# How can I build up an arbitrary quantum circuit given a certain unitary matrix operation?

Suppose I want to put a qubit whose initial state is $$|0\rangle$$ to the final state $$\frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac{2}{3}}|1\rangle$$.

Well, in that case, the unitary matrix that performs such operation is given by: $$U = \frac{1}{\sqrt{3}}\begin{pmatrix}1&-\sqrt{2}\\ \sqrt{2} & 1 \end{pmatrix}$$ So the question is, how can I build a quantum circuit with the usual quantum gates (X, Y, Z, etc) which reproduces this behavior?

• You can implement this using a U3 gate, see how to do this in this question – met927 Jan 20 '20 at 14:27

You can use $$Ry$$ (y-rotation gate). Its general matrix is

$$Ry(\theta) = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}$$

Applying the gate on $$|0\rangle$$ state, you get a state

$$|\psi\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle$$

Hence $$\cos(\theta/2) = \frac{1}{\sqrt{3}}$$ and

$$\theta = 2\arccos\Big(\frac{1}{\sqrt{3}}\Big) = 1.910633.$$

You can validate that such setting of $$\theta$$ gets your matrix $$U$$.