# 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). Jun 12 '18 at 21:51
• @AHusain that sounds like something worthy of being an answer!
– glS
Jun 13 '18 at 9:22
• Isn't the statement you quote basically tautological? (I mean: Either they generate the full algebra, or a subalgebra.) Dec 29 '21 at 15:32
• @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
– glS
Dec 30 '21 at 8:49

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).