Spectral Theory III: Applications

Abstract

In this chapter we examine applications of the theory of unbounded operators in Hilbert spaces, where spectral theory, as developed in Chaps. 8 and 9, plays a paramount technical role during the proofs. The final part of the chapter presents a series of classical results about certain operators of interest in Quantum Mechanics, in particular regarding self-adjointness and spectral lower bounds.

Particles are solutions to differential equations.

Werner Karl Heisenberg

Exercises

10.1

Referring to Example 10.16(1), assume $\gamma >0$. Prove the solution to the Klein–Gordon equation with source $e^{i\omega } \psi$ and dissipative term has the form:

$$u_t = e^{-\gamma t} \left[ \cos \left( t \sqrt{A-\gamma I}\right) v +\sin \left( t \sqrt{A-\gamma I}\right) (A-\gamma I)^{-1/2}(v'+ \gamma v)\right] $$

$$+ e^{-\gamma t} C_t\psi + e^{i\omega t} (A^2 - \omega ^2 I + 2i\gamma \omega I)^{-1} \psi \,,$$

where $||C_t||\le 1$.

Hint. Apply the definition of $\int _a^b L_\tau \psi _\tau d\tau$ given in (10.10). Then pass to the spectral measures of A and use the Fubini–Tonelli theorem, carefully verifying the assumptions.

10.2

If $A_k \in {\mathfrak B}({\mathsf H}_k)$, $k=1,\ldots , N$, prove

$$A_1 \otimes \cdots \otimes A_k \in {\mathfrak B}({\mathsf H}_1 \otimes \cdots \otimes {\mathsf H}_N)\,.$$

Solution. Consider $N=2$, the general case being similar. If $\psi = \{f_i\}_{i \in I}$ and $\{g_j\}_{j \in J}$ are bases of ${\mathsf H}_1$ and ${\mathsf H}_2$, take the finite sum $\psi := \sum _{ij} c_{ij} f_i \otimes g_j$. Then $||(A_1\otimes I) \psi ||^2 = \sum _j ||\sum _i c_{ij}A_1 f_i ||^2 \le \sum _{j} ||A_1||^2 \sum _i |c_{ij}|^2 = ||A_1||^2 ||\psi ||^2$. A density argument allows to conclude $||A_1\otimes I||\le A$, and therefore $||A_1\otimes A_2|| \le ||A_1\otimes I|| \, ||I \otimes A_2|| \le ||A_1||\, ||A_2||$.

10.3

If $A_k \in {\mathfrak B}({\mathsf H}_k)$, $k=1,\ldots , N$, prove that

$$|| A_1 \otimes \cdots \otimes A_k|| = ||A_1|| \cdots ||A_N||\,.$$

Solution. We already know that $A_1 \otimes \cdots \otimes A_k \in {\mathfrak B}({\mathsf H}_1 \otimes \cdots \otimes {\mathsf H}_N)$ from Exercise 10.2. Take $n=2$, the generalisation being similar. If $A_1=0$ or $A_2=0$ the claim is obvious, so assume $||A_1||, ||A_2|| >0$. When solving (2) we found $||A_1\otimes A_2|| \le ||A_1||\, ||A_2||$, so it is enough to obtain the opposite inequality. By definition of $||A_1||$ and $||A_2||$, for any $\varepsilon >0$ there are $\psi ^{(\varepsilon )}_1\in {\mathsf H}_1$ and $\psi ^{(\varepsilon )}_2 \in {\mathsf H}_2$, $||\psi ^{(\varepsilon )}_1||, ||\psi ^{(\varepsilon )}_2|| =1$, such that $|\,||A_i \psi ^{(\varepsilon )}_i|| - ||A_i|| \,| < \varepsilon$. In particular, $||\psi ^{(\varepsilon )}_i || \ge ||A_i|| -\varepsilon$. With these choices

$$||(A_1\otimes A_2) (\psi ^{(\varepsilon )}_1\otimes \psi ^{(\varepsilon )}_2) || = ||A_1\psi ^{(\varepsilon )}_1||\, ||A_2 \psi ^{(\varepsilon )}_2|| \ge ||A_1||\, ||A_2|| - \varepsilon (||A_1|| + ||A_2||) +\varepsilon ^2\,.$$

Since $||\psi ^{(\varepsilon )}_1\otimes \psi ^{(\varepsilon )}_2||=1$, and from

$$||A_1\otimes A_2|| = \sup _{||\psi ||=1} ||A_1 \otimes A_2 \psi || \ge ||(A_1\otimes A_2) (\psi ^{(\varepsilon )}_1\otimes \psi ^{(\varepsilon )}_2) ||\,,$$

for any $\varepsilon >0$ we have $||A_1\otimes A_2|| \ge ||A_1||\, ||A_2|| - \varepsilon (||A_1|| + ||A_2||) +\varepsilon ^2$, where $-\varepsilon (||A_1|| + ||A_2||) +\varepsilon ^2 <0$. That value tends to 0 as $\varepsilon \rightarrow 0^+$. Eventually, $||A_1\otimes A_2|| \ge ||A_1||\, ||A_2||$ as required.

