# Unbounded operators in a Hilbert space and the Trotter product formula

Jordan Bell
August 25, 2015

## 1 Unbounded operators

Let $H$ be a Hilbert space with inner product $\left\langle\cdot,\cdot\right\rangle$. We do not assume that $H$ is separable. By an operator in $H$ we mean a linear subspace $\mathscr{D}(T)$ of $H$ and a linear map $T:\mathscr{D}(T)\to H$. We define

 $\mathscr{R}(T)=\{Tx:x\in\mathscr{D}(T)\}.$

If $\mathscr{D}(T)$ is dense in $H$ we say that $T$ is densely defined.

Write

 $\mathscr{G}(T)=\{(x,y)\in H\times H:x\in\mathscr{D}(T),y=Tx\}.$

When $\mathscr{G}(T)\subset\mathscr{G}(S)$, we write

 $T\subset S,$

and say that $S$ is an extension of $T$. If $\mathscr{G}(T)$ is a closed linear subspace of $H\times H$, we say that $T$ is closed.

We say that an operator $T$ in $H$ is closable if there is a closed operator $S$ in $H$ such that $T\subset S$. If $T$ is closable, one proves that there is a unique closed operator $\overline{T}$ in $H$ with $T\subset\overline{T}$ and such that if $S$ is a closed operator satisfying $T\subset S$ then $\overline{T}\subset S$.

Suppose that $T$ is a densely defined operator in $H$. We define $\mathscr{D}(T^{*})$ to be the set of those $y\in H$ for which

 $x\mapsto\left\langle Tx,y\right\rangle,\qquad x\in\mathscr{D}(T),$

is continuous. For $y\in\mathscr{D}(T^{*})$, by the Hahn-Banach theorem there is some $\lambda_{y}\in H^{*}$ such that

 $\lambda_{y}x=\left\langle Tx,y\right\rangle,\qquad x\in\mathscr{D}(T).$

Next, by the Riesz representation theorem, there is a unique $x_{y}\in H$ such that

 $\lambda_{y}x=\left\langle x,x_{y}\right\rangle,\qquad x\in H,$

and hence

 $\left\langle x,x_{y}\right\rangle=\left\langle Tx,y\right\rangle,\qquad x\in% \mathscr{D}(T).$

If $v\in H$ satisfies

 $\left\langle x,v\right\rangle=\left\langle Tx,y\right\rangle,\qquad x\in% \mathscr{D}(T),$

then

 $\left\langle x,v\right\rangle=\left\langle x,x_{y}\right\rangle,\qquad x\in% \mathscr{D}(T),$

and because $\mathscr{D}(T)$ is dense in $H$ this implies that $v=x_{y}$. We define $T^{*}:\mathscr{D}(T^{*})\to H$ by $T^{*}y=x_{y}$, which satisfies

 $\left\langle Tx,y\right\rangle=\left\langle x,T^{*}y\right\rangle,\qquad x\in% \mathscr{D}(T).$

$T^{*}$ is called the adjoint of $T$. One checks that $\mathscr{D}(T^{*})$ is a linear subspace of $H$ and that $T^{*}:\mathscr{D}(T^{*})\to H$ is a linear map. We say that $T$ is self-adjoint when $T=T^{*}$.

For operators $S$ and $T$ in $H$ we define

 $\mathscr{D}(S+T)=\mathscr{D}(S)\cap\mathscr{D}(T)$

and

 $\mathscr{D}(ST)=\{x\in\mathscr{D}(T):Tx\in\mathscr{D}(S)\}.$

One checks that

 $(R+S)+T=R+(S+T),\qquad(RS)T=R(ST),$

and

 $RT+ST=(R+S)T,\qquad TR+TS\subset T(R+S).$

We now determine the adjoint of products of densely defined operators.11 1 Walter Rudin, Functional Analysis, second ed., p. 348, Theorem 13.2.

###### Theorem 1.

If $S$, $T$, and $ST$ are densely defined operators in $H$, then

 $T^{*}S^{*}\subset(ST)^{*}.$

If $S\in\mathscr{B}(H)$, then

 $T^{*}S^{*}=(ST)^{*}.$
###### Proof.

Let $y\in\mathscr{D}(T^{*}S^{*})$ and let $x\in\mathscr{D}(ST)$. Then $S^{*}y\in\mathscr{D}(T^{*})$ and $x\in\mathscr{D}(T)$, so

 $\left\langle Tx,S^{*}y\right\rangle=\left\langle x,T^{*}S^{*}y\right\rangle.$

On the other hand, $y\in\mathscr{D}(S^{*})$, so

 $\left\langle STx,y\right\rangle=\left\langle Tx,S^{*}y\right\rangle.$

Hence

 $\left\langle STx,y\right\rangle=\left\langle x,T^{*}S^{*}y\right\rangle,$

which implies that $(ST)^{*}y=T^{*}S^{*}y$ for each $y\in\mathscr{D}(T^{*}S^{*})$, that is, $T^{*}S^{*}\subset(ST)^{*}$.

