Stone's theorem

 Stone's theorem is the basic result describing group-like unitary flows on Hilbert space. If the map $t \mapsto U(t)$ is continuous in a sense we will make precise later, and each $U(t)$ is a unitary map on a Hilbert space $\mathcal{H},$ and we have $U(t+s)=U(t)U(s),$ then Stone's theorem asserts the existence of a (self-adjoint, positive definite, unbounded) operator $\Delta$ satisfying $U(t) = \Delta^{it}.$ This reduces the study of group-like unitary flows to the study of (self-adjoint, etc etc) operators. Quantum mechanically, it tells us that every group-like unitary evolution is generated by a time-independent Hamiltonian. This lets us study very general symmetry transformations in terms of Hamiltonians.

The standard proof of Stone's theorem, which you'll see if you look at Wikipedia, involves trying to make sense of a limit like $\lim_{t \to 0} (U(t) - 1)/t$. However, I have recently learned of a beautiful proof of Stone's theorem that works instead by studying analytic continuations of the map $it \mapsto U(t)$ away from the imaginary axis into a strip in the complex plane. This proof is extremely informative, as it teaches us how to think about analytic continuations of unitary flows, which are a useful tool in functional analysis. Furthermore, many of the techniques used in the proof (e.g. the Riesz representation theorem, Morera's theorem, the Phragmen-Lindelof principle) are broadly applicable in mathematical physics. The purpose of the post is to explain that proof.

The outline is:

1. In section 1, I will briefly state some essential definitions having to do with unbounded operators, including the definitions of adjoints, of closed operators, and of positive definite, self-adjoint operators.
2. In section 2, I will explain the analytic structure of the map $z \mapsto \Delta^{z}$ for a positive, self-adjoint operator $\Delta.$ Defining analyticity requires being able to take limits of quantities like $\frac{\Delta^{z+h} - \Delta^{z}}{h},$ but the operators $\Delta^{z+h}$ and $\Delta^{z}$ cannot be compared using a norm. Instead, they must be compared in terms of how they act on a fixed vector. I will show that whenever $|\xi\rangle$ is in the domain of $\Delta,$ it is in the domain of $\Delta^z$ for $z$ in the strip $0 \leq \text{Re}(z) \leq 1,$ and that the function $z \mapsto \Delta^z |\xi\rangle$ is analytic in the interior of this strip and continuous on its boundary. I will also show that this is a complete characterization of the domain of $\Delta$: if $|\xi\rangle$ is a vector for which the map $it \mapsto \Delta^{it} |\xi\rangle$ has an analytic continuation to the strip, then $|\xi\rangle$ is in the domain of $\Delta.$
3. In section 3, I will show that if $t \mapsto U(t)$ is a group-like unitary flow satisfying an appropriate continuity condition, then $U(t)$ is of the form $\Delta^{it}$ for some positive definite, self-adjoint operator $\Delta.$ The operator is constructed using intuition from section 2: its domain is defined as the set of all vectors $| \xi \rangle$ for which $it \mapsto U(t) |\xi\rangle$ admits an analytic continuation to a strip in the complex plane, and $\Delta |\xi\rangle$ is defined by evaluating that analytic continuation at the point $z=1.$ The hard work is all in showing that this definition does in fact give rise to a positive definite, self-adjoint operator.

I learned all of this material from Chapters 2 and 9 of Stratila and Zsido. All of the details that I skim over in this post can be found there.

Prerequisites: It will be very helpful to know some things about unbounded operators and spectral theory, but I will try to state the important prerequisite theorems as needed.

Table of Contents

1. Brief review of unbounded operators
2. Analyticity of positive operator flows
3. Stone's theorem

1. Brief review of unbounded operators

All of the material in this section can be found in any book on functional analysis. I will not provide proofs.

Let $\mathcal{H}$ be a Hilbert space. In general terms, an operator $T$ on $\mathcal{H}$ is a linear map from some subspace $D_T \subseteq \mathcal{H}$ to $\mathcal{H}$. The subspace $D_T$ is called the domain of $T$. Commonly, $D_T$ is taken to be dense in $\mathcal{H}$, and $T$ is said to be densely defined.

If $T$ is densely defined and bounded on its domain, then it can be extended to a bounded linear operator everywhere on $\mathcal{H}$ by continuity. The interesting application of operator domains comes therefore in the study of unbounded operators. The purpose of this section is to state some definitions concerning unbounded operators, and to describe their spectral theory.

Every bounded operator $T$ has the property that if $|x_n\rangle$ converges to $|x\rangle$, then $T |x_n\rangle$ converges to $T |x\rangle.$ An unbounded operator $T$ on $\mathcal{H}$ is said to be closed if whenever a sequence $|x_n\rangle \in D_T$ converges to $|x\rangle$ and $T |x_n \rangle$ converges to something, then $|x\rangle$ is in $D_T$ and we have $T |x_n \rangle \to T |x\rangle.$ So the action of a closed operator on a convergent sequence need not converge, but if it does, then it converges in the expected way.

Given a densely defined operator $T$, one can define an adjoint operator $T^{\dagger}.$ We would like to define it in the usual way via the equation

$$\langle x | T y \rangle = \langle T^{\dagger} x | y \rangle.$$

This is a little complicated by the fact that this expression can only make sense if $|y\rangle$ is in the domain of $T$ and $|x\rangle$ is in the domain of $T^{\dagger}.$ In fact, we define the domain of $T^{\dagger}$ by demanding that this equation makes sense. We define $D_{T^{\dagger}}$ to be the set of vectors $|x\rangle$ such that the map $|y\rangle \mapsto \langle x | T y \rangle$ is bounded on $D_T.$ The Riesz representation theorem, together with density of $D_T$ in $\mathcal{H}$, then guarantees the existence of a unique vector $T^{\dagger} |x\rangle$ satisfying the defining equation

$$\langle x | T y \rangle = \langle T^{\dagger} x | y \rangle.$$

One can show that every adjoint is automatically closed.