10.4

If $A_k \in {\mathfrak B}({\mathsf H}_k)$, $k=1,\ldots , N$ prove

$$(A_1 \otimes \cdots \otimes A_k)^*= A^*_1 \otimes \cdots \otimes A^*_N\,.$$

Hint. Check $A^*_1 \otimes \cdots \otimes A^*_N$ satisfies the properties of the adjoint to a bounded operator (Proposition 3.36).

10.5

If $P_k \in {\mathfrak B}({\mathsf H}_k)$, $k=1,\ldots , N$ are orthogonal projectors, show that $P_1 \otimes \cdots \otimes P_k$ is an orthogonal projector.

10.6

Suppose that $A: D(A) \rightarrow {\mathsf H}$ is a closed, densely defined, normal operator ($A^*A=AA^*$). Prove that $D(A)=D(A^*)$.

Solution. Since $A^*A = AA^*$ can be written $A^*A = (A^*)^*A^*$, Theorem 10.37(c) implies $D(A)=D(A^*)$.

10.7

Suppose that $A: D(A) \rightarrow {\mathsf H}$ is a closed, densely defined, normal operator ($A^*A=AA^*$). Referring to the polar decompositions $A=UP$ and $A^*=VS$, show the following facts hold

• (i) $V=U^*$ and $S=P$,

• (ii) U is normal,

• (iii) $UA \subset AU$, $U^*A \subset AU^*$, $UP\subset PU$, $U^*P\subset PU^*$, $AP \subset PA$, $A^*P\subset PA^*$.

Solution. Start from $UPPU^*= AA^*= A^*A= PU^*UP$. Since $U^*U|_{Ran(P)}=I|_{Ran(P)}$, we have $UP^2U^* = P^2$ and so $P^2U^*=U^*P^2$. Therefore $U^*$ commutes with the PVM of $P^2$, and hence with every function of it, in particular $P=\sqrt{P^2}$: $U^*P\subset PU^*$. Taking the adjoint we also have $UP\subset PU$. Taking the adjoint of both sides of $A=UP$ we get $A^*= PU^*$, and then $A^*x =U^*Px$ provided $x\in D(P)$. However $D(P)= D(A^*)$ because $D(P)=D(A)=D(A^*)$ (Theorem 10.37), so that $A^* =U^*P$. Now, notice that $U^*$ is an isometry on Ran(P) because U is an isometry on Ran(U) and $(U^*Px|U^*Px) = (x|PUU^*Px) =(x|AA^*x)= (x|A^*Ax)= (x|PU^*UPx) = (x|PPx)= (Px|Px)$. A similar argument proves that $(Px| (U^*U-UU^*) Px) =0$. Actually, since $Ker (P) \perp Ran(P)$, $Ker(P)\oplus \overline{Ran(P)}= {\mathsf H}$ and finally $U(Ker(P))\subset { Ker}(P)$ and $U^*(Ker(P))\subset { Ker}(P)$ (for U and $U^*$ commute with P), then $(Px| (U^*U-UU^*) Px) =0$ can be extended to $(y| (U^*U-UU^*) y) =0$ for every $y \in {\mathsf H}$. We conclude that $UU^*=U^*U$ as requested, but also $Ker(U^*)= { Ker}(U)$. The latter contains Ker(P) by hypothesis, so $Ker(U) \supset { Ker}(P)$. In summary, $A^*=U^*P$ satisfies all requirements defining the polar decomposition of $A^*$. Uniqueness concludes the proof of the fact that $A^* = U^*P$ is the polar decomposition of $A^*$. Finally observe that, since U commutes with P, we have $UA = UUP \subset UPU =AU$. A similar argument proves $UA^* \subset A^*U$. Regarding the last inclusions: $AP= UPP \subset PUP = PA$ and $A^*P = U^*PP \subset PU^*P = PA^*$.

10.8

Suppose that $A: D(A) \rightarrow {\mathsf H}$ is a closed, densely defined, self-adjoint (or anti-self-adjoint) operator and assume that $Ker(A)= \{\mathbf{0}\}$. Referring to the polar decomposition $A=UP$ prove that $U=U^*=U^{-1}$ (respectively, $U=-U^*=U^{-1}$).

Hint. Use the spectral decomposition theorem.

10.9

Study the polar decomposition $\overline{A} = U P$ for the operator A of Sect. 9.1.4. Prove that U satisfies

$$U \psi _n = \psi _{n-1}$$

if $n\ge 1$ and $\{\psi _n\}_{n\in {\mathbb N}}$ is the basis of $L^2({\mathbb R}, dx)$ defined in Sect. 9.1.4.

