No-Cloning and No-Signaling

Two restrictions organize quantum communication before any protocol is built.

No-cloning says that one physical device cannot make exact, deterministic copies of every state in a family containing distinct nonorthogonal pure states. No-signaling says that a party cannot encode a message into another party’s unconditioned local statistics merely by choosing a local operation on a shared state. The first limits replication; the second limits controllable remote influence.

They are often mentioned together because a hypothetical perfect cloner could turn different ensemble decompositions of the same density operator into a superluminal communication scheme. They are nevertheless different statements. No-cloning follows from the linear, inner-product-preserving structure of quantum dynamics. No-signaling is an operational constraint on observable marginal probabilities. Neither slogan should be used as a substitute for its assumptions.

Required background. Tensor products describes the input and copy registers. Partial trace gives local states, and quantum channels supplies physical maps and their dilations.

Exact cloning of pure states

Work with finite-dimensional registers and ordinary completely positive, trace-preserving quantum channels. Let $\mathcal S=\{|\psi_i\rangle\}$ be a possible input family. An exact deterministic cloner would be one physical channel $\mathcal C$, independent of $i$, such that, with $\rho_i=|\psi_i\rangle\langle\psi_i|$,

$$\mathcal C(\rho_i\otimes|0\rangle\langle0|) =\rho_i\otimes\rho_i$$

for every $|\psi_i\rangle\in\mathcal S$. The blank state $|0\rangle$ and the device are fixed. The input label is not supplied as side information.

The theorem of Wootters and Zurek (1982) and Dieks (1982) can be stated sharply:

A single physical process can exactly and deterministically clone a family of pure states only when every pair is either orthogonal or represents the same ray. In particular, no universal cloner exists for all pure states of a nontrivial quantum system.

The common word “unknown” is operational shorthand. It does not impose a mysterious property on the state. It means that the device must work without being told which member of the promised family arrived. If a complete classical description of one fixed state is already known, another instance can simply be prepared.

Inner-product proof

Every channel admits a Stinespring dilation. Absorb the fixed blank and initial environment state into an isometry $V$. If exact cloning were possible, its action on the unknown input would be

$$\begin{aligned} V|\psi\rangle &= |\psi\rangle|\psi\rangle|e_\psi\rangle, \\ V|\phi\rangle &= |\phi\rangle|\phi\rangle|e_\phi\rangle. \end{aligned}$$

Isometries preserve inner products. Writing $s=\langle\psi|\phi\rangle$ gives

$$s = s^2 \langle e_\psi|e_\phi\rangle.$$

Taking absolute values,

$$|s| = |s|^2 \left| \langle e_\psi|e_\phi\rangle \right| \leq |s|^2.$$

For distinct nonorthogonal states, $0<|s|<1$, so $|s|^2<|s|$. The two inequalities contradict each other. Exact cloning is possible only for $|s|=0$, meaning orthogonal alternatives, or $|s|=1$, meaning the same physical pure state up to phase.

This proof includes noisy open-system implementations: any environment, ancilla, measurement, and discarded record can be absorbed into the dilation. Allowing a larger laboratory does not evade the theorem.

Linearity made visible

Suppose a controlled-NOT gate copies the computational-basis alternatives:

$$\begin{aligned} |0\rangle|0\rangle&\longmapsto|0\rangle|0\rangle,\\ |1\rangle|0\rangle&\longmapsto|1\rangle|1\rangle. \end{aligned}$$

For a superposition $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, linearity requires

$$|\psi\rangle|0\rangle \longmapsto \alpha|00\rangle+\beta|11\rangle.$$

This output is generally entangled. Two copies would instead be

$$|\psi\rangle|\psi\rangle ={} \alpha^2|00\rangle +\alpha\beta|01\rangle + \alpha\beta|10\rangle +\beta^2|11\rangle.$$

The cross terms expose the contradiction. A gate that copies one orthogonal basis copies the classical label represented by that basis; it does not clone arbitrary superpositions.

What can still be copied?

No-cloning has exact qualifications, not loopholes.

