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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.

glS
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2 Answers2

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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.

Sanchayan Dutta
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user1271772 No more free time
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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).

Alexey Uvarov
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