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I have a central model of this form $$H = \sum_{i=1}^{N} S^z_0\otimes S^z_i$$ where the $S^z_i$ acts on the $i$th element of the environment, i.e. the Hilbert space is of the following form $\mathcal{H}_0 \otimes_{i=1}^N\mathcal{H}_{E_i}$.

I want to study the integrability of this model, more specifically I'm interested in the conserved charges. The model is very basic and I'd like us it as a test bed.

I found a set of $2N_E + 1$ conserved charges, i.e. $$\{S_0^z +S^z_i,S_0^2,S_i^2\}$$ Since the number of charge scale as the dimension of the system, does that prove that the system is integrable and these are the conserved charges?

My interest is to study Generalised Gibbs Ensemble (GGE) of this system and see if I can approximate the state locally by this GGE.

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Actually, since by construction $S_k^2$ ($k=0…N$) are proportional to identity they don’t really count. More simply, you have $N+1$ conserved quantities $S_k^z$. They for a complete set of observables commuting with the Hamiltonian, i.e. you can uniquely label a basis with their simultaneous eigenvalues.

Btw, you can rewrite the Hamiltonian to be: $$ H=S_0^z\sum_{i=1}^NS_i^z $$

Hope this helps.

Answer to comment

Talking about integrability in your case is a bit overkill. Indeed, your system is closer to a independent particle system. The dynamics are reduced to independent one body dynamics since the energy eigenstates are tensor products of $S^z$ eigenvectors. There is no need for the Bethe ansatz. Things are typically more interesting when considering 1D lattice models or when you mix up the spin components.

I guess that the generalized Gibbs ensemble in your case is: $$ \rho = \frac{1}{Z}e^{\sum_{k=0}^N h_k S_k^z} $$

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  • $\begingroup$ Does this mean that these are the conserved charge that make the system integrable? Do you also know if the GGE can be written in terms of these conserved quantities? $\endgroup$ Commented Jun 1, 2023 at 16:19
  • $\begingroup$ This answer is correct and complete. The model suggested in the question is integrable in the trivial sense of being exactly solvable. But the eigenstates correspond to a free model, and are 1-body reducible (not 2-body reducible like the Heisenberg chain, no BA required). The only GGE is the one written down. The "charges" listed in the original question are not linearly independent if the spins are two-level. For spin-1 and higher, one can also include powers of $S^z_k$. $\endgroup$ Commented Jun 9, 2023 at 23:04

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