Wednesday, 16 January 2019

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


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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