In Ref. [1] absolutely maximally entangled (AME) states are defined as:

An $\textrm{AME}(n,d)$ state (absolutely maximally entangled state) of $n$ qudits of dimension $d$, $|\psi\rangle \in \mathbb{C}^{\otimes n}_d$, is a pure state for which every bipartition of the system into the sets $B$ and $A$, with $m = |B| \leq |A| = n − m$, is strictly maximally entangled such that $$ S(\rho_B) = m \log_2 d. $$

As the name would have you believe, does this mean that an $\textrm{AME}(n,d)$ state is maximally entangled across all entanglement monotones (for fixed $n$ and $d$)?

[1]: Helwig, Wolfram, et al. "Absolute maximal entanglement and quantum secret sharing." Physical Review A 86.5 (2012): 052335.

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    $\begingroup$ You could define a monotone to give an extra high value to your favourite maximally entangled state, just to make it the unique ME state for that monotone. But barring such absurdities, I think the answer is 'yes'. $\endgroup$ Jun 21, 2018 at 13:56

1 Answer 1


There is no a priori reason why an entanglement monotone should have this property. I think the particular issue is that, restricting to pure states, when you start to look at multiparticle states, you get inequivalent states. For example, for 3 qubits, the W state and GHZ state are inequivalent, and both are maximally entangled representatives for their SLOCC equivalence class (stochastic local operations and classical communication). There's no particular reason that an entanglement monotone should always favour GHZ (the AME(3,2)) over W-state. Indeed, they don't. Looking through the list of common multipartite entanglement monotones on Wikipedia, the one that stands out to me is the Schmidt Measure. It has value $\log_2(3)$ for a W state and one a value 1 for the GHZ state, so the AME is not maximally entangled by this measure.


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