I How to understand the Bloch sphere in the quantum computation?

  • Thread starter Haorong Wu
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I've read that ##\left | \psi \right > =cos \frac \theta 2 \left | 0 \right > + e^{i \phi} sin \frac \theta 2 \left | 1 \right >##, and the corresponding point in the Bloch sphere is as the fig below shows.

无标题.jpg


I think ##\left | 0 \right >## and ##\left | 1 \right >## are orthonormal vectors, then why they seem to apear parallel in the Bloch sphere?

Also, I can understand the ##cos \frac \theta 2 \left | 0 \right >## part, but I cannot understand how ##e^{i \phi} sin \frac \theta 2 \left | 1 \right >## part can match the fig.

Thanks!
 

DrClaude

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Consider a spin-1/2 particle. The two eigenvalues of ##S_z## are ##\pm \hbar/2##, which correspond to opposite points along ##z##. The fact that ##|+z\rangle## and ##|-z\rangle## are orthogonal states must not be confused with the orthogonality of the cartesian axes.

Likewise, consider that
$$
|\pm x \rangle = \frac{1}{\sqrt{2}} \left( |+z \rangle \pm |-z \rangle \right) \\
|\pm y \rangle = \frac{1}{\sqrt{2}} \left( |+z \rangle \pm i |-z \rangle \right)
$$
This is exactly what is transposed to the Bloch sphere.
 
I think ##\left | 0 \right >## and ##\left | 1 \right >## are orthonormal vectors, then why they seem to apear parallel in the Bloch sphere?
Spin half particles take 720 degrees to return to their original state.

Cheers
 

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