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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)

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I don't think there is any such general method to decompose a general many-body state in this manner. If you were to discover one that would be huge. But awesome idea. –  user346 Apr 15 '11 at 16:15
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I like the idea. But are you talking about the wave-function, or about an ensemble represented by a density matrix? For the wave-function, the (Anti-)symmetrized products $\prod_i |\phi_i\rangle$ already form a complete set of basis states, so every wavefunction can be written as a linear combination of these. For Fermions, e.g., it would be a linear combination of Slater Determinants. –  Lagerbaer Apr 15 '11 at 16:54
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In classical statistical mechanics you use the cluster expansion to derive the Van der Waals equation. This is a expansion in powers of the number of interacting particles. I know nothing about its application to quantum statistical mechanics. –  Johannes Apr 15 '11 at 17:19
    
@Johannes, exactly! this is what i'm looking for; i didn't knew the term, a quantum version of the virial expansion; however i think this becomes subtle because interactions and entanglement are mutually related, but have a not very clear overlap –  lurscher Apr 15 '11 at 17:41
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@Lurscher the question of whether or not a state is separable is neatly answered in Miyake's paper on the classification of multipartite entanglement. One has to realize that the state space for a vector living in a $k+1$ dimensional Hilbert space is the complex projective space $CP^k$. The space of the composite system of two "particles" in a $k_1+1$ and $k_2+1$ dimensional hilbert spaces respectively is $CP^{(k_1+1)(k_2+1)-1}$. The subspace consisting of separable states is of the form $CP^{k_1} \times CP^{k_2}$. The latter has an embedding in the –  user346 Apr 17 '11 at 18:27

1 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.

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@wsc nice answer +1. Yes a BEC has long-range correlations, but the ideal BEC is still a product state!! ... there is a paper by C. Simon which claims that every BEC is highly entangled. I'm not sure if I follow his line of reasoning. So which one is it? or equally, which one should it be? for a BEC. –  user346 Apr 16 '11 at 4:54
    
Ah, this is why I said the OP had to be more clear. If you look at Simon's paper, what he is demonstrating is an entanglement between two spatially separated regions, not any entanglement between the internal degrees of freedom (the bosons). It's a subtle thing, but if you think about it real hard (and do a little math!) you can also show that even a simple plane wave on a 1D lattice can "entangle" two halves of the lattice across some cut –  wsc Apr 16 '11 at 5:08
    
And actually as I read further, Simons clarifies very well that his result is a peculiarity of writing his BEC wavefunction in the fixed-number basis [and here comes the part where I admit that I think of BECs almost exclusively in terms of coherent states...] –  wsc Apr 16 '11 at 5:12
    
@wsc I LOL as I read your comments. True about the plane wave example. Also following Srednicki's line of argument isn't any subsystem of a QFT entangled with its complement via their common boundary? He only looks at the ground state of a non-interacting massless scalar. But if there is such a result for a massless scalar then one should be able to extend it ... shows how much catching up I have to do to get to the present day understanding of this material! –  user346 Apr 16 '11 at 5:53
    
I'm not familiar with Srednicki's work on the topic, but even in the paper you linked there was a footnote about the vacuum of relativistic QFTs generally being entangled in the sense that he describes. Anyway, I make no claim to expertise, but I think there is a lot being learned about the nature of entanglement both in relativistic and non-relativistic QFT and we do ourselves no favors by mixing up our words and elementary QM intuitions when there's so much math to be done! –  wsc Apr 17 '11 at 0:15

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