Unbounded Operators

The position operator in quantum mechanics is unbounded. Specifically, consider \newcommand{\H}{\mathcal{H}}\H=L^2(\R) and let $D(T)$ be the set of all functions $f\in L^2(\mathbb{R})$ such that ${\int }_{\mathbb{R}}{x}^2|f(x){|}^2\text{dd}x<\infty$. For $f\in D(T)$, the position operator $T$ acts on $f$ such that $(Tf)(x)=xf(x).$Then $T$ is clearly unbounded because we can consider $f_n\in L^2(\mathbb{R})$ satisfying $f_n(x):=\{\begin{array}{l}1,\text{if }x\in [n,n+1],\\ 0,\text{otherwise}.\end{array}$Then $\Vert f_n\Vert =1$, but $\Vert T{f}_n\Vert ={\int }_{\mathbb{R}}{x}^2|f_n(x){|}^2\text{dd}x=\frac{1}{3}((n+1{)}^3-n^3)$, which diverges to $\infty$ as $n\to \infty$.

Domain of an Operator

Closed Operator

An operator T\colon \H_{1}\to \H_{2} between two Hilbert spaces is called closed if for any sequence $\{f_n\}\subset D(T)$, $f_n\to f$ and $T{f}_n\to g$ implies $f\in D(T)$ and $Tf=g$. Equivalently, $T$ is closed if its graph is closed in \H_{1}\times \H_{2}.

e.g. Clearly, any bounded linear operator is closed.

Extension

Let T_{1},T_{2}\colon \H_{1}\to \H_{2} be operators between two Hilbert spaces. We say that $T_1$ is an extension of $T_2$ if their graphs satisfy $\Gamma (T_1)\supset \Gamma (T_2)$ and we write $T_1\supset T_2$.

Proposition

$T_1\supset T_2$ if and only if $D(T_1)\supset D(T_2)$ and $T_{1}f=T_{2}f$ for all $f\in D(T_2)$.

Proof

Closable

An operator is closable

e.g. Let $\stackrel{˝}{=}{L}^2(\mathbb{R})$, two operators $T_1:=i\text{ddf}x$ and $T_2:=i\text{ddf}x$ but defined on two different domains: $D(T_1)=C_0^1(\mathbb{R})$, $D(T_2)=C_0^{\infty }(\mathbb{R})$. Then $T_2\subset T_1$.

Adjoints of Densely Defined Operators

Adjoint of Densely Defined Operator

Let T\colon \H_{1}\to \H_{2} be a densely defined (i.e. $D(T)$ is dense in \H_{1}) operator between Hilbert spaces. Define the adjoint operator T^{*}\colon \H_{2}\to \H_{1} on the domain D(T^*):=\left\{f\in \H_{2} \mid \exists g\in\H_{2}. \langle T\varphi,f\rangle =\langle \varphi,g\rangle \text{ for all } \varphi\in D(T) \right\} such that $⟨Tf,g⟩=⟨f,T^{\ast }g⟩$ for all $f\in D(T)$ and $g\in D(T^{\ast })$.

The following proposition guarantees the unique existence of adjoints for unbounded operators under certain conditions:

Proposition

The adjoint of a densely defined operator is unique if exits.

Proof Fix some $g\in D(T^{\ast })$, suppose $h_1$ and $h_2$ satisfy $⟨Tf,g⟩=⟨f,h_1⟩$ and $⟨Tf,g⟩=⟨f,h_2⟩$ for all $f\in D(T)$. Then, we have $⟨f,h_1-h_2⟩=0$ for all $f\in D(T)$. Since $D(T)$ is dense in \H_{1}, there exists a sequence $\{x_n\}\subset D(T)$ converges to $h_1-h_2$, which implies that $⟨x_n,h_1-h_2⟩\to ⟨h_1-h_2,h_1-h_2⟩=0,$hence $h_1=h_2$. $\square$

Proposition

Suppose T\colon \H_{1}\to \H_{2} and S\colon \H_{1}\to \H_{2} are operators such that $S\subset T$, then $T^{\ast }\subset S^{\ast }$.

Proof If $S\subset T$, then $D(S)\subset D(T)$, and $S{|}_{D(S)}=T{|}_{D(S)}$. For all $\psi \in D(T^{\ast })$, we have $⟨T\phi ,\psi ⟩=⟨\phi ,T^{\ast }\psi ⟩$ for all $\phi \in D(T)$. So it also holds in $D(S)$, hence $S^{\ast }\psi =T^{\ast }\psi$ for all $\psi \in D(T^{\ast })$. This shows that $\psi \in D(S^{\ast })$, so $D(T^{\ast })\subset D(S^{\ast })$. Therefore, $T^{\ast }\subset S^{\ast }$. $\square$

Theorem

Let T\colon \H_{1}\to \H_{2} be a densely defined operator, then

• $T^{\ast }$ is closed;
• $T$ is closable if and only if $D(T^{\ast })$ is dense, in which case $\bar{T}=T^{\ast \ast }$;
• If $T$ is closable, then $(\bar{T}{)}^{\ast }=T^{\ast }$.

Proof For the first statement, suppose $\{f_n\}\subset D(T^{\ast })$ such that $f_n\to f$ and $T^{\ast }{f}_n\to g$, then for all $\phi \in D(T)$, we have $⟨T\phi ,f⟩={\lim }_{n\to \infty }⟨T\phi ,f_n⟩={\lim }_{n\to \infty }⟨\phi ,T^{\ast }{f}_n⟩=⟨\phi ,g⟩,$so $f\in D(T^{\ast })$ and $T^{\ast }f=g$. This shows that $T^{\ast }$ is closed. Now we prove the second statement. Suppose $D(T^{\ast })$ is dense, then

Resolvent Set & Resolvent Operator

Let $T$ be a closed operator on a Hilbert space \H. Then the resolvent set of $T$ is defined as $\rho (T):=\{\lambda \in \text{C}:\lambda \cdot 1-T:D(T)\to \stackrel{˝}{\text{ is bijective with bounded inverse}}\}.$For any $\lambda \in \rho (T)$, the resolvent of $T$ at $\lambda$ is defined as $R_{\lambda }(T):=(\lambda \cdot 1-T{)}^{-1}.$

Remark

Note that