Antecedentes

We will now start to discuss entropy-production and reversibility in the time-average model. Let us first discuss a very simple toy-example that shows that there can be entropy pro- duction in the quasi-static limit. Consider the following Hamiltonian trajectory, which is

continuous but not differentiable: u∈ [−1, 1] 7→H(u) = ( uσx, −1≤u<0, uσz, 0≤u≤1. (10.30)

Now suppose the initial state is an eigenstate of σx, e.g. ρ(0) = |+ ih+ |. Then foru≤0,

time-averaging does not alter the change at all and we have ωt.a.(u) = |+ ih+ |for all

u ≤ 0. For any u > 0 the eigenbasis of the Hamiltonian H(u) is constant orthogonal to |+ i. Taking the time-average of the state |+ ih+ |in this new eigenbasis therefore completely mixes the state and we have ωt.a.(u) = 1/2 for any u > 0. We conclude

that the entropy of the effective description changes fromS(−1) =0 to S(1) = log(2). This dissipation can be attributed to the fact that while the eigenvalues of the Hamiltonian can be described by smooth functions, the eigenbasis cannot. It should be clear that in principlea similar effect can occur in interacting many-body systems (even though it may be unlikely, e.g., due to the phenomenon of avoided crossings). Thus, the example shows that we cannot expect to prove the absence of entropy production in general.

But the example also hints at the sufficient condition for absent of entropy production, namely the smoothness of the Hamiltonian trajectory. This is indeed the case, as shown by the following lemma.

Lemma 10.2 (Absence of entropy production in time-average model). Letu ∈ [0, 1] 7→

H(u) be a differentiable trajectory of Hamiltonians, in the sense that the eigenvectors ofH(u) are continuous and the eigenbasis ofH(u) is differentiable. Then there is no entropy-production in the time-average equilibration model: For any initial equilibrium state ωt.a.(0)its spectrum is conserved during the process and thereforeS(0) =S(1).

Proof. Let the eigenbasis ofH(u)be given by|Ek(u)i. Then the eigenvaluespk(u+δu)

of the equilibrium state at parameter valueu+δu can be written as

pk(u+δu) = hEk(u+δu)|ωt.a.(u)|Ek(u+δu)i (10.31)

=

∑

l

pl(u)|hEk(u+δu)|El(u)i|2, (10.32)

because they are simply the diagonal elements of ωt.a.(u)in the new eigenbasis|Ek(u+δu)i.

Using differentiability of the eigenbasis,|Ek(u)iwe can write them as

|Ek(u+δu)i = |Ek(u)i + |Xk(u)iδu+O(δu2). (10.33)

Since the bases are ortho-normalized, we haveRe(hEk(u)|Xk(u)i) =0. This implies

|hEk(u+δu)|El(u)i|2=δl,k+O(δu2).

Hence we havepk(u+δu) = pk(u) +O(δu2). Taking the limit δu→0 we see that the

eigenvaluespk(u)remain constant.

Lemma 10.2 is essentially a consequence of the adiabatic theorem of quantum mechan- ics. While in the quantum mechanical context, the qualifier "adiabatic" simply means "very slow", in this context we see that it also indeed acquires the meaning of vanishing entropy- production.

From the discussion of the time-average model we already learn that in the context of GGEs, vanishing entropy production in the quasi-static limit requires smoothness condi- tions on the thermodynamic protocols.

In the general case of arbitrary GGEs, we should expect that such conditions can be phrased in terms of the Lagrange-multipliers λj(u) determining the equilibrium states.

We will now provide such a characterization. To understand this characterization, first recall that in the case of a discrete thermodynamic protocol, the Lagrange-multipliers are implicitly determined by the equation

Trω_GGE(i) Q(i)_j



=Trω_GGE(i−1)Q(i)_j



, j=1, . . . , m. (10.34) We then take a continuum limit to obtain the functions λj(u) in the quasi-static limit.

The following Lemma shows that the entropy production vanishes as long as the functions

A Q UA N T U M O F T H E R M O DY N A M I C S 123

Lemma 10.3 (Absence of entropy production in GGEs). Consider a quasi-static process along a trajectory of Hamiltonians u ∈ [0, 1] 7→ H(u)(with associated m conserved quantities Qj(u)) giving rise to the equilibrium states ωGGE(u). Then the entropy of

ωGGE(u)in the quasi-static limit if the functionsu 7→ λj(u)determined by(10.16) are

smooth.

Proof. If the functions λj(u)are smooth, we can take the continuum limit of equation (10.16)

to obtain

Tr dωGGE(u) du Qj(u)



=0, j=1, . . . , m. (10.35) But, due to normalization of the states ωGGE(u), the derivative ofS(u)is given by

dS(u) du = −Tr  dωGGE(u) du log(ωGGE(u))  = m

∑

Thus the entropy remains constant.

To illustrate this Lemma, let us discuss a simply counter-example in which the quasi- static limit does not give rise to smooth Lagrange-multipliers. To do this, we will work in the Gibbs-equilibration model, which is a special case of GGEs and consider again a two-level system with Hamiltonian trajectory

u∈ [0, 1] 7→H(u) = (1−u)|1ih1|. (10.37)

Let the initial state be given by ω_β(0)(H(0))with β(0) > 0. We will now see that this example has the following properties:

1. The inverse temperature β(u)diverges asu→1, therefore it is not smooth on the whole interval[0, 1].

2. The entropy does not remain constant.

3. The Gibbs model does not provide a good description of the system.

First note that all HamiltoniansH(u)have the same eigenbasis. Therefore the actual quan- tum state of the system remains constant and equal to the initial state ω_β(0)(H(0)). This in turn implies that the effective description in terms of a Gibbs-state requires that the effective inverse temperature and the HamiltonianH(u)fulfill

β(u)H(u) =β(0)H(0). (10.38)

Since H(u) = H(0)(1−u) we then deduce β(u) = β(0)/(1−u), which proves the

first claim. To see that the entropy increases, note that the final Hamiltonian is fully de- generate. Hence any Gibbs-state of the final Hamiltonian is equal to the maximally mixed state and has entropylog(2), which shows the second claim. Note that the entropy pro- duction only happens if we follow the trajectory all the way to u = 1, which is the only point at which β(u)is not smooth (since it is ill-defined). The reason for this behaviour is that atu=1, the Hamiltonian is degenerate and cannot "resolve" the two different eigen- levels of the actual quantum state of the system. This also explains the last claim: Any additional, independent observable that commutes with H(1) would distinguish the two different eigen-levels of the quantum state and would lead to smooth Lagrange-multipliers and vanishing entropy-production.

Summarizing, we find that quasi-static processes result in a constant entropy if and only if the process is sufficiently smooth and all Lagrange-multipliers are well-behaved. Let us now turn to the Gibbs-equilibration model.