# How to show that the GHZ state is absolutely maximally entangled?

A multipartite state is called absolutely maximally entangled if for its any bipartition the reduced density matrix of smaller part is maximally mixed. Show that GHZ state has this property.

• Hi and welcome to Quantum Computing SE. This looks like a homework. What did you try so far? Nov 26, 2023 at 7:34
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I assume that you talk about the standard case of a $$3$$-qubit GHZ state (the generalized versions of $$n$$-qubit GHZ states are not absolutely maximally entangled):

$$|GHZ\rangle = \frac{1}{\sqrt{2}} \big( |000\rangle + |111\rangle \big)$$

Then its density matrix is:

$$|GHZ\rangle \langle GHZ| = \begin{bmatrix} \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \end{bmatrix}$$

If you trace over any of the possible pairs of qubits ({0, 1}, {1, 2}, {0, 2}) you are left with the same reduced density matrix:

$$\rho_1 = \text{Tr}_{\{0, 1\}} \big(|GHZ\rangle \langle GHZ| \big) = \text{Tr}_{\{1, 2\}} \big(|GHZ\rangle \langle GHZ| \big) = \text{Tr}_{\{0, 2\}} \big(|GHZ\rangle \langle GHZ| \big) = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}$$

And we can see right away that it's maximally mixed ($$\rho_1 = I / 2$$ is a known case). We can also check the purity of $$\rho_1$$ to confirm:

$$\text{Tr}(\rho_1^2) = \text{Tr}\Bigg(\begin{bmatrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{bmatrix} \Bigg) = \frac{1}{2} = \frac{1}{N_1}$$

Where $$N_1 = 2$$ is the Hilbert space's dimension of the (smaller) single-qubit subsystem. This indicates that $$\rho_1$$ is maximally mixed.