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No-Cloning Theorem Explained: Why You Cannot Copy a Qubit

No, you cannot copy an unknown qubit state. Linearity of unitary operations strictly forbids cloning. See the 3-step algebraic proof & CNOT simulation →

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Written by

Malcolm Low
Part of the Quantum Computing: A Complete Learning Path series.
Quick Answer & Key Insight

Can You Make an Identical Copy of an Arbitrary Qubit?

No, you cannot clone an unknown qubit state. The mathematical linearity of unitary quantum operators strictly forbids cloning arbitrary quantum states. See the 3-step algebraic proof and quantum teleportation alternative below →

QUANTUM SERIES 2026
The no-cloning theorem: why an unknown quantum state cannot be copied, and what that buys us.

Copying is so ordinary in classical computing that we never think about it. Every assignment, every memory write, every Ctrl+C duplicates a bit string perfectly and at will. Quantum mechanics flatly refuses to do the same for an unknown state. There is no machine, no clever circuit, no sequence of gates that takes an arbitrary qubit and hands you back two identical copies of it. This is the no-cloning theorem, and far from being a limitation we tolerate, it is the structural reason quantum key distribution is secure, the reason entanglement cannot be used to signal faster than light, and the reason quantum error correction had to be reinvented from scratch.

The proof is short. It falls out of two properties we have already relied on throughout this series: linearity of quantum evolution, and unitarity. Below we prove it both ways, clear up the very common confusion about CNOT “copying” a qubit, walk through the consequences, and then watch a cloning attempt fail in Qiskit.

1  ·  What “cloning” would actually mean

A cloning machine would be a single fixed unitary U together with a standard blank register |e⟩, such that for every possible input state |ψ⟩ it produces:

U  ( |ψ⟩ ⊗ |e⟩ )  =  |ψ⟩ ⊗ |ψ⟩

Three pieces of that statement deserve unpacking, because the whole theorem turns on them.

What |e⟩ is, and what “blank” means. The register |e⟩ is simply the second slot, the one that will end up holding the copy. We call it “blank” by analogy with an empty sheet of paper or an uninitialised memory cell: before the operation it carries no information whatsoever about the input |ψ⟩. The conventional choice is |0⟩, or |0…0⟩ when the copy needs several qubits, but |e⟩ does not have to be the zero vector specifically. Any fixed pure state works equally well for the theorem. The property that actually matters is not its literal value but that it is independent of the unknown input. A register that already secretly equalled |ψ⟩ would not be blank, and you would not have copied anything, you would have been handed the answer.

What “fixed” means, for both U and |e⟩. Fixed means chosen in advance and identical on every run, never allowed to depend on which |ψ⟩ happens to come in. This is the condition that keeps the problem honest. If you were permitted to pick |e⟩ or U based on the input, you would first have to know what |ψ⟩ is, and at that point there is nothing left to clone: you could just prepare a second copy yourself. So the cloner gets one unitary and one starting register, and they must work for an arbitrary, unknown input. The claim of the theorem is that no such pair (U, |e⟩) exists.

Why cloning a known state is allowed, and how you do it. The theorem forbids copying an unknown arbitrary state. It says nothing about states you already understand, and there are two everyday ways such a state can be “known”:

You know… How you get two copies
The preparation recipe (the amplitudes, or the circuit that builds it) Just run the preparation twice on two fresh |0⟩ qubits. You are re-preparing from a known description, not copying an unknown state.
That the state is one of a known orthogonal set Measure in the basis that distinguishes the set. Orthogonal states are eigenstates of that measurement, so they survive it undisturbed. Read off the label, then re-prepare as many copies as you like.

As a concrete example of the first route, if you are told the state is |ψ⟩ = Ry(θ)|0⟩ for a known angle θ, you simply apply Ry(θ) to two separate qubits. No universal device is involved, and no rule is broken. The contradiction in the next two sections appears only when no such side information exists, that is, when the same machine must handle a continuum of unknown inputs at once.

The one-line distinction: you can always re-prepare a state you know how to build, and you can copy states that act like classical labels. What you cannot do is build one fixed machine that duplicates an arbitrary, unknown state handed to you blind.

