Convert state Vectors to Bloch Sphere angles

Question

I think this question is a bit low brow for the forum. I want to take a state vector $ \alpha |0\rangle + \beta |1\rangle $ to the two bloch angles. What's the best way? I tried to just factor out the phase from $\alpha$, but then ended up with a divide by zero when trying to compute $\phi$ from $\beta$.

Answer 1

You are probably dividing by $\alpha$ at some point to eliminate a global phase, leading to your divide by zero in some cases. It would be better to get the phase angles of $\alpha$ and $\beta$ with $\arg$, and set the relative phase $\phi=\arg(\beta)-\arg(\alpha)$. Angle $\theta$ is now simply extracted as $\theta = 2\cos^{-1}(|\alpha|)$ (note that the absolute value of $\alpha$ is used). This is all assuming that you want to get to

$$|\psi\rangle = \cos(\theta/2)|0\rangle + \mathrm{e}^{i\phi}\sin(\theta/2)|1\rangle\,,$$

which neglects global phase.

Answer 2

$\phi$ is the relative phase between $\alpha$ and $\beta$ (so the phase of $\alpha/\beta$). You will only get zero or divide-by-zero when $\alpha=0$ or $\beta=0$. But in that case, $\phi$ is arbitrary. And when $\alpha$ or $\beta$ are close to zero, you are near the poles of the Bloch sphere, and $\phi$ doesn't really matter that much.

Answer 3

To those who may be having a similar question: make sure you are normalizing the qubit before trying to find the Bloch angles.

Take for example the eigenvectors of the $X$ operator (Pauli matrix $\sigma^x$), in the $\{|0\rangle, |1\rangle\}$ basis:

$$ -|0\rangle + |1\rangle \\ |0\rangle + |1\rangle $$

If you try to use those coefficients directly, you will try for example $\cos \theta/2 = \pm 1$, which will lead to the inconsistent equation that the question mentions. But the right coefficients, after normalizing, should be:

$$ -\frac{ 1}{ \sqrt{2}}|0\rangle + \frac{ 1}{ \sqrt{2}}|1\rangle \\ \frac{ 1}{ \sqrt{2}}|0\rangle + \frac{ 1}{ \sqrt{2}}|1\rangle $$

These give you the correct positions in the Bloch sphere: $(\pi/2,\pi),(\pi/2,0)$