An operator $T$ is said to be symmetric if we have $D_{T^{\dagger}} \supseteq D_{T}$ and the operators $T^{\dagger}$ and $T$ agree on their common domain. It is said to be self-adjoint if it is symmetric and furthermore we have $D_{T^{\dagger}} = D_T.$ Note that because we only define adjoints for densely defined operators, an operator that is symmetric or self-adjoint is automatically densely defined. Note also that because adjoints are automatically closed, self-adjoint operators are closed.

An operator is said to be positive if we have $\langle x | T | x \rangle \geq 0$ for all $|x\rangle$ in $D_T.$

Given any self-adjoint operator $T$, there is an associated spectral theory. There are a few ways to think about this theory. One is in terms of a projection-valued measure, which is a map that takes Borel subsets of the real line to projections in an appropriately additive way. We denote this map by $\omega \mapsto \chi_{\omega}(T).$ Given any vectors $|x\rangle, |y\rangle,$ there is some associated complex measure on the real line given by the map $\mu_{x, y} : \omega \mapsto \langle x | \chi_{\omega}(T) | y \rangle.$ Furthermore, when $|y\rangle$ is in the domain of $T$, then the matrix elements of $T$ can be expressed as

$$\langle x | T | y \rangle = \int d\mu_{x, y}\, \text{id},$$

i.e., the integral of the identity operator on the real line with respect to the measure $\mu_{x, y}.$

The existence of a projection-valued measure corresponding to the spectral decomposition of $T$ lets us define certain functions of $T$. If $f : \mathbb{R} \to \mathbb{C}$ is any measurable function, then there is a corresponding densely defined operator $f(T)$ given as follows. (This is called the "functional calculus" for a self-adjoint unbounded operator, because it lets us calculate functions of operators.)

• The domain of $f(T)$ is the set of all vectors $|x\rangle$ for which we have
  $$\int d\mu_{x, x} |f|^2 < \infty.$$
  Note that since we have $\int d\mu_{x, x} = \langle x | x \rangle < \infty,$ this implies that if $f$ is a bounded function, then $f(T)$ is defined on all of Hilbert space and is therefore bounded.
• Given any $|y \rangle$ in $D_{f(T)}$ and any $|x\rangle \in \mathcal{H},$ the matrix elements of $f(T)$ are defined by
  $$\langle x| f(T) | y \rangle = \int d\mu_{x, y} f.$$
• If $f, g : \mathbb{R} \to \mathbb{C}$ are measurable, then we have $D_{f(T) g(T)} \subseteq D_{(fg)(T)}$ and the operators $f(T) g(T)$ and $(fg)(T)$ agree on their common domains. Explicitly, we have
  $$D_{f(T) g(T)} = D_{(fg)(T)} \cap D_{g(T)}.$$
• We have $\bar{f}(T) = f(T)^{\dagger}.$

From these properties it is easy to see that if $f : \mathbb{R} \to \mathbb{C}$ is real, then $f(T)$ is self-adjoint. If $f : \mathbb{R} \to \mathbb{C}$ is nonnegative, then $f(T)$ is self-adjoint and positive. Note also that the notation $\chi_{\omega}(T)$ for the spectral projection onto $\omega$ is consistent with the functional calculus for $\chi_{\omega}$ being the indicator function on $\omega.$

There is one last important point, which is that the spectral measure $\omega \mapsto \chi_{\omega}(T)$ is actually supported on the spectrum of $T$, which is the set of points $\lambda \in \mathbb{R}$ such that $T - \lambda$ cannot be inverted as a bounded operator. Furthermore, if $\omega \subseteq \mathbb{R}$ has zero spectral measure in the sense that $\chi_{\omega}(T)$ vanishes, then $f(T)$ is independent of the behavior of $f$ on the set $\omega.$

One specific consequence of this is as follows. An unbounded, self-adjoint, positive operator $\Delta \geq 0$ is said to be positive definite if it has trivial kernel. It is easily seen that $\chi_{\{0\}}(\Delta)$ is the projector onto the kernel of $\Delta,$ so for a positive definite operator, $f(\Delta)$ is independent of the behavior of $f$ at zero.

2. Analyticity of positive operator flows

Let $\Delta \geq 0$ be a positive definite, self-adjoint, unbounded operator with domain $D_{\Delta}.$ Consider the map $t \mapsto \Delta^{it}.$ We ultimately want to show that every unitary flow satisfying certain properties can be written in this form. To reach that goal, it will be helpful to understand the properties of the map $t \mapsto \Delta^{it}$.

First, let us be a little precise about what we mean by the operator $\Delta^{it}.$ We mean the operator $f_t(\Delta),$ defined via the functional calculus of the previous section, for the function $f_t : [0, \infty) \to \mathbb{C}$ given by

$$f(x) = \begin{cases} e^{i t \log{x}} & x > 0 \\ 0 & x = 0\end{cases}.$$

This function is bounded, so $\Delta^{it}$ is bounded; in fact, since we have assumed that $\Delta$ has trivial kernel, the behavior of $f$ at $x=0$ is irrelevant. From the functional calculus described in the previous section, we easily see $(\Delta^{it})^{\dagger} = \Delta^{-i t}$ and $\Delta^{i t} \Delta^{-i t} = 1,$ so each $\Delta^{it}$ is unitary. The functional calculus also gives $\Delta^{it} \Delta^{is} = \Delta^{i(t+s)},$ so the map $t \mapsto \Delta^{it}$ is group-like.

