# What is the Schmidt number of generalized GHZ and W states?

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?

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 $$|0\rangle^{\otimes cn}\left(\sum_{i=1}^{(1-c)n}|0\rangle^{\otimes(i-1)}|1\rangle|0\rangle^{\otimes((1-c)n-i)}\right)+\left(\sum_{i=1}^{cn}|0\rangle^{\otimes(i-1)}|1\rangle|0\rangle^{\otimes(cn-i)}\right)|0\rangle^{\otimes (1-c)n}.$$