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?

  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Commented Mar 27, 2022 at 18:09
  • $\begingroup$ Related : Determine the state $|\psi\rangle$. $\endgroup$
    – Frobenius
    Commented Mar 28, 2022 at 13:45

1 Answer 1


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.

  • $\begingroup$ So I know that means $\phi = -\frac{\pi}{2}$ and $\theta = \frac{\pi}{2}$. Could you explain your first sentence a bit more? $\endgroup$ Commented Mar 27, 2022 at 18:18
  • $\begingroup$ @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. $\endgroup$
    – Kurt G.
    Commented Mar 27, 2022 at 18:36
  • $\begingroup$ @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 $\theta$... $\endgroup$ Commented Mar 27, 2022 at 18:38
  • $\begingroup$ @KurtG. Why should I make the first component of $|q\rangle$ imaginary? My understanding by first component is the component of $|0\rangle$. $\endgroup$ Commented Mar 27, 2022 at 19:03
  • $\begingroup$ @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. $\endgroup$
    – J.G.
    Commented Mar 27, 2022 at 19:04

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