No-Cloning and Reversible Copying
Classical reversible computing has a perfectly respectable way to copy: fan-out with a zero
ancilla. A CNOT with a fresh 0 bit as target sends
(x, 0) \mapsto (x, x \oplus 0) = (x, x) — a flawless duplicate, and a
flawless bijection (run it again and the copy vanishes). Nothing in the reversibility rulebook forbids
copying; you have used this move in half the circuits of this course. So here is the natural bet: the
same
gate, being unitary, should copy a qubit just as happily. Place your wager, apply
the CNOT to a superposition — and watch quantum mechanics decline, politely but absolutely. The failure
is not a technicality. It is the no-cloning theorem, and it marks the precise boundary
between classical and quantum reversibility.
Run the classical copier on a superposition
Take an unknown qubit |\psi\rangle = \alpha|0\rangle + \beta|1\rangle and a
fresh ancilla |0\rangle, and apply CNOT
(|\psi\rangle as control). Unitaries are linear, so the gate acts branch by
branch — on |0\rangle|0\rangle it does nothing, on
|1\rangle|0\rangle it flips the target:
\mathrm{CNOT}\,\big(\alpha|0\rangle + \beta|1\rangle\big)|0\rangle \;=\; \alpha|00\rangle + \beta|11\rangle.
Now write down what a genuine copy would have been — two independent instances of
|\psi\rangle, i.e. the product state:
|\psi\rangle \otimes |\psi\rangle \;=\; \big(\alpha|0\rangle + \beta|1\rangle\big)\otimes\big(\alpha|0\rangle + \beta|1\rangle\big) \;=\; \alpha^2|00\rangle + \alpha\beta|01\rangle + \alpha\beta|10\rangle + \beta^2|11\rangle.
Compare the two, coefficient by coefficient. The CNOT output has no
|01\rangle or |10\rangle terms at all, and its
diagonal amplitudes are \alpha, \beta rather than
\alpha^2, \beta^2. The two expressions agree only when
\alpha\beta = 0 — that is, only when |\psi\rangle
was a plain basis state to begin with. For any true superposition, what the CNOT produced is not two
copies of anything: it is the entangled state
\alpha|00\rangle + \beta|11\rangle, a single shared coin that cannot be
factored into "this qubit's state" times "that qubit's state". The copier didn't copy — it
correlated.
One picture of the boundary
The circuit below is the same in both rows; only the input changes. Feed it a basis state and it is the
honest classical fan-out you have trusted all course. Feed it a superposition and the identical gate
manufactures entanglement instead of copies — the cross terms a real clone would need simply never
appear.
No machine can do better: the no-cloning theorem
Perhaps CNOT is just the wrong gadget? The
no-cloning
theorem closes every loophole: no unitary whatsoever clones an unknown state.
The proof is two lines of linearity. Suppose a unitary U cloned both basis
states into a blank register:
U|0\rangle|0\rangle = |0\rangle|0\rangle and
U|1\rangle|0\rangle = |1\rangle|1\rangle. Then on a superposition,
linearity leaves no freedom:
U\big(\alpha|0\rangle + \beta|1\rangle\big)|0\rangle = \alpha\,U|0\rangle|0\rangle + \beta\,U|1\rangle|0\rangle = \alpha|00\rangle + \beta|11\rangle.
But cloning |\psi\rangle demands the product
|\psi\rangle|\psi\rangle, whose expansion (above) contains
\alpha\beta|01\rangle and \alpha\beta|10\rangle
terms. Equating the two forces \alpha\beta = 0. So any machine that
copies the basis states is thereby forbidden from copying their superpositions — the linear
structure of quantum mechanics makes "copies everything" self-contradictory. Cloning is not hard; it is
inconsistent.
Notice what the theorem does not say. Copying known states is fine (prepare another
one from the recipe). Copying basis states is fine — that is classical fan-out. What cannot
exist is a single fixed unitary that duplicates an arbitrary unknown state. Classical
reversible copying survives in the quantum world exactly as far as one chosen basis, and not one
superposition further.
Measurement: the one door out of unitarity
This lesson completes a ledger the course has been keeping. In quantum mechanics every gate is
unitary — reversible without exception. The lone non-unitary operation in the formalism is
measurement: it takes \alpha|0\rangle + \beta|1\rangle to
|0\rangle or |1\rangle, many states to
one outcome — a genuine funnel, irreversible in exactly the truth-table sense of this course. Even here,
modern physics softens the verdict: in the decoherence picture a measurement is just a unitary that
entangles the system with a vast environment — the information is not destroyed but scattered beyond
recall, an accounting you will recognise from the thermodynamic half of this course. Practically,
though, the rule of thumb stands: unitaries rewind; measurements do not.
- classical reversible fan-out is CNOT with a zero ancilla:
(x, 0) \mapsto (x, x) — a legal bijection, and a legal quantum gate;
- on a superposition it entangles instead of copying:
\mathrm{CNOT}(\alpha|0\rangle + \beta|1\rangle)|0\rangle = \alpha|00\rangle + \beta|11\rangle \neq |\psi\rangle|\psi\rangle,
the product state having extra \alpha\beta cross terms;
- no unitary clones an unknown state: cloning the basis states plus linearity forces
\alpha\beta = 0 — copying is consistent only within one fixed basis;
- measurement is the sole irreversible operation of the theory — and even it is unitary evolution in
disguise once the environment is counted.
Classical error correction leans on copying — store the bit three times, take majority votes. No-cloning
seems to slam that door for qubits, and for decades sceptics assumed quantum computing would die of
noise. The escape is beautiful: a quantum code entangles instead of copying. The three-qubit
repetition code encodes \alpha|0\rangle + \beta|1\rangle as
\alpha|000\rangle + \beta|111\rangle — built, in fact, by exactly the two
CNOT fan-outs of this lesson. Look closely: that state is not
|\psi\rangle|\psi\rangle|\psi\rangle (no cross terms — the very signature you
learned to check today). No qubit individually carries \alpha or
\beta; the information lives only in the correlations, where local noise
cannot reach all of it at once. The "failure mode" of the classical copier — entangling rather than
duplicating — turns out to be precisely the resource that protects quantum information.
The state \alpha|00\rangle + \beta|11\rangle looks like two copies —
measure both qubits and you even get matching answers every time! But run the checks. A true pair of
copies is a product state: measuring one qubit would tell you nothing new about the other, and each
qubit alone would behave exactly like |\psi\rangle (interference and all).
Here neither holds: the outcomes are perfectly correlated (measure one, the other is decided), and each
qubit alone acts like a classical random bit — its interference is gone, washed out by entanglement with
its partner. The infallible test is the one from this lesson: expand the would-be copy
|\psi\rangle \otimes |\psi\rangle and look for the
\alpha\beta cross terms on |01\rangle and
|10\rangle. If they are missing, you are holding correlation, not copies.