Lecture 0’s cradle did not thermalise, and Lecture 18 solved a chain with a conserved occupation for every mode. So the question is

The answer: local observables, when they relax at all, to the average in the ensemble fixed by a complete family of conserved charges (Definition 19.5); for the magnetisation after a field quench of the Ising chain this is proved (Theorem 19.7), and the Gibbs ensemble gets it wrong (Figure 19.2). The worked example is that quench.

From here on the system is a ring of $N$ sites, each carrying a spin $\tfrac12$: the Hilbert space is $(\C^{2})^{\otimes N}$, site $N+1$ is site $1$, and $\tau$ is the translation that moves every site one step, $n\mapsto n+1$. An operator acts on a set of sites if it is the identity on every other factor.

The fixed range is the whole content: with $r=N$ allowed, any translation-invariant operator $A$ would qualify, as the sum of the translates of $A/N$. The total spin $S^{z}=\sum_n S^{z}_n$ is a local charge of range $1$, and so is a nearest-neighbour Hamiltonian such as Lecture 18’s Ising chain, of range $2$. Local charges are extensive: $Q$ is a sum of $N$ terms of norm $\norm{q}$, so $\norm{Q}\le N\norm{q}$, and in a translation-invariant state $\langle Q\rangle=N\langle q\rangle$ — proportional to the length, as energy and particle number are. And local charges are rare.

The chain is the XXZ chain with a next-nearest-neighbour coupling, $$H=\sum_{n=1}^{N}\big(S^{x}_nS^{x}_{n+1}+S^{y}_nS^{y}_{n+1}+\XXZ\,S^{z}_nS^{z}_{n+1}\big) +J_2\sum_{n=1}^{N}\big(S^{x}_nS^{x}_{n+2}+S^{y}_nS^{y}_{n+2}+\XXZ\,S^{z}_nS^{z}_{n+2}\big).$$(19.1) At $J_2=0$ it is the XXZ chain, which the algebraic Bethe ansatz solves (Lecture 15’s reading); for $J_2\ne0$ it is not known to be integrable. Figure 19.1 compares $J_2=0$ with $J_2=0.5$. The anisotropy is $\XXZ=0.55$, away from the values $\cos(\pi p/q)$ with small $q$ at which XXZ acquires extra degeneracies.

Prepare a system in a pure state $\ket{\psi_0}$ that is not an eigenstate of $H$ and let it evolve. Nothing about $\ket{\psi(t)}$ relaxes: the state stays pure and, on a finite system, almost periodic. The question is about a local observable $\mathcal O$ (Definition 19.1), in the thermodynamic limit taken first. Expanding in eigenstates $H\ket n=E_n\ket n$, $$\bra{\psi(t)}\mathcal O\ket{\psi(t)} \;=\;\sum_{n}\abs{c_n}^{2}\bra n\mathcal O\ket n \;+\;\sum_{m\ne n}c_m^{*}c_n\,e^{-\ii(E_n-E_m)t}\bra m\mathcal O\ket n , \qquad c_n=\braket{n}{\psi_0} .$$(19.2)

So whatever a local observable settles down to is fixed by $\rho_{\mathrm{d}}$ — exponentially many numbers, and no ensemble worth the name. (Not by the $\abs{c_n}^{2}$ alone: inside a degenerate eigenspace the relative phases survive dephasing, and a spectator spin prepared in $\ket{{\pm}x}$ makes that visible.) Every relaxation statement is the claim that far fewer of those numbers matter.

If (19.3) holds and $\ket{\psi_0}$ lies in one symmetry sector with subextensive energy fluctuations, Proposition 19.3 averages a smooth function over a narrow window and the answer is the microcanonical one: the system thermalises, in the only sense a closed quantum system allows. Nothing has been proved — ETH is a hypothesis about a Hamiltonian, checked numerically model by model, and (19.3), sector by sector, is exactly the statement a figure can test (Figure 19.1).

The left panel is not an accident of the observable. A chain with about $N$ local conserved charges, all commuting, can have its eigenstates labelled by all of them at once. What is expected to replace (19.3) there is a generalised eigenstate thermalisation: $\bra n\mathcal O\ket n$ a smooth function of the densities of a complete family of conserved charges — the strictly local ones for the Ising chain, quasi-local ones as well for the XXZ chain (Lecture 20’s reading) cited: Caux and Essler 2013. That reduces to a function of the energy alone only if the other densities are fixed by it, and they are not: two eigenstates at the same energy density generically carry different values of the other charges. The charges remember which state the system was prepared in, and a local observable can see the memory. The example this lecture proves is the Ising chain of Lecture 18, whose local charges are those of Proposition 19.8.

A quadratic Hamiltonian makes the Heisenberg equations linear: $\dd\gamma_k/\dd t=\ii\comm{H}{\gamma_k}=-\ii\varepsilon(k)\gamma_k$, so each mode turns at its own frequency and ignores the others — the quantum form of the linear flow $\dot\varphi=\freq$ on a torus, and the reason there is no scattering.

(Lecture 18’s caveats say why this is special: the Heisenberg chain’s fermions interact.)

With one charge, $Q_1=H$, this is the Gibbs state and $\beta_1=1/T$; the question is what the others do. The claim to be tested — that a local observable relaxes to its $\rho_{\mathrm{GGE}}$ average — can be settled by calculation on Lecture 18’s chain.

