How to get the Bloch sphere angles given an arbitrary qubit?

Question

I understand that given a pure state $ |\psi\rangle$, we can express it in terms of two angles $\theta$ and $\varphi$ such that $|\psi\rangle = \cos(\theta/2)|0\rangle + \mathrm{e}^{i\varphi}\sin(\theta/2)|1\rangle $, and this is derived by converting from $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ into their representations in terms of $r\mathrm{e}^{i\theta}$, and then factoring and rearranging that.

But how do I convert between the two representations given arbitrary states? I know that $|0\rangle = (\theta,\varphi) = (0,0), |1\rangle = (\pi,0), |+\rangle = (\pi/2, 0)$ etc, but how do I get it for an arbitrary state $|\psi\rangle$?

So far, I have:

1. If $\alpha$ is complex, shift the entire state by phase $\bar{\alpha}$, where $\bar\alpha$ is the complex conjugate of $\alpha$, to end up with $\alpha\bar\alpha |0\rangle + \bar\alpha\beta|1\rangle$
2. Use the formulas:

$$ \theta = 2 \arccos(\alpha\bar\alpha) \\ \varphi = ??? $$

Answer 1

As in if given $ |\psi\rangle=\alpha |0\rangle+\beta |1\rangle$, and you want it in the form $\cos(\theta/2) |0\rangle+e^{i\phi}\sin(\theta/2) |1\rangle$?

Assume the state is normalised $|\alpha|^2+|\beta|^2=1$.

I'd first start by multiplying by $\frac{\alpha^*}{|\alpha|}$ which is a phase (complex number unit length), and use $\alpha \alpha^*=|\alpha|^2$.

$$\frac{\alpha^*}{|\alpha|}|\psi\rangle=|\alpha| |0\rangle+\frac{\beta\alpha^*}{|\alpha|}|1\rangle =\cos(\theta/2) |0\rangle+e^{i\phi}\sin(\theta/2) |1\rangle$$

So $\theta=2\arccos(|\alpha|)$ or $\theta=2\arcsin(|\beta|)$.

Then $\frac{\beta\alpha^*}{|\alpha|}=e^{i\phi}|\beta|$ or $\frac{\beta\alpha^*}{|\alpha||\beta|}=e^{i\phi}$,

so that $\phi=\arg\left(\frac{\beta\alpha^*}{|\alpha||\beta|}\right)$ and it depends on how you want to calculate that, which branch of the loagarithm to take/where to measure angles from.

You could do something like $\phi=-i\ln\left(\frac{\beta\alpha^*}{|\alpha||\beta|}\right)$ etc.

In summary try

$$ \theta = 2 \arccos(|\alpha|) \\ \varphi = \arg\left(\frac{\beta\alpha^*}{|\alpha||\beta|}\right) $$

Answer 2

I found that the accepted answer does not allow $ |\alpha| $ or $ |\beta| $ to be zero. To allow this, I first find the phase/argument of $ \alpha $ and then rotate $ \beta $ with a factor $ \exp(-i\arg(\alpha))$.

Apologies for the Python, but then I get:

import cmath
import math

alpha = 0.5 + 0.5j
beta = 0.5 - 0.5j

magn = math.sqrt(abs(alpha)**2 + abs(beta)**2)
alpha /= magn
beta /= magn

theta = cmath.acos(abs(alpha)) * 2
if theta:
    phi = cmath.log(beta * cmath.exp(-1j*cmath.phase(alpha)) / cmath.sin(theta / 2)) / 1j
else:
    phi = 0

theta = round(theta.real, 2)
phi = round(phi.real, 2)

print(f"{theta=}, {phi=}")