\input amstex \documentstyle{amsppt} \Monograph \TagsOnRight \NoBlackBoxes \topmatter \title Spectral Analysis of Rank One Perturbations and Applications \endtitle \author Barry Simon \endauthor \affil Division of Physics, Mathematics, and Astronomy \\ California Institute of Technology, 253-37 \\ Pasadena, CA 91125 \endaffil \abstract A review of the general theory of self-adjoint operators of the form $A+\alpha B$ where $B$ is rank one is presented. Applications include proofs of localization for Schr\"odinger operators, results on inverse spectral theory, and examples of operators with singular continuous spectrum. \endabstract \endtopmatter \document \vskip 3.0in \flushpar Lecture given at the Vancouver Summer School in Mathematical Physics, August 10--14, 1993. This material is based upon work supported by the U.S.~National Science Foundation under Grant No.~DMS-9101715. The Government has certain rights in this material. \newpage \title\chapter{1} Introduction, Borel Transforms,\\ the Krein Spectral Shift, and All That \endtitle \leftheadtext{Introduction, Borel Transforms, the Krein Spectral Shift} \endtopmatter \vskip 0.30in \head I.0 Introduction \endhead \rightheadtext{Introduction} \bigpagebreak Our goal here is to review the spectral theory (with applications) of rank one perturbations of positive self-adjoint operators on a separable complex Hilbert space, $\Cal H$: $$A_{\alpha}=A+\alpha B \qquad B=(\varphi, \cdot)\varphi. \tag I.1$$ The cynic might feel that I have finally sunk to my proper level. I started with quantum field theory, analysis in infinitely many variables. That was too hard so I switched to the $N$-body Schr\"odinger equation; but that was too hard so I switched to one-body, then one-dimensional, then discrete one-dimensional. Finally to rank one perturbations---maybe something so easy that I can say something useful! Alas, we'll see even this is hard and exceedingly rich. I should warn you that (I.1) is somewhat more general than you might think. First, I'll bet you thought that $\varphi$ was a unit vector. It need not be. Big deal, you think---normalize $\varphi$ and renormalize $\alpha$. But $\varphi$ need not be a normalizable vector. We'll consider $B$'s which are rank one but only form bounded perturbations of $A$. In one fell swoop, we've absorbed the theory of variation of boundary conditions for Sturm-Liouville problems on a half-line! Moreover, as I'll discuss in \S I.5, $\alpha$ may be infinite. In some ways, my recent relationship with rank one perturbations reminds me of my relationship with trace ideals fifteen years ago. Then it turned out three distinct problems: Yukawa field theories, scattering theory, and semi-classical bounds on the number of bound states led me to similar mathematics, viz estimates on the trace ideal properties of certain pseudo-differential operators. This past year, three distinct research projects: \roster \item my work, in part with del Rio, Jitomirskaya, and Makarov, on singular spectrum; \item my trying to understand the work of Aizenman and Molchanov; \item my work with Gesztesy on trace formulas \endroster all led to rank one perturbations and enriched my own knowledge of the subject. (1)--(3) are the themes of Chapters II--IV of these notes. In this chapter I'll discuss the general theory in a new systematic way; many of the ideas were developed in discussions with Fritz Gesztesy, to whom I am grateful. A key role will be played by $$F_{\alpha}(z)=(\varphi, (A_{\alpha}-z)^{-1}\varphi). \tag I.2$$ In terms of a suitable spectral measure $d\mu_{\alpha}$: $$F_{\alpha}(z)=\int\frac{d\mu_{\alpha}(\lambda)}{\lambda -z} \tag I.3$$ so we'll begin with an exhaustive study of the Borel transform of a measure and its relation to the measure in \S I.1. That section and \S I.2 are the technical core behind most of what follows. \vskip 0.3in \head I.1 Borel Transforms of Measures \endhead \rightheadtext{Borel Transforms of Measures} \bigpagebreak Throughout, we let $d\mu$ be a (positive) measure on $[a, \infty)$ for some $a>-\infty$ with $$\int\frac{d\mu(\lambda)}{|\lambda|+1} <\infty. \tag I.4$$ \remark{Remarks} 1. We'll suppose that $\text{supp}(\mu)$ is bounded from below and the operator $A$ in (I.1) is positive. Much of our theory works without those restrictions but the details are a little simpler with them and they hold in the applications I want to make, so I've made them. 2. For some results when one only knows that $\int\frac{d\mu(\lambda)}{1+\lambda^{2}}<\infty$, see Aronszajn-Donoghue [4]. \endremark \medpagebreak When I.4 holds, we can define for $z\in\Bbb C\backslash(-\infty, a)$: $$F(z)=\int\frac{d\mu(\lambda)}{\lambda - z} \tag I.5$$ the {\it Borel transform} of $\mu$ (also called the Stieltjes transform or the Borel-Stieltjes transform). The key issues concern boundary values of $F$ as $z=x+i\epsilon\downarrow x\in\text{supp}(d\mu)$. We begin with the simple stuff away from $\text{supp}(d\mu)$. \proclaim{Theorem I.1} \roster\runinitem"\rom{(i)}" $F$ is an anlytic function on $\Bbb C\backslash(a, \infty)$. \item"\rom{(ii)}" $F$ is positive on $(-\infty, a)$ and $\text{\rom{Im }} F>0$ if $\text{\rom{Im }} z>0$. \item"\rom{(iii)}" $\lim\limits_{\gamma\to\infty} F(-\gamma)=0$. \item"\rom{(iv)}" $\lim\limits_{\gamma\to\infty} F(-\gamma)(\gamma) =\int d\mu(\lambda)$ where $\int d\mu=\infty$ is allowed. Similarly, $$\lim\limits_{\gamma\to\infty}F'(-\gamma)(\gamma)^{2}=\int d\mu(\lambda).