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I am trying to decompose a set of two qubit gates repecting parity, though I am not sure whether parity is the right word for this. The gate has the following structure:

$$\begin{bmatrix} u_{11}&0&0&u_{12} \\ 0&v_{11}&v_{12}&0 \\ 0&v_{21}&v_{22}&0 \\ u_{21}&0&0&u{22} \end{bmatrix} .$$

This symmetry requirement is required for quantum gates acting on two fermions. In the spin language it is preserving a $Z_2$ symmety.

I want to decompose this gate into a gate consisting CNOT gate and arbitrary single qubit gates. I adopt the optimal decomposition method for general two qubit gates. It generally takes three CNOT gates. Since CNOT gates are especially prone to error due to the longer evolution time times compared with single qubit gates, I wonder if I can reduce the number of CNOT gates further.

What structure in the two qubit gates would determine the number of CNOT gates used? If I want to preserve parity while having nontrivial unitary gates (other than identity), is three CNOT gates the least I can get?

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  • $\begingroup$ C-Z can do with less. $\endgroup$ – Norbert Schuch Sep 10 at 15:27
  • $\begingroup$ @NorbertSchuch That can't be right; the CZ differs from the CNOT only in single qubit rotations and in this context those are free. $\endgroup$ – Craig Gidney 2 hours ago
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You can't use fewer than three CNOTs for this task.

Consider the operation $U(a, b) = e^{ia(X\otimes X+Y\otimes Y)} \cdot e^{ibZ\otimes Z}$

This operation preserves $Z_2$ symmetry. In fact it preserves the total number of set bits, so it's even more restrictive.

When you take the Cartan KAK decomposition of this operation, you just get the same operation back. That is to say, the interaction coefficients are $(x=a, y=a, z=b)$. Whenever the three interaction coefficients of an operation are all not zero, you need three CNOTs in order to implement it.

Therefore, except for a measure zero subset where $a=0$ or $b=0$, $Z_2$ preserving operations require three CNOTs.

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