2  ·  Proof I: linearity forbids it

Assume, for contradiction, that a universal cloner U exists. Pick two arbitrary states |ψ⟩ and |φ⟩. By assumption U clones both of them:

U |ψ⟩|e⟩  =  |ψ⟩|ψ⟩
U |φ⟩|e⟩  =  |φ⟩|φ⟩

Now feed the machine a superposition of the two inputs, |χ⟩ = (1/√2)(|ψ⟩ + |φ⟩). Quantum evolution is linear, so U acts term by term on the input:

U |χ⟩|e⟩  =  (1/√2) ( U|ψ⟩|e⟩  +  U|φ⟩|e⟩ )
         =  (1/√2) ( |ψ⟩|ψ⟩  +  |φ⟩|φ⟩ )

But if U is genuinely a cloner, then by definition it must also clone |χ⟩ itself, producing |χ⟩|χ⟩. Expanding that out:

|χ⟩|χ⟩  =  (1/2) ( |ψ⟩ + |φ⟩ )( |ψ⟩ + |φ⟩ )
      =  (1/2) ( |ψ⟩|ψ⟩ + |ψ⟩|φ⟩ + |φ⟩|ψ⟩ + |φ⟩|φ⟩ )

The two results disagree. Linearity gave us only the “diagonal” terms |ψ⟩|ψ⟩ and |φ⟩|φ⟩, whereas true cloning of |χ⟩ demands the cross terms |ψ⟩|φ⟩ and |φ⟩|ψ⟩ as well. They can only match if the cross terms vanish, which happens precisely when |ψ⟩ and |φ⟩ are identical or orthogonal. For general, non-orthogonal states the equality is impossible. The cloner cannot exist.

The contradiction is the same one that powers most quantum no-go results: a device defined by its action on a basis is forced, by linearity, into a behaviour on superpositions that contradicts the original specification.

3  ·  Proof II: unitarity forbids it

The linearity argument shows cloning fails for superpositions. A second proof, often considered the cleaner one, uses the fact that unitary operations preserve inner products. Suppose again that U clones two states:

U |ψ⟩|e⟩  =  |ψ⟩|ψ⟩
U |φ⟩|e⟩  =  |φ⟩|φ⟩

The inner product is taken over the full two-register states that U acts on, not over |ψ⟩ and |φ⟩ on their own. Label the two inputs and their images under U:

Input A   = |ψ⟩|e⟩        Output UA  = |ψ⟩|ψ⟩
Input B   = |φ⟩|e⟩        Output UB  = |φ⟩|φ⟩

If U is unitary, then   ⟨A|B⟩  =  ⟨UA|UB⟩

That identity, ⟨A|B⟩ = ⟨UA|UB⟩, is the definition of a unitary: it preserves inner products. U never acts on the inner product itself, which is only a number; it acts on the states, and unitarity says doing so cannot change their overlap. Now evaluate both sides, using the tensor-product factorisation ⟨a|⟨b| · |c⟩|d⟩ = ⟨a|c⟩ ⟨b|d⟩ and ⟨e|e⟩ = 1:

⟨A|B⟩   = ⟨ψ|⟨e| · |φ⟩|e⟩  =  ⟨ψ|φ⟩ ⟨e|e⟩  =  ⟨ψ|φ⟩
⟨UA|UB⟩ = ⟨ψ|⟨ψ| · |φ⟩|φ⟩  =  ⟨ψ|φ⟩ ⟨ψ|φ⟩  =  ⟨ψ|φ⟩²

so   ⟨ψ|φ⟩  =  ⟨ψ|φ⟩²

Write x = ⟨ψ|φ⟩. The condition x = x² rearranges to x(x − 1) = 0, whose only solutions are x = 0 or x = 1. In other words, cloning is consistent only when the two states are orthogonal (x = 0) or identical (x = 1). Any pair of distinct, non-orthogonal states, the generic case, breaks the equality. No universal cloner can exist.

