Unbounded operators in a Hilbert space and the Trotter product formula

1 Unbounded operators

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

$$ℛ(T)=\{Tx:x\in 𝒟(T)\}.$$

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

Write

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

When $𝒢(T)\subset 𝒢(S)$, we write

$$T\subset S,$$

and say that $S$ is an extension of $T$. If $𝒢(T)$ 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 $𝒟(T^{*})$ to be the set of those $y\in H$ for which

$$x\mapsto ⟨Tx,y⟩,x\in 𝒟(T),$$

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

$${\lambda }_{y}x=⟨Tx,y⟩,x\in 𝒟(T).$$

Next, by the Riesz representation theorem, there is a unique $x_y\in H$ such that

$${\lambda }_{y}x=⟨x,x_y⟩,x\in H,$$

and hence

$$⟨x,x_y⟩=⟨Tx,y⟩,x\in 𝒟(T).$$

If $v\in H$ satisfies

$$⟨x,v⟩=⟨Tx,y⟩,x\in 𝒟(T),$$

then

$$⟨x,v⟩=⟨x,x_y⟩,x\in 𝒟(T),$$

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

$$⟨Tx,y⟩=⟨x,T^{*}y⟩,x\in 𝒟(T).$$

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

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

$$𝒟(S+T)=𝒟(S)\cap 𝒟(T)$$

and

$$𝒟(ST)=\{x\in 𝒟(T):Tx\in 𝒟(S)\}.$$

One checks that

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

and

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

We now determine the adjoint of products of densely defined operators.

Theorem 1.

If $S$, $T$, and $S\mathit{}T$ are densely defined operators in $H$, then

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

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

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

Proof.

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

$$⟨Tx,S^{*}y⟩=⟨x,T^{*}{S}^{*}y⟩.$$

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

$$⟨STx,y⟩=⟨Tx,S^{*}y⟩.$$

Hence

$$⟨STx,y⟩=⟨x,T^{*}{S}^{*}y⟩,$$

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

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

$$⟨Tx,S^{*}y⟩=⟨STx,y⟩=⟨x,{(ST)}^{*}y⟩.$$

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

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

∎

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

$$⟨Tx,y⟩=⟨x,Ty⟩,x,y\in 𝒟(T).$$

Theorem 2.

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

Proof.

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

$$|⟨Tx,y⟩|=|⟨x,Ty⟩|\le \parallel x\parallel \parallel Ty\parallel ,$$

hence $x\mapsto ⟨Tx,y⟩$ is continuous on $𝒟(T)$, i.e. $y\in 𝒟(T^{*})$. For $x\in 𝒟(T)$, on the one hand,

$$⟨Tx,y⟩=⟨x,T^{*}y⟩,$$

and on the other hand,

$$⟨Tx,y⟩=⟨x,Ty⟩.$$

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

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

$$⟨Tx,y⟩=⟨x,T^{*}y⟩=⟨x,Ty⟩,$$

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

$$⟨(a,b),(c,d)⟩=⟨a,c⟩+⟨b,d⟩.$$

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

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

which belongs to $ℬ(H\times H)$. It is immediate that $V{V}^{*}=I$ 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.

Theorem 3.

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

$$𝒢(T^{*})={\left(V𝒢(T)\right)}^{⟂}.$$

Theorem 4.

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

Proof.

$V𝒢(T)$ 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 $𝒢(T^{*})$ 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

$$H\times H=V𝒢(T)\oplus 𝒢(T^{*})$$

is an orthogonal direct sum.

Proof.

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

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

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

$$H\times H=V𝒢(T)\oplus {(V𝒢(T))}^{⟂}.$$

By Theorem 3, this is

$$H\times H=V𝒢(T)\oplus 𝒢(T^{*}),$$

proving the claim. ∎

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

The following theorem establishes several properties of symmetric densely defined operators. We remind ourselves that if $T$ is an operator in $H$, the statement $𝒟(T)=H$ 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. If $𝒟(T)=H$ then $T$ is self-adjoint and $T\in ℬ(H)$.

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

3. If $ℛ(T)$ is dense in $H$, then $T$ is one-to-one.

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

If $T\in ℬ(H)$ then $T^{**}=T$. The following theorem says that this is true for closed densely defined operators.

Theorem 7.

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

The following theorem gives statements about $I+T^{*}T$ when $T$ is a closed densely defined operator.

Theorem 8.

