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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!

Zhengyan Shi
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3 Answers3

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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.

Wolpertinger
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  • 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? – Zhengyan Shi Jul 08 '16 at 15:57
  • @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 – Wolpertinger Jul 08 '16 at 16:09
  • How about a theory with unitary evolution operators and nonlinear, non-commuting observables? Do we still have HUP in that case? – Zhengyan Shi Jul 08 '16 at 16:15
  • You probably still have HUP (see e.g. http://physics.stackexchange.com/questions/247394/heisenbergs-uncertainty-principle-for-mean-deviation 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. – Wolpertinger Jul 09 '16 at 18:33
  • Thanks for the link! But what had me confused is that: 1, You claimed that neither implication is correct. 2, Rococo said that Aaronson's claims are right. But Aaronson's claim is precisely the equivalence between the two things (Quoting himself: "let me show you how the No-Cloning Theorem implies a rough-and-ready, information theoretic version of the Uncertainty Principle, and vice versa." p.126 of the book I mentioned). Although he did not explicitly say HUP, he means some kind of HUP. That's why I asked about alternative HUPs that do not assume standard hermitian operators. – Zhengyan Shi Jul 10 '16 at 02:56
  • I'm not quite sure how you can make a physical theory without hermitian operators in the first place. They would potentially have imaginary eigenvalues... – Wolpertinger Jul 10 '16 at 09:30
  • Yes you need Hermitian operators to ensure real eigenvalues! Sorry for the confusion. But you can have Hermitian operators that are nonlinear (therefore not standard). Is there a generalization of HUP to nonlinear Hermitian operators? – Zhengyan Shi Jul 10 '16 at 14:25
  • @ZhengyanShi does udrv's post below answer your question? It contains a lot of detail on non-linear Hermitian operators, that I did not know about. If it doesn't I really don't know what you're on about. – Wolpertinger Jul 10 '16 at 22:46
  • udrv's post does answer my question! To appreciate it fully is a different story haha. Should go read carefully. – Zhengyan Shi Jul 11 '16 at 03:07
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    @ 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 (https://arxiv.org/abs/1503.00238). – XXDD Jul 19 '16 at 13:51
  • @X.Dong fair point, i didnt know that!! thsnks for the paper :) – Wolpertinger Jul 19 '16 at 13:55
  • @XXDD: I do not quite understand your point. Why does the non-commutativity of some two self-adjoint operators have to do with the unitarity of some other operators? If there is such a implicational relationship, say in the paper you referenced, could you please point to the exactly theorem that states this? – Hans Mar 30 '22 at 05:11
  • @Wolpertinger: It seems you agree with XXDD. If so, could you please read my comment above and tell me your opinion? – Hans Mar 30 '22 at 05:15
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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.

Rococo
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    +1 on this one. My answer is possibly is possibly only scratching the surface. – Wolpertinger Jul 09 '16 at 18:34
  • 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? – Zhengyan Shi Jul 10 '16 at 03:02
  • @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. – Rococo Jul 10 '16 at 17:55
  • I think the key thing to keep in mind here is that Aaronson is specifically interested in what it takes to get all of the information about the quantum state. If the HUP did not exist, and all measurements commuted, you could do this by repeated measurements of different observables, while if cloning were possible you could do it in the way you describe above. However, this is only one consequence of HUP, for example. and other consequences (like that a particle cannot be in a simultaneous eigenstate of position and momentum) are certainly not related to no-cloning in any obvious way. – Rococo Jul 10 '16 at 18:06
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    Ok. I think things are much more clear now. Thank you!! – Zhengyan Shi Jul 11 '16 at 02:53
  • I do not see how "Aaronson's claims are true". The uncertainty principle describes the non-commutativity of some two self-adjoint operators, while the no-cloning theorem depicts the behavior of a tensor vectors under a unitary operator. They are independent of each other. If you still think they are equivalent, could you please write out a mathematical proof of the proposition? – Hans Mar 31 '22 at 06:56
  • Hi @Hans: in the first sentence of my answer, what I am saying is that neither I nor Aaronson (as far as I can tell) are claiming that the two are equivalent. Only the OP uses this word, and in my answer I claim that it is an incorrect interpretation of the point that Aaronson is trying to make, and then I go on to try to clarify what he was actually after. Hope that helps. – Rococo Mar 31 '22 at 13:58
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    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." – Hans Mar 31 '22 at 20:23
  • @Hans- ah, indeed he does. Forgive me; it's been six years since I answered this question, after all. Upon rereading the relevant passage, I interpret his statement as being about the qualitative content of the no-cloning theorem: the statement that one cannot deterministically duplicate a single copy of an unknown quantum state, rather than the mathematical statement about unitary evolution. – Rococo Apr 01 '22 at 01:16
  • To wit: if the HUP could be violated, this qualitative statement would not be true, as one could copy a quantum state by measuring it enough to fully characterize it. Since this process would involve measurements it would presumably be non-unitary and evade this particular mathematical formulation of the no-cloning theorem. That said, only Aaronson could give you a definitive answer here as to his intent. – Rococo Apr 01 '22 at 01:16
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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 $$ \langle \psi |(\Delta A)^2|\psi\rangle \langle \psi |(\Delta B)^2|\psi\rangle 0 \le |\langle \psi |\frac{1}{2i}[A,B]|\psi\rangle|^2 = \frac{1}{4} \;\;\;!! $$

udrv
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