No-Cloning and Uncertainty: Connections or Misconception
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.
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.
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:
- 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.
- 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 $$ \langle \psi |(\Delta A)^2|\psi\rangle \langle \psi |(\Delta B)^2|\psi\rangle 0 \le |\langle \psi |\frac{1}{2i}[A,B]|\psi\rangle|^2 = \frac{1}{4} \;\;\;!! $$