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

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

The following statements are true:

1. $Q:𝒟(Q)\to H$ is a bijection, and there are $B,C\in ℬ(H)$ with $\parallel B\parallel \le 1$, $B\ge 0$, $\parallel C\parallel \le 1$, $C=TB$, and

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

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

2. Let $T_0$ be the restriction of $T$ to $𝒟(T^{*}T)$. Then $𝒢(T_0)$ is dense in $𝒢(T)$.

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.

The following theorem is about $T+iI$ when $T$ is a symmetric operator in $H$.

Theorem 9.

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

1. ${\parallel Tx+jx\parallel }^2={\parallel x\parallel }^2+{\parallel Tx\parallel }^2$ for $x\in 𝒟(T)$.

2. $T$ is closed if and only if $ℛ(T+jI)$ is a closed subset of $H$.

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

4. If $ℛ(T+jI)=H$ then $T$ is maximally symmetric.

3 The Cayley transform

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

$$𝒟(U)=ℛ(T+iI).$$

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

$$𝒟(T-iI)=𝒟(T)=𝒟(T-iI)$$

and $𝒟({(T+iI)}^{-1})=ℛ(T+iI)$,

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

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

$=ℛ(T+iI)$

$=𝒟(U).$

We define

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

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

We have

$$ℛ(U)=U𝒟(U)=Uℛ(T+iI)=(T-iI){(T+iI)}^{-1}ℛ(T+iI)=(T-iI)𝒟(T+iI),$$

and $𝒟(T+iI)=𝒟(T)=𝒟(T-iI)$ so

$$ℛ(U)=(T-iI)𝒟(T-iI)=ℛ(T-iI).$$

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

$${\parallel (T+iI)x\parallel }^2={\parallel Tx+ix\parallel }^2={\parallel x\parallel }^2+{\parallel Tx\parallel }^2={\parallel Tx-ix\parallel }^2={\parallel (T-iI)x\parallel }^2,$$

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

$$\parallel Ux\parallel =\parallel (T-iI){(T+iI)}^{-1}x\parallel =\parallel (T+iI){(T+iI)}^{-1}x\parallel =\parallel x\parallel ,$$

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.

Theorem 10.

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

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

2. $ℛ(I-U)=𝒟(T)$, $I-U$ is one-to-one, and

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

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

If $V$ is an operator in $H$ that is an isometry and $I\mathrm{-}V$ 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 $\lambda \in ℂ$ such that $T-\lambda I:𝒟(T)\to H$ is a bijection and ${(T-\lambda I)}^{-1}\in ℬ(H)$. That is, $\lambda \in \rho (T)$ if and only if there is some $S\in ℬ(H)$ such that

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

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

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

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

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

and

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

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

5 Resolutions of the identity

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

$$E:𝒮\to ℬ(H)$$

satisfying:

1. $E(\mathrm{\varnothing })=0$, $E(\mathrm{\Omega })=I$.

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

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

4. If $a\cap b=\mathrm{\varnothing }$, then $E(a\cup b)=E(a)+E(b)$.

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

   $$E_{x,y}(a)=⟨E(a)x,y⟩,a\in 𝒮,$$

   is a complex measure on $𝒮$.

We check that if $a_n\in 𝒮$ and $E(a_n)=0$ for each $n=1,2,\mathrm{\dots }$, then for $a={\bigcup }_{n=1}^{\mathrm{\infty }}{a}_n$, $E(a)=0$.

Let $\{D_i\}$ be a countable collection of open discs that is a base for the topology of $ℂ$, i.e., $\bigcup D_i=ℂ$ 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 $f:(\mathrm{\Omega },𝒮)\to (ℂ,{ℬ}_{ℂ})$ 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 $ℂ\setminus V$, and we say that $f$ is essentially bounded if the essential range of $f$ is a bounded subset of $ℂ$. We define the essential supremum of $f$ to be

$$\parallel f{\parallel }_{\mathrm{\infty }}=sup\{|\lambda |:\lambda \in ℂ\setminus V\}.$$

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

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

for which

$$N=\{f\in B:{\parallel f\parallel }_{\mathrm{\infty }}=0\}$$

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

$${\parallel f+N\parallel }_{\mathrm{\infty }}={\parallel f\parallel }_{\mathrm{\infty }}.$$

The unity of $L^{\mathrm{\infty }}(E)$ is $1+N$. Because $L^{\mathrm{\infty }}(E)$ is a Banach algebra, it makes sense to speak about the spectrum of an element of $L^{\mathrm{\infty }}(E)$. For $f+N\in L^{\mathrm{\infty }}(E)$, the spectrum of $f+N$ is the set of those $\lambda \in ℂ$ for which there is no $g+N\in L^{\mathrm{\infty }}(E)$ 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 $ℬ(H)$ is said to be normal when $ST=TS$ for all $S,T\in A$ and $T\in A$ implies that $T^{*}\in A$. (To say that $T\in ℬ(H)$ is normal means that $T{T}^{*}=T^{*}T$, and this is equivalent to the statement that the set $\{T,T^{*}\}$ is normal.)

