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The protocol is: We start with a supply of identically prepared bipartite non-Gaussian states. The overall protocol then amounts to an iteration of the following basic steps.

  1. The states will be mixed pairwise locally at 50:50 beam splitters.
  2. On one of the outputs of each beam splitter, a photon detector distinguishes between the absence and presence of photons. It should be noted that we do not require photon counters that can discriminate between different photon numbers.
  3. In case of absence of photons at both detectors for a particular pair, one keeps the remaining modes as an input for the next iteration, otherwise the state is discarded. This is one iteration of the protocol which we will continue until we finally end up with a small number of states that closely resemble Gaussian states. enter image description here

This protocol is presented in the paper: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.67.062320

Specifically, what I am trying to understand is: why does vacuum detection at both outputs lead to distillation? What is the motivation for this step?

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  • $\begingroup$ what is the question? it seems like you are just asking for a review of the paper, which is too broad a question. Can you edit the question to be more specific on what you don't find clear in the paper/protocol? $\endgroup$ – glS Jun 27 at 15:54
  • $\begingroup$ I hope this version of the question is more specific. $\endgroup$ – Devjyoti Tripathy Jun 28 at 12:31
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To see intuitively why this protocol increases the entanglement after each iteration, we can work out an example where our initial state is say $\lvert 00\rangle+\lvert 11\rangle$. Upon passing through a 50:50 Beam Splitter, we get: $$|00\rangle+|11\rangle = |00\rangle+ a_{0}^{+}a_{1}^{+}|00\rangle \hspace{3mm}transforms\hspace{3mm}|00\rangle + \frac{1}{2}(a_{2}^{+} + ia_{3}^{+})(ia_{2}^{+} + a_{3}^{+})|00\rangle \\ =|00\rangle + \frac{i}{\sqrt2}(|20\rangle+|02\rangle)$$ (Ignore normalisation in the above equations) Now, upon detecting vacuum in this state and another copy of this state, we clearly see that the remaining modes now occur in a form of superposition very similar to the two mode squeezed sates which are entangled. And since the recurrence happens by the action of the same beam splitter on similar kind of states, we can intuitively see that we get closer and closer to the superposition $$|\psi>=\frac{1}{coshr}\sum^{\infty}_{n=0} (-1)^{n}e^{in\phi}tanh^{n}r |n,n>$$

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