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.

  • $\begingroup$ Unitaries are bijective, hence onto $\endgroup$ – Phoenix87 Feb 2 '15 at 9:25
  • $\begingroup$ so you mean a partial isometry then :) $\endgroup$ – Phoenix87 Feb 2 '15 at 9:48
  • $\begingroup$ if you talk about kernel of a map on a vector space to itself I understand that the map is linear. Moreover, in finite dimension, trivial kernel imply bijective, hence onto again $\endgroup$ – Phoenix87 Feb 2 '15 at 11:10
  • $\begingroup$ As Timaeus said, MEs and Fs don't form vector subspaces, and it doesn't really make that much sense to consider non-linear mappings in the setting of vector spaces. $\endgroup$ – Phoenix87 Feb 2 '15 at 12:08
  • $\begingroup$ @Phoenix87 The example I gave was just an example. I understand that the sets are not linear subspaces. But it still makes sense to ask what the image of a general subset under a linear map looks like (and sometimes interesting to: as an unrelated example: unitaries map spheres to spheres), which is what all this is aimed at. $\endgroup$ – Selene Routley Feb 2 '15 at 12:21

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).

  • $\begingroup$ Many thanks Norbert. A wonderful paper, exactly what I am looking for! I am also interested in the question of when $f(MES)\subset PS$ as a proper subset, but this paper certainly gives me a great deal to think about. $\endgroup$ – Selene Routley Feb 2 '15 at 21:29

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

  • $\begingroup$ Thanks Māris, there are some great, simple and useful points here. "Product" is what I mean by "factorizable". I also hadn't thought about mixed (globally) states, but you show rather well here broadening the discussion to include them makes the problem simpler. Those look like some great notes too. Many thanks again. $\endgroup$ – Selene Routley Feb 4 '15 at 9:40

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.

  • $\begingroup$ I guess I'm wording it badly focussing on bijectivity. I mean a unitary restricted to, say ME states / Factorisables: is this restricted unitary's range either within the other set or onto the other set? Without further discussion, the unitary image of one could be a proper subset of the other: or there (as I suspect) is no linear relationship at all between the two sets or even between one and a proper subset of the other (as you say - neither are linear subspaces - and they are very NONlinear!) - this is the point of my question. $\endgroup$ – Selene Routley Feb 2 '15 at 9:45

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