 Deterministic reduction of maximally entangled Schmidt rank

Let $|{\Phi }_d⟩=d^{-1/2}{\sum }_{j=0}^{d-1}|j,j⟩$ and $1\le m\le d$. It can be converted deterministically by LOCC to $|{\Phi }_m⟩$. For each $m$-element subset $S$ of the local basis labels, let ${\Pi }_S$ be its projector and let Alice's Kraus operator be $K_S={\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)}$. Every label occurs in $\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)$ subsets, so ${\sum }_S{K}_S^{†}{K}_S=I$. Each outcome has probability $m/[d(\genfrac{}{}{0ex}{}{d-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d}{m}\right)$ and leaves the maximally entangled state $m^{-1/2}{\sum }_{j\in S}|j,j⟩$. Alice reports $S$, and both parties relabel these $m$ basis states to $0,\dots ,m-1$ by local unitary operations.

For $d=6$, $m=4$, only three outcomes are needed: partition the six labels into disjoint pairs $T_0,T_1,T_2$ and use $K_r=(I-{\Pi }_{T_r})/\sqrt{2}$. Each label is retained twice, proving completeness, and each outcome has probability $1/3$. Every branch leaves four equal Schmidt coefficients, hence two independent Bell pairs after binary relabelling.