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.
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?
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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