Correction to: Thermalization and Canonical Typicality in Translation-Invariant Quantum Lattice Systems

1 Correction to: Commun. Math. Phys. 340, 499–561 (2015) https://doi.org/10.1007/s00220-015-2473-y

In our result about dynamical thermalization, the proof of the upper bound on the time average of the distance between the local evolved state $\rho ^{(n)}(t)$ and the time-averaged state $\rho _\mathrm{avg}^{(n)}$ is wrong. While it is correct that this distance tends to zero for block size $|\Lambda _n|\rightarrow \infty$ (see corrected proof below), it is unclear whether it can be shown that this happens exponentially fast in $|\Lambda _n|$. This affects Theorem 31, and hence also Theorem 3 (the summary of Theorem 31) and Theorem 33 (a small modification of Theorem 31).

This mistake is due to an error in Ref. [3] which we have used in our proof of Lemma 30. Ref. [3] claims that the Rényi entropy $H_q$ is convex in its parameter q, which is incorrect. This claim has been corrected in an erratum published on the author’s homepage [4], but we became aware of this only recently.

We give a corrected version of Theorem 31 of our paper [1] in Theorem 4 below. Its summary (and hence the correction of Theorem 3 of our paper) reads as follows.

Theorem 1

(Correction of [1, Theorem 3]). If there is a unique equilibrium state around inverse temperature $\beta :=\lim _{n\rightarrow \infty }\beta _n$, if the (possibly pure) initial state has close to maximal population entropy, in the sense that

$$\begin{aligned} {\bar{S}}(\rho _0^{(n)})\ge S(\gamma _{\Lambda _n}^p (\beta _n))-o(|\Lambda _n|), \end{aligned}$$

and if each $H_{\Lambda _n}^p$ is non-degenerate with uniformly bounded gap degeneracy $\sup _n D_G(H_{\Lambda _n}^p)<\infty$, then unitary time evolution thermalizes the subsystem $\Lambda$ for most times t:

$$\begin{aligned} \left\langle \left\| \mathrm{Tr}_{\Lambda _n\setminus \Lambda } \rho ^{(n)}(t)-\mathrm{Tr}_{\Lambda _n\setminus \Lambda } \frac{\exp (-\beta _n H_{\Lambda _n}^p)}{Z_n}\right\| _1\right\rangle&{\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}&0. \end{aligned}$$

The gap degeneracy [5] is defined as $D_G(H_{\Lambda _n}^p):=\max _E|\{(i,j)\,\,|\,\, i\ne j, E_i-E_j=E\}|$, with $E_i$ the eigenvalues of $H_{\Lambda _n}^p$.

This formulation differs from the old one in the following two ways. First, it does not give concrete bounds on the time-averaged distance between $\rho ^{(n)}(t)$ and its time average (it only says that this distance tends to zero for $n\rightarrow \infty$); second, it presumes that the gap degeneracy is uniformly bounded.

To prove its formal version (Theorem 4 below), we need two elementary lemmas.

Lemma 2

Let $\Phi$ be a translation-invariant finite-range interaction which is not physically equivalent to zero, and let ${\bar{u}}$ be some energy density for which there is a unique Gibbs state at inverse temperature $\beta ({\bar{u}})$. Then the real function $u\mapsto s(u)$ defined in [1, Lemma 9] is strictly concave at ${\bar{u}}$ in the following sense: If ${\bar{u}} =\lambda u_0+(1-\lambda )u_1$ for some $u_0<u_1$ and $\lambda \in (0,1)$ then $s({\bar{u}})>\lambda s(u_0)+(1-\lambda )s(u_1)$.

