ETH states that for a system, all of its eigenstates thermalize. To be more specific, consider an energy eigenstate of the full system $H|n\rangle=E_n|n\rangle$. If the full system is in this eigenstate, $\rho=\rho^{(n)}=|n\rangle\langle n |$. Denote the system we are interested in as $A$ and the rest of the system as $B$, which is the environment. The reduced density matrix of $A$ is $\rho_A^{(n)}=\mathrm{Tr}_B\left(|n\rangle\langle n |\right)$. ETH states that $\rho_A^{(n)}$ looks thermal: $\rho_A^{(n)}=\rho_A^{\mathrm{eq}}(T_n)$, where $\rho^{\mathrm{eq}}(T) = Z^{-1}\exp(-H/k_BT)$ is the thermal equilibrium density matrix. $T_n$ is deternmined by the energy (density) of the eigenstate $E_n$.

Now, consider the ground state of the full system $|0\rangle$.

  • Since it is a ground state, the full system must be the lowest temperature possible, which is the zero temperature. Hence $\rho_A^{(0)}=\rho_A^{\mathrm{eq}}(T_0)$ is a thermal equilibrium at $T=0$, which is a pure state with only the ground state.

  • However, suppose $|0\rangle$ is not a product state (which is often the case, for example, the superfluid ground state of Bose-Hubbard model) so that the reduced density matrix of $A$ is a mixed state.

Since $\rho_A^{(0)}$ cannot be both pure and mixed, a contradiction occurs. What is wrong with the previous argument?


1 Answer 1


You need to be careful about how you go from the full system to the subsystem $A$. You define $\rho^\text{eq}(T) = Z^{-1} \exp(-H/T)$ as the thermal state of the whole system, but then you use $\rho_A^\text{eq}(T)$ without defining how you are reducing the density matrix of the whole system onto just the subsystem. There are two reasonable ways to do so:

(1) You could project $\rho^\text{eq}(T)$ onto the subsystem: $\rho_A^\text{eq}(T) = \text{Tr}_B\ \rho^\text{eq}(T)$. In this case, at zero temperature $\rho^\text{eq}(T) = |0\rangle \langle 0 |$ is indeed a pure state, but its partial trace over $B$ is not (necessarily). In this case, ETH is trivially true at zero temperature because the (whole system) pure state $|0 \rangle \langle 0 |$ and the thermal state are exactly the same.

(2) You could first project the Hamiltonian into a truncated Hamiltonian $H_A$ supported only on subsystem $A$, then exponentiate to get an approximate thermal state $\rho_A^\text{approx}(T) = \exp(-H_A/T)\ /\ \text{Tr}(\exp(-H_A/T))$. The order of projection and exponentiation matters: this is not the exact reduced density matrix $\rho_A^\text{eq}(T)$ that ETH uses, because you've lost $H_B$ and the bonds connecting subsystems $A$ and $B$. In fact, the projection procedure to take $H$ down to $H_A$ is not even fully defined as stated. Usually for high temperatures the $\rho_A^\text{approx}(T) \approx \rho_A^\text{eq}(T)$, but at low temperatures the approximation fails.

See http://arxiv.org/abs/1503.00729 eqs. (2a) and (2b) for more details.


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