We would like to understand the topological properties of the map $t \mapsto \Delta^{it}.$ In particular, we would like to know in what sense it is continuous. There are a few natural notions of continuity on the space of bounded operators on a Hilbert space. One is norm continuity; we would say the map $t \mapsto \Delta^{it}$ is norm continuous if $s \to t$ implies

$$\lVert \Delta^{i s} - \Delta^{it} \rVert \to 0.$$

Another is strong continuity; we would say the map $t \mapsto \Delta^{it}$ is strongly continuous if $s \to t$ implies

$$\Delta^{is} |\xi\rangle \to \Delta^{it} |\xi \rangle$$

for every vector $|\xi\rangle \in \mathcal{H}.$ A final possibility is weak continuity; we would say that hte map $t \mapsto \Delta^{it}$ is weakly continuous if $s \to t$ implies

$$\langle \eta| \Delta^{is} |\xi \rangle \to \langle \eta | \Delta^{it} | \xi \rangle$$

for every $|\eta\rangle, |\xi\rangle \in \mathcal{H}.$

It is easy to check that if a map is norm continuous then it is automatically strongly continuous, and if it is strongly continuous then it is automatically weakly continuous. So to make the most general statement about the map $t \mapsto \Delta^{it},$ we should first check if it is norm continuous; if not, we should check if it is strongly continuous; and so on. It turns out that $t \mapsto \Delta^{it}$ is not always norm continuous, but it is always strongly continuous. We will only check continuity in the limit $t \to 0,$ but using the group structure of $t \mapsto \Delta^{it},$ it is easy to see that this is equivalent to checking continuity $t \to s$ for any $s.$

In particular, $t \mapsto \Delta^{it}$ is not norm continuous if the spectrum of $\Delta$ contains points arbitrarily close to zero. In that case, for any $t > 0$ one can find an open subset $O$ of the spectrum of $\Delta$ such that for $x \in O,$ the number $t \log{x}$ is arbitrarily close to $2 \pi n + \pi$ for some integer $n.$ For any vector $|\xi\rangle$ supported in that spectral range, we have

$$\lVert (\Delta^{it} - 1) | \xi\rangle \rVert^2 = \int d\mu_{\xi, \xi} |e^{i t \log{x}} - 1|^2 \geq \inf_{x \in O} |e^{i t \log{x}} - 1|^2,$$

and this is bounded away from zero. So as $t$ goes to zero, we can keep changing $|\xi\rangle$ to find a vector on which $\Delta^{it} |\xi\rangle$ is not close to $|\xi\rangle.$ This means that $\Delta^{it}$ cannot converge to the identity operator in norm as $t$ goes to zero. However, when we fix a vector $|\xi\rangle$ once and for all, then take the limit $t \to 0,$ we will show that we have $\Delta^{it} |\xi \rangle \to |\xi \rangle,$ which means that the map $t \mapsto \Delta^{it}$ is strongly continuous.

To see strong continuity, let $|\xi\rangle$ be some fixed vector, and let $|\xi_{\epsilon}\rangle = \chi_{(\epsilon, 1/\epsilon)}(\Delta) |\xi\rangle$ be the projection of $|\xi\rangle$ onto the spectral range $(\epsilon, 1/\epsilon).$ Let $\Delta_{\epsilon}$ denote the spectral restriction $\chi_{(\epsilon, 1/\epsilon)}(\Delta) \Delta \chi_{(\epsilon, 1/\epsilon)}(\Delta).$ For each $\epsilon,$ the function $t \mapsto \Delta^{it}_{\epsilon}$ is strongly continuous, because $\Delta_\epsilon$ is a bounded operator with spectrum bounded away from zero, so we can write $\Delta_{\epsilon}^{it}$ in terms of the exponential $e^{i t \log \Delta_{\epsilon}}.$ Furthermore, we have

$$\lVert \Delta_{\epsilon}^{it} |\xi\rangle - \Delta^{it} |\xi\rangle \rVert \leq \lVert \Delta^{it} \rVert \lVert |\xi_{\epsilon}\rangle - |\xi\rangle \rVert = \lVert |\xi_{\epsilon} \rangle - |\xi\rangle \rVert.$$

In the limit $\epsilon \to 0,$ the right-hand side goes to zero. So the continuous functions $t \mapsto \Delta^{i t}_{\epsilon} |\xi \rangle$ converge uniformly to the function $t \mapsto \Delta^{it} |\xi \rangle.$ Therefore this function is continuous.

So far, we know that $t \mapsto \Delta^{it}$ is a group-like unitary flow such that every map of the form $t \mapsto \Delta^{it} |\xi \rangle$ is continuous. Let us study what happens if we change $it$ to another complex number; can we say something about the map $z \mapsto \Delta^{z},$ or about maps of the form $z \mapsto \Delta^{z} |\xi\rangle$?

For $z$ not purely imaginary, $\Delta^{z}$ is not a unitary operator, and in fact it is generally not even bounded. So we shouldn't really expect to be able to say much about the map $z \mapsto \Delta^z.$ But we do expect to be able to say something about maps $z \mapsto \Delta^z |\xi\rangle$ for fixed $|\xi\rangle,$ so long as we work in a region of the complex plane where $|\xi\rangle$ is in the domain of $\Delta^z$ for every $z.$ This leads us to the following observation, which I will explain with a figure and elaborate further:

• If $|\xi\rangle$ is in the domain of $\Delta,$ then $|\xi \rangle$ is in the domain of $\Delta^z$ for every $z$ in the strip $0 \leq \text{Re}(z) \leq 1.$ On the left edge of the strip, which is the imaginary axis, the map $z \mapsto \Delta^z |\xi\rangle$ reduces to the continuous unitary flow $it \mapsto \Delta^{it}|\xi\rangle.$ In fact, the map $|\xi\rangle \to \Delta^{z} |\xi \rangle$ is analytic in the whole strip $0 < \text{Re}(z) < 1,$ and continuous on the boundary.

Note that when I say a Hilbert-space valued function is analytic, I mean it in the sense of the fourth section of my post on vector integration. Namely, I mean that limits like $\lim_{h \to 0} \frac{\Delta^{z + h} |\xi\rangle - \Delta^z |\xi\rangle}{h}$ exist in the Hilbert space topology. I showed in that post that many of the basic theorems of complex analysis hold for analytic functions with target space given by a Hilbert space.