Proof

Let $u_0<u_1$ and $u=\lambda u_0+(1-\lambda )u_1$ for some $\lambda \in (0,1)$. Let $\omega _{\beta (u_0)}$ be an arbitrary Gibbs state with energy density $u_0$ at inverse temperature $\beta (u_0)$, and similarly $\omega _{\beta (u_1)}$. Set $\omega :=\lambda \omega _{\beta (u_0)}+(1-\lambda )\omega _{\beta (u_1)}$, a translation-invariant state. Since the entropy density is affine on the translation-invariant states ([2, Thm. IV.2.4]), we have

$$\begin{aligned} s(\omega )=\lambda \, s(\omega _{\beta (u_0)})+(1-\lambda )s(\omega _{\beta (u_1)})=\lambda s(u_0)+(1-\lambda ) s(u_1). \end{aligned}$$

By construction, $u(\omega )=u$. Thus, due to [1, Lemma 9], we have $s(\omega )\le s(u)$, hence $u\mapsto s(u)$ is concave.

Let us now apply the previous argumentation to the special case $u:={\bar{u}}$, an energy density with a unique Gibbs state. Suppose that $s({\bar{u}})=s(\omega )$. Then the variational principle ([1, Definition 6]) implies that $\omega$ is a Gibbs state at inverse temperature $\beta ({\bar{u}})$. But the set of Gibbs states at inverse temperature $\beta ({\bar{u}})$ is a face of the set of all translation-invariant states [2, p. 348], hence $\omega _{\beta (u_0)}$ and $\omega _{\beta (u_1)}$ must both be Gibbs states at inverse temperature $\beta ({\bar{u}})$, too. But these are distinct states, since they have different energy densities, contradicting the uniqueness of the Gibbs state at $\beta ({\bar{u}})$. Therefore $s({\bar{u}})>s(\omega )$, and we get the statement of strict concavity as claimed. $\quad \square$

Lemma 3

Let $\Phi$ be a translation-invariant finite-range interaction which is not physically equivalent to zero. Suppose that the maximal energy degeneracy of $H_{\Lambda _n}^p$ grows at most subexponentially in $|\Lambda _n|$, i.e. $\log \max \{{\mathrm {tr}}(\pi _i^{(n)})\}=o(|\Lambda _n|)$, where $(\pi _i^{(n)})_i$ denotes the eigenprojectors of $H_{\Lambda _n}^p$. Let $(\rho ^{(n)})_{n\in {\mathbb {N}}}$ be any sequence of $\Lambda _n$-translation-invariant states with

$$\begin{aligned} {[}\rho ^{(n)},H_{\Lambda _n}^p]=0,\quad S(\rho ^{(n)})\ge s\cdot |\Lambda _n|+o(|\Lambda _n|), \quad \mathrm{tr}(\rho ^{(n)} H_{\Lambda _n}^p)=u\cdot |\Lambda _n|+o(|\Lambda _n|), \end{aligned}$$

where $u\in (u_{\min }(\Phi ),u_{\max }(\Phi ))$ is an energy density such that there is a unique Gibbs state at inverse temperature $\beta (u)$, and $s=s(u)$. Then $\max _i {\mathrm {tr}}(\rho ^{(n)}\pi _i^{(n)}){\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}0$.

Proof

We can write u as some convex combination of two distinct energy densities in a small neighborhood of u, and then Lemma 2 implies that $s=s(u)>0$. Let us now argue by contradiction. Suppose that $\lambda ^{(n)}:=\max _i {\mathrm {tr}}(\rho ^{(n)}\pi _i^{(n)})$ does not converge to zero. Decompose the state $\rho ^{(n)}$ as follows:

$$\begin{aligned} \rho ^{(n)}=\lambda ^{(n)}\tau ^{(n)}+(1-\lambda ^{(n)})\sigma ^{(n)}, \end{aligned}$$

(1)

where $\tau ^{(n)}=\pi _i^{(n)}\rho ^{(n)}\pi _i^{(n)}/\lambda ^{(n)}$ (note that $\lambda ^{(n)}>0$), with $\pi _i^{(n)}$ the maximizing projector. If $\lambda ^{(n)}\ne 1$, define $\sigma ^{(n)}:={\bar{\pi }}_i^{(n)}\rho ^{(n)}{\bar{\pi }}_i^{(n)}/(1-\lambda ^{(n)})$, where ${\bar{\pi }}_i^{(n)}:={\mathbf {1}}-\pi _i^{(n)}$; if $\lambda ^{(n)}=1$, set $\sigma ^{(n)}={\bar{\pi }}_i^{(n)}/{\mathrm {tr}}({\bar{\pi }}_i^{(n)})$ (if n is large enough, then $\pi _i^{(n)}\ne {\mathbf {1}}$, hence this is well-defined). It follows that $\tau ^{(n)}$ and $\sigma ^{(n)}$ are mutually orthogonal $\Lambda _n$-translation-invariant states that commute with $H_{\Lambda _n}^p$.

