0

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
  • 11

1 Answers1

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.

J.G.
  • 25,615