Right. But when you build a quantum computer, you want to have a certain set of gates that you want to implement, and all other gates (unitary matrices) can be built from that set of gates. This is known as a universal set. Surprisingly (or not), it is quite small. One example of a universal set is: $G = \{H, S, T, CNOT \}$ where

$$H = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}, \hspace{.25 cm} T =\begin{pmatrix} 1 & 0\\ 0 & e^{i\pi/4} \end{pmatrix}, \hspace{.25 cm} S = \begin{pmatrix} 1 & 0\\ 0 & i \end{pmatrix}, \hspace{.25 cm} CNOT = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$

This is just an example of one of many universal sets. These are the native gates you can say on your quantum hardware. For instance, on IBM hardware, the native gates are: { CNOT, I, RZ, SX, X }. The point is every unitary matrix can be approximated to an arbitrary, $\epsilon$, error using those gates. This formed your quantum circuit.

The matrix example you gave is the matrix representation of the *SWAP* gate. It is not a native gate on a quantum hardware but this is because you can simply implement it as a sequence of CNOT gates as follows:

1In addition to the other, fuller, answers, your SWAP matrix is a permutation, which is also

sufficientfor being a quantum gate. Permutations happen to benecessaryfor being implementable as a classical reversible gate, which are subsets of quantum gates. – Mark S – 2021-02-03T03:08:24.2101yes, effectively any universal quantum device can implement any unitary matrix up to arbitrary precision. – Condo – 2021-02-10T14:12:26.793