1
$\begingroup$

enter image description here

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?

$\endgroup$
2
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ This? commons.wikimedia.org/wiki/File:Cartesian_planes_and_axis.jpg $\endgroup$ – guest Mar 30 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). $\endgroup$ – Davit Khachatryan Mar 30 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.