The No-Cloning Theorem

Derivation

The No-Cloning Theorem

D-402 Home PU-404 Threads chance · symmetry · matter Depends on quantum-superposition-principle, unitary-time-evolution

Statement

There exists no unitary operator \( \hat{U} \) acting on a composite system (data register plus blank target register) that copies an arbitrary unknown pure state, i.e. no fixed \( \hat{U} \) and fixed blank \( \lvert b \rangle \) can satisfy \( \hat{U}\lvert \psi \rangle \otimes \lvert b \rangle = \lvert \psi \rangle \otimes \lvert \psi \rangle \) for every state \( \lvert \psi \rangle \) in a Hilbert space of dimension \( \ge 2 \).

Why it matters

Cloning would let you make many copies of one qubit, measure each in a different basis, and thereby learn an unknown state to arbitrary precision without disturbing it. That would collapse the Heisenberg-style trade-offs at the heart of quantum mechanics and permit superluminal signalling through entanglement. The theorem forbids all of this with a two-line argument, so its consequences are structural rather than technological.

Constructively, no-cloning is the security foundation of quantum key distribution (an eavesdropper cannot copy the carrier qubits) and the reason quantum error correction must protect information without ever reading or duplicating the logical state. It is the sharpest single statement of how quantum information differs from classical bits.

Assumptions

Evolution is unitary (linear and inner-product preserving).If evolution could be nonlinear, linearity could not be invoked in Step 3 and the contradiction dissolves; approximate or probabilistic "cloners" of low fidelity then become possible.

The copier is a single fixed device: one operator \( \hat{U} \) and one blank \( \lvert b \rangle \), independent of the input \( \lvert \psi \rangle \).If \( \hat{U} \) were allowed to depend on \( \lvert \psi \rangle \) you would already have to know the state, which defeats the purpose; state-specific copiers of known states are of course allowed.

The target register starts in a fixed pure blank state, uncorrelated with the data.If the blank were entangled with the input or drawn from a mixed ensemble, the "copy" relation is ill-posed and the inner-product bookkeeping in Step 4 changes; the clean theorem statement no longer applies.

The two states to be copied are non-orthogonal and distinct.Orthogonal states can be cloned (they are perfectly distinguishable, hence effectively classical); the impossibility is specifically about copying arbitrary, possibly overlapping, superpositions.

Derivation

\[ \hat{U}\bigl(\lvert \psi \rangle \otimes \lvert b \rangle\bigr) = \lvert \psi \rangle \otimes \lvert \psi \rangle \]

Suppose, for contradiction, that a perfect cloner exists. Write down what it must do to an input \( \lvert \psi \rangle \) placed against the fixed blank \( \lvert b \rangle \). A

\[ \hat{U}\bigl(\lvert \phi \rangle \otimes \lvert b \rangle\bigr) = \lvert \phi \rangle \otimes \lvert \phi \rangle \]

The same device, with the same \( \hat{U} \) and same \( \lvert b \rangle \), must also copy any second state \( \lvert \phi \rangle \); this is just the "arbitrary state" hypothesis applied twice. A

\[ \hat{U}\Bigl(\bigl(\alpha\lvert \psi \rangle + \beta\lvert \phi \rangle\bigr)\otimes\lvert b\rangle\Bigr) = \alpha\,\lvert \psi \rangle\otimes\lvert \psi \rangle + \beta\,\lvert \phi \rangle\otimes\lvert \phi \rangle \]

Apply \( \hat{U} \) to the superposition \( \alpha\lvert\psi\rangle+\beta\lvert\phi\rangle \). Because \( \hat{U} \) is linear it distributes over the sum and acts term by term as in Steps 1–2. This is the crux: linearity fixes the output. B

\[ \hat{U}\Bigl(\bigl(\alpha\lvert \psi \rangle + \beta\lvert \phi \rangle\bigr)\otimes\lvert b\rangle\Bigr) \overset{!}{=} \bigl(\alpha\lvert \psi \rangle + \beta\lvert \phi \rangle\bigr)\otimes\bigl(\alpha\lvert \psi \rangle + \beta\lvert \phi \rangle\bigr) \]

