I have looked at the following:
What is the difference between a relative phase and a global phase? In particular, what is a phase?
Global and relative phases of kets in QM
Global phases and indistinguishable quantum states, mathematical understanding
If two states differ by a scalar of magnitude of 1, then they are indistinguishable. Consider: \begin{align} \vert \psi_1 \rangle &= \dfrac{1}{\sqrt{2}} \vert 0 \rangle + \dfrac{i}{\sqrt{2}} \vert 1 \rangle\\ \vert \psi_2 \rangle &= \color{red}{i}\left(\dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle\right). \end{align}
Which of the following is true about $\vert \psi_1 \rangle$ and $\vert \psi_2 \rangle$?
- $\vert \psi_1 \rangle = \vert \psi_2 \rangle$
- $\vert \psi_1 \rangle \neq \dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle$
- $\vert \psi_1 \rangle = \dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle$ up to global phase.
- If we just ignore the global phase in $\vert \psi_2 \rangle$ and only deal with $\dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle$ , do we still have the state vector on a Bloch sphere yields the same projection as $\vert \psi_1 \rangle$?
Lastly, since the global phase is not physically observable, is it mathematically evident?