CNOT on Superpositions

When a CNOT gate's control qubit is in a definite basis state, its behavior is purely classical: flip the target or don't. Things change dramatically when the control is in superposition — the gate acts on both branches at once, and the result has no classical counterpart.

Linearity turns superposition into entanglement

Suppose the control starts in $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and the target in $|0\rangle$. The two-qubit state before the CNOT is

CNOT is a linear operator, so it acts on each term separately:

Putting the pieces back together:

This is the Bell state $|\Phi^+\rangle$. It cannot be written as a product of two independent single-qubit states — the two qubits are entangled. A measurement of qubit 0 yields 0 or 1 with equal probability, but whichever outcome appears, qubit 1 is instantly found in the same state.

The recipe: H then CNOT

Preparing $|\Phi^+\rangle$ takes exactly two gates:

1. H on the control — turns $|0\rangle$ into $|+\rangle$.
2. CNOT(control, target) — entangles the pair.

This two-gate sequence is the most common entry point to quantum entanglement and underlies protocols such as quantum teleportation and superdense coding.

Try it

Build the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)$. Apply $H$ to qubit 0 first, then use CNOT with control 0 and target 1. The grader checks the full state vector.

After running, the Amplitudes panel should show $\approx 0.707$ for $|00\rangle$ and $|11\rangle$ and zero for the other two basis states — the signature of a maximally entangled pair.

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