Suppose that $S\in\mathscr{B}(H)$, hence $S^{*}\in\mathscr{B}(H)$, for which $\mathscr{D}(S^{*})=H$. Let $y\in\mathscr{D}((ST)^{*})$. For $x\in\mathscr{D}(ST)$,

 $\left\langle Tx,S^{*}y\right\rangle=\left\langle STx,y\right\rangle=\left% \langle x,(ST)^{*}y\right\rangle.$

This implies that $S^{*}y\in\mathscr{D}(T^{*})$ and hence $y\in\mathscr{D}(T^{*}S^{*})$, showing

 $\mathscr{D}((ST)^{*})\subset\mathscr{D}(T^{*}S^{*}).$

If $T$ is an operator in $H$, we say that $T$ is symmetric if

 $\left\langle Tx,y\right\rangle=\left\langle x,Ty\right\rangle,\qquad x,y\in% \mathscr{D}(T).$
###### Theorem 2.

Let $T$ be a densely defined operator in $H$. $T$ is symmetric if and only if $T\subset T^{*}$.

###### Proof.

Suppose that $T$ is symmetric and let $(y,Ty)\in\mathscr{G}(T)$. For $x\in\mathscr{D}(T)$,

 $|\left\langle Tx,y\right\rangle|=|\left\langle x,Ty\right\rangle|\leq\left\|x% \right\|\left\|Ty\right\|,$

hence $x\mapsto\left\langle Tx,y\right\rangle$ is continuous on $\mathscr{D}(T)$, i.e. $y\in\mathscr{D}(T^{*})$. For $x\in\mathscr{D}(T)$, on the one hand,

 $\left\langle Tx,y\right\rangle=\left\langle x,T^{*}y\right\rangle,$

and on the other hand,

 $\left\langle Tx,y\right\rangle=\left\langle x,Ty\right\rangle.$

Therefore $\left\langle x,T^{*}y\right\rangle=\left\langle x,Ty\right\rangle$ for all $x\in\mathscr{D}(T)$, and because $\mathscr{D}(T)$ is dense in $H$ we get that $T^{*}y=Ty$, i.e. $(y,Ty)\in\mathscr{G}(T^{*})$. Therefore $\mathscr{G}(T)\subset\mathscr{G}(T^{*})$.

Suppose that $\mathscr{G}(T)\subset\mathscr{G}(T^{*})$. Let $x,y\in\mathscr{D}(T)$. We have $(y,Ty)\in\mathscr{G}(T^{*})$, i.e. $y\in\mathscr{D}(T^{*})$ and $T^{*}y=Ty$. Hence

 $\left\langle Tx,y\right\rangle=\left\langle x,T^{*}y\right\rangle=\left\langle x% ,Ty\right\rangle,$

showing that $T$ is symmetric. ∎

One proves that if $T$ is a symmetric operator in $H$ then $T$ is closable and $\overline{T}$ is symmetric. An operator $T$ in $H$ is said to be essentially self-adjoint when $T$ is densely defined, symmetric, and $\overline{T}$ (which is densely defined) is self-adjoint.

## 2 Graphs

For $(a,b),(c,d)\in H\times H$, we define

 $\left\langle(a,b),(c,d)\right\rangle=\left\langle a,c\right\rangle+\left% \langle b,d\right\rangle.$

This is an inner product on $H\times H$ with which $H\times H$ is a Hilbert space. We define $V:H\times H\to H\times H$ by

 $V(a,b)=(-b,a),\qquad(a,b)\in H\times H,$

which belongs to $\mathscr{B}(H\times H)$. It is immediate that $VV^{*}=I$ and $V^{*}V=I$, namely, $V$ is unitary. As well, $V^{2}=-I$, whence if $M$ is a linear subspace of $H\times H$ then $V^{2}M=M$. The following theorem relates the graphs of a densely defined operator and its adjoint.22 2 Walter Rudin, Functional Analysis, second ed., p. 352, Theorem 13.8.

###### Theorem 3.

Suppose that $T$ is a densely defined operator in $H$. It holds that

 $\mathscr{G}(T^{*})=\left(V\mathscr{G}(T)\right)^{\perp}.$
###### Theorem 4.

If $T$ is a densely defined operator in $H$, then $T^{*}$ is a closed operator.

###### Proof.

$V\mathscr{G}(T)$ is a linear subspace of $H\times H$. The orthogonal complement of a linear subspace of a Hilbert space is a closed linear subspace of the Hilbert space, and thus Theorem 3 tells us that $\mathscr{G}(T^{*})$ is a closed linear subspace of $H\times H$, namely, $T^{*}$ is a closed operator. ∎

Let $T$ be a densely defined operator in $H$. If $T$ is self-adjoint, then the above theorem tells us that $T$ is itself a closed operator.

###### Theorem 5.

Suppose that $T$ is a closed densely defined operator in $H$. Then

 $H\times H=V\mathscr{G}(T)\oplus\mathscr{G}(T^{*})$

is an orthogonal direct sum.

###### Proof.

Generally, if $M$ is a linear subspace of $H\times H$,

 $H\times H=\overline{M}\oplus M^{\perp}=\overline{M}\oplus(\overline{M})^{\perp}$

is an orthogonal direct sum. For $M=V\mathscr{G}(T)$, because $\mathscr{G}(T)$ is a closed linear subspace of $H\times H$, so is $M$. Thus

 $H\times H=V\mathscr{G}(T)\oplus(V\mathscr{G}(T))^{\perp}.$

By Theorem 3, this is

 $H\times H=V\mathscr{G}(T)\oplus\mathscr{G}(T^{*}),$

proving the claim. ∎

If $T$ is an operator in $H$ that is one-to-one, we define $\mathscr{D}(T^{-1})=\mathscr{R}(T)$, and $T^{-1}$ is a densely defined operator with domain $\mathscr{D}(T^{-1})$.

