This video is the best derivation I found: https://www.youtube.com/watch?v=a-dIl1Y1aTs.

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A single complex number z = a + ib has two degrees of freedom, the two real numbers a and b.

A qubit, or a two-level quantum state can be represented with two complex numbers, the amplitudes of the $$\ket{\psi} = \alpha\ket{0} + \beta\ket{1}$$

But $$\alpha$$ and $$\beta$$ are not unconstrained parameters.

First, we have the normalization constraint, which means that $$\lvert \alpha \rvert^2 + \lvert \beta \rvert^2 = 1$$.

So we may write $$\ket{\psi}$$ as:

$\ket{\psi} = {\rm e}^{i\phi_0}\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\phi_1}\sin{\frac{\theta}{2}}\ket{1}$

with $$0 \le \theta \le \pi$$ (since only positive real parts of the amplitude suffice), and $$0 \le \phi_0, \phi_1 \lt 2\pi$$ (since $$2\pi$$ is the same phase as $$0$$). Using $$\frac{\theta}{2}$$ is more convenient than $$\theta$$ for visualization.

Secondly, the global phase of a qubit does not matter – only the relative phase between the amplitudes is physically detectable. This means:

$\ket{\psi} = {\rm e}^{i\phi_0}\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\phi_1}\sin{\frac{\theta}{2}}\ket{1} = {\rm e}^{i\phi_0}\left(\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\left(\phi_1-\phi_0\right)}\sin{\frac{\theta}{2}}\ket{1}\right)$

Ridding the global phase factor of $${\rm e}^{i\phi_0}$$, and writing $$\phi = \phi_1 - \phi_0$$, this state is equivalent to:

$\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\phi}\sin{\frac{\theta}{2}}\ket{1}$

So now we are able to describe a two-level quantum state with only two real numbers, $$0 \le \theta \le \pi$$ and $$0 \le \phi \lt 2\pi$$.

For $$\ket{\psi} = \ket{0}$$, $$\theta = 0$$, and for $$\ket{\psi} = \ket{1}$$, $$\theta = \pi$$ and $$\phi = 0$$.

While we could scatter plot qubit states in a 2-D $$\theta {\rm vs.} \phi$$ chart, visualizing using spherical co-ordinates gives a richer and more useful picture. In this spherical co-ordinate system since we have only unit vectors, we can render qubit states as points on the unit sphere with $$\ket{0}$$ on the north pole, and $$\ket{1}$$ on the south pole of the sphere.