# Why are the number of one qubit quantum gates uncountably infinite?

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?

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.
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.
• $\aleph_1,$ if $\theta \in \mathbf{R}.$ Commented Dec 29, 2022 at 8:56
• @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). Commented Dec 30, 2022 at 18:30