13.5 Spectral theory

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Eigenvectors

As in Chapter 4 a complex number $\alpha$ is an eigenvalue of a bounded linear operator $A$ : $\mathcal{H}\to \mathcal{H}$ if there exists a non-zero vector $u\in \mathcal{H}$ such that

$$Au=\alpha u.$$

$u$ is called the eigenvector of $A$ corresponding to the eigenvalue $\alpha$.

Theorem 13.15 · Eigenvalues and orthogonality for Hermitian operators

All eigenvalues of a hermitian operator $A$ are real, and eigenvectors corresponding to different eigenvalues are orthogonal.

Proof

If $Au=\alpha u$ then

$$⟨u\mid Au⟩=⟨u\mid \alpha u⟩=\alpha \Vert u{\Vert }^2.$$

Since $A$ is hermitian

$$⟨u|Au⟩=⟨Au|u⟩=⟨\alpha u|u⟩=\overline{\alpha }\Vert u{\Vert }^2.$$

For a non-zero vector $\Vert u\Vert \ne 0$ , we have $\alpha =\overline{\alpha };$ the eigenvalue $\alpha$ is real.

If $Av=\beta v$ then

$$⟨u|Av⟩=⟨u|\beta v⟩=\beta ⟨u|v⟩$$

and

$$⟨u\mid Av⟩=⟨Au\mid v⟩=⟨\alpha u\mid v⟩=\overline{\alpha }⟨u\mid v⟩=\alpha ⟨u\mid v⟩.$$

If $\beta \ne \alpha$ then $⟨u|v⟩=0$

A hermitian operator is said to be complete if its eigenvectors form a complete o.n. set.

Example 13.12

The eigenvalues of a projection operator $P$ are always 0 or 1, for

$$Pu=\alpha u⟹P^{2}u=P(\alpha u)=\alpha Pu={\alpha }^{2}u$$

and since $P$ is idempotent,

$$P^{2}u=Pu=\alpha u.$$

Hence ${\alpha }^2=\alpha$ , so that $\alpha =0$ or 1. If $P=P_M$ then the eigenvectors corresponding to eigenvalue 1 are the vectors belonging to the subspace $M$, while those having eigenvalue 0 belong to its orthogonal complement $M^{\perp }$ . Combining Theorem 13.8 and Theorem 13.2, we see tha every projection operator is complete.

Theorem 13.16 · Eigenvalues and orthogonality for unitary operators

The eigenvalues of a unitary operator $U$ are of theform $\alpha ={\mathrm{e}}^{ia}$ where $a$ is a real number, and eigenvectors corresponding to different eigenvalues are orthogonal.

Proof

Since $U$ is an isometry, if $Uu=\alpha u$ where $u\ne 0$ , then

$$\Vert u{\Vert }^2=⟨u|u⟩=⟨Uu|Uu⟩=⟨\alpha u|\alpha u⟩=\overline{\alpha }\alpha \Vert u{\Vert }^2.$$

Hence $\overline{\alpha }\alpha =|\alpha {|}^2=1$ , and there exists a real $a$ such that $\alpha ={\mathrm{e}}^{ia}$

If $Uu=\alpha u$ and $Uv=\beta v$ , then

$$⟨u|Uv⟩=\beta uv.$$

But $U^{\ast }U=I$ implies $u=U^{\ast }Uu=\alpha U^{\ast }u.$ , so that

$$U^{\ast }u={\alpha }^{-1}u=\overline{\alpha }u\text{ since }|\alpha {|}^2=1.$$

Therefore

$$⟨u\mid Uv⟩=⟨U^{\ast }u\mid v⟩=⟨\overline{\alpha }u\mid v⟩=\alpha ⟨u\mid v⟩.$$

Hence $(\alpha -\beta )⟨u\mid v⟩=0.\mathrm{If}\alpha \ne \beta$ then $u$ and $v$ are orthogonal, $⟨u\mid v⟩=0$

Spectrum of a bounded operator

In the case of a finite dimensional space, the set of eigenvalues of an operator is known as its spectrum. The spectrum is non-empty (see Chapter 4), and forms the diagonal elements in the Jordan canonical form. In infinite dimensional spaces, however, operators may have no eigenvalues at all.

Example 13.13

In ${\ell }^2$ , the right shift operator $S$ has no eigenvalues, for suppose

$$S(x_1,x_2,\dots )=(0,x_1,x_2,\dots )=\lambda (x_1,x_2,\dots ).$$

If $\lambda \ne 0$ then $x_1=0,x_2=0,...$ , hence $\lambda$ is not an eigenvalue. But $\lambda =0$ also implies $x_1=x_2=\cdot \cdot \cdot =0.$ , so this operator has no eigenvalues at all.

