Two Different Qubit States

There are two different qubit states for each pair of measurement probabilities of $|0\rangle$ and $|1\rangle$. For instance, the states $\frac{\alpha|0\rangle+\beta|1\rangle}{2}$ and $\frac{\alpha|0\rangle-\beta|1\rangle}{2}$ have the identical measurement probabilities, as does any pair of states, such as $|\psi\rangle$ and $|\psi'\rangle$ whose state vector is mirrored at the Z-axis. This is depicted in the following image.

Regarded as waves, the two states $|\psi\rangle$ and $|\psi'\rangle$ denote two waves shifted by half their wavelength. One’s crest is the other’s trough.

This notion of a qubit lets us distinguish two opposite phases, but what about all the other possible phases a qubit can be in? Similar to waves, the phase can be any arbitrary value. The only meaningful restriction we can formulate is that the phase repeats once it exceeds the wavelength. It may remind you of the angle $\theta$ we used to rotate the qubit state vector and change its amplitudes.

We can represent the qubit phase as an angle $\phi$ (the Greek letter “phi”) that spans a circle around the center, and that is orthogonal to the circle of the amplitudes. This circle uses another dimension.

In the following figure, the angle $\theta$ describes the probability of the qubit to result in $|0\rangle$ or $|1\rangle$ and the angle $\phi$ describes the phase the qubit is in.

These two circles form a sphere around the center. This sphere is known as the Bloch sphere.

The Bloch sphere offers a visual reference of both the phase and the probabilities of measuring a qubit as either of the basis states $|0\rangle$ or $|1\rangle$. In this sphere, the angle $\theta$ that determines the measurement amplitudes revolves around the Y-axis. Correspondingly, the $R_Y$-gate we’ve used so far rotates the qubit state vector around this axis. It cuts the Z-axis in the basis states $|0\rangle$ and $|1\rangle$. If we don’t apply a phase shift, it “lies” flat on the plane the X-axis spans, but once the qubit state has a different phase, it rises from this plane.

The angle $\varphi$ that determines the phase revolves around the Z-axis. So, any change in the phase of a qubit does not affect the proximity to the Z-axis or any point on it, such as the basis states $|0\rangle$ and $|1\rangle$ that denote the top and the bottom of the Z-axis.

Therefore, the phase does not affect the measurement probabilities.

Mathematically, we could create a three-dimensional sphere using three-dimensional vectors, such as $v=\begin{bmatrix}v_0 \\ v_1 \\ v_2\end{bmatrix}$ ...