Now, for a figure, I have in mind something like this:

We have an infinite strip of unit width in the complex plane. There is some Hilbert-space valued function defined in this strip, which is analytic in the interior and continuous on the boundary. At the midpoint of the strip's right edge, the function takes the value $\Delta |\xi\rangle.$ Along its left edge, the function takes the values $\Delta^{it} |\xi\rangle.$ In fact, at every point in the strip, the function takes the value $\Delta^{z} | \xi \rangle$. Let's prove the claim that such an analytic function exists.

First, we'll prove that this function makes sense. That is, for $|\xi\rangle$ in the domain of $\Delta,$ I need to prove that $|\xi\rangle$ is in the domain of $\Delta^z$ for every $z$ in the strip. This is easy to show using the rules of the functional calculus that I described in the previous section. Given $\Delta$ and $|\xi\rangle,$ there is an associated spectral measure $\mu_{\xi, \xi},$ and $|\xi \rangle$ is in the domain of $f(\Delta)$ if and only if we have

$$\int d\mu_{\xi, \xi} |f|^2 < \infty.$$

In particular, we are considering the function $f_z(\Delta)$ given on the positive real axis by

$$f_{z}(x) = \begin{cases} e^{z \log{x}} & x > 0, \\ 0 & x = 0 \end{cases}.$$ 

It is easy to see that for $z$ in the strip, we have $|f|^2 < x^2$ for $x > 1.$ So since the integral $\int d\mu_{\xi, \xi} x^2$ is finite by assumption that $|\xi\rangle$ is in the domain of $\Delta,$ basic domination arguments imply that the integral $\int d\mu_{\xi,\xi} |f_z(x)|^2$ is finite as well. (The portion of the integral between $x=0$ and $x=1$ might get bigger, but it of course remains finite.)

Second, let's show that the map $z \mapsto \Delta^{z} |\xi\rangle$ is analytic in the strip and continuous on the boundary. If you take a look at the fourth section of my post on vector integration, you'll see that it suffices to find a sequence of functions $f_n : \text{strip} \to \mathcal{H}$ that are holomorphic on the strip and continuous on the boundary, such that the sequence converges uniformly to $z \mapsto \Delta^z |\xi\rangle$ on the strip.

This is easy. We're working with an unbounded operator $\Delta,$ so there's really only one thing we can do to regulate it: project onto a bounded subset of its spectrum. We'll do the same thing we did above to prove that $t \mapsto \Delta^{it}$ is strongly continuous: we'll denote by $\Delta_n$ the projection of $\Delta$ onto the spectral range $(1/n, n).$ For each such $\Delta_n,$ the map $z \mapsto \Delta_n^{z} |\xi\rangle$ is holomorphic not just in the strip, but in the entire complex plane, since it can be written as the exponential $z \mapsto e^{z \log \Delta_n} |\xi \rangle.$ So we need only show that the maps $z \mapsto \Delta_n^{z} |\xi\rangle$ converge uniformly on the strip to the map $z \mapsto \Delta^z |\xi\rangle.$ But this is easy. Using the spectral measure $\mu_{x,x},$ we have

$$\lVert (\Delta_n^z - \Delta^z) |\xi \rangle \rVert^2 = \int_{0}^{1/n} d\mu_{\xi, \xi} |x^z|^2 + \int_{n}^{\infty} d\mu_{\xi, \xi} |x^z|^2.$$ 

In the first integral, we have $0 < x < 1,$ so $|x^z|$ is upper bounded by 1. In the second integral, we have $1 < x < \infty,$ and $|x^z|$ for $z$ in the strip is upper bounded by $x.$ So we have

$$\lVert (\Delta_n^z - \Delta^z) |\xi \rangle \rVert^2 \leq \langle \xi| \chi_{(0, 1/n)}(\Delta) |\xi\rangle + \int_{n}^{\infty} d\mu_{\xi, \xi} |x|^2.$$

Both terms are independent of $z,$ and both can be shown to vanish in the limit $n \to \infty.$ This establishes uniform convergence, and therefore establishes the desired continuity and analyticity properties of the map $z \mapsto \Delta^{z} |\xi\rangle.$

So far, we know that if $|\xi\rangle$ is in the domain of $\Delta,$ then it is in the domain of $\Delta^z$ for all $z$ in the strip, and the maps $z \mapsto \Delta^z |\xi\rangle$ are continuous on the strip and holomorphic in its interior. We will now show a converse: that if $|\xi\rangle$ is a vector such that the map $i t \mapsto \Delta^{it} |\xi\rangle$ admits an analytic/continuous extension to the strip, then $|\xi\rangle$ must be in the domain of $\Delta.$

To see this, suppose that $z \mapsto |F(z)\rangle$ is a vector valued function that is analytic in the strip and continuous on its edges, and that equals $\Delta^{it} |\xi \rangle$ on the edge. For any $|\eta\rangle$ in the domain of $\Delta,$ our preceding discussion tells us that the map

$$it \mapsto \langle \eta | F(it) \rangle = \langle \Delta^{-it} \eta | \xi \rangle$$

is the left-edge restriction of the holomorphic map $z \mapsto \langle \Delta^{\bar{z}} \eta | \xi \rangle$ in the strip. By usual complex analysis arguments, there is only one holomorphic extension into the strip, so we must have

$$\langle \eta | F(z) \rangle = \langle \Delta^{\bar{z}} \eta | \xi \rangle.$$

This equation tells us that the map $|\eta\rangle \mapsto \langle \xi | \Delta^{\bar{z}} \eta \rangle$ is bounded, so $|\xi\rangle$ is in the domain of the adjoint of $\Delta^{\bar{z}},$ which we know is $\Delta^z.$ We conclude that $|\xi\rangle$ must be in the domain of $\Delta^z$ for every $z$ in the strip.

