In a paper I am reading, it states:

For open-loop coherent controllability a quantum system with Hamiltonian $H$ is open-loop controllable by a coherent controller if and only if the algebra $\mathcal{A}$ generated from $\{ H, H_i \}$ by commutation is the full algebra of Hermitian operators for the system.

How would you produce an algebra from the set $\{ H, H_i \}$ using commutation? What is the basic idea in this regard?


1 Answer 1


In general, an algebra $\mathcal{A}$ generated from a set $\{H_1, H_2,..., H_n\}$ by commutation refers to the algebra whose generators are $H_1,H_2,...,H_n$, all their first-order commutators $C_{ij} = [H_i,H_j]$, and all their second-order commutators $C_{ijk} = [[H_i, H_j],H_k]$ and so on.

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    $\begingroup$ So what you stated produces a set of new operators generated from the original set. Is the underlying algebraic structure (of addition, multiplication and scalar multiplication etc) defined in a pointwise way? $\endgroup$
    – John Doe
    Commented Jun 19, 2018 at 11:18
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    $\begingroup$ @JohnDoe In such cases, the elements of the algebra $\mathcal{A}$ can be represented as polynomials of the generators, with coefficients in an underlying field $K$. $\endgroup$ Commented Jun 19, 2018 at 11:23
  • $\begingroup$ Take the Free Lie Algebra L on the set $X = {x_j}$. Then send each $x_j \to i H_j$ in the set of anti-Hermitian matrices. (Math convention so different factors of $i$). $\endgroup$
    – AHusain
    Commented Jul 15, 2018 at 19:00

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