Checkpoint: Register Manipulation

Module 1 built up a toolkit for multi-qubit registers: allocating $n$ -qubit state vectors, applying single-qubit gates to individual qubits inside a register, encoding integers, building uniform superpositions with Hadamards, and controlling gates on multiple qubits. This checkpoint combines those ideas in a single state-preparation problem.

The target state

Your goal is to prepare

$|\text{target}\rangle = \frac{1}{\sqrt{2}}\bigl(|010\rangle + |101\rangle\bigr).$

Qubit 0 is the most significant bit, so $|010\rangle$ means $q_0 = 0$ , $q_1 = 1$ , $q_2 = 0$ , and $|101\rangle$ means $q_0 = 1$ , $q_1 = 0$ , $q_2 = 1$ . Each of the two terms has amplitude $1/\sqrt{2}$ , so the Born-rule probability of each outcome is $|1/\sqrt{2}|^2 = 1/2$ .

Strategy: branch on the most significant qubit

A useful design pattern for state preparation is to create a superposition on one qubit and then fill in the remaining qubits conditioned on which branch you are in.

Step 1 — Hadamard on $q_0$ . Apply $H$ to qubit 0. The register becomes

$\frac{1}{\sqrt{2}}\bigl(|000\rangle + |100\rangle\bigr).$

The two branches are now $q_0 = 0$ (top) and $q_0 = 1$ (bottom).

Step 2 — Handle the $q_0 = 0$ branch. In this branch, $q_1$ must become $1$ . We need a CNOT that fires when $q_0 = 0$ rather than when $q_0 = 1$ . To invert the polarity of a control qubit, sandwich the controlled gate between two $X$ gates on the control:

$X_0 \cdot CX_{0 \to 1} \cdot X_0 \equiv \text{``CNOT controlled on } q_0 = 0\text{''}.$

The first $X_0$ temporarily flips $q_0$ , so the $CX$ fires on the originally-zero branch. The second $X_0$ restores $q_0$ . After this three-gate block the state is

$\frac{1}{\sqrt{2}}\bigl(|010\rangle + |100\rangle\bigr).$

Step 3 — Handle the $q_0 = 1$ branch. In this branch, $q_2$ must become $1$ . A standard $CX(0 \to 2)$ fires exactly when $q_0 = 1$ , turning $|100\rangle$ into $|101\rangle$ :

$\frac{1}{\sqrt{2}}\bigl(|010\rangle + |101\rangle\bigr) = |\text{target}\rangle. \checkmark$

The total circuit depth is five gates: $H, X, CX, X, CX$ . Each step uses gates from the standard universal set, and the control-on-zero trick is exactly the technique introduced in the multi-controlled gates lesson.

Try it

Build the circuit that prepares $|\text{target}\rangle$ . The grader checks the complete state vector against the reference solution, so every amplitude — including its sign and phase — must be correct. Leaving the circuit empty will fail.

Run your code to see the quantum state.

After a successful run you should see two amplitudes of $\approx 0.707 = 1/\sqrt{2}$ at positions $|010\rangle$ and $|101\rangle$ , with all other amplitudes equal to zero.