$$ \item"\rom{(v)}" $\lim\limits_{\gamma\to\infty} F'(-\gamma)/F^{2}(- \gamma)=(\int d\mu(\lambda))^{-1}$ where $\int d\mu=\infty$ is allowed with $\infty^{-1}\mathbreak =0$. \endroster \endproclaim \demo{Proof} (i) and (ii) are elementary. (iii) uses: $$\text{Im } F(x+i\beta)=\int\frac{\beta}{(x-y)^{2}+\beta^{2}}\,d\mu(y). \tag I.6$$ (iii) and (iv) follow from the monotone convergence theorem. Actually, by a slightly more involved argument if $|z|\to\infty$ with $|\text{arg}(-z)|<\epsilon$ with $\epsilon >0$ and small, it is not hard to see that $(-z)F(z)\to\int d\mu$. If $\int d\mu <\infty$, a similar use of the monotone convergence theorem shows that $$\gamma^{2}F'(-\gamma)=\gamma^{2}\int\frac{d\mu(\lambda)}{(\lambda + \gamma)^{2}} \longrightarrow\int d\mu(\lambda)$$ and (iv) follows. If $\int d\mu=\infty$, we use the fact that $|1/[z F(z)]|\to 0$ as $|z|\to\infty$ with $\text{Re } z<0$. Thus, writing $$\biggl(\frac1F\biggr)'(z)=(2\pi i)^{-1}\int\limits_{|w-z|=|z|/2} (w-z)^{-2}\frac1{F(w)}\, dw$$ we see that $(\frac1F)'\to 0$ but $(\frac1F)'=-\frac{F'}{F^2}$ implying (v). \qed \enddemo Next we talk about boundary values of $F$. The function $$G(x)=\int\frac{d\mu(y)}{(x-y)^{2}}$$ which is defined for $x\in (-\infty, \infty)$ and takes values in $(0, \infty]$ (note $\infty$ is allowed), will play an important role. \proclaim{Theorem I.2} \roster\runinitem"\rom{(i)}" $G(x)<\infty$ implies that $$\int\frac{d\mu(y)}{|x-y|}<\infty. \tag I.7$$ \item"\rom{(ii)}" {\rom{[12,27]}} $\{x\mid G(x)=\infty\}$ is a dense $G_{\delta}$ in $\text{\rom{supp}}(d\mu)$. \item"\rom{(iii)}" If {\rom{(1.7)}} holds, and, in particular if $G(x)<\infty$, then $\lim\limits_{\epsilon\downarrow 0} F(x+i\epsilon)$ exists and is real. \item"\rom{(iv)}" $\lim\limits_{\epsilon\downarrow 0} \epsilon^{-1} \text{\rom{Im }}F(x+i\epsilon)\to G(x)$. \item"\rom{(v)}" If $G(x)<\infty$, then $\lim\limits_{\epsilon\downarrow 0}(i\epsilon)^{-1}[F(x+i\epsilon)- F(x+i0)]=G(x)$. \endroster \endproclaim \remark{Remark} A $G_\delta$ is a countable intersection of open sets. \endremark \demo{Proof} (i) is immediate if one notes that $$\frac1{|x-y|}\leq\frac1{|x-y|^2} + \frac{|x|+2}{|y|+1}$$ since $$|y|+1\leq |x|+|x-y|+1\leq (|x-y|)(|x|+2)$$ if $|x-y|\geq 1$. (ii) By the lemma below, if $\mu([0, 1])\geq 1$, there is an $x_{0}\in [0, 1]$ with $\int\frac{d\mu(y)}{|x_{0}-y|}=\infty$. So, by scaling, if $\mu([a, b])>0$, there is an $x_{0}\in [a, b]$ with $\int\frac{d\mu(y)}{|x_{0}-y|}=\infty$. It follows that if $a_{0}\in\text{supp}(d\mu)$, then there exists $a_{n}\to a_{0}$ with $\int\frac{d\mu(y)}{|a_{n}-y|}=\infty$. So, by (i), $\{x\mid G(x)=\infty\}$ is dense in $\text{supp}(d\mu)$. It is a $G_\delta$ because if we define $G_{n}(x)=\int (|x-y|^{2}+n^{-1})^{-1}\,d\mu(y)$, then $G_n$ is continuous and $$\{x\mid G(x)=\infty\}=\bigcap_{m}\bigcup_{n}\{x\mid G_{n}(x)>m\}.$$ (iii) By (I.6), for any $\delta$ $$\text{Im }F(x+i\beta)\leq\beta\int\limits_{|x-y|>\delta} |x-y|^{-2}\,d\mu(y)+ \int\limits_{|x-y|\leq\delta} |x-y|^{-1}\,d\mu(y)$$ so $$\overline{\lim_{\beta\downarrow 0}}\, \text{Im }F(x+i\beta) \leq\overline{\lim_{\delta\downarrow 0}}\, \int\limits_{|x-y|\leq \delta} |x-y|^{-1}\,d\mu(y)=0$$ if (I.7) holds. (I.7) also implies $$\lim_{\beta\downarrow 0}\, \text{Re }F(x+i\beta)=\int (y-x)^{-1}\,d\mu(y)$$ by the dominated convergence theorem. (iv) follows from (I.6) and the monotone convergence theorem. (v) If $G(x)<\infty$, then, by the proof of (iii) $$F(x+i0)=\int\frac{d\mu(y)}{y-x}$$ so $$(i\epsilon)^{-1}[F(x+i\epsilon)-F(x-i0)]=\int\frac{d\mu(y)}{(y-x) (y-x-i\epsilon)}\to G(x)$$ by the dominated convergence theorem if $G(x)<\infty.$ \qed \enddemo \proclaim{Lemma 1.3} Let $\mu([0, 1])\geq 1$. Then, there exists $x_{0}\in [0, 1]$ with $\int\frac{d\mu(y)}{|x_{0}-y|}=\infty$. \endproclaim \demo{Proof} Define $I_n$ inductively as follows. Let $I_1$ be one of $[0, \frac12], [\frac12, 1]$ with $\mu(I_{1})\geq \frac12$. Now subdivide $I_1$ in half and choose $I_2$ to be one of those halves with $\mu(I_{2})\geq\frac14$. By induction $$I_{1}\supset I_{2}\supset\cdots \quad \text{all closed}; \quad\mu(I_{n})\geq 2^{-n}; \quad |I_{n}|=2^{-n}.$$ By compactness, $\cap I_{n}=\{x_{0}\}$ for some $x_0$. Obviously $$\int\limits_{I_n} |x-x_{0}|^{-1}\,d\mu(x)\geq (2^{-n})^{-1} 2^{-n}=1.$$ But if $\int |x-x_{0}|^{-1}\,d\mu(x)<\infty$, $\int\limits_{I_n}|x- x_{0}|^{-1}\,d\mu(x)\downarrow 0$ by monotone convergence, so we must have $\int |x-x_{0}|^{-1}\,d\mu(x)=\infty$. \qed \enddemo \proclaim{Theorem I.4} $\lim\limits_{\beta\downarrow 0} F(x+i\beta)$ exists and is finite for a.e.~$x$. \endproclaim \demo{Proof} This is a standard harmonic analysis result on non-tangential boundary values of analytic maps, see [31], so we'll only sketch the details. Let $g(z)=\frac{z-i}{z+i}$, the fractional linear map of the upper half plane to the unit disc. Thus $\tilde{F}=g\circ F\circ g^{-1}$ is a map of the unit disc to itself. Such maps have boundary values by a standard maximal function argument, see Katznelson [31]. By the same argument as in the next theorem, $\{\theta\mid \lim\limits_{r\uparrow 1} \tilde{F}(re^{i\theta})=-1\}$ has measure zero, so $\lim\limits_{\beta\downarrow 0}F(x+i\beta)=\infty$ on a set of measure 0 and so $F$ is finite a.e. \qed \enddemo \proclaim{Theorem I.5} Let $F_{1}, F_{2}$ be the Borel transforms of two \rom(positive\rom) measures \rom(one may be zero\rom) and fix $\alpha\in\Bbb C$. Then $$|\{x\mid F_{1}(x+i0)-F_{2}(x+i0)=\alpha\}|>0$$ only if $\alpha=0$ and $F_{1}=F_2$. \endproclaim \demo{Proof} Let $g$ be defined as in the proof of Theorem I.4 and let $$h=e^{iF_{1}\circ g^{-1}} - (e^{iF_{2}\circ g^{-1}}) e^{i\alpha}.$$ $h$ is a bounded analytic function on the disc and the theorem asserts that under $h\not\equiv 0$, $\{\theta\mid\lim\limits_{r\uparrow 1} h(re^{i\theta})=0\}$ has Lebesgue measure zero. This is a standard theorem; see Katznelson [31]. \qed \enddemo Let $d\mu_{\text{\rom ac}}, d\mu_{\text{\rom sc}}, d\mu_{\text{\rom pp}}$ be the absolutely continuous, singular continuous and pure point parts of a measure $d\mu$. Let $d\mu_{\text{\rom sing}}= d\mu_{\text{\rom sc}}+d\mu_{\text{\rom pp}}$. The main theorem on Borel transforms is \proclaim{Theorem I.6} Let $F(z)$ be the Borel transform of a measure $d\mu$ obeying {\rom{(I.4)}}. Then \roster \item"\rom{(i)}" $\pi^{-1}\,\text{\rom{Im }} F(E+i\epsilon) dE\to \,d\mu$ weakly in the sense that $$\int\pi^{-1} f(E)\, \text{\rom{Im }} F(E+i\epsilon)\, dE\longrightarrow\int f(E)\,d\mu(E) \tag I.8$$ for all continuous functions, $f$, of compact support. \item"\rom{(ii)}" $\mu_{\text{\rom sing}}$ is supported on $\{E\mid\lim\limits_{\epsilon\downarrow 0}\, \text{\rom{Im }} F(E+i0)=\infty\}$. \item"\rom{(iii)}" $\mu(\{E_{0}\})=\lim\limits_{\epsilon\downarrow 0} \epsilon \text{\rom{ Im }}F(E_{0}+i\epsilon)$. Moreover, for any $E_0\in\Bbb R$: $$\lim_{\epsilon\downarrow 0}\epsilon\text{\rom{ Re }}F(E_{0}+i0)=0.