Take the transverse-field Ising chain (18.1) at coupling $J$, start in the ground state at field $h_0$, and at $t=0$ change the field to $h$. Both Hamiltonians are quadratic in the same fermions $c_k$: only the Bogoliubov angle moves. Write $\theta_k$ for the post-quench angle of (18.15), $\theta^{0}_k$ for the pre-quench one, and $$\chi_k=\theta_k-\theta^{0}_k ,\qquad \cos2\chi_k=\frac{a_ka^{0}_k+b_k^{2}}{\varepsilon_k\varepsilon^{0}_k},\qquad \sin2\chi_k=\frac{4J(h_0-h)\sin k}{\varepsilon_k\varepsilon^{0}_k} ,$$(19.5) with $a_k=2(h-J\cos k)$, $b_k=2J\sin k$, $\varepsilon_k=\sqrt{a_k^{2}+b_k^{2}}$ as in §18.5, and $a^{0}_k$, $\varepsilon^{0}_k$ the same at $h_0$. The rotation taking pre-quench modes to post-quench ones is the rotation by the difference angle, $\gamma_k=\cos\chi_k\,\gamma^{0}_k-\ii\sin\chi_k\,(\gamma^{0}_{-k})^{\dagger}$, and the pre-quench ground state is the $\gamma^{0}$ vacuum.

The difference angle is not a bookkeeping convention: the sum angle gives a well-formed, plausible and wrong $n_k$, off by as much as $0.48$ for $J=1$, $h_0=0.4\to h=2.1$.

The transverse magnetisation $m^{z}=\tfrac1N\sum_n\sigma^{z}_n$ is the density of a one-site operator, and in a translation-invariant state it is $\langle\sigma^{z}_1\rangle$ exactly; (18.4) makes it $1-\tfrac2N\sum_kc_k^{\dagger}c_k$.

The hypotheses are spent there: at $h_0=J$ the zero of (19.9) is simple and the decay is $t^{-1}$, at $h=0$ the dispersion is flat and nothing decays, and at $h_0=0$ there is no single ground state.

Theorem 19.7 takes $N\to\infty$ first and $t\to\infty$ second, and the two do not commute: at finite $N$ the sum (19.7) revives, as the pale curve of Figure 19.2 does, and only its time average converges.

Definition 19.5 asks for charges, and a single $n_k$ is not an observable of the chain: it is a sum over all $N$ sites whose coefficients never decay. Lecture 18’s reading built the local charges by weighting the $k$-block of (18.13); this is that weighting.

The $\varepsilon$ weight is not decoration. Drop it, and $Q[\cos mk]$ is not local: its coefficients decay exponentially, at the rate $\abs{\ln(h/J)}$ — the branch points of $1/\varepsilon$ sit at $z=e^{\ii k}=h/J$ and $z=J/h$, and whichever is nearer the unit circle sets the rate — without ever vanishing. Such a charge is quasi-local, the class that Lecture 20’s reading shows the XXZ chain needs. Both ends of the range are special: at $h=J$ the two branch points meet on the circle and the coefficients decay only like $1/d$, while at $h=0$, where $\varepsilon=2J$ is constant, the weight is a number and every trigonometric polynomial is strictly local again.

None of this weakens Theorem 19.7: knowing every $Q[\varepsilon P]$ is knowing every Fourier coefficient of $\varepsilon(k)n(k)$, and $\varepsilon>0$ almost everywhere, so the strictly local charges fix $n_k$ almost everywhere and with it every local observable’s relaxed value. Numerically, keeping $Q[\varepsilon\cos mk]$ for $m\le M$ and maximising the entropy gives $m^{z}$ to $1.7\times10^{-3}$ at $M=4$ and $8\times10^{-8}$ at $M=16$ — and at $M=0$, where only the energy is kept, that is the Gibbs ensemble, with exactly the error of Figure 19.2.

• Lecture 0’s question, answered in one model: after a quench the Ising chain’s magnetisation relaxes, and to the generalised Gibbs value, not the Gibbs one (Theorem 19.7, Figure 19.2).

• Why: the occupations $n_k$ are conserved (Proposition 19.6), and the local charges fix them (Proposition 19.8); a local observable remembers them.

• The ensemble that Lecture 20 carries to the interacting gas, where one function, the rapidity density, plays the part of the $n_k$.

• Status. Proposition 19.3 (proof in a reading), Proposition 19.6 and Theorem 19.7 are proved; Proposition 19.8 is sketched. ETH is a hypothesis, checked numerically; generalised ETH is cited; the GGE’s definition is cited.

• The hypotheses of Theorem 19.7 are spent. At $h_0=J$ the decay is slower, at $h=0$ it does not happen, at $h_0=0$ there is no single ground state, and at $h=J$ the observable $m^{z}$ cannot tell the Gibbs and generalised ensembles apart.

• The order of the limits: $N\to\infty$ first, then $t\to\infty$. At finite $N$ the magnetisation revives, and only its time average converges.

• Which charges. For the Ising chain the strictly local charges fix the occupations almost everywhere (Proposition 19.8); for the XXZ chain the ultralocal charges are not enough, and quasi-local ones are needed (Lecture 20’s reading).

• Every statement about a non-integrable chain is numerical, and a state that carries a thermal state’s charge densities relaxes to thermal values even in an integrable chain.