3. Stone's theorem

The preceding section told us that whenever $\Delta \geq 0$ is a positive definite, self-adjoint operator, then we have the following results.

• The map $t \mapsto \Delta^{it}$ is a strongly continuous group-like unitary flow.
• $|\xi \rangle$ is in the domain of $\Delta$ if and only if the map $it \mapsto \Delta^{it} |\xi\rangle$ admits an analytic continuation to the strip $0 \leq \text{Re}(z) \leq 1$ that is continuous on the boundaries of the strip. In this case, $|\xi\rangle$ is in the domain of every $\Delta^z$ for $z$ in the strip, and the analytic continuation to the strip is given explicitly by $z \mapsto \Delta^{z} |\xi\rangle.$

We would now like to look at the same kind of object from a different angle. Let us consider a strongly continuous, group-like unitary flow $t \mapsto U(t).$ We will define an operator $\Delta$ by saying that $|\xi\rangle$ is in the domain of $\Delta$ if the map $it \mapsto U(t) |\xi\rangle$ admits an analytic continuation to the strip, and define $\Delta |\xi \rangle$ by evaluating this analytic continuation at $z=1$. We will check that $\Delta$ is in fact a linear operator, that it is densely defined, that it is positive definite and closed, and that the analytic continuation of $it \mapsto U(t) |\xi\rangle$ is given by $z \mapsto \Delta^z |\xi\rangle.$ This will let us write $U(t) = \Delta^{it},$ which is the essential content of Stone's theorem.

Before we begin, though, let me give some context about an unconventional aspect of the proof. Namely, that we won't actually use strong continuity; we'll use a weaker assumption. The basic reason for this is that whenever a function $z \mapsto |F(z)\rangle$ is analytic in the sense of Hilbert space, all functions of the form $z \mapsto \langle \eta | F(z) \rangle$ are analytic in the ordinary sense of complex analysis. Consequently, it is easier to check analyticity for the functions $z \mapsto \langle \eta | F(z) \rangle$ than for the function $z \mapsto | F(z) \rangle$. It turns out that we can prove Stone's theorem in a way that only requires us to use this weaker notion of analyticity. What's cool about this is that we can make a weaker assumption than strong continuity of $t \mapsto U(t)$ and still recover the formula $U(t) = \Delta^{it}$, which implies that $t \mapsto U(t)$ was actually strongly continuous all along.

The weaker condition we will use is that $t \mapsto U(t)$ is weakly continuous, i.e., that all maps of the form $t \mapsto \langle \eta| U(t) | \xi\rangle$ are continuous.

Here's a rough plan of how we're going to proceed. Note that if I say a vector-valued function $z \mapsto |F(z)\rangle$ is weakly continuous/analytic, that means all functions of the form $z \mapsto \langle \eta | F(z) \rangle$ are continuous/analytic.

• Assume that $t \mapsto U(t)$ is a weakly continuous, group-like unitary flow.
• We will show that for a strip of any width, $0 \leq \text{Re}(z) \leq \alpha,$ there is a dense set of vectors $|\xi \rangle$ for which $U(t) |\xi\rangle$ admits a weakly continuous/analytic extension to $0 \leq \text{Re}(z) \leq \alpha$.
• For any such $|\xi \rangle$ and any $z$ in the strip $0 \leq \text{Re}(z) \leq \alpha,$ we will define the vector $\Delta^z |\xi \rangle$ by setting it equal to the corresponding weakly analytic continuation evaluated at $z.$
• We will show that the vector $\Delta^z |\xi \rangle$ is well defined (i.e., independent of the choice of $\alpha$), and that $\Delta^z$ is a linear operator.
• We will show that the operators $\Delta^z$ have an appropriate group-like property: $\Delta^{z+w} |\xi \rangle = \Delta^z (\Delta^w |\xi\rangle)$ whenever this equation makes sense. (Whether it makes sense depends on $z, w$ and $|\xi\rangle,$ due to domain issues for unbounded operators.)
• We will show that the operators $\Delta^z$ are closed and satisfy $(\Delta^z)^{\dagger} = \Delta^{\bar{z}},$ and that if $z$ is a positive real number then $\Delta^z$ is a positive definite operator.
• We will show that the operators $\Delta^z$ are in fact the complex powers of the operator $\Delta.$
• We will conclude that the weakly analytic extensions of the maps $it \mapsto U(t) | \xi \rangle$ to $0 \leq \text{Re}(z) \leq 1$ are given by $\Delta^z | \xi \rangle$, and therefore that we have $U(t) = \Delta^{it}.$

There are a couple of steps below that involve funny iterated analytic continuations; they can be a little confusing to read in a text-only format, so I've included figures where I think they will be helpful. If you get stuck, scroll ahead a bit and see if there's a figure that helps you orient yourself.

To start, assume that $t \mapsto U(t)$ is a weakly continuous, group-like unitary flow. I would like to show that there is a dense set of vectors $|\xi \rangle$ such that the map $it \mapsto U(t) |\xi\rangle$ admits a weakly analytic extension to the full complex plane; in particular, this map admits a weakly analytic/continuous extension to strips of any width.

