I was reading some group theory notes and I am familiar with the concept of a Lie algebra, but I cannot imagine what the following formulation means:

What is more, not only states, but also the operators come in representations of the symmetry group. Clearly, the operators on a Hilbert space form an algebra, hence in particular a vector space.

Does this refer to the eigenspace of the operator?

  • 6
    $\begingroup$ It's an algebra, ie it forms a vector space (you can add and scale operators), and there exists a distributive product on them (performing successive applications of operators on the Hilbert vector). $\endgroup$
    – Slereah
    Sep 28, 2021 at 10:05
  • 2
    $\begingroup$ Wikipedia $\endgroup$
    – jacob1729
    Sep 28, 2021 at 10:15
  • $\begingroup$ Ah ok thanks thant makes sense! $\endgroup$ Sep 28, 2021 at 10:18

1 Answer 1


As per the comments, an algebra $(\mathcal A, \circ)$ consists of a vector space $\mathcal A$ and a bilinear operation $\circ:\mathcal A\times \mathcal A \rightarrow \mathcal A$ which provides some notion of vector multiplication. Examples include $\mathbb R^3$ equipped with the cross product and the space of $2\times 2$ skew-hermitian matrices equipped with the commutator bracket.

If $\mathcal H$ is some Hilbert space, then the bounded linear operators $\mathfrak B(\mathcal H)$ also constitute an algebra, with the multiplication operation being given by composition. In other words, for $\alpha,\beta\in\mathfrak B(\mathcal H)$ and $\psi\in \mathcal H$,

$$(\alpha \circ \beta)(\psi):= \alpha \big( \beta(\psi)\big)$$


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