Training a Quantum Classifier

Previous lessons introduced parameterized quantum circuits and feature maps. Now we ask: how do we actually train such a circuit to separate classes?

The classification task

Suppose we have a dataset of labeled points, each a real number $x \in [0, \pi]$. Points with $x \leq \pi/2$ belong to class 0; points with $x > \pi/2$ belong to class 1. We want a quantum circuit $U(x, \boldsymbol{\theta})$ — parameterized by both the data $x$ and trainable weights $\boldsymbol{\theta}$ — whose measurement on a designated label qubit tells us the class.

A natural circuit structure for one data point $x$ on two qubits is:

Qubit 0 is the data qubit (encoding $x$) and qubit 1 is the label qubit (read out after the circuit). The trainable parameter $\theta_w$ sits on the label qubit after the entangling gate.

How training works

Starting from $|00\rangle$, the circuit produces the state

After the $R_y(x)$ on qubit 0:

The CNOT flips qubit 1 only when qubit 0 is $|1\rangle$, entangling the two qubits:

Finally, $R_y(\theta_w)$ on qubit 1 rotates the $|1\rangle$ component of each term. The probability of measuring qubit 1 in $|1\rangle$ — the "class 1" output — becomes a function of both $x$ and $\theta_w$.

Training means choosing $\theta_w$ so that this probability is small for class-0 points and large for class-1 points. A classical optimizer (gradient descent, COBYLA, etc.) adjusts $\theta_w$ by evaluating the circuit repeatedly and minimizing a cost function that penalizes misclassified points. Gradients can be computed exactly with the parameter-shift rule (covered in the next lessons).

The trained classifier

For our toy dataset, suppose the optimizer has already run and found the optimal weight $\theta_w = \pi/4$. At this setting the probability of measuring $|1\rangle$ on the label qubit grows monotonically with $x$ over $[0, \pi]$, correctly ranking class-0 inputs below the classification threshold and class-1 inputs above it.

We can verify this concretely for the boundary point $x = \pi/2$:

After the encoding step: $\cos(\pi/4)|00\rangle + \sin(\pi/4)|11\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$.

After $R_y(\pi/4)$ on qubit 1, the amplitude on $|11\rangle$ picks up a rotation by $\pi/8$. Using the general result $P(\text{qubit }1 = |1\rangle) = \tfrac{1}{2}\bigl(1 - \cos x\cdot\cos\theta_w\bigr)$, substituting $x = \pi/2$ and $\theta_w = \pi/4$ gives $\tfrac{1}{2}\bigl(1 - \cos(\pi/2)\cdot\cos(\pi/4)\bigr) = \tfrac{1}{2}$ — exactly at the 50% threshold, as expected for the class boundary: inputs with $x < \pi/2$ produce $P < \tfrac{1}{2}$ (class 0) and inputs with $x > \pi/2$ produce $P > \tfrac{1}{2}$ (class 1).

Try it

Implement the trained circuit: encode the data point $x = \pi/2$ on qubit 0 with $R_y(\pi/2)$, entangle the two qubits with a CNOT, then apply the trained weight $R_y(\pi/4)$ on the label qubit. The grader checks the exact statevector produced by your circuit.

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