Theorem 11.

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

$$⟨\mathrm{\Psi }(f)x,y⟩={\int }_{\mathrm{\Omega }}f𝑑E_{x,y},f\in L^{\mathrm{\infty }}(E),x,y\in H.$$

Furthermore,

$${\parallel \mathrm{\Psi }(f)x\parallel }^2={\int }_{\mathrm{\Omega }}{|f|}^2𝑑E_{x,x},f\in L^{\mathrm{\infty }}(E),x\in H.$$

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

$${\int }_{\mathrm{\Omega }}f𝑑E=\mathrm{\Psi }(f).$$

For $L^{\mathrm{\infty }}(E)$, $\sigma (\mathrm{\Psi }(f))$ is equal to the essential range of $f$.

6 The spectral theorem

The following is the spectral theorem for self-adjoint operators.

Theorem 12.

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

$$E:{ℬ}_{ℝ}\to ℬ(H)$$

such that

$$⟨Tx,y⟩={\int }_{ℝ}\lambda 𝑑E_{x,y}(\lambda ),x\in 𝒟(T),y\in H.$$

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

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 $ℬ(H)$ and a unique isometric ${}^{*}$-isomorphism $\mathrm{\Psi }:L^{\mathrm{\infty }}(E)\to A$ such that

$$⟨\mathrm{\Psi }(f)x,y⟩={\int }_{\sigma (T)}f(\lambda )𝑑E_{x,y}(\lambda ),f\in L^{\mathrm{\infty }}(E),x,y\in H.$$

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

$$e^{itT}=\mathrm{\Psi }(f_t)={\int }_{\sigma (T)}{e}^{it\lambda }𝑑E(\lambda ).$$

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

$$\mathrm{\Psi }{(f_t)}^{*}\mathrm{\Psi }(f_t)=\mathrm{\Psi }(\overline{f_t})\mathrm{\Psi }(f_t)=\mathrm{\Psi }(f_{-t})\mathrm{\Psi }(f_t)=\mathrm{\Psi }(f_{-t}{f}_t)=\mathrm{\Psi }(f_0)=I,$$

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

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

$$I=\mathrm{\Psi }(f_0)=\mathrm{\Psi }(f_t{f}_{-t})=\mathrm{\Psi }(f_t)\mathrm{\Psi }(f_{-t})=e^{itT}{e}^{i(-t)T},$$

and for $s,t\in ℝ$ we have

$$e^{isT}{e}^{itT}=\mathrm{\Psi }(f_s)\mathrm{\Psi }(f_t)=\mathrm{\Psi }(f_s{f}_t)=\mathrm{\Psi }(f_{s+t})=e^{i(s+t)T},$$

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

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

$${\parallel {\mathrm{\Psi }}_{t}x-x\parallel }^2={\parallel \mathrm{\Psi }(f_t-1)x\parallel }^2={\int }_{\sigma (T)}{|f_t-1|}^2𝑑E_{x,x}={\int }_{\sigma (T)}{|e^{it\lambda }-1|}^2𝑑E_{x,x}(\lambda ).$$

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

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

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

$$\parallel e^{itT}x-x\parallel \to 0$$

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

Conversely, Stone’s theorem on one-parameter unitary groups states that if $\{U_t:t\in ℝ\}$ 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 $t\in ℝ$.

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

$$⟨iTx,y⟩=i⟨Tx,y⟩=i{\int }_{ℝ}\lambda 𝑑E_{x,y}(\lambda )$$

and by Theorem 11,

$$⟨\mathrm{\Psi }(g_t)x,y⟩={\int }_{\sigma (T)}{g}_t𝑑E_{x,y}={\int }_{\sigma (T)}\frac{e^{it\lambda }-1}{t}𝑑E_{x,y}(\lambda ),$$

so

$$⟨\mathrm{\Psi }(g_t)x-iTx,y⟩={\int }_{\sigma (T)}\left(\frac{e^{it\lambda }-1}{t}-i\lambda \right)𝑑E_{x,y}(\lambda ).$$

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

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

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

$$⟨\mathrm{\Psi }(g_t)x-iTx,y⟩={\int }_{\sigma (T)}\left(\frac{e^{it\lambda }-1}{t}-i\lambda \right)𝑑E_{x,y}(\lambda )\to 0$$

as $t\to 0$. In particular,

$${\parallel \mathrm{\Psi }(g_t)x-iTx\parallel }^2\to 0$$

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

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

as $t\to 0$. In other words, $iT$ is the infinitesimal generator of the one-parameter group $e^{itT}$. We remark that because $T^{*}=T$, the adjoint of $iT$ is ${(iT)}^{*}=\overline{i}{T}^{*}=-i{T}^{*}=-iT=-(iT)$.

