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Because we know the state of a qubit can be described as: $$ |q\rangle=\cos{\frac{\theta}{2}}|0\rangle+e^{i\phi}\sin{\frac{\theta}{2}}|1\rangle\\\ \\ \theta, \phi \in \mathbb{R} $$

How do I find the values of $\theta$ and $\phi$ when the qubit is in the state below? $$ \frac{1}{\sqrt{2}}\begin{bmatrix}i\\1\end{bmatrix} $$

What I've done so far:

$$ |q\rangle = \frac{i}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle $$ Therefore $$ \cos{\frac{\theta}{2}} = \frac{i}{\sqrt{2}}\\ e^{i\phi}\sin{\frac{\theta}{2}} = \frac{1}{\sqrt{2}} $$

But I don't know where to go from here. Could anyone give me some guidance?

Henry Hudson
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1 Answers1

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Your parameterization assumes we rescale $|q\rangle$ by a unit complex factor so $\langle q|0\rangle\ge0$. In this case, you need to multiply by $-i$ first. So you actually want to solve $\cos\frac{\theta}{2}=\frac{1}{\sqrt{2}},\,e^{i\phi}\sin\frac{\theta}{2}=\frac{-i}{\sqrt{2}}$. I leave you to solve that.

Edit: in the comments below, @KurtG. has noted the alternative (which is to multiply $|q\rangle$ by $+i$) $$\tfrac{i}{\sqrt{2}}|0\rangle+\tfrac{1}{\sqrt{2}}|1\rangle=i\cos\tfrac{\theta}{2}|0\rangle+ie^{i\phi}\sin\tfrac{\theta}{2}|1\rangle,$$ which works the same way.

J.G.
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  • So I know that means $\phi = -\frac{\pi}{2}$ and $\theta = \frac{\pi}{2}$. Could you explain your first sentence a bit more? – Henry Hudson Mar 27 '22 at 18:18
  • @HenryHudson . Unit complex factor is the phase $e^{ia}$ you are free to multiply your $|q\rangle$ with before solving for $\phi$ and $\theta$. Use a phase that makes the first component of $|q\rangle$ purely imaginary instead of real. – Kurt G. Mar 27 '22 at 18:36
  • @HenryHudson Put differently: your vector $|q\rangle$ (in the first equation) cannot be equal to $[i,1]^T$ simply because $\langle 0|q\rangle \in \mathbb R$ and thus $\neq i$, for all $\theta$... – Tobias Fünke Mar 27 '22 at 18:38
  • @KurtG. Why should I make the first component of $|q\rangle$ imaginary? My understanding by first component is the component of $|0\rangle$. – Henry Hudson Mar 27 '22 at 19:03
  • @HenryHudson I think the suggestion was to write your original $|q\rangle$ as $e^{i\alpha}$ times your Ansatz. Edit: as of the comment below, that is indeed the suggestion. – J.G. Mar 27 '22 at 19:04
  • Because we know that there is no real $\theta$ that solves $\cos\frac{\theta}{2}=\frac{i}{\sqrt{2}}$. You will have more luck when you try to solve $i\cos\frac{\theta}{2}=\frac{i}{\sqrt{2}}$. – Kurt G. Mar 27 '22 at 19:05
  • Okay, so I have $|q\rangle = e^{i\phi}\frac{i}{\sqrt{2}}|0\rangle + e^{i\phi}\frac{1}{\sqrt{2}}|1\rangle$ which equals (when $\phi = -\frac{\pi}{2}$) $\frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle$. Then I can solve for $\theta$ as I already know $\phi = -\frac{\pi}{2}$? – Henry Hudson Mar 27 '22 at 19:16
  • @HenryHudson I've updated my answer to show how it works, as you've not quite done it right. – J.G. Mar 27 '22 at 19:19