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What is the assumption for Boltzmann H-theorem? One can derive it just from the unitarity of quantum mechanics, so this should be generally true, does it imply a closed system will always thermalize eventually? Does it apply for many-body localized states?

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    $\begingroup$ Could you specifically reference the derivation you mention. The assumptions should be in there. But just out of the top of my head, you're going to have to assume some specific initial conditions (they have to be "typical"). The way Boltzmann did it was with the Stosszahlansatz. But more modern derivations use more explicit assumptions on the initial conditions. $\endgroup$ – Raskolnikov Jan 31 '14 at 17:51
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    $\begingroup$ Boltzmann's H-theorem is not for unitary quantum mechanics, but for classical (nearly ideal) gases. For the quantum mechanical H-theorem, see von Neumann's work from 1929. For an English translation, see arxiv.org/abs/1003.2133 $\endgroup$ – Zoltan Zimboras Jan 31 '14 at 18:05
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I would like to share my thoughts and questions on the issue. The Boltzmann H theorem based on classical mechanics is well discussed in various literatures, the irreversibility comes from his assumption of molecular chaos, which cannot be justified from the underlying dynamical equation. Here I will try to say something on quantum H theorem, the point I want to make is that, although seemingly H theorem can be derived from unitarity, the true entropy increase in fact comes from the non-unitary part of quantum mechanics. Let me first recap the derivation using unitarity $^{1,2}$.

H theorem as a consequence of unitarity

Denote by $P_k$ the probability of a particle appearing on the state $|k\rangle$, $A_{kl}$ the transition rate from state $|k\rangle$ to state $|l\rangle$, then by the master equation

$${\frac {dP_{k}}{dt}}=\sum _{l}(A_{{kl }}P_{l }-A_{{l k}}P_{k})=\sum _{{l\neq k}}(A_{{kl }}P_{l }-A_{{l k}}P_{k})\cdots\cdots(1).$$ Then we take the derivative of entropy

$$S=-\sum_k P_k\ln P_k\cdots\cdots(2),$$

we obtain

$$\frac{dS}{dt}=-\sum_k\frac{dP_k}{dt}\left(1+\ln P_k\right)\cdots\cdots(3).$$ Together with (1) we have $$\frac{dS}{dt}=-\sum_{kl}\left\{(1+\ln P_k)A_{{kl }}P_{l }-(1+\ln P_k)A_{{l k}}P_{k}\right\}\cdots(4).$$ For the seond second term let us interchange the dummy indices $k$ and $l$, we get $$\frac{dS}{dt}=\sum_{kl}(\ln P_l-\ln P_k)A_{kl}P_l\cdots\cdots(5)$$ Now use the mathematical identity $(\ln P_l-\ln P_k)P_l\geq P_l- P_k$, we obtain

$$\frac{dS}{dt}\geq \sum_{kl}(P_l-P_k)A_{kl}= \sum_{kl}P_l(A_{kl}-A_{lk})\\=\sum_{l}P_l\big\{\sum_{k}(A_{kl}-A_{lk})\big\}\cdots\cdots(6)$$

Now unitarity ensures $\sum_{k}A_{kl}$ and $\sum_{k}A_{lk}$ are both 0, because as transition rates, $$\sum_{k}A_{kl}=\frac{d}{dt}\sum_{k}|\langle l|S|k\rangle|^2=\frac{d}{dt}\sum_{k}\langle l|S|k\rangle\langle k|S^{\dagger}|l\rangle\\=\frac{d}{dt}\langle l|SS^{\dagger}|l\rangle=\frac{d}{dt}\langle l|l\rangle=0\cdots\cdots(7),$$ where $S$ is the unitary time evolution operator describing the system. This is nothing but saying the total transition probability from one state to all states must be 1. It is clear (6) and (7) imply the H theorem: $$\frac{dS}{dt}\geq 0.$$ Where does the irreversibility come from?

