# What does the phase $\phi_1$ in a state $|\psi\rangle=a_1|0\rangle+a_2|1\rangle$ with $a_j=r_j e^{i\phi_j}$ say about state $|1\rangle$?

I'm a beginner in quantum computing and this question has been bugging me for quite some time. I have seen in various articles that a qubit is a device whose state can be represented by a unit vector in a 2-dimensional "complex" vector space. That is $$|\Psi> = a_1 |0> + a_2 |1>$$ where $$a_1$$ and $$a_2$$ are complex numbers that can be represented as $$a_1 = r_1 e^{i\Phi_1}$$ where $$r_1$$ the amplitude of the complex number $$a_1$$, it represents the square root of the probability of the qubit to be in "state 1" on measurement.

But what confuses me is that what does the phase of $$a_1$$, $$\Phi_1$$ represent about the "state 1". Does it even represent anything with respect to "state 1"? Or does the qubit just have a global phase (is it the phase difference between the coefficients? ) which has nothing to do with state |0> or |1> individually?

The phase $$\phi_1$$ by itself does not mean much (because of this global phase issue). However, the phase difference $$\phi_2-\phi_1$$ means a lot. You'll never see it if all you're doing is interpreting your qubit in a classical way (probability of getting 0 or 1), but that's not the only thing you can do with your qubit. For example, I could perform an X measurement, and ask "what is the probability of getting one of the answers $$(|0\rangle\pm|1\rangle)/\sqrt{2}$$?". (Alternatively, perform a Hadamard rotation on your qubit, and then perform a standard measurement.)
\begin{align} |\psi\rangle &= a_1|0\rangle + a_2|1\rangle \\ &= r_1e^{i\phi_1}|0\rangle + r_2e^{i\phi_2}|1\rangle\\ &= e^{i\phi_1}\left(r_1|0\rangle + r_2e^{i\left(\phi_2-\phi_1\right)}|1\rangle\right) \end{align}
So now you can see that you have a global phase of $$\phi_1$$ which isn't so important (or as you said has nothing to do with the states $$|0\rangle$$ or $$|1\rangle$$ individually) but you do have a relative phase of $$\phi_2-\phi_1$$ between states $$|0\rangle$$ and $$|1\rangle$$ (or in other words, the state $$|1\rangle$$ has a phase of $$\phi_2-\phi_1$$ after disregarding the global phase).