The following theorem establishes several properties of symmetric densely defined operators.33 3 Walter Rudin, Functional Analysis, second ed., p. 353, Theorem 13.11. We remind ourselves that if $T$ is an operator in $H$, the statement $\mathscr{D}(T)=H$ means that $T$ is a linear map $H\to H$, from which it does not follow that $T$ is continuous.

###### Theorem 6.

Suppose that $T$ is a densely defined symmetric operator in $H$. Then the following statements are true:

1. 1.

If $\mathscr{D}(T)=H$ then $T$ is self-adjoint and $T\in\mathscr{B}(H)$.

2. 2.

If $T$ is self-adjoint and one-to-one, then $\mathscr{R}(T)$ is dense in $H$ and $T^{-1}$ is densely defined and self-adjoint.

3. 3.

If $\mathscr{R}(T)$ is dense in $H$, then $T$ is one-to-one.

4. 4.

If $\mathscr{R}(T)=H$, then $T$ is self-adjoint and $T^{-1}\in\mathscr{B}(H)$.

If $T\in\mathscr{B}(H)$ then $T^{**}=T$. The following theorem says that this is true for closed densely defined operators.44 4 Walter Rudin, Functional Analysis, second ed., p. 354, Theorem 13.12.

###### Theorem 7.

If $T$ is a closed densely defined operator in $H$, then $\mathscr{D}(T^{*})$ is dense in $H$ and $T^{**}=T$.

The following theorem gives statements about $I+T^{*}T$ when $T$ is a closed densely defined operator.55 5 Walter Rudin, Functional Analysis, second ed., p. 354, Theorem 13.13.

###### Theorem 8.

Suppose that $T$ is a closed densely defined operator in $H$ and let $Q=I+T^{*}T$, with

 $\mathscr{D}(Q)=\mathscr{D}(T^{*}T)=\{x\in\mathscr{D}(T):Tx\in\mathscr{D}(T^{*}% )\}.$

The following statements are true:

1. 1.

$Q:\mathscr{D}(Q)\to H$ is a bijection, and there are $B,C\in\mathscr{B}(H)$ with $\left\|B\right\|\leq 1$, $B\geq 0$, $\left\|C\right\|\leq 1$, $C=TB$, and

 $B(I+T^{*}T)\subset(I+T^{*}T)B=I.$

$T^{*}T$ is self-adjoint.

2. 2.

Let $T_{0}$ be the restriction of $T$ to $\mathscr{D}(T^{*}T)$. Then $\mathscr{G}(T_{0})$ is dense in $\mathscr{G}(T)$.

Let $T$ be a symmetric operator in $H$. We say that $T$ is maximally symmetric if $T\subset S$ and $S$ being symmetric imply that $S=T$. One proves that a self-adjoint operator is maximally symmetric.66 6 Walter Rudin, Functional Analysis, second ed., p. 356, Theorem 13.15.

The following theorem is about $T+iI$ when $T$ is a symmetric operator in $H$.77 7 Walter Rudin, Functional Analysis, second ed., p. 356, Theorem 13.16.

###### Theorem 9.

Suppose that $T$ is a symmetric operator in $H$ and let $j$ be $i$ or $-i$. Then:

1. 1.

$\left\|Tx+jx\right\|^{2}=\left\|x\right\|^{2}+\left\|Tx\right\|^{2}$ for $x\in\mathscr{D}(T)$.

2. 2.

$T$ is closed if and only if $\mathscr{R}(T+jI)$ is a closed subset of $H$.

3. 3.

$T+jI$ is one-to-one.

4. 4.

If $\mathscr{R}(T+jI)=H$ then $T$ is maximally symmetric.

## 3 The Cayley transform

Let $T$ be a symmetric operator in $H$ and define

 $\mathscr{D}(U)=\mathscr{R}(T+iI).$

Theorem 9 tells us that $T+iI$ is one-to-one. Because

 $\mathscr{D}(T-iI)=\mathscr{D}(T)=\mathscr{D}(T-iI)$

and $\mathscr{D}((T+iI)^{-1})=\mathscr{R}(T+iI)$,

 $\displaystyle\mathscr{D}((T-iI)(T+iI)^{-1})$ $\displaystyle=\{x\in\mathscr{R}(T+iI):(T+iI)^{-1}x\in\mathscr{D}(T)\}$ $\displaystyle=\{x\in\mathscr{R}(T+iI):(T+iI)^{-1}x\in\mathscr{D}(T+iI)\}$ $\displaystyle=\mathscr{R}(T+iI)$ $\displaystyle=\mathscr{D}(U).$

We define

 $U=(T-iI)(T+iI)^{-1}.$

$U$ is called the Cayley transform of $T$.

We have

 $\mathscr{R}(U)=U\mathscr{D}(U)=U\mathscr{R}(T+iI)=(T-iI)(T+iI)^{-1}\mathscr{R}% (T+iI)=(T-iI)\mathscr{D}(T+iI),$

and $\mathscr{D}(T+iI)=\mathscr{D}(T)=\mathscr{D}(T-iI)$ so

 $\mathscr{R}(U)=(T-iI)\mathscr{D}(T-iI)=\mathscr{R}(T-iI).$

Also, for $x\in\mathscr{D}(T)$, Theorem 9 tells us

 $\left\|(T+iI)x\right\|^{2}=\left\|Tx+ix\right\|^{2}=\left\|x\right\|^{2}+\left% \|Tx\right\|^{2}=\left\|Tx-ix\right\|^{2}=\left\|(T-iI)x\right\|^{2},$

hence for $x\in\mathscr{D}(U)$, for which $(T+iI)^{-1}x\in\mathscr{D}(T+iI)=\mathscr{D}(T)$,

 $\left\|Ux\right\|=\left\|(T-iI)(T+iI)^{-1}x\right\|=\left\|(T+iI)(T+iI)^{-1}x% \right\|=\left\|x\right\|,$

showing that $U$ is an isometry in $H$.

