The No-Cloning Theorem
Why an unknown quantum state can never be copied
By the end of this topic you'll be able to
Wootters and Zurek proved in 1982 that a single quantum cannot be cloned: no unitary operation can take an arbitrary, unknown state |ψ⟩ and produce two independent copies of it. This is not an engineering limitation awaiting a cleverer piece of hardware — it is a short, direct consequence of linearity, provable in a few lines.
The claim to rule out is a single unitary U that works for every unknown input: U(|ψ⟩|0⟩) = |ψ⟩|ψ⟩ for all |ψ⟩, where the second qubit starts blank and ends up as a copy of the first. Note the word 'unitary' is doing real work here — cloning a state you already know classically (just prepare a second copy from scratch) is trivial and not what's at stake.
Suppose such a U existed, and that it correctly clones two particular states, |0⟩ and |1⟩: U(|0⟩|0⟩) = |0⟩|0⟩ and U(|1⟩|0⟩) = |1⟩|1⟩. Now apply U to the superposition |ψ⟩ = (|0⟩+|1⟩)/√2 as its input.
Because U is required to be a linear operator (every valid quantum gate is), it must act on the superposition as the same weighted sum of its action on each piece: U(|ψ⟩|0⟩) = (1/√2)(U(|0⟩|0⟩) + U(|1⟩|0⟩)) = (1/√2)(|0⟩|0⟩ + |1⟩|1⟩).
But a faithful clone of |ψ⟩ would instead have to be |ψ⟩|ψ⟩ = (1/2)(|0⟩+|1⟩)⊗(|0⟩+|1⟩) = (1/2)(|00⟩+|01⟩+|10⟩+|11⟩) — a state with all four basis terms present, not just two. Linearity forces one answer, faithful copying demands a different one, and the two disagree: no U can satisfy both at once.
The no-cloning theorem is why quantum error correction cannot simply 'make backup copies' the way a classical repetition code does — a code has to spread information across entangled qubits instead, never literally duplicating an unknown state — and it's part of why quantum key distribution protocols like BB84 can detect an eavesdropper: intercepting and re-transmitting an unknown qubit without disturbing it would require exactly the cloning this theorem forbids.
Try It Yourself
Suppose a proposed cloning machine U correctly copies |+⟩ = (|0⟩+|1⟩)/√2 and |−⟩ = (|0⟩−|1⟩)/√2 (i.e. U(|+⟩|0⟩)=|+⟩|+⟩ and U(|−⟩|0⟩)=|−⟩|−⟩). Show it fails to correctly clone |0⟩.
- 1Write |0⟩ in terms of |+⟩ and |−⟩: since |+⟩+|−⟩ = √2|0⟩, we have |0⟩ = (|+⟩+|−⟩)/√2.
- 2By linearity, U(|0⟩|0⟩) = U((|+⟩+|−⟩)/√2 ⊗ |0⟩) = (1/√2)(U(|+⟩|0⟩) + U(|−⟩|0⟩)) = (1/√2)(|+⟩|+⟩ + |−⟩|−⟩).
- 3Expand that result back in the {|0⟩,|1⟩} basis: (1/√2)(|+⟩|+⟩+|−⟩|−⟩) works out to (1/√2)(|00⟩+|11⟩) — the Bell state Φ⁺, not the product state |0⟩|0⟩ a faithful clone requires.
- 4So a machine tuned to clone |+⟩ and |−⟩ correctly produces an entangled Bell state, not a clean copy, when fed |0⟩ — exactly the contradiction the general proof predicts.
U(|0⟩|0⟩) comes out as the entangled state (|00⟩+|11⟩)/√2, not the faithful copy |0⟩|0⟩ — confirming no single U can clone every state correctly, even one that happens to clone two particular states just fine.
Reference
| No-cloning theorem | Unknown quantum states cannot be copied | |
| Cause | A linear U's action on a superposition is forced, and disagrees with faithful copying |
Quick Check
Why can't a linear unitary U clone an arbitrary unknown quantum state?
What does the no-cloning theorem rule out for quantum error correction?