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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.

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  • $\begingroup$ How would you produce a state such as $(|000\rangle+|111\rangle)/\sqrt{2}$? $\endgroup$
    – DaftWullie
    Nov 16 at 12:55
  • $\begingroup$ 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. $\endgroup$
    – Physkid
    Nov 16 at 13:10
  • $\begingroup$ Not ccnot. Two controlled-nots. So now, if you change Hadamard to a rotation that produces different amplitudes? $\endgroup$
    – DaftWullie
    Nov 16 at 13:27
  • $\begingroup$ @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? $\endgroup$
    – Physkid
    Nov 16 at 13:30
  • $\begingroup$ Yes, but you might equivalently think of $R_Y(\pi/2)$ has having the same effect as Hadamard on $|0\rangle$. $\endgroup$
    – DaftWullie
    Nov 16 at 13:42

1 Answer 1

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One way to achieve this is the following way.

Initialise 1 qubit in the $|0\rangle$-state and rotate along the $Y$-axis to \begin{equation} \frac{1}{2}|0\rangle+\frac{\sqrt{3}}{2}|1\rangle. \end{equation} 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 \begin{equation} \frac{1}{2}|000\rangle+\frac{\sqrt{3}}{2}|111\rangle. \end{equation} Apply an $X$-rotation to the 2nd qubit and you will find \begin{equation} \frac{1}{2}|010\rangle+\frac{\sqrt{3}}{2}|101\rangle. \end{equation}

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  • $\begingroup$ I managed to work out a solution following Daftwullie's hint. $\endgroup$
    – Physkid
    Nov 16 at 14:12

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