But the cloning hypothesis of Step 1, applied directly to the input state \( \lvert\chi\rangle=\alpha\lvert\psi\rangle+\beta\lvert\phi\rangle \), demands the tensor square of that whole superposition. We now have two expressions for the same left-hand side. B

\[ \alpha\,\lvert \psi\psi \rangle + \beta\,\lvert \phi\phi \rangle = \alpha^2\lvert \psi\psi \rangle + \beta^2\lvert \phi\phi \rangle + \alpha\beta\bigl(\lvert \psi\phi \rangle + \lvert \phi\psi \rangle\bigr) \]

Set the Step 3 output equal to the Step 4 output and expand the right-hand tensor square. Here \( \lvert\psi\psi\rangle \equiv \lvert\psi\rangle\otimes\lvert\psi\rangle \), etc. The two sides can match only for very special coefficients. B

\[ \alpha = \alpha^2,\qquad \beta = \beta^2,\qquad \alpha\beta = 0 \]

For generic non-orthogonal \( \lvert\psi\rangle,\lvert\phi\rangle \) the kets \( \lvert\psi\psi\rangle,\lvert\phi\phi\rangle,\lvert\psi\phi\rangle,\lvert\phi\psi\rangle \) are linearly independent, so their coefficients must match term by term. C

\[ \{\alpha,\beta\} = \{0,1\}\ \text{only} \quad\Rightarrow\quad \text{no genuine superposition can be cloned} \]

\( \alpha=\alpha^2 \) and \( \alpha\beta=0 \) force \( \alpha\in\{0,1\} \) with the other coefficient zero. Any true superposition (\( \alpha,\beta\neq0 \)) violates the equations. Contradiction with the assumption of a universal cloner. C

\[ \langle \psi \vert \phi \rangle = \langle \psi \vert \phi \rangle^2 \quad\Longrightarrow\quad \langle \psi \vert \phi \rangle \in \{0,1\} \]

An equivalent, basis-free proof: take the inner product of the two Step 1/2 outputs. Unitarity preserves inner products, so \( \langle\psi\vert\phi\rangle\langle b\vert b\rangle = \langle\psi\vert\phi\rangle\langle\psi\vert\phi\rangle \). With \( \langle b\vert b\rangle=1 \) this forces \( x=x^2 \), whose only solutions are orthogonal (\(x=0\)) or identical (\(x=1\)) states. C

Result

\[ \boxed{\;\nexists\ \hat{U},\lvert b\rangle:\ \ \hat{U}\lvert\psi\rangle\lvert b\rangle=\lvert\psi\rangle\lvert\psi\rangle\ \ \forall\,\lvert\psi\rangle\;}\qquad\text{equivalently}\qquad \langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2 \]

Reading. A single fixed unitary can perfectly duplicate at most a set of mutually orthogonal states. The moment two inputs overlap without coinciding (\( 0<\lvert\langle\psi\vert\phi\rangle\rvert<1 \)), no unitary can copy both, and hence none can copy an arbitrary unknown state. Linearity alone kills universal cloning; the inner-product form (Step 8) shows the same conclusion follows from unitarity without ever choosing a basis.

Units check. Every quantity is a dimensionless amplitude or inner product; states are normalised, \( \langle\psi\vert\psi\rangle=1 \), so \( \langle\psi\vert\phi\rangle \) is a pure (dimensionless) complex number and the equation \( x=x^2 \) is dimensionally consistent. \( \hat{U} \) is unitary hence dimensionless, and the tensor product \( \lvert\psi\rangle\otimes\lvert\psi\rangle \) carries no units. Consistent throughout.

Limiting cases

\( \langle\psi\vert\phi\rangle=0 \) (orthogonal): the constraint \( x=x^2 \) is satisfied; orthogonal states are perfectly clonable, recovering the classical bit that can be freely copied.
\( \langle\psi\vert\phi\rangle=1 \) (identical): trivially "clonable" because there is only one state; no information content to duplicate.
Known state (\( \lvert\psi\rangle \) given in advance): you may build a state-specific preparation device \( \hat{U}_\psi \); this is not cloning because \( \hat{U}_\psi \) already encodes the answer.
Approximate cloning: dropping "perfect", the optimal universal cloner reaches fidelity \( F=5/6 \) for a qubit — high but strictly below \(1\), the smooth remnant of the theorem.
Classical limit (macroscopically distinguishable pointer states): overlaps become effectively zero, copying is unobstructed, and everyday photocopying is restored.

