Framework of 'Thermodynamic cost of creating correlations' by Huber et al

Question

Currently, I am reading Thermodynamic cost of creating correlations by Huber et al.

I am having hard time understanding the framework of the paper (Section 2). I am sure I am missing something very obvious.

The Section 2 starts as follows.

We consider a global system comprised of $n$ initially uncorrelated $d$-dimensional quantum systems. Each system is taken to have the same (arbitrary) local Hamiltonian $H={{\sum }_{i}}{{E}_{i}}|i\rangle \langle i|$, and the same temperature ${{k}_{{\rm B}}}T=1/\beta $. Hence the initial state of the global system is

and $\mathcal{Z}={\rm Tr}\left( {{{\rm e}}^{-\beta H}} \right)$ is the partition function. When discussing qubits we will denote by $E$ the energy of the excited state and

the ground state probability.

My questions:

1. Should I assume that $H={{\sum }_{i}}{{E}_{i}}|i\rangle \langle i|$ is actually $H={{\sum }^n_{i = 1}}{{E}_{i}}|i\rangle \langle i|$?
2. The authors say that the initial state of the global system is $\rho_i$. Why do we need the index $i$ if it is global? Or is it a different $i (initial)$ than the index in question $i$?

Answer 1

1 - yes

2 - $\rho_i$ it is the initial probability of the system being in state $| i \rangle $. Each state has its own initial probability, so that you have $n$ different $\rho_k$ for $1\le k\le n$.

Answer 2

1. No. n is the number of systems. The n systems are initially uncorrelated. These n systems are allowed to interact which makes them entangled. Each system is a d dimensional system meaning it can be described by a d-dimensional Hilbert space. So we can find d energy eigenvectors $\lvert i \rangle $ and expand the Hamiltonian in terms of projection onto these d eigenvectors (so the sum is from 1 to d).
2. Yes. The i here is a different i standing for "initial state" before the interaction, with f the "final state" after the interaction. If this was a pure quantum system we would describe it by a state vector $\lvert \psi_i \rangle $ which would then evolve by some unitary operation to $\lvert \psi_f \rangle = U \lvert \psi_i \rangle $.
   However here we want to consider a system in a mixed state so we describe the quantum state by a density operator $\rho$ (in the special case of a pure state $\rho = \lvert \psi \rangle \langle \psi \rvert $ ). Since $\rho$ is an operator it will evolve under U according to $ \rho_f = U \rho_i U^\dagger$.
   Since initially the n systems are uncorrelated we can write the density operator for the complete n systems together as the tensor product of the density matrix for each individual system (which are assumed to be identical), hence $\rho_i$ has the form $ \rho_i = \tau^{\otimes n}$.