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Consider generalizations of the GHZ state and the W state to $n$ qubits.

What is the Schmidt number of these two states for any bipartition $ c n $ and $(1 - c) ~n $, for $c < 1$?

Does it depend on the bipartition chosen?

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The Schmidt number for both of these states, for any $c$ such that the bipartition is non-trivial, is 2.

For the GHZ state, it always divides up as $$ |0\rangle^{\otimes cn}|0\rangle^{\otimes (1-c)n}+|1\rangle^{\otimes cn}|1\rangle^{\otimes (1-c)n} $$ so it should be trivial to see.

For the W state, because all basis elements involved are all-zero except for a single 1, then for every basis element, the one must be on one side of the bipartition. On the other side, the state is the all-zero state. Hence, you get a term of the form $$ |0\rangle^{\otimes cn}\left(\sum_{i=1}^{(1-c)n}|0\rangle^{\otimes(i-1)}|1\rangle|0\rangle^{\otimes((1-c)n-i)}\right)+\left(\sum_{i=1}^{cn}|0\rangle^{\otimes(i-1)}|1\rangle|0\rangle^{\otimes(cn-i)}\right)|0\rangle^{\otimes (1-c)n}. $$

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