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I came across an unusual multi-partite generalization of the Schmidt decomposition in my work, which I describe below. Usually, when people say "a multi-partite Schmidt decomposition", they mean a sum of product states which is in some sense minimal, and there are different decomposition depending on what metric is used to measure minimality. In contrast, I'm looking at a decomposition of a state into a maximal sum of locally-orthogonal (but not necessarily product) states. I would like to know if this structure has been studied before so I can read more about it.

Suppose we are given a normalized quantum state $\vert \psi \rangle$ from a Hilbert space $\mathcal{H}$ composed of $N$ smaller subsystems $\mathcal{H}^{(n)}$ that are tensored together:

$\vert \psi \rangle \in \mathcal{H} = \mathcal{H}^{(1)} \otimes \cdots \otimes \mathcal{H}^{(N)}$.

Now I am looking for a preferred decomposition

$\vert \psi \rangle = \sum_i \sqrt{p_i} \vert \psi_i \rangle$

of $\vert \psi \rangle$ expressed as a sum (weighted by the positive values $\sqrt{p_i}$) of orthonormal vectors $\vert \psi_i \rangle$ such that the $\vert \psi_i \rangle$ live on orthogonal subspaces of each subsystem. More precisely, we require

$\vert \psi_i \rangle \in \bigotimes_n \mathcal{H}^{(n)}_i$


$\mathcal{H}^{(n)} = \bigoplus_i \mathcal{H}^{(n)}_i$

is a decomposition of $\mathcal{H}^{(n)}$ into orthogonal subspaces.

This is interesting because it means that the $i$-conditional reduced density matrices of any subsystem $\mathcal{H}^{(n)}$,

$\rho^{(n)}_i = \mathrm{Tr}_{\overline{\mathcal{H}^{(n)}}} \left[ \vert \psi_i \rangle \langle \psi_i \vert \right]$

are restricted to the same orthogonal subspaces. (Here, the trace is over all spaces except $\mathcal{H}^{(n)}$.) Therefore, observers can make local measurements on any of the systems $\mathcal{H}^{(n)}$ and determine which "branch" $\vert \psi_i \rangle$ they are on.

This structure is special for the following reason. (a) There exists a unique decomposition which maximizes the number of vectors in the sum while still satisfying the requirement of orthogonal records in each of the subsystems. (b) This maximal decomposition can be obtained through a mechanical (i.e. algorithmic) fine-graining procedure; the same decomposition is obtained regardless of the order in which the fine-graining is done. (c) The quantity $E = - \sum_i p_i \ln p_i$ is maximized for the maximal decomposition and defines a measure of the global entanglement (with respect to the particular choice of subsystems); if $v$ is such that any of the subsystems are unentangled with the rest, this quantity vanishes. (d) This structure reduces to the Schmidt decomposition (and $E$ reduces to the traditional entropy of entanglement) when $N=2$. (e) This structure, and the corresponding value of $E$, is insensitive to pairwise entangling interactions between subsystems (so long as they do not destroy the orthogonality of the conditional local states $\rho_i^{(n)}$) even when the strength/type of these entangling interactions are conditional on the branch $i$.

So: has this been explored before?

(I also submitted essentially the same question in math notation here.)

share|cite|improve this question
Dear Jess Riedel. In general, it is frown upon to cross-post simultaneously, because it may waste potential answerer's time. As a minimum OP should mention the cross-post (on both sites!). The preferred procedure is to not cross-post, and if the post hasn't received an acceptable answer after, say, a couple of days, then OP could flag for migration. – Qmechanic Sep 28 '13 at 16:34
OK, I will follow that advice in the future. If I can get anyone (anyone!) to look at one of these, I'll close the other one and have it redirect. – Jess Riedel Sep 29 '13 at 1:03
Spooky! I'd never heard this kind of thing called the Schmidt decomposition before, but right this minute I finished writing a post on Glauber's work in showing that a general quantum state (of the EM field) is a superposition of tensor products of coherent states - not quite the same thing, but eerily alike in flavour. Now I know the general abstraction of this idea, thanks to your question. – WetSavannaAnimal aka Rod Vance Sep 29 '13 at 10:30

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