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.
 A: 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
\begin{equation}
\rho^A = \text{Tr}_{\mathcal{H}_B}\,\rho.
\end{equation}
Then the entanglement entropy of subsystem $A$ is the von Neumann entropy of the reduced density matrix,
\begin{equation}
S^A = - \text{Tr}_{\mathcal{H}_A}\,\rho^A\log\rho^A.
\end{equation}
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,
\begin{equation}
\rho^A = \text{Tr}_{BC}\, \rho,
\end{equation}
with $\text{Tr}_{BC} = \text{Tr}_{\mathcal{H}_B \otimes \mathcal{H}_C}$. We obtain the reduced density matrices
\begin{equation}
\rho^A_{\text{GHZ}} = \frac{1}{2}\left(|0\rangle\langle 0| + |1\rangle\langle1|\right),
\end{equation}
\begin{equation}
\rho^A_{\text{W}} = \frac{1}{3}\left(|0\rangle\langle 0| + |0\rangle\langle 0| + |1\rangle\langle1|\right),
\end{equation}
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."
The reader is encouraged to double-check my back-of-the-envelope math.