Task	Possible?	Reason
prepare another instance of a known state	yes	the classical preparation description is available
copy states from one known orthogonal basis	yes	measure the basis label and reprepare it, or copy it coherently
clone an arbitrary nonorthogonal family exactly and deterministically	no	inner products cannot be preserved
exactly clone a finite linearly independent family with a heralded flag	yes	every input can have nonzero success probability; failures remain possible
produce imperfect copies	yes	fidelity remains below one for some inputs
broadcast a commuting mixed-state family	yes	the family is classical in one common eigenbasis

A probabilistic exact cloner has a success flag. On success it produces exact copies, but a distinct nonorthogonal pair cannot succeed deterministically. For a finite family of distinct pure-state rays, requiring a nonzero success probability for every member gives the linear-independence criterion of Duan and Guo (1998). Orthogonal families can of course succeed with probability one. The successful branch and the failed branch together form the physical channel. An approximate cloner is deterministic but sacrifices fidelity. For the optimal symmetric universal $1\!\to\!2$ qubit cloner (Bužek and Hillery, 1996; Gisin and Massar, 1997), each marginal output has state fidelity

$$F = \langle\psi|\rho_{\mathrm{clone}}|\psi\rangle = \frac56.$$

That value is an optimum for a specific universal, symmetric, single-copy task. It is not a generic fidelity bound for every state-dependent or asymmetric cloning problem.

Cloning and broadcasting are not synonyms

For a mixed input $\rho_i$, cloning would demand the product output

$$\rho_i \longmapsto \rho_i\otimes\rho_i.$$

Broadcasting asks only for an output $\omega_i^{AB}$ whose two marginals are both $\rho_i$:

$$\operatorname{Tr}_B\omega_i^{AB} = \operatorname{Tr}_A\omega_i^{AB} = \rho_i.$$

The output may be correlated, so broadcasting is weaker. A family of density operators is exactly broadcastable by one channel if and only if its members commute (Barnum et al., 1996). For example, copying the eigenbasis label of $\rho_p=p|0\rangle\langle0|+(1-p)|1\rangle\langle1|$ produces the correlated state $p|00\rangle\langle00|+(1-p)|11\rangle\langle11|$. Both marginals equal $\rho_p$, but the output is generally not $\rho_p\otimes\rho_p$. For pure inputs, a pure marginal cannot be correlated with another system, so broadcasting reduces to cloning. The full theorem and its mixed-state interpretation are in the No-Broadcasting Theorem.

No-signaling as a probability constraint

Consider two separated parties. Alice chooses a setting $x$ and records an outcome $a$; Bob chooses $y$ and records $b$. Their experiment defines

$$p(a,b|x,y).$$

No-signaling from Alice to Bob means that Bob’s marginal does not depend on Alice’s setting:

$$\sum_a p(a,b|x,y) = p(b|y)$$

for every $b,x,y$. No-signaling from Bob to Alice similarly requires

$$\sum_b p(a,b|x,y) = p(a|x).$$

These equations do not require factorization. The joint distribution may contain strong correlations, and learning Alice’s setting and outcome may change Bob’s conditional distribution. The restriction is that Bob cannot detect Alice’s freely chosen setting from his local, unconditioned data.

The setting choice must be independent of the source preparation in this comparison. A difference caused by a source that was itself prepared differently for different settings does not establish an effect of Alice’s later local operation.

Trace preservation keeps the remote state unchanged

Let $\rho_{AB}$ be an arbitrary shared state. Alice chooses a local trace-preserving channel $\Lambda_A^{x}$ with Kraus operators $\{K_\mu^{x}\}$:

$$\begin{aligned} \Lambda_A^{x}(X)&=\sum_\mu K_\mu^{x}XK_\mu^{x\dagger},\\ \sum_\mu K_\mu^{x\dagger}K_\mu^{x}&=I_A. \end{aligned}$$

After Alice’s operation, the joint state is

$$\rho_{AB}' = \sum_\mu \left(K_\mu^{x}\otimes I_B\right) \rho_{AB} \left(K_\mu^{x\dagger}\otimes I_B\right).$$

To compare Bob’s states, test them against an arbitrary observable $O_B$. Cyclicity of the full trace and trace preservation give

$$\begin{aligned} \operatorname{Tr} \left[ (I_A\otimes O_B)\rho_{AB}' \right] &= \sum_\mu \operatorname{Tr} \left[ (K_\mu^{x\dagger}K_\mu^{x}\otimes O_B) \rho_{AB} \right]\\ &= \operatorname{Tr} \left[ (I_A\otimes O_B)\rho_{AB} \right]. \end{aligned}$$

Because this equality holds for every $O_B$, Bob’s reduced state is unchanged:

$$\rho_B' = \operatorname{Tr}_A\rho_{AB}' = \operatorname{Tr}_A\rho_{AB} = \rho_B.$$

The result holds for every initial state, including entangled states, and for every local trace-preserving quantum channel acting on that fixed shared state. Alice may rotate, dephase, measure and forget the outcome, attach an ancilla, or discard her subsystem. Without a system or classical record reaching Bob, his local statistics do not reveal $x$.

