In quantum mechanics, given a Hermitian operator $A$, it gives rise to a symmetry/unitary operator by exponentiation $e^{i\lambda A}$, which can be properly defining using the eigenvector expansion, i.e. we define this exponentiation in the basis of $A$ as

$$e^{i\lambda A } = \sum_{n} e^{i\lambda a_n} |a_n\rangle \langle a_n|.$$

On the other hand, by Stone's theorem, given a strongly continuous one-parameter family of unitary operators $U(\lambda)$ we get a Hermitian operator $A$ such that

$$U(\lambda)=e^{i\lambda A},$$

where $A = U'(0)$.

Given, then, a Lie group $G$ and a smooth curve $\gamma : I\subset \mathbb{R}\to G$, we get an element of the Lie algebra as $\gamma'(0)$.

It turns out though that if $G$ acts unitarily on the state space of a system by $U : G\to \mathcal{L}(\mathcal{H})$, we can consider a curve $\gamma$ on $G$ and get a one-parameter family of unitary operators $U(\gamma(\lambda))$ on $\mathcal{H}$.

In the examples I looked at, it turns out that the Hermitian operators derived from Stone's theorem from these one-parameter families "$(U\circ\gamma)'(0)$" correspond to the Lie algebra elements $\gamma'(0)$.

The examples I saw of this are basically the Poincare group $G = P(1,3)$ acting by $U(a,\Lambda)$ so that if we consider the one-parameter families obtained by the coordinate lines of the usual coordinate system on $G$ we get that the unitary one-parameter families are for example translations in each of the $4$ directions. The associated observables by Stone's theorem are the $4$-momentum components. They seem related to the Lie algebra elements of the group, I just don't know how to make this connection really precise.

My question is: Given Lie groups, Lie algebras and one-parameter families of unitary operators in a quantum mechanical state space, is there really a connection between the observables and the Lie algebra? How does one make this connection precise?


1 Answer 1


Let $G$ be a Lie group, $\mathfrak{g}$ its Lie algebra, and $H$ a Banach space with a Banach representation $\pi : G \to \mathrm{GL}(H)$. The analytic vectors are defined to be $$ H^\omega := \{h\in H \mid G\to H, x\mapsto \pi(x)h \text{ is an analytic function}\}.$$ For the derived representation $\mathrm{d}\pi : \mathfrak{g}\to\mathrm{Hom}(H^\infty,H),A\mapsto \frac{\mathrm{d}}{\mathrm{d}t}(\pi(\mathrm{e}^{tA})h)\lvert_{t=0}$, where $H^\infty$ are the smooth vectors defined analogously to $H^\omega$, one can then show that $$ \pi(\mathrm{e}^{tA})h = \mathrm{e}^{t\mathrm{d}\pi(A)}h$$ for all $A\in\mathfrak{g}$, all $h\in H^\omega$ and all $t\in \mathbb{R}$. This is precisely the equation you're looking for: $\mathrm{d}\pi(A)$ is the generator of a one-parameter group by Stone's theorem, and $\mathrm{e}^{tA}$ is the equivalent of that one-parameter group in the Lie group. The equation says these two groups are mapped onto each other by the representation $\pi$.

You can find a more elaborate development of these notions in these lecture notes on representations of the Poincaré group by Eberhard Freitag, from whom I learnt this.

  • $\begingroup$ Thanks for the answer, but the link isn't working. What would be this derived representation? I mean,if $GL(H)$ were a smooth manifold, then $d\pi$ would be defined by $d\pi(\gamma'(0))=(\pi\circ\gamma)'(0)$ for all $\gamma(t)$. But here $H$ is a Banach space which might be infinite dimensional, and $GL(H)$ are all linear maps on $H$. I'm not sure of what is this derived representation in this context. $\endgroup$
    – Gold
    Apr 8, 2017 at 22:11
  • $\begingroup$ @user1620696 You're right, the servers seem to be down at the moment. I added the definition of the derived representation, it's really the same as what you wrote. The point is that it's defined only on $H^\infty$, where the notion of differentiation makes sense by definition, and one can show that $H^\infty$ is dense in $H$ for Banach representations $\pi$ (i.e. $\pi(G)\subset B(H)$ is bounded and $g\mapsto \pi(g)h$ is continuous on a dense subset of $H$, and perhaps some other technical condition I'm missing) $\endgroup$
    – ACuriousMind
    Apr 8, 2017 at 22:31

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