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I keep running into this statement everywhere I go, and the source is never quoted. Is it because the entries for the matrix representing the gate are complex numbers and hence uncountably infinite? Where do I find a source for this?

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The most general transformation implemented by a unitary gate is of the form $$ U=e^{i\xi}\left(\begin{array}{cc} a&b\\ -b^*&a^*\end{array}\right)\, , \tag{1} $$ where $a,b\in\mathbb{C}$ and $\vert a\vert^2+\vert b\vert^2=1$. A convenient way to reparametrize this is with $a=e^{i\varphi}\cos\theta$ and $b=e^{i\gamma}\sin\theta$, with $0\le \xi,\varphi,\gamma \le 2\pi$ and $0\le \theta\le \pi$.

As there is no other constraint on an arbitrary $2\times 2$ unitary, there are clearly infinitely many possible 4-tuples $(\xi,\varphi,\theta,\gamma)$, and moreover any range of the parameter is dense on its range. That's enough to show what you want.

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All phase-shifting operators, i.e. matrices of the form $\begin{pmatrix}1&0\\ 0&e^{i\theta}\\ \end{pmatrix}$, with arbitrary $\theta \in [0, 2\pi)$, are gates. There are $2^{\aleph_0}$ of these.

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    $\begingroup$ $\aleph_1,$ if $\theta \in \mathbf{R}.$ $\endgroup$
    – Mark H
    Dec 29, 2022 at 8:56
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    $\begingroup$ @MarkH We know $|[0, 2\pi)| = |\mathbf{R}| = 2^{\aleph_{0}}$, but it's unsettled/unknown if $2^{\aleph_{0}}=\aleph_{1}$ (this is continuum hypothesis). $\endgroup$ Dec 30, 2022 at 18:30
  • $\begingroup$ Sorry, Mark H was pointing out a mistake in my answer (I had written aleph_0), then I edited my answer. $\endgroup$
    – Plop
    Dec 31, 2022 at 11:27

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