The sequences of real numbers $S(\sigma ^{(n)})/|\Lambda _n|$, $\mathrm{tr}(\sigma ^{(n)}H_{\Lambda _n}^p)/|\Lambda _n|$, $\mathrm{tr}(\tau ^{(n)}H_{\Lambda _n}^p)/|\Lambda _n|$ and $\lambda ^{(n)}$ are all bounded (the latter sequence bounded away from zero by assumption). Thus, we can find a subsequence $(n_k)_{k\in {\mathbb {N}}}$ such that

$$\begin{aligned}&\lambda ^{(n_k)}{\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }}\delta >0,\quad \frac{1}{|\Lambda _{n_k}|} S(\sigma ^{(n_k)}) {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }}s_1,\quad \frac{1}{|\Lambda _{n_k}|} {\mathrm {tr}}(\tau ^{(n_k)}H_{\Lambda _{n_k}}^p){\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }}u_0,\quad \\&\quad \frac{1}{|\Lambda _{n_k}|} {\mathrm {tr}}(\sigma ^{(n_k)}H_{\Lambda _{n_k}}^p){\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }}u_1, \end{aligned}$$

where $s_1$, $u_0$, $u_1$ are real numbers, and $0<\delta \le 1$. Due to (1), computing the von Neumann entropy, we have $S(\rho ^{(n_k)})=\lambda ^{(n_k)} S(\tau ^{(n_k)})+(1-\lambda ^{(n_k)})S(\sigma ^{(n_k)})+{\mathcal {O}}(1)$. Since $S(\tau ^{(n_k)})\le \log {\mathrm {tr}}(\pi _i^{(n_k)})=o(|\Lambda _{n_k}|)$, this implies $s\le (1-\delta )s_1$. Thus, $s>0$ yields $\delta <1$. Similarly, computing the energy expectation value, we obtain $u=\delta u_0+(1-\delta )u_1$.

Suppose that $s_1\ge s(u_1)$, then $s_1-\beta u_1\ge p(\beta ,\Phi )$ for $\beta :=\beta (u_1)$, hence [1, Lemma 8] implies that we must have equality, i.e. $s_1=s(u_1)$. In summary, we conclude that $s_1\le s(u_1)$. Therefore

$$\begin{aligned} s(u)=s\le (1-\delta )s_1 \le \delta \, s(u_0)+(1-\delta )s(u_1). \end{aligned}$$

Since s is strictly concave at u due to Lemma 2 above, this is only possible if $u_0=u_1=u$. Hence

$$\begin{aligned} 0<s(u)\le (1-\delta )s_1 \le (1-\delta )s(u_1)=(1-\delta )s(u) \end{aligned}$$

which is a contradiction. $\quad \square$

This allows us to obtain a corrected version of [1, Theorem 31].

Theorem 4

(Correction of [1, Theorem 31]: Thermalization, periodic boundary conditions). Let $\Phi$ be a translation-invariant finite-range interaction which is not physically equivalent to zero. Suppose that the maximal energy degeneracy of $H_{\Lambda _n}^p$ grows at most subexponentially in $|\Lambda _n|$, i.e. $\log \max \{{\mathrm {tr}}(\pi _i^{(n)})\}=o(|\Lambda _n|)$, where $(\pi _i^{(n)})_i$ denotes the eigenprojectors of $H_{\Lambda _n}^p$, and $\sup _n D_G(H_{\Lambda _n}^p)<\infty$. Let $(\rho _0^{(n)})_{n\in {\mathbb {N}}}$ be some sequence of initial states on $\Lambda _n$ which have energy expectation value $U_n:={\mathrm {tr}}(\rho _0^{(n)}H_{\Lambda _n}^p)$ with density $U_n/|\Lambda _n|$ converging to some value $u\in (u_{\min }(\Phi ),u_{\max }(\Phi ))$ as $n\rightarrow \infty$, such that there is a unique Gibbs state around inverse temperature $\beta (u)$.

