In chapter 9 of Scott Aaronson's book "Quantum Computing Since Democritus", he make interesting but peculiar claims relating the no-cloning theorem and the Heisenberg Uncertainty Principle (HUP). Here is his statement:

"On the one hand, if you could measure all properties of a quantum state to unlimited accuracy, then you could produce arbitrarily-accurate clones. On the other hand, if you could copy a state $|\psi\rangle$ an unlimited number of times, then you could learn all its properties to arbitrary accuracy."

In his interpretation, the first sentence proves that no-cloning theorem implies HUP, while the second sentence proves that HUP implies no-cloning theorem. So he is suggesting that the no-cloning theorem is EQUIVALENT to HUP.

However, I cannot feel comfortable about this statement. There are two questions that I have in mind:

  1. To my understanding, HUP is the claim that a quantum state cannot be a simultaneous eigenstate of momentum and position operators. Therefore, suppose I can clone two copies of a quantum state, I could collapse copy 1 to a position eigenstate, and copy 2 to a momentum eigenstate, and never collapse either copy into a simultaneous eigenstate, therefore never violating the HUP. So is Aaronson wrong on his arguments here?

  2. I am a little unsure about the assumptions that go into the proofs of no-cloning theorem and HUP. To my understanding, No-Cloning follows from Linearity and Unitarity of quantum operators, while HUP follows from unitarity and non-commutativity of quantum operators. My question, which might be stupid, is this: Is it possible to derive a generalized HUP from a nonlinear modification of quantum theory (Invent some nonlinear observables that is neither position nor momentum)? If so, I believe that no-cloning theorem would really have nothing to do with HUP whatsoever.

Any comments or insights about the connection or implication of no-cloning and HUP are also welcomed!


3 Answers 3


If you go through the proof of the no-cloning theorem (see e.g. wikipedia) you will notice that only the following properties are used:

  • Unitarity of the evolution operator
  • Hilbert space axioms for a composite state $|\phi \rangle | 0 \rangle$ (which imply the Cauchy-Schwartz inequality)

The no-cloning theorem than says that no unitary operator exists that does the transformation $|\phi \rangle | 0 \rangle \rightarrow |\phi \rangle | \phi \rangle$ for arbitrary $|\phi \rangle$.

The Heisenberg uncertainty principle on the other hand is a statement about how the uncertainties for two different operators on the Hilbert space are related, namely that they can not simultaneously be small for non-commuting operators.

But for the no-cloning theorem you didn't even need observables, it is a property of the Hilbert space and the unitary time evolution in Quantum Mechanics alone.

Summary: They are completely different things and not related.

  • $\begingroup$ Hi Numrok, thanks for the answer. I have read through the proofs before. I lean toward your conclusion but not the way you arrived at it. Just because HUP and No-Cloning Theorem involve different mathematical objects, we cannot conclude that they are not equivalent. If they are derivable from each other, then they are equivalent! My question is, do HUP and No-Cloning Theorem make the same assumption about quantum mechanical axioms? Or is it that No-Cloning Theorem requires linearity of all quantum operators while HUP doesn't? $\endgroup$ Jul 8, 2016 at 15:57
  • $\begingroup$ @ZhengyanShi In Quantum Mechanics we define certain operators on the Hilbert space, which are the observables. We also have the Schrödinger equation which is tells you how things evolve in time and it is unitary. So in QM we have both no-cloning and HUP. But if you want to make a theory with only commuting observables you don't get HUP (you can still have Schrödinger equation) and if you want to make a theory with no unitary time evolution (which doesn't really make sense, but why not), you can do that too. So how would one imply the other? I think my argument in the answer is complete $\endgroup$ Jul 8, 2016 at 16:09
  • $\begingroup$ How about a theory with unitary evolution operators and nonlinear, non-commuting observables? Do we still have HUP in that case? $\endgroup$ Jul 8, 2016 at 16:15
  • $\begingroup$ You probably still have HUP (see e.g. physics.stackexchange.com/questions/247394/… for HUP on unusual operators). I don't see what this has to do with the question though. My point is that you neither have $HUP \Rightarrow No-cloning$ nor $No-Cloning \Rightarrow HUP$, which I thought was the question and which I demonstrated in my answer why it is the case. Of course they both follow from the axioms of Quantum mechanics, but that does not make one imply the other. You need to invoke the other axioms. $\endgroup$ Jul 9, 2016 at 18:33
  • 1
    $\begingroup$ @ Zhengyan Shi About the conclusion of Numrok to say they are not related, I may hold my opinion. From a geometrical point of view, the uncertainty relationship is determined by the Kahler structure of the Hilbert space, where the fibre bundle is determined by the unitary operation assumption. So they ARE REALLY closely related. If we eliminate the linear unitary operation assumption, then the geometrical structure will be different and then the uncertainty relation will not hold any more. This is exactly what udrv showed above. May this paper can help (arxiv.org/abs/1503.00238). $\endgroup$
    – XXDD
    Jul 19, 2016 at 13:51

Aaronson's claims are true, but your statement about what he means is not correct.