Instruments, branches, and postselection

A measurement is more finely described by an instrument $\{\mathcal E_{a|x}\}_a$. Bob’s unnormalized conditional state is

$$\sigma_B^{a|x} = \operatorname{Tr}_A \left[ \left(\mathcal E_{a|x}\otimes\mathcal I_B\right) (\rho_{AB}) \right].$$

Its trace is the branch probability,

$$p(a|x) = \operatorname{Tr}\sigma_B^{a|x},$$

and, when that probability is nonzero, the normalized conditional state is

$$\rho_B^{a|x} = \frac{\sigma_B^{a|x}}{p(a|x)}.$$

Different settings and outcomes can produce different conditional states. That is remote steering at the level of an ensemble decomposition. But if Bob does not know which outcome occurred, he must average:

$$\sum_a \sigma_B^{a|x} = \rho_B$$

for every $x$. Sorting joint data by Alice’s particular outcome record requires access to that record; delivering it is ordinary classical communication. A trace- nonincreasing branch by itself is therefore not a counterexample to no-signaling.

Two ensembles with the same local state

Suppose Alice and Bob share

$$|\Phi^+\rangle_{AB} = \frac{|00\rangle+|11\rangle}{\sqrt2}.$$

Bob’s reduced state is maximally mixed:

$$\rho_B = \operatorname{Tr}_A |\Phi^+\rangle\langle\Phi^+| = \frac{I_B}{2}.$$

If Alice measures in the $Z$ basis, Bob’s unnormalized conditional states are

$$\begin{aligned} \sigma_B^{0|Z}&=\frac12|0\rangle\langle0|,\\ \sigma_B^{1|Z}&=\frac12|1\rangle\langle1|. \end{aligned}$$

If she measures in the $X$ basis, where $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$, they are

$$\begin{aligned} \sigma_B^{+|X}&=\frac12|+\rangle\langle+|,\\ \sigma_B^{-|X}&=\frac12|-\rangle\langle-|. \end{aligned}$$

The ensembles differ, but their averages do not:

$$\begin{aligned} \sigma_B^{0|Z}+\sigma_B^{1|Z} &= \frac{I_B}{2}, \\ \sigma_B^{+|X}+\sigma_B^{-|X} &= \frac{I_B}{2}. \end{aligned}$$

No measurement performed only on Bob’s system can determine whether Alice chose $Z$ or $X$. Once Alice communicates her setting and outcome, Bob can sort his records into different conditional ensembles and observe the correlations.

For a shared Bell state, Alice’s $Z$ or $X$ measurement prepares different conditional states for Bob. Each outcome has probability $1/2$. Both ensembles average to the center of the Bloch ball, $I/2$, so Bob cannot identify her setting from his local data before a message arrives. The connecting diameters show possible Bloch vectors, not physical trajectories.

Why a perfect cloner would distinguish remote ensembles

The worked example also shows why cloning and signaling are historically linked. Imagine, contrary to quantum mechanics, that Bob could perfectly clone the pure state arriving in each run. He could make many replicas, estimate that run’s state with arbitrary precision, and decide whether Alice prepared a $Z$-basis or $X$-basis ensemble. Alice’s setting would then become readable before her classical message arrived.

The legitimate density operator $I/2$ has no observable tag identifying its ensemble decomposition. A universal cloner that acted differently on those decompositions would fail to be a linear, well-defined channel on density operators. This counterfactual explains a shared mathematical pressure: linearity protects both ensemble equivalence and no-signaling.

It does not make the two propositions logically identical. General no-signaling theories can have state spaces and correlations unlike quantum theory. Conversely, the no-cloning theorem by itself is not a complete statement of relativistic causality.

No-signaling is weaker than Bell locality

A Bell-local model with settings independent of its shared variable $\lambda$ has the factorized form

$$p(a,b|x,y) = \int d\lambda\, q(\lambda) p(a|x,\lambda) p(b|y,\lambda).$$

This condition implies no-signaling, but the converse is false. Quantum correlations from entangled states can violate Bell inequalities while their local marginals remain independent of the remote settings.

For the CHSH expression $S$, the familiar hierarchy is

$$\begin{aligned} |S|_{\mathrm{local}}&\leq2,\\ |S|_{\mathrm{quantum}}&\leq2\sqrt2,\\ |S|_{\mathrm{no\text{-}signaling}}&\leq4. \end{aligned}$$

Thus no-signaling alone permits hypothetical correlations stronger than quantum mechanics, such as Popescu–Rohrlich boxes. Quantum theory occupies a proper subset of the no-signaling correlation set (Popescu and Rohrlich, 1994). The CHSH Inequality develops the bounds, while Entanglement in Foundations separates entanglement, steering, Bell nonlocality, and causal language.

