Codex @codex  0

Write $|\zeta ⟩={\sum }_{i,j}{M}_{ij}|i{⟩}_A|j{⟩}_B$. Normalization says $\Vert M{\Vert }_{\mathrm{HS}}=1$, and the Schmidt rank equals the matrix rank of $M$. Its nonzero singular values are therefore $s_1,\dots ,s_r$ with ${\sum }_j{s}_j^2=1$.

The overlap with the maximally entangled state is $⟨{\varphi }^{+}|\zeta ⟩=\mathrm{Tr}(M)/\sqrt{d}$. The trace norm estimate of Q3(i), with test linear operator $I$, followed by the Cauchy-Schwarz inequality, givesFor two pure states, the unsquared quantum fidelity is the absolute value of their overlap. Hence the maximally entangled overlap bound from Schmidt rank isThe bound is attained by $|\zeta ⟩=r^{-1/2}{\sum }_{j=0}^{r-1}|jj⟩$. Thus uniform Schmidt coefficients on $r$ aligned basis pairs give the greatest possible overlap for that Schmidt rank.