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I'm trying to solve the two equations to solve for: $\gamma',\beta'$: \begin{align} \cos\left(\frac{\gamma'}{2}\right) = g_0\qquad e^{i\beta'}\sin\left(\frac{\gamma'}{2}\right) = g_1 \end{align} Where $g_0$ is real, and $g_1$ is some complex number. I tried \begin{align*} \gamma'=2\cos^{-1}(g_0),\qquad\cos(\beta')\sin\left(\frac{\gamma'}{2}\right)=Re(g_1),\ \text{(or)}\ \sin(\beta')\sin\left(\frac{\gamma'}{2}\right)=Im(g_1) \end{align*} Then for each $\gamma'$, there should be two corresponding $\beta'$s. However, it turned out that my solution doesn't quite work. I'm wondering is there another way I can solve this?

Thanks!

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    $\begingroup$ If you make that "or" an "and" you get one solution for $\beta'$. What makes you say this doesn't work? $\endgroup$
    – anon
    Commented Mar 18, 2021 at 22:07

3 Answers 3

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$$e^{i\beta}=\pm\frac{g_1}{\sqrt{1-g_0^2}}$$ and $$\beta=-i\log\frac{g_1}{\sqrt{1-g_0^2}}+k\pi.$$

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  • $\begingroup$ Thanks for the answer! Should the sign of the first term of $\beta$ be $\pm$? $\endgroup$
    – ZR-
    Commented Mar 18, 2021 at 22:19
  • $\begingroup$ @ZR-: why should it ? $\endgroup$
    – user65203
    Commented Mar 19, 2021 at 8:18
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Solve one for $\sin{\frac{\gamma '}{2}}$, solve the other for $\cos{\frac{\gamma '}{2}}$, and then use $\sin^2 x + \cos^2 x = 1$.

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Clearly $\gamma^\prime\pm2\arccos g_0(\mod4\pi)$, which is consistent with $\sin\tfrac{\gamma^\prime}{2}=\pm|g_1|$ iff $g_0^2+|g_1|^2=1$, in which case $e^{i\beta^\prime}=\operatorname{sgn}g_1$ (except for the case $g_0=\pm1,\,g_1=0$, where $\beta^\prime$ is arbitrary). This constrains $\beta^\prime$ to either the even or odd multiples of $\pi$.

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