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Take an injective, translation invariant MPS with transfer matrix $E = \sum_\sigma \overline{A^\sigma} \otimes A^\sigma$ (i am using the terminology of https://arxiv.org/abs/quant-ph/0410227 , eq. (6)).

Then (by injectivity), $E$ will have one eigenvalue $1$ and all other eigenvalues $\lambda_i$ have $|\lambda_i|<1$. This corresponds to exponential clustering of correlation functions, in particular, the smallest correlation length in the system is given by $1/\xi = - \log |\lambda|$, with $\lambda$ the second largest eigenvalue.

The point is that thus this part of the spectrum of $E$ has direct physical meaning: given a state with a given correlation length, we know that we should use a MPS with some given gap between the first and the second largest eigenvalue.

From there i would be interested if there is more information with direct physical interpretation in the spectrum of $E$. I will state my problem with decreasing generality:

  1. Assume $|\lambda_1| \leq |\lambda_2| \leq \cdots$ is the spectrum of $|E|$, and assume we have a matrix product state of large bond dimension with a given correlation length, gap above the ground state, Renyi entropy etc. Can we say sth about the decay properties of $|\lambda_n|$?

  2. More explicitely i am wondering if it is possible to find an $N$ depending only on a "thermodynamic" quantities (like correlation lengths, energy gaps etc.) s.t. $$ \sum_{n > N} |\lambda_n| < 1 \ .$$

  3. I want to do this for the following reason: given an injective matrix product state $\rho$ on a subsystem $X$, it holds that $$ \lim_{X \rightarrow \mathbb{Z}} \text{Tr}(\rho_X^2) = C = \text{const.}$$ Is it possible to quantify how fast this limit is approached, again in terms of correlation lengths, gaps etc? i.e. can one find some decaying function $f(x) > 0$ s.t. $$ |\text{Tr}(\rho_X^2) - C | \leq f(|X|) \ ,$$ where the form of $f$ depends on properties of the state alone (and not the tensor network chosen to represent it)?

I can get more explicit on the last point. Assume the MPS has bond dimension $D$ and correlation length $\xi_D$, then:

$$ |\text{Tr}(\rho_X^2) - C | \leq D e^{-|X|/\xi_D} \ , \quad (1) $$

but this explicitely depends on the bond dimension of the chosen matrix product representation.

EDIT: I have a sketchy idea of what i want to do... so here is said sketch.

I interpret the results of this paper https://arxiv.org/abs/cond-mat/0505140 as: Given a state $\psi$ satisfying an area law for the Renyi-entropy $S_t$ for some $t < 1$, i can find a translation invariant MPS state $\psi_D$ which approximates $\psi$ on any subset. Denote the reduced density matrices to a subset $X:=\{1,2,\cdots L\}$ by

$$ \rho_L := \text{Tr}_{X^c}(|\psi\rangle \langle \psi|) \ , \quad \rho_{L,D} := \text{Tr}_{X^c}(|\psi_D\rangle \langle \psi_D|) \ .$$

Then the statement is that there are $r \in (1, \infty)$ and constants $C(r,\psi)$ such that

$$ \| \rho_L - \rho_{L,D} \|_1 := \text{Tr}| \rho_L - \rho_{L,D} | \leq C(r,\psi) L^2 D^{-r} \ . \quad (2) $$

I think this should be true; in the paper i mention this is similar to equation (5), only that there only the case $L=2$ is considered. In any case, by blocking, i can reduce the general $L$ case to the $L=2$ case, and that gives the $L^2$ dependence on the r.h.s. of the above equation. Then furthermore i used the estimate of Lemma 2 to upper bound what in the paper is called $\epsilon(D)$, and i just assume that by choosing some appropriate prefactors the r.h.s. of equation (5) can be brought in such a form. It seems that in this case, $C(r,\psi)$ depends on $r$ and on the upper bound of the Renyi entropy, also the dimension of the on-site Hilbert spaces. In particular it should not depend on $D$. But i am stressing that this part i am not very sure about.

However, if this is indeed true, i can do the following:

$$ | \text{Tr}(\rho_L^2) - \text{Tr}(\rho_{L+1}^2)| \leq \\ \leq | \text{Tr}(\rho_L^2) - \text{Tr}(\rho_{L,D}^2)| + | \text{Tr}(\rho_{L+1}^2) - \text{Tr}(\rho_{L+1,D}^2)| +| \text{Tr}(\rho_{L,D}^2) - \text{Tr}(\rho_{L+1,D}^2)| \leq \\ \leq \| \rho_L - \rho_{L,D}\|_2 + \| \rho_L - \rho_{L,D}\|_2 + | \text{Tr}(\rho_{L,D}^2) - \text{Tr}(\rho_{L+1,D}^2)|$$

Now for the first two terms i use $(2)$ and for the last term i use (1):

$$ | \text{Tr}(\rho_L^2) - \text{Tr}(\rho_{L+1}^2)| \leq 2 C(r,\psi)(L+1)^2 D^{-r} + 2 D e^{-L/\xi_D} \ .$$

Now assume that the state $\psi$ has correlation length $\xi$. Then pick some $\zeta > \xi$, and now let $D$ grow with $L$ in the following way:

$$ D(L) = e^{L/\zeta} \ . $$

Since $\psi_D$ converges to $\psi$ for $D\rightarrow \infty$, the correlation length $\xi_D$ has to converge to $\xi$. Hence there is only a finite number of instances of $L$ s.t. $1/\xi_{D(L)} - 1/\zeta < 0$. Thus, we can find some constants $C',\zeta'$ s.t.

$$ | \text{Tr}(\rho_L^2) - \text{Tr}(\rho_{L+1}^2)| \leq C' e^{-L/\zeta'} $$

Hence the limit (which will be denoted by $k$) exists and

$$ | \text{Tr}(\rho_L^2) - k| \leq C'' e^{-L/\zeta'} $$

for some additional constant $C''$.

So this is how far i got; i am not sure whether equation (2) holds. Note that unless the bound on the right hand side does not grow exponentially with $L$, the argument still works, though.

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  • $\begingroup$ In arxiv.org/abs/cond-mat/0701055, Hastings gives an example for an MPS where the entanglement is exponentially larger than the correlation length. Whether this also implies a counterexample to your point 3, I don't know. (I would however suspect yes. In fact, I suspect that there are correlation length $0$ MPS where the entropy has a finite length scale.) $\endgroup$ – Norbert Schuch Apr 3 at 20:59
  • $\begingroup$ Update: Isn't your point 3 a triviality? Basically, isn't this just rephrasing the definition of the limit in terms of f(|X|)? As long as you don't talk about convergence rates, I don't see what your (ultimate) question is $\endgroup$ – Norbert Schuch Apr 3 at 21:04
  • $\begingroup$ My question is about convergence rates... Saying that a certain quantity converges to another obviously implies that their difference decays, but i was wondering what can be said about the speed of this convergence. I mean i want to know the form of f independently from the left hand side. I will look into the Hastings example, thanks! $\endgroup$ – Lorenz Mayer Apr 3 at 21:19
  • $\begingroup$ Well, the way you state it in point 3 exactly does not ask about convergence rates! If it were, f(x) had to be independent of the state (that it is indep. of the representation is obvious). $\endgroup$ – Norbert Schuch Apr 3 at 21:45
  • $\begingroup$ ... the way it is currently put it is unclear what you are after, since f(x) will have to depend on sth. (e.g. the correlation length etc.) -- what do you have in mind, what would, and what wouldn't, make you happy? $\endgroup$ – Norbert Schuch Apr 3 at 21:49

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