The Bloch sphere uses half angles to map quantum state probabilities.
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Reference discussions on physics discussion forums demonstrate that the standard representation of a qubit state on the Bloch sphere utilizes half angles (theta over 2) to correctly map polar coordinates to quantum state probability amplitudes.
# Why is $\theta \over 2$ used for a Bloch sphere instead of $\theta$?
Tags: quantum-information, hilbert-space, group-theory, representation-theory, bloch-sphere
- Score: 32
- Views: 12028
- Answers: 7
- Answered: yes
- Asked by: Mike Wong (469 rep)
- Asked: 2015-04-06
- Edited: 2017-03-10
- Site: physics
## Question
I'm a beginner in studying quantum info, and I'm a little confused about the representation of a qubit with a Bloch Sphere. Wikipedia says that we can use $$\lvert\Psi\rangle=\cos\frac{\theta}{2} \lvert 0\rangle + e^{i\phi}\sin\frac{\theta}{2} \lvert 1\rangle$$
to represent a pure state, and map it to the polar coordinates of the sphere.
What I'm not sure about is, where does the "$\frac{\theta}{2}$" come in?
I mean, in polar coordinate, the vector equals $\cos{\theta}\ \hat{z} + e^{i\phi}\sin{\theta}\ \hat{x}$, but even if we use $\hat{z}=\lvert 0\rangle$ and $\hat{x}=\lvert 0\rangle + \lvert 1\rangle$, it's still different from above. How could this be transformed into the formula above?
Or... does this mean that the sphere is simply a graphical representation of $\theta$ and $\phi$, while $\lvert 0\rangle$ and $\lvert 1\rangle$ do not geometrically correspon
# Definition of points on Bloch sphere
Tags: quantum-mechanics, quantum-information, hilbert-space, bloch-sphere
- Score: 3
- Views: 1053
- Answers: 3
- Answered: yes
- Asked by: anonymous (398 rep)
- Asked: 2016-09-04
- Edited: 2016-09-04
- Site: physics
## Question
In the definition of the Bloch sphere, one demands that $\theta \in [0, \pi]$ ans $\phi \in [0, 2\pi)$ so that any state on the Bloch sphere can be represented by
$$|\phi \rangle= \cos(\theta/2)|0 \rangle+ e^{i \phi} \sin(\theta/2)|1 \rangle.$$
But I was wondering why the representation is chosen to be like this since in my opinion the natural way to choose this representation would be
$$|\phi \rangle=\cos(\theta)|0 \rangle+ e^{i \phi} \sin(\theta)|1 \rangle,$$
with $\theta \in [0, \pi], \phi \in [0, 2\pi)$.
If one chooses this representation, you would get in trouble since for example the states $|\phi_1\rangle$ with $\theta_1=\pi/4$ and $\phi_1=0$ and $|\phi_2\rangle$ with $\theta_2=3\pi/4$ and $\phi_2=\pi$ would both lead (when neglecting an irrelevant phase) to the representation
$$|\phi_1 \rangle=|\phi_2\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle).$$
But imagine that the Axiom of Quantum mechanics, t
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