I have variously heard people describe the no-cloning theorem as an essential feature of "quantum physics", akin to saying "we cannot copy arbitrary quantum information to arbitrary precision". However, this question is about the interpretation of the quantum result only insofar as this sheds light on the classical result presented in the course of the question.

The basic formal statement of the theorem is that given a Hilbert space $H$, there is no unitary operator $U : H \otimes H \to H \otimes H$ such that $$ U (\lvert a \rangle \otimes \lvert b\rangle) = \lvert a\rangle \otimes \lvert a\rangle$$ for all $\lvert a\rangle \in H$ and a "blank" state $\lvert b\rangle$. The question is, is anything about this result surprising or "quantum" from the viewpoint of classical mechanics?

Here's an argument that "no-cloning" also holds in classical mechanics, taken almost verbatim from "There's no cloning in symplectic mechanics" by Fenyes:

Let $(M,\omega)$ be a symplectic manifold, the $\omega$ is the symplectic form that encodes what we physicists call the Poisson bracket by $\omega(X_f,X_g) = \{f,g\}$ where $X_f$ is the vector field defined by $\mathrm{d} f = \omega(X_f,-)$. Then all physical motions on $M$ are symplectomorphisms, i.e. functions $M\to M$ that preserve $\omega$, because they are integral flows of the Hamiltonian vector field $X_H$ which is a symplectic vector field by construction.

The combined phase space of two systems $(M,\omega),(M',\omega')$ is $(M\times M',\omega + \omega')$, where $\times$ is the Cartesian product of manifolds. Now, the classical analogue to the no-cloning theorem would clearly be the statement that there is no symplectomorphism $\phi : M\times M\to M\times M$ such that $$ \phi(a,b) = \phi(a,a)$$ for all $a\in M$ and a blank state $b\in M$. And, indeed, this is true:

Let $u,v\in T_{(b,b)} (M\times M)$ be tangent vectors at $(b,b)$. Since $\phi(x,b) = (x,x)$ by assumption, curves $(\gamma(t),b)$ starting at $(b,b)$ get mapped to curves $(\gamma(t),\gamma(t))$ starting at $(b,b)$, and so $\mathrm{d} \phi_{(b,b)} (w,0) = (w,w)$ for all $w\in T_{(b,b)}(M\times M)$. Therefore, \begin{align} & (\omega + \omega)((u,0),(v,0)) = (\omega + \omega)((u,u),(v,v)) \\ \implies & \omega(u,v) + \omega(0,0) = \omega(u,v) + \omega(u,v) \\ \implies & \omega(u,v) = 0\end{align} which is a contradiction because symplectic forms are non-degenerate by definition. Therefore, no classical Hamiltonian cloning map exists.

So, what does this result actually show? Are the assumptions of the no-cloning theorem silly and the desired cloning map does not actually reflect what we mean by being able to copy arbitrary information in either case? Is there a subtle difference between the classical and the quantum setting which makes the assumptions silly in the classical, but not in the quantum setting? If the assumptions are not silly, then what is the significance of the classical result?

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    $\begingroup$ Hi @ACuriousMind, have you considered that people simply use quantum mechanics as the most fundamental tool available to say that in the nature (in the real world) there could be no duplication of information? I mean - yes, as you've demonstrated, this feature isn't unique to QM, but once you want to demonstrate that cloning is forbidden, doesn't it make sense to use QM to do so over anything else? I know that it probably doesn't help, just want to share my point of view. $\endgroup$ Commented Dec 5, 2016 at 14:24
  • $\begingroup$ @SolenodonParadoxus Well, the point is that people seem to think that we can copy "classical" information, which seems to contradict the theorem in the question, so what exactly does this version of "no-cloning" mean? That's what I want to know when I ask whether the cloning map does not actually reflect what we mean by being able to copy arbitrary information. $\endgroup$
    – ACuriousMind
    Commented Dec 5, 2016 at 18:06
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    $\begingroup$ Well, its mostly used in information theory. Classical bits can be cloned and quantum qubits can't. The point is that classical circuits are in constant interaction with the outside world, thus Hamiltonian description is inapplicable and cloning is possible. Quantum circuits are isolated, because we want to avoid decoherence. $\endgroup$ Commented Dec 6, 2016 at 7:25
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    $\begingroup$ @SolenodonParadoxus So, you're saying the significance of the classical theorem is "classical cloning requires dissipation", but no one ever cared about that because dissipation is not a problem for classical information storage? That would then mean that "no-cloning vs. cloning" is not quantum vs. classical, but closed systems vs. open systems. $\endgroup$
    – ACuriousMind
    Commented Dec 6, 2016 at 15:54
  • $\begingroup$ Pretty much, yes. $\endgroup$ Commented Dec 6, 2016 at 16:08

3 Answers 3


The answer seems to be given in "Limitations on Cloning in Classical Mechanics", also by Fenyes.

