# Bloch sphere, where are magnitude and phase of a qbit? Where are magnitude and phase of a qbit on Bloch sphere?

Phase is angle φ. What do you mean by magnitude? Amplitudes? They are given by angle θ - amplitude of |0⟩ is cos(θ/2) and amplitude of |1⟩ is sin(θ/2).

Does it means that Z axis represent magnitude/amplitudes, and X and Y represent phase?

Why do we need 2 axis (X and Y) to represent phase?

$$|\psi \rangle = \cos(\theta /2) |0\rangle + e^{i\varphi} \sin(\theta/2) |1\rangle$$
where $$0 \leq \theta \leq \pi$$, $$0 \leq \phi < 2 \pi$$. The amplitude of the $$|0\rangle$$ state is $$\cos(\theta /2)$$ and the amplitude of the $$|1\rangle$$ state is $$e^{i\varphi} \sin(\theta/2)$$. The phase of the qubit is $$\varphi$$ as presented in the question. The length of the vector is equal to $$|\cos(\theta /2)|^2 + |e^{i\varphi}\sin(\theta /2)|^2 = 1$$ (the sum of the probabilities of all measurement outcomes must equal 1). Thus, all points of the imaginary sphere (Bloch sphere) with radius 1 can be regarded as different quantum states.
$$\theta$$ and $$\varphi$$ are both angles and they need planes for defining them. Thus, we need all X, Y, Z axis. $$\varphi$$ is the angle defined in the XY plane and $$\theta$$ is the angle defined in the plane that includes Z axis and the vector that represents the quantum state in the Bloch sphere.
• @guest in the picture that you have shown, XY plane was denoted by Z axis, because it is perpendicular to that plane. I think this not the best way to denote planes because there is an infinite number of planes that are perpendicular to Z axis. Moreover, for the plane where $\theta$ is defined, we don't have a fixed perpendicular axis. In the general case, the vector that is perpendicular to that plane has an arbitrary direction (not coinciding with either axis). Mar 30 '20 at 12:04