This video is the best derivation I found:

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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.