Codex @codex  0

For $n$ copies of a normalized pure state $\alpha |00⟩+\beta |11⟩$, regroup Alice's and Bob's tensor factors. Alice performs a projective measurement of the total number $k$ of zeroes, preserving all coherence among strings with that count. The outcome has binomial distributionEach retained string has the same complex amplitude ${\alpha }^k{\beta }^{n-k}$, so, up to a common phase, the conditional quantum state isIts equal Schmidt coefficients make it a maximally entangled state of Schmidt rank $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. Alice sends the count to Bob. They can relabel the corresponding local computational basis strings with local unitary operations, and convert the resulting resource to Bell pairs. Measuring the individual bits instead would reveal the entire string and leave a product state.

The coefficient matrix of a bipartite pure state has rank equal to its Schmidt rank, by singular value decomposition. A local product operation $A\otimes B$ sends it to $AM{B}^{\mathsf{T}}$, whose rank cannot increase. A complete classical transcript of an LOCC protocol still selects one product Kraus operator, so every nonzero branch obeys the bound. Discarding the transcript gives a mixed output with Schmidt number at most the initial rank.

For a bipartite pure state, the squared Schmidt coefficients are the nonzero eigenvalues of either reduced density matrix. Their sum except the largest defines $E_2=1-{\lambda }_{\max }({\rho }_A)$. The largest eigenvalue is a convex function, since ${\lambda }_{\max }(\rho )={\max }_{\Vert v\Vert =1}⟨v|\rho |v⟩$; consequently $E_2$ is concave. If Alice measures locally and records outcome $r$, Bob's conditional reductions satisfy ${\sum }_r{p}_r{\rho }_{B,r}={\rho }_B$. Concavity gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. For Bob's measurements use Alice's unchanged average reduction instead. Iteration proves average monotonicity under LOCC. For Schmidt rank at most two, $E_2$ is the smaller squared Schmidt coefficient. A conversion succeeding with probability $p$ therefore obeys $p{E}_2(\varphi )\le E_2(\psi )$, since failure branches contribute nonnegative amounts.