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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.

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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.

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  • $\begingroup$ 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 $| \Psi ^ { + } \rangle = ( | 0,1 \rangle + | 1,0 \rangle ) / \sqrt { 2 }$ ? $\endgroup$ – user159729 Dec 20 '18 at 17:53

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