Connection between bonddimension of a matrix product state and entanglement
The entanglement of any region in a matrix product state of bond dimension $D$ is bounded by $S\le 2\log D$. Thus, in order to simulate a system with a lot of entanglement, the bond dimension (and thus the memory and time of the computation) will grow exponentially with the entropy.
Conversely, we know that if for a state $\vert\psi\rangle$ the $\alpha$Renyi entropy (for $\alpha<1$) of any block is bounded by a constant, then an efficient approximation of $\vert\psi\rangle$ by an MPS exists (i.e., where $D$ scales polynomially in the system size and the inverse precision). This is proven in https://arxiv.org/abs/condmat/0505140. The general connection between entropy scaling and the approximability by MPS is discussed in https://arxiv.org/abs/0705.0292, where in particular examples of states with an area law for $\alpha\ge1$ are given which cannot be approximated efficiently by MPS. (Note that for $\alpha=1$, there is (afaik) no translation invariant example.)
Joshuah Heath
Updated on August 29, 2020Comments

Joshuah Heath about 3 years
The bond dimension is the dimension of the truncated matrix product state (MPS). Let us assume that I am simulating some manybody system with high entanglement via the density matrix renormalization group (DMRG). At a conference I was attending a few days ago, someone told me that the bonddimension increases with the amount of entanglement in the system. Therefore, simulating a highly entangled system with DMRG requires a huge amount of computational time. However, how, exactly is the bonddimension connected to a system's entanglement?

Meng Cheng over 8 yearsLarger bond dimension means more entanglement. Think of a 1D MPS as starting as many maximally entangled pairs between "virtual particles" on sites, the number of which is roughly the bond dimension. Then project the virtual particles back to physical states. Apparently the more virtual particles you have, the more entangled the virtual states can get.


Everett You over 8 yearsWhy is there a factor 2 in front of $\ln D$? Given the bond dimension $D$, there are at most $D$ Schmidt eigenvalues of $D^{1}$, which lead to an entropy $S=D(D^{1}\ln D^{1})=\ln D$. I am wondering why your formula has an additional factor 2.

Norbert Schuch over 8 years@EverettYou If you cut a region in the middle of an MPS, you get a factor of $2$ since you have bonds on both ends of the region. If you cut an infinite/OBC MPS in two parts, the bound is indeed $D$.

Abhinav about 7 yearsThank you for this answer. Could you add a reference for the converse statement? And are there counterexamples for $\alpha=1$?

Norbert Schuch about 7 years@Abhinav I have added the reference, as well as a reference to the paper which contains the corresponding counterexamples.