Breaks when

Nonlinear quantum mechanics. If the Schrödinger evolution acquired a nonlinear term, Step 3 fails: the map need not distribute over superpositions, and Gisin–Polchinski-type constructions permit both cloning and superluminal signalling. The theorem is exactly as strong as linearity.
Non-orthogonal set collapses to orthogonal. Restricting the input alphabet to a mutually orthogonal (or classical) set removes the overlap term \( \alpha\beta(\lvert\psi\phi\rangle+\lvert\phi\psi\rangle) \) in Step 5; a copier exists and no contradiction arises. The theorem only bites for genuinely quantum, overlapping inputs.
Approximate / probabilistic cloning. Allowing fidelity \( F<1 \), or success only on a subset with some failure probability, evades the exact-equality constraints; imperfect cloners are physically realisable and are studied precisely because perfect ones are forbidden.

Failure modes

"But measurement copies information." Measurement yields one classical outcome and destroys the superposition; it does not produce a second copy of the pre-measurement amplitude \( \alpha,\beta \).
Confusing cloning with entangling. \( \hat{U}_{\text{CNOT}} \) maps \( \lvert x\rangle\lvert 0\rangle\to\lvert x\rangle\lvert x\rangle \) on basis states, which students misread as a cloner. It clones the computational basis but entangles superpositions, giving \( \alpha\lvert00\rangle+\beta\lvert11\rangle \), not \( (\alpha\lvert0\rangle+\beta\lvert1\rangle)^{\otimes2} \).
Assuming \( \hat{U} \) may depend on \( \lvert\psi\rangle \). A device tuned to the unknown state presupposes knowledge of it; this smuggles the answer in and is not universal cloning.
Forgetting the blank must be fixed. Letting \( \lvert b\rangle \) vary with the input is the same illegitimate move dressed differently.
Claiming no-cloning forbids teleportation. Teleportation moves a state (destroying the original); it does not duplicate it, so it is fully consistent with the theorem.
Treating \( \lvert\psi\psi\rangle,\lvert\psi\phi\rangle \) as identical vectors. Order and slot matter in the tensor product; dropping the distinction hides the cross terms that drive the contradiction.

Discussion

The theorem is remarkable for how little it needs: only that state evolution is linear (Step 3) or, equivalently, that it preserves inner products (Step 8). No dynamics, no Hamiltonian, no measurement postulate is invoked. This is why no-cloning is often called a "structural" theorem — it is a property of Hilbert-space linearity itself, and any physical theory built on a linear state space inherits it. The two proofs illuminate the same fact from complementary angles: the linearity proof shows the output is over-determined, while the inner-product proof shows the geometry of state space simply has no room for a universal copier.

The counting version sharpens the intuition. A perfect universal cloner would let you convert one copy of \( \lvert\psi\rangle \) into \( N \) copies, measure them, and estimate the two real parameters of a qubit (its point on the Bloch sphere) to precision \( \sim 1/\sqrt{N} \), all from a single physical system. Since you started with one qubit — two dimensions, one bit of extractable information — this would be information from nothing. No-cloning is the guardrail that keeps the accessible (Holevo) information of one qubit bounded by one classical bit.

Its constructive payoff is enormous. In BB84 quantum key distribution, an eavesdropper who intercepts carrier qubits cannot copy them to measure later; any attempt disturbs the states and reveals itself in the error rate. In quantum error correction, the logical state must be protected without being read or duplicated, forcing the elegant machinery of stabiliser codes that detect errors via syndromes rather than by looking at the data. Even the linearity of quantum computation — the impossibility of "backing up" a mid-computation state — traces to the same root.

At the deepest level, no-cloning is dual to no-deleting (you cannot unitarily erase an unknown copy against a blank either) and both are faces of the reversibility of unitary dynamics. They connect to the no-signalling principle: were cloning possible, Alice could encode a bit in her choice of measurement basis on an entangled pair and Bob, cloning his half, could read it instantly, violating relativistic causality. Thus no-cloning sits at the intersection of quantum linearity, thermodynamic reversibility, and relativistic locality — three pillars that a consistent physical theory must simultaneously respect, and which mutually reinforce one another through exactly this result.

Common misconceptions. No-cloning does not forbid copying known states, does not forbid transmitting or teleporting states, and does not forbid approximate copies (fidelity up to \(5/6\) for a qubit). It forbids only the perfect, universal, deterministic duplication of an arbitrary unknown state by a fixed device.

