9
$\begingroup$

When I encounter the definition of the mathematical definition of quantum entanglement. System composed by many parts $A$, $B$,.., $N$ can be described by a density matrix operator $\hat{\rho}$ acting on a Hilbert space of the tensorial product structure: $$H = H_A \otimes H_B \otimes \ldots \otimes H_N$$ The quantum state $\hat{\rho}$ is said to be at least $m$-separable if there exists splitting of the $N$ parties into $m$ parts $P_1, P_2, \ldots P_m$ such that [1] $$\hat{\rho}=\sum\limits_k p_k \hat{\rho}_k^{(1)} \otimes \hat{\rho}_k^{(2)} \otimes \ldots \hat{\rho}_k^{(m)}.$$ Otherwise the state is at least $m$-partite entangled.

All the quantum information science and quantum metrology (see Fisher information) is based on this definition and most importantly people apply this definition to indistinguishable particles (the parties are the collection of particles) which is not correct because the parties in the above definition are distinguishable (we are not dealing for example with the symmetrized Hilbert space). Maybe if the distinguishable particles are entangled according to the above definition they are also entangled in the indistinguishable case (some kind of lower bound)? How does this definition of entanglement applies to particle entanglement which are indistinguishable?

Here are some news phys.org.

UPDATE

Since my interest in the quantum metrology using atoms (e.g. atomic clocks) I wonder why the link between Quantum Fisher information (QFI) and multi-particle entanglement is correct [2]? Without doubt the QFI sets the achievable precision of interferometers, but the proof that it is connected to multi-particle entanglement is not convincing to me (as long as indistinguishable particles are concerned). It follows from standard definition of entanglement (above formula) where distinguishable particles are considered. Is it actually correct? In real experiments atoms are the same.

$\endgroup$
  • $\begingroup$ You might want to try reading the article you cite, and look up the paper mentioned, which can be found here: arxiv.org/abs/1511.03445. $\endgroup$ – alanf Apr 19 '17 at 13:14
8
$\begingroup$

Many people have noticed this problem.

Indeed, for identical particles, it is artificial to split them into groups and treat them as if they were distinguishable.

A fair way to treat identical particles identically is not to split.

For example, for identical fermions, the total wave function must be antisymmetric and hence can never be separable by the usual definition. But is it appropriate to say that fermions are always entangled? The simplest wave function of a collection of identical fermions is a Slater determinant. Is it appropriate to say that a Slater determinant contains entanglement? This kind of entanglement is always in the background, and should be get rid of.

Therefore, a proposed measure of entanglement for fermions is the so-called geometric measure of entanglement. Given a fermionic wave function, try to find a Slater determinant having the largest overlap with it. The larger the overlap, the closer the wave function to a Slater determinant, and the less the entanglement.

We have carried out this in some tentative papers:

Optimal multiconfiguration approximation of an N-fermion wave function

Optimal Slater-determinant approximation of fermionic wave functions

Geometric entanglement in the Laughlin wave function

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.