I read a Paper about quantum error corrections. I don't know why this is a CNOT gate. How to calculate this kind of CNOT gate as a topology form?

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1 Answer 1


There are a variety ways of showing this, depending on your prior knowledge. The simplest is to know that you can convert a braiding diagram into a ZX calculus graph by changing each ring into a connected component of nodes of the same type and putting edges between components whose rings are of opposite type and linked:



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which is the same as (by contracting adjacent nodes of the same color and discarding self-edges, which doesn't change the operation of the graph):


which is the same as (by discarding degree 2 nodes, which has no effect on the operation):

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Which, if you're familiar with the ZX calculus, is the standard CNOT graph. You can evaluate as a tensor network to see that it corresponds to the unitary matrix of the CNOT.

Alternatively, what you can do is show that the topological construction supports parity sheets corresponding to the stabilizer tableau of a CNOT operation:

$$\text{Tableau}(CNOT) = \begin{array}{r|cc|cc} & X_c & Z_c & X_t & Z_t \\ \hline c & X & Z & & Z \\ t & X & & X & Z \\ \end{array}$$

For example here is the parity sheet of the $X_c \rightarrow X_c X_t$ entry:

parity sheet

The main rule you need to know, in order to find these sheets, is that they can be extended through space arbitrarily but can only terminate on boundaries of the same type. So in this quick sketch above the rule I was following is that the "red" sheet could only terminate on the dark boundary, and I found a way to link a top-left tube to a bottom-right and bottom-left tube while respecting that rule.

I've given a talk going into this in more detail: https://www.youtube.com/watch?v=1ojXEEm_JiI

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  • $\begingroup$ Thank you for your rich reply. I watched your video, really interesting. But I am still thinking about how to use a mathematical method to get this CNOT gate... It is my first time to see this kind of topology for gates. $\endgroup$ Apr 1, 2021 at 8:29

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