$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\Ker}{\operatorname{Ker}}$ $\newcommand{\Ran}{\operatorname{Ran}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$

13.4. Elements of Spectral Theory

1. General theory
2. Sel-Adjoint theory

General theory

Definitions

Let $\mathsf{H}$ be a Hilbert space (see Definition 4.3.3)

Definition 1. Linear operator $L:\mathsf{H}\to\mathsf{H}$ is bounded if \begin{equation} | Lu|\le M|u|\qquad \forall u\in \mathsf{H}; \label{eq-13.4.1} \end{equation} the smallest constant $M$ for which it holds is called operator norm of $L$ and denoted $\|L\|$.

However one needs to consider also unbounded operators. Such operators not only fail (\ref{eq-13.4.1}) but they are not defined everywhere.

Definition 2. Consider a linear operator $L:\ D(L)\to \mathsf{H}$ where $D(L)$ is a linear subset in $\mathsf{H}$ (i.e.~it is a linear subspace but we do not call it this way because it is not closed) which is dense in $\mathsf{H}$ (i.e.~for each $u\in \mathsf{H}$ there exists a sequence $u_n \in D(L)$ converging to $u$ in $\mathsf{H}$). Then operator $L$ is closed if $u_n\to u$, $Lu_n\to f$ imply that $u\in D(L)$ and $Lu=f$.

Definition 3.

1. $\lambda \in \mathbb{C}$ belongs to resolvent set of $L$ $\rho(L)$ if there exist bounded operator $R(\lambda) := R_L (\lambda)=(\lambda -L)^{-1}$, that is \begin{gather} R(\lambda) (\lambda-L) = (\lambda-L) R(\lambda)=I. \label{eqn-13.4.1} \end{gather}
2. $\sigma (L):= \mathbb{C}\setminus \rho(L)$ is the Spectrum of $L$.
3. $r(L)$, the smallest $r\ge 0$ such that $\sigma (L)\subset\{\lambda\colon |\lambda|\le r\}$ is the spectral radius of $L$.

Remark 1. Obviously $\rho(L)$ is an open set and $\sigma(L)$ is a closed set and $r(L)\le |L|$. Even in Linear algebra one can find examples of operator $L\ne 0$ with $r(L)=0$.

Classification of spectra

Definition 4. Let $L$ be a bounded linear operator and $\lambda \in \sigma(L)$. Then

1. $\lambda $ belongs to point spectrum of $L$ if $\Ker (\lambda-L)\ne {0}$.
2. $\lambda $ belongs to continuous spectrum of $L$ if $\Ker (\lambda-L)= {0}$ and exists $u_n$ with $\|u_n\|=1$ such that $\|(\lambda -L)u_n\|\to 0$.
3. $\lambda \in \sigma(L)$ belongs to residue spectrum of $L$ if it belongs neither to point spectrum, nor continuous spectrum, that means $\Ran (\lambda -L)\ne H$.

Remark 2. So, in the general theory spectrum of operator consists of three non-overlapping sets. Not so in self-adjoint theory!

Function of operator

Definition 5. Let $L$ be a bounded operator and $f$ analytic in the vicinity of $\sigma(L)$ function. Then \begin{gather} f(L)= \frac{1}{2\pi i}\int_\gamma f(\lambda)(\lambda -L)^{-1}\,d\lambda \label{eqn-13.4.3} \end{gather} where $\gamma\subset \rho(L)$ is a contour going once in the counter-clockwise direction around $\sigma (L)$.

Remark 3. One can prove easily (requires some knowledge of Complex Variables) that $f(L)$ does not depend on the choice of $\gamma$.

Self-adjoint theory

Definition 6. Let $L:\mathsf{H}\to\mathsf{H}$ be a bounded linear operator.

1. Adjoint operator $L^*$ is defined as \begin{equation} (Lu, v)= (u,L^*v) \qquad \forall u,v\in \mathsf{H}; \label{eq-13.4.2} \end{equation}
2. Operator $L$ is self-adjoint if $L^*=L$.

However one needs to consider also unbounded operators.

