# Landauer's Principle and the balance equation

## February 29, 2016*

Tags: Landauer, physics, math

I’ve been working on my thesisEdit: this was my master’s thesis

over reading week, and I think I’ve finished my introduction to Landauer’s Principle. I ended up writing a pretty detailed derivation of the balance equation, and thus Landauer’s bound, so I thought it might be useful to post here.

Landauer's principle states that there is a minimal energetic cost for a state transformation $$\rho^\text{i}\to \rho^\text{f}$$ on a system $$\mathcal{S}$$ via the action of a thermal reservoir $$\mathcal{E}$$ at temperature $$(k_B\beta)^{-1}$$ $$k_B \approx 1.38 \times 10^{-23}$$ Joules per Kelvin is Boltzmann's constant.

. In particular, if $$\Delta S_\mathcal{S}$$ is the change of entropy of the system $$\mathcal{S}$$, and $$\Delta Q_\mathcal{E}$$ is the change in energy of the reservoir $$\mathcal{E}$$, then $\Delta Q_\mathcal{E}\geq \beta^{-1}\Delta S_\mathcal{S}. \qquad(1)$ This principle has generated interest since its inception in 1961; see [Section 1, RW14] [RW14]: An improved Landauer Principle with finite-size corrections, D. Reeb and M. Wolf, 2014 (v3).

for a recent summary. First, the bound has allusions to practicality: perhaps the energy efficiency of our computers will be limited. For changing the state of a classical or quantum bit however, the bound is at most $\Delta Q_\mathcal{E}\geq k_B\cdot T \log 2 \approx (9.6 \times 10^{-24} J/K)\cdot$ which is extremely small for reasonable temperatures $$T$$; yet, modern processors are within several orders of magnitude of this limit, as shown in Figure 1.

Moreover, in 1973 Bennett showed that any Turing machine program may be implemented in a reversible manner, so that $$\Delta S_\mathcal{S}=0$$. Reversible computing is an area of considerable practical interest and continuing theoretical work.

More fundamentally, Landauer's bound is a direct relationship between energy and information (entropy). From now on, we will use natural units so that $$k_B=\hbar=1$$.

In fact, Landauer's principle follows from the entropy balance equation $\Delta S_\mathcal{S}+ \sigma = \beta \Delta Q_\mathcal{E} \qquad(2)$ where $$\sigma$$ is the entropy production.

We will define $$\sigma$$ and prove (eq. 2), in a finite dimensional quantum unitary setup, following [Section 3, RW14] and [JP14] [JP14]: A note on the Landauer principle in quantum statistical mechanics, V. Jaksic and C. Pillet, 2014 (v1).

. We assume the system $$\mathcal{S}$$ is described by a finite dimensional Hilbert space $$\mathcal{H}_\mathcal{S}$$, with self-adjoint Hamiltonian $$h_{\mathcal{S}}$$. The initial state on the system is given by a density matrixnon-negative trace-one operator on $$\mathcal{H}_\mathcal{S}$$

$$\rho^\text{i}$$. Likewise, we assume the environment is described by a finite dimensional Hilbert space $$\mathcal{H}_\mathcal{E}$$ with self-adjoint Hamiltonian $$h_{\mathcal{E}}$$, and initial state $\xi^\text{i}= \frac{\exp(-\beta h_{\mathcal{E}})}{\operatorname{tr}(\exp(-\beta h_{\mathcal{E}}))} \qquad(3)$ the Gibbs stateGibbs states on $$\mathcal{E}$$ are invariant under the free dynamics $$h_{\mathcal{E}}$$; in this finite dimensional context, they are uniquely so. They thus have the interpretation of thermal equilibrium states.

at temperature $$\beta^{-1}$$. The system and environment start uncoupled, so the joint initial state is $$\rho^\text{i}\otimes \xi^\text{i}$$. The evolution of the joint system is given by a unitary operator $$U \in \mathcal{B}(\mathcal{H}_\mathcal{S}\otimes \mathcal{H}_\mathcal{E})$$, leading to the final joint state $$U\rho^\text{i}\otimes \xi^\text{i}U^*$$. We decouple the systems, yielding \begin{aligned} \rho^\text{f}= \operatorname{tr}_\mathcal{E}(U \rho^\text{i}\otimes \xi^\text{i}U^*), \qquad \xi^\text{f}= \operatorname{tr}_\mathcal{S}(U\rho^\text{i}\otimes \xi^\text{i}U^*) \end{aligned} as the final state on the system, environment, respectively.