$$ \item"\rom{(iv)}" $d\mu_{\text{\rom ac}}(E)=\pi^{-1}\, \text{\rom{Im }} F(E+i0)\, dE$. \endroster \endproclaim \demo{Proof} (i) It is a well-known calculation that for $f$, continuous of compact support: $$\lim_{\epsilon\downarrow 0}\frac1\pi \int f(E') \frac{\epsilon} {(E-E')^{2}+\epsilon^2}\, dE'\to f(E).$$ This and the dominated convergence theorem easily imply the result. (ii) We will prove the weaker fact that $\mu_{\text{\rom sing}}$ is supported on $\{E\mid\overline{\lim} \text{ Im }F(E+i2^{-n})= \infty\}$. The stronger fact (theorem of de Vall\'ee Poussin) can be found in Saks ([40]), but for most applications, this weaker fact suffices. Since the stronger result is true and more elegant, we'll use it. For the theorem of Aronszajn on mutual singularity (Theorem II.2(iv)), one does need the stronger fact. We will prove that for any $a$ $$\bigl\{E\mid\sup_{n}\, \text{Im }F(E+i2^{-n})\leq a\bigr\}=A_a$$ has $\mu_{\text{\rom sing}}$-measure 0 from which the result follows. Notice that \align \text{Im }F(x+i\epsilon) &=\int\frac{\epsilon}{(x-y)^{2}+\epsilon^2} \,d\mu(y) \\ &\geq \frac1{2\epsilon}\int\limits_{|x-y|\leq\epsilon}\, d\mu(y). \endalign So for fixed $x\in A_{a}$ $$\sup_{n} 2^{n}\mu\left([x-2^{-n}, x+2^{-n}]\right)\leq 2a$$ and so, since any $2^{n}\leq\epsilon^{-1}\leq 2^{n+1}$ for some $n$ $$\sup_{n}\epsilon^{-1}\mu\left([x-\epsilon, x+\epsilon]\right)\leq 4a$$ and thus looking at the two subintervals: $$\sup_{x\in [\alpha, \beta]} |\beta -\alpha|^{-1}\mu ([\alpha, \beta])\leq 8a.$$ By regularity of measures, for any $\epsilon$, we can find a closed set $C$ and open $O$ with $C\subset O$, so $\mu_{\text{\rom sing}} (\Bbb R\backslash C)\leq\epsilon$ and $|O|\leq\epsilon$. We write $O= \operatornamewithlimits{\cup}\limits^{\infty}_{n=1}(\alpha_{n}, \beta_{n})$ as a disjoint union of intervals (union of connected components). Then \align \mu_{\text{\rom sing}}(C\cap A_{a}) &\leq \mu_{\text{\rom sing}} \biggl(\bigcup_{\{n\mid (\alpha_{n}, \beta_{n})\cap A_{a}\neq\emptyset\}} (\alpha_{n}, \beta_{n})\biggr) \\ &\leq \sum_{\{n\mid (\alpha_{n}, \beta_{n})\cap A_{a}\neq\emptyset\}} \mu (\alpha_{n}, \beta_{n}) \\ &\leq 8a \sum_{n}|\beta_{n}-\alpha_{n}|=8a |O|\leq 8a\epsilon \endalign so $$\mu_{\text{\rom sing}}(A_{a})\leq\mu_{\text{\rom sing}} (C\cap A_{a}) + \mu_{\text{\rom sing}} (\Bbb R\backslash C)\leq \epsilon+8a\epsilon.$$ Since $\epsilon$ is arbitrary, $\mu_{\text{\rom sing}}(A_{a})=0$ as was to be proven. (iii) The first part is a trivial consequence of the monotone convergence theorem, if we note that $\frac{\epsilon^2}{x^{2}+\epsilon^{2}}$ converges monotonically to 0 (resp.~1) if $x\neq 0$, (resp.~$x=0$). For the second: $$\epsilon\int\limits_{|x-y|\geq 1} \frac{|x-y|}{|x-y|^{2}+\epsilon^{2}} \,d\mu(y)\leq \epsilon\int\limits_{|x-y|\geq 1}\frac1{|x-y|}\, d\mu(y) \longrightarrow 0$$ while $$\lim_{\epsilon\downarrow 0} \int\limits_{|x-y|<1} \frac{\epsilon|x-y|}{(x-y)^{2} +\epsilon^{2}}\, d\mu(y)=0$$ by the dominated convergence theorem since $\alpha\beta/(\alpha^{2}+\beta^{2})\leq \frac12$. (iv) Write $d\mu_{\text{\rom ac}}(E)=h(E)\,dE$ and let $F_{\text{\rom ac}}(z)$ be the Borel transform of $d\mu_{\text{\rom ac}}$. By the Lebesgue theorem on the derivatives of the integral (that says if $h\in L^{1}$, for a.e.~$x$, $h(x)=\lim\limits_{a\to 0}(2a)^{-1} \int\limits^{x+a}_{x-a} h(y)\,dy$; see Katznelson [31]), one sees that $\frac1{\pi}\text{ Im }F_{\text{\rom ac}}(E+i0)=h(E)$ for a.e.~$E$. By Fatou's lemma, for $f\geq 0$ and continuous: \align \pi^{-1}\int [\text{Im }F(E+i0)]f(E)\,dE &\leq \underline{\lim}\,\pi^{-1} \int\text{Im } F(E+i\epsilon)f(E)\,dE \\ &= \int f(E)\,d\mu(E). \endalign Thus $\pi^{-1}\text{Im }F(E+i0)\,dE\leq d\mu$ and so for a.e.~$E$: $$\frac1\pi \text{ Im }F(E+i0)\leq h(E)=\frac1\pi \text{ Im } F_{\text{\rom ac}} (E+i0)\leq\frac1\pi \text{ Im }F(E+i0)$$ since $d\mu\geq d\mu_{\text{\rom ac}}$ implies $\text{Im }F\geq \text{Im }F_{\text{\rom ac}}$. Thus $$h(E)=\frac1\pi \text{ Im }F(E+i0)$$ as was to be proven. \qed \enddemo \vskip 0.3in \head I.2 Rank One Perturbations: The Set-Up and Basic Formulas \endhead \rightheadtext{Rank One Perturbations: Set-Up and Basic Formulas} \bigpagebreak Let $A\geq 0$ be a possibly unbounded self-adjoint operator on a Hilbert space $\Cal H$. One defines the scale of spaces $\Cal H_{s}(A)$ associated to $A$ as follows. For $s\geq 0$, $\Cal H_{s}(A)$ is $D(A^{s/2})$ with the norm $$\|\varphi\|_{s}=\|(A+1)^{s/2}\varphi\| \tag I.9$$ in which $\Cal H_{s}$ is easily seen to be complete. For $s<0$, take $\Cal H$ with the norm given by (I.9) and complete it. In a natural way, $\Cal H_{s}$ and $\Cal H_{-s}$ are duals in such a way that $\varphi\in\Cal H$ is associated to the functional on $\Cal H_{s}$ given by $\ell_{\varphi}(\eta)=(\varphi, \eta)$, so $|\ell_{\varphi}(\eta)|\leq\|\varphi\|_{-s}\|\eta\|_{s}$ and $\Cal H_{s}\subset\Cal H_{t}$ if $s>t$. $(A+1)^{-t/2}$ is an isometry from $\Cal H_{s}$ to $\Cal H_{s+t}$ so for $s>0$, if $\Delta$ is a bounded subset of $\Bbb R$, $E_{\Delta}(A)$, the spectral projection of $A$, maps $\Cal H_{-s}$ to $\Cal H_{s}$ since $(A+1)^{+s/2}E_{0}(A)(A+1)^{s/2}$ is bounded. In particular, we can define a spectral measure $d\mu^{\varphi}_{A}$ for any $\varphi\in \Cal H_{-s}(A)$ by $$\mu^{\varphi}_{A}(\Delta)=(\varphi, E_{\Delta}(A)\varphi).$$ It is easy to see that $$\int\frac{d\mu^{\varphi}_{A}(x)}{(|x|+1)^{s}} < \infty \tag I.10$$ and, in particular, the spectral measures of $\varphi\in \Cal H_{-1}(A)$ have Borel transforms. Now let $\varphi\in\Cal H_{-1}(A)$ and let $b$ be the quadratic form on $\Cal H_{+1}(A)$ given by $$b_{\varphi}(\eta, \eta')=\overline{\ell_{\varphi}(\eta)}\, \ell_{\varphi}(\eta').