that $A(\rho ) > 0$ if and only if $\rho \ne \mathcal {D}(\rho )$.

The above reasoning shows that protocols based on energy measurements cannot reach the optimal average work extraction, since they lose the possibility of extracting work from the coherence of the quantum state. This intuition can also be grounded in the non equilibrium free energy. As we discussed above, $\langle W \rangle _{max}(\rho ) = \Delta F (\rho ) := F(\rho ) - F(\tau )$. Summing and subtracting $\Delta F(\mathcal {D}(\rho )) = F(\mathcal {D}(\rho )) - F(\tau )$ one obtains, using the definition of $A(\rho )$,

$$\begin{aligned} \Delta F(\rho ) = \Delta F(\mathcal {D}(\rho )) + kT A(\rho ), \end{aligned}$$

(11.28)

i.e. the non-equilibrium free energy neatly decomposes into a contribution coming from the diagonal part of the state and a contribution coming from coherence [61, 62]. As discussed above, $\langle W \rangle _{meas} = \Delta F(\mathcal {D}(\rho ))$, i.e. the diagonal non-equilibrium free energy captures the component that can be converted into work by the energy measurement protocol, whereas the coherent contribution is lost. One has $\Delta F(\rho )>\Delta F(\mathcal {D}(\rho ))$ whenever $[\rho ,H_S]\ne 0$.