The No-Cloning Basis

The previous lessons asserted that an eavesdropper cannot simply copy the qubits and measure later. That assertion is the no-cloning theorem, and it is not a practical limitation but a structural consequence of the linearity of quantum mechanics. Here we prove it and trace exactly how it underwrites QKD security.

Statement and proof

No-cloning theorem. There is no unitary $U$ that copies an arbitrary unknown state: no $U$ satisfies $U(|\psi\rangle \otimes |b\rangle) = |\psi\rangle \otimes |\psi\rangle$ for all $|\psi\rangle$ , where $|b\rangle$ is a fixed blank register.

Proof. Suppose such a $U$ existed. Apply it to two states $|\psi\rangle$ and $|\phi\rangle$ :

U(|\psi\rangle|b\rangle) = |\psi\rangle|\psi\rangle, \qquad U(|\phi\rangle|b\rangle) = |\phi\rangle|\phi\rangle.

Take the inner product of the two left-hand sides and the two right-hand sides. Because $U$ is unitary it preserves inner products, so the left side gives

\langle\psi|\phi\rangle\,\langle b|b\rangle = \langle\psi|\phi\rangle,

while the right side gives $\langle\psi|\phi\rangle^2$ . Hence

\langle\psi|\phi\rangle = \langle\psi|\phi\rangle^2 \;\Longrightarrow\; \langle\psi|\phi\rangle \in \{0, 1\}.

So cloning is possible only if every pair of states is either identical or orthogonal. A device cannot clone two states like $|0\rangle$ and $|+\rangle$ whose overlap is $\tfrac{1}{\sqrt2}$ — exactly the non-orthogonal pair BB84 relies on. $\blacksquare$

The same argument rules out a probabilistic or approximate cloner from doing better than chance on non-orthogonal inputs; optimal approximate cloning has a hard fidelity ceiling strictly below 1.

Why the two BB84 bases matter

A known basis is clonable: if Eve knew every qubit was a $Z$ -basis state, she would measure in $Z$ , learn the bit perfectly, and resend an identical copy — no disturbance, no detection. Security requires mixing in a second, conjugate basis. The four BB84 states

|0\rangle, \ |1\rangle, \ |+\rangle, \ |-\rangle

form two orthonormal pairs that are pairwise non-orthogonal across the bases ( $\langle 0|+\rangle = \tfrac{1}{\sqrt2}$ , etc.). No-cloning therefore forbids any device that copies all four. This is the precise sense in which BB84 lives in a "no-cloning basis set": the encoding states are chosen so that perfect copying is provably impossible.

Information–disturbance: the same coin

No-cloning is one face of a deeper principle. Its operational twin is the information–disturbance tradeoff: extracting any information distinguishing two non-orthogonal states necessarily disturbs at least one of them. If Eve could gain information without disturbance, she could repeat the procedure to amass arbitrarily many independent records of $|\psi\rangle$ — effectively cloning it. The contrapositive is BB84's security guarantee: every bit Eve learns about the conjugate-coded key costs her a measurable disturbance, which Alice and Bob see as QBER.

What no-cloning does not forbid

It is worth dispelling confusions. No-cloning does not forbid copying a known state (just prepare a fresh one), nor copying classical (orthogonal, distinguishable) information, nor copying a known member of a known orthonormal set. It also does not contradict measurement or teleportation: teleportation moves a state and destroys the original, consistent with there being only ever one copy. The theorem bites precisely on the one thing QKD needs — duplicating an unknown qubit drawn from non-orthogonal possibilities.