7 Trotter product formula

We remind ourselves that for an operator $T$ in $H$ to be closed means that $𝒢(T)$ 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 $\mathrm{D}\mathit{}\mathrm{(}T\mathrm{)}$ with the norm

$${\parallel x\parallel }_T=\parallel x\parallel +\parallel Tx\parallel .$$

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 $n\to \mathrm{\infty }$.

Theorem 14.

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

$$e^{it(A+B)}\psi =\underset{n\to \mathrm{\infty }}{lim}\left({(e^{itA/n}{e}^{itb/n})}^n\psi \right).$$

Proof.

The claim is immediate for $t=0$, and we prove the claim for $t>0$; it is straightforward to obtain the claim for using the truth of the claim for $t>0$. Let $D=𝒟(A+B)=𝒟(A)\cap 𝒟(B)$. Because $A+B$ is self-adjoint, $A+B$ is closed (Theorem 4), so by Theorem 13, the linear space $D$ with the norm ${\parallel \varphi \parallel }_{A+B}=\parallel \varphi \parallel +\parallel (A+B)\varphi \parallel$ is a Banach space. Because $D$ is a Banach space, the uniform boundedness principle tells us that if $\mathrm{\Gamma }$ is a collection of bounded linear maps $D\to H$ and if for each $\varphi \in D$ the set $\{\gamma \varphi :\gamma \in \mathrm{\Gamma }\}$ is bounded in $H$, then the set $\{\parallel \gamma \parallel :\gamma \in \mathrm{\Gamma }\}$ is bounded, i.e. there is some $C$ such that $\parallel \gamma \varphi \parallel \le C{\parallel \varphi \parallel }_{A+B}$ for all $\gamma \in \mathrm{\Gamma }$ and all $\varphi \in D$.

For $s\in ℝ$, 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 $ℬ(H)$. For $n\ge 1$,

$$\sum _{j=0}^{n-1}{U}_{t/n}^j(S_{t/n}-U_{t/n})S_{t/n}^{n-j-1}=U_{t/n}^n-S_{t/n}^n=U_{t/n}^n-S_t,$$

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

$\parallel (S_t-U_{t/n}^n)\xi \parallel$ $=\parallel {\displaystyle \sum _{j=0}^{n-1}}{U}_{t/n}^j(S_{t/n}-U_{t/n})S_{t/n}^{n-j-1}\xi \parallel$

$\le {\displaystyle \sum _{j=0}^{n-1}}\parallel (S_{t/n}-U_{t/n})S_{t/n}^{n-j-1}\xi \parallel$

$={\displaystyle \sum _{j=0}^{n-1}}\parallel (S_{t/n}-U_{t/n})S_{t-\frac{j+1}{n}}\xi \parallel$

$\le {\displaystyle \sum _{j=0}^{n-1}}\underset{0\le s\le t}{sup}\parallel (S_{t/n}-U_{t/n})S_s\xi \parallel .$

That is,

$$\parallel (S_t-U_{t/n}^n)\xi \parallel \le n\underset{0\le s\le t}{sup}\parallel (S_{t/n}-U_{t/n})S_s\xi \parallel ,\xi \in H,n\ge 1.$$ (1)

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

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

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

$$V_s(iB\varphi )+V_s\left(\frac{W_s-I}{s}-iB\right)\varphi +\frac{V_s-I}{s}\varphi =\frac{U_s-I}{s}\varphi ,$$

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

$$V_s(iB\varphi )+V_s\left(\frac{W_s-I}{s}-iB\right)\varphi +\frac{V_s-I}{s}\varphi \to iB\varphi +iA\varphi$$

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

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

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

$$\frac{S_s-U_s}{s}\varphi \to 0,s\downarrow 0.$$

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

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

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel n(S_{t/n}-U_{t/n})\varphi \parallel =0,\varphi \in D.$$ (4)