The Cayley transform of a symmetric operator in $H$ (which we do not presume to be densely defined) has the following properties.88 8 Walter Rudin, Functional Analysis, second ed., p. 385, Theorem 13.19.

###### Theorem 10.

Suppose that $T$ is a symmetric operator in $H$. Then:

1. 1.

$U$ is closed if and only if $T$ is closed.

2. 2.

$\mathscr{R}(I-U)=\mathscr{D}(T)$, $I-U$ is one-to-one, and

 $T=i(I+U)(I-U)^{-1}.$
3. 3.

$U$ is unitary if and only if $T$ is self-adjoint.

If $V$ is an operator in $H$ that is an isometry and $I-V$ is one-to-one, then there is a symmetric operator $S$ in $H$ such that $V$ is the Cayley transform of $S$.

## 4 Resolvents

Let $T$ be an operator in $H$. The resolvent set of $T$, denoted $\rho(T)$, is the set of those $\lambda\in\mathbb{C}$ such that $T-\lambda I:\mathscr{D}(T)\to H$ is a bijection and $(T-\lambda I)^{-1}\in\mathscr{B}(H)$. That is, $\lambda\in\rho(T)$ if and only if there is some $S\in\mathscr{B}(H)$ such that

 $S(T-\lambda I)\subset(T-\lambda I)S=I.$

We call $R:\rho(T)\to\mathscr{B}(H)$ defined by

 $R(\lambda)=(T-\lambda I)^{-1}$

the resolvent of $T$. The spectrum of $T$ is $\sigma(T)=\mathbb{C}\setminus\rho(T)$. It is a fact that $\rho(T)$ is open, that $\sigma(T)$ is closed, and that if $\sigma(T)\neq\mathbb{C}$ then $T$ is a closed operator, that

 $R(z)-R(w)=(z-w)R(z)R(w),\qquad z,w\in\rho(T),$

and

 $\frac{d^{n}R}{dz^{n}}(z)=n!R^{n+1}(z),\qquad z\in\rho(T).$

If $T$ is a self-adjoint operator in $H$, one proves that $\sigma(T)\subset\mathbb{R}$.

## 5 Resolutions of the identity

Let $(\Omega,\mathscr{S})$ be a measurable space. A resolution of the identity is a function

 $E:\mathscr{S}\to\mathscr{B}(H)$

satisfying:

1. 1.

$E(\emptyset)=0$, $E(\Omega)=I$.

2. 2.

For each $a\in\mathscr{S}$, $E(a)$ is a self-adjoint projection.

3. 3.

$E(a\cap b)=E(a)E(b)$.

4. 4.

If $a\cap b=\emptyset$, then $E(a\cup b)=E(a)+E(b)$.

5. 5.

For each $x,y\in H$, the function $E_{x,y}:\mathscr{S}\to\mathbb{C}$ defined by

 $E_{x,y}(a)=\left\langle E(a)x,y\right\rangle,\qquad a\in\mathscr{S},$

is a complex measure on $\mathscr{S}$.

We check that if $a_{n}\in\mathscr{S}$ and $E(a_{n})=0$ for each $n=1,2,\ldots$, then for $a=\bigcup_{n=1}^{\infty}a_{n}$, $E(a)=0$.

Let $\{D_{i}\}$ be a countable collection of open discs that is a base for the topology of $\mathbb{C}$, i.e., $\bigcup D_{i}=\mathbb{C}$ and for each $i,j$ and for $z\in D_{i}\cap D_{j}$, there is some $k$ such that $x\in D_{k}\subset D_{i}\cap D_{j}$. Let $f:(\Omega,\mathscr{S})\to(\mathbb{C},\mathscr{B}_{\mathbb{C}})$ be a measurable function and let $V$ be the union of those $D_{i}$ for which $E(f^{-1}(D_{i}))=0$. Then $E(f^{-1}(V))=0$. The essential range of $f$ is $\mathbb{C}\setminus V$, and we say that $f$ is essentially bounded if the essential range of $f$ is a bounded subset of $\mathbb{C}$. We define the essential supremum of $f$ to be

 $\left\|f\right\|_{\infty}=\sup\{|\lambda|:\lambda\in\mathbb{C}\setminus V\}.$

Now define $B$ to be the collection of bounded measurable functions $(\Omega,\mathscr{S})\to(\mathbb{C},\mathscr{B}_{\mathbb{C}})$, which is a Banach algebra with the norm

 $\sup\{|f(\omega):\omega\in\Omega\},$

for which

 $N=\{f\in B:\left\|f\right\|_{\infty}=0\}$

is a closed ideal. Then $B/N$ is a Banach algebra, denoted $L^{\infty}(E)$, with the norm

 $\left\|f+N\right\|_{\infty}=\left\|f\right\|_{\infty}.$

The unity of $L^{\infty}(E)$ is $1+N$. Because $L^{\infty}(E)$ is a Banach algebra, it makes sense to speak about the spectrum of an element of $L^{\infty}(E)$. For $f+N\in L^{\infty}(E)$, the spectrum of $f+N$ is the set of those $\lambda\in\mathbb{C}$ for which there is no $g+N\in L^{\infty}(E)$ satisfying $(g+N)(f+N-\lambda(1+N))=1+N$. Check that the spectrum of $f+N$ is equal to the essential range of $g$, for any $g\in f+N$.

