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In [1], the problem of simulating a Hamiltonian using repeated applications of a different set of Hamiltonians is discussed.

In particular, let $A$ and $B$ be a pair of Hermitian operators, and let $\mathcal L$ be the algebra generated from $A, B$ through repeated commutation $^{\mathbf{(\dagger)}}$.

The author then asks (first paragraph of third page) what is $\mathcal L$ for an arbitrary pair of observables $A$ and $B$, and argues that $\mathcal L$ is the space of all Hermitian matrices, unless (quoting from the paper) both $e^{iA t}$ and $e^{iB t}$ lie in an $n$-dimensional unitary representation of some Lie group other than $U(n)$.

I'm not too familiar with the theory of Lie algebras, so this statement is quite cryptic for me. How can this be shown more explicitly? Equivalently, is there a more direct way to show this fact?


$(\dagger)$: More explicitly, this is the vector space spanned by $A, B, i[A,B], [A,[A,B]], ...$

[1] Lloyd 1995, Almost Any Quantum Logic Gate is Universal, Link to PRL.

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    $\begingroup$ For anyone who likes Lie algebras more: You can just take those two symbols A, B and make the free Lie algebra $Free_2$. So no relations on the $\dagger$ besides those that make sure it is still a Lie algebra. Then let $\rho$ be the representation that goes down to actual matrices (so $\rho (A)$ is what you're calling A above). From here there are some very powerful theorems that go under the name of Kashiwara-Vergne. These are useful in understanding that long Baker-Campbell-Hausdorff formula (stronger formula than Trotter). $\endgroup$
    – AHusain
    Jun 12, 2018 at 21:51
  • $\begingroup$ @AHusain that sounds like something worthy of being an answer! $\endgroup$
    – glS
    Jun 13, 2018 at 9:22
  • $\begingroup$ Isn't the statement you quote basically tautological? (I mean: Either they generate the full algebra, or a subalgebra.) $\endgroup$ Dec 29, 2021 at 15:32
  • $\begingroup$ @NorbertSchuch I think the nontriviality is in the relation between algebra and group. The statement boils down to, I believe, asking whether given $\mathfrak g\equiv\langle A,B\rangle$, algebra generated by $A,B$, the Lie group elements $e^{tA},e^{tB}$ generate the full $G$. Or at least the version of this statement specialised to $G=\mathbf U(N)$. In other words, assuming compactness of $G$, this should amount to asking whether $e^{tC}\in G$ for arbitrary $C\in\mathfrak g$ can be written as product of $e^{tA}$ and $e^{tB}$. Which looks like a variation of BCH's statement $\endgroup$
    – glS
    Dec 30, 2021 at 8:49

2 Answers 2

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I'm not too familiar with the theory of Lie algebras, so this statement is quite cryptic for me. How can this be shown more explicitly? Equivalently, is there a more direct way to show this fact?

At around the same time, David Deutsch et al. proved the same thing in this paper: Universality in Quantum Computation (1995), but without ever using the word "algebra" or "Lie" in the whole paper. The proof starts on page 3 and the main point is at Eq. 9, which is the same equation that appears in Seth Lloyd's paper, but here it is explained without reference to "Lie algebras". Eq. 9 is an application of what in physics we often just call the "Trotter splitting". It was written down almost 100 years earlier by Sophus Lie, but you do not need to know anything about Lie Algebras or even vector spaces in order to apply the formula as done in Eq. 9.

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  • $\begingroup$ Why would this answer the question? In the paper, H1 and H2 are related (by a swap), so they seem exactly NOT independent as asked in the question! $\endgroup$ Nov 19, 2018 at 22:37
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If $e^{iAt}$ and $e^{iBt}$ lie in a unitary representation of a group, this means that they belong to some matrix subgroup $G \subset U(n)$. This means that all possible compositions of these operators belong to $G$. If $G \neq U(n)$, then there is some unitary $U$ that is not generated by these unitaries. There is a Hermitian matrix $F$ such that $U = e^{iF}$. Since $U$ is not generated by $e^{iAt}$ and $e^{iBt}$, $F$ cannot belong to the Lie algebra generated by $A$ and $B$.

EDIT: The last statement probably needs more explanation. It is known that $e^{iAt}e^{iBt}$ can be expressed using the Baker-Campbell-Hausdorff formula. A nice corollary is called the Suzuki-Trotter theorem. By cleverly applying the latter, we can approximate any $e^{iC}$ for any $C \in \mathcal{L}$ (I think).

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