Checkpoint: Amplify an Amplitude

This is the module checkpoint. You now know every moving part of amplitude amplification: the geometric interpretation, the oracle, the diffuser, and how to adapt all three to a custom subroutine. The challenge is to assemble the complete iteration from scratch on a concrete example and drive a target amplitude from $\tfrac{1}{4}$ to $\approx 1$ .

Setting up the problem

Consider the two-qubit subroutine

$\mathcal{A} = R_y\!\left(\frac{\pi}{3}\right) \otimes I,$

which acts only on qubit 0 and leaves qubit 1 in $|0\rangle$ :

$\mathcal{A}|00\rangle = \cos\!\frac{\pi}{6}\,|00\rangle + \sin\!\frac{\pi}{6}\,|10\rangle = \frac{\sqrt{3}}{2}|00\rangle + \frac{1}{2}|10\rangle.$

The target (good) state is $|10\rangle$ . Its initial probability is $\sin^2(\pi/6) = (1/2)^2 = \tfrac{1}{4}$ , so $\sin\theta = \tfrac{1}{2}$ and $\theta = \tfrac{\pi}{6}$ .

One amplitude-amplification iteration rotates by $2\theta = \tfrac{\pi}{3}$ , landing the state at angle $(2 \cdot 1 + 1)\theta = 3\theta = \tfrac{\pi}{2}$ : probability 1 on the good state.

What the iteration looks like

The iteration is the composition of two reflections.

Oracle $\mathcal{O}$ : phase-flip every state in the good subspace.

$\mathcal{O}|10\rangle = -|10\rangle, \qquad \mathcal{O}|00\rangle = |00\rangle.$

Because the good state has qubit 0 equal to 1 and qubit 1 equal to 0, you build the oracle by temporarily flipping qubit 1 with $X$ , applying $CZ$ (which marks $|11\rangle$ ), and undoing the flip:

$\mathcal{O} = (I \otimes X)\;CZ\;(I \otimes X).$

Diffuser $\mathcal{D}$ : reflect about $|\psi\rangle = \mathcal{A}|00\rangle$ .

$\mathcal{D} = \mathcal{A}\,(2|00\rangle\langle 00| - I)\,\mathcal{A}^\dagger.$

The inner factor $2|00\rangle\langle 00| - I$ phase-flips everything except $|00\rangle$ . Circuit realization: $X^{\otimes 2}\;CZ\;X^{\otimes 2}$ (which flips the sign of $|00\rangle$ up to global phase), sandwiched between $\mathcal{A}^\dagger$ and $\mathcal{A}$ :

$\mathcal{D} = R_y\!\left(\tfrac{\pi}{3}\right) \;(X \otimes X)\;CZ\;(X \otimes X)\; R_y\!\left(-\tfrac{\pi}{3}\right).$

Try it

The starter code applies subroutine $\mathcal{A}$ . Add the oracle and then the diffuser. The grader checks the output distribution — after one full iteration the probability on $|10\rangle$ should be $1$ .

Run your code to see the quantum state.