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

Question

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.

Answer 1

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.