Is there a unitary, linear bijection between (1) Maximally Entangled and (2) Factorizable States? Pretty much as the title says. I am interested in the two particle system, each particle having two dimensional quantum states; naturally if there is a generalisation I'd be interested in that too. So, the particular problem I am thinking of is: can one find a $4\times 4$ unitary matrix $U$ that always maps a factorizable state to a maximally entangled one and contrariwise?.
By "maximally entangled" I mean that the von Neumann entropy of one particle alone (which in general seems to be in a mixed state owing to its entanglement with the other) is maximal  and equal to one bit (I think this is a standard definition: this is not my field).
I am asking this question as part of trying to find ways to visualise the set of maximally entangled and factorizable states in the bipartite system state space.
PS: I am having trouble deciding which tags to put on my question; I'd appreciate help here too.
 A: Following http://arxiv.org/abs/quant-ph/0110082, for two qudits the manifold of all product states is $4(d-1)$ dimensional, while the manifold of maximally entangled states has $d^2-1$ dimensions.  Thus, with the possible exception of $d=3$, these two sets cannot be mapped onto each other by any kind of "nice" mapping (and in particular not by a linear map).
A: If you are interested in how these sets look like, I can provide their characterization.
A state on two $d$-dimensional systems is maximally entangled if and only if it can be written as follows:
$$\frac{1}{\sqrt{d}} \sum_{i=1}^d |u_i\rangle |v_i\rangle$$
where $|u_i\rangle$ is the $i$-th column of some $d \times d$ unitary matrix $U$ and similarly $|v_i\rangle$ is the $i$-th column of some $d \times d$ unitary matrix $V$. You can assume without loss of generality that either $U$ or $V$ is the $d \times d$ identity matrix. Note that all maximally entangled states are pure globally and completely mixed locally.
There are two related notions: product and separable states (I'm not sure which of them you mean by "factorizable"). A bipartite state is product if and only if it is of the form
$$\rho \otimes \sigma$$
for some $d \times d$ density matrices $\rho$ and $\sigma$. Note that most product states are mixed, so they cannot be unitarily mapped to maximally entangled states (a pure product state is of the form $|u\rangle |v\rangle$).
A quantum state is separable if and only if it can be written as
$$\sum_i p_i \sigma_i \otimes \rho_i$$
for some probability distribution $p$ and some $d \times d$ density matrices $\rho_i$ and $\sigma_i$. Here the index set over which $i$ ranges can be arbitrarily large but you can put bounds on it (using Carathéodory’s Theorem you can show that $d^4$ terms always suffice). You can also assume without loss of generality that the states $\rho_i$ and $\sigma_i$ are pure. Note that every product state is separable. Also, note that most separable states are mixed, so such states cannot be mapped to a maximally entangled states by a global unitary.
You can find more about these three sets of states in John Watrous' lecture notes: https://cs.uwaterloo.ca/~watrous/CS766/LectureNotes/all.pdf
A: Any 4x4 unitary matrix is onto, so onto is going to happen one you insist on a unitary linear map.
Neither the set of maximally entangled, nor the set of separable states is a linear subspace.  However, if you insist that every (nonzero) separable state map to a maximally entangled state, then you would have to send $|++\rangle$, $|+-\rangle$, $|-+\rangle$, and $|--\rangle$ to maximally entangled states.  And since they are orthogonal and unit length, you'd have to map them to four mutually orthogonal unit length maximally entangled states.  For a moment I didn't think there were four such states, but  $\frac{|++\rangle+|--\rangle}{\sqrt{2}}$, $\frac{|+-\rangle+|-+\rangle}{\sqrt{2}}$,$\frac{|+-\rangle-|-+\rangle}{\sqrt{2}}$, and $\frac{|++\rangle-|--\rangle}{\sqrt{2}}$ respectively might do it.
But the separable states naturally form a torus within the unit ball modulo overall phase, and a linear unitary map would have to respect that, so if it is also bijection, then the maximally entangled states would have to likewise be exactly such a configuration.  I'm not sure they are.