Exercise

Show that every $\lambda$ such that $|\lambda |<1$ is an eigenvalue of the left shift operator $S^{\prime }=S^{\ast }$ . Note that the spectrum of $S$ and its adjoint $S^{\ast }$ may be unrelated in the infinite dimensional case.

Example 13.14

Let $\alpha (x)$ be a bounded integrable function on a measure space $X$, and let $A_{\alpha }:g\mapsto \alpha g$ be the multiplication operator defined in Example 13.9. There is no normalizable function $g\in L^2(X)$ such that $\alpha (x)g(x)=\lambda g(x)$ unless $\alpha (x)$ has the constant value $\lambda$ on an interval $E$ ofnon-zero measure. For example, if $\alpha (x)=x$ on $X=[a,b]$ , then $f(x)$ is an eigenvector of $A_x$ iff there exists $\lambda \in \mathbb{C}$ such that

$$xf(x)=\lambda f(x),$$

which is only possible through $[a,b]\mathrm{if}f(x)=0$ . In quantum mechanics (see Chapter 14) this problem is sometimes overcome by treating the eigenvalue equation as a distributional equation. Then the Dirac delta function $\delta (x-x_0)$ acts as a distributional eigenfunction, with eigenvalue $a<\lambda =x_0<b$ ,

$$x\delta (x-x_0)=x_0\delta (x-x_0).$$

Examples such as 13.14 lead us to consider a new definition for the spectrum of an operator. Every operator $A$ has a degeneracy at an eigenvalue $\lambda$, in that $A-\lambda I$ is not an invertible operator. For, if $(A-\lambda I{)}^{-1}$ exists then Au $\ne \lambda u$ , for if $Au=\lambda u$ then

$$u=(A-\lambda I{)}^{-1}(A-\lambda I)u=(A-\lambda I{)}^{-1}0=0.$$

We say a complex number $\lambda$ is a regular value of a bounded operator $A$ on a Hilbert space $\mathcal{H}$ if $A-\lambda I$ is invertible – that is, $(A-\lambda I{)}^{-1}$ exists and is bounded. The spectrum $\Sigma (A)$ of $A$ is defined to be the set of $\lambda \in \mathbb{C}$ that are not regular values of $A$. If $\lambda$ is an eigenvalue of $A$ then, as shown above, it is in the spectrum of $A$ but the converse is not true. The eigenvalues are often called the point spectrum. The other points of the spectrum are called the continuous spectrum. At such points it is conceivable that the inverse exists but is not bounded. More commonly, the inverse only exists on a dense domain of $\mathcal{H}$ and is unbounded on that domain. We will leave discussion of this to Section 13.6.

Example 13.15

$\mathrm{If}\alpha (x)=x$ then the multiplication operator $A_{\alpha }$ on $L^2([0,1])$ has spectrum consisting of all real numbers $\lambda$ such that $0\le \lambda \le 1.\mathrm{If}\lambda >1$ or $\lambda <0$ or has non-zero imaginary part then the function $\beta =x-\lambda$ is clearly invertible and bounded on the interval [0, 1]. Hence all these are regular values of the operator $A_x$ . The real values $0\le \lambda \le 1$ form the spectrum of $A_x$ . From Example 13.14 none of these numbers are eigenvalues, but they do lie in the spectrum of $A_x$ since the function $\beta$ is not invertible. The operator $A_{\beta }$ is then defined, but unbounded, on the dense set $[0,1]-\{\lambda \}$

Theorem 13.17 · Bounds and compactness of the spectrum of a bounded operator

Let $A$ be a bounded operator on a Hilbert space $\mathcal{H}.$

(i) Every complex number $\lambda \in \Sigma (A)$ has magnitude $|\lambda |\le \Vert A\Vert$

(ii) The set ofregular values of $A$ is an open subset of $\mathbb{C}$

(iii) The spectrum of $A$ is a compact subset of $\mathbb{C}$.

Proof

(i) Let $|\lambda |>\Vert A\Vert$ . The operator $A/\lambda$ then has norm $<1$ and by Theorem 13.12 the operator $I-A/\lambda$ is invertible and

$$(A-\lambda I{)}^{-1}=-{\lambda }^{-1}{\left(I-\frac{A}{\lambda }\right)}^{-1}=-{\lambda }^{-1}\sum _{n=0}^{\infty }{\left(\frac{A}{\lambda }\right)}^n.$$

Hence $\lambda$ is a regular value. Spectral values must therefore have $|\lambda |\le \Vert A\Vert$