We identify two quantities of interest during this process: $$\Delta S_\mathcal{S}$$, the change of entropy of the system of interest, and $$\Delta Q_\mathcal{E}$$, the change of energy of the environment, defined as Note the sign convention.

\begin{aligned} \Delta S_\mathcal{S}:\!&= S(\rho^\text{i}) - S(\rho^\text{f}), & \Delta Q_\mathcal{E}:\!&= \operatorname{tr}(h_{\mathcal{E}}\xi^\text{f})- \operatorname{tr}(h_{\mathcal{E}}\xi^\text{i}), \end{aligned} where $$S(\rho) := - \operatorname{tr}\rho\log\rho$$ is the von Neumann entropy. Recall the relative entropy $$S(\eta|\nu) = \operatorname{tr}(\eta\log\eta-\log \nu))$$ of two faithful states $$\eta$$ and $$\nu$$ has $$S(\eta|\nu)\geq 0$$ with equality if and only if $$\eta=\nu$$. See sections 2.5-2.6 of Entropic Fluctuations in Quantum Statistical Mechanics. An Introduction by Jaksic et al (2011) for a review of entropy functions in finite dimensional quantum mechanics.

With this function, we define the entropy production $\sigma := S(U \rho^\text{i}\otimes \xi^\text{i}\,U^* | \rho^\text{f}\otimes \xi^\text{i}).$ By definition of relative entropy, the entropy production may be written \begin{aligned} \sigma = \operatorname{tr}\big(U \rho^\text{i}\otimes \xi^\text{i}U^* \, \log (U \rho^\text{i}\otimes \xi^\text{i}U^*)\big) -\operatorname{tr}\big(U \rho^\text{i}\otimes \xi^\text{i}U^* \, \log(\rho^\text{f}\otimes \xi^\text{i})\big). \end{aligned} We then recognize the first term as an entropy, and expand the second term using the following claim.

If $$A$$ and $$B$$ are positive (and thus self-adjoint) on a finite dimensional Hilbert space, then

$\log ( A\otimes B) = \log (A)\otimes \operatorname{id}+ \operatorname{id}\otimes \log(B).$

If $$A,B$$ have spectral decompositions $$A = \sum_i \mu_i P_i$$ and $$B= \sum_j \lambda_j Q_j$$, then $$A\otimes B= \sum_{ij} \mu_i \lambda_j P_i \otimes Q_j$$. With this,

\begin{aligned} \log(A\otimes B)&= \sum_{ij} \log(\mu_i \lambda_j) P_i \otimes Q_j \\&= \sum_{ij} (\log \mu_i + \log \lambda_j) P_i\otimes Q_j \\&= \sum_{ij} \log \mu_i P_i\otimes Q_j + \sum_{ij} \log \lambda_j P_i\otimes Q_j \\&= \sum_i \log \mu_i P_i\otimes \operatorname{id}+ \sum_j \log \lambda_j \operatorname{id}\otimes Q_j\\&= \log(A)\otimes \operatorname{id}+ \operatorname{id}\otimes \log(B). \end{aligned}

This yields

\begin{aligned}\sigma &= - S(U \rho^\text{i}\otimes \xi^\text{i}U^*) - \operatorname{tr}\big( U \rho^\text{i}\otimes \xi^\text{i}U^* \, (\log \rho^\text{f}\otimes \operatorname{id})\big) \\&\qquad- \operatorname{tr}\big( U \rho^\text{i}\otimes \xi^\text{i}U^* \, ( \operatorname{id}\otimes \log \xi^\text{i})\big).\end{aligned}

Since entropy is invariant under a unitary transformationAs may immediately be seen by the spectral theorem: if $$\rho$$ has spectral decomposition $$\rho = \sum_i \mu_i P_i$$, then $$S(\rho) = -\sum_i \mu_i\log \mu_i= S(U\rho U^*)$$, since eigenvalues are invariant under unitary transformations (change of basis).