A subset $A$ of $\mathscr{B}(H)$ is said to be normal when $ST=TS$ for all $S,T\in A$ and $T\in A$ implies that $T^{*}\in A$.99 9 Walter Rudin, Functional Analysis, second ed., p. 319, Theorem 12.21. (To say that $T\in\mathscr{B}(H)$ is normal means that $TT^{*}=T^{*}T$, and this is equivalent to the statement that the set $\{T,T^{*}\}$ is normal.)

###### Theorem 11.

If $(\Omega,\mathscr{S})$ is a measurable space and $E:\mathscr{S}\to H$ is a resolution of the identity, then there is a closed normal subalgebra $A$ of $\mathscr{B}(H)$ and a unique isometric ${}^{*}$-isomorphism $\Psi:L^{\infty}(E)\to A$ such that

 $\left\langle\Psi(f)x,y\right\rangle=\int_{\Omega}fdE_{x,y},\qquad f\in L^{% \infty}(E),\quad x,y\in H.$

Furthermore,

 $\left\|\Psi(f)x\right\|^{2}=\int_{\Omega}|f|^{2}dE_{x,x},\qquad f\in L^{\infty% }(E),\quad x\in H.$

For $f\in L^{\infty}(E)$, we define

 $\int_{\Omega}fdE=\Psi(f).$

For $L^{\infty}(E)$, $\sigma(\Psi(f))$ is equal to the essential range of $f$.1010 10 Walter Rudin, Functional Analysis, second ed., p. 366, Theorem 13.27.

## 6 The spectral theorem

The following is the spectral theorem for self-adjoint operators.1111 11 Walter Rudin, Functional Analysis, second ed., p. 368, Theorem 13.30.

###### Theorem 12.

If $T$ is a self-adjoint operator in $H$, then there is a unique resolution of the identity

 $E:\mathscr{B}_{\mathbb{R}}\to\mathscr{B}(H)$

such that

 $\left\langle Tx,y\right\rangle=\int_{\mathbb{R}}\lambda dE_{x,y}(\lambda),% \qquad x\in\mathscr{D}(T),\quad y\in H.$

This resolution of the identity satisfies $E(\sigma(T))=I$.

If $T$ is a self-adjoint operator in $H$ applying the spectral theorem and then Theorem 11, we get that there is a closed normal subalgebra $A$ of $\mathscr{B}(H)$ and a unique isometric ${}^{*}$-isomorphism $\Psi:L^{\infty}(E)\to A$ such that

 $\left\langle\Psi(f)x,y\right\rangle=\int_{\sigma(T)}f(\lambda)dE_{x,y}(\lambda% ),\qquad f\in L^{\infty}(E),\quad x,y\in H.$

For $t\in\mathbb{R}$ and $f_{t}:\sigma(T)\to\mathbb{C}$ defined by $f_{t}(\lambda)=e^{it\lambda}$, this defines

 $e^{itT}=\Psi(f_{t})=\int_{\sigma(T)}e^{it\lambda}dE(\lambda).$

Because $\Psi$ is a ${}^{*}$-homomorphism, for $t\in\mathbb{R}$ we have

 $\Psi(f_{t})^{*}\Psi(f_{t})=\Psi(\overline{f_{t}})\Psi(f_{t})=\Psi(f_{-t})\Psi(% f_{t})=\Psi(f_{-t}f_{t})=\Psi(f_{0})=I,$

and likewise $\Psi(f_{t})\Psi(f_{t})^{*}=I$, showing that $e^{itT}=\Psi(f_{t})$ is unitary. We denote by $\mathscr{U}(H)$ the collection of unitary elements of $\mathscr{B}(H)$. $\mathscr{U}(H)$ is a subgroup of the group of invertible elements of $\mathscr{B}(H)$.

Furthermore, because $\Psi$ is a ${}^{*}$-homomorphism, for $t\in\mathbb{R}$ we have

 $I=\Psi(f_{0})=\Psi(f_{t}f_{-t})=\Psi(f_{t})\Psi(f_{-t})=e^{itT}e^{i(-t)T},$

and for $s,t\in\mathbb{R}$ we have

 $e^{isT}e^{itT}=\Psi(f_{s})\Psi(f_{t})=\Psi(f_{s}f_{t})=\Psi(f_{s+t})=e^{i(s+t)% T},$

showing that $t\mapsto e^{itT}$ is a one-parameter group $\mathbb{R}\to\mathscr{B}(H)$.

For $t\in\mathbb{R}$ and $x\in H$, by Theorem 11 we have

 $\left\|\Psi_{t}x-x\right\|^{2}=\left\|\Psi(f_{t}-1)x\right\|^{2}=\int_{\sigma(% T)}|f_{t}-1|^{2}dE_{x,x}=\int_{\sigma(T)}|e^{it\lambda}-1|^{2}dE_{x,x}(\lambda).$

For each $\lambda\in\sigma(T)$, $|e^{it\lambda}-1|^{2}\to 0$ as $t\to 0$, and thus we get by the dominated convergence theorem

 $\int_{\sigma(T)}|e^{it\lambda}-1|^{2}dE_{x,x}(\lambda)\to 0,\qquad t\to 0.$

That is, for each $x\in H$,

 $\left\|e^{itT}x-x\right\|\to 0$

as $t\to 0$, showing that $t\mapsto e^{itT}$ is strongly continuous, i.e. $t\mapsto e^{itT}$ is continuous $\mathbb{R}\to\mathscr{B}(H)$ where $\mathscr{B}(H)$ has the strong operator topology.