$$ $b_\varphi$ is positive but if $\varphi\notin\Cal H$, it is neither closed nor closable. See [30,38] for discussion of forms and their perturbation theory. \proclaim{Proposition I.7} For any $\varphi$, $b_{\varphi}$ is a form bounded perturbation of $A$ with relative bound zero. \endproclaim \demo{Proof} Given $\epsilon$, find $\varphi_{\epsilon}\in\Cal H_{-1} (A)$ so $\|\varphi_{\epsilon}\|^{2}_{-1}\leq\frac12 \epsilon$ and $\psi_{\epsilon}\equiv\varphi - \varphi_{\epsilon}\in\Cal H$. Then \align b_{\varphi} &= b_{\varphi_{\epsilon}+\psi_{\epsilon}}\leq b_{\varphi_{\epsilon}+\psi_{\epsilon}}+b_{\varphi_{\epsilon}- \psi_{\epsilon}} \\ &= 2 b_{\varphi_{\epsilon}}+2 b_{\psi_{\epsilon}}. \endalign So, for any $\eta\in\Cal H_{+1}(A)$ \align b_{\varphi}(\eta, \eta) &\leq 2 \|\varphi_{\epsilon}\|^{2}_{-1} \|\eta\|^{2}_{+1} + 2\|\psi_{\epsilon}\|^{2} \|\eta\|^{2} \\ &\leq \epsilon (\eta, A\eta) + (2\|\psi_{\epsilon}\|^{2}+\epsilon) \|\eta\|^{2} \endalign as was to be proven. \qed \enddemo The perturbation theory of forms [30,38] thus implies that for any $\alpha\in\Bbb R$, $A+\alpha b$ is the quadratic form of a self- adjoint operator $A_{\alpha}$ with $\Cal H_{s}(A_{\alpha})=\Cal H_{s} (A)$ if $|s|\leq 1$. We will write $$A_{\alpha}=A+\alpha B=A+\alpha(\varphi, \cdot)\varphi \tag I.11$$ although there is no operator $B$ or vector $\varphi$ in the usual sense if $\varphi\notin\Cal H$. We will occasionally write $\|\varphi\|^{2}$ which may be infinite if $\varphi\notin\Cal H$. Since $\Cal H_{-1}(A_{\alpha})=\Cal H_{-1}(A)$ and so $(A_{\alpha}- z)^{-1}\varphi\in\Cal H_{+1}(A_{\alpha})=\Cal H_{+1}(A)\subset\Cal H$. The strong resolvent formula holds, viz $$(A_{\alpha}-z)^{-1}-(A-z)^{-1}=-\alpha ((A_{\alpha}-\bar z)^{-1} \varphi, \cdot)(A-z)^{-1}\varphi. \tag I.12$$ Don't blink now because in just about one page we're going to present four of the five critical formulas in the subject (the fifth is (I.17) below). Define $$F_{\alpha}(z)=(\varphi, (A_{\alpha}-z)^{-1}\varphi)\equiv\int \frac{d\mu_{\alpha}(x)}{x-z}$$ where $d\mu_{\alpha}\equiv d\mu^{\varphi}_{A_{\alpha}}$ is the spectral measure for $\varphi$ which has a Borel transform by (I.10) and $\Cal H_{-1}(A_{\alpha})=\Cal H_{-1}(A)$. We define $F(z)\equiv F_{\alpha=0}(z)$. Taking matrix elements of (I.12) with $\varphi$ on both sides, we get $F_{\alpha}(z)-F(z)=-\alpha F_{\alpha}(z)\mathbreak F(z)$, or solving for $F_{\alpha}$, we get the Aronszajn-Krein formula: $$F_{\alpha}(z)=\frac{F(z)}{1+\alpha F(z)}, \tag I.13$$ the first critical formula. Applying (I.12) to $\varphi$ and using (I.13) to see that $1-\alpha F_{\alpha}=[1+\alpha F(z)]^{-1}$, we get the second key formula: $$(A_{\alpha}-z)^{-1}\varphi=(1+\alpha F(z))^{-1} (A-z)^{-1}\varphi. \tag I.14$$ Plugging this into (I.12) we get the third key formula: $$(A_{\alpha}-z)^{-1}=(A-z)^{-1}-\frac{\alpha}{1+\alpha F(z)} ((A-\bar{z})^{-1}\varphi, \cdot)(A-z)^{-1}\varphi. \tag I.15$$ Notice that I.14 says for $\eta, z$ fixed, $(\eta, (A_{\alpha}-z)^{-1} \varphi)=a/(\alpha + b)$, the key to the Aizenman-Molchanov theory. For the final formula, (I.12) says that $(A_{\alpha}-z)^{-1}-(A-z)^{- 1}$ is rank one, so trace class, and by (I.15) we have $$\text{Tr}\left[(A-z)^{-1}-(A_{\alpha}-z)^{-1}\right]=\frac{\alpha} {1+\alpha F(z)} (\varphi, (A-z)^{-2}\varphi).$$ Notice that $(\varphi, (A-z)^{-2}\varphi)=\frac{d}{dz} F(z)$. Thus, we have what I'll call the trace formula, the key to our discussion of the Krein spectral shift in \S I.4 below: $$\text{Tr}\left[(A-z)^{-1}-(A_{\alpha}-z)^{-1}\right] = \frac{d}{dz} \ln(1+\alpha F(z)) \tag I.16$$ where we take the branch of the log which is positive for $z$ real and very negative (recall by Theorem I.1 (iii), that even for $\alpha<0$, $1+\alpha F(z)>0$ for $z$ real and very negative). \vskip 0.3in \head I.3 The Integral Formula \endhead \rightheadtext{The Integral Formula} \bigpagebreak The critical last formula of the general theory is \proclaim{Theorem I.8} $$\int\limits^{\infty}_{-\infty}[d\mu_{\alpha}(E)]\,d\alpha=dE \tag I.17$$ in the sense that if $f$ is in $L^{1}(\Bbb R, dE)$, then $f\in L^{1}(\Bbb R, d\mu_{\alpha})$ for a.e.~$\alpha$, $\int f(E)\,d\mu_{\alpha}(E)\mathbreak \in L^{1}(\Bbb R, d\alpha)$ and $$\int \biggl(\int f(E)\,d\mu_{\alpha}(E)\biggr)\,d\alpha = \int f(E)\,dE. \tag I.18$$ \endproclaim \demo{Proof} By an elementary argument, it suffices to prove this result for $f_{z}(E)=(E-z)^{-1}-(E+i)^{-1}$ for any $z\in\Bbb C\backslash \Bbb R$ (use analyticity in $z$ and then Stone-Weierstrass approximation). By closing the contour in the upper half plane \alignat2 \int\limits^{\infty}_{-\infty}f_{z}(E)\,dE &= 0 &&\qquad \text{if \, Im z<0} \\ &= 2\pi i &&\qquad \text{if \, Im z>0}. \endalignat On the other hand, by (I.13) \align \int f_{z}(E)\,d\mu_{\alpha}(E) &= F_{\alpha}(z)-F_{\alpha} (-i)=\frac{1}{\alpha+F(z)^{-1}} - \frac{1}{\alpha+F(-i)^{-1}} \\ &\equiv h_{z}(\alpha). \endalign Now, if $\pm\text{Im }z>0$, then $\pm\text{Im }F(z)>0$ so $\pm\text{Im }F(z)^{-1}<0$. Thus, $h_{z}(\alpha)$ has either two poles in the lower half plane if $\text{Im }z<0$ or one in each half plane if $\text{Im }z>0$. The same contour integral as for $E$ (but now for $\alpha$!) implies that \alignat2 \int\limits^{\infty}_{-\infty} d\alpha\, h_{z}(\alpha) &= 0 &&\qquad \text {if \, Im z<0} \\ &\equiv 2\pi i &&\qquad \text{if \, Im z>0} \endalignat so I.18 is proven for $f_{z}(E)$. \qed \enddemo The above implies a regularity result for the density of states in $n$-dimensional Jacobi matrices (see Theorem I.19 below) due to Wegner [49]. Wegner proved his result by tracking eigenvalues in finite-dimensional approximations, but his underlying result is similar to (I.17) so this is sometimes called a Wegner estimate. The Russian literature, however, has similar results much earlier in the case of boundary condition dependence; see Javrjan [28]. \newpage \head I.4 The Krein Spectral Shift \endhead \rightheadtext{The Krein Spectral Shift} \bigpagebreak In this section, we make a presentation of the Krein spectral shift [33] due to Gesztesy and me. By the general theory of Borel transforms, $F(z)$ has a boundary value $F(\lambda+i0)$ for a.e.~$\lambda$ in $\Bbb R$. Moreover, by Theorem I.5, if $\alpha$ is fixed, $F(\lambda+i0)\neq -\alpha^{-1}$ for a.e.~$\alpha$. Thus we can define $\text{Arg}(1+\alpha F(\lambda +i0))$ for a.e.~$\lambda$. Since $\text{Im }F(\lambda +i0)\geq 0$, if $\alpha >0$ (resp.~$\alpha <0$), this arg lies in $[0, \pi]$ (resp.~$[-\pi, 0]$). If $1+\alpha F (\lambda+i0)\in (-\infty, 0)$ that sets the argument to (sgn $\alpha)\pi$. \definition{Definition} The Krein spectral shift, $\xi_{\alpha}(\lambda)$, is defined by $$\xi_{\alpha}(\lambda)=\frac{1}{\pi} \text{ Arg}(1+\alpha F(\lambda +i0)).