# Does the Lie closure of a set of Hamiltonians describe all unitaries you can generate with them?

Suppose we have $$k$$ Hermitian matrices $$H_1, ..., H_k \in M_d(\mathbb{C})$$. Define $$\mathfrak{g} = \langle iH_1, ... iH_k \rangle_{\text{Lie}}$$ to be the Lie algebra generated by these Hermitian matrices (i.e. the vector space spanned by these matrices and all possible nested commutators). Assume also that $$\operatorname{dim} \mathfrak{g} < d^2$$, so that these matrices don't generate the entire $$U(d)$$.

Denote $$G \subset U(d)$$ as the group generated by matrices of the type $$e^{itH_j}$$. Denote also as $$e^{\mathfrak{g}}$$ the set of all unitary matrices that are exponents of some elements in $$\mathfrak{g}$$.

My question is, how do $$G$$ and $$e^{\mathfrak{g}}$$ relate to each other?

1. $$e^{\mathfrak{g}} \subseteq G$$. This direction is apparently shown by Lloyd (1995) by showing that $$e^{[iA, iB]t}$$ can be performed to arbitrary precision.
2. $$G \subseteq e^{\mathfrak{g}}$$. This is where it gets tricky. For small enough $$\epsilon$$, we know that $$e^{i\epsilon H_l} e^{i\epsilon H_m}$$ will belong to $$e^\mathfrak{g}$$ from the Baker-Campbell-Hausdorff formula, but for general case it's unclear. In other words, is it possible, by repeated multiplication of matrices of the type $$e^{iH_j t}$$, to obtain a unitary $$U = e^{iF}$$ such that $$iF \notin \mathfrak{g}$$?
• – glS
Commented Jun 23, 2022 at 13:26
• I found a related question on math.SE: math.stackexchange.com/questions/3847555. Will come back to this question when I make sense of it Commented Jun 27, 2022 at 14:03

Theorem. Given $$H_1, \ldots, H_k\in\mathbb C^{d\times d}$$ Hermitian let $$\mathfrak{g} := \langle iH_1, ... iH_k \rangle_{\text{Lie}}$$ denote the associated Lie algebra and let $$e^{\mathfrak g}$$ be the corresponding Lie group. Moreover, let $$G \subseteq U(d)$$ be the group generated by all individual one-parameter groups $$\{e^{itH_j}:t\in\mathbb R\}$$. Then $$G=e^{\mathfrak g}$$.
Proof. $$e^{\mathfrak g}\subseteq G$$: Note that $$e^{\mathfrak g}$$ is a connected Lie group so the inclusion in question follows from Lemma 6.2 in Jurdjevic, Sussmann, "Control Systems on Lie Groups", Journal of Differential Equations 12, p.313-329 (1972). As a side note, in the qubit case such decompositions are known as Z-Y decomposition (and its generalizations), refer to Theorem 4.1 ff. in Nielsen & Chuang.
$$G\subseteq e^{\mathfrak g}$$: Given any $$x\in G$$, by definition of $$G$$ there exist $$m\in\mathbb N$$ and real numbers $$\{t_{ab}:a=1,\ldots,k, b=1,\ldots,m\}$$ such that $$x=\prod_{j=1}^m e^{it_{1j}H_1}\cdot\ldots\cdot e^{it_{kj}H_k}$$. Re-writing $$x=\prod_{j=1}^m e^{i|t_{1j}|{\rm sgn}(t_{1j})H_1}\cdot\ldots\cdot e^{i|t_{kj}|{\rm sgn}(t_{kj})H_k}$$ shows that $$x$$ lies in the reachable set of the bilinear control system $$\dot x(t)=\sum_{j=1}^ku_j(t)H_jx(t)$$, $$x(0)={\bf1}$$ where the $$u_j$$ are piecewise constant functions which only take values $$1$$ and $$- 1$$. But this reachable set is equal to $$e^{\mathfrak g}$$, cf. Theorem 5.1 ff. in the previously cited Jurdjevic & Sussmann paper; hence $$x\in e^{\mathfrak g}$$ which concludes the proof. $$\square$$
Finally let me stress that your first inclusion $$e^{\mathfrak g}\subseteq G$$ does not follow from the arguments made in the paper of Lloyd you cited. The reason for this is that he implicitly requires either $$e^{\mathfrak g}$$ or $$G$$ to be closed (which need not be true in general).
• His first argument---the Lie product formula for commutators (Eq.(2) in his paper)---implies $$e^{\mathfrak g}\subseteq\overline{G}$$, where the closure $$\overline{G}$$ of $$G$$ comes into play because we have to allow for approximations = the limit $$n\to\infty$$. If $$G$$ is closed, then this does indeed show $$e^{\mathfrak g}\subseteq G$$. However, $$G$$ need not be closed: he common counterexample here are irrational windings on a torus, e.g., $$H={\rm diag}(1,\sqrt2)$$ in which case $$G=\{e^{itH}:t\in\mathbb R\}\subsetneq\overline{\{e^{itH}:t\in\mathbb R\}}=\{{\rm diag}(e^{i\theta_1},e^{i\theta_2}),\theta_1,\theta_2\in\mathbb R\}\,,$$ refer also to Proposition 2.5 in Elliott's book "Bilinear Control Systems", Springer (2009).
• In his second argument---the "noninfinitesimal" construction---Lloyd argues that "if $$\mathfrak g$$ has finite dimension, [the space $$e^{\mathfrak g}$$] is compact: As a result, at most, a number of transformations proportional to the number of generators of $$\mathfrak g$$ is required to reach any desired transformation in the space". The mistake made here is that $$\dim\mathfrak g<\infty$$ does in general not imply compactness of $$e^{\mathfrak g}$$; again, irrational windings on a torus show that $$e^{\mathfrak g}$$ need not be closed.