KMS States and Thermal States

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?

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.