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For a pure state $\rho_{AB}$, the entropy of entanglement of subsystem $A$ is

\begin{equation} S( \rho_A) = -tr (\rho_A \log \rho_A) \end{equation}

where $\rho_A$ is the reduced density matrix of A.

For a single site of a spin chain, $\rho_A$ can be written in terms of single site correlation functions $\langle \sigma_l^\alpha \rangle$ where $\alpha = x,y,z$.

Are there any entanglement measures for mixed states that use only the same correlation functions, $\langle \sigma_l^\alpha \rangle$ where $\alpha = x,y,z$?

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  • $\begingroup$ @PiotrMigdal Thanks, but no, that's not what I'm asking. I know that the density matrix can be written in terms of the Pauli matrices. I want to know whether there are any entanglement measures for mixed states which only use the 'single site' correlation functions (i.e. the expectation values of the Pauli matrices). I hope that's clearer. $\endgroup$
    – Calvin
    Commented Jan 25, 2012 at 21:45
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    $\begingroup$ I guess one would not call that "correlation functions", so maybe that's what caused the confusion. With this clarification, the answer to your question is no: The maximally entangled and the (unentangled) maximally mixed state have the same reduced single site density matrices (and thus the same expectation values for any local operator), but completely different entanglement. $\endgroup$
    – Norbert Schuch
    Commented Jan 26, 2012 at 4:42
  • $\begingroup$ @NorbertSchuch Slightly off topic, but why wouldn't you call them correlation functions? $\endgroup$
    – Calvin
    Commented Jan 26, 2012 at 16:52
  • $\begingroup$ @Calvin: Wikipedia says "A correlation function is the correlation between random variables at two different points in space or time". $\endgroup$
    – Norbert Schuch
    Commented Jan 26, 2012 at 20:00

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It seems that such a measure for mixed states is fundamentally impossible, since you can have both entangled and separable states which have exactly the local expectation values. For pure states, monogamy of entanglement ensures that the impurity of a reduced density matrix (which can be infered from the expectation values of local Pauli operators) is directly related to entanglement. However for mixed states, this is not the case, as the following example will homefully make clear:

Consider a two qubit system, in which the two reduced density matrices are maximally mixed. In this case, it is possible the system is separable, composed of two copies of the maximally mixed state, or it is maximally entangled, composed of a single EPR pair, or anything in between.

Thus no function of local expectation values can distinguish separable from entangled states in general.

However, purity (which is a function of single site correlation functions), can indeed be used as a bound on the entanglement of a system, again due to monogomy of entanglement. If the local system is not maximally mixed, then it is not maximally entangled, and hence maximum amount of entanglement possible for a system is a monotonic function of its (im)purity.

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  • $\begingroup$ Thank you. Just to check, by local expectation value, you mean one made up of just one Pauli matrix, while something like $\langle \sigma_l^\alpha \sigma_{l+m}^\beta \rangle$ with $\alpha,\beta = x,y,z$ is not. $\endgroup$
    – Calvin
    Commented Jan 26, 2012 at 16:50
  • $\begingroup$ @Calvin: Exactly, although there is no need to restrict to just Pauli operators. $\endgroup$
    – Joe Fitzsimons
    Commented Jan 26, 2012 at 20:10
  • $\begingroup$ As Joe says, the local spectrum will still tell you something about entanglement if you have some kind of promise/information about the global purity. For example, if $S(\rho_{AB})<S(\rho_A)$, then the state is $A:B$-entangled $\endgroup$
    – Marco
    Commented Jan 26, 2012 at 21:06
  • $\begingroup$ Joe: Sorry to reply here, but the website won't let me edit your post (not enough characters); you've made a typo with 'homefully' (hopefully) and also 'monogomy' (monogamy). Please delete this comment, if you can, as it's otherwise useless. $\endgroup$
    – Noon Silk
    Commented Jan 26, 2012 at 22:45
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    $\begingroup$ @calvin: As long as the expectation values are all within the same part of the system, they all won't tell you anything about the entanglement (unless you are given additional information). $\endgroup$
    – Norbert Schuch
    Commented Jan 27, 2012 at 5:42
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For any multipartite state $\rho_{ABC...}$ the product state $\rho_A\otimes\rho_B\otimes\rho_C\otimes\ldots$, with $\rho_A$, $\rho_B$, $\rho_C$, ... the local reduced density matrices of $\rho_{ABC...}$, has exactly the same single-site expectation values. So, with only single-site expectation values one cannot say anything about any kind of correlation, not only about entanglement. On the other hand, thanks to the Schmidt decomposition and the invariance under local unitaries of entanglement measures, it is obvious that all bipartite entanglement measures on pure states can only depend on the spectrum of the reduced density matrix.

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