Global phase is a property of a quantum state that comes out of mathematics description of a quantum state. However it is physically completely unobservable and we often ignore it.

To gain an intuition what global phase looks like, let us start from a single qubit written using Dirac notation. Probability amplitudes $a\in\mathbb{C}$ and $b\in\mathbb{C}$ are complex numbers which are respectively associated with the states $\lvert 0\rangle$ and $\lvert 1\rangle$. According to the Born rule, their modulus squared $|a|^2$ and $|b|^2$ are the real probabilities the qubit would collapse to either state $\lvert 0\rangle$ or $\lvert 1\rangle$

$$\lvert\psi\rangle = a\lvert 0\rangle + b\lvert 1\rangle$$

Because $a$ and $b$ are complex numbers, we can rewrite them as $Ae^{i\alpha}$ and $Be^{i\beta}$ respectively $A, B\in\mathbb{R}$ and $\alpha,\beta\in(-\pi, \pi]$.

$$\lvert\psi\rangle = Ae^{i\alpha}\lvert 0\rangle + Be^{i\beta}\lvert 1\rangle$$

What would happen if created a new qubit $\lvert\psi'\rangle$ by multiplying the qubit $\lvert\psi\rangle$ by some arbitrary phase $e^{i\phi}$?

$$\lvert\psi'\rangle = e^{i\phi}\lvert\psi\rangle = Ae^{i(\alpha + \phi)}\lvert 0\rangle + Be^{i(\beta + \phi)}\lvert 1\rangle$$

It looks like both probability amplitudes accrued the same phase $\phi$. So are qubits $\lvert\psi\rangle$ and $\lvert\psi'\rangle$ different in any way? No. In fact, they are completely indistinguishable. Global phase has no effect on probabilities and no effect on interference. There is no experiment we can do to distinguish observe global phase change. Thus, we call these two qubits identical up to global phase.

$$\lvert\psi\rangle \sim e^{i\phi}\lvert\psi\rangle$$

Because global phase is unobservable, we can even simplify the expression for the original qubit $\lvert\psi\rangle$. We rotate both probability amplitudes $a\in\mathbb{C}$ and $b\in\mathbb{C}$ in the complex plane by an equal amount such that one of them becomes real. In other words, we can factor either $e^{i\alpha}$ or $e^{i\beta}$ and then just ignore it. By convention the probability amplitude $a$ associated with $\lvert 0\rangle$ is modified to be real, while the probability amplitude $b$ associated with state $\lvert 1\rangle$ is tasked with carrying information relative phase $e^{i\beta - \alpha}$

$$\lvert\psi\rangle = e^{i\alpha}\left(A\lvert 0\rangle + Be^{i(\beta - \alpha)}\lvert 1\rangle\right)$$

Here we can identify $\lvert\psi\rangle \sim e^{i\alpha}\lvert\psi\rangle$ and let $\varphi = \beta - \alpha$ be the relative phase between probability amplitudes. Unlike global phase, relative phase is actually important for entanglement.

$$\lvert\psi\rangle = A\lvert 0\rangle + Be^{i\var}\lvert 1\rangle$$

First let's focus on $\lvert\psi\rangle$.

$$\lvert\psi\rangle = Ae^{i\alpha}\lvert 0\rangle + Be^{i\beta}\lvert 1\rangle$$

Let's apply the Born's rule. Only thing to remember is that $|e^{ix}| = 1$ because it represents a complex number on the unit circle.

$$P_0 = |a|^2 = |Ae^{i\alpha}|^2 = |A|^2|e^{i\alpha}|^2 = |A|^2\cdot 1 = |A|^2 $$ $$P_1 = |b|^2 = |Be^{i\beta}|^2 = |B|^2|e^{i\beta}|^2 = |B|^2\cdot 1 = |B|^2 $$

Now let's turn to $\lvert\psi'\rangle = e^{i\varphi}\lvert\psi\rangle$ $$\lvert\psi'\rangle = e^{i\varphi}\lvert\psi\rangle = e^{i\varphi}\left(Ae^{i\alpha}\lvert 0\rangle + Be^{i\beta}\lvert 1\rangle\right) = Ae^{i(\alpha + \varphi)}\lvert 0\rangle + Be^{i(\beta + \varphi)}\lvert 1\rangle$$

From this, we can tell that both probability amplitudes accrued a phase shift $\varphi$. This is the global phase.

$$Ae^{i\alpha}\rightarrow Ae^{i(\alpha + \varphi)}$$ $$Be^{i\beta}\rightarrow Be^{i(\beta + \varphi)}$$

So now let's reapply Born rule $$P_0 = |Ae^{i(\alpha + \varphi)}|^2 = |A|^2|e^{i(\alpha + \varphi)}|^2 = |A|^2\cdot 1 = |A|^2 $$ $$P_1 = |Be^{i(\beta + \varphi)}|^2 = |B|^2|e^{i(\beta + \varphi)}|^2 = |B|^2\cdot 1 = |B|^2 $$

The probabilities are identical

At this point, we can factor out either $e^{i\alpha}$ or $e^{i\beta}$ let's do the former.