# 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.

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Did you mean to put $\sqrt{3}/2$ in the first sentence? – axblount Jul 23 '12 at 19:04
yes... sorry about that – fosho Jul 23 '12 at 19:07
Given that it is $\left(\frac{\sqrt{3}}2\right)+x^2=1$ there are two real solutions. Please add the first sentence back into the post, as it explains the question. – Ross Millikan Jul 23 '12 at 20:40
PLEASE, don't post and answer the questions from www.coursera.org. This one is taken from www.coursera.org/course/qcomp (First assignment, problems 8 and 9). See www.coursera.org/about/terms for more information. Moderators, please, delete this question and the answer to it. Thanks. – Physicsworks Jul 24 '12 at 18:07

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.

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and In C2, how many complex (unit) vectors are there whose projection onto |1⟩ has length 3√2? – fosho Jul 23 '12 at 19:22
@Daniel: Every $\beta$ such that $|\beta|=1/2$ will make $\frac{\sqrt{3}}{2}|1\rangle+\beta|2\rangle$ such a unit vector. (I assume you mean $\sqrt{3}/2$ again.) So: how many complex numbers $\beta$ are there such that $|\beta|=1/2$? Infinity! – anon Jul 23 '12 at 19:24
@Daniel: Picture the Bloch sphere (space that the qubits lie in $\mathbb{C}^2$). The set of complex vectors with the above described projection onto $|1 \rangle$ can be represented by a circle that is $\sqrt(3)/2$ away from the origin. – tatterdemalion Jul 23 '12 at 20:39