Spectral Theory II: Unbounded Operators on Hilbert Spaces

Abstract

In this second chapter on spectral theory we shall examine a number of mathematical issues concerning typically unbounded self-adjoint operators.

The language of mathematics turns out to be unreasonably effective in natural sciences, a wonderful gift that we don’t understand nor deserve.

Eugene Paul Wigner

Exercises

9.1

Consider a spectral measure $P:\varSigma ({\mathsf X}) \ni E \mapsto P(E)\in {\mathfrak B}({\mathsf H})$ and a unitary operator (isometric and onto) $V: {\mathsf H}\rightarrow {\mathsf H}'$, where ${\mathsf H}$ is a complex Hilbert space. Prove

$$P': \varSigma ({\mathsf X}) \ni E \mapsto P'(E) := VP(E)V^{-1} \in {\mathfrak B}({\mathsf H}')$$

is a PVM.

9.2

In relationship to Exercise 9.1, prove the following facts.

(i) If $f: {\mathsf X}\rightarrow {\mathbb C}$ is measurable then $\psi \in \varDelta _{f}$ $\;\Leftrightarrow \;$ $V\psi \in \varDelta '_f$, where $\varDelta '_f$ is the domain of the integral of f in $P'$.

(ii) $V\int _{{\mathsf X}} f(x) dP(x) V^{-1} = \int _{{\mathsf X}} f(x) dP'(x)$.

9.3

Prove that (iv) in Theorem 9.13(b) can be strengthened as follows: let $T: D(T) \rightarrow {\mathsf H}$ be self-adjoint on the Hilbert space ${\mathsf H}$. Then $\lambda \in \sigma _c(T)$ is equivalent to asking that $0<||T\phi -\lambda \phi ||$, $\forall \,\phi \in D(T)$ with $||\phi ||=1$, and that for any $\varepsilon >0$ there exists $\phi _\varepsilon \in D(T)$, $||\phi _\varepsilon ||=1$, such that

$$||T\phi _\varepsilon - \lambda \phi _\varepsilon || \le \varepsilon \,.$$

Hint. The second condition amounts to saying $\lambda$ does not belong to $\sigma _p(T)$, so $(T-\lambda I)^{-1}: Ran(T-\lambda I) \rightarrow D(T)$ exists. Then $\lambda \in \sigma (T) = \sigma _p(T) \cup \sigma _c(T)$. Can $(T-\lambda I)^{-1}$ be bounded?

9.4

Consider the space $L^2({\mathsf X},\mu )$ with $\mu$ positive and finite on the Borel $\sigma$-algebra of a space ${\mathsf X}$. Let $f: {\mathsf X}\rightarrow {\mathbb R}$ be an arbitrary real, measurable, and locally $L^2$ map (i.e. $f\cdot g \in L^2({\mathsf X}, \mu )$ for any $g\in C_c({\mathsf X})$). Consider the operator on $L^2({\mathsf X},\mu )$:

$$T_f : h \mapsto f\cdot h$$

where $D(T_f) := \{h \in L^2({\mathsf X},\mu )\,|\, f\cdot h \in L^2({\mathsf X},\mu )\}$. Prove $T_f$ is self-adjoint. Without using Proposition 9.17 explicitly, show

$$\sigma (T_f) = \text {ess ran}(f)\,.$$

For $f: {\mathsf X}\rightarrow {\mathbb R}$, ess ran(f) is the essential rank of the measurable map f, defined by ${\mathbb R}\ni v\in \mathrm{ran~ess} (f)$ $\;\Leftrightarrow \;$ $\mu \left( f^{-1}(v-\varepsilon , v+ \varepsilon ) \right) >0$ for any $\varepsilon >0$.

Hint. The domain of $T_f$ is dense because f is locally $L^2$, and the self-adjointness comes from computing $T_f^*=T_f$. The second part goes along these lines. Observe that $\lambda \in \rho (T_f)$ $\;\Leftrightarrow \;$ the resolvent $R_\lambda (T_f)$ exists on $L^2({\mathsf X},\mu )$ and is bounded, i.e. there is $M>0$ such that $||R_\lambda (T_f) h|| \le M$ for any unit map $h \in L^2({\mathsf X},\mu )$. That is to say, $\lambda \in \rho (T_f)$ if and only if:

$$\int _{{\mathsf X}} \frac{|h(x)|^2}{|f(x) -\lambda |^2} d\mu (x) < M \quad \text {for any unit}\, h \in L^2({\mathsf X},\mu )\,.$$

If $\lambda \not \in {ess~ran} (f)$, by definition of essential rank and $\mu ({\mathsf X}) < +\infty$ we see that the condition holds, so $\lambda \not \in {ess~ran} (f)$ $\Rightarrow$ $\lambda \not \in \sigma (T_f)$. If $\lambda \in {ess~ran} (f)$, still by definition of essential rank we may build a sequence of unit vectors $h_n$ such that $\int _{{\mathsf X}} \frac{|h(x)|^2}{|f(x) -\lambda |^2} d\mu (x) > 1/n^2$ for any $n=1,2,\ldots$. Hence $\lambda \in {ess~ran} (f)$ $\Rightarrow$ $\lambda \in \sigma (T_f)$.