Now we are in a position to question ourselves with Loschmidt's paradox, analogously to its classical version: There are many unitary and time-reversible quantum mechanical systems, if we have just derived H theorem using unitarity alone, how can it be reconciled with time-reversibility of the underlying dynamics?

What sneaked into the above derivation?

The crucial thing to notice is that, in the quantum regime, the definition of entropy using equation (2) is inherently an impossible one: the value of the entropy in (2) depends on the basis we choose to describe the system!

Consider a two-level system with two choices of orthogonal basis $\{|1\rangle, |2\rangle\}$ and $\{|a\rangle, |b\rangle\}$ related by $$|1\rangle=\frac{1}{\sqrt2}(|a\rangle+|b\rangle),\\|2\rangle=\frac{1}{\sqrt2}(|a\rangle-|b\rangle).$$ Suppose the system is in the state $|1\rangle$, then the entropy formula gives $S=0$ in the first choice of basis since it has 100% chance to appear in $|1\rangle$, while in the other basis $S=\ln2$ since it has 50%-50% chance to appear in either $|a\rangle$ or $|b\rangle$.

Now we may argue, it is one thing that to say the system is in $\frac{1}{\sqrt2}(|a\rangle+|b\rangle)$ and have the potential 50%-50% chance to transit into $|a\rangle$ and $|b\rangle$ after a measurement, but a different thing to say the transition has been realized by some measurement. Two situations must be described differently. If we look back to our derivation, it is not hard to see what we really did was, after a basis state evolves to a new state which is a superposition of the basis states, we assumed transitions to original basis states have happened instead of just staying in that superposition state, and in fact the original definition of entropy is not capable of describing such situation, as explained just now.

A plausible definition of quantum entropy is the Von Neumann entropy, which is a basis-independent definition of entropy, and in this description, the entropy of a unitarily evolving system is constant in time, while a (projective) measurement can increase the entropy.

Based on the above comparison, we see the irreversibility really comes as an assumption, the assumption that a measurement/decoherence has happened, and as we know, a (projective) measurement is a non-unitary, irreversible process, no paradox anymore.

My own question on the issue is, what to make of the fact that von Neumann entropy is constant in time? Does it mean it is incapable of describing a closed system evolving from non-equilibrium to equilibrium, or should we just reverse the argument and say any non-equilibrium to equilibrium evolution must be described by some non-unitary process?


1.Rephrased from section 3.6 of The Quantum Theory of Fields, Vol1, S. Weinberg

2.If I remember correctly(which I'm not quite confident on), such derivation was first given by Pauli, and he correctly spotted the origin of irreversibility, which he called the "random phase assumption".

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  • $\begingroup$ The von Neumann entropy is not the same thing as thermodynamic entropy, for the same reason the information entropy $\int -\rho\ln \rho\, dq\,dp$ is not thermodynamic entropy in classical statistical physics. Both von Neumann and information entropy are constant for isolated Hamiltonian systems. Thermodynamic ( Clausius ) entropy has to be derived from statistical physics differently - it is proportional to logarithm of the accessible phase space(classical statistical physics) or logarithm of the number of accessible microstates (quantum statistical physics). $\endgroup$ – Ján Lalinský Feb 8 '14 at 15:42
  • $\begingroup$ The latter entropies are defined only for equilibrium states. There is no law saying that the entropy has to be defined for non-equilibrium states or that it has to increase in time. $\endgroup$ – Ján Lalinský Feb 8 '14 at 15:45
  • $\begingroup$ @JánLalinský: thanks for pointing out my misunderstanding. So in the quantum case what would be the proper definition of thermodynamic entropy? Or what would be a basis-independent way of counting microstates? $\endgroup$ – Jia Yiyang Feb 8 '14 at 17:20
  • $\begingroup$ I am not sure about this, but the closest thing is this: let the system be isolated with energy somewhere in $\langle E,E+\Delta E\rangle$ (the interval is large enough to contain zillion of Hamiltonian eigenstates). Define statistical entropy as $S = k_B \ln ( D(E)\Delta E )$ where $D(E)$ is density of the Hamiltonian eigenfunctions on the energy line. This is statistical entropy for microcanonical ensemble and for macro-systems (such ideal gas with many particles) can be treated as quantity similar to thermodynamic entropy. $\endgroup$ – Ján Lalinský Feb 8 '14 at 17:43
  • $\begingroup$ @JánLalinský: ok, I'll try to look into this, thanks. $\endgroup$ – Jia Yiyang Feb 9 '14 at 0:57
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Contrary to claims made in the question, we cannot derive the H-theorem "just from the unitarity of quantum mechanics". In fact, the theorem requires a non-unitary extension of quantum mechanics. I will first comment on this misunderstanding about unitarity that one finds in Weinberg textbook and latter I will address other questions made in the question.

