# How entangled are the $|W\rangle$ and the $|GHZ\rangle$ states?

The two states $$| GHZ \rangle = \frac { 1 } { \sqrt { 2 } } ( | 0,0,0 \rangle + | 1,1,1 \rangle )$$ and $$| \mathrm { W } \rangle = \frac { 1 } { \sqrt { 3 } } ( | 0,0,1 \rangle + | 0,1,0 \rangle + | 1,0,0 \rangle )$$ are considered unequivalent entangled quantum states.

I read on wikipedia : "The W state is, in a certain sense "less entangled" than the GHZ state; however, that entanglement is, in a sense, more robust against single-particle measurements"

The Von Neumann entropy measures the degree of mixing of the system's state: $$S ( \rho ) = - { tr } ( \rho \ln \rho )=- \sum _ { j } M _ { j j } \ln \left( M _ { j j } \right)$$ with $$M_{jj}$$ the diagonal coefficient and density matrix $$\rho$$.

How can we compare the two entropies $$S ( \rho_{GHZ} )$$ and $$S ( \rho_{W} )$$ to clarify in what sense is one state more entangled ?What's the difference with tri-partite entanglement?

Thank you.

Entanglement entropy is one measure of how entangled a state is. If we divide our system into two subsystems $$A$$ and $$B$$, we write the Hilbert space $$\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B$$, and the total density matrix is $$\rho$$, the reduced density matrix for subsystem $$A$$ is $$$$\rho^A = \text{Tr}_{\mathcal{H}_B}\,\rho.$$$$ Then the entanglement entropy of subsystem $$A$$ is the von Neumann entropy of the reduced density matrix, $$$$S^A = - \text{Tr}_{\mathcal{H}_A}\,\rho^A\log\rho^A.$$$$ In particular, the entanglement entropy for a product state is 0.
To apply this to the current case, we have a three-qubit Hilbert space $$\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C$$. We can calculate the reduced density matrix for the $$A$$ qubit by tracing over the latter two factors, $$$$\rho^A = \text{Tr}_{BC}\, \rho,$$$$ with $$\text{Tr}_{BC} = \text{Tr}_{\mathcal{H}_B \otimes \mathcal{H}_C}$$. We obtain the reduced density matrices $$$$\rho^A_{\text{GHZ}} = \frac{1}{2}\left(|0\rangle\langle 0| + |1\rangle\langle1|\right),$$$$ $$$$\rho^A_{\text{W}} = \frac{1}{3}\left(|0\rangle\langle 0| + |0\rangle\langle 0| + |1\rangle\langle1|\right),$$$$ and the corresponding von Neumann entropies $$S_{\text{GHZ}}^A = \log 2 \approx 0.69$$ and $$S_{\text{W}}^A = \log 3 - \frac{2}{3}\log 2\approx0.64$$. So by this measure, the GHZ state is "more entangled."
• Isn't $\hat { \rho _ {A } } = { Tr } _ { { H } _ { B} \otimes {H } _ { C } } ( \hat { \rho } )$ so $\rho_ { ( A) } = { tr } _ {A} | { GHZ } \rangle \left\langle { GHZ } \left| = \frac { 1 } { 2 } ( | 0,0 \rangle \langle 0,0 | + | 1,1 \rangle \langle 1,1 | )\right. \right.$ and ${ tr } _ { A } | { W } \rangle \left\langle { W } \left| = \frac { 1 } { 3 } \right| 0,0 \right\rangle \left\langle 0,0 \left| + \frac { 2 } { 3 } \right| \Psi ^ { + } \right\rangle \left\langle \Psi ^ { + } |\right.$ with $| \Psi ^ { + } \rangle = ( | 0,1 \rangle + | 1,0 \rangle ) / \sqrt { 2 }$ ? – user159729 Dec 20 '18 at 17:53