Unbounded generators of unitary groups

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
2. Proving the main theorem

1. Smooth vectors

Let $G$ be a Lie group and $\rho : G \to \mathcal{U}(\mathcal{H})$ a strongly continuous unitary representation. Because $G$ is a Lie group, it always has a left Haar measure $\mu.$ To obtain vectors in Hilbert space that are "well behaved" with respect to $G$, we will start with a generic vector $|\psi\rangle$ and smear it with the action of $G$ using the Haar measure. For any compactly supported, smooth function $f : G \to \mathbb{C},$ we will define

$$|\psi_f\rangle = \int_G d\mu(g)\, f(g)\, \rho(g) |\psi\rangle.$$
This integral is defined in the sense of the Bochner integral.

Any vector constructed this way has the property that the map $g \mapsto \rho(g) |\psi_f\rangle$ is a smooth map from $G$ to $\mathcal{H}.$ Any vector $|\phi\rangle$ for which the map $g \mapsto \rho(g) |\phi\rangle$ is smooth is called a 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.

Now consider an arbitrary Lie algebra element $X$, and let $H_X$ be the generator of the unitary group $e^{i H_X t} = U_X(t) = \rho(\exp[t X]).$ While this is an unbounded operator, we will show that every smooth vector is in its domain.

To see this, we note from Stone's theorem that the domain of $H_X$ consists of all those vectors $|\psi\rangle$ for which the limit

$$H_X |\psi \rangle = \lim_{t \to 0} \frac{U_X(t) |\psi \rangle - |\psi \rangle}{it}$$

exists. If $|\psi\rangle$ is a smooth vector, then obviously the map $t \mapsto U_X(t) |\psi\rangle$ is smooth, so this limit exists and is just the derivative of that map.

So far we have shown that the space of smooth vectors is simultaneously in the domain of every $H_X.$ We also know that the space of smooth vectors is dense in Hilbert space. This makes it sound like to understand each $H_X,$ it is sufficient to study the restriction of $H_X$ to the space of smooth vectors. But this is not necessarily true!

For an unbounded operator $T$ with domain $D$, we say its 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

$$\overline{T}|\psi\rangle = \lim_n T |\psi_n\rangle.$$
In other words, the closure of an operator is obtained by studying limits of sequences for which the image sequence is convergent.

Given a closed operator $S$ with domain $D$, and a dense sub-domain $\tilde{D},$ the identity

$$S = \overline{S|_{\tilde{D}}}$$

may or may not be true depending on the details of $\tilde{D}.$ If it is true, then $\tilde{D}$ is said to be a core for $S$, since anything you want to know about $S$ can be understood by taking limits of its restriction to $\tilde{D}.$

What we will now show is that for each generator $H_X$, the space of smooth vectors is a core. To see this, let $f_n$ be a sequence of compactly supported smooth functions of $G$ that limit to a delta function on the identity element. As mentioned above, we have $|\psi_{f_n} \rangle \to |\psi \rangle$ for any $|\psi\rangle$, and each $|\psi_{f_n}\rangle$ is a smooth vector. The action of $H_X$ on this sequence is given by
$$H_X |\psi_{f_n}\rangle = \lim_{t \to 0} \frac{U_X(t) |\psi_{f_n}\rangle - |\psi_{f_n}\rangle}{i t},$$

or

$$H_X |\psi_{f_n}\rangle = \lim_{t \to 0} \frac{1}{i t} \left( \int_G d\mu(g)\, f_n(g) (\rho(\exp[t X]) - 1) |\psi\rangle \right).$$

When $|\psi\rangle$ is in the domain of $H_X$, the limit can be taken inside the integral to obtain

$$H_X |\psi_{f_n}\rangle = \int_G d\mu(g)\, f_n(g) H_X |\psi\rangle,$$

from which we can deduce

$$\lim_n H_X |\psi_{f_n} \rangle = H_X |\psi \rangle.$$

We also had $|\psi_{f_n} \rangle \to |\psi\rangle,$ and each $|\psi_{f_n}\rangle$ is smooth, so this completes the claim that the smooth vectors form a core for $H_X.$

2. Proving the main theorem

Continuing with the notation of the previous section, suppose that $X$ and $Y$ are two Lie algebra elements for $G$, with $H_X$ and $H_Y$ the corresponding self-adjoint generators. We want to show $H_{X+Y} = \overline{H_X + H_Y}.$ We know that the space of smooth vectors is a core for each of these operators individually, so we have

$$H_{X+Y} = \overline{H_{X+Y}|_{\text{smooth}}}.$$

For a smooth vector $|\psi\rangle$, we clearly have

$$H_{X+Y}|\psi\rangle = \lim_{t \to 0} \frac{\rho(\exp[t(X+Y)])-1}{i t}|\psi\rangle.$$

This is the derivative of the smooth function $t \mapsto \rho(\exp[t(X+Y)]) |\psi \rangle,$ which can be computed using continuity of $\rho$ and the derivative of the exponential map to obtain

$$H_{X+Y}|\psi\rangle = H_X |\psi\rangle + H_Y |\psi\rangle.$$

So the fundamental identity we want to prove holds on smooth vectors, and we have

$$H_{X+Y} = \overline{H_{X+Y}|_{\text{smooth}}} = \overline{(H_X + H_Y)|_{\text{smooth}}}.$$

So $H_{X+Y}$ is the closure of a restriction of $H_X + H_Y$; to show that it is the closure of $H_X + H_Y,$ it suffices to show that the domain of $H_{X+Y}$ contains the full domain of $H_X + H_Y.$

To show this, we note that for $|\phi\rangle$ in the domain of $H_X + H_Y$ and $|\psi\rangle$ a smooth vector, we can use individual self-adjointness of $H_X$ and $H_Y$ to obtain

$$\langle \phi | (H_X + H_Y) |\psi \rangle = \langle (H_X + H_Y) \phi | \psi \rangle.$$

This implies that $|\phi\rangle$ is in the domain of the adjoint of $(H_X + H_Y)|_{\text{smooth}},$ but since this operator has a self-adjoint closure $H_{X+Y},$ the domain of its adjoint is the domain of its closure, and we have that $|\phi\rangle$ is in the domain of $H_{X+Y},$ as desired.