Global phase is a property of a quantum state that emerges from the mathematical description but is physically completely unobservable. Multiplying an entire quantum state by a complex number of unit magnitude $e^{i\phi}$ produces a state that is physically identical to the original in every possible measurement.
To gain intuition, start from a single qubit with probability amplitudes $a = Ae^{i\alpha}$ and $b = Be^{i\beta}$ where $A, B \in \mathbb{R}$ and $\alpha, \beta \in (-\pi, \pi]$.
$$\lvert\psi\rangle = Ae^{i\alpha}\lvert 0\rangle + Be^{i\beta}\lvert 1\rangle$$
If we create a new qubit $\lvert\psi'\rangle$ by multiplying by some global phase $e^{i\phi}$, both amplitudes accrue the same shift $\phi$.
$$\lvert\psi'\rangle = e^{i\phi}\lvert\psi\rangle = Ae^{i(\alpha+\phi)}\lvert 0\rangle + Be^{i(\beta+\phi)}\lvert 1\rangle$$
The states $\lvert\psi\rangle$ and $\lvert\psi'\rangle$ are physically indistinguishable. We write this equivalence as $\lvert\psi\rangle \sim e^{i\phi}\lvert\psi\rangle$.
The Born rule gives measurement probabilities $|a|^2$ and $|b|^2$. Since $|e^{i\phi}| = 1$ for any real $\phi$, multiplying by a global phase leaves all probabilities unchanged.
$$P_0 = |Ae^{i(\alpha+\phi)}|^2 = A^2|e^{i(\alpha+\phi)}|^2 = A^2$$ $$P_1 = |Be^{i(\beta+\phi)}|^2 = B^2|e^{i(\beta+\phi)}|^2 = B^2$$
These are identical to the probabilities for $\lvert\psi\rangle$. No experiment can detect the global phase $\phi$.
Because global phase is unobservable, we can always choose to factor out $e^{i\alpha}$ and ignore it, placing the full phase difference onto the $\lvert 1\rangle$ amplitude.
$$\lvert\psi\rangle = e^{i\alpha}\!\left(A\lvert 0\rangle + Be^{i(\beta-\alpha)}\lvert 1\rangle\right) \sim A\lvert 0\rangle + Be^{i\varphi}\lvert 1\rangle \qquad \varphi = \beta - \alpha$$
The quantity $\varphi$ is the relative phase between the two probability amplitudes. Unlike global phase, relative phase is physically observable — it determines interference patterns and measurement outcomes in rotated bases. For example, $\lvert +\rangle = (\lvert 0\rangle + \lvert 1\rangle)/\sqrt{2}$ and $\lvert -\rangle = (\lvert 0\rangle - \lvert 1\rangle)/\sqrt{2}$ are physically distinct states even though they give the same probabilities in the computational basis: their relative phases differ by $\pi$.
Using the equivalence $\lvert\psi\rangle \sim A\lvert 0\rangle + Be^{i\varphi}\lvert 1\rangle$, we can write the most general qubit state using only three real numbers: the magnitudes $A, B \geq 0$ (with $A^2 + B^2 = 1$) and the relative phase $\varphi \in (-\pi, \pi]$. Substituting $A = \cos(\theta/2)$ and $B = \sin(\theta/2)$ gives the standard Bloch sphere parameterization.
$$\lvert\psi\rangle = \cos\frac{\theta}{2}\lvert 0\rangle + e^{i\varphi}\sin\frac{\theta}{2}\lvert 1\rangle$$