Questions on Matrix Product States (Tensor Networks)

Question

I just have a few questions on Matrix Product States. I have learnt them from the point of view of performing cuts via Schmidt Decompositions, as seen in Chubb and Bridgeman 2017.

1. Is the bond dimension for a certain cut equal to the Schmidt Number across that cut? I have seen authors mention the fact that we can 'truncate' the diagonal matrix of Schmidt weights if a finite number of them are nonzero, but this confuses me, as (a) Can a Schmidt Weight even be zero? Surely at that point it is not part of the Schmidt Decomposition as it provides zero contribution? (b) Is the bond dimension not equal to the number of nonzero Schmidt weights, meaning truncating the matrix to discard nonzero values would not change the bond dimension?

2. Why is the complexity of describing strong-area law entanglement states $\mathcal{O}(dNc^2)$? It would be great to gain some intuition on how that is derived. $d$ is the local dimension, $N$ the number of sites and $c$ a constant such that the entanglement entropy along any bipartition is bounded by $S \leq \log(c) $

3. Would be great to understand how the Schmidt Number can be unbounded? Is this unbounded in the sense that as the number N of qubits increases the Schmidt Number keeps increasing? How do I visualise the physical differences between states where Schmidt Numbers are all bounded and states where they are not?

Would appreciate answers on any of these questions. Thanks a lot and apologies if some of them are a little imprecise: I have not got a lot of experience with MPS states.

Answer 1

Let me answer point by point.

1. The bond dimension between two sites is exactly the Schmidt number of the bipartite state with the two subsystems, which the bond connects. To understand this, let's isolate the cut look at the Schmidt decomposition of some state $$\psi = \sum_{i=1}^r c_i \phi^{(1)}_i \otimes \phi^{(2)}_i $$ You only need to store $2r$ many states $\phi^{(1/2)}_i$ to recover the full state $\psi$. Assume there are $N_1$ sites in subsystem 1 and $N_2$ sites in subsystem 2 each with a local dimension $d$, you thus need two tensors $A_1$ with $r\cdot d^{N_1}$ coefficients and $A_2$ with $r\cdot d^{N_2}$ coefficients, or you just give them each two indices $(i,\mu_{1/2})$ with $i\in \{1, ..., r\}$ and $\mu_{1/2} \in \{1, ..., d^{N_{1/2}}\}$. Iterate this over all sites and you get to the matrix product structure $$\psi = \sum_{\mu_1, ..., \mu_N} {\rm Tr}(A_1^{\mu_1} \cdot A_2^{\mu_2} \cdots A_N^{\mu_N}) e^{(1)}_{\mu_1} \otimes e^{(2)}_{\mu_2} \otimes ... \otimes e^{(N)}_{\mu_N}$$ with some local basis $e_{\mu}$. Going reverse, you can take the above structure and contract all indices except one bond and retrieve the Schmidt-decomposition with respect to a certain bond.

2. Having understood 1., it should be clear that a bounded entanglement entropy $S<\log(c)$ can be exactly represented by a state whose Schmidt rank is maximally $c$, or in MPS language, with a bond dimension maximally $c$. If you have $N$ matrices each with local dimension $d$ and bond dimension $c$, you got $dNc^2$ coefficients. Since some matrices might have smaller bond dimensions, because some bipartitions admit less entanglement, you could say that you need $\mathcal{O}(dNc^2)$ coefficients.

3. If $N$ is finite, of course also the Hilbert space dimension and thus the bond dimension is finite. The bond dimension is thus always trivially bounded by $d^N$. What people probably mean when they say (un)bounded is whether there exists a bound that is constant in $N$, i.e. that stays constant in the "thermodynamic limit", whatever that is. Examples for states with unbounded bond dimension are MPS admitting so-called "volume law entanglement", in that case the bond dimension grown linear in $N$.