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