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In the type I fusion gate of the Rudolph-Browne protocol, I fail to understand how the detection of a photon means the entanglement of the two Bell pairs.

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If you consider two Bell pairs (in polarization), the initial state is $$ \frac{1}{2}\big(|HH\rangle + |VV\rangle\big) \otimes \big(|HH\rangle + |VV\rangle\big) $$ Now consider applying a Type-I fusion gate between one qubit from each pair. The original paper describes the projection operator corresponding to detection of a single-photon as $$ \frac{1}{2}\big(|H\rangle\langle HH| \pm |V\rangle\langle VV|\big) $$ with the sign determined according to whether a horizontal or vertical polarized photon is detected. If you apply this projection to the initial state you will obtain a 3-qubit entangled state equivalent to a graph state whose underlying graph is a line. This is what is described in Fig 3a of the paper.

Note this is not the same as "entanglement of the two Bell pairs" - there are no more Bell pairs, there is a 3-qubit state. When they succeed, Type-I and Type-II fusions create larger entangled states from smaller ones.

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  • $\begingroup$ thank you. it helped! Could I ask you another question on this fusion gate. The authors talk about creating 2-tree cluster states where the central qubit is redundantly encoded from a 4-qubit GHZ state by locally acting with a Hadamard on two qubits of that GHZ state. I fail to see how that gate would do that? $\endgroup$
    – physics22
    Nov 25, 2023 at 13:41
  • $\begingroup$ where does it talk about 2-tree cluster states? $\endgroup$
    – ChrisD
    Nov 28, 2023 at 4:34
  • $\begingroup$ Sorry, I should have linked to this paper arxiv.org/pdf/quant-ph/0702044.pdf (on page 3). And I'm also confused because they say're going to show that LOQC is possible as long as the product of photon detector and source is larger that 2/3, but I don't see them doing that in the paper. $\endgroup$
    – physics22
    Nov 28, 2023 at 20:15
  • $\begingroup$ It's probably easiest to see in a circuit diagram. Prepare the 4-GHZ in the usual way ($|+\rangle|0\rangle|0\rangle|0\rangle$ followed by three CNOTs). Then apply Hadamards to qubits 3 and 4, then propagate them backwards. This will give you a circuit that you can think of as preparing $|\overline{+}\rangle |+\rangle |+\rangle$ where the $|\overline{+}\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$ is a redundantly encoded $|+\rangle$, followed by CPHASE gates between this and the other two (unencoded) qubits. This is the definition of a three-qubit tree cluster state. $\endgroup$
    – ChrisD
    Nov 30, 2023 at 2:08
  • $\begingroup$ Which they then call a "2-tree" for some reason. The result that efficient LOQC is possible if $\eta_D\eta_S > 2/3$ depends on a result from their earlier paper (Ref[15]), so yes they are only showing part of it in this one $\endgroup$
    – ChrisD
    Nov 30, 2023 at 2:26

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