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 $|1\rangle$?

Question

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?

Answer 1

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.)

Answer 2

Nice answer by DaftWullie! Just to add a little bit, you can see it by writing out the state explicitly:

$\begin{align} |\psi\rangle &= a_1|0\rangle + a_2|1\rangle \\ &= r_1e^{i\phi_1}|0\rangle + r_2e^{i\phi_2}|1\rangle\\ &= e^{i\phi_1}\left(r_1|0\rangle + r_2e^{i\left(\phi_2-\phi_1\right)}|1\rangle\right) \end{align}$

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).