A qubit state \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\) has two complex numbers (\(\alpha\) and \(\beta\)), which means 4 real numbers. But two of those are redundant:
The two surviving numbers become \(\theta\) and \(\phi\) on the Bloch sphere. Here's the full process with a concrete example.
$$|\psi\rangle = \frac{1+i}{2}|0\rangle + \frac{1-i}{2}|1\rangle$$
\(\alpha = \frac{1+i}{2}\) — real part \(\frac{1}{2}\), imaginary part \(\frac{1}{2}\)
\(\beta = \frac{1-i}{2}\) — real part \(\frac{1}{2}\), imaginary part \(-\frac{1}{2}\)
That's 4 real numbers: \((\frac{1}{2},\, \frac{1}{2},\, \frac{1}{2},\, -\frac{1}{2})\). We want to reduce this to just \(\theta\) and \(\phi\).
Goal: make \(\alpha\) real and positive by rotating both amplitudes.
\(\alpha = \frac{1+i}{2}\) points at \(45°\) in the complex plane. To land it on the positive real axis, rotate everything by \(-45°\) — multiply the entire state by \(e^{-i\pi/4}\).
First, turn \(e^{-i\pi/4}\) into a concrete number using Euler's formula:
$$e^{-i\pi/4} = \cos(-45°) + i\sin(-45°) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$$
Now multiply each amplitude by this number:
$$\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) \cdot \frac{1+i}{2} = \frac{1}{2\sqrt{2}}\Big[(1-i)(1+i)\Big]$$
$$= \frac{1}{2\sqrt{2}}\Big[1 + i - i - i^2\Big] = \frac{1}{2\sqrt{2}}\Big[1+1\Big] = \frac{1}{\sqrt{2}}$$
The \(+i\) and \(-i\) cancelled, and \(-i^2 = +1\). Result: real and positive ✓
$$\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) \cdot \frac{1-i}{2} = \frac{1}{2\sqrt{2}}\Big[(1-i)(1-i)\Big]$$
$$= \frac{1}{2\sqrt{2}}\Big[1 - i - i + i^2\Big] = \frac{1}{2\sqrt{2}}\Big[-2i\Big] = \frac{-i}{\sqrt{2}}$$
The real parts cancelled, leaving pure imaginary. β absorbed the relative phase.
$$|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{-i}{\sqrt{2}}|1\rangle$$
\(\alpha = \frac{1}{\sqrt{2}}\) — one real number.
\(\beta = \frac{-i}{\sqrt{2}}\) — still complex (length + angle) = two numbers.
3 real numbers remain.
\(\beta\) has a length and an angle. But the length isn't free — normalization forces it:
$$|\beta| = \sqrt{1 - |\alpha|^2} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$$
So \(\beta\)'s length is determined by \(\alpha\). The only free thing left in \(\beta\) is its angle.
The two surviving free parameters:
The Bloch sphere formula is a template that packages these two parameters:
$$|\psi\rangle = \underbrace{\cos\frac{\theta}{2}}_{\alpha}|0\rangle \;+\; \underbrace{e^{i\phi}}_{\beta\text{'s angle}} \cdot \underbrace{\sin\frac{\theta}{2}}_{\beta\text{'s length}}|1\rangle$$
We have two unknowns (\(\theta\) and \(\phi\)) and two equations to solve them.
$$\cos\frac{\theta}{2} = \alpha = \frac{1}{\sqrt{2}}$$
Reverse lookup on the unit circle: \(\cos(45°) = \frac{1}{\sqrt{2}}\), so \(\frac{\theta}{2} = \frac{\pi}{4}\).
$$\boxed{\theta = \frac{\pi}{2}} \quad\text{(equator of the Bloch sphere)}$$
\(\beta\) is made of angle × length. We know the length (\(|\beta| = \frac{1}{\sqrt{2}}\)), so divide it out to isolate the angle:
$$e^{i\phi} = \frac{\beta}{|\beta|} = \frac{-i\,/\,\sqrt{2}}{1\,/\,\sqrt{2}} = -i$$
Reverse lookup on the unit circle: where is \(-i\)? It's the point \((0, -1)\) — the bottom of the circle.
Verify: Euler's formula says \(e^{i\phi} = \cos\phi + i\sin\phi\). For this to equal \(-i = 0 + i(-1)\):
$$\cos\phi = 0, \quad \sin\phi = -1$$
$$\boxed{\phi = -\frac{\pi}{2}} \quad\text{(−90° = pointing −Y on the Bloch sphere)}$$
Before: \(\quad\frac{1+i}{2}|0\rangle + \frac{1-i}{2}|1\rangle\) — 4 real numbers, messy
After: \(\quad\frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle\) — clean form after global phase removal
Bloch sphere: \(\quad\theta = \frac{\pi}{2},\; \phi = -\frac{\pi}{2}\) — the \(|-i\rangle\) state (equator, pointing −Y)
Same qubit. Same probabilities. Same gate behavior. Just progressively simpler representations.