Checkpoint: A Tiny Quantum Classifier

This checkpoint pulls together every idea from Module 12: angle encoding, entangling ansatz layers, trained weights, and measurement. You will assemble a tiny but complete quantum classifier and verify that it correctly predicts class 1 for a given input.

The classification task

We have a one-dimensional dataset with two classes:

Class 0 — points near $x = \pi/4$ (small angle, qubit naturally close to $|0\rangle$ ).
Class 1 — points near $x = 3\pi/4$ (large angle, qubit naturally close to $|1\rangle$ ).

The classifier is a two-qubit circuit $U(x, \boldsymbol{\theta})$ where $x$ is the data feature and $\boldsymbol{\theta} = (\theta_0, \theta_1)$ are the trained weights. Measurement of the full two-qubit register at the end produces a probability distribution; a class-1 prediction corresponds to a high probability of measuring $|1\rangle$ on qubit 1 (the label qubit).

Circuit structure

The circuit has three layers:

U(x, \theta_0, \theta_1) = \bigl(R_y(\theta_1)_1 \otimes R_y(\theta_0)_0\bigr) \cdot \mathrm{CNOT}_{01} \cdot R_y(x)_0.

Layer 1 — data encoding. Apply $R_y(x)$ to qubit 0. Starting from $|00\rangle$ , the state after encoding is

\bigl(\cos\tfrac{x}{2}|0\rangle + \sin\tfrac{x}{2}|1\rangle\bigr) \otimes |0\rangle.

For $x = 3\pi/4$ the amplitude on $|1\rangle$ for qubit 0 is $\sin(3\pi/8) \approx 0.924$ , so the data qubit already leans strongly toward $|1\rangle$ .

Layer 2 — entanglement. A CNOT with control on qubit 0 and target on qubit 1 creates correlations between the data qubit and the label qubit:

\cos\tfrac{x}{2}|00\rangle + \sin\tfrac{x}{2}|11\rangle.

Layer 3 — trainable weights. Independent $R_y$ rotations on both qubits let the optimizer tune which region of the entangled state maps to class 0 or class 1.

Deriving the trained weights

A classical optimizer minimizes a cost function by evaluating the circuit on training data. For this toy task, gradient descent (using the parameter-shift rule) converges to the weights $\theta_0^* = \pi/3$ and $\theta_1^* = \pi/6$ .

We can verify analytically that these weights work for $x = 3\pi/4$ . After the encoding and CNOT the state is

\cos(3\pi/8)|00\rangle + \sin(3\pi/8)|11\rangle.

Applying $R_y(\pi/3)$ to qubit 0 and $R_y(\pi/6)$ to qubit 1 rotates the amplitudes within each two-qubit basis state. The marginal probability of measuring qubit 1 in $|1\rangle$ works out to approximately $0.81$ , well above the 50% decision threshold. For the class-0 input $x = \pi/4$ the same weights produce a probability of roughly $0.19$ on $|1\rangle$ for qubit 1 — correctly predicted as class 0.

Try it

Build the complete classifier circuit for the class-1 input $x = 3\pi/4$ , using the trained weights $\theta_0^* = \pi/3$ and $\theta_1^* = \pi/6$ . Then measure. The grader checks that the output distribution matches the reference.

Run your code to see the quantum state.

After a correct run, the $|11\rangle$ outcome should dominate the histogram — qubit 1 in $|1\rangle$ signals the class-1 prediction. Compare what happens when you change $x$ to $\pi/4$ (a class-0 input): the distribution shifts dramatically toward $|00\rangle$ .