Worked examples

Example A — Why CNOT is not a cloner. Attempt to copy \( \lvert+\rangle=\tfrac{1}{\sqrt2}(\lvert0\rangle+\lvert1\rangle) \) with CNOT and a blank \( \lvert0\rangle \).

Set up the state and the gate before inserting numbers. A

\[ \text{CNOT}\,\lvert+\rangle\lvert0\rangle = \text{CNOT}\,\tfrac{1}{\sqrt2}\bigl(\lvert00\rangle+\lvert10\rangle\bigr) = \tfrac{1}{\sqrt2}\bigl(\lvert00\rangle+\lvert11\rangle\bigr) \]

Apply CNOT term by term: it flips the target iff the control is \(1\). A

\[ \text{Desired clone: } \lvert+\rangle\otimes\lvert+\rangle = \tfrac12\bigl(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle+\lvert11\rangle\bigr) \]

Write the state a true copier would have produced, for comparison. A

\[ \bigl\lvert\langle +{+}\vert \Phi^+\rangle\bigr\rvert^2 = \Bigl\lvert \tfrac12\cdot\tfrac{1}{\sqrt2}(1+0+0+1)\Bigr\rvert^2 = \Bigl\lvert\tfrac{1}{\sqrt2}\Bigr\rvert^2 = \tfrac12 \]

Overlap of CNOT output \( \lvert\Phi^+\rangle \) with the true clone. Fidelity is only \( 1/2 \). B

\[ \boxed{\ F = \tfrac12\ \text{for }\lvert+\rangle,\quad\text{while }F=1\text{ for }\lvert0\rangle,\lvert1\rangle\ } \]

Reading. CNOT perfectly copies the basis states \( \lvert0\rangle,\lvert1\rangle \) but produces an entangled Bell state, not a product of copies, for the superposition \( \lvert+\rangle \). Its state-dependent fidelity is exactly what the theorem forbids from being uniformly \(1\). Units: fidelity is a dimensionless probability, \( 0\le F\le1 \).

Example B — Optimal approximate cloning fidelity for a qubit. The optimal universal \( N\to M \) qubit cloner (Bužek–Hillery / Gisin–Massar) has per-copy fidelity \[ F_{N\to M} = \frac{N}{M} + \frac{(M-N)(N+1)}{M\,(N+2)}. \] Evaluate the \(1\to2\) case.

Quote the closed-form fidelity symbolically first. B

\[ N=1,\quad M=2 \quad\Rightarrow\quad F = \frac{1}{2} + \frac{(2-1)(1+1)}{2\,(1+2)} \]

Substitute the copy numbers \( N=1,\ M=2 \). A

\[ F = \frac{1}{2} + \frac{2}{6} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\approx 0.833 \]

Arithmetic: \( (M-N)(N+1)=1\cdot2=2 \) over \( M(N+2)=2\cdot3=6 \), added to the trivial-transfer term \( N/M=1/2 \). B

\[ F_{1\to2}^{\text{qubit}} = \frac{5}{6}\approx 0.8\overline{3} \qquad(\text{per-copy fidelity, }d=2) \]

This \( 5/6 \) is the proven optimum; no physical device beats it. It is the sharp ceiling on how well one unknown qubit can be duplicated, the quantitative residue of the exact-cloning ban. C

\[ \boxed{\ F_{1\to2}=\tfrac{5}{6}\approx0.833\ } \]

Reading. Even the best possible universal machine copies an unknown qubit with fidelity \( 5/6 \), safely below perfect. The \( 1/6 \) deficit is the quantitative shadow of no-cloning: you can approach but never reach exact duplication. Units: \( F \) is a dimensionless overlap probability.

