Pure-State Entanglement Across a Tensor Cut

Choose ordered orthonormal bases of $H_A,H_B$ using [F5], and let $C$ be the coefficient matrix of $\psi$ in their product basis [L2]. For any other ordered orthonormal bases, let $A,B$ have the new basis vectors as columns in the old coordinates. Their columns are orthonormal, so the finite row-by-column identities $A^{\ast }A=B^{\ast }B=I$ hold; because they are basis changes, both matrices are invertible and hence unitary. Expanding the new product basis vectors in the old one and using conjugate-linearity in the second argument gives $c_{pq}^{\mathrm{\prime }}={\sum }_{i,k}\overline{A_{ip}}{c}_{ik}\overline{B_{kq}}$, so the new coefficient matrix is $C^{\mathrm{\prime }}=A^{\ast }C\bar{B}$. The factors $A^{\ast }$ and $\bar{B}$ are invertible, so the map represented by $C^{\mathrm{\prime }}=A^{\ast }C\bar{B}$ is $A^{\ast }\circ T_C\circ \bar{B}$, where $T_C$ is the map represented by $C$. Since $\bar{B}$ is onto, its image is $A^{\ast }(\mathrm{im}{T}_C)$; since $A^{\ast }$ is injective, this image has the same dimension as $\mathrm{im}{T}_C$. By [F2], the matrix rank is therefore unchanged, so it is basis-independent. Multiplying $\psi$ by any nonzero scalar multiplies $C$ by that scalar and also preserves its rank.

In any Schmidt decomposition $\psi ={\sum }_{j=1}^r{s}_j{u}_j\otimes v_j$ with all $s_j>0$ and orthonormal factor families, tensor coordinates multiply, so its coefficient matrix is $C=UD{V}^{\mathsf{T}}$, where the columns of $U,V$ are the coordinates of the $u_j,v_j$ and $D=\mathrm{diag}(s_1,\dots ,s_r)$. The orthonormal columns give $U^{\ast }U=V^{\ast }V=I_r$, so $U$ is injective; conjugating $V^{\ast }V=I_r$ gives $V^{\mathsf{T}}\bar{V}=I_r$, so $V^{\mathsf{T}}$ is surjective; and $D$ is invertible. Thus $\mathrm{rank}(C)=r$. Conversely, [L3] constructs such a Schmidt decomposition from the SVD, and [F3] and [F4] show that its number of positive coefficients is the rank of $C$. Therefore the number of terms is the same for every Schmidt decomposition and equals the defined Schmidt rank.

If $\psi =c(a\otimes b)$ for nonzero $a,b,c$, its coefficient matrix is $c\alpha {\beta }^{\mathsf{T}}$ for the coordinate columns $\alpha ,\beta$, so its rank is one. Conversely, if $C$ has rank one, choose a nonzero column $\alpha$; every column is a scalar multiple ${\beta }_k\alpha$, so $C=\alpha {\beta }^{\mathsf{T}}$ with $\beta \mathrm{\ne }0$. The coefficient expansion then gives $\psi =a\otimes b$ for the vectors with coordinates $\alpha ,\beta$. Thus the ray is a product ray exactly when the Schmidt rank is one.

Since $\psi \mathrm{\ne }0$, its coefficient matrix is nonzero and has positive rank. By the product-ray equivalence in step 2.1, failure to be a product ray is equivalent to Schmidt rank greater than one, which is precisely entanglement by the Definition.