The point will be, given a vector $|y \rangle,$ to write it as the limit of a sequence $|y_n\rangle$ of vectors  for which $it \mapsto U(t) |y_n\rangle$ admits a weakly analytic extension to the full complex plane. To do this, I will formally write

$$|y\rangle = \int_{-\infty}^{\infty} ds\, \delta(s) U(s) |y\rangle,$$

then mollify the delta function by Gaussians, to write

$$|y_n \rangle = \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n s^2} U(s) |y \rangle.$$

We can't actually guarantee that this integral converges using only the assumption that $U(t)$ is weakly continuous. But I can try to formally define the overlap of $|y_n\rangle$ with a generic vector $|\eta \rangle$ via the integral

$$\int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n s^2} \langle \eta | U(s) | y \rangle.$$

Because $U(s)$ is weakly continuous, this integral is well defined if its absolute value is finite. We have

$$\left| \int_{-\infty}^{\infty} dt\, \sqrt{\frac{n}{\pi}} e^{-n s^2} \langle \eta | U(s) | y \rangle \right|\leq \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n s^2} |\langle \eta | U(s) | y \rangle |.$$

Using the Cauchy-Schwarz inequality and unitarity of $U(s),$ we obtain

$$\left| \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{-n s^2} \langle \eta | U(s) | y \rangle \right| \leq  \lVert \eta \rVert \lVert y \rVert.$$

In particular, this means that the map

$$|\eta \rangle \mapsto \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{-n s^2} \langle \eta | U(s) | y \rangle$$

is antilinear and bounded, so by the Riesz representation theorem, there does exist a unique vector $|y_n\rangle$ satisfying 

$$\langle \eta | y_n \rangle = \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n s^2} \langle \eta | U(s) | y \rangle.$$

We can now show that the map $it \mapsto U(t) |y_n \rangle$ admits a weakly analytic extension to the complex plane. To see this, we write

$$\langle \eta | U(t) y_n \rangle = \langle U(-t) \eta | y_n \rangle = \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n s^2} \langle \eta | U(t+s) | y \rangle.$$

Making the change of variables $s \mapsto s - t$ gives

$$\langle \eta | U(t) y_n \rangle = \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n (s+i (it))^2} \langle \eta | U(s) | y \rangle.$$

So a candidate analytic continuation for the map $it \mapsto \langle \eta | U(t) y_n \rangle$ is 

$$z \mapsto \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n (s+i z)^2} \langle \eta | U(s) | y \rangle.$$

By a similar argument as given above using the Riesz representation theorem, we are guaranteed the existence of a vector-valued map $z \mapsto |y_n(z)\rangle$, defined everywhere on the complex plane, whose overlaps are given by

$$\langle \eta | y_n(z) \rangle = \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n(s+iz)^2} \langle \eta | U(s) | y \rangle.$$

So the vector-valued function $z \mapsto |y_n(z)\rangle$ certainly agrees with $it \mapsto \langle \eta | U(t) | y_n \rangle$ on the imaginary axis; we only need to show that $|y_n(z)\rangle$ is weakly analytic in $z.$ This isn't too hard to do, but it's a little annoying to write out explicitly, so I'll leave it as an exercise. The most straightforward way I know to prove analyticity is an application of Morera's theorem.

So now we know that each $|y_n\rangle$ has the property that the map $it \mapsto U(t) |y_n\rangle$ admits a weakly analytic extension to the full complex plane. Furthermore, the sequence $|y_n\rangle$ converges weakly to $|y\rangle,$ as we have

$$| \langle \eta | y_n \rangle - \langle \eta | y \rangle | = \int_{-\infty}^{\infty} ds\, \sqrt{\frac{n}{\pi}} e^{- n s^2} \langle \eta | (U(s) - 1) | y \rangle,$$

and one can easily show that this goes to zero in the limit $n \to \infty$ using the weak continuity of $U(s)$ near $s=0,$ and the fact that the Gaussian $e^{- n s^2}$ becomes arbitrarily peaked near $s=0$ in the limit $n\to \infty.$

From this, we conclude that there is a weakly dense set of vectors $|\xi\rangle$ with the property that $it \mapsto U(t) | \xi\rangle$ admits a weakly entire extension. Furthermore, this set is a linear subspace of $\mathcal{H}$.  But it is easy to check that if a subspace of $\mathcal{H}$ is weakly dense, then it is dense in the usual sense. To see this, let $V \subseteq \mathcal{H}$ be a subset of $\mathcal{H}$ that is weakly dense in $\mathcal{H}.$ Then for any $|x\rangle \in \mathcal{H}$ that is orthogonal to $V$, by writing $|x\rangle$ as the weak limit of a sequence $|v_n\rangle,$ we have

$$\langle x | x \rangle = \lim_n \langle x | v_n \rangle = 0,$$

so $|x\rangle$ is the zero vector. Since the only vector orthogonal to $V$ is the zero vector, we conclude that $V$ is dense in $\mathcal{H}.$

This post is getting a bit long, so let's check back in with our outline. Here's what we've done so far:

• Assume that $t \mapsto U(t)$ is a weakly continuous, group-like unitary flow.
• We will show that for a strip of any width, $0 \leq \text{Re}(z) \leq \alpha,$ there is a dense set of vectors $|\xi \rangle$ for which $U(t) |\xi\rangle$ admits a weakly continuous/analytic extension to $0 \leq \text{Re}(z) \leq \alpha$.

And here's what we still need to do:

• For any such $|\xi \rangle$ and any $z$ in the strip $0 \leq \text{Re}(z) \leq \alpha,$ we will define the vector $\Delta^z |\xi \rangle$ by setting it equal to the corresponding weakly analytic continuation evaluated at $z.$
• We will show that the vector $\Delta^z |\xi \rangle$ is well defined (i.e., independent of the choice of $\alpha$), and that $\Delta^z$ is a linear operator.
• We will show that the operators $\Delta^z$ have an appropriate group-like property: $\Delta^{z+w} |\xi \rangle = \Delta^z (\Delta^w |\xi\rangle)$ whenever this equation makes sense. (Whether it makes sense depends on $z, w$ and $|\xi\rangle,$ due to domain issues for unbounded operators.)
• We will show that the operators $\Delta^z$ are closed and satisfy $(\Delta^z)^{\dagger} = \Delta^{\bar{z}},$ and that if $z$ is a positive real number then $\Delta^z$ is a positive definite operator.
• We will show that the operators $\Delta^z$ are in fact the complex powers of the operator $\Delta.$
• We will conclude that the weakly analytic extensions of the maps $it \mapsto U(t) | \xi \rangle$ to $0 \leq \text{Re}(z) \leq 1$ are given by $\Delta^z | \xi \rangle$, and therefore that we have $U(t) = \Delta^{it}.$

Now, fix $z \in \mathbb{C}$ with $\text{Re}(z) \geq 0.$ We define a set $D_{\Delta^z}$ consisting of all vectors $|\xi\rangle$ for which $it \mapsto U(t) |\xi\rangle$ admits a weakly analytic/continuous continuation into the strip $0 \leq \text{Re}(w) \leq \text{Re}(z)$. (If this map can be analytically continued further, that's fine; it's none of our business.) The following figure may be helpful.

