# Why does (almost) every pair of Hamiltonians generate, through repeated commutation, the whole space of Hermitian matrices?

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.

• 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). – AHusain Jun 12 '18 at 21:51
• @AHusain that sounds like something worthy of being an answer (not one easily understandable by me, but nonetheless..)! – glS Jun 13 '18 at 9:22