Quantum Kernel Methods

qc/ml

An entirely different way to build a “quantum ML model” from QNNs: instead of training circuit parameters, use a fixed feature map $|\Psi (x)⟩$ to compute a fidelity kernel — a similarity score between two data points — and hand that kernel matrix to an ordinary classical support-vector machine.

$$K(x,x^{\prime })=|⟨\Psi (x)|\Psi (x^{\prime })⟩{|}^2$$

Key insight: no quantum circuit parameters are ever optimized here. The quantum computer’s only job is to compute $K(x,x^{\prime })$ for every pair of data points once, producing a kernel/similarity matrix; all actual learning — finding the decision boundary — happens classically afterward via a standard SVM (QSVC = quantum kernel + classical SVC). This is a fundamentally different division of labor than a QNN, where the circuit is the trainable model. Here the quantum part is a fixed similarity-measuring subroutine, and the “quantumness” of the model lives entirely in the choice of feature map — a different encoding circuit gives a different notion of similarity, and therefore a different classifier, without any quantum-side training at all.

# schematic — sQUlearn-style
from squlearn.encoding_circuit import YZ_CX_EncodingCircuit
from squlearn.kernel import FidelityKernel, QSVC
 
fmap = YZ_CX_EncodingCircuit(num_qubits=4, num_layers=2, num_features=2)
kernel = FidelityKernel(fmap)
model = QSVC(kernel)
model.fit(X_train, y_train)   # kernel matrix computed once, SVM optimizes classically

Related

• Data Encoding Circuits (Feature Maps) — the only quantum-side choice that matters here
• Quantum Neural Networks (QNN) — contrast: circuit parameters are trained there, not here

Self-Check

• What role does the quantum computer actually play in a quantum kernel method, versus a QNN?
• If no circuit parameters are trained, what quantum-side choice still determines how good the classifier is?
• Why might two different feature maps produce two different classifiers from the exact same classical SVM training procedure?