4
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.