Conversely, Stone’s theorem on one-parameter unitary groups1212 12 cf. Walter Rudin, Functional Analysis, second ed., p. 382, Theorem 38. states that if $\{U_{t}:t\in\mathbb{R}\}$ is a strongly continuous one-parameter group of bounded unitary operators on $H$, then there is a unique self-adjoint operator $A$ in $H$ such that $U_{t}=e^{itA}$ for each $t\in\mathbb{R}$.

For $t\neq 0$, define $g_{t}:\sigma(T)\to\mathbb{C}$ by $g_{t}(\lambda)=\frac{e^{it\lambda}-1}{t}$. By Theorem 12, for $x\in\mathscr{D}(T)$ and $y\in H$,

 $\left\langle iTx,y\right\rangle=i\left\langle Tx,y\right\rangle=i\int_{\mathbb% {R}}\lambda dE_{x,y}(\lambda)$

and by Theorem 11,

 $\left\langle\Psi(g_{t})x,y\right\rangle=\int_{\sigma(T)}g_{t}dE_{x,y}=\int_{% \sigma(T)}\frac{e^{it\lambda}-1}{t}dE_{x,y}(\lambda),$

so

 $\left\langle\Psi(g_{t})x-iTx,y\right\rangle=\int_{\sigma(T)}\left(\frac{e^{it% \lambda}-1}{t}-i\lambda\right)dE_{x,y}(\lambda).$

For each $\lambda\in\sigma(T)$, $\frac{e^{it\lambda}-1}{t}-i\lambda\to 0$ as $t\to 0$, and for each $t$,

 $\left|\frac{e^{it\lambda}-1}{t}-i\lambda\right|\leq\left|\frac{e^{it\lambda}-1% }{t}\right|+|\lambda|\leq 2|\lambda|,$

and as $x\in\mathscr{D}(T)$, by Theorem 12 we have that $\lambda\mapsto|\lambda|$ belongs to $L^{1}(E_{x,y})$. Thus by the dominated convergence theorem,

 $\left\langle\Psi(g_{t})x-iTx,y\right\rangle=\int_{\sigma(T)}\left(\frac{e^{it% \lambda}-1}{t}-i\lambda\right)dE_{x,y}(\lambda)\to 0$

as $t\to 0$. In particular,

 $\left\|\Psi(g_{t})x-iTx\right\|^{2}\to 0$

as $t\to 0$. That is, for each $x\in\mathscr{D}(T)$,

 $\frac{e^{itT}x-x}{t}\to iTx$

as $t\to 0$. In other words, $iT$ is the infinitesimal generator of the one-parameter group $e^{itT}$.1313 13 cf. Walter Rudin, Functional Analysis, second ed., p. 376, Theorem 13.35. We remark that because $T^{*}=T$, the adjoint of $iT$ is $(iT)^{*}=\overline{i}T^{*}=-iT^{*}=-iT=-(iT)$.

## 7 Trotter product formula

We remind ourselves that for an operator $T$ in $H$ to be closed means that $\mathscr{G}(T)$ is a closed linear subspace of $H\times H$.

###### Theorem 13.

Let $T$ be an operator in $H$. $T$ is closed if and only if the linear space $\mathscr{D}(T)$ with the norm

 $\left\|x\right\|_{T}=\left\|x\right\|+\left\|Tx\right\|.$

is a Banach space.

The following is the Trotter product formula, which shows that if $A$, $B$, and $A+B$ are self-adjoint operators in a Hilbert space, then for each $t$, $(e^{itA/n}e^{itB/n})^{n}$ converges strongly to $e^{it(A+B)}$ as $n\to\infty$.1414 14 Barry Simon, Functional Integration and Quantum Physics, p. 4, Theorem 1.1; Konrad Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, p. 122, Theorem 6.4.

###### Theorem 14.

Let $H$ be a Hilbert space, not necessarily separable. If $A$ and $B$ are self-adjoint operators in $H$ such that $A+B$ is a self-adjoint operator in $H$, then for each $t\in\mathbb{R}$ and for each $\psi\in H$,

 $e^{it(A+B)}\psi=\lim_{n\to\infty}\left((e^{itA/n}e^{itb/n})^{n}\psi\right).$
###### Proof.

The claim is immediate for $t=0$, and we prove the claim for $t>0$; it is straightforward to obtain the claim for $t<0$ using the truth of the claim for $t>0$. Let $D=\mathscr{D}(A+B)=\mathscr{D}(A)\cap\mathscr{D}(B)$. Because $A+B$ is self-adjoint, $A+B$ is closed (Theorem 4), so by Theorem 13, the linear space $D$ with the norm $\left\|\phi\right\|_{A+B}=\left\|\phi\right\|+\left\|(A+B)\phi\right\|$ is a Banach space. Because $D$ is a Banach space, the uniform boundedness principle1515 15 Walter Rudin, Functional Analysis, second ed., p. 45, Theorem 2.6. tells us that if $\Gamma$ is a collection of bounded linear maps $D\to H$ and if for each $\phi\in D$ the set $\{\gamma\phi:\gamma\in\Gamma\}$ is bounded in $H$, then the set $\{\left\|\gamma\right\|:\gamma\in\Gamma\}$ is bounded, i.e. there is some $C$ such that $\left\|\gamma\phi\right\|\leq C\left\|\phi\right\|_{A+B}$ for all $\gamma\in\Gamma$ and all $\phi\in D$.