$$ \enddefinition \definition{Definition} $i(\alpha)\equiv\min(\inf\text{ spec}(A), \inf\text{ spec}(A_{\alpha}))$. \enddefinition \proclaim{Theorem I.9} \roster\runinitem"\rom{(i)}" $0\leq\pm\xi_{\alpha}(\lambda)\leq 1$ if $\pm\alpha>0$; $\xi_{\alpha}(\lambda)=0$ if $\lambda 0$, $i(\alpha)=\inf\text{ spec}(A)$ and $F(\lambda+i0)>0$ for $\lambda 0$. (ii) Let $H_{\alpha}(z)=\ln(1+\alpha F(z))$ on $(-\infty, i(\alpha))$ analytically continued to \linebreak $\Bbb C\backslash [i(\alpha), \infty)$.Consider a contour, $C$, which goes from $i(\alpha)$ just above the real axis to $R$, then directly up to $R+iR$, then on a circle of radius $R\sqrt2$ to $R-iR$, then up to just above the real axis, then along the bottom of the real axis to $i(\alpha)$. Consider for $z_{0}\in (-\infty, i(\alpha))$: $$\frac{1}{2\pi i} \oint\limits_{C} \frac{H_{\alpha}(z)}{(z-z_{0})^{2}} \,dz\equiv I_{\alpha}(z_{0}).$$ By Cauchy's formula, once $R\sqrt2 > |z_{0}|$, the integral is just $dH_{\alpha}/dz$, which is the left side of (I.19) by (I.16). Since $|\text{Im }H_{\alpha}|\leq \pi$, and the real parts of the contour from $R$ to $R+iR$ and from $R$ to $R-iR$ cancel, and since $F(z)\to 0$ as $z\to\infty$ with $|\text{Arg }z|>\frac{\pi}{2}$, the infinite part of the contour contributes 0 as $R\to\infty$. Since $|\text{Im }H_{\alpha}|\leq\pi$, dominated convergence lets us take the contour to the real axis and get $\text{Im }\ln(1+\alpha F(z))=\text{Arg}(1+\alpha F(z))$. Thus, we get the right side of (I.19). (iii) By Theorem I.1(iii) and (iv) and the fact that $\int d\mu_{0}(z)=\|\varphi\|^{2}$, we see that $$\lim_{\gamma\to\infty} \gamma^{2} \frac{d}{dz} H_{\alpha} (-\gamma)=\alpha \|\varphi\|^{2}$$ so, by (ii) $$\lim_{\gamma\to\infty}\int \frac{\gamma^{2}}{(x+\gamma)^{2}} \xi_{\alpha}(x)\, dx=\alpha\|\varphi\|^{2}.$$ By monotone convergence we obtain (I.20). \qed \enddemo The next result extends (I.19) to more general functions than $f(x)=(x- z)^{-1}$. The formula we want is: $$\text{Tr}(f(A)-f(A_{\alpha}))=-\int f'(x)\xi_{\alpha}(x)\, dx. \tag I.21$$ Formally this holds for any reasonable $f$ since it holds for $f(x)=(x-z)^{-1}$ and those $f$'s are dense'' in reasonable $f$. One issue is that we don't even know that $f(A)-f(A_{\alpha})$ is trace class! Our hypotheses on $f$ aren't optimal but include the critical example $f(x)=e^{-sx}$ (cutoff near $-\infty$). \proclaim{Theorem I.10} Let $f$ be a $C^2$ on $\Bbb R$ with $(1+|x|)^{2}f^{(j)}\in L^{2}(0, \infty)$ for $j=1, 2$. Then for any $\alpha$, $f(A)-f(A_{\alpha})$ is trace class and {\rom{(I.21)}} holds. \endproclaim \demo{Proof} We only sketch the details supposing w.l.o.g.~ that $\alpha >0$ and $A\geq 1$ (otherwise, reverse the roles and/or add a constant to $A$). And we write $B\equiv A_{\alpha}$ for notational simplicity. 1. $B^{-1}-A^{-1}$ is rank 1 and so trace class. It follows that $B^{-2}e^{isB}-A^{-2}e^{isA}$ is trace class with trace norm bounded by $c(1+|s|)$ since $$\multline B^{-2}e^{isB}-A^{-2}e^{isA}=B^{-1}e^{isB}(B^{-1}-A^{-1}) \\ - i\int\limits^{s}_{0} e^{iuB}(B^{-1}-A^{-1})e^{i(s-u)A}+ (B^{-1}-A^{-1})e^{isA}A^{-1}. \endmultline$$ 2. Let $\hat f$ be the Fourier transform of $f$. Under the hypotheses $$\int (1+|k|)\left[\biggl|\frac{d\hat f}{dk}\biggr| + \biggl|\frac{d^{2}\hat f}{dk^2}\biggr|\right] dk<\infty \tag I.22$$ and that implies that we can integrate by parts twice to find that $$f(B)-f(A)=-(2\pi)^{-1/2}\int\biggl[\frac{d^{2}\hat f}{dk^{2}}(k)\biggr] [B^{-2}e^{ikB}-A^{-2}e^{ikA}]\, dk.$$ It follows by step 1 that $f(B)-f(A)$ is trace class. 3. From (I.19) one deduces (I.21) for $f(x)=x^{-2}e^{ixs}$ by writing $$f(x)=\lim_{n\to\infty} [x^{-2}(1-ixs/n)^{-n}]$$ for one can use analyticity to get (I.21) for $f(x)=(x-z)^{-n}$ and then for this special $f$. 4. (I.21) for general $f$ follows from the formula in step 2 and the equality proven in step 3. \qed \enddemo \remark{Remark} This does not capture all $f$ for which (I.21) holds. For example, $f(x)=(x+1)^{-\beta}$ with $0<\beta<1$ does not obey the hypotheses, but using $$(A+1)^{-\beta}=\frac{\sin\pi\beta}{\pi}\int\limits^{\infty}_{0} (A+1+x)^{-1} |x|^{-\beta}\, dx$$ and $\|(A+x)^{-1}-(A_{\alpha}+x)^{-1}\|_{1}\leq C(|x|+1)^{-2}$ for $x$ large, one can prove (I.21) for such $f$. \endremark \proclaim{Proposition I.11} Suppose that $(a, b)$ is an interval disjoint from $\text{\rom{spec}}(A)\cup\text{\rom{spec}}(A_{\alpha})$. Then \roster \item"\rom{(i)}" $\xi_{\alpha}(x)$ is constant on $(a, b)$. \item"\rom{(ii)}" Its value is either $0$ or $1$. \item"\rom{(iii)}" $P_{(-\infty, a)}(A_{\alpha})-P_{(-\infty, a)}(A)$ is trace class and its trace is the common value of $\xi_{\alpha}(x)$ in $(a, b)$. \endroster \endproclaim \demo{Proof} (i) Let $f\in C^{\infty}_{0}(a, b)$. Then, $f(A)=f(A_{\alpha})=0$ so $\int f'(x)\xi_{\alpha}(x)\, dx=0$ by Theorem I.10. Thus the distributional derivative $d\xi_{\alpha}/dx=0$ on $(a, b)$ so $\xi_{\alpha}$ is constant there. (iii) Let $f$ be a function in $C^{\infty}(\Bbb R)$ which is 1 on $(i(\alpha), a)$, 0 on $[b, \infty)$. Then $f(A)=P_{(-\infty, a)}(A), f(A_{\alpha})=P_{(-\infty, a)}(A_{\alpha})$, and $-\int f'(x)\xi_{\alpha}(x)=-\int\limits^{b}_{a} f'(x)\xi_{\alpha}(x)= -\xi_{\alpha}(x_{0})\mathbreak\int\limits^{b}_{a}f'(x)\, dx=\xi_{\alpha} (x_{0})$, the common value of $\xi_{\alpha}(x)$ in $(a, b)$. (ii) It is a general fact about the trace of a difference of projections that is trace class, that it is an integer [16,5]. Since $0\leq\xi_{\alpha}(x)\leq 1$, it must be 0 or 1. [There is a simple argument: Since $(a, b)\cap\text{spec}(A)=\emptyset$, $F(x+i0)$ is real and continuous on $(a, b)$. If $F(x_{0}+i0)=-\alpha^{-1}$ anywhere on $(a, b)$, then $F_{\alpha}(z)=F(z)/[1+\alpha F(z)]$ has a pole at $x_0$ and so $x_0$ is an eigenvalue of $A_\alpha$. Since $(a, b)\cap \text{spec}(A_{\alpha})=\emptyset$, this cannot happen; that is, $1+\alpha F(x+i0)$ has no zeros and so its argument is constant at either 0 or 1.] \qed \enddemo \remark{Remark} Because of (iii), $\xi_{\alpha}(x)$ is called a spectral shift. \endremark There is an interesting extension of Theorem I.8, an abstraction of a result of Javrjan [28]: \proclaim{Theorem I.12} Let $\beta<\gamma$. Then $$\int\limits^{\gamma}_{\beta} (d\mu_{\alpha}(E))\,d\alpha = (\xi_{\gamma}(E)-\xi_{\beta}(E))\,dE. \tag I.23$$ \endproclaim \remark{Remark} If $c\neq 0$, $\lim\limits_{\alpha\to\infty}\text{Arg} (1+\alpha c)-\text{Arg}(1-\alpha