Stone's theorem tells us when a one-parameter unitary group has a self-adjoint generator. If $U(t)$ is a group --- i.e., it satisfies $U(t_1 + t_2) = U(t_1) U(t_2)$ --- then we can write $U(t) = e^{i H t}$ for some unbounded $H$ if and only if we have $\lim_{t \to t_0} U(t) |\psi \rangle = U(t_0) |\psi\rangle$ for every $|\psi\rangle$ in Hilbert space. This is the condition that $U(t)$ is **strongly continuous**.

What if we have a unitary group with multiple parameters? Within any one-parameter subgroup we should be able to find a generator. But is there a connection between the generators of different subgroups? Concretely, imagine we have a two-parameter subgroup that forms a representation of $\mathbb{R}^2.$ I.e., assume we have

$$U(t_1 + t_2, s_1 + s_2) = U(t_1, s_1) U(t_2, s_2).$$

Assume further that the map from $\mathbb{R}^2$ to unitary operators is strongly continuous. Then for any one-parameter subgroup we have a generator, in particular, we have

$$U(t,0) = e^{i H_1 t}$$

and

$$U(0, t) = e^{i H_2 t}.$$

But what about the generator of $U(t, t)$? We have some operator $K$ with

$$U(t, t) = e^{i K t}.$$

What is the connection between $K$ and $H_1$ and $H_2$? By looking at the structure of $\mathbb{R}^2$ we expect to be able to write something like $K = H_1 + H_2,$ but is this really correct? These operators are all unbounded; how do we know that we can add them together?

The correct statement is, in fact, $K = \overline{H_1 + H_2},$ where the overline denotes a closure of unbounded operators. This means that $K$ can be obtained from the sum of $H_1$ and $H_2$ by taking limits in the shared domain. The fact that $H_1$ and $H_2$ have *any* vectors in their shared domain is highly nontrivial; it is even stronger to claim that one can learn everything about $K$ by studying its restriction to this shared domain. More generally, if $G$ is a Lie group and $\rho : G \to \mathcal{U}(\mathcal{H})$ is a strongly continuous representation, then for any Lie algebra element $X$ we can study the one-parameter unitary group

$$U_{X}(t) = \rho(\exp[t X]).$$

This group has a self-adjoint generator $H_X.$ The theorem we will prove in this post is that generators compose the way you expect, in that we have

$$H_{X + Y} = \overline{H_{X} + H_{Y}}.$$

The fact that $H_X,$ $H_Y$, and $H_{X+Y}$ all live together in some representation of a Lie group is crucial in establishing this result.

To proceed, we will first (i) find a dense subspace of Hilbert space on which all of the generators $H_X$ can be studied simultaneously, then (ii) establish that the expected relation holds within this subspace, and (iii) show that the main theorem can be proved by taking appropriate limits.

I learned this material by studying chapter 10 of Konrad Schmudgen's book Unbounded Operator Algebras and Representation Theory, and filling in some gaps with my own calculations.

**Prerequisites: **Basics of unbounded operators; in particular, it will be very helpful to know what is the closure of an operator before trying to read this post, though I will explain it as it arises.

__Table of contents__

## 1. Smooth vectors

This integral is defined in the sense of the Bochner integral.

**smooth vector**; we have constructed a broad class of examples above. In fact, by taking a sequence of functions $f_n$ that approaches a delta function on the identity of $G$, one can show $|\psi_{f_n}\rangle \to |\psi\rangle.$ From this we see that every vector in Hilbert space is arbitrarily well approximated by smooth vectors, so the space of smooth vectors is dense in Hilbert space.

**But this is not necessarily true!**

**closure**(if it exists) is the operator $\overline{T}$ with domain $\overline{D}$, where $\overline{D}$ consists of all vectors $|\psi\rangle$ that can be obtained as the limit of a sequence $|\psi_n\rangle \in D$ such that $T |\psi_n\rangle$ also converges. For such a vector, $\overline{T}$ is defined by

In other words, the closure of an operator is obtained by studying limits of sequences for which the image sequence is convergent.

**core**for $S$, since anything you want to know about $S$ can be understood by taking limits of its restriction to $\tilde{D}.$

$$H_X |\psi_{f_n}\rangle = \lim_{t \to 0} \frac{U_X(t) |\psi_{f_n}\rangle - |\psi_{f_n}\rangle}{i t},$$

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