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I am studying how to decompose a $4 \times 4$ unitary matrix into multiple 2-level unitary matrices.

I have found a total of six 2-level unitary matrices, and they are as follows.

enter image description here

At this point, I realized that (1,2) corresponds to $0V$, (3,4) corresponds to $1V$, (1,3) is $V0$, and (2,4) is $V1$, similar to a controlled-$V$ gate. enter image description here

When considering the decomposition using six 2-level unitary matrices, it is known that while the controlled V gate can be further decomposed using single-qubit gates and $\text{CNOT}$ gates, the (1,4) and (2,3) combinations are more challenging to decompose because they simultaneously affect two qubits.

My question is, when you have an arbitrary $4 \times 4$ unitary matrix and decompose it using these six 2-level unitary matrices, there seems to be no issue with not using (1,4) and (2,3).

Does that mean you don't need to use them at all? I'm curious about the reason behind this.

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Yes, you do not need to use all types of 2-level unitaries. Controlled NOT gates can be used to exchange rows/columns 1 and 2, or 3 and 4. For example, you can exclude the type (2,3) by using $(2,3)=\mathrm{CNOT}\cdot (2,4)\cdot \mathrm{CNOT}$ where $\mathrm{CNOT}=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}$ is the CNOT gate, which itself is of type (3,4).

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  • $\begingroup$ The reason I wanted to know is accurate! What do you think about this idea? $\endgroup$
    – junghyunHa
    Dec 28, 2023 at 3:21

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