c)=\pi$ so $\lim\limits_{\alpha\to\infty} [\xi_{\alpha}(E)-\xi_{-\alpha}(E)]=1$ and (I.23) implies (I.17). \endremark \demo{Proof} By (I.13), if $\text{Im }z\neq 0$: \align \int\limits^{\gamma}_{\beta}\biggl[\int\frac{d\mu_{\alpha}(E)} {(E-z)^{2}}\biggr] d\alpha &= \int\limits^{\gamma}_{\beta}\frac{d}{dz} \biggl[\frac{F(z)}{1+\alpha F(z)}\biggr]d\alpha \\ &= \frac{d}{dz}\int\limits^{\gamma}_{\beta}\frac{F(z)}{1+\alpha F(z)} \,d\alpha \\ &= \frac{d}{dz}\,[\ln(1+\gamma F(z))-\ln(1+\beta F(z))] \\ &= \int \frac{\xi_{\gamma}(E)-\xi_{\beta}(E)}{(E-z)^{2}}\, dE \endalign proving (I.23) smeared with $(E-z)^{2}$. Such functions have linear combinations which are dense in a big enough space to imply (I.23) as an equality of measures. \qed \enddemo Since $d\mu_{\alpha}$ obeys $\int d\mu_{\alpha}(\lambda) (|\lambda|+1)^{-1}<\infty$, with bounds uniform in $\alpha$ on compacts, this result implies a similar estimate for $\xi_{\alpha}$. Here's another proof: \proclaim{Theorem I.13} \roster\runinitem"\rom{(i)}" If $\alpha<\infty$, then $$\int |\xi_{\alpha}(E)|(1+|E|)^{-1}\,dE<\infty \tag I.24$$ and $$\int \xi_{\alpha}(E)(E-z)^{-1}\,dE=\ln(1+\alpha F(z)). \tag I.25$$ \item"\rom{(ii)}" $\int |\xi_{\alpha}(E)|\,dE\equiv\infty$ if and only if $\varphi\notin\Cal H$. \endroster \endproclaim \demo{Proof} (i) Integrating $$\frac{d}{dz}\ln(1+\alpha F(z))=\int \xi_{\alpha}(y)(y-z)^{-2}\,dy.$$ >From $-w$ to $-x$ for $x\|A\|$ so the integral diverges logarithmically exactly as $\ln x$. This is typical of the case when $\|\varphi\|<\infty$. \proclaim{Theorem I.17 ([25])} If $\|\varphi\|<\infty$, then $\int (1+|x|)^{-1}[1-\xi_{\infty}(x)]\,dx<\infty$ and $$\ln\left[(-z)F(z)/\|\varphi\|^{2}\right]=\int\limits^{\infty}_{0} \frac{[\xi_{\infty}(x)-1]}{x-z}\,dx. \tag I.30$$ \endproclaim \demo{Proof} We have that $$\frac{d}{dz}\ln F(z)=\int \frac{\xi_{\infty}(x)}{(x-z)^{2}}\,dx \tag I.31$$ while clearly $$\frac{d}{dz}\ln (-z^{-1})=\int\limits^{\infty}_{0}\frac{1} {(x-z)^{2}}\,dx \tag I.32$$ since for $z<0$, both sides are just $(-z)^{-1}$. (I.32) is a direct calculation but can also be regarded as (I.31) for the case $F(z)=- 1/z$. By (I.31--32): $$\frac{d}{dz}\ln((-z)F(z)/\|\varphi\|^{2}) = \int\limits^{\infty}_{0} \frac{(\xi_{\infty}(x)-1)}{(x-z)^{2}}\,dx.$$ By Theorem I.1(iv), $\ln ((-z)F(z)/\|\varphi\|^{2})\to 0$ as $z\to - \infty$ along the real axis. The same monotone convergence theorem argument that we used in the proof of Theorem I.13(i) and the fact that $1-\xi_{\alpha}(x)\geq 0$ implies that $(1+|x|)^{-1}(1- \xi_{\infty})$ is in $L^1$ and that (I.30) holds. \qed \enddemo In \S IV.1, we will extend this idea to some cases where $\|\varphi\|=\infty$. How can one get a handle on a spectral measure for $A_{\infty}$? At first sight, it seems a hard job since $$\lim_{\alpha\to\infty} \int \frac{d\mu_{\alpha}(x)}{x-z}= \lim_{\alpha\to\infty} \frac{F(z)}{1+\alpha F(z)}=0$$ so $d\mu_{\alpha}(x)\to 0$ weakly. The key is to renormalize! Define $$d\rho_{\alpha}(x)=(1+\alpha^{2})\,d\mu_{\alpha}(x).$$ One can prove the following \proclaim{Theorem I.18 ([25])} {\rom{(i)}} There exists a measure $d\rho_{\infty}$ \rom(which is non-zero except for the trivial case where $\varphi$ is an eigenvector of $A$\rom) so that $$\int\frac{d\rho_{\infty}(x)}{(1+|x|)^{2}}<\infty$$ and $$\lim_{\alpha\to\infty} \int f(x)\, d\rho_{\alpha}(x)=\int f(x) \,d\rho_{\infty}(x) \tag I.33$$ for any $f\in C^{\infty}_{0}(\Bbb R)$. {\rom{(ii)}} There exists a vector $\eta\in\Cal H_{-2}(A_{\infty})$ so that $d\rho_{\infty}$ is the spectral measure $d\mu^{\eta}_{A_{\infty}}$. If $\varphi$ is cylic for $A$, then $\eta$ is cyclic for $A_{\infty}$. {\rom{(iii)}} {\rom{(I.33)}} holds for $f(x)=(x-z)^{-3}$ but it only holds for $f(x)=(x-z)^{-2}$ if $\|\varphi\|=\infty$; explicitly, $$\lim_{\alpha\to\infty} \int \frac{d\rho_{\alpha}(x)}{(x-z)^{2}}= \|\varphi\|^{-2} + \int \frac{d\rho_{\infty}(x)}{(x-z)^{2}}.$$ {\rom{(iv)}} $d\rho_{\infty}$ is the boundary value of $-F(z)^{-1}$ in the sense that for $f\in C^{\infty}_{0}(\Bbb R)$ $$\int f(x)\, d\rho_{\infty}(x)=\lim_{\epsilon\downarrow 0} \frac 1\pi \int f(x) \text{\rom{ Im}}[-F(x+i\epsilon)^{-1}]\, dx.$$ \endproclaim We refer the reader to [25] for a complete proof but will make a series of remarks that illuminate the result and partly prove it. \remark{Remarks} 1. Notice that $\eta$ is only stated to be in $\Cal H_{-2}(A_{\infty})$, not $\Cal H_{-1}(A_{\infty})$. Indeed, the differential equation case of the next section when $V=0$ is one where $d\rho_{\infty}(x)\sim cx^{1/2}\,dx$ and $\eta\notin\Cal H_{-1} (A_{\infty})$. Indeed [25], for any $\theta<2$, there are examples where $\int d\rho_{\infty}(x)/(1+|x|)^{\theta}=\infty$. 2. If $A$ is bounded, $\eta=(1-P)A\varphi$. Otherwise, one has that $\eta$ obeys $$(A_{\infty}-z)^{-1}\eta = (1-P) F(z)^{-1} (A-z)^{-1}\varphi$$ (with the convention that $P=0$ if $\|\varphi\|=\infty$). 3. $$\int\frac{d\mu_{\alpha}(x)}{(x-z)^{2}}=\frac{d}{dz}\,\frac{F(z)} {1+\alpha F(z)} = \frac{F'(z)}{(1+\alpha F(z))^{2}}$$ so $$\lim_{\alpha\to\infty} \int \frac{d\rho_{\alpha}(x)}{(x-z)^{2}}= \frac{F'}{F^2} \tag I.34$$ so the $\|\varphi\|^{-2}$ term in (iii) is related to Theorem I.1(v). 4. Notice that the right side of (I.34) is $\frac{d}{dz}(- 1/F)$. This is the key to proving (iv). \endremark There is an interesting reparametrization of the $\rho$'s that is exactly what is used in the differential equation case. For $\theta\in (0, \pi)$, we can define $\alpha=-\cot(\theta)$ so $\alpha$ runs from $- \infty$ to $\infty$ as $\theta$ runs from 0 to $\pi$. Let $$d\tilde{\rho}_{\theta}(x)=d\rho_{-\cot(\theta)}(x)= [\sin^{2}(\theta)]^{-1}\,d\mu_{-\cot(\theta)}(x).$$ Then, $d\tilde{\rho}_{\theta}$ has a limit as $\theta\to 0, \pi$ which is just what we called $d\rho_{\infty}(x)$. Moreover, we claim that $$\int\limits^{\pi}_{\theta} (d\tilde{\rho}_{\theta}(E))\,d\theta=dE \tag I.35$$ for $$\int\limits^{\pi}_{0} d\tilde{\rho}_{\theta}(E)\, d\theta = \int\limits^{\infty}_{-\infty} (1+\alpha^{2})\, d\mu_{\alpha}(E) \,\frac{d\alpha}{1+\alpha^{2}}=dE.