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:

- 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. - 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 $\Delta.$ - 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

**densely defined**.

**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.

*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

**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.

**positive**if we have $\langle x | T | x \rangle \geq 0$ for all $|x\rangle$ in $D_T.$

**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

- 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 $\bar{f}(T) = f(T)^{\dagger}.$

**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

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

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

**weak continuity**; we would say that hte map $t \mapsto \Delta^{it}$ is weakly continuous if $s \to t$ implies

*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.

## 3. Stone's theorem

- 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 $z \mapsto \Delta^{z} |\xi\rangle.$

- 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 $U(t) = \Delta^{it}.$

*does*exist a unique vector $|y_n\rangle$ satisfying

- 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 $U(t) = \Delta^{it}.$

- 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 $U(t) = \Delta^{it}.$

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