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I am recently reading a paper about entanglement entropy. It mentions that if we consider a 1D spin chain and write a pure state in the matrix product state:

\begin{align} |\psi\rangle = A^{\sigma_1}A^{\sigma_2}\cdots A^{\sigma_i}A^{\sigma_{i+1}}\cdots A^{\sigma_N}\left|\sigma_1\cdots\sigma_N\right\rangle \end{align}

And we treat spins $1,\dots,i$ to be subsystem A and spins $i+1,\dots,N$ to be subsystem B, then we can define the $n$-th Renyi entropy between A and B:

\begin{align} \rho &= \left|\psi\right\rangle\left\langle \psi\right|\\ \rho_B &= \text{Tr}_A\rho\\ S_n(B) &= \frac{1}{1-n}\log\text{Tr}\rho_B^n \end{align}

For the zeroth Renyi entropy, it equals

\begin{align} S_0(B) =\log r \end{align}

Here $r$ is the number of nonzero eigenvalues of $\rho_B$ and it is also the Schmidt rank. But the paper also says that $r$ equals $d_{i,i+1}$, the local dimension between $A^{\sigma_i}$ and $A^{\sigma_{i+1}}$ in MPS. Why is it true? How to prove it?

References:

  1. https://doi.org/10.1103/PhysRevX.7.031016 Appendix A Eqn.A11.
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It is only true that $r\le d_{i,i+1}$.

This is easy to see: First, write the matrix products using explicit sums over the matrix indices. Then, the only summation index which couples the subsystems A and B is the index from the matrix product $A^{\sigma_i}A^{\sigma_{i+1}}$, which can take $d_{i,i+1}$ values.

Thus, the state can be written as $$ \sum_{k=1}^{d_{i,i+1}} |a_k\rangle|b_k\rangle\ . $$ The number of terms in the sum upper bounds the number of terms in the Schmidt decomposition. Thus, $r\le d_{i,i+1}$.

Of course, if you pose additional conditions on the MPS (like a "nice" canonical form), it might also hold that $r=d_{i,i+1}$.

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  • $\begingroup$ Yes, this is what I believe. If you are interested the paper, it is link Appendix A. I guess they are talking about the possible minimal d_{i,i+1}? Thanks for the answer. $\endgroup$
    – jisutich
    Jul 3 at 23:07
  • $\begingroup$ Hm, that's two pages. You don't seriously expect anyone to read two full pages just to spot what you are after? $\endgroup$ Jul 4 at 7:34
  • $\begingroup$ Sorry, it is my fault. In fact, I just mean eqn. A11. I apologize for it. $\endgroup$
    – jisutich
    Jul 4 at 14:32
  • $\begingroup$ Ideally, you should edit your question to include this information. $\endgroup$ Jul 4 at 21:26
  • $\begingroup$ I guess that's what "In an efficient representation" right before (A11) is supposed to indicate. $\endgroup$ Jul 4 at 21:27

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