Let me summarize it: the point is that the notion of copying that is usually seen (such as in the OP) is actually not what we mean by copying. It only allows for the object that we want to copy, and the new object. In that case it is indeed impossible to clone. But the linked article shows that if we also include a copier, then it becomes possible to clone in the classical case (but remains impossible in the quantum case).

(Note: he actually doesn't prove that it's always possible classically, but at least he gives some explicit examples where it is.)


You're right that no-cloning isn't inherently quantum. The classical analogue of quantum cloning is "probability cloning".

Here's the probability cloning challenge: create a machine that takes one coin prepared to be heads up with probability $p$, but outputs two coins that are heads up with probability $p$. The catch is that you don't know $p$ (the machine should work for all $p$) and the two output coins are supposed to be independent. The statistics of the output should satisfy $P(HH) = p^2$, $P(HT) = P(TH) = p (1-p)$, and $P(TT) = (1-p)^2$.

I hope you see immediately why this machine can't possibly exist (or, at least, must perform very poorly). The coin's state doesn't tell you enough about the probability that generated it. One sample just isn't enough information to get a good idea of the underlying distribution. So the best you can do is make the two coins match the input, maybe throw in a bit of noise, and hope the statistical tests miss the obvious.

Another way to think about it is that there's no stochastic matrix $S$ that satisfies:

$$S \begin{bmatrix} p \\ 1-p \end{bmatrix} = \begin{bmatrix} p \\ 1-p \end{bmatrix}^{\otimes 2} =\begin{bmatrix} p^2 \\ p (1-p) \\ p(1-p) \\ (1-p)^2 \end{bmatrix}$$

This is obvious because the output requires components of the input to be multiplied and squared. You can't go from $p$ to $p^2$ using a linear operation.

See also: the 2002 paper 'Classical No-Cloning Theorem' by A. Daffertshofer et al., which is cited by the paper you cited.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Commented Dec 9, 2016 at 2:15

Comments to the post (v1):

  1. So there are at least 4 related theorems:

    1. The quantum No-cloning theorem for pure states by Wootters, Zurek & Dieks.

    2. The quantum No-broadcast theorem for density operators.

    3. The classical symplectic No-cloning theorem by Refs. 1-2. This is what OP asks about.

    4. The classical probabilistic No-cloning theorem using Kullback-Leibler divergence by Ref. 3. We will have nothing more to say about item 4.

  2. Note that both Thm. 1 & Thm. 3 have elementary one-line proofs. It might therefore not be very constructive/productive/useful/meaningful to try to compare them. [E.g. such a 'detail' of whether we consider (i) a Hilbert space or (ii) a projective Hilbert space/ray space (as we do in quantum mechanics) is already more subtle, cf. e.g. Wigner's theorem & its non-trivial proof.]

  3. In geometric quantization, under the correspondence principle between classical & quantum mechanics, the Hilbert space ${\cal H}$ is identified with a Lagrangian submanifold via polarization. Therefore we cannot use the Hilbert space ${\cal H}$ to probe non-zero directions of the symplectic structure. Hence we may argue that the cloning obstruction in Thm. 3 does not reflect the cloning obstruction in Thm. 1.

  4. Rather Thm. 1 is of a different nature, possibly closer related to Thm. 2. If time permits, we may elaborate further on this point in the future.


  1. A. Fenyes, There’s no cloning in symplectic mechanics, 2010.

  2. A. Fenyes, J. Math. Phys. 53 (2012) 012902, http://arxiv.org/abs/1010.6103.

  3. A. Daffertshofer, A. R. Plastino, & A. Plastino, Phys. Rev. Lett. 88 (2002) 210601.

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    $\begingroup$ Indeed, upon reading Ref. 2 as also pointed out more explicitly by RubenVerresen's answer, it turns out the classical obstruction vanishes upon including the state of the copying machine, which the quantum obstruction doesn't. The superficial similarity of the statements and their proofs is misleading as you suspect in point 2. $\endgroup$
    – ACuriousMind
    Commented Dec 7, 2016 at 1:01
  • $\begingroup$ $\uparrow$ Point 2? $\endgroup$
    – Qmechanic
    Commented Dec 7, 2016 at 7:49

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