Quantum Kernels

Kernel methods are a classical machine-learning idea: instead of operating on raw feature vectors, compare data points through a kernel function $k(x, y)$ that measures similarity in a high-dimensional feature space. Quantum computers can evaluate kernels that are classically hard to compute, by using the quantum state itself as the feature vector.

The quantum kernel

Given a data-encoding unitary $U_\phi(x)$ (a feature map), the corresponding quantum state is

|\phi(x)\rangle = U_\phi(x)|0\rangle.

The quantum kernel between two data points $x$ and $y$ is the squared overlap of their feature states:

k(x, y) = |\langle\phi(y)|\phi(x)\rangle|^2.

Because the inner product lives in a Hilbert space whose dimension grows exponentially with the number of qubits, the feature space is vastly larger than anything a classical computer can represent explicitly.

The overlap circuit

The key insight is that $k(x,y)$ is a measurement probability. Consider the circuit

U_\phi(y)^\dagger \, U_\phi(x) |0\rangle.

The probability that this state measures as $|0\rangle$ is

P(|0\rangle) = |\langle 0|\, U_\phi(y)^\dagger U_\phi(x)\,|0\rangle|^2 = |\langle\phi(y)|\phi(x)\rangle|^2 = k(x, y).

So the recipe is: encode $x$ , then invert the encoding of $y$ , then measure. The fraction of shots landing on the all-zero outcome estimates the kernel.

Angle encoding and the cosine kernel

For single-qubit angle encoding, $U_\phi(x) = R_y(x)$ , the feature state is

|\phi(x)\rangle = \cos\!\tfrac{x}{2}|0\rangle + \sin\!\tfrac{x}{2}|1\rangle.

The inner product is

\langle\phi(y)|\phi(x)\rangle = \cos\!\tfrac{x}{2}\cos\!\tfrac{y}{2} + \sin\!\tfrac{x}{2}\sin\!\tfrac{y}{2} = \cos\!\tfrac{x - y}{2},

so the kernel is the squared cosine:

k(x, y) = \cos^2\!\!\tfrac{x - y}{2}.

For $x = \pi/3$ and $y = 2\pi/3$ :

k\!\left(\frac{\pi}{3},\frac{2\pi}{3}\right) = \cos^2\!\!\left(\frac{\pi/3 - 2\pi/3}{2}\right) = \cos^2\!\!\left(-\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}.

The kernel circuit combines both rotations into a single net rotation $R_y(\pi/3 - 2\pi/3) = R_y(-\pi/3)$ , and $P(|0\rangle) = \cos^2(\pi/6) = 3/4$ as expected.

From a kernel to a classifier

Once a kernel matrix $K_{ij} = k(x_i, x_j)$ is available, any classical kernel method (support vector machine, kernel ridge regression) can use it directly. The quantum computer evaluates the matrix entries; the classical optimizer does the rest.

Try it

Build the overlap circuit for $x = \pi/3$ and $y = 2\pi/3$ . Apply the encoding $R_y(\pi/3)$ , then the inverse encoding $R_y(-2\pi/3)$ , and measure. The grader checks the output distribution: $P(0)$ should be $3/4$ .

Run your code to see the quantum state.