Define the ‘population entropy” ${\bar{S}}(\rho _0^{(n)}):=S(\lambda _1,\ldots ,\lambda _N)$, where S is Shannon entropy, and $\lambda _i:={\mathrm {tr}}(\rho _0^{(n)}\pi _i^{(n)})$ is the probability that the i-th level is populated. Suppose that for every n large enough, either $H_{\Lambda _n}^p$ is non-degenerate or every $\pi _i^{(n)}\rho _0^{(n)}\pi _i^{(n)}$ is $\Lambda _n$-translation-invariant. Then, determine the inverse temperature $\beta _n$ for which

$$\begin{aligned} {\mathrm {tr}}(H_{\Lambda _n}^p \gamma _{\Lambda _n}^p(\beta _n))=U_n,\quad \text{ where } \gamma _{\Lambda _n}^p(\beta _n):=\frac{\exp (-\beta _n H_{\Lambda _n}^p)}{Z_n}. \end{aligned}$$

If the initial states have close to maximal population entropy in the sense that

$$\begin{aligned} {\bar{S}}(\rho _0^{(n)})\ge S(\gamma _{\Lambda _n}^p(\beta _n))-o(|\Lambda _n|), \end{aligned}$$

then unitary time evolution $\rho ^{(n)}(t):=\exp (-itH_{\Lambda _n}^p)\rho _0^{(n)}\exp (it H_{\Lambda _n}^p)$ thermalizes the subsystem $\Lambda _m$ for most times t:

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\langle \left\| \mathrm{Tr}_{\Lambda _n\setminus \Lambda _m} \rho ^{(n)}(t)-\mathrm{Tr}_{\Lambda _n\setminus \Lambda _m} \frac{\exp (-\beta _n H_{\Lambda _n}^p)}{Z_n}\right\| _1\right\rangle =0, \end{aligned}$$

where $Z_n={\mathrm {tr}}(\exp (-\beta _n H_{\Lambda _n}^p))$, and $\langle \cdot \rangle$ denotes the average over all times $t\ge 0$. Furthermore, in this statement, $\beta _n$ can be replaced by $\beta :=\beta (u)$.

Proof

The only ingredient in the proof of [1, Theorem 31] that has to be corrected is the argument that lower-bounds the “effective dimension” $d_{\mathrm{eff}}$. The old proof erroneously claimed that $d_{\mathrm{eff}}$ grows exponentially in $|\Lambda _n|$, but this relied on a wrong claim about the Rényi entropy of Ref. [3]. We now give a simple alternative argument which makes use of the Rényi entropy $S_\infty (\lambda _1,\ldots ,\lambda _N)=-\log \max _i\lambda _i$ and the inequality $S_2\ge S_\infty$ [4]. Namely,

$$\begin{aligned} d_{\mathrm{eff}}=\exp (S_2(\lambda _1,\ldots ,\lambda _N))\ge \exp (S_\infty (\lambda _1,\ldots ,\lambda _N))=\left( \max _i\lambda _i\right) ^{-1} {\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}\infty \end{aligned}$$

according to Lemma 3 above, applied to the sequence of states ${\bar{\rho }}_0^{(n)}=\sum _i \pi _i^{(n)}\rho _0^{(n)}\pi _i^{(n)}$. Since we have assumed that the gap degeneracy is uniformly bounded, this is enough to show that $\rho ^{(n)}(t)$ is close to its time average for most times t if n is large. The rest of the proof works without modification (note that $\rho (\beta _n)$ should read $\gamma _{\Lambda _n}^p(\beta _n)$). $\quad \square$

Finally, [1, Theorem 33] has to be corrected analogously. We omit the obvious details.