By the preceding arguments, the set $D_{\Delta^z}$ is dense in $\mathcal{H}.$ It is also easily seen to be a linear subspace of $\mathcal{H},$ since the sum of two weakly analytic functions is weakly analytic. For $|\xi \rangle \in D_{\Delta^z},$ we will denote by $w \mapsto |F_{\xi}(w) \rangle$ the weakly analytic extension of the map $it \mapsto U(t) |\xi\rangle.$ Note that if $|\xi\rangle$ is in both $D_{\Delta^{z_1}}$ and $D_{\Delta^{z_2}}$ for distinct values of $z$ with $\text{Re}(z_1) \neq \text{Re}(z_2)$, then there is no ambiguity in what we mean by $|F_{\xi}(w)\rangle$ for $w$ in the strips $0 \leq \text{Re}(w) \leq \text{Re}(z_1)$ or $0 \leq \text{Re}(w) \leq \text{Re}(z_2)$, due to the uniqueness of analytic continuations. The following figure may be helpful.

For $|\xi \rangle$ in $D_{\Delta^z},$ we define the vector $\Delta^z |\xi\rangle$ by $|F_{\xi}(z)\rangle$. This is well defined by the above comment about uniqueness of analytic continuations. It is also not hard to show that for each $z,$ the map $|\xi \rangle \mapsto \Delta^z |\xi\rangle$ is linear. This is because the maps $w \mapsto |F_{\xi_1}(w)\rangle + |F_{\xi_2}(w)\rangle$ and $w \mapsto |F_{\xi_1 + \xi_2}(w) \rangle$ are weakly analytic functions that agree on the imaginary axis, so they agree everywhere.

We would now like to argue that it makes sense to think of all the operators we have defined as being powers of the operator $\Delta.$ To show this, it will first be helpful to show that these operators satisfy an appropriate grouplike multiplication property. Namely, if $|\xi\rangle$ is in the domain of $\Delta^{z+w},$ then $|\xi\rangle$ is in the domain of $\Delta^w,$ $\Delta^w |\xi\rangle$ is in the domain of $\Delta^z,$ and we can write

$$\Delta^{z+w}|\xi\rangle = \Delta^{z} (\Delta^{w} |\xi\rangle).$$

The definitions given above make it obvious that if $|\xi\rangle$ is in the domain of $\Delta^{z+w}$ then it is in the domain of $\Delta^w,$ as $w$ lies in the strip bounded on the right by $\text{Re}(z+w).$ To show that $\Delta^w |\xi\rangle$ is in the domain of $\Delta^z,$ we must show that the map $it \mapsto U(t) |F_{\xi}(w)\rangle$ admits an analytic continuation to the strip bounded on the right by $\text{Re}(z).$ But this is clear; that analytic continuation is simply $\zeta \mapsto \Delta^{\zeta+w} |\xi\rangle.$ For fixed $w,$ we know this is analytic for $0 \leq \text{Re}(\zeta) \leq \text{Re}(z)$. All we need to show is that on the imaginary axis, we have

$$\Delta^{it+w}|\xi\rangle = U(t) |F_{\xi}(w) \rangle.$$

But that can be shown by showing that for fixed $t,$ these are both analytic continuations of the same function $is \mapsto U(s+t)|\xi\rangle.$

The following figure may be helpful.

Note that because $\Delta^{z} (\Delta^w |\xi\rangle)$ and $\Delta^{z+w} |\xi\rangle$ are analytic continuations of the same function, they are equal.

Alright, we're making a lot of progress. The last few steps will go by very quickly, but it may be helpful to take stock before proceeding. So far we've shown:

• Assume that $t \mapsto U(t)$ is a weakly continuous, group-like unitary flow.
• We will show that for a strip of any width, $0 \leq \text{Re}(z) \leq \alpha,$ there is a dense set of vectors $|\xi \rangle$ for which $U(t) |\xi\rangle$ admits a weakly continuous/analytic extension to $0 \leq \text{Re}(z) \leq \alpha$.
• For any such $|\xi \rangle$ and any $z$ in the strip $0 \leq \text{Re}(z) \leq \alpha,$ we will define the vector $\Delta^z |\xi \rangle$ by setting it equal to the corresponding weakly analytic continuation evaluated at $z.$
• We will show that the vector $\Delta^z |\xi \rangle$ is well defined (i.e., independent of the choice of $\alpha$), and that $\Delta^z$ is a linear operator.
• We will show that the operators $\Delta^z$ have an appropriate group-like property: $\Delta^{z+w} |\xi \rangle = \Delta^z (\Delta^w |\xi\rangle)$ whenever this equation makes sense. (Whether it makes sense depends on $z, w$ and $|\xi\rangle,$ due to domain issues for unbounded operators.)