Problems

Show explicitly that two orthogonal states \( \lvert0\rangle,\lvert1\rangle \) satisfy the constraint \( \langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2 \), and hence can be cloned by a unitary.
Solution
With \( \langle0\vert1\rangle=0 \), the equation reads \( 0=0^2=0 \), satisfied. A cloner is the CNOT: \( \lvert0\rangle\lvert0\rangle\to\lvert0\rangle\lvert0\rangle \) and \( \lvert1\rangle\lvert0\rangle\to\lvert1\rangle\lvert1\rangle \). Both computational basis states are perfectly duplicated because they carry no relative superposition to be mangled. No contradiction arises since the overlap term \( \alpha\beta \) is absent.
For two states with real overlap \( \langle\psi\vert\phi\rangle = s \), the equation \( s=s^2 \) has solutions \( s=0 \) and \( s=1 \). Explain physically why any intermediate \( 0<s<1 \) is forbidden for a perfect cloner, and compute the "gap" \( s-s^2 \) at \( s=\tfrac12 \).
Solution
Intermediate overlap means the states are neither perfectly distinguishable nor identical; a cloner would have to both preserve inner products (unitarity gives output overlap \( s \)) and square them (cloning gives \( s^2 \)), which is impossible unless \( s=s^2 \). The gap \( s-s^2 \) measures the inconsistency: at \( s=\tfrac12 \), gap \( =\tfrac12-\tfrac14=\tfrac14 \). The maximum gap over \( [0,1] \) occurs at \( s=\tfrac12 \), where the theorem bites hardest.
A student proposes cloning \( \lvert\psi\rangle=\cos\theta\lvert0\rangle+\sin\theta\lvert1\rangle \) using CNOT. Compute the fidelity \( F(\theta)=\lvert\langle\psi\psi\vert\text{CNOT}\,\psi0\rangle\rvert^2 \) and find where it is worst.
Solution
CNOT gives \( \cos\theta\lvert00\rangle+\sin\theta\lvert11\rangle \). The desired clone is \( (\cos\theta\lvert0\rangle+\sin\theta\lvert1\rangle)^{\otimes2}=\cos^2\theta\lvert00\rangle+\cos\theta\sin\theta(\lvert01\rangle+\lvert10\rangle)+\sin^2\theta\lvert11\rangle \). Overlap \( =\cos\theta\cdot\cos^2\theta+\sin\theta\cdot\sin^2\theta=\cos^3\theta+\sin^3\theta \). So \( F=(\cos^3\theta+\sin^3\theta)^2 \). At \( \theta=0 \) or \( \pi/2 \), \( F=1 \) (basis states). Worst at \( \theta=\pi/4 \): \( \cos^3+\sin^3=2\cdot(1/\sqrt2)^3=2/(2\sqrt2)=1/\sqrt2 \), so \( F=1/2 \), matching Worked Example A.
Prove the inner-product form of the theorem in full: assuming \( \hat{U}\lvert\psi\rangle\lvert b\rangle=\lvert\psi\rangle\lvert\psi\rangle \) and \( \hat{U}\lvert\phi\rangle\lvert b\rangle=\lvert\phi\rangle\lvert\phi\rangle \) with \( \hat{U} \) unitary and \( \langle b\vert b\rangle=1 \), derive \( \langle\psi\vert\phi\rangle\in\{0,1\} \).
Solution
Take the inner product of the two output equations. The left side is \( \langle\psi\vert\langle b\vert\,\hat U^\dagger\hat U\,\lvert\phi\rangle\lvert b\rangle=\langle\psi\vert\phi\rangle\langle b\vert b\rangle=\langle\psi\vert\phi\rangle \) using \( \hat U^\dagger\hat U=\mathbb{1} \). The right side is \( \langle\psi\vert\phi\rangle\langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2 \). Hence \( \langle\psi\vert\phi\rangle=\langle\psi\vert\phi\rangle^2 \), i.e. \( x=x^2 \) with roots \( x=0,1 \). Any pair with \( 0<\lvert x\rvert<1 \) cannot be simultaneously cloned. QED.
Explain, using no-cloning, why an eavesdropper on BB84 cannot copy the qubits to measure them later, and estimate the induced error rate if she instead measures in a random basis and resends.
Solution
No-cloning forbids duplicating the unknown carrier qubit (prepared in one of the four BB84 states, which are pairwise non-orthogonal), so "store a copy, measure later" is impossible. Forced to measure-and-resend, Eve guesses the basis; she picks the correct basis half the time (no error) and the wrong basis half the time. When wrong, her resent state gives Bob the correct bit only \( 1/2 \) the time. Induced error probability \( =\tfrac12\cdot0+\tfrac12\cdot\tfrac12=\tfrac14=25\% \). This large disturbance is exactly what makes eavesdropping detectable, and it is a direct operational consequence of the theorem.