No-signaling and relativistic causality

The tensor-product theorem proved above is an operational statement about local subsystem operations. It should not be silently promoted into every notion of spacetime causality.

In a laboratory model, “local” means that Alice’s channel acts as $\Lambda_A\otimes\mathcal I_B$ on a declared tensor factor. The theorem then says that Bob’s reduced state is unchanged. It does not by itself assign spacetime positions, establish a finite propagation speed, or prove microcausality for quantum fields.

Relativistic quantum theory adds spacetime structure. Observables associated with spacelike separated regions commute in standard local quantum field theory, and admissible operations must respect the corresponding causal organization. General bipartite channels can also be classified by whether they signal in neither direction, one direction, or both; being non-signaling does not automatically mean that an operation can be implemented by separated parties using only local operations and shared entanglement (Beckman et al., 2001).

The Causality, Support, and Interpretation page treats finite-speed propagation and explains why instantaneous support of a nonrelativistic wavefunction is a different issue.

Consequences for communication protocols

Teleportation needs the classical message

Before Alice’s two classical bits arrive in qubit teleportation, Bob’s state averaged over her Bell-measurement outcomes is

$$\frac14 \sum_{a,b\in\{0,1\}} Z^aX^b |\psi\rangle\langle\psi| X^bZ^a = \frac{I}{2}.$$

It contains no locally accessible dependence on $|\psi\rangle$. Alice’s bits identify the Pauli frame and make the transfer usable. The input is consumed, so Quantum Teleportation respects both no-cloning and no-signaling.

Repeaters cannot copy and amplify

A classical repeater can read a symbol and regenerate many copies. A quantum repeater cannot measure an unknown qubit, clone it, and forward perfect instances. It instead distributes entanglement over elementary links, stores heralded successes, improves entanglement when allowed, and connects links by entanglement swapping. The success probabilities, pair consumption and waiting times determine its final output rate.

Quantum error correction encodes rather than clones

An encoding isometry can map

$$\alpha|0\rangle+\beta|1\rangle \longmapsto \alpha|0_L\rangle+\beta|1_L\rangle$$

across many physical systems. Those systems are not independent copies of the logical qubit. Their reduced states generally do not equal the input state, and syndrome measurements reveal error information without revealing the logical amplitudes. Why Quantum Error Correction Is Possible explains this distinction.

QKD needs more than no-cloning

No-cloning prevents a perfect “copy now, measure later” attack on arbitrary nonorthogonal signals. It does not bound Eve’s optimal approximate attack, authenticate the classical channel, estimate finite-sample leakage, model detector flaws, or produce a composable secrecy parameter. Those requirements enter a complete Quantum Key Distribution security proof. The BB84 protocol shows concretely how incompatible-basis sampling makes intercept–resend disturbance visible while keeping that diagnostic distinct from a proof against general attacks.

Entanglement does not add a hidden message channel

Pre-shared entanglement can change communication resources: it enables teleportation, superdense coding, entanglement-assisted capacities, and distributed protocols. It cannot transmit a freely chosen message on its own. Every usable protocol includes an actual quantum transmission, classical communication, or both. Communication with Quantum Systems makes those carriers explicit.

Interpreting an apparent violation

For a copying experiment, identify the promised input family, the information supplied to the device, and whether success is deterministic, approximate or heralded. A successful branch alone does not describe the device’s unconditional action.

For a signaling experiment, compare locally available data from all declared trials. Selecting coincidences using a remote detector record can change the sample in a setting-dependent way. Source drift, shared control lines, setting-dependent detection and finite-sample fluctuations can also create apparent differences. A measured marginal difference needs uncertainty and a model of which events entered the sample before it can be interpreted as a failure of the theorem’s assumptions.

Exercises

1. Basis copying is not universal cloning

Apply a controlled-NOT gate to $|+\rangle|0\rangle$. Compare the output with $|+\rangle|+\rangle$ and explain the difference.

Solution

Since $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$, linearity gives

$$\begin{aligned} \operatorname{CNOT}|+\rangle|0\rangle &=\frac{|00\rangle+|11\rangle}{\sqrt2}\\ &=|\Phi^+\rangle. \end{aligned}$$

Two copies would be

$$|+\rangle|+\rangle={}\tfrac12(|00\rangle+|01\rangle) +\tfrac12(|10\rangle+|11\rangle).$$

The first state is entangled and has maximally mixed one-qubit marginals. The second is a product state with pure $|+\rangle$ marginals. Controlled-NOT copies the computational-basis label, not an arbitrary input state.