If cloning were possible, the HUP would still exist but would pose a non-absolute limit on how much you could learn about a single copy of an unknown quantum state- that is, any limit on how much you learned about a state could be attributed to not doing a very good measurement.

On the other hand, with no-cloning the HUP provides a fundamental limit on how much you can learn about a single copy of an unknown quantum state.

So they are certainly not identical. Maybe a better way to think about them is that they act in concert to limit the amount of information one may extract from a general state.

There may be a deeper way of looking at the connection between them. What Numrok says about the structural difference between the two statements is true, but it is also true that it is difficult to modify just one part of quantum mechanics. There is another nice paper by Aaronson that discusses this point (1). I think it is possible that there is a deeper reason that the structure of quantum theories fits together so neatly, which we do not fully understand as of yet.

  • 1
    $\begingroup$ +1 on this one. My answer is possibly is possibly only scratching the surface. $\endgroup$ Jul 9, 2016 at 18:34
  • $\begingroup$ Thanks for the paper, the abstract seems very interesting! "Act in concert to limit the amount of information" is a cool perspective, although I don't quite understand it yet... Could you clarify your first point a bit more? Why is it that "if cloning were possible, the HUP would... pose a non-absolute limit on how much you could learn about a single copy of an unknown quantum state" ? Are you just saying that you produce arbitrarily many copies, measure them all, and map out the original wavefunctions to arbitrary accuracy? $\endgroup$ Jul 10, 2016 at 3:02
  • $\begingroup$ @ZhengyanShi "Are you just saying that you produce arbitrarily many copies, measure them all, and map out the original wavefunctions to arbitrary accuracy?" Yes, that's all that I (and presumably Aaronson) mean. You can already do this if you can prepare many copies of a given quantum state, of course, but cloning would allow you to do this for any state. $\endgroup$
    – Rococo
    Jul 10, 2016 at 17:55
  • 1
    $\begingroup$ Ok. I think things are much more clear now. Thank you!! $\endgroup$ Jul 11, 2016 at 2:53
  • 1
    $\begingroup$ I checked Scott Aaronson's book Quantum Computing Since Democritus, p.126. He did say "I claim that the No-Cloning Theorem basically implies the Uncertainty Principle and vice versa." $\endgroup$
    – Hans
    Mar 31, 2022 at 20:23

Wrote this to address the 2nd question in the OP: "Is it possible to derive a generalized HUP from a nonlinear modification of quantum theory (Invent some nonlinear observables that is neither position nor momentum)?", following the discussion on the topic in comments. Although it is not a clear-cut answer, perhaps it may help.

But first a couple of remarks adding to the other answers:

  1. It is no simple historical coincidence that the HUP was discovered first as essential to the initial development of quantum theory, but not the no-cloning theorem.
  2. There may be many corollaries to the no-cloning theorem, but the main one, which brought it to light in the first place, is that it forbids a faster-than-light communication loophole through an exploitation of entanglement. So regardless of whether the existence of cloning would allow or not some sort of by-pass on the limits set by the HUP, it would definitely raise the much more troubling issue of open conflict with relativity. HUP does nothing of the sort, to the contrary. Hence again, no, HUP cannot be equivalent to the no-cloning theorem.

On the issue of nonlinear observables:

We already know of some useful nonlinear observables, see the entropy $-k_B \text{Tr}(\rho \ln\rho)$ and related entropic UPs, hence the question is appropriate. The real problem is that the issue is quite complex. Even if the Hilbert space structure is left in place, true nonlinear observables would modify the theory drastically and the usual HUP definitely does not apply in all nonlinear cases.

This is because in general nonlinear operators can no longer be characterized in terms of action on bases or even in terms of matrix elements. The simplest example is one that appears precisely in the algebraic proof of the HUP. If $A$ is an arbitrary linear and hermitic operator, define a nonlinear application ${\mathcal \not A}$ through $$ {\mathcal \not A}|\psi\rangle = A|\psi\rangle - \frac{\langle \psi |A|\psi\rangle}{\langle \psi |\psi\rangle} |\psi\rangle $$ Application ${\mathcal \not A}$ is homogeneous, $$ {\mathcal \not A}|a\psi\rangle = a {\mathcal \not A} |\psi\rangle $$ but not linear, $$ {\mathcal \not A}|(a\psi + b\phi)\rangle \neq a {\mathcal \not A} |\psi\rangle + b {\mathcal \not A} |\phi\rangle $$ and therefore has some unusual properties:

  • Its action on any eigenvector $|\lambda\rangle$ of $A$ vanishes identically: $$ A|\lambda\rangle = \lambda |\lambda\rangle \;\;\;\Rightarrow \;\;\; {\mathcal \not A}|\lambda\rangle = A|\lambda\rangle - \frac{\langle \lambda |A|\lambda\rangle}{\langle \lambda|\lambda\rangle} |\lambda\rangle = 0 $$ An entire basis set is in its kernel, but ${\mathcal \not A}$ is not a null application!

