How to implement a random selection of one of the 3 qubits on the IBM Q (composer)?

Question

I would like to implement a quantum program on the IBM Composer with following characteristics:

1. The output is the observed value of 3 qubits
2. Only one of the 3 qubits should be observed as $1$, the 2 others should be $0$
3. The probability that a qubit is observed as $1$ is $1/3$

So the 3 possible outputs of this circuit are: $|100\rangle$,$|010\rangle$,$|001\rangle$ (all 3 having equality probability of being outputted)

So how can we best implement such a circuit on the IBM Composer ?

Answer 1

The pure quantum state that satisfies your conditions is the W state in three qubits, $$ \frac{1}{\sqrt{3}} \left(|001\rangle + |010\rangle + |100\rangle \right) $$

You can look at this answer for a high level circuit to construct this. The first gate in that circuit is a single qubit gate that effects the transformation, $$ |0 \rangle \rightarrow \frac{1}{\sqrt{3}} |1 \rangle + \sqrt{\frac{2}{3}} | 0\rangle .$$

This you can implement in the composer as a $U_3$ gate with an appropriate value of theta.

Next you will need a controlled H gate between the first and second qubits, and a Toffoli gate. To implement them in the composer you can use the circuits given here . The control gates in the answer have the control qubits flipped (the controls are triggered by $0$ and not $1$ ). So you will need to sandwich your control qubits in the composer between $X$ gates to get the desired circuit. As you can see, constructing this from scratch in the composer is rather tedious.