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TL/DR: Can unitary operators:

$$U_a=e^{-it(H_{a1}+H_{a2}+\cdots)}$$

and

$$U_b=e^{-it(H_{b1}+H_{b2}+\cdots)}$$

commute, even though $[H_{aj},H_{ak}]\ne 0$ and $[H_{bj},H_{bk}]\ne 0$ for all $j,k$?


When stated as above I'm pretty sure the answer is yes. But, for background I'm motivated by some confusion on two separate problems.

Initially Kane and additionally Kane, Sharif, and Silverberg have a couple of schemes for quantum money based on modular forms and quaternion algebras. Quoting from one of the abstracts:

... In order to instantiate this protocol, one needs to find a cryptographically complicated system of computable, commuting, unitary operators...

Separately, Irani and Jiang, among many others, have explored the computational complexity of finding the ground state of a system of commuting local Hamiltonians. Quoting from their abstract:

We study the complexity of local Hamiltonians in which the terms pairwise commute. Commuting local Hamiltonians (CLHs) provide a way to study the role of non-commutativity in the complexity of quantum systems... (Emphasis added).

Given my general ignorance I originally thought that quantum money based on commuting unitaries was somehow related to the commuting Hamiltonian problem, but now I don't think that's so?

That is, I think in the commuting local Hamiltonian problem, the Hamiltonian simulation is classically efficient as there's no need to trotterize different local terms (because, by definition, we always have $[H_j,H_k]=0$ for all terms). While, the quantum money schemes actually do involve Hamiltonian simulation with trotterization, etc. based on random walks generated by Hecke operators or Brandt operators (whatever those are), but in the above quantum money schemes although $[H_{aj},H_{ak}]\ne 0$ and $[H_{bj},H_{bk}]\ne 0$, we can nonetheless have $[U_a,U_b]=0$.

Is this understanding correct?

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    $\begingroup$ I'm not familiar with quantum money schemes, but the answer to your TL;DR question is trivially "yes". It's trivial because an example would be to set $H_{aj} = H_{bj}$ for all $j$, so that $U_a = U_b$. In general, $[\sum_j H_{aj}, \sum_j H_{bj}] = 0$ would imply $[U_a(t), U_b(t)] = 0 \forall t$. $\endgroup$ Commented May 3 at 2:22
  • $\begingroup$ Yeah I’d buy that… $\endgroup$ Commented May 3 at 2:27
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    $\begingroup$ @Solarflare0 if you post that as an answer I’d likely accept it, although I’m also interested when $U_a\ne U_b$, which I think is just as trivial $\endgroup$ Commented May 4 at 1:34

1 Answer 1

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Set $H_a = \sum_j H_{aj}$ and $H_b = \sum_j H_{bj}$. Then, $U_a = e^{-iH_a t}$ and $U_b = e^{-iH_bt}$. Then, $[H_a, H_b] = 0 \implies [U_a, U_b] = 0$, which can be proved by the following Taylor expansion argument: \begin{align} [U_a, U_b] &= \left[\sum_{k=0}^{\infty} \frac{(-iH_at)^k}{k!},\sum_{l=0}^{\infty} \frac{(-iH_bt)^l}{l!}\right]\\ &= \sum_{k=0}^{\infty} \sum_{l=0}^{\infty} \frac{(-it)^k(-it)^l}{k! l!}[H_a^k, H_b^l]\\ &= 0. \end{align} Importantly, this is true regardless of the commutation relations within the $\{H_{aj}\}$ set and within the $\{H_{bj}\}$ set.

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