# Finding $\theta$ and $\phi$ when qubit state is $\frac{1}{\sqrt 2}[i ,1]^T$

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?

• I assume $\phi,\theta$ are real. The representation in your first formula allows only real coefficients for $|0\rangle$ but we know that $|q\rangle$ can be multiplied by a phase $e^{ia}$ without changing the physics. Commented Mar 27, 2022 at 18:09
• Related : Determine the state $|\psi\rangle$. Commented Mar 28, 2022 at 13:45

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.
• So I know that means $\phi = -\frac{\pi}{2}$ and $\theta = \frac{\pi}{2}$. Could you explain your first sentence a bit more? Commented Mar 27, 2022 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. Commented Mar 27, 2022 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 $\theta$... Commented Mar 27, 2022 at 18:38
• @KurtG. Why should I make the first component of $|q\rangle$ imaginary? My understanding by first component is the component of $|0\rangle$. Commented Mar 27, 2022 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.