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What is the intuition behind the following entanglement distillation protocol for continuous variable systems?

Quantum Computing Asked by Devjyoti Tripathy on July 25, 2020

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?

One Answer

To see intuitively why this protocol increases the entanglement after each iteration, we can work out an example where our initial state is say $lvert 00rangle+lvert 11rangle$. Upon passing through a 50:50 Beam Splitter, we get: $$|00rangle+|11rangle = |00rangle+ a_{0}^{+}a_{1}^{+}|00rangle hspace{3mm}transformshspace{3mm}|00rangle + frac{1}{2}(a_{2}^{+} + ia_{3}^{+})(ia_{2}^{+} + a_{3}^{+})|00rangle \ =|00rangle + frac{i}{sqrt2}(|20rangle+|02rangle)$$ (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 $$|psi>=frac{1}{coshr}sum^{infty}_{n=0} (-1)^{n}e^{inphi}tanh^{n}r |n,n>$$

Answered by Devjyoti Tripathy on July 25, 2020

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