9.5

Consider a PVM $P: \mathscr {B}({\mathbb C}) \rightarrow {\mathsf H}$ with ${\mathsf H}$ separable. Prove $A \in {\mathfrak B}({\mathsf H})$ has the form $A= \int _{\mathbb C}f dP$ for some $f\in M_b({\mathbb C})$ if and only if it commutes with every $B\in {\mathfrak B}({\mathsf H})$ satisfying $BP(E)=P(E)B$ for any $E\in \mathscr {B}({\mathbb C})$.

Solution. The sufficient implication is known, so we just prove the necessary part of the equivalence. Divide supp(P) in a disjoint collection, at most countable, of bounded sets $E_n$, and ${\mathsf H}$ in the corresponding orthogonal sum ${\mathsf H}= \oplus _{n} {\mathsf H}_n$, ${\mathsf H}_n := P(E_n)({\mathsf H})$. Every ${\mathsf H}_n$ is A-invariant, since $AP(E_n) = P(E_n)A$ by assumption. If $A_n := A\upharpoonright _{{\mathsf H}_n}: {\mathsf H}_n \rightarrow {\mathsf H}_n$, then $A\psi = \sum _n A_n\psi$ for any $\psi \in {\mathsf H}$. Moreover (see Corollary 9.42) $A_n$ commutes with any operator in ${\mathfrak B}({\mathsf H}_n)$ that commutes with the bounded normal operator $T_n := \int _{E_n} z dP(z)$ and its adjoint. By Theorem 9.11, $A_n = \int _{E_n} f_n dP$ for some $f_n \in M_b(E_n)$. Define $f(z) := f_n(z)$ on $z\in E_n$, for any $z\in {\mathbb C}$. Then $f_n \rightarrow f$ (the $f_n$ are null outside $E_n$) in $L^2({\mathbb C}, \mu _\psi )$ by dominated convergence if $\psi \in \varDelta _f$. Therefore $A\psi = \int _{\mathbb C}f dP\psi$ for $\psi \in \varDelta _f$, by definition of $\int _{\mathbb C}f dP$. As A is bounded, Corollary 9.5 implies f must be bounded, $\varDelta _f = {\mathsf H}$ and $A= \int _{\mathbb C}f dP$.

9.6

Let ${\mathsf H}$ be separable and $T: {\mathscr {D}}(T) \rightarrow {\mathsf H}$ self-adjoint on ${\mathsf H}$ (not necessarily bounded). Prove that $A\in {\mathfrak B}({\mathsf H})$ has the form $A=f(T)$, for some $f: {\mathbb R}\rightarrow {\mathbb C}$ measurable and bounded, if and only if A commutes with every $B\in {\mathfrak B}({\mathsf H})$ such that $BT \psi = TB\psi$ for any $\psi \in {\mathscr {D}}(T)$.

Solution. If $P^{(T)}$ is the PVM of T, $BP^{(T)}(E) = P^{(T)}(E)B$ $\;\Leftrightarrow \;$ $BT\psi = TB\psi$ for any $\psi \in {\mathscr {D}}(T)$. The claim boils down to proving $A = \int f dP^{(T)}$, f bounded, iff A commutes with any $B\in {\mathfrak B}({\mathsf H})$ commuting with $P^{(T)}$. Exercise 9.5 does exactly that.

9.7

If A is the self-adjoint generator of a strongly continuous one-parameter unitary group $U_t=e^{itA}$, prove that A is bounded (and hence it is defined on the whole Hilbert space) if and only if $||U_t -I|| \rightarrow 0$, as $t\rightarrow 0$.

Hint. Passing to the spectral representation of A, we have $||U_t-I|| = ||f_t||_\infty$ where $f_t(\lambda ) = |e^{it\lambda } -1 |$. Since $(a, b) \ni \lambda \mapsto f_t(\lambda )$ tends to 0 uniformly in $\lambda$, as $t \rightarrow 0$, if and only if a, b are finite, the claim follows.

9.8

Consider the operators A, $A^\star$ of Sect. 9.1.4. Prove they are closable, and

$$\sigma _p(\overline{A}) = {\mathbb C}$$

so that $\sigma (\overline{A})= {\mathbb C}$ while $\sigma _c(\overline{A})= \sigma _r(\overline{A})= \varnothing$.

