# 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."

The reader is encouraged to double-check my back-of-the-envelope math.

• 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 $| \Psi ^ { + } \rangle = ( | 0,1 \rangle + | 1,0 \rangle ) / \sqrt { 2 }$ ? Dec 20, 2018 at 17:53

The difference between pure 2-way and 3-way entanglement can be seen by considering the GHZ and W-states (...)

The essential difference between these states becomes obvious if we perform a measurement on one of the three qubits. In the case of the GHZ state, if we measure an arbitrary qubit and obtain the result 0, the system collapses into the state |000>. Clearly, this is no longer an entangled state. Measuring any one of the qubits therefore completely destroys the entanglement. This is therefore “essential three-way entanglement”.

In contrast, if we measure the third qubit of the W state and obtain the result 0, we are left with the state |010> + |100>, in which the first two qubits are still maximally entangled. This type of entanglement is therefore called pairwise entanglement.

The different types of entanglement are complementary: If a system is strongly three-way entangled, its bipartite entanglement cannot be large.