# Qubit projections

Given the qubit: $$\frac{|0\rangle+i|1\rangle}{\sqrt{2}}$$

What is the corresponding point on the extended complex plane and Bloch sphere?

How to perform calculations and get the point representing the state of qubit?

If you read the wikipedia article about the Bloch sphere, you will see that any pure state has the form

$|\psi\rangle = \cos\left(\tfrac{\theta}{2}\right) |0 \rangle \, + \, ( \cos \phi + i \sin \phi) \, \sin\left(\tfrac{\theta}{2}\right) |1 \rangle$

with $0 \leq \theta \leq \pi$ and $0 \leq \phi < 2 \pi$. Notice that if you have a complex coefficient for $|0\rangle$, you have to factor it and forget about the global phase of the qbit.

As the basis ($| 0 \rangle$, $|1\rangle$) is orthonormal, we have $\langle 1 | 0 \rangle = 0$, and we can identify the coefficients of the decomposition. You have

$|\psi\rangle = (1/\sqrt{2}) |0\rangle + i (1/\sqrt{2}) |1\rangle$

so

$\cos(\theta/2) = \frac{1}{\sqrt{2}}$

and thus (with inverse functions, if you want) $\theta=\pi/2$. So we have also

$\sin(\theta/2) = \frac{1}{\sqrt{2}}$

and we must have $\sin(\phi) = 1$ and $\cos(\phi) = 0$, which gives $\phi=\pi/2$ too.

After some time, you will see where the qbit goes on the Bloch sphere without tedious calculations. For example, here you can see that the coefficients of $| 0 \rangle$ and $|1\rangle$ have the same magnitude (it happens when $\theta=\pi/2$), so the state lies on the equator. Now, you can see that the $\phi$ coordinate is just the cylindrical coordinate on the circle. It means that the relative phase between $|0\rangle$ and $|1\rangle$ gives the angular position of the qbit on the $xy$ plane. Conversely, the relative magnitude between $|0\rangle$ and $|1\rangle$ gives you the angular position in the plane which contains $z$ and your state.

You can obtain the coordinates in 3-space corresponding to the Bloch sphere, by taking expectation values of the Pauli spin operators: \begin{align*} x &= \langle\psi | \sigma_x |\psi\rangle \\ y &= \langle\psi | \sigma_y |\psi\rangle \\ z &= \langle\psi | \sigma_z |\psi\rangle \end {align*} and in the case of the state you describe, we have $(x,y,z) = (0,1,0)$, i.e. your state is the one on the positive $y$ axis.