Two-qubit gates and CNOT

Everything so far moved one qubit at a time, which can never make two qubits depend on each other. The gate that finally couples them is the simplest imaginable: an if-statement across two wires. Look at the control qubit; if it is 1, flip the target; if it is 0, do nothing. That gate is CNOT, and running it on a superposition is where quantum computing stops looking like classical logic.

The tempting wrong answer

Run CNOT on plain bits and it behaves like a copier. With the target starting at $|0⟩$: a control of $|0⟩$ leaves the target at $|0⟩$, and a control of $|1⟩$ flips the target to $|1⟩$. Either way the target ends up equal to the control. So it is natural to think CNOT copies the control, always.

Watch the control decide

Step the circuit for each input. For the four basis inputs the truth table fills in and the control simply decides whether the target flips. Then pick the $|+⟩$ control and watch the output refuse to factor into two separate qubits.

CNOT as a 4x4 matrix

Two qubits have four basis states, so a two-qubit gate is a $4\times 4$ matrix. The CNOT gate, with the top wire as control, permutes the basis states: it leaves $|00⟩$ and $|01⟩$ alone (control 0) and swaps $|10⟩\leftrightarrow |11⟩$ (control 1, so the target flips).

It is a permutation matrix: one 1 in every row and column, so it is unitary (reversible and length-preserving) just like the single-qubit gates.^[2] On the four basis states it does exactly what the truth table said, and there it really does copy a definite control into the target.

The move that is not copying

Now feed it a control in superposition. Start with $|+⟩$ on the control and $|0⟩$ on the target, and expand with the tensor product:

CNOT is linear, so it acts on each piece separately. It leaves $|00⟩$ untouched and sends $|10⟩$ to $|11⟩$:

This is the ${\Phi }^{+}$ Bell state. Look at what it is not: it is not $|+⟩\otimes |+⟩$, which would expand to $\frac{1}{2}(|00⟩+|01⟩+|10⟩+|11⟩)$, all four terms present. Two copies of $|+⟩$ would have the crossed terms $|01⟩$ and $|10⟩$; the Bell state has zero weight on both. The output cannot be split into a state for qubit one times a state for qubit two. That inseparability is entanglement, and it is the next lesson.^[1]

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector

qc = QuantumCircuit(2)
qc.h(0)          # control -> |+>
qc.cx(0, 1)      # CNOT: control 0, target 1

Statevector(qc)
# (|00> + |11>) / sqrt(2)   -> the Bell state, amplitudes only on 00 and 11
# NOT (|00> + |01> + |10> + |11>)/2, which two copies of |+> would give