# Rotation of qubit - Pauli Gates XYZ

I don't understand how to apply a Pauli Gate on a qubit.
Lets say 8 got a qubit with in state:

$$|\psi\rangle = 0.891 |0\rangle+ 0.454i |1\rangle$$

How would I compute e.g. rotating it 90 degrees about the z-axis? And how would I get the angles theta and azimuth from the result vector? So for the qubit above the angles are:

$$\theta=3\pi/10$$ and $$\phi=6\pi/12$$

as we can also write $$|\psi\rangle = \cos{(3\pi/20)}|0\rangle + \sin{(3\pi/20)}\exp(i 6\pi/12)|1\rangle$$

derived by $$\psi = \cos{(\theta/2)}|0\rangle + \sin{(\theta/2)}\exp(i \phi)|1\rangle$$

I don't know what do you mean by pauli gatter but if you want to apply a $$Z$$-gate to your state, or any gate, then first write the the state as vector in a particular basis. Your state in computational basis in vector form will be

$$\left|\psi\right\rangle = \left(\begin{array}{cc} 0.891\\ 0.454𝑖\\ \end{array}\right)$$

Now, to perform rotation, just multiply this vector to its left by a rotation matrix (a unitary matrix). To flip it around $$Z$$-axis, apply $$Z$$-gate.

Say your new state is $$|\phi\rangle$$.

$$\therefore |\phi\rangle = Z \left|\psi\right\rangle$$ $$\therefore |\phi\rangle = \left(\begin{array}{cc} 1&0\\ 0&-1\\ \end{array}\right)\left(\begin{array}{cc} 0.891\\ 0.454𝑖\\ \end{array}\right)$$ $$\therefore |\phi\rangle =\left(\begin{array}{cc} 0.891\\ -0.454𝑖\\ \end{array}\right)$$

writing it back in computational basis $$\{|0\rangle, |1\rangle\}$$,

$$|\phi\rangle = 0.891|0\rangle -0.454𝑖|1\rangle$$

Now, if you want to find bloch sphere angles $$\phi$$ and $$\theta$$ for any state, again, first write that state in the vector form and then compare it with

$$\left(\begin{array}{cc} \cos\frac{\theta}{2}\\ e^{i\phi}\sin\frac{\theta}{2} \end{array}\right)$$

and you will have two equations to solve. First will give you $$\theta$$ and second will give you $$\phi$$ .

Remember, general state you can write where $$\phi$$ and $$\theta$$ are bloch sphere angles is

$$\left|{\psi}\right\rangle = \cos\frac{\theta}{2}\left|0\right\rangle+ e^{i\phi}\sin\frac{\theta}{2}\left|1\right\rangle$$

• Ahh makes sense! And how would i now get the angles theta and azimuth from that? May 2, 2023 at 6:43
• what do you mean by theta and azimuth? can you explain what exactly are you saying? May 2, 2023 at 7:25
• i edited the question above: So i can also articulate a qubit by its angles theta and phi and i would like to know what are the values for the angles after the rotation May 2, 2023 at 9:01
• Okay, I think I get what you are asking. I have edited my answer. Check if this what you are asking for. May 2, 2023 at 15:35