$$ \vskip 0.3in \head I.6 Boundary Condition Variation of ODE's \endhead \rightheadtext{Boundary Condition Variation of ODE's} \bigpagebreak For applications to Schr\"odinger operators, there are two main categories of the general theory: variation of potential in the discrete (Jacobi) case, which we'll discuss in the next section, and the subject of this section, variation of boundary condition. We begin with a lightning review of the standard Weyl $m$-function approach (see [48]) and then rephrase the set-up into the context of these lectures. We will consider a Schr\"odinger operator on $(0, \infty)$ (and later on $(-\infty, \infty)$) of the form $$H_{\theta}u=-u''(x)+V(x)\,u(x) \tag I.36$$ where, for simplicity, we suppose that $V(x)$ is bounded from below but continuous. This makes $H$ limit point at infinity so we only need a boundary condition at zero, which is where the parameter $\theta$ enters. The boundary condition at 0 is $$u(0)\cos\theta+u'(0)\sin\theta=0 \qquad 0\leq\theta<\pi \tag I.37$$ so that $\theta=0$ is the Dirichlet boundary condition and $\theta=\pi/2$ is the Neumann boundary condition. Explicitly, $H_\theta$ defines a self-adjoint operator with domain $$\split D(H_{\theta})&= \{u\in L^{2}(0, \infty)\mid u, u'\text{ are locally absolutely continuous functions}, \\ &\qquad u \text{ obeys (I.37) and } H_{\theta}u\in L^{2}\}. \endsplit$$ We will often be interested in solutions of the differential equation $$-u''+Vu=zu \tag I.38$$ and will consider the Wronskian $W(f, g)$ of two functions $$W(f, g)(x) = f(x) g'(x)-f'(x)g(x).$$ As usual, if $f,g$ solve (I.38) for the same $z$, then $W(f, g)(x)$ is independent of $x$. Given $\theta$, three solutions of (I.38) will concern us: $\phi$ which obeys the boundary condition (I.37), $\eta$ defined with a dual'' boundary condition, and $\psi_+$ for $\text{Im }z\neq 0$, the solution which is $L^2$ near $x=\infty$. Explicitly, $\phi, \eta$ are defined by boundary values at $x=0$: \alignat2 &\phi_{\theta}(z, 0)=-\sin(\theta) &&\qquad\phi'_{\theta}(z, 0) =\cos(\theta) \\ &\eta_{\theta}(z, 0)=\cos\theta &&\qquad\eta'_{\theta}(z, 0)=\sin(\theta) \endalignat so $$W(\eta_{\theta}, \phi_{\theta})(x)=1. \tag I.39$$ The third solution $\psi_+$ is defined with mixed boundary conditions; it is in \linebreak $L^{2}((0, \infty))$, which determines it up to a multiplicative constant, and it is normalized by $$W(\psi_{+,\theta}, \phi_{\theta})=\psi_{+,\theta}(0)\cos\theta + \psi'_{+,\theta}(0)\sin\theta = 1. \tag I.40$$ Since $\phi, \eta$ are independent solutions of (I.38), we have $\psi_{+}=a\eta + b\phi$. By (I.39/40), $a=1$. We therefore define the Weyl $m$-function, $m_{\theta}(z)$, by $$\psi_{+,\theta}(z, x)=\eta_{\theta}(z, x)+m_{\theta}(z) \phi_{\theta}(z, x) \tag I.41a$$ or equivalently: $$m_{\theta}(z)=W(\eta_{\theta}, \psi_{+, \theta}) \tag I.41b$$ or $$m_{\theta}(z)=-\lim_{R\to\infty} \frac{\eta_{\theta}(z, R)} {\phi_{\theta}(z, R)}. \tag I.41c$$ The last (which follows from $\psi_{+}\to 0$ at infinity) is the usual definition in the limit point case because it is what Weyl used, although the first two are more useful. It can be seen that $$\text{Im }z>0 \Longrightarrow \text{Im }m_{\theta}(z)>0. \tag I.42$$ We won't prove this now, since it will follow from the general theory of Borel transforms once we have set up the connection. The spectral measure $d\rho_{\theta}$ is defined by $$d\rho_{\theta}(x)=\lim_{\epsilon\downarrow 0} \frac{1}{\pi} \,[\text{Im }m_{\theta}(x+i\epsilon)]\, dx.$$ For $\theta\neq 0$, as we'll see $$m_{\theta}(z)=\cot(\theta)+\int \frac{d\rho_{\theta}(x)}{x-z}. \tag I.43$$ The Green's function, $G_{\theta}(x, x'; z)$ is the integral kernel of $(H_{\theta}-z)^{-1}$, that is $$((H_{\theta}-z)^{-1}u(x)=\int G_{\theta}(x, x'; z)u(x')\,dx'.$$ Because of the normalization (I.40), one has $$G_{\theta}(x, x'; z)=\phi_{\theta}(z, x_{<})\psi_{+,\theta}(z, x_{>})$$ where $x_{<}=\min(x, x'), x_{>}=\max(x, x')$. In particular, by (I.41a) $$G_{\theta}(0, 0; z)=+\sin^{2}\theta(-\cot\theta+m_{\theta}(z)). \tag I.44$$ As a final general formula, we note that Wronskian gymnastics (we'll get it other ways below) shows that $$m_{\kappa}(z)=\frac{-\sin(\kappa-\theta)+\cos(\kappa-\theta)m_{\theta}(z)} {\cos(\kappa-\theta)+\sin(\kappa-\theta)m_{\theta}(z)}. \tag I.45a$$ In particular, $$m_{0}(z)=-\frac{1}{m_{\pi/2}(z)}. \tag I.45b$$ We want to show how to think of this in terms of rank one perturbations. Let $u\in D(H_{\theta})$ be real-valued. Then, integrating by parts \align (u, H_{\theta}u) &= \int\limits^{\infty}_{0} (u(x))[-u''+Vu](x)\,dx \\ &= \int\limits^{\infty}_{0} [u'(x)^{2}+V(x)u^{2}(x)]\,dx+u'(0)u(0) \\ &\equiv \tau(u, u)-\cot(\theta)u(0)^{2} \tag I.46 \endalign because of the boundary condition. Let $A$ be the $\theta=\pi/2$ Neumann boundary condition operator. A simple Sobolev estimate shows that for any $u\in\Cal H_{+1}(A)$ is continuous in $x$ and $$|u(0)|^{2}\leq c(u,(A+1)u),$$ that is, $\delta(x)$, the Dirac delta function, lies in $\Cal H_{-1}(A)$. We can form $$A_{\alpha}=A+\alpha\delta(x).$$ (I.46) says that $H_{\theta}=A_{-\cot(\theta)}$. Moreover, $$F_{\alpha}(z)=(\delta, (A_{\alpha}-z)^{-1}\delta)$$ is exactly $G_{\theta}(0, 0; z)$. Since $G$ is related to $m$ by (I.44), equation (I.45) is just the relation $F_{\alpha}(z)=F(z)/[1+\alpha F(z)]$. In addition, because of (I.44) and (I.43), we have $$d\rho_{\theta}(E)=\sin^{2}\theta\, d\mu_{-\cot(\theta)}(E),$$ that is, the spectral measure $d\rho$ is precisely the measure we called $d\tilde{\rho}_{\theta}$ in the last section. The Dirichlet spectral measure, $d\tilde{\rho}_{0}$, is an example of a $d\rho_{\infty}$ with $\int d\rho_{\alpha}(E)/[|E|+1]=\infty$. (I.45b) is just Theorem I.18(iv). The vector $\eta$ of \S I.5 is just $\delta'$; see [25]. The point of this section is the spectral analysis of rank one perturbations in the next chapter will apply to boundary condition variation directly (rather than by analogy). \vskip 0.3in \head I.7 Jacobi Matrices \endhead \rightheadtext{Jacobi matrices} \bigpagebreak Jacobi matrices are operators on $\ell^{2}(\Bbb Z^{\nu})$ which are discrete analogs of Schr\"odinger operators. We write $u\in\ell^{2}(\Bbb Z^{\nu})$ as a function on $\Bbb Z^\nu$. Given a function, $V$, on $\Bbb Z^\nu$, we define the associated Jacobi matrix, $h$, by $$(hu)(n)=\sum_{|j|=1} u(n+j)+V(n)u(n) \tag I.47a$$ which we write as $$h=h_{0}+V. \tag I.47b$$ Let $\delta_{0}$ be the vector in $\ell^2$ which is given by $$\delta_{0}(n)=\delta_{n0}.