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Use the Schmidt decomposition of the bipartite pure state:The two reduced density matrices, obtained by partial trace, areTheir nonzero eigenvalues are identical, even if the two ambient dimensions differ. Zero eigenvalues contribute no Von Neumann entropy, soThis common Von Neumann entropy is the entanglement entropy of the bipartite pure state.

For $|J,J⟩$, tracing out subsystem 2 gives the reduced density matrixwhich is pure, so its entanglement entropy is zero.

For $|J,J-1⟩$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and givesIts nonzero eigenvalues are $j_1/J$ and $j_2/J$, so the Von Neumann entropy isWhen $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log 2$, the maximal entropy for a rank-two reduced state.

Cutting an open-boundary matrix product state across one virtual bond of dimension $\chi$ gives Schmidt rank at most $\chi$, so its entanglement entropy obeysA periodic interval cuts two bonds and obeys $S(A)\le 2\log \chi$. Since the boundary of a one-dimensional interval has a constant number of points, this is an area law.

For a projected entangled pair state, a bipartition crossing $n_{\partial }$ virtual bonds has Schmidt rank at most ${\chi }^{n_{\partial }}$. The tensor-network area-law bound is thereforeBecause $n_{\partial }$ is proportional to the lattice boundary area, every fixed-bond-dimension PEPS satisfies an area law.

A bipartite density operator is a product state when it factors as ${\rho }_A\otimes {\rho }_B$. A pure product state factors as $|{\psi }_A⟩\otimes |{\psi }_B⟩$; its reduced density matrices have rank one and zero entanglement entropy. Mixed product states can have nonzero marginal Von Neumann entropy.

If a bipartition cuts $n_{\partial }$ tensor-network bonds of dimension at most $\chi$, its Schmidt rank is at most ${\chi }^{n_{\partial }}$ and its entanglement entropy is at most $n_{\partial }\log \chi$.