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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?


1 Answer 1


An arbitrary 1 qubit pure state can be represented as:

$$|\psi \rangle = \cos(\theta /2) |0\rangle + e^{i\varphi} \sin(\theta/2) |1\rangle$$

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.

  • $\begingroup$ This? commons.wikimedia.org/wiki/File:Cartesian_planes_and_axis.jpg $\endgroup$
    – guest
    Mar 30, 2020 at 4:48
  • $\begingroup$ @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). $\endgroup$ Mar 30, 2020 at 12:04

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