⟨ψ|φ⟩ Relationship Cloneable?
0 Orthogonal (a known basis set) Yes, this is just classical copying
1 Identical (already known) Yes, trivially
anything else Distinct, non-orthogonal No

Both proofs converge on the same boundary. Orthogonal states carry distinguishable classical labels and can be copied; overlapping quantum states cannot.

4  ·  “But doesn’t CNOT copy a qubit?”

This is the single most common point of confusion, and it is worth resolving carefully. A CNOT with the input qubit as control and a fresh |0⟩ as target looks exactly like a copy operation. On computational basis states it behaves like one:

CNOT |0⟩|0⟩  =  |0⟩|0⟩
CNOT |1⟩|0⟩  =  |1⟩|1⟩

So far it really is copying: the second qubit ends up matching the first. The trouble appears the moment the control is a superposition. Take |ψ⟩ = a|0⟩ + b|1⟩ as control and |0⟩ as target:

CNOT ( a|0⟩ + b|1⟩ )|0⟩  =  a|00⟩ + b|11⟩

Compare that with what a true clone would have produced, namely |ψ⟩|ψ⟩:

|ψ⟩|ψ⟩  =  a²|00⟩ + ab|01⟩ + ab|10⟩ + b²|11⟩

These are different states. The CNOT output a|00⟩ + b|11⟩ is an entangled pair, not two independent copies. If you discard the original and look at the “copy” on its own, you do not find |ψ⟩; you find a mixed state that has lost the phase relationship between a and b entirely. CNOT copies the value in the computational basis but destroys the superposition structure that made the state quantum in the first place. That is exactly the gap between copying classical information, which is always allowed, and cloning a quantum state, which is not.

Rule of thumb: CNOT copies basis labels, not quantum states. The instant the input is in superposition, the “copy” is entanglement, not duplication.

5  ·  Why this single theorem matters so much

No-cloning is not a footnote. Three of the most important results in quantum information rest directly on it.

Quantum key distribution. In BB84, Alice sends qubits encoded in randomly chosen bases. An eavesdropper, Eve, would love to intercept each qubit, keep a copy, and forward the original untouched so she can measure her copies later once the bases are announced. No-cloning makes that impossible. Eve cannot duplicate an unknown qubit, so she is forced to measure in transit, and measuring in the wrong basis disturbs the state and injects a detectable error rate. The security of the protocol is the no-cloning theorem wearing a different hat.

No faster-than-light signalling. Entanglement correlates distant measurement outcomes, which has always tempted people into superluminal-communication schemes. Many of those schemes secretly assume Bob can clone his half of an entangled pair, make many copies, and read off the statistics to learn which basis Alice measured in. No-cloning closes that loophole and is one of the reasons quantum mechanics and relativity coexist without paradox.

Quantum error correction. Classical error correction leans on the simplest trick imaginable: copy the bit several times and take a majority vote. That is illegal in the quantum setting, because you cannot copy an unknown logical state. Quantum codes had to be built on a different principle, spreading logical information across many physical qubits using entanglement and measuring error syndromes without ever reading or duplicating the encoded state. The whole architecture of fault tolerance is shaped by the fact that the obvious classical fix is off the table.

A useful way to hold it in your head: no-cloning is what stops quantum information from behaving like classical information. Almost everything that makes quantum protocols both powerful and secure traces back to that one restriction.

6  ·  Watching a clone fail in Qiskit

The abstract proof is convincing, but it is satisfying to watch the failure happen numerically. We attempt the naive CNOT “clone” of the state |+⟩ = H|0⟩, then measure how badly it falls short by two yardsticks: fidelity against the intended |+⟩|+⟩, and the purity of the copy qubit on its own.

Three terms in plain language:

Fidelity is a similarity score between two quantum states, from 0 to 1. Think of it as a match percentage: 1.0 means identical, 0 means completely unrelated.

Reduced state is the honest description of one qubit when you ignore its partner. Picture two coins glued so they always land the same way: cover one and look at just the other, and that single coin looks like a plain 50/50 flip. The “match” information lived in the pair, not in either coin alone. Discarding the partner mathematically is called tracing it out.

