# Calculating the state of the qubit when $\alpha$ and $\beta$ are given [closed]

I am new to quantum computing. Can someone help me in calculating state of the qubit when $$\alpha = \frac{3}{5}e^{i\pi/7}$$ and $$\beta = -\frac{4i}{5}$$.

Thanks

• Assuming the qubit is given by $\alpha\vert 0\rangle + \beta\vert 1\rangle$, this question is trivial right? However, from your question it does not become clear what you are really asking. There seems to be little calculation involved here. – nippon May 11 '20 at 7:48
• @nippon are you able to see deleted posts? A deleted answer by Mariia has a comment by the original author of the question, that clarifies a little bit, what they want. – user1271772 May 11 '20 at 14:07

The state is usually just expressed as $$\alpha|0\rangle+\beta|1\rangle,$$ which is straightforward given that you're stating $$\alpha$$ and $$\beta$$.
However, if what you actually want is a geoemtric interpretation, that means mapping your system to something of the form $$e^{i\gamma}(\cos\frac{\theta}{2}|0\rangle+\sin\frac{\theta}{2}e^{i\phi}|1\rangle)$$ where you want the values of $$\theta$$ and $$\phi$$. In this case, you immediately have $$\gamma=\pi/7$$ and $$\cos\frac{\theta}{2}=\frac{3}{5}$$. You want to be a little careful in your choice of $$\sin\frac{\theta}{2}$$ - you probably want it to be positive (as was also true of cos), so $$\sin\frac{\theta}{2}=\frac{4}{5}$$. Hence, $$\sin\theta=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}.$$ You can numerically evaluate this of you want, but it's probably better to leave it just like that.
Now we need to take care of $$\phi$$. We have $$e^{i\phi}=-ie^{-i\pi/7}=e^{-i\pi/7-i\pi/2},$$ so we conclude that $$\phi=2\pi-\frac{\pi}{7}-\frac{\pi}{2}=\frac{19}{14}\pi,$$ since you're probably expected to give a value of $$\phi$$ in the range 0 to $$2\pi$$.