In the algebraic approach to QM and specially to QFT one defines KMS states, for example, as Wald does in his lecture notes:

Let $\mathscr{A}$ be one $\ast$-algebra describing a system and $\alpha_t : \mathscr{A}\to \mathscr{A}$ a one-parameter family of automorphisms. In this situation, a state $\omega : \mathscr{A}\to \mathbb{C}$ is called a KMS-state at inverse temperature $\beta$ with respect to $\alpha_t$ if the following two conditions are satisfied:

  1. For any collection of $a_i\in \mathscr{A}$, the function $\bar{t} = (t_1,\dots, t_n) \mapsto F_{a_1\dots a_n}(\bar{t})$ defined by $$F_{a_1\dots a_n}(\bar{t})=\omega(\alpha_{t_1}(a_1)\cdots \alpha_{t_n}(a_n))$$ has an analytic continuation to the strip $$\mathfrak{T}^\beta_n = \{(z_1,\dots,z_n)\in \mathbb{C}^n | 0 < \operatorname{Im}(z_j)-\operatorname{Im}(z_i) < \beta, 1\leq i\leq j\leq n\},$$ this function is required to be bounded and continuous at the boundary.

  2. On the boundary, we have $$F_{a_1\dots a_n}(t_1,\dots, t_{k-1},t_k+i\beta,\dots, t_n+i\beta)= F_{a_k\dots a_n a_1\dots a_{k-1}}(t_k,\dots, t_n,t_1,\dots, t_{k-1})$$

The KMS states are used as a generalization of thermal states for situations on which no density matrix exists.

Now I simply can't understand why this is a good way to characterize thermal states. In QM we can define a thermal state by a density matrix $\rho$ of the form

$$\rho = \dfrac{1}{Z}e^{-\beta H}$$

where $H$ is the Hamiltonian operator and $Z$ the partition function. Now there seems to be a huge gap between this and the definition of KMS states.

On Wikipedia's page the author tried to give some explanation, basically by saying that in a thermal state we have

$$\langle \alpha_t(A)B\rangle =\langle B\alpha_{t+i\beta}(A)\rangle$$

where $\alpha_t$ is the time-evolution operator for a time interval $t$. He also states that the function $z\mapsto \langle \alpha_z(A)B\rangle$ is analytic on $-\beta < \operatorname{Im}(z)\rangle < 0$ while $z\mapsto \langle B\alpha_z(A)\rangle$ is analytic on $0 < \operatorname{Im}(z)<\beta$.

In other words, a usual thermal state is a KMS state.

Now, these are not obvious properties from a physics point of view. Futhermore, it is not clear why: "a thermal state satisfies (1) and (2)" be enough justification for "(1) and (2) characterize thermal states".

So what is the intuition behind KMS states? How can one arrive at this definition, if trying to describe thermal states without density matrices? Why the KMS condition implies thermal behavior for the system under consideration?


2 Answers 2


The KMS condition (which you consider not intuitive) is equivalent to the Gibbs condition $e^{-\beta H}$ on matrix algebras (finite-dimensional quantum systems). It seems that only implication of the KMS condition from the Gibbs state has been emphasized in the literature. I see that such quick introduction of the KMS condition is not so persuasive.

However, the KMS condition is considered as a well-founded characterization of equilibrium states for large quantum systems of infinite number of degrees of freedom. For example, it is equivalent to the energy-entropy balance EEB criterion (which essentially says minimum free-energy) and a form of the second law of thermodynamics as in Passive states and KMS states for general quantum systems by W. Pusz and S. L. Woronowicz which is available from projecteuclid.org/euclid.cmp for free.

There are some books. A famous one is a monograph by Bratteli-Robinson. However it has a reputation of being mathematical oriented, though its writing style is solid and the coverage is rather comprehensive. For physics oriented readers I may suggest

  • Quantum Mechanics and Its Emergent Macrophysics by G. L. Sewell.
  • Many-Body Boson Systems: Half a Century Later by André F. Verbeure

The following is cited from the book by Verbeure:

On the other hand, the quantum mechanical KMS conditions preach such a self-evident physical interpretation of equilibrium -- Nevertheless we are convinced that there are good reasons popularize and prompt a number of general basic ideas of this algebraic approach,---

I hope that my answer will be some help to your question.


The problem with the Gibbs state $e^{-\beta H}/tr(e^{-\beta H})$ is that it does not exist for infinite numbers of degrees of freedom (which you need for example if you want to see a phase transition): Not only that the trace of $e^{-\beta H}$ diverges, also $H$ itself is ill defined (consider for example an infinite Ising model where it would be an infinite sum of $\pm 1$ which does not converge and is typically infinite).

The physicists' typical reaction is to regularize both infinities and to hope that details will not matter. But in this case they typically do: The formula for the Gibbs state suggests that given $\beta$ and $H$ there is a unique equilibrium state. But at a phase transition that is not the case: For example, at zero Centigrade water can both be solid or liquid (or any mixture). Both state have the same $\beta$ and $H$ so the difference can only be "in the details of the regularization".

If you consider KMS states (maybe not for water but for model systems like quantum spin systems or free particles Fermions or Bosons, the latter having a Bose-Einstein condensate phase transition) you find exactly that: There is more than one KMS state for a given $\beta$.

Besides the properties mentioned above (Gibbs states being KMS, KMS being essentially equivalent to passive i.e. states with a second law and the energy-entropy balance inequality) you can also show certain stability properties of KMS state (that you would expect of thermal equilibrium states). Those are typically that you study what happens when you start with a KMS state but then apply a perturbation to time evolution (think: Hamiltonian). Under some technical assumptions (that are supposed to guarantee the analogue of ergodicity) you can show that your state approaches again a KMS state over time for the perturbed time evolution (note that the time evolution is in the definition of KMS).


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