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Quantum gates are basically matrices belonging to $C_{2\times2}$. Now, what are the properties of these matrices? We know that to preserve the normalization factor of the qubits these are unitary. All good. But then again we know that every unitary matrix is represented by rotation around Z, Y, and Z-axis (I am mentioning the Z-Y decomposition here).

So that means any unitary matrix is a rotation?

One funny thing, is this similar to rotation matrices similar to $R^{3}$ with determinant 1 etc?

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  • $\begingroup$ I'm sure you mean single-qubit gates. Note that $n$-qubit gates are elements of $SU(2^n)$ and their properties are summarized here: special unitary group. $\endgroup$ – keisuke.akira Nov 10 '20 at 11:51
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Yes, unitary matrix is a natural generalization of orthogonal matrix (which is a rotation with a possible sign flip) in the case of complex numbers. It's defined as a linear map that doesn't change inner products between vectors in $\mathbb{C}^n$.

A unitary matrix with a real numbers is exactly an orthogonal matrix. We have $U^\dagger = (\overline{U})^T$ for any matrix $U$, for unitary we have $UU^\dagger=I$, and for real we have $U=\overline{U}$. Hence $UU^T=I$.

Their determinant can be any complex number with modulus $1$.

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  • $\begingroup$ Thanks for confirming. $\endgroup$ – user27286 Nov 10 '20 at 9:22
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Since I think question 1 is solved well by the upper answer, I will skip to question 2.

They have some similar property with the rotation in $R^3$, but from a group theory point of view, the quantum gates(2D unitary matrix) belong to a larger group, $SU(2)$, while 3D rotations belong to $SO(3)$, and there is a two-to-one map between $SU(2)$ and $SO(3)$ ($SU(2)$ contains $SO(3)$). For a detailed explanation, see Wikipedia.

And there is a more direct way to see the difference, if you draw the Bloch sphere of a quantum state sometimes you need to denote their phase while in the general $R^3$ case, this is nonsense.

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  • $\begingroup$ Thanks for clarifying. $\endgroup$ – user27286 Nov 10 '20 at 16:46

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