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Where are magnitude and phase of a qubit on the Bloch sphere?

I've read that

Phase is angle $\varphi$. What do you mean by magnitude? Amplitudes? They are given by angle $\theta$ - amplitude of $|0\rangle$ is $\cos(\theta/2)$ and amplitude of $|1\rangle$ is $\sin(\theta/2)$.

Does this mean that the $z$ axis represents magnitude/amplitudes, and the $x$ and $y$ axes represent phase?

Why do we need 2 axes ($x$ and $y$) to represent phase?

glS
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guest
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  • See this thread https://quantumcomputing.stackexchange.com/questions/10116/how-to-get-bloch-sphere-angles-given-arbitrary-qbit-as-linear-combination-of-bas on how to convert qubit description from $\alpha|0\rangle+\beta|1\rangle$ to representation on Bloch sphere. Maybe it helps. – Martin Vesely Mar 29 '20 at 06:55

2 Answers2

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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.

Davit Khachatryan
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  • This? https://commons.wikimedia.org/wiki/File:Cartesian_planes_and_axis.jpg – guest Mar 30 '20 at 04:48
  • @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). – Davit Khachatryan Mar 30 '20 at 12:04
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Maybe infinite planes can be taken through z but perforce it is a closed surface, so I don’t think it really needs to be specified. That normal z plane is usually used to illustrate that the rest of the surface is similarly represented as a closed, consistent metric, riemann submanifold. It’s closed, has to be, that’s how we are assured of the constant metric and the phase rotation on some field. (So division and multiplication is everywhere convergent via modulo math.)

Psmith Boston
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