10.10

Study the polar decomposition $\overline{A}^* = V P_1$ for the operator A of Sect. 9.1.4. Prove that V satisfies

$$V \psi _n = \psi _{n+1}$$

if $n\ge 0$ and $\{\psi _n\}_{n\in {\mathbb N}}$ is the basis of $L^2({\mathbb R}, dx)$ defined in Sect. 9.1.4.

10.11

Referring to Exercises (10.9) and (10.10), prove that

$$P= \sqrt{N}\quad \text{ and }\quad P_1= \sqrt{N+I}$$

where N is the unique self-adjoint extension of the symmetric operator defined on the finite span of the Hilbert basis $\{\psi _n\}_{n \in {\mathbb N}}$, introduced in Sect. 9.1.4, such that $N \psi _n = n \psi$ for $n \in {\mathbb N}$.

10.12

Consider the operator A of Sect. 9.1.4 and prove that $\sigma _p(A^*)= \sigma _c(A^*) = \emptyset$ and $\sigma _r(A^*)= {\mathbb C}$.

Solution. We already know from Exercise 9.8 that $\sigma _p(\overline{A}) = {\mathbb C}$, while Exercises 10.10 and 10.11 tell us that $A^* = V\sqrt{N+1}$. Consider $\psi \in {\mathsf H}$ such that $A^*\psi = \lambda \psi$ for some $\lambda \in {\mathbb C}$. Since $\sqrt{N+1}$ is invertible, this equation is equivalent to $V\phi = \lambda \sqrt{N+1}^{-1}\phi$ where both V and $\sqrt{N+1}^{-1}$ are bounded and everywhere defined. Expanding $\phi = \sum _{n\in {\mathbb N}} c_n \psi _n$, with reference to the Hilbert basis $\{\psi _n\}_{n \in {\mathbb N}}$ introduced in Sect. 9.1.4, we find the identity

$$\sum _{n} c_n \psi _{n+1} = \lambda \sum _{n} \frac{c_n}{\sqrt{n+1}} \psi _{n}$$

and hence

$$\lambda c_n = \sqrt{n+1} c_{n-1}\quad n=1,2, \ldots \,.$$

If $\lambda =0$ the only solution is $c_n=0$ for every n, and so $\psi = \sqrt{N+1}^{-1}{} \mathbf{0}=\mathbf{0}$. If $\lambda \ne 0$, we have $c_n = \frac{\sqrt{(n+1)!}}{\lambda ^n} c_0$. Unless $c_0=0$, giving $\psi =\mathbf{0}$ again, we have $\sum _n |c_n|^2 = +\infty$, and we conclude $\psi$ cannot exist. We have proved that $\sigma _p(A^*)= \emptyset$. Next observe that, as $\overline{A}$ is closed and the domain of $A^*= \overline{A}^*$ is dense, Theorem 5.10(c) implies $Ran(A^*-{\lambda } I)^{\perp } = { Ker}(\overline{A}- \overline{\lambda }I) {\ne } \{\mathbf{0}\}$ (the last inequality was proved in Exercise 9.8). Hence $Ran(A^*-\lambda I)$ cannot be dense for every $\lambda \in {\mathbb C}$, and therefore $\sigma _r(A^*)= {\mathbb C}$ because $A^*-\lambda I$ is injective for every $\lambda \in {\mathbb C}$, as we proved above.

10.13

Prove the statement in Remark 10.43(2).

10.14

Suppose $V: {\mathbb R}^3 \rightarrow {\mathbb R}$ makes the symmetric operator $H_1$, given by the differential operator $-\varDelta _{x} + V(x)$, essentially self-adjoint on ${\mathscr {S}}({\mathbb R}^3)$, where $\varDelta _x := \sum _{k=1}^3 \frac{\partial ^2}{\partial x_i^2}$ is the Laplacian. Prove that the symmetric operator H on $L^2({\mathbb R}^3\times {\mathbb R}^3, dx\otimes dy)$ defined by the differential operator $-\varDelta _{x} + V(x) -\varDelta _{y} + V(y)$ is essentially self-adjoint on the span of finite products of a map in x in ${\mathscr {S}}({\mathbb R}^3)$ and a map in y in ${\mathscr {S}}({\mathbb R}^3)$. Then show

$$\sigma (\overline{H}) = \overline{\sigma (H_1) + \sigma (H_1)}\,.$$

10.15

Prove that the attractive Coulomb potential in ${\mathbb R}^3$:

$$V(x) = \frac{eQ}{|x|}\,,$$

with $e<0$, $Q>0$ constants, $|x| := \sqrt{x_1^2+x_2^2+x_3^2}$, satisfies Kato’s Theorem 10.50. What happens if we increase or decrease the dimension of ${\mathbb R}^3$?