# Constructing a two 3-qubit state involving either X, Y or Z rotation gate

The goal is to construct the state

$$|\psi\rangle = \frac{1}{2}|010\rangle + \frac{\sqrt{3}}{2}|101\rangle$$

Two issues I am facing:

1. What is a good choice for initial state?
2. What is a good choice for rotation operator?

Hints or assistance would be helpful.

• How would you produce a state such as $(|000\rangle+|111\rangle)/\sqrt{2}$? Nov 16 at 12:55
• I apply a H gate to the first qubit of an initial state $|000\rangle$. Then apply a CCNOT gate centered on first qubit where left-most is the first qubit. Nov 16 at 13:10
• Not ccnot. Two controlled-nots. So now, if you change Hadamard to a rotation that produces different amplitudes? Nov 16 at 13:27
• @DaftWullie Yes two CNOTs. A hadamad is produced by $\frac{1}{\sqrt{2}}[X+Z]$ and $X = R_{X}(\pi), Z = R_{Z}(\pi)$. Is this correct? Nov 16 at 13:30
• Yes, but you might equivalently think of $R_Y(\pi/2)$ has having the same effect as Hadamard on $|0\rangle$. Nov 16 at 13:42

Initialise 1 qubit in the $$|0\rangle$$-state and rotate along the $$Y$$-axis to $$$$\frac{1}{2}|0\rangle+\frac{\sqrt{3}}{2}|1\rangle.$$$$ Append two qubits in the $$|0\rangle$$-state, and apply a multi-CNOT gate between the 3 qubits: the control is the first qubit in the superposition state, the targets are the other 2 fresh qubits.This gives $$$$\frac{1}{2}|000\rangle+\frac{\sqrt{3}}{2}|111\rangle.$$$$ Apply an $$X$$-rotation to the 2nd qubit and you will find $$$$\frac{1}{2}|010\rangle+\frac{\sqrt{3}}{2}|101\rangle.$$$$