Here's what we still need to do:

• We will show that the operators $\Delta^z$ are closed and satisfy $(\Delta^z)^{\dagger} = \Delta^{\bar{z}},$ and that if $z$ is a positive real number then $\Delta^z$ is a positive definite operator.
• We will show that the operators $\Delta^z$ are in fact the complex powers of the operator $\Delta.$
• We will conclude that the weakly analytic extensions of the maps $it \mapsto U(t) | \xi \rangle$ to $0 \leq \text{Re}(z) \leq 1$ are given by $\Delta^z | \xi \rangle$, and therefore that we have $U(t) = \Delta^{it}.$

Showing the identity $(\Delta^{z})^{\dagger} = \Delta^{\bar{z}}$ is sufficient to show the operators are closed, because as I mentioned in section 1, all adjoints are closed operators. To show $(\Delta^{z})^{\dagger} = \Delta^{\bar{z}}$, first note that by the definitions given above, the domains $D_{\Delta^z}$ and $D_{\Delta^{\bar{z}}}$ are the same, since $z$ and $\bar{z}$ have the same real part. Next note that for $|\xi\rangle,$ $|\eta\rangle$ in this common domain, we must have

$$\langle \Delta^{\bar{z}} \xi | \eta \rangle = \langle \xi | \Delta^z \eta \rangle,$$

since both of these can be written in terms of analytic functions that agree on the imaginary axis. So we have $D_{(\Delta^z)^{\dagger}} \supseteq D_{\Delta^{\bar{z}}},$ with the two operators agreeing on their common domain. All we need to show is that the domains are equal; i.e., that if a vector is in the domain of $(\Delta^z)^{\dagger},$ then it is in the domain of $\Delta^{\bar{z}}.$ Concretely, if $(\Delta^z)^{\dagger}$ can act on $|\eta\rangle,$ then we must show that the map $it \mapsto U(t) |\eta\rangle$ admits a weakly analytic extension to the strip $0 \leq \text{Re}(w) \leq \text{Re}(z).$

We will construct the extension using the Riesz representation theorem. Our extension is supposed to be a function $w \mapsto |F_{\eta}(w)\rangle,$ defined in the strip $0 \leq \text{Re}(w) \leq \text{Re}(z),$ that satisfies

$$|F_{\eta}(it)\rangle = U(t) |\eta\rangle$$

and

$$\langle F_{\eta}(\bar{w}) | \xi \rangle = \langle \eta | \Delta^{w} \xi \rangle.$$

The Riesz representation theorem guarantees the existence and uniqueness of such a vector $|F_{\eta}(\bar{w})\rangle$ if the map $|\xi \rangle \mapsto \langle \eta | \Delta^{w} \xi \rangle$ is bounded.

For $w = x + i y$ in the strip $0 \leq \text{Re}(w) \leq \text{Re}(z),$ we have

$$|\langle \eta | \Delta^{w} \xi \rangle| = |\langle \eta | U(y) \Delta^{x} \xi \rangle| \leq \lVert |\eta\rangle \rVert \lVert | F_{\xi}(x) \rVert.$$

Since $x$ lies in the compact range $0 \leq x \leq \text{Re}(z),$ we conclude that the function $w \mapsto \langle \eta | \Delta^{w} \xi \rangle$ is uniformly bounded on the strip. By the Phragmen-Lindelof principle, its maximum is attained on the boundary of the strip. So for any $w$ in the strip, we have

$$|\langle \eta | \Delta^{w} \xi \rangle| \leq \lVert |\xi\rangle \rVert \text{max}\left\{ \lVert |\eta\rangle \rVert, \lVert (\Delta^{z})^{\dagger} |\eta\rangle \rVert \right\},$$

where I have used the assumption that $|\eta\rangle$ is in the domain of $(\Delta^{z})^{\dagger}.$ We conclude that the map $|\xi\rangle \mapsto \langle \eta | \Delta^w \xi \rangle$ is bounded, so by the Riesz theorem there exists a vector $|F_{\eta}(\bar{w})\rangle$ satisfying

$$\langle F_{\eta}(\bar{w}) \eta | \xi \rangle = \langle \eta | \Delta^{w} \xi\rangle.$$

So $|F_{\eta}(w)\rangle$ furnishes some extension of $it \mapsto U(t) |\eta\rangle$, and weak continuity/analyticity in the strip are easy to show using the above formulas. We conclude $(\Delta^{z})^{\dagger} = \Delta^{\bar{z}}.$

Finally, we must show that when $z$ is real, the operator $\Delta^z$ is positive. But this follows from self-adjointness and the group multiplication laws we showed earlier. Given $x \geq 0$ and $|\xi\rangle$ in the domain of $\Delta^x$, we have $|\xi\rangle \in D_{\Delta^{x/2}},$ and

$$\langle \xi | \Delta | \xi \rangle = \langle \xi | \Delta^{x/2} \Delta^{x/2} | \xi \rangle = \langle \Delta^{x/2} \xi| \Delta^{x/2} \xi \rangle \geq 0.$$

To see that $\Delta^z$ has trivial kernel, note that if we have $\Delta^{z} |\xi \rangle = 0,$ then we have $\Delta^{z+it} |\xi \rangle = U(t) \Delta^z |\xi \rangle = 0,$ so $w \mapsto \Delta^w | \xi \rangle$ vanishes on the full vertical line $\{z + i t\}_{t \in \mathbb{R}},$ and by analyticity it vanishes everywhere. Evaluating at $w=0$ gives $|\xi \rangle = 0$.

We are now ready to conclude by showing that the operators $\Delta^{z}$ are in fact the complex powers of the operator $\Delta,$ at least for $0 \leq \text{Re}(z) \leq 1.$ Let's change our notation to make this argument clear. We will now use the notation $\Delta^z$ to denote an actual power of $\Delta,$ and the notation $\Delta_z |\xi \rangle$ to denote what we have previously been calling the vector $\Delta^{z} |\xi\rangle.$ We must show that in the strip $0 \leq \text{Re}(z) \leq 1,$ we have

$$\Delta^{z} |\xi\rangle = \Delta_z |\xi \rangle.$$

By analyticity, it suffices to show this for a set of points $z$ in the strip that accumulate. But using the group multiplication laws we have already shown, it is easy to show this for a dense set of rational numbers in the range $0 \leq z \leq 1$ by subdividing this interval into smaller and smaller fractions. We conclude that for any $|\xi\rangle$, we have $\Delta^{it} |\xi\rangle = U(t) |\xi\rangle.$ This completes the proof.