For each $n\ge 1$, 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 $\varphi \in D$,

$$\parallel {\gamma }_n\varphi \parallel \le n\parallel S_{t/n}\varphi \parallel +n\parallel U_{t/n}\varphi \parallel \le n\parallel \varphi \parallel +n\parallel \varphi \parallel \le 2n{\parallel \varphi \parallel }_{A+B},$$

showing that each ${\gamma }_n$ is a bounded linear map $D\to H$, where $D$ is a Banach space with the norm ${\parallel \varphi \parallel }_{A+B}=\parallel \varphi \parallel +\parallel (A+B)\varphi \parallel$. Moreover, (4) shows that for each $\varphi \in D$, there is some $C_{\varphi }$ such that

$$\parallel {\gamma }_n\varphi \parallel \le C_{\varphi },n\ge 1.$$

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

$$\parallel {\gamma }_n\varphi \parallel \le C{\parallel \varphi \parallel }_{A+B},$$

i.e.

$$\parallel n(S_{t/n}-U_{t/n})\varphi \parallel \le C{\parallel \varphi \parallel }_{A+B},n\ge 1,\varphi \in D.$$ (5)

Let $K$ be a compact subset of $D$, where $D$ is a Banach space with the norm ${\parallel \varphi \parallel }_{A+B}=\parallel \varphi \parallel +\parallel (A+B)\varphi \parallel$. Then $K$ is totally bounded, so for any $ϵ>0$, there are ${\varphi }_1,\mathrm{\dots },{\varphi }_M\in K$ such that $K\subset {\bigcup }_{m=1}^M{B}_{ϵ/C}({\varphi }_m)$. By (4), for each $m$, $1\le m\le M$, there is some $n_m$ such that when $n\ge n_m$,

$$\parallel n(S_{t/n}-U_{t/n}){\varphi }_m\parallel \le ϵ.$$

Let $N=\max \{n_1,\mathrm{\dots },n_M\}$. For $n\ge N$ and for $\varphi \in D$, there is some $m$ for which , and using (5), as $\varphi -{\varphi }_m\in D$, we get

$\parallel n(S_{t/n}-U_{t/n})\varphi \parallel$ $\le \parallel n(S_{t/n}-U_{t/n})(\varphi -{\varphi }_m)\parallel +\parallel n(S_{t/n}-U_{t/n}){\varphi }_m\parallel$

$\le C{\parallel \varphi -{\varphi }_m\parallel }_{A+B}+ϵ$

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

(6)

Let $\varphi \in D$, let $s_0\in ℝ$, and let $ϵ>0$. Because $s\mapsto S_s$ is strongly continuous $ℝ\to ℬ(H)$, there is some ${\delta }_1>0$ such that when , , and there is some ${\delta }_2>0$ such that when , , and hence with $\delta =\min \{{\delta }_1,{\delta }_2\}$, when we have

${\parallel S_s\varphi -S_{s_0}\varphi \parallel }_{A+B}$ $=\parallel S_s\varphi -S_{s_0}\varphi \parallel +\parallel (A+B)(S_s\varphi -S_{s_0}\varphi )\parallel$

$=\parallel S_s\varphi -S_{s_0}\varphi \parallel +\parallel S_s(A+B)\varphi -S_{s_0}(A+B)\varphi )\parallel$

showing that $s\mapsto S_s\varphi$ is continuous $ℝ\to D$. Therefore $\{S_s\varphi :0\le s\le t\}$ is a compact subset of $D$, so applying (6) we get that for any $ϵ>0$, there is some $n_{ϵ}$ such that if $n\ge n_{ϵ}$ and $0\le s\le t$, then

and therefore if $n\ge n_{ϵ}$ then

$$\underset{0\le s\le t}{sup}\parallel n(S_{t/n}-U_{t/n})S_s\varphi \parallel \le ϵ.$$ (7)

Finally, let $ϵ>0$. The statement that $A+B$ is self-adjoint in $H$ entails the statement that $D$ is dense in $H$, so there is some $\varphi \in D$ such that . For $n\ge 1$,

$\parallel (S_t-U_{t/n}^n)\psi \parallel$ $\le \parallel (S_t-U_{t/n}^n)(\psi -\varphi )\parallel +\parallel (S_t-U_{t/n}^n)\varphi \parallel$

$\le 2\parallel \psi -\varphi \parallel +\parallel (S_t-U_{t/n}^n)\varphi \parallel$

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

$$\parallel (S_t-U_{t/n}^n)\varphi \parallel \le n\underset{0\le s\le t}{sup}\parallel (S_{t/n}-U_{t/n})S_s\varphi \parallel \le ϵ.$$

Therefore for $n\ge n_{ϵ}$,

proving the claim. ∎