(i) By a well-known theorem, entropy is conserved by unitary evolutions. This theorem is proven in many statistical mechanics books. The Schrödinger equation conserves entropy $dS/dt=0$. Yes, we can define a coarse-grained entropy $S = S_\mathrm{cg}+ \Delta S$ and force it to evolve via unitary evolution, but it will do in ways that violate both the second law of thermodynamics and experiments/observation, not to mention the existence of paradoxes associated to the arbitrariness in the coarse-graining.

(ii) Above point is the reason why Weinberg doesn't start from the Schrödinger equation but from a master equation he introduces in an ad hoc way. As is well-known, the master equation is incompatible with the Schrödinger equation (yes I know some authors pretend to derive the former from the latter, but those 'derivations' are completely wrong). Thus Weinberg is proving the H-theorem by starting from an irreversible master equation that already contains the essence of the theorem.

(iii) We have known since Boltzmann's epoch that unitarity relates the direct process $(i\rightarrow j)$ with the inverse process $(j\rightarrow i)$, but the master equation describes a superposition of both processes $(i\rightleftharpoons j)$ and this superposition breaks unitarity. We can check it by rewriting the master equation from the usual gain-loss form used by Weinberg to the kinetic form $dP_i/dt = K_{ij} P_j$ and then observing that the propagator $\exp(Kt)$ for the whole evolution is not unitary, predicting a final equilibrium state with $P_i^\mathrm{eq} = P_j^\mathrm{eq}$ independently of the initial state. This equilibrium state maximizes entropy (minimizes the H-function)

$$ S^\mathrm{eq} = -k_\mathrm{B} \sum_j^W P_j^\mathrm{eq} \ln P_j^\mathrm{eq} = -k_\mathrm{B} \ln P_j^\mathrm{eq} \sum_j^W P_j^\mathrm{eq} =-k_\mathrm{B} \ln P_j^\mathrm{eq} = -k_\mathrm{B} \ln \frac{1}{W} = k_\mathrm{B} \ln W $$

(iv) Weinberg claims in his book that his discussion of the H-theorem is more general than in "statistical mechanics textbooks". Well, that is only true when comparing with introductory textbooks. As explained in more advanced textbooks, the general master equation contains three terms: a flow term, a memory term, and a noise term. By taking a special initial state --sometimes named the assumption of initial random phase-- the noise term can be eliminated. By taking for the density matrix a base that diagonalizes the unperturbed Hamiltonian, the flow term identically vanish. Then only the memory term in the master equation remains. Applying a further diagonal within the kernel of the memory term, $\rho_{kl} \rightarrow \rho_{kl}\delta_{kl}$, and finally applying a Markovian approximation we obtain the simple master equation used by Weinberg.


About the other queries in the question, let me say that the H-theorem has the same limited validity than the approximated master equation used for his derivation. More general irreversible master equations are required for many phenomena in condensed phase systems, and those equations predicts all kind of complex evolutions for the entropy (or the H-function). In the general case the entropy of a 'closed' (I guess you mean isolated) system doesn't increase monotonically, and it is possible that there is no local extrema. It is needed to study each system and experimental situation individually.

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protected by ACuriousMind Dec 13 '16 at 21:20

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