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Starting from the zero pure qubit, how can I get the normalized qubit $$ \alpha\left|0\right\gt + \beta\left|1\right\gt $$ such that $$ |\alpha|^2 + |\beta|^2 = 1 $$ using the spherical coordinates $(\theta, \phi)$ on the Bloch Sphere. I suppose that those coordinates are calculated as

$$ (\theta, \phi) = \left(2 * \arctan\left(\frac{\text{module}(\beta)}{\text{module}(\alpha)}\right), \arg(\beta)- \arg(\alpha)\right) $$

and then I need to use the rotation matrices over the Bloch sphere axes.

\begin{align} R_x(\omega) &= \left( \begin{matrix} \cos\left(\frac{\omega}{2}\right) & -i\sin\left(\frac{\omega}{2}\right) \\ -i\sin\left(\frac{\omega}{2}\right) & \cos\left(\frac{\omega}{2}\right)\\ \end{matrix} \right) \\ R_y(\omega) &= \left(\begin{matrix} \cos\left(\frac{\omega}{2}\right) & -\sin\left(\frac{\omega}{2}\right)\\ \sin\left(\frac{\omega}{2}\right) & \cos\left(\frac{\omega}{2}\right)\\ \end{matrix}\right) \\ R_z(\omega) &= \left(\begin{matrix} e^{-i{\omega}/{2}} & 0\\ 0 & e^{i{\omega}/{2}}\\ \end{matrix}\right) \end{align}

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closed as off-topic by Norbert Schuch, AccidentalFourierTransform, Jon Custer, honeste_vivere, Gert Jan 23 '17 at 2:05

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Start with $\vert 0\rangle= \left(\begin{array}{c} 1 \\ 0\end{array}\right)$ and apply the sequence \begin{align} R_z(\phi)R_y(\theta)\left(\begin{array}{c} 1 \\ 0\end{array}\right) &= \left( \begin{array}{cc} e^{-\frac{i \phi }{2}} \cos \left(\frac{\theta }{2}\right) & -e^{-\frac{i \phi }{2}} \sin \left(\frac{\theta }{2}\right) \\ e^{\frac{i \phi }{2}} \sin \left(\frac{\theta }{2}\right) & e^{\frac{i \phi }{2}} \cos \left(\frac{\theta }{2}\right) \\ \end{array} \right)\left(\begin{array}{c} 1 \\ 0\end{array}\right)\, ,\\ &=\left(\begin{array}{c} e^{-\frac{i \phi }{2}} \cos \left(\frac{\theta }{2}\right) \\ e^{\frac{i \phi }{2}} \sin \left(\frac{\theta }{2}\right)\end{array}\right)=\left(\begin{array}{c} \alpha \\ \beta\end{array}\right) \end{align} from which it immediately follows that $$ \alpha=e^{-\frac{i\phi}{2}}\cos\left(\frac{\theta}{2}\right)\, ,\qquad \beta=e^{\frac{i\phi}{2}}\sin\left(\frac{\theta}{2}\right)\, . $$

Suppose now $$ \alpha= \frac{2+45 i}{3\sqrt{574}}=\sqrt{\frac{2029}{5166}}e^{i\varphi_a}\, ,\qquad \beta= \frac{56+I}{3\sqrt{574}}=\sqrt{\frac{3137}{5166}}e^{i\varphi_b} $$ where $$ \varphi_a=\arctan(45/2)\, ,\qquad \varphi_b=\arctan(1/56) $$ Using $$ \frac{\beta}{\alpha}=e^{i\phi}\tan\left(\frac{\theta}{2}\right) $$ we obtain $\phi=-\arctan\left(\frac{2518}{157}\right)$ and \begin{align} R_z(\phi)R_y(\theta)\left(\begin{array}{c}1 \\ 0 \end{array}\right)&= \left( \begin{array}{cc} 0.456729\, +0.429138 i & -0.567903-0.533597 i \\ 0.567903\, -0.533597 i & 0.456729\, -0.429138 i \\ \end{array} \right)\left(\begin{array}{c}1 \\ 0 \end{array}\right)\, ,\\ &=\left(\begin{array}{c} 0.456729\, +0.429138 i\\ 0.567903\, -0.533597 i \end{array}\right)\ \end{align} which differs from your target vector $(\alpha,\beta)$ by an unimportant overall phase $e^{2.02367 i}$.

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  • $\begingroup$ I try this to get the qubit $$ (2 + 45i) |0 \gt + (56 + i) |1 \gt $$ that has norm 71.8748, so the normalized qubit is $$ (0.02782612904 + 0.6260879034i) |0\gt + (0.7791316131 + 0.01391306452i) |1\gt $$ The spherical coordinates are $\theta = 1.786955007 $ and $\phi = -1.5085258$ The result of the rotations proposed are (0.45673-0.5679i) |0> + (-0.42914-0.5336i) |1> that it's different from the desired qubit. $\endgroup$ – CodeSniffer Jan 20 '17 at 11:14
  • $\begingroup$ @CodeSnifer: you were out by an overall phase. $\endgroup$ – ZeroTheHero Jan 20 '17 at 13:35
  • $\begingroup$ What do you mean? Sorry, I'm a beginner in the subject. How can I correct that overall phase to get exactly the same amplitudes? $\endgroup$ – CodeSniffer Jan 20 '17 at 13:48
  • $\begingroup$ In QM, two vectors like $(\alpha,\beta)$ and $e^{i\xi}(\alpha,\beta)$ are physically equivalent (describe the same physics). Your target vector is just $\endgroup$ – ZeroTheHero Jan 20 '17 at 13:57
  • $\begingroup$ Your target vector is just ....? $\endgroup$ – CodeSniffer Jan 20 '17 at 14:02

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