Maximally entangled mixed states for a fixed spectrum do not always exist

Entanglement is a resource under local operations assisted by classical communication (LOCC). Given a set of states $S$, if there is one state in $S$ that can be transformed by LOCC into all other states in $S$, then this state is maximally entangled in $S$. It is a well-known result that the $d$-dimensional Bell state is the maximally entangled state in the set of all bipartite states of local dimension $d$. Since in practical applications noise renders every state mixed, it is interesting to study whether sets of mixed states of relevance enable the notion of a maximally entangled state. A natural choice is the set of all states with the same spectrum. In fact, for any given spectrum distribution on two-qubit states, previous work has shown that several entanglement measures are all maximized by one particular state in this set. This has led to consider the possibility that this family of states could be the maximally entangled states in the set of all states with the same spectrum, which should then maximize all entanglement measures. In this work I answer this question in the negative: there are no maximally entangled states for a fixed spectrum in general, i.e. for every possible choice of the spectrum. In order to do so, I consider the case of rank-2 states and show that for particular values of the eigenvalues there exists no state that can be transformed to all other isospectral states not only under LOCC but also under the larger class of non-entangling operations. This in particular implies that in these cases the state that maximizes a given entanglement measure among all states with the same spectrum depends on the choice of entanglement measure, i.e. it cannot be that the aforementioned family of states maximizes all entanglement measures.