2. Why pure-state broadcasting becomes cloning

Let $\omega_{AB}$ have a pure marginal $\operatorname{Tr}_B\omega_{AB}=|\psi\rangle\langle\psi|$. Show that $\omega_{AB}=|\psi\rangle\langle\psi|\otimes\tau_B$ for some $\tau_B$.

Solution

Choose a basis whose first vector is $|\psi\rangle$. The probability for the $A$ system to lie in the orthogonal subspace is zero:

$$\begin{aligned} Q&=(I-|\psi\rangle\langle\psi|)\otimes I_B,\\ \operatorname{Tr}(Q\omega_{AB})&=0. \end{aligned}$$

Positivity then implies that $\omega_{AB}$ has support only on the subspace $\operatorname{span}\{|\psi\rangle\}\otimes\mathcal H_B$. Every operator on that subspace has the form $|\psi\rangle\langle\psi|\otimes\tau_B$. If the $B$ marginal is also $|\psi\rangle\langle\psi|$, the output is the cloned product state.

3. No-signaling for a discarded measurement

Alice performs a projective measurement $\{P_a\}$ and discards the result. Prove directly that Bob’s reduced state remains $\rho_B$.

Solution

The nonselective joint state is

$$\rho_{AB}' = \sum_a (P_a\otimes I) \rho_{AB} (P_a\otimes I).$$

For every $O_B$,

$$\begin{aligned} \operatorname{Tr} \left[ (I\otimes O_B)\rho_{AB}' \right] &= \sum_a \operatorname{Tr} \left[ (P_a^2\otimes O_B)\rho_{AB} \right]\\ &= \operatorname{Tr} \left[ (I\otimes O_B)\rho_{AB} \right], \end{aligned}$$

where $\sum_aP_a=I$. Equality of all local expectation values implies $\rho_B'=\rho_B$.

4. Conditional change without signaling

For $|\Phi^+\rangle$, Alice measures $Z$ and obtains outcome $0$. What state does Bob assign before and after learning that outcome? Why is the change not a signal?

Solution

Without Alice’s outcome, Bob assigns

$$\begin{aligned} \rho_B&=\tfrac12|0\rangle\langle0|+\tfrac12|1\rangle\langle1|\\ &=\frac I2. \end{aligned}$$

Conditioned on outcome $0$, he assigns $|0\rangle\langle0|$. The conditional state changes because the classical record selects a subensemble. Alice cannot choose her random outcome to encode a message. If Bob already knows she measured $Z$, his own $Z$ measurement can reveal the correlated outcome, but its random distribution carries no controllable signal. His unconditioned state is unchanged.

5. A no-signaling correlation beyond quantum theory

Let binary outcomes satisfy $a\oplus b=xy$ with probability one, and assign probability $1/2$ to each of the two outcome pairs satisfying that equation. Show that the distribution is no-signaling and reaches the algebraic CHSH value $4$.

Solution

For every fixed $x,y$, exactly one allowed pair has $a=0$ and one has $a=1$, each with probability $1/2$. Thus $p(a|x,y)=1/2$, independent of $y$. Likewise $p(b|x,y)=1/2$, independent of $x$, so the distribution is no-signaling.

Writing outcomes as signs, the relation gives correlations

$$E_{00} = E_{01} = E_{10} = 1, \qquad E_{11} = -1.$$

Hence

$$S = E_{00}+E_{01}+E_{10}-E_{11} = 4.$$

This exceeds the quantum Tsirelson bound $2\sqrt2$. No-signaling is therefore strictly weaker than belonging to the quantum correlation set.

6. Error correction versus copying

The repetition-code map sends $\alpha|0\rangle+\beta|1\rangle$ to $\alpha|000\rangle+\beta|111\rangle$. Show from one-qubit reduced states that the three output qubits are not three copies of the input.

Solution

Tracing out any two output qubits removes the coherence between $|000\rangle$ and $|111\rangle$:

$$\rho_{\mathrm{one}} = |\alpha|^2|0\rangle\langle0| +|\beta|^2|1\rangle\langle1|.$$

The original pure-state density operator is

$$\rho_{\mathrm{in}}={}|\alpha|^2|0\rangle\langle0| +|\beta|^2|1\rangle\langle1| +\alpha\beta^*|0\rangle\langle1| +\alpha^*\beta|1\rangle\langle0|.$$

For a genuine superposition, the reduced output lacks the off-diagonal terms and is not equal to $\rho_{\mathrm{in}}$. The logical amplitudes reside in nonlocal correlations of the codeword, not in independent physical copies.

References

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