Significance of angle $\phi$ on bloch sphere

Question

So far I learned that a qubit can be written as $| \psi \rangle = \alpha | 0 \rangle + \beta | 1\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$ and reparametrized as $| \psi \rangle = cos( \theta / 2) + e^{i \phi} sin (\theta / 2)$ to visualize it on the bloch sphere: https://en.wikipedia.org/wiki/Bloch_sphere. Moreover, the source I am reading says that the probability to measure $0$ is $|\alpha|^2$ and the probability to measure $1$ is $|\beta|^2$.

Which role does the angle $\phi$ play since changing it does not influence the probabilities $|\alpha|^2$ and $|\beta|^2$ (due to $|e^{i \phi}| = 1$).

While thinking about this question it came to my mind that one could measure the qubit with different basis as well. Is this assumption true? If yes, are there some examples of application?

Answer 1

You got it exactly; it matters when measuring in different bases.

One point to clarify is that $0\leq \theta \leq \pi$, so the only difference between the $X$-eigenvectors $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ is $\phi = \frac{\pi}{2}$ for $|+\rangle$ and $\frac{3\pi}{2}$ for $|-\rangle$ since $\theta = \frac{\pi}{2}$ for both.

In terms of applications, many of the advantageous quantum computing algorithms, including Shor's, use interference of these phases (or analogous phases for larger quantum states) to select one desired state/answer from a large input superposition of all possible answers. Outside of computing, many quantum sensing schemes use a measurement of the phase $\phi$ to say something about the signal of interest.