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W state


A W-state of $ d $ qubits is definied as

\[\ket{\mathrm{W}} = \frac{1}{\sqrt{d}} \left( \ket{10\ldots 0} + \ket{01\ldots 0} + \ldots + \ket{00\ldots 1} \right).\]


To focus attention, let us consider a three qubit setup: $\ket{\mathrm{W}} = \frac{1}{\sqrt{3}} \left( \ket{100} + \ket{010} + \ket{001} \right)$. There is no generic three qubit entanglement in W-states and all entanglement monotones as well as their normal form will be equal to zero. However tracing out one of the parties leaves the remaing two entangled. In this sense three qubit W-states contain maximal amount of sum of two qubit entanglement.