For $s\in\mathbb{R}$, let $S_{s}=e^{is(A+B)}$, $V_{s}=e^{isA}$, $W_{s}=e^{isB}$, $U_{s}=V_{s}W_{s}$, which each belong to $\mathscr{B}(H)$. For $n\geq 1$,

 $\sum_{j=0}^{n-1}U_{t/n}^{j}(S_{t/n}-U_{t/n})S_{t/n}^{n-j-1}=U_{t/n}^{n}-S_{t/n% }^{n}=U_{t/n}^{n}-S_{t},$

so, because a product of unitary operators is a unitary operator and a unitary operator has operator norm $1$ and also using the fact that $S_{t/n}^{n-j-1}=S_{t-\frac{j+1}{n}}$, for $\xi\in H$ we have

 $\displaystyle\left\|(S_{t}-U_{t/n}^{n})\xi\right\|$ $\displaystyle=\left\|\sum_{j=0}^{n-1}U_{t/n}^{j}(S_{t/n}-U_{t/n})S_{t/n}^{n-j-% 1}\xi\right\|$ $\displaystyle\leq\sum_{j=0}^{n-1}\left\|(S_{t/n}-U_{t/n})S_{t/n}^{n-j-1}\xi\right\|$ $\displaystyle=\sum_{j=0}^{n-1}\left\|(S_{t/n}-U_{t/n})S_{t-\frac{j+1}{n}}\xi\right\|$ $\displaystyle\leq\sum_{j=0}^{n-1}\sup_{0\leq s\leq t}\left\|(S_{t/n}-U_{t/n})S% _{s}\xi\right\|.$

That is,

 $\left\|(S_{t}-U_{t/n}^{n})\xi\right\|\leq n\sup_{0\leq s\leq t}\left\|(S_{t/n}% -U_{t/n})S_{s}\xi\right\|,\qquad\xi\in H,\quad n\geq 1.$ (1)

Let $\phi\in D$. On the one hand, because $i(A+B)$ is the infinitesimal generator of $\{S_{s}:s\in\mathbb{R}\}$, we have

 $\frac{S_{s}-I}{s}\phi\to i(A+B)\phi,\qquad s\downarrow 0.$ (2)

On the other hand, for $s\neq 0$ we have, because an infinitesimal generator of a one-parameter group commutes with each element of the one-parameter group,

 $V_{s}(iB\phi)+V_{s}\left(\frac{W_{s}-I}{s}-iB\right)\phi+\frac{V_{s}-I}{s}\phi% =\frac{U_{s}-I}{s}\phi,$

and as $V_{s}$ converges strongly to $I$ as $s\downarrow 0$ and as $iB$ is the infinitesimal generator of the one-parameter group $\{W_{s}:s\in\mathbb{R}\}$ and $iA$ is the infinitesimal generator of the one-parameter group $\{V_{s}:s\in\mathbb{R}\}$,

 $V_{s}(iB\phi)+V_{s}\left(\frac{W_{s}-I}{s}-iB\right)\phi+\frac{V_{s}-I}{s}\phi% \to iB\phi+iA\phi$

as $s\downarrow 0$, i.e.

 $\frac{U_{s}-I}{s}\phi\to i(A+B)\phi,\qquad s\downarrow 0.$ (3)

Using (2) and (3), we get that for each $\phi\in D$,

 $\frac{S_{s}-U_{s}}{s}\phi\to 0,\qquad s\downarrow 0.$

Therefore, for each $\phi\in D$, with $s=t/n$ we have

 $\frac{n}{t}(S_{t/n}-U_{t/n})\phi\to 0,\qquad n\to\infty,$

equivalently ($t$ is fixed for this whole theorem),

 $\lim_{n\to\infty}\left\|n(S_{t/n}-U_{t/n})\phi\right\|=0,\qquad\phi\in D.$ (4)

For each $n\geq 1$, define $\gamma_{n}:D\to H$ by $\gamma_{n}=n(S_{t/n}-U_{t/n})$. Each $\gamma_{n}$ is a linear map, and for $\phi\in D$,

 $\left\|\gamma_{n}\phi\right\|\leq n\left\|S_{t/n}\phi\right\|+n\left\|U_{t/n}% \phi\right\|\leq n\left\|\phi\right\|+n\left\|\phi\right\|\leq 2n\left\|\phi% \right\|_{A+B},$

showing that each $\gamma_{n}$ is a bounded linear map $D\to H$, where $D$ is a Banach space with the norm $\left\|\phi\right\|_{A+B}=\left\|\phi\right\|+\left\|(A+B)\phi\right\|$. Moreover, (4) shows that for each $\phi\in D$, there is some $C_{\phi}$ such that

 $\left\|\gamma_{n}\phi\right\|\leq C_{\phi},\qquad n\geq 1.$

Then applying the uniform boundedness principle, we get that there is some $C>0$ such that for all $n\geq 1$ and for all $\phi\in D$,

 $\left\|\gamma_{n}\phi\right\|\leq C\left\|\phi\right\|_{A+B},$

i.e.

