# 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?

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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.

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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.

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