Entanglement is a purely quantum mechanical phenomena that e.q. increase precision of quantum interferometers, atomic clocks or have application in quantum computing. There are many "measures" of entanglement and one of the most recognized is the von Neumann entropy

$$S[\hat{\rho}] = -Tr[\hat{\rho}\ln\hat{\rho}],$$

which is exactly $0$ for separable states and grater than $0$ for entangled states. In order to determine the value of Entropy one needs to know the density matrix operator $\hat{\rho}$. Experimentally, this can be done using quantum tomography methods. There are also different measures of entanglement that also require the knowledge of density matrix operator.

I was wondering if there is some indirect method (response to weak perturbations, transport properties, noise etc.) that could potentially be connected to entanglement measure, but does not destroy the quantum state or requires just one destructive measurement (a few)? Recently, Hauke et al. showed that quantum Fisher information - measure of mutlipartite entanglement - can be determined via dynamic susceptibilities. Do you know of any other possibilities?

  • $\begingroup$ Your question is quite broad: Do you want e.g. to assume purity of the state? (Your discussion of the von Neumann entropy suggests so, but this is a rather strong assumption.) Are you allowed to measure the joint state or only individual parts. Are non-desctructive measurements "good", and if yes, do you care about the pre- or post-measurement state? What do you mean by "a few"? A single measurement will certainly not suffice (unless you repeat it on many copies, allow for non-projective measurements, or allow to measure the joint state). $\endgroup$ Sep 23, 2015 at 20:37
  • $\begingroup$ This probably isn't quite what you had in mind, but it may be of interest: measuring entanglement entropy by site-resolved interference of two copies of the same system. arxiv.org/abs/1509.01160 . It's not a generic method, though. $\endgroup$
    – Rococo
    Sep 24, 2015 at 0:14
  • 1
    $\begingroup$ Did you try to read up on entanglement witnesses? $\endgroup$ Sep 24, 2015 at 11:51
  • 2
    $\begingroup$ What you can measure in an experiment are exactly linear functionals of the density operator. The great thing is that by measuring a single such functional and finding that it is negative, you know that your state is entangled (and the number even gives you a bound on how much). Note that people have developed witnesses which can be measured in a simple way experimentally. -- BTW, if you don't use @NorbertSchuch in your replies, I don't get notified and only find out that you replied by accident by revisiting the question. $\endgroup$ Sep 25, 2015 at 6:55
  • 2
    $\begingroup$ E.g., arxiv.org/abs/0811.2803 contains an overview of both theory and some experiments. $\endgroup$ Sep 25, 2015 at 12:00

1 Answer 1


Entanglement witnesses are a way to probe entanglement through measurements.

An entanglement witness is a linear operator $W$ which is constructed such that $\mathrm{tr}[W\rho_s]\ge0$ for any unentangled state $\rho_s$. Thus, if we find that $\mathrm{tr}[W\rho]<0$, we know that $\rho$ is entangled. Moreover, for any entangled state $\rho$ we can construct a corresponding witness $W_\rho$ such that $\mathrm{tr}[W_\rho\rho]<0$.

Note that $\mathrm{tr}[W\rho]$ is just the expectation value of $W$ in the state $\rho$, this is, it can be determined by repeatedly measuring $W$. Moreover, we can expand $W$ in a product basis, this is, we can determine $\mathrm{tr}[W\rho]$ by doing separate measurements on the individual qubits. Also, witnesses can be optimized to be easily measurable.

An overview, including experimental implementations of witness measurements, can be found e.g. in this review paper by Gühne and Toth.

  • 1
    $\begingroup$ Note that, as Norbert said, for any entangled state one can create a witness that will separate it from separable states, but it is not possible to construct one that will work for all states. Any given witness will fail to distinguish some entangled states from separable ones. This is due to the Hahn-Banach theorem. $\endgroup$
    – Rococo
    Oct 2, 2015 at 1:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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