# Type I fusion Gate

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.

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.
• 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. Nov 30, 2023 at 2:08
• 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 Nov 30, 2023 at 2:26