Purity tells you whether a state is sharp and definite or a blurry random mixture. Purity 1 is a clean, definite state. For a single qubit the lowest value is 0.5, called maximally mixed, which is total randomness, the quantum version of a fair coin holding no information.

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector, partial_trace, state_fidelity

qc = QuantumCircuit(2)
qc.h(0)         # prepare |+> on qubit 0 (the “original”)
qc.cx(0, 1)    # naive “cloning” attempt via CNOT

state = Statevector(qc)
print(state)   # (|00> + |11>)/sqrt(2), a Bell state

target = Statevector.from_label(‘++’)   # what we wanted: |+>|+>
print(state_fidelity(state, target))   # 0.5

rho_copy = partial_trace(state, [0])   # trace out the original
print(rho_copy.data)   # I/2, maximally mixed
print(rho_copy.purity().real)   # 0.5

Running it gives:

Output state:  (√2/2)|00⟩ + (√2/2)|11⟩
Fidelity to |+⟩|+⟩:  0.5
Reduced state of copy:  [[0.5, 0], [0, 0.5]]
Purity of copy:  0.5

Every number tells the same story. The output is a maximally entangled Bell state, not the product |+⟩|+⟩ we were hoping for, and the fidelity to the intended clone is only 0.5. Trace out the original and the supposed “copy” is the maximally mixed state I/2, with purity 0.5 rather than 1. The copy retains no memory of the |+⟩ superposition at all; it is as good as a fair coin. The CNOT did precisely what the theorem says it must: it produced correlation, not duplication.

Try changing qc.h(0) to any other single-qubit rotation, say qc.ry(0.7, 0). The fidelity stays stubbornly below 1 for every non-basis input, which is the no-cloning theorem reproduced one statevector at a time.

Frequently Asked Questions

What is the no-cloning theorem?

It states that no unitary operation can take an arbitrary, unknown quantum state and produce two independent copies of it. It follows directly from linearity and unitarity of quantum evolution, both proven above.

Can quantum entanglement beat the no-cloning theorem?

No. Entanglement correlates two qubits, it does not duplicate a state onto each of them independently. A CNOT “clone” attempt produces an entangled Bell pair, not two copies — tracing out either qubit leaves a maximally mixed state carrying none of the original superposition.

Does CNOT copy a qubit?

Only on computational basis states (|0⟩ or |1⟩), where it behaves like classical bit-copying. The moment the control qubit is in superposition, CNOT produces entanglement instead of a copy — the output state differs from a true clone, as shown in the algebra above.

Why does the no-cloning theorem matter for security?

It’s what makes quantum key distribution protocols like BB84 secure. An eavesdropper cannot copy an unknown qubit in transit to measure later, so intercepting it forces a measurement that disturbs the state and reveals the eavesdropper’s presence through detectable errors.

Quantum Series 2026  ·  Built with Qiskit 1.x

✦ This article was generated with the assistance of Claude by Anthropic ✦

About the author

Malcolm Low is an Associate Professor at the Singapore Institute of Technology, writing on quantum computing, programming, and applied computing from Singapore.

Website: malcolmlow.com  ·  Singapore

Related interactive experiment: The no-cloning theorem limits what quantum information can do; entanglement reveals what it can do. See our guide to Two-Qubit Entanglement & The 4 Bell States, or explore the contrast with this Bell’s inequality simulator: Play and learn with the CHSH game.
Continue the Quantum Series
Quantum Series Recommended Next Reading in Quantum Computing

Deepen your quantum intuition with our step-by-step mathematical and circuit guides:

FAQ Frequently Asked Questions

What is the No-Cloning Theorem in quantum mechanics?

The No-Cloning Theorem, proven by Wootters, Zurek, and Dieks in 1982, states that it is impossible to create an identical copy of an arbitrary unknown quantum state using unitary transformations.

If qubits cannot be cloned, how does quantum teleportation work?

Quantum teleportation transmits quantum information by destroying the original state through Bell-basis measurement and reconstructing it at the receiver using shared entanglement, strictly respecting no-cloning because the original is erased.

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