 $\left\|n(S_{t/n}-U_{t/n})\phi\right\|\leq C\left\|\phi\right\|_{A+B},\qquad n% \geq 1,\quad\phi\in D.$ (5)

Let $K$ be a compact subset of $D$, where $D$ is a Banach space with the norm $\left\|\phi\right\|_{A+B}=\left\|\phi\right\|+\left\|(A+B)\phi\right\|$. Then $K$ is totally bounded, so for any $\epsilon>0$, there are $\phi_{1},\ldots,\phi_{M}\in K$ such that $K\subset\bigcup_{m=1}^{M}B_{\epsilon/C}(\phi_{m})$. By (4), for each $m$, $1\leq m\leq M$, there is some $n_{m}$ such that when $n\geq n_{m}$,

 $\left\|n(S_{t/n}-U_{t/n})\phi_{m}\right\|\leq\epsilon.$

Let $N=\max\{n_{1},\ldots,n_{M}\}$. For $n\geq N$ and for $\phi\in D$, there is some $m$ for which $\left\|\phi-\phi_{m}\right\|_{A+B}<\frac{\epsilon}{C}$, and using (5), as $\phi-\phi_{m}\in D$, we get

 $\displaystyle\left\|n(S_{t/n}-U_{t/n})\phi\right\|$ $\displaystyle\leq\left\|n(S_{t/n}-U_{t/n})(\phi-\phi_{m})\right\|+\left\|n(S_{% t/n}-U_{t/n})\phi_{m}\right\|$ $\displaystyle\leq C\left\|\phi-\phi_{m}\right\|_{A+B}+\epsilon$ $\displaystyle<\epsilon+\epsilon.$

This shows that any compact subset $K$ of $D$ and $\epsilon>0$, there is some $n_{\epsilon}$ such that if $n\geq n_{\epsilon}$ and $\phi\in K$, then

 $\left\|n(S_{t/n}-U_{t/n})\phi\right\|<\epsilon.$ (6)

Let $\phi\in D$, let $s_{0}\in\mathbb{R}$, and let $\epsilon>0$. Because $s\mapsto S_{s}$ is strongly continuous $\mathbb{R}\to\mathscr{B}(H)$, there is some $\delta_{1}>0$ such that when $|s-s_{0}|<\delta_{1}$, $\left\|S_{s}\phi-S_{s_{0}}\phi\right\|<\epsilon$, and there is some $\delta_{2}>0$ such that when $|s-s_{0}|<\delta_{2}$, $\left\|S_{s}(A+B)\phi-S_{s_{0}}(A+B)\phi\right\|<\epsilon$, and hence with $\delta=\min\{\delta_{1},\delta_{2}\}$, when $|s-s_{0}|<\delta$ we have

 $\displaystyle\left\|S_{s}\phi-S_{s_{0}}\phi\right\|_{A+B}$ $\displaystyle=\left\|S_{s}\phi-S_{s_{0}}\phi\right\|+\left\|(A+B)(S_{s}\phi-S_% {s_{0}}\phi)\right\|$ $\displaystyle=\left\|S_{s}\phi-S_{s_{0}}\phi\right\|+\left\|S_{s}(A+B)\phi-S_{% s_{0}}(A+B)\phi)\right\|$ $\displaystyle<\epsilon+\epsilon,$

showing that $s\mapsto S_{s}\phi$ is continuous $\mathbb{R}\to D$. Therefore $\{S_{s}\phi:0\leq s\leq t\}$ is a compact subset of $D$, so applying (6) we get that for any $\epsilon>0$, there is some $n_{\epsilon}$ such that if $n\geq n_{\epsilon}$ and $0\leq s\leq t$, then

 $\left\|n(S_{t/n}-U_{t/n})S_{s}\phi\right\|<\epsilon,$

and therefore if $n\geq n_{\epsilon}$ then

 $\sup_{0\leq s\leq t}\left\|n(S_{t/n}-U_{t/n})S_{s}\phi\right\|\leq\epsilon.$ (7)

Finally, let $\epsilon>0$. The statement that $A+B$ is self-adjoint in $H$ entails the statement that $D$ is dense in $H$, so there is some $\phi\in D$ such that $\left\|\phi-\psi\right\|<\epsilon$. For $n\geq 1$,

 $\displaystyle\left\|(S_{t}-U_{t/n}^{n})\psi\right\|$ $\displaystyle\leq\left\|(S_{t}-U_{t/n}^{n})(\psi-\phi)\right\|+\left\|(S_{t}-U% _{t/n}^{n})\phi\right\|$ $\displaystyle\leq 2\left\|\psi-\phi\right\|+\left\|(S_{t}-U_{t/n}^{n})\phi\right\|$ $\displaystyle<\epsilon+\left\|(S_{t}-U_{t/n}^{n})\phi\right\|.$

Using (1) with $\xi=\phi$ and then using (7), there is some $n_{\epsilon}$ such that when $n\geq n_{\epsilon}$,

 $\left\|(S_{t}-U_{t/n}^{n})\phi\right\|\leq n\sup_{0\leq s\leq t}\left\|(S_{t/n% }-U_{t/n})S_{s}\phi\right\|\leq\epsilon.$

Therefore for $n\geq n_{\epsilon}$,

 $\left\|(S_{t}-U_{t/n}^{n})\psi\right\|<2\epsilon,$

proving the claim. ∎