Are GHZ states absolutely maximally entangled?

Question

The definition of an absolutely maximally entangled state is as follows:

"A multipartite state $|\psi\rangle$ of a system S is called absolutely maximally entangled if for any bipartition A|B of S, the reduced density operator is maximally mixed $\rho_A=\rho_B=I/d$"

Any bipartition of the GHZ state is as follows:

$$\rho = \left( \begin{array}{cccc} \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} \\ \end{array} \right)\,.$$

This is not a maximally mixed state $I/d$ but somehow GHZ state is an absolutely maximally entangled state.

Answer 1

I think the issue here is the definition of Absolutely Maximally Entangled (AME) states on Wikipedia, which you are quoting. I had not heard of AME states before, but after coming across your question, I went into a rabbit hole of reading some papers.

In the paper Goyeneche et al.(2015)[arXiv:1506.08857] they define AME states in the Section $\text{II(A)}$ as

Entangled states of $N$-party systems, such that tracing out arbitrary $N − k$ subsystems, the remaining $k$ subsystems have associated a maximally mixed state. Such states are often called $k$-uniform and by construction, the integer number $k$ cannot exceed $N/2$.

So you can see here, for GHZ state, $N=3$ and hence, maximum value of $k$ would be $$\text{max(k)} = \left\lfloor \frac{N}{2} \right\rfloor = \left\lfloor \frac{3}{2} \right\rfloor = 1\,.$$

So, for tracing out any $N-k = 3-1 = 2$ qubits from 3 qubits, it should leave the remaining $k=1$ qubit in the maximally entangled state. You can trace over any of the possible pairs of qubits ({0, 1}, {1, 2}, {0, 2}), and you are left with the same reduced density matrix, $I/2$.

Hence, we can confirm that the GHZ state is an AME state according to the definition mentioned above. Goyeneche et al. (2015) also confirms this in Section $\text{III(A)}$:

In the case of three qubits the well-known GHZ state is an AME.

Answer 2

I suppose the term "Reduced Density Matrix" makes sense here. As we have a three qubits system GHZ state such the following:

$$|\psi⟩ = \frac{1}{\sqrt{2}} (|000⟩ + |111⟩)$$

If we build the density matrix up from above state and then take a partial trace from from any possible bipartition A|B, B|A, A|C, C|A, C|B, B|C, then what we end up is a maximally mixed state.

Answer 3

For higher order GHZ states, like as you said for four qubits case, we can write the density matrix of this state like the following:

$$\rho_{ABCD} = \frac{1}{2}(|0000⟩⟨0000|+|0000⟩⟨1111|+|1111⟩⟨0000|+|1111⟩⟨1111|$$

Then like as you have mentioned in the theorem, we must search for the bipartitions in the Hilbert Space of the whole system which we have the same again: A|B, A|C, A|D, etc.

For instance for finding the density matrix of in A|B, then we must take into account the following partial trace:

$$\rho_{AB} = Tr_{CD}(\rho_{ABCD}) = \frac{1}{2}(|00⟩⟨00|+ |11⟩⟨11|)$$

this will give us the AB subsystem. Then by continuing to have $\rho_{A} = \rho_{B} = I/d $, we must take another partial trace which gives us the following:

$$\rho_{A} = Tr_{B}(\rho_{AB}) = \frac{1}{2}(|0⟩⟨0|+ |1⟩⟨1|)$$

This is exactly what we expected. I hope this can be helpful.