I am stuck in simplyfing the following cNOT teleportation in ZX-calculus. I don't know how to proceed further. The circuit I start from is taken from this thesis (Fig 2.14, page 22).

Which property can I use to simplify the circuit further? Am I forced to sit down and compute the matrix product?

In case the image is not super clear, in d), the element in the bottom green Z spider is $(-1)^{b_1} \pi b_2$ (I could have removed the $(-1)^{b_1}$ already as it is multiplied by $\pi$ but I kept it to be sure you see my step-by-step calculation).

enter image description here


1 Answer 1


I think I see where you went wrong. In ZX-Calculus, only connectivity matters. So in your final stage, you can connect the middle two phaseless nodes with a vertical line, instead of the cup. Then, removing the middle two phaseless nodes using the identity rule, you are left with a CNOT gate.

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    $\begingroup$ Ooh right. It was actually very simple. Thanks $\endgroup$ Sep 27, 2023 at 9:29
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