Expansion of multi-particle state vector as a sum of n-entangled states

Question

Physically, quantum entanglement is ranged from full long-range entanglement (Bose-Einstein condensate), described by a basis of states that look like this:

$$ |\Psi\rangle = |\phi_{i_{0} i_{1} ... i_{N}}\rangle $$

to full-decoherence (a Maxwell-Boltzmann ideal gas) which a basis of states that look like this:

$$ |\Psi\rangle = \prod_{i}{ |\phi_{i}\rangle } $$

And in the middle of this range we have 2-particle entanglement terms, 3-particle entanglement, etc.

So it seems natural to arrange the wavefunction as a series looking like:

$$ |\Psi\rangle = \sum{ \prod_{i}{ |\phi_{i}\rangle } } + \sum{ \lbrace \prod_{i_{0} ,i_{1} > i_{0}}{ |\phi_{i_{0} i_{1}}\rangle } \rbrace } + \sum{ \lbrace \prod_{i_{0} ,i_{1} > i_{0} , i_{2} > i_{1}}{ |\phi_{i_{0} i_{1} i_{2}}\rangle } \rbrace } + \dotsb $$

BEGIN EDIT I feel that i need to put in a bit more clear footing the mathematics behind this separation. Let's take an arbitrary state vector $|\Psi\rangle$, we might write it like this:

$$ |\Psi\rangle = \prod_{i}{ c^{0}_{i} |\phi_{i}\rangle } + |\Psi_{Remainder}\rangle $$

That is, we write the state vector as a vector that is completely separable and a remainder that is not. This decomposition is unique. Proof: take another decomposition with $\widehat{c^{0}_{i}}$, take the difference between both decompositions and verify that both remainders are completely separable, which is against the definition

The idea is that one should be able to further this decomposition of the remainder state vector, for instance lets take a vector with zero completely separable part (that is, we are on the equivalence class of our remainder above) and attempt to write it like:

$$ |\Psi_{R_0}\rangle = \prod_{i_{0} ,i_{1} > i_{0}}{ c^{1}_{i_0 i_1} |\phi_{i_{0} i_{1}}\rangle } + |\Psi_{R_1}\rangle $$

Analogously, one can prove that the $c^{1}_{i_0 i_1}$ are unique and depend only on the $|\Psi_{R_0}\rangle$ vector. And we don't need to do any symmetrization operation to get this result, so this is an universal decomposition of the state vector (for bosons and fermions)

(note: i'm aware that this leaves out a lot of products with mixed entanglement, i.e: some single-particle states multiplied by two-particle entangled states, but since they don't add anything to this particular argument i choosed to leave them aside)

END EDIT

BEGIN 2ND EDIT

separability of states is a property that is invariant under unitary transformations, so a separable state is not equivalent to a separable one in any basis you choose.

I've investigated a bit more and the problem in general of knowing if a state is separable or not is known as QSP (quantum separability problem). For a definition please look at this paper

END 2ND EDIT

Question: Where can i read more about this sort of expansion and do you know if there is a computational framework to estimate the relative magnitudes of each term (for instance, i would expect that for Bose-Einstein condensates you need to keep all the terms of the expansion, while for relatively high-temperature solids you would be able to get away with 3 or 4 terms)

Answer 1

Lagerbaer gets it right in the comments:

The set of all (properly [anti-]symmetrized) product states is a complete basis. That's it, game over. So the first sum in your series represents every possible state.

We could also make (generally overcomplete) bases that consist entirely of products of maximally entangled pairs -- for example the valence bond basis is the natural description for resonating valence bond (or valence bond solid) states. Thus the second term in your series, by itself, could also represent a complete many-body basis.

However there is a flaw implicit in the notation, and this has to do with the monogamy of entanglement -- if two particles in my system are maximally entangled, they cannot be entangled with anything else! My point being: you have to be more clear with what you mean by "entanglement in a BEC." Yes a BEC has long-range correlations, but the ideal BEC is still a product state!!

In any case, as far as estimating relative contributions, writing down the best possible product ground state for a certain interacting hamiltonian is essentially mean-field theory. The true ground state will have corrections that will be superpositions of states, but still in the product basis. As a side-note, for a particularly cute way of going slightly beyond MFT in a language that helps to understand what entanglment entails in many-body states, look up Steve White's "Minimally Entangled Typical Thermal States" algorithm.

EDIT: after edit in the OP... Okay, so the point is that non-separable states are not such different objects from separable states (at least when talking about pure states). The structure of Hilbert space is such that when I take two objects and combine them, I get a complete basis from all possible tensor products of the two original bases! As an example, suppose I mix two qubits: I can write any state at all as the vector $$|\Psi\rangle=\sum_{ij}c_{ij}|i\rangle|j\rangle$$ Nonseparable states are states like Bell states with, for example, $c_{01}=c_{10}=0, c_{11}=c_{00}=1/\sqrt{2}$. There is no separable part versus non-separable part, the state is a superposition of product states, and is non-separable because it cannot be written as a single product without the sum. From your edit it seems like you understand this, but I feel the need to be explicit.

Additionally we might imagine writing states in the Bell basis; no one can stop me from writing $$|\Psi\rangle=c_{1}(|00\rangle+|11\rangle)+c_{2}(|00\rangle-|11\rangle)+c_{3}(|10\rangle+|01\rangle)+c_{4}(|10\rangle-|01\rangle)$$ It seems like this is what you imagine as the second term in your series, but it's nothing more than a rotation of my original basis. I don't gain anything at all by mixing my descriptions -- the original basis already provided a complete description of nonseparable states.