  • Moreover, its average also vanishes on any $|\psi\rangle$, $$ \langle \psi | {\mathcal \not A}|\psi\rangle = \langle \psi |A|\psi\rangle - \frac{\langle \psi |A|\psi\rangle}{\langle \psi |\psi\rangle} \langle \psi |\psi\rangle = 0 $$

In other words, many of the tools used liberally with linear observables fly out the window even for this modest example.

In particular, the action of observables, and operators at large, can no longer be defined in terms of action on a basis set, but has to be defined for each state vector individually. Unfortunately when basis sets become insufficient for characterization, so do matrix representations. And once this happens, the concepts of hermitian conjugate, self-adjoint observable, and eigenbasis, all loose their celebrated significance.

Suppose though that observable averages would still be given by real diagonal matrix elements $\langle \psi |{\mathcal O}|\psi\rangle \in {\mathbb R}$, a condition that already sets boundaries on the set of acceptable nonlinear applications $\mathcal O$. Then here is a simple example of nonlinear "observables" $A$, $B$ that break the HUP as applied in the usual form $$ \langle \psi |(\Delta A)^2|\psi\rangle \langle \psi |(\Delta B)^2|\psi\rangle \ge |\langle \psi |\frac{1}{2i}[A,B]|\psi\rangle|^2 $$ Let $A$, $B$ be such that $\langle \psi |A|\psi\rangle \in {\mathbb R}$, $\langle \psi |B|\psi\rangle \in {\mathbb R}$ for any $|\psi\rangle$, and in addition such that for some given $|\psi_0\rangle$ $$ A|\psi_0\rangle = 0 \;\;\; \Rightarrow \;\;\; \langle \psi_0 |A|\psi_0 \rangle = 0 $$ $$ B|\psi_0\rangle = a |\psi_0\rangle \neq 0 \;\;\; \Rightarrow \;\;\; \langle \psi_0 |B|\psi_0 \rangle = a = a^* \neq 0 $$ and for any $|\psi_\bot\rangle$, $\langle\psi_\bot|\psi_0\rangle = 0$, $$ A|\psi_\bot \rangle = |\psi_0\rangle \;\;\; \Rightarrow \;\;\; \langle \psi_\bot |A|\psi_\bot \rangle = 0 \\ B|\psi_\bot \rangle = b |\psi_\bot \rangle + c |\psi_0\rangle \;\;\; \Rightarrow \;\;\; \langle \psi_\bot |B|\psi_\bot \rangle = b = b^* \neq 0 \\ A|b\psi_\bot + c\psi_0\rangle = |\psi_\bot\rangle \;\;\; \Rightarrow \;\;\; \langle b\psi_\bot + c \psi_0 |A| b\psi_\bot + c\psi_0 \rangle = b \langle \psi_\bot | \psi_\bot \rangle $$ Then we have that for any $|\psi_\bot\rangle$, $\langle\psi_\bot|\psi_\bot\rangle = 1$, $\langle\psi_\bot|\psi_0\rangle = 0$, $$ \langle \psi_\bot |(\Delta A)^2|\psi_\bot \rangle = \langle\psi_\bot |A^2|\psi_\bot \rangle - \langle\psi_\bot |A|\psi_\bot \rangle^2 = 0 $$ but $$ |\langle \psi |[A,B]|\psi\rangle|^2 = |\langle\psi_\bot |AB|\psi_\bot \rangle - \langle\psi_\bot |BA|\psi_\bot \rangle|^2 = |\langle\psi_\bot |A(b\psi_\bot + c\psi_0)\rangle - \langle\psi_\bot |B|\psi_0 \rangle|^2 = $$ $$ = |\langle\psi_\bot |A(b\psi_\bot + c\psi_0)\rangle - a \langle\psi_\bot |\psi_0 \rangle|^2 = |\langle\psi_\bot |A(b\psi_\bot + c\psi_0)\rangle|^2 = |\langle \psi_\bot |\psi_\bot \rangle|^2 =1 $$ and so $$ \langle \psi |(\Delta A)^2|\psi\rangle \langle \psi |(\Delta B)^2|\psi\rangle 0 \le |\langle \psi |\frac{1}{2i}[A,B]|\psi\rangle|^2 = \frac{1}{4} \;\;\;!! $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.