, we have $$S(U \rho^\text{i}\otimes \xi^\text{i}U^*) = S(\rho^\text{i}\otimes \xi^\text{i})$$. Furthermore, by definition of the partial trace, \begin{aligned}\operatorname{tr}\big( U \rho^\text{i}\otimes \xi^\text{i}U^* \, (\log \rho^\text{f}\otimes \operatorname{id})\big) = \operatorname{tr}\big(\operatorname{tr}_\mathcal{S}(U \rho^\text{i}\otimes \xi^\text{i}U^*) \log \rho^\text{f}\big),\end{aligned} which is simply $$\operatorname{tr}(\rho^\text{f}\log \rho^\text{f}) = -S(\rho^\text{f})$$. Using this argument for the third term as well, we are left with \begin{aligned}\sigma =-S(\rho^\text{i}\otimes \xi^\text{i}) + S(\rho^\text{f}) - \operatorname{tr}(\xi^\text{f}\log \xi^\text{i}).\end{aligned} But \begin{aligned}S(\rho^\text{i}\otimes \xi^\text{i}) &= -\operatorname{tr}\big(\rho^\text{i}\otimes \xi^\text{i}\log(\rho^\text{i}\otimes \xi^\text{i}) \big) \\&= -\operatorname{tr}\big(\rho^\text{i}\otimes \xi^\text{i}(\log\rho^\text{i}\otimes \operatorname{id})\big) - \operatorname{tr}\big(\rho^\text{i}\otimes \xi^\text{i}(\operatorname{id}\otimes\log \xi^\text{i})\big)\\&=-\operatorname{tr}\big(\rho^\text{i}\log\rho^\text{i}\otimes \xi^\text{i}\big) - \operatorname{tr}\big(\rho^\text{i}\otimes \xi^\text{i}\log \xi^\text{i}\big) \\&=- \operatorname{tr}(\rho^\text{i}\log \rho^\text{i}) \operatorname{tr}(\xi^\text{i}) - \operatorname{tr}(\xi^\text{i}\log \xi^\text{i}) \operatorname{tr}(\rho^\text{i})\\&= S(\rho^\text{i}) + S(\xi^\text{i}),\end{aligned} using that $$\operatorname{tr}(\rho^\text{i}) = \operatorname{tr}(\xi^\text{i})=1$$. Then, \begin{aligned}\sigma &= -S(\rho^\text{i}) +S(\rho^\text{f})- S(\xi^\text{i}) - \operatorname{tr}(\xi^\text{f}\log \xi^\text{i}) \\&= -\Delta S_\mathcal{S}- \operatorname{tr}\big((\xi^\text{f}- \xi^\text{i}) \log \xi^\text{i}\big).\end{aligned} Additionally, using (eq. 3), \begin{aligned}\log \xi^\text{i}= -\beta h_{\mathcal{E}}- \log(\operatorname{tr}(\exp(-\beta h_{\mathcal{E}})),\end{aligned} so \begin{aligned}\sigma &= -\Delta S_\mathcal{S}+ \beta\operatorname{tr}\big((\xi^\text{f}- \xi^\text{i}) h_{\mathcal{E}}\big) + \log(\operatorname{tr}(\exp(-\beta h_{\mathcal{E}})))\operatorname{tr}(\xi^\text{f}- \xi^\text{i}) \\&= -\Delta S_\mathcal{S}+ \beta \Delta Q_\mathcal{E},\end{aligned} using that $$\operatorname{tr}(\xi^\text{f}- \xi^\text{i}) = 1-1=0$$.

We thus have the balance equation (eq. 2). Landauer’s Principle (eq. 1) follows by simply noting $$\sigma\geq 0$$ as it is a relative entropy.

We may interpret (eq. 2) as a microscopic Clausius formulation of the Second Law of Thermodynamics [BHN+14] [BHN+14]: The second laws of quantum thermodynamics F. Brandao et al, 2013 (v4).

. More specifically, we may interpret $$\beta \Delta Q_\mathcal{E}=\int_\text{i}^\text{f}\frac{\mathrm{d} Q_\mathcal{E}}{T}= \Delta S_\mathcal{E}^\text{Clausius}$$ as the Clausius entropy change of the environment. Note: We’re making an analogy to the Clausius formulation of the Second Law of Thermodynamics, but not a formal relationship. The balance equation presented here and the 2nd law are part of two different frameworks.

Then, with a minus sign to account for our sign convention, we will interpret $$\Delta S_\mathcal{S}^\text{Clausius} =-\Delta S_\mathcal{S}$$, and the Second Law is \begin{aligned}\Delta S^\text{Clausius}_\mathcal{E}+\Delta S^\text{Clausius}_\mathcal{S}=\text{entropy production}\geq 0.\end{aligned} In this language then, $$\sigma$$ serves as the entropy production, which gives it its name. The classical Second Law, however, is a statement about macroscopic quantities obtained from the behavior of $$\gtrsim 10^{23}$$ particles. Within the theory of quantum mechanics and our assumptions, however, the balance equation (eq. 2) is exact on a microscopic level.