(ii) If ${\lambda }_0$ is a regular value, then for any other complex number $\lambda$

$$\begin{array}{r}I-(A-{\lambda }_{0}I{)}^{-1}(A-\lambda I)=(A-{\lambda }_{0}I{)}^{-1}((A-{\lambda }_{0}I)-(A-\lambda I))\\ =(A-{\lambda }_{0}I{)}^{-1}(\lambda -{\lambda }_0).\end{array}$$

Hence

$$\Vert I-(A-{\lambda }_{0}I{)}^{-1}(A-\lambda I)\Vert =|\lambda -{\lambda }_0|\Vert (A-{\lambda }_{0}I{)}^{-1}\Vert <1$$

if

$$\left|\lambda -{\lambda }_0\right|<\frac{1}{\Vert {\left(A-{\lambda }_{0}I\right)}^{-1}\Vert }.$$

By Theorem 13.12, for $\lambda$ in a small enough neighbourhood of$\lambda$ the operator $I-(I-(A-{\lambda }_{0}I{)}^{-1}(A-\lambda I))=(A-{\lambda }_{0}I{)}^{-1}(A-\lambda I)$ is invertible. If $B$ is its inverse, then

$$B(A-{\lambda }_{0}I{)}^{-1}(A-\lambda )=I$$

and $A-\lambda I$ is invertible with inverse $B(A-{\lambda }_{0}I{)}^{-1}$ . Hence the regular values form an open set.

(iii) The spectrum $\Sigma (A)$ is a closed set since it is the complement ofan open set (the regular values). By part (i), it is a subset of a bounded set $|\lambda |\le \Vert A\Vert$ , and is therefore a compact set. -

Spectral theory of hermitian operators

Of greatest interest is the spectral theory of hermitian operators. This theory can become quite difficult, and we will only sketch some of the proofs.

Theorem 13.18 · Reality of the spectrum of a Hermitian operator

The spectrum $\Sigma (A)$ of a hermitian operator $A$ consists entirely of real numbers.

Proof

Suppose $\lambda =a+ib$ is a complex number with $b\ne 0$ . Then $\Vert (A-\lambda I)u{\Vert }^2=$ $\Vert (A-aI)u{\Vert }^2+b^2\Vert u{\Vert }^2$ , and

$$\Vert u\Vert \le \frac{1}{|b|}\Vert (A-\lambda I)u\Vert .\text{\hspace{1em}(13.12)}$$

The operator $A-\lambda I$ is therefore one-to-one, for if $(A-\lambda I)u=0$ then $u=0$

The set $V=\{(A-\lambda I)u\mid u\in \mathcal{H}\}$ is a subspace of $\mathcal{H}$. To show closure (the vector sub space property is trivial), let $v_n=(A-\lambda I)u_n\to v$ be a convergent sequence of vectors in $V$. From the fact that it is a Cauchy sequence and the inequality Eq. (13.12), it follows that $u_n$ is also a Cauchy sequence, having limit $u$. By continuity of the operator $A-\lambda I$ , it follows that $V$ is closed, for

$$(A-\lambda I)u=\underset{n\to \infty }{\lim }(A-\lambda I)u_n=\underset{n\to \infty }{\lim }{v}_n=v.$$

Finally, $V=\mathcal{H},$ , for if $w\in V^{\perp }$ , then $⟨(A-\lambda I)u|w⟩=⟨u|(A-\overline{\lambda }I)w⟩=0$ for all $u\in$ $\mathcal{H}.$ . Setting $u=(A-\overline{\lambda }I)w$ gives $(A-\overline{\lambda }I)w=0$ . Since $A-\overline{\lambda }I$ is $\mathtt{one}-\mathtt{to}-\mathtt{one},w=0$ Hence $V^{\perp }=\{0\}$ , the subspace $V=\mathcal{H}$ and every vector $u\in \mathcal{H}$ can be written in the form $u=(A-\lambda I)v$ . Thus $A-\lambda I$ is invertible, and the inequality Eq. (13.12) can be used to show it is bounded. -

The full spectral theory of a hermitian operator involves reconstructing the operator from its spectrum. In the finite dimensional case, the spectrum consists entirely of eigenvalues, making up the point spectrum. From Theorem 13.15 the eigenvalues may be written as a non-empty ordered set of real numbers ${\lambda }_1<{\lambda }_2<\cdots <{\lambda }_k$ . For each eigenvalue ${\lambda }_i$ there corresponds an eigenspace $M_i$ of eigenvectors, and different spaces are orthogonal to each other. A standard inductive argument can be used to show that every hermitian operator on a finite dimensional Hilbert space is complete, so the eigenspaces span the entire Hilbert space. In terms of projection operators into these eigenspaces $P_i=P_{M_i}$ , these statements can be summarized as