Definition 7. Consider a linear operator $L:\ D(L)\to \mathsf{H}$ where $D(L)$ is a dense linear subset in $\mathsf{H}$

1. Operator $L$ is symmetric if \begin{gather} (Lu, v)= (u,L v) \qquad \forall u,v\in D(L); \label{eqn-13.4.5} \end{gather} then \begin{gather} D(L)\subset D(L^*) := \{v\colon (Lu,v)\le C\| u\| \quad \forall u\in D(L), \ C=C(v)\}. \label{eqn-13.4.6} \end{gather}
2. Symmetric operator $L$ is self-adjoint if $L^*=L$ that is $D(^*)=D(L)$. \end{enumerate}

Remark 4. It is known that

1. Symmetric operator $L$ is self-adjoint if and only if $\sigma (L)\subset \mathbb{R}$;.

2. Co-dimension of $\Ran (\lambda -L)$ is the same in all points $\lambda\in \mathbb{C}_\pm:= \{\lambda\colon \Im (\lambda)\gtrless 0\}$; those are called deficiency indices of $L$.

3. Symmetric operator $L$ can be extended to self-adjoint operator $K$ (that is $L \subset K =K^*\subset L^* $ if and only if these deficiency indices coincide.

4. Let symmetric operator be semi-bounded from below \begin{gather*} (Lu,u)\ge M\|u\|^2\quad \forall u\in D(L). \end{gather*} Then it can be extended to self-adjoint operator with the same $M$.

Remark 5.

1. For bounded operators symmetric'' equalsself-adjoint'';
2. Not so for unbounded operators. F.e. $Lu=-u''$ on $(0,l)$ with $D(L)=\{u(0)=u'(0)=u(l)=u'(l)=0\}$ is symmetric but not self-adjoint;
3. Self-adjoint operators have many properties which symmetric but not self-adjoint operators do not have;
4. In Quantum Mechanics observables are self-adjoint operators.

Theorem 1. The following statements are equivalent:

1. $L$ is self-adjoint;
2. $L$ generates unitary group $e^{itL}$ ($t\in \mathbb{R}$: $\| e^{itL} u\|=\|u\|$, $e^{i(t_1+t_2)L}= e^{it_1L}e^{it_2L}$, $u\in D(L)\implies e^{itL}u\in D(L)$, $\frac{d\ }{dt}e^{itL} u= L e^{itL}u$ for all $u\in D(L)$ (and conversely, if $e^{itL}u$ is differentiable by $t$ then $u\in D(L)$);
3. Exist spectral projectors -- operators $\theta (\tau -L)$ ($\theta(\tau)=0$ as $\tau\le 0$ and $\theta(\tau)=1$ as $\tau>0$) such that $\theta(\tau -L)$ are orthogonal projectors, $\theta (\tau_1-L)\theta (\tau_2-L)=\theta (\tau-L)$ with $\tau=\min (\tau_1,\tau_2)$, $\theta (\tau-L)u\to 0$ as $\tau\to -\infty$; $\theta (\tau-L)u\to u$ as $\tau\to +\infty$; $\theta (\tau-L)u\to \theta (\tau^*-L)$ as $\tau\to \tau^*-0$ and \begin{gather} L =\int \tau\, d \theta (\tau-L) \label{eqn-13.4.7} \end{gather}

Function of operator

Definition 8. For real function $f$ \begin{gather} f(L) =\int f(\tau)\, d \theta (\tau-L) \label{eqn-13.4.8} \end{gather}

Classification of spectra

Consider spectrum of self-adjoint operator. Then there is no residue spectrum but spectral decomposition allows us more refined classification of continuous spectrum:

For simplicity we assume that $\mathsf{H}$ is separable Hilbert space (that is it has enumerable basis). Let $\mathsf{H}_p$ be a closed span of all eigenvectors and $\mathsf{H}_c=\mathsf{H}_p^\perp$ be its orthogonal complement.

Then one can prove that $D(L)\cap \mathsf{H}_p$ is dense in $\mathsf{H}_p$, $D(L)\cap \mathsf{H}_c$ is dense in $\mathsf{H}_c$ and $L$ is a self-adjoint operator on $\mathsf{H}_p$ and $\mathsf{H}_c$ and $\theta (\lambda -L)$ are projectors on $\mathsf{H}_p$ and $\mathsf{H}_c$.

