Qubits and vector projections In $\Bbb C^2$, how many real unit vectors are there whose projection onto $|1\rangle$ has length $\sqrt{3}/2$?
I would think zero as $\bigl(\frac{\sqrt{3}}{2}\bigr)^2 + x^2 = 1$, therefore there are no real values of $x$ to satisfy the equation. Please help.
 A: Write $\Bbb C^2$ as $\Bbb C|1\rangle\oplus \Bbb C|2\rangle$ and the canonical projection $P$ down to the subspace spanned by $|1\rangle$,
$$P:\alpha|1\rangle+\beta|2\rangle\mapsto\alpha|1\rangle.$$
Then $P(\cdot)=\frac{\sqrt{3}}{2}|1\rangle$ means $\alpha=\frac{\sqrt{3}}{2}$. Furthermore, the condition of being a unit vector means
$$\left\|\frac{\sqrt{3}}{2}|1\rangle+\beta|2\rangle\right\|=1\iff \left|\frac{\sqrt{3}}{2}\right|^2+|\beta|^2=1\iff |\,\beta|=\frac{1}{2}.$$
The only two real solutions for $\beta$ are $\pm1/2$, so the answer is two. The two real-coordinate unit vectors are given by $\frac{\sqrt{3}}{2}|1\rangle\pm\frac{1}{2}|2\rangle$. Note that losing the restriction of real-coordinates means that $\beta$ could be chosen anywhere on the circle of radius $1/2$ about the origin in the complex plane $\Bbb C$.

As an anonymous user points out in a suggested edit to my answer, I did not really pay attention to the first comment below. The question doesn't say the projection has coordinate $\sqrt{3}/2$, it says this is the length of the projection, so $\alpha$ may be $\pm\sqrt{3}/2$. This yields four possible solutions total.