Outline of solution. The operators are closable because they admit closed extensions, for $A \subset (A^\star )^*$ and $A^\star \subset A^*$. Using the Hilbert basis $\{\psi _n\}_{n \in {\mathbb N}}$ of Sect. 9.1.4, construct explicitly an eigenvector $\psi \in {\mathsf H}\setminus \{\mathbf{0}\}$ of $\overline{A}$ (i.e. $\overline{A}\psi = \lambda \psi$) for every $\lambda \in {\mathbb C}\setminus \{0\}$. Supposing $\psi = \sum _{n\in {\mathbb N}} c_n \psi _n$ we may heuristically assume that $\overline{A}\psi = \lambda \sum _{n\in {\mathbb N}} c_n \sqrt{n}\psi _{n-1}$, so that

$$c_{n+1}= \frac{c_n}{\lambda \sqrt{n+1}}\,.$$

and thus the candidate eigenvector reads

$$\psi = \sum _{n\in {\mathbb N}} \frac{c_0 \lambda ^{-n}}{ \sqrt{(n+1)!}}\psi _n\,.$$

It is easy to prove that, for $c_0\ne 0$, the series converges to a non-zero element of ${\mathsf H}$ which belongs to $D(\overline{A})$ and satisfies $\overline{A} \psi = \lambda \psi$. We already know that $\psi _0$ satisfies $\overline{A}\psi _0 =A\psi _0=\mathbf{0}$, so $0\in \sigma _p(\overline{A})$.

9.9

Consider the operators A, $A^\star$ and the Hilbert basis $\{\psi _n\}_{n\in {\mathbb N}}$ of Sect. 9.1.4. Prove that $A^* = \overline{A}^* =\overline{A^\star }$, that

$$\overline{A} \left( \sum _{n=0}^{+\infty } c_n \psi _n \right) = \sum _{n=0}^{+\infty } \sqrt{n+1}c_{n+1} \psi _n$$

and that

$$A^* \left( \sum _{n=0}^{+\infty } c_n \psi _n \right) = \sum _{n=1}^{+\infty } \sqrt{n}c_{n-1} \psi _n\,,$$

where

$$D(\overline{A})= \left\{ \psi \in {\mathsf H}\,\left| \, \sum _{n=0}^{+\infty }(n+1)|(\psi |\psi _{n+1})|^2 <+\infty \right. \right\} $$

and

$$D(A^*)= \left\{ \psi \in {\mathsf H}\,\left| \, \sum _{n=1}^{+\infty }n|(\psi | \psi _{n-1})|^2 <+\infty \right. \right\} \,.$$

Conclude that $D(\overline{A})=D(A^*)$.

Outline of solution. The solution mostly relies on Proposition 5.17 and on the very definition of adjoint. Apply the definition of adjoint of A and prove that $D(A^*)$ and $A^*$ take the form written above. Next observe that $\overline{A} = (A^*)^*$. Then, applying the definition of adjoint, prove that $D(\overline{A})$ and $\overline{A}$ have the form claimed. Finally, again exploiting the definition of adjoint, demonstrate that $(A^\star )^*= \overline{A}$ and conclude that $\overline{A^\star }= ((A^\star )^*)^* = \overline{A}^*= A^*$. The last statement is quite evident if one simply rearranges the expressions of $D(\overline{A})$ and $D(A^*)$ and uses $\psi \in H$.

9.10

Consider the operators A, $A^\star$ of Sect. 9.1.4. Prove that

$$N:= A^* \overline{A} =\overline{A^\star A}$$

is the unique self-adjoint extension of the symmetric operator N defined on the span of the vectors $\psi _n$ satisfying $N \psi _n = n\psi _n$ for $n\in {\mathbb N}$. The Hilbert basis $\{\psi _n\}_{n \in {\mathbb N}}$ is the one in Sect. 9.1.4.

9.11

Consider the operators A, $A^\star$ of Sect. 9.1.4, prove that $A + A^\star$ and $i(A - A^\star )$ are essentially self-adjoint on ${\mathscr {S}}({\mathbb R})$. Next, study the relation of the closures of those operators and the self-adjoint position and momentum operators.

9.12

Consider the operators A and $A^\star$ of Sect. 9.1.4. Compute $e^{\overline{\alpha A + \overline{\alpha } A^\star }} \psi _n$ with $\alpha \in {\mathbb C}$ given.

