= Exact dilution obstruction from Schmidt rank
{title2=$m\geq n\quad\text{for exact }\mathrm{GHZ}^{\otimes m}\to\Psi_p^{\otimes n}$}
Let three parties share the normalized <quantum state>
$$
|\Psi_p\rangle=\sqrt p\,|000\rangle+\sqrt{1-p}\,|111\rangle,\qquad 0<p<1.
$$
For $n$ copies, group each party's $n$ <qubits> together. Across any nontrivial division of these three parties, the <Schmidt decomposition> has one nonzero coefficient for every binary string of length $n$, so its <Schmidt rank> is $2^n$. Similarly, $m$ copies of the standard <GHZ state> have <Schmidt rank> $2^m$ across the same division.
Every recorded <LOCC> branch acts by a product of local <Kraus operators>. <Schmidt-rank contraction under product operators> therefore rules out an exact successful conversion when $m<n$, even if success need only have nonzero <probability>. This is a rank obstruction for the full target, not merely a bound on its <entanglement entropy>.
Approximation changes the question. In the expansion of $|\Psi_p\rangle^{\otimes n}$, squared coefficients have the binary-product distribution with parameter $p$. A <quantum typical subspace> retains weight tending to one while using only $2^{nh_2(p)+o(n)}$ strings, where $h_2$ is <binary entropy>. The normalized truncation approaches the full target in <quantum fidelity>, although its <Schmidt rank> can be far smaller than $2^n$. Consequently, the exact finite-copy rank argument does not forbid an asymptotic <entanglement dilution> protocol with vanishing approximation error.
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