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It is well known (Geometry of quantum states by Bengtsson and Życzkowski) that the set of $N$-dimensional density matrices is stratified by the adjoint action of $U(N)$, where each stratum corresponds to orbits with a fixed type of degeneracy structure of the density matrix spectrum.

These orbits are flag manifolds. For example the orbit of density matrices of spectrum $(1, 0, 0, 0)$ (pure states) is $\mathbb{C}P^3$ and the orbit of density matrices of spectrum $(0.5, 0.5, 0, 0)$ (and also $(0.4, 0.4, 0.1, 0.1)$) is the complex Grassmannian $Gr(4,2, \mathbb{C})$.

These spaces - being coadjoint orbits - are known to be Kahler-Einstein.

An observation by Ingemar Bengtsson which was stated and proved in this article, asserts that in the composite system of two $N-$ state quantum systems $\mathcal{H}^N \otimes \mathcal{H}^N$, whose orbit of pure states is $\mathbb{C}P^{N^2-1}$, the orbit of maximally entangled pure states is $\mathbb{R}P^{N^2-1}$, which is a minimal Lagrangian submanifold.

My questions:

  1. Does this result generalize to non-pure states? For example, the orbit of maximally entangled states in $Gr(N^2,2, \mathbb{C})$ (the orbit of density matrices of the type $(0.5, 0.5, 0, 0, . . .)$ in $\mathcal{H}^N \otimes \mathcal{H}^N$) isomorphic to $Gr(N^2,2, \mathbb{R})$.

  2. Is there a physical interpretation of this result? (this is also a question left open by Bengtsson).

Update:

This is a clarification following Peter Shor's remark:

Among the biparticle pure states, the maximally entangled states have the property that their partial trace with respect to one system is mixed (see the discussion following equation 22 in the second reference). As a generalization, I wish to know the local orbits of the states within a fixed biparticle density matrix orbit whose partial trace has the maximal von Neumann entropy relative to all other states in the same biparticle orbit. My motivation is that if the local orbits will happen to be the real flag Lagrangian submanifolds of the complex flag manifolds defining the biparticle state orbits (which is the case for the pure states), these manifolds have well known geometries, and this can contribute to our understanding of mixed state entanglement.

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    $\begingroup$ I don't understand your example. Maximally entangled states are pure states. I don't know what you mean by maximally entangled states in Gr(N^2,2,C). If you mean the states of that type with maximal entanglement, then don't you have to specify the entanglement measure? (There's a unique natural entanglement measure for pure states, but not for mixed states.) $\endgroup$
    – Peter Shor
    Oct 9, 2011 at 18:51
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    $\begingroup$ Thanks for the remark. The criterion for maximal entanglement that I have in mind (and didn't spell out explicitely in the question) is the one given by Bengtsson (in the second reference) following equation (22), namely,a state within a given orbit (of the biparticle density matrix) is to be called maximally entangled if its partial trace with respect to system 2 is maximally mixed (maximal Von Neumann entropy) relative to all states within the same orbit. I'll write an update to the question for clarification. $\endgroup$
    – David Bar Moshe
    Oct 10, 2011 at 9:31
  • $\begingroup$ I read somewhere that the mathematical definition of two maximally entangled two-dimensional particles $a$ and $b$ is $$\frac 1 {\sqrt 2}\big(|0,b\rangle_1|0,b^{'}\rangle_2+|1,b\rangle_1|1,b^{'}\rangle_2\big),$$ but shoudn't also products of states between o and 1 (and vice-versa) be involved? $\endgroup$
    – Deschele Schilder
    Dec 8, 2017 at 0:12
  • $\begingroup$ @descheleschilder: No, one of the definitions of a maximally entangled state of two systems is that its reduced density matrix is maximally mixed, i.e., proportional to the unit matrix. If the state contains mixed products, its reduced density matrix will not be diagonal; hence it is not maximally entangled. (The question however, is about mixed states) $\endgroup$
    – David Bar Moshe
    Dec 14, 2017 at 10:50
  • $\begingroup$ @PeterShor I'm very confused by your claim that "Maximally entangled states are pure states." I though that maximally entangled states have density matrices proportional to the identity and pure states have rank-one density matrices, which are mutually exclusive possibilities. Could you clarify? $\endgroup$
    – tparker
    Jun 4, 2018 at 17:19

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