9.13

Prove Stone’s formula, valid for a self-adjoint operator $T: D(T) \rightarrow {\mathsf H}$ with spectral measure $P^{(T)}$ and $b > a$. Use the weak operator topology:

$$\frac{1}{2}(P^{(T)}(\{a\}) + P^{(T)}(\{b\})) + P^{(T)}((a, b))= w-\lim _{\varepsilon \rightarrow 0^+}\frac{1}{2\pi i}\int _a^b \frac{1}{T-\lambda -i\varepsilon } - \frac{1}{T-\lambda +i\varepsilon } d \lambda \,.$$

The integral is understood in the sense of Proposition 9.31. Is the identity still valid for $a=b$?

Outline of solution. Define $S(\varepsilon ) := \frac{1}{2\pi i}\int _a^b \frac{1}{T-\lambda -i\varepsilon } - \frac{1}{T-\lambda +i\varepsilon } d \lambda$. Next take $\psi , \phi \in {\mathsf H}$ and prove that

$$(\psi |S(\varepsilon ) \phi ) = \frac{1}{2\pi i} \int _{{\mathbb R}} \left[ \int _a^b \frac{2i\varepsilon }{(\ell -\lambda )^2+ \varepsilon ^2} d\lambda \right] d\mu _{\psi ,\phi }(\ell )$$

$$= \frac{1}{\pi } \int _{{\mathbb R}}\left[ \tan ^{-1}((b -\ell )/\varepsilon ) - \tan ^{-1}((a -\ell )/\varepsilon )\right] d\mu _{\psi ,\phi }(\ell )\,. $$

Prove that, taking the limit when $\varepsilon \rightarrow 0^+$ one obtains

$$\lim _{\varepsilon \rightarrow 0^+} (\psi |S(\varepsilon ) \phi ) = \int _{{\mathbb R}}\left[ \chi _{(a, b)}(\ell ) + \frac{1}{2} (\chi _{\{b\}}(\ell ) + \chi _{\{a\}}(\ell ))\right] d\mu _{\psi ,\phi }(\ell )$$

and conclude. The identity is generally not valid for $a=b$: the right-hand side always vanishes while the left-hand side may not.

9.14

Prove that the result of Exercise 9.13 is valid if we use the strong operator topology:

$$ \frac{1}{2}(P^{(T)}(\{a\}) + P^{(T)}(\{b\})) + P^{(T)}((a, b))= s-\lim _{\varepsilon \rightarrow 0^+}\frac{1}{2\pi i}\int _a^b \frac{1}{T-\lambda -i\varepsilon } - \frac{1}{T-\lambda +i\varepsilon } d \lambda \,.$$

The integral is understood in the sense of Proposition 9.31.

Outline of solution. Since we already know that the convergence is weak, it suffices to show that, if $\phi \in {\mathsf H}$,

$$\lim _{\varepsilon \rightarrow 0^+}||S(\varepsilon ) \phi ||^2 = \left| \left| \frac{1}{2}(P^{(T)}(\{a\})\phi + P^{(T)}(\{b\}))\phi + P^{(T)}((a, b))\phi \right| \right| ^2\,. $$

The left-hand side can be written as

$$\lim _{\varepsilon \rightarrow 0^+}\frac{1}{\pi } \int _{{\mathbb R}}\left| \tan ^{-1}((b -\ell )/\varepsilon ) - \tan ^{-1}((a -\ell )/\varepsilon )\right| ^2 d\mu _{\phi }(\ell )\,,$$

and the limit produces the result we want.

9.15

Consider the operator H in formula (9.66), example Sect. 9.1.4. Show that $\rho _\beta = e^{-\beta H}$ is a well-defined trace-class operator for every $\beta \in {\mathbb C}$, $Re(\beta )>0$. Compute $tr \rho _\beta$ for these values of $\beta$. For $A\in {\mathfrak B}({\mathsf H})$, define

$$\alpha _z(A) := e^{iz H}A e^{-iz H}\,,\quad z \in {\mathbb C}$$

$$\langle A \rangle _\beta := tr(e^{-\beta H}A)\,.$$

and finally

$$F^{(\beta )}_{AB}(z) := \langle B\alpha _z(A)\rangle _\beta \,, \quad G^{(\beta )}_{AB}(z) := \langle \alpha _z(A)B\rangle _\beta \,.$$

Prove that (i) $F^{(\beta )}_{AB}(z)$ is an analytic function on the strip $0< Im(z) < \beta$ and $G^{(\beta )}_{AB}(z)$ is analytic on the strip $-\beta< Im(z) < 0$; (ii) $F_{AB}^{(\beta )}$ and $G_{AB}^{(\beta )}$ are bounded, continuous functions, and they can be extended continuously to the boundaries of their strips; (iii) along the boundaries the KMS condition

$$G_{AB}^{(\beta )}(t) = F_{AB}^{(\beta )}(t+i\beta )\,$$

holds.