$$A={\lambda }_1{P}_1+{\lambda }_2{P}_2+\cdots +{\lambda }_k{P}_k$$

where

$$P_1+P_2+\cdots +P_k=I,P_i{P}_j=P_j{P}_i={\delta }_{ij}{P}_i.$$

Essentially, this is the familiar statement that a hermitian matrix can be ‘diagonalized with its eigenvalues along the diagonal. If we write, for any two projection operators, $P_M\le P_N\mathrm{iff}M\subseteq N$ , we can replace the operators $P_i$ with an increasing family ofprojection operators $E_i=P_1+P_2+\cdot \cdot \cdot +P_i$ . These are projection operators since they are clearly hermitian and idempotent, $(E_i{)}^2=E_i$ , and project into an increasing family of subspaces, $V_i=L(M_1\cup M_2\cup \cdot \cdot \cdot \cup M_i)$ , having the property $V_i\subset V_j\mathrm{if}i<j$ . Since $P_i=E_i-E_{i-1}$ where $E_0=0$ , we can write the spectral theorem in the form

$$A=\sum _{i=1}^n{\lambda }_i(E_i-E_{i-1}).$$

For infinite dimensional Hilbert spaces, the situation is considerably more complicated, but the projection operator language can again be used to effect. The full spectral theorem in arbitrary dimensions is as follows:

Theorem 13.19 · Spectral theorem for bounded Hermitian operators

Let $A$ be a hermitian operator on a Hilbert space $\mathcal{H}$, with spectrum $\Sigma (A)$ By Theorem 13.17 this is a closed bounded subset of $\mathbb{R}$. There exists an increasingfamily ofprojection operators $E_{\lambda }\left(\lambda \in \mathbb{R}\right)$ , with $E_{\lambda }\le P_{{\lambda }^{\prime }}\text{for}\lambda \le {\lambda }^{\prime }$ , such that

$$E_{\lambda }=0\text{for}\lambda <\inf (\Sigma (A)),E_{\lambda }=I\text{for}\lambda >\sup (\Sigma (A))$$

and

$$A={\int }_{-\infty }^{\infty }\lambda \mathrm{d}{E}_{\lambda }.$$

The integral in this theorem is defined in the Lebesgue–Stieltjes sense. Essentially it means that if $f(x)$ is a measurable function, and $g(x)$ is of the form

$$g(x)=c+{\int }_0^{x}h(x)\mathrm{d}x$$

for some complex constant $c$ and integrable function $h(x)$ , then

$${\int }_a^{b}f(x)\mathrm{d}(g(x))={\int }_a^{b}f(x)h(x)\mathrm{d}x.$$

A function $g$ of this form is said to be absolutely continuous; the function $h$ is uniquely defined almost everywhere by $g$ and we may write it as a kind of derivative of $g$, $h(x)=g^{\prime }(x)$ For the finite dimensional case this theorem reduces to the statement above, on setting $E_{\lambda }$ to have discrete jumps by $P_i$ at each of the eigenvalues ${\lambda }_i$ . The proof of this result is not easy. The interested reader is referred to [3, 6] for details.

Problems

Problem 13.24

Show that a non-zero vector $u$ is an eigenvector of an operator $A$ if and only if $|⟨u|Au⟩|=\Vert Au\Vert \Vert u\Vert$

Problem 13.25

For any projection operator $P_M$ show that every value $\lambda \ne 0$ , 1 is a regular value, by showing that $(P_M-\lambda I)$ has a bounded inverse.

Problem 13.26

Show that every complex number $\lambda$ in the spectrum of a unitary operator has $|\lambda |=1$

Problem 13.27

Prove that every hermitian operator $A$ on a finite dimensional Hilbert space can be written as

$$A=\sum _{i=1}^k{\lambda }_i{P}_i\text{ where }\sum _{i=1}^k{P}_i=I,P_i{P}_j=P_j{P}_i={\delta }_{ij}{P}_i.$$

Problem 13.28

For any pair of hermitian operators $A$ and $B$ on a Hilbert space $\mathcal{H}$, define $A\le B$ iff $⟨u\mid Au⟩\le ⟨u\mid Bu⟩$ for all $u\in \mathcal{H}$ . Show that this is a partial order on the set of hermitian operators – pay particular attention to the symmetry property, $A\le B$ and $B\le A$ implies $A=B$

(a) For multiplication operators on $L^2(X)$ show that $A_{\alpha }\le A_{\beta }\mathrm{iff}\alpha (x)\le \beta (x)$ a.e. on $X.$

(b) For projection operators show that the definition given here reduces to that given in the text, $P_M\le P_N\mathrm{iff}M\subseteq N.$