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?


  1. https://doi.org/10.1103/PhysRevX.7.031016 Appendix A Eqn.A11.

1 Answer 1


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 $r=d_{i,i+1}$.

  • $\begingroup$ 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
    Commented Jul 3, 2021 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? $\endgroup$ Commented Jul 4, 2021 at 7:34
  • $\begingroup$ Sorry, it is my fault. In fact, I just mean eqn. A11. I apologize for it. $\endgroup$
    – jisutich
    Commented Jul 4, 2021 at 14:32
  • $\begingroup$ Ideally, you should edit your question to include this information. $\endgroup$ Commented Jul 4, 2021 at 21:26
  • $\begingroup$ I guess that's what "In an efficient representation" right before (A11) is supposed to indicate. $\endgroup$ Commented Jul 4, 2021 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.