On $\mathsf{H}_p$ operator $L$ can have a continuous spectrum (consisting of all limit points of the point spectrum which are not eigenvalues, but on $\mathsf{H}_c$ operator $L$ has only continuous spectrum.

Then $\theta (\lambda -L)$ is a continuous function on $\mathsf{H}_c$. We know from Real Analysis that continuous monotone functions could be decomposed in absolutely continuous and singular continuous components.

Namely, one can decompose $\mathsf{H}_c=\mathsf{H}_{ac}\oplus \mathsf{H}_{sc}$, such that $D(L)\cap \mathsf{H}_{ac}$ is dense in $\mathsf{H}_{ac}$, $D(L)\cap \mathsf{H}_{sc}$ is dense in $\mathsf{H}_{sc}$ and $L$ is a self-adjoint operator on both $\mathsf{H}_{ac}$ and $\mathsf{H}_{sc}$, and $\theta(\tau -L)$ are projectors on both $ \mathsf{H}_{ac}$ and $ \mathsf{H}_{sc}$ and the former is absolutely continuous and the second is singular continuous.

Therefore, finally \begin{gather} \mathsf{H} = \mathsf{H}_{p} \oplus \mathsf{H}_{ac}\oplus \mathsf{H}_{sc}. \label{eqn-13.4.9} \end{gather}

Definition 9. Spectrum of $L$ on On $\mathsf{H}_{ac}$ is called absolutely continuous spectrum and spectrum of $L$ on On $\mathsf{H}_{sc}$ is called singular continuous spectrum of $L$; these spectra are denoted as $\sigma_{ac}(L)$ and $\sigma_{sc}(L)$.

Remark 6.

1. $\sigma_{ac}(L) $ and $\sigma_{sc}(L)$ are closed sets. $\sigma_{sc}(L)$ has Lebesgue measure $0$.
2. $\sigma_p(L)$ is not necessary closed set. $\sigma_{pp}(L):= \overline{\sigma_p(L)}$ is called pure point spectrum of $L$.
3. All these sets are not necessarily disjoint and can intersect.

Definition 10.

1. $\lambda $ belongs discrete spectrum of $L$ ($\lambda \in \sigma_d(L)$) if it is isolated from the rest of $\sigma (L)$ and is eigenvalue of final multiplicity.
2. Essential spectrum of $L$ is $\sigma(L)$ with the exception of discrete spectrum.

Remark 7. $\sigma_d(L)$ is not necessarily closed and such eigenvalues can accumulate to points of essential spectrum.

Families of commuting operators

What can be simpler? Operators $L$ and $M$ commute if $LM=ML$! Not so fast: for unbounded operators crucial role is played by domains and with such ``definition'' neither resolvents, nor unitary groups, nor spectral projectors are necessarily commuting.

Example 1. $L_1\colon L_1u=-u''$ on $[0,1]$ with $D(L_1)=\{u(0)=u(1)=0\}$ and $L_2\colon L_2u=-u''$ on $[0,1]$ with $D(L_2)=\{u'(0)=u'(1)=0\}$ do not commute as properly defined below.

Definition 11. Consider self-adjoint operators $L_1,\ldots, L_n$. These operators commute if their spectral projectors $\theta (\tau_1-L_1), \ldots , \theta (\tau_n-L_n)$ commute.

Then $\theta (\tau_1-L_1)\cdots \theta (\tau_n-L_n)$ generate joint spectral measure on $\mathbb{R}^n$.

Definition 12.

1. Support of this spectral measure is the joint spectrum of $L_1,\ldots,L_n$.
2. For family of commuting operators we define function: \begin{gather} f(L_1,\ldots, L_n) :=\int_{\mathbb{R}^n} f(\tau_1,\ldots,\tau_n) \, d \theta (\tau_1-L_1)\cdots d\theta (\tau_n-L_n). \label{eqn-13.4.10} \end{gather}

$\Leftarrow$  $\Uparrow$  $\Rightarrow$

__