$$ Let $P$ be the projection on $\delta_{0}: (\delta_{0}, \cdot)\delta_{0}$. Fix $V$ and let $A$ be the Jacobi matrix associated to $V-\delta_{\cdot 0}(V(0))$. Then $$h=A+V(0)P$$ and variations of $V$ are precisely rank one perturbations. As an application of this idea, let $V_{\omega}(n)$ be a random process where the $V_{\omega}(n)$ are independent, identically distributed random variables with distribution $p(v)\,dv$ where $p$ is bounded. There is a fundamental quality called the integrated density of states, $k(\lambda)$, which is given by the following: Let $d\mu^{(0)}_{\omega}$ be the spectral measure for $h_{0}+V_{\omega}$ with vector $\delta_{0}$. Then $$k(\lambda)=E_{\omega}(\mu^{(0)}_{\omega}(-\infty, \lambda)). \tag I.48$$ This is not the basic definition but a derived formula for $k$ (see, e.g., [6,9]). We claim \proclaim{Theorem I.19 (Wegner's Estimate [49])} $$|k(\lambda)-k(\lambda')|\leq c|\lambda-\lambda'|$$ where $c=\sup\limits_{\gamma}p(\gamma)$. \endproclaim \demo{Proof} The result is equivalent to saying that $dk$ is absolutely continuous with respect to $d\lambda$ with $dk=q(\lambda)\,d\lambda$ with $\|q\|_{\infty}\leq c$. Fix $(V_{\omega}(n))_{n\neq 0}$. Then, by Theorem I.8 $$\int [d\mu^{(0)}_{\omega}(\lambda)] p(v(0))\,dv(0)\leq c \int d\mu^{(0)}_{\omega}(\lambda)\,dv(0)=c\,d\lambda.$$ Averaging over $(V_{\omega}(n))_{n\neq 0}$, we get $$E_{\omega}(d\mu_{\omega})\leq c\,d\lambda$$ which is exactly what we want. \qed \enddemo \newpage \title\chapter{2} Spectral Theory of Rank One Perturbations \endtitle \leftheadtext{Spectral Theory of Rank One Perturbations} \endtopmatter \vskip 0.3in \head II.0 Introduction \endhead \rightheadtext{Introduction} \bigpagebreak We begin our analysis with the study of the absolutely continuous spectrum, obtaining a result that also follows from the trace class theory of scattering, but without wave operators. We next discuss the fundamental work of Aronszajn [3] and Donoghue [4], which identifies spectral information for $A_\alpha$ in terms of $F(z)$. Next, we discuss the Simon-Wolff [47] work on localization and apply it to one-dimensional random Jacobi matrices. Finally, we discuss recent results of Gordon [27] and del Rio, et al.~[12] on singular spectrum. {\it Henceforth, we assume that $\varphi$ is a cyclic vector for $A$, that is, $\{(A-z)^{-1}\varphi\mid z\in\Bbb C\}$ is a total set for $\Cal H$.} In general, if $\Cal H_{0}$ is the closed subspace generated by these vectors, then $\Cal H^{\perp}_{0}$ is an invariant space for each $A_\alpha$ and $A_{\alpha}=A$ on $\Cal H^{\perp}_{0}$. Thus, the extension from the cylic to general case is trivial. We make the assumption to be able to focus on the real issues and not have to state the theorems in convoluted forms to give the general case. \vskip 0.3in \head II.1 Invariance of the Absolutely Continuous Spectrum \endhead \rightheadtext{Invariance of the Absolutely Continuous Spectrum} \bigpagebreak Let $f(x)\,dx$ and $g(x)\,dx$ be two absolutely continuous (positive) measures on $\Bbb R$. Recall these measures are equivalent if and only if $\{x\mid f(x)\neq 0\}$ and $\{x\mid g(x)\neq 0\}$ agree up to sets of Lebesgue measure zero. Moreover, if $A$ is multiplication by $x$ on $L^{2}(\Bbb R, f\,dx)$ and $C$ is multiplication by $x$ on $L^{2} (\Bbb R, g\,dx)$, then $A$ and $C$ are unitarily equivalent if and only if $f\,dx$ and $g\,dx$ are equivalent. With these preliminaries \proclaim{Theorem II.1} For all $\alpha\neq\beta$, $(A_{\alpha})_ {\text{\rom ac}}$ and $(A_{\beta})_{\text{\rom ac}}$, the absolutely continous parts of $A_\alpha$ and $A_\beta$ are unitarily equivalent. \endproclaim \demo{Proof} Since $A_{\alpha}=A_{\beta}+(\alpha-\beta)(\varphi, \cdot)\varphi$, we can suppose for $\beta=0$ w.l.o.g. By the above discussion and Theorem I.6(iv), it suffices to show that $S_{1}\equiv \{E\mid\text{Im }F(E+i0)\neq 0\}$ and $S_{2}\equiv\{E\mid \text{Im } F_{\alpha}(E+i0)\neq 0\}$ agree up to sets of measure zero. By the Aronszajn-Krein formula $F_{\alpha}=\frac{F}{1+\alpha F}$, which implies that $$\text{Im }F_{\alpha}(z)=\frac{\text{Im }F(z)}{|1+\alpha F(z)|^{2}}. \tag II.1$$ It follows that $S_1$ and $S_2$ agree up to the sets where $F(E+i0)$ fails to exist, the set where $F(E+i0)=\infty$, and the set where $F(E+i0)=-\alpha^{-1}$. These sets all have measure zero, so $S_1$ and $S_2$ agree up to sets of measure zero. \qed \enddemo \remark{Remarks} 1. The result extends to the non-cyclic case and so extends to the finite rank case. It should be possible to accommodate the trace class case. This is an area under present study. 2. Since $(A_{\infty}+1)^{-1}-(A_{0}+1)^{-1}$ is rank one in the usual sense, we can apply the theorem to $(A_{\infty}-1)^{-1}$ and $(A_{0}+1)^{-1}$ and so, by a spectral mappping theorem, extend the result to allow $\alpha=\infty$. One can also get $\alpha=\infty$ by using $d\rho_{\infty}$ and $\text{Im }(-1/F)=\text{Im }F/|F|^{2}$. \endremark \vskip 0.3in \head II.2 The Aronszajn-Donoghue Theory \endhead \rightheadtext{The Aronszajn-Donoghue Theory} \bigpagebreak Recall that $G(x)=\int\frac{d\mu(y)}{(x-y)^{2}}$. The following basic theorem was proven by Aronszajn [3] for boundary condition dependence of spectrum for Sturm-Liouville operators and then extended to (bounded) rank one perturbations by Donoghue [14]. \proclaim{Theorem II.2} For $\alpha\neq 0$ \rom($\alpha=$ infinity allowed with $\infty^{-1}=0$\rom), define \alignat2 &S_{\alpha} &&= \{x\in\Bbb R\mid F(x+i0)=-\alpha^{-1}; G(x)=\infty\} \\ &P_{\alpha} &&= \{x\in\Bbb R\mid F(x+i0)=-\alpha^{-1}; G(x)<\infty\} \\ &L &&= \{x\in\Bbb R\mid\text{Im }F(x+i0)\neq 0\}. \endalignat Then \roster\runinitem"\rom{(i)}" $\{S_{\alpha}\}_{\alpha\neq 0; |\alpha|\leq\infty}, \{P_{\alpha}\}_{\alpha\neq 0;|\alpha|\leq\infty}$ and $L$ are mutually disjoint. \item"\rom{(ii)}" $P_\alpha$ is the set of eigenvalues of $A_\alpha$. In fact \alignat2 (d\mu_{\alpha})_{\text{\rom pp}}(x) &= \sum_{x_{n}\in P_{\alpha}} \frac{1}{\alpha^{2}G(x_{n})}\, \delta(x-x_{n}) &&\qquad \alpha<\infty \\ (d\rho_{\infty})_{\text{\rom pp}}(x)&= \sum_{x_{n}\in P_{\infty}} \frac{1}{G(x_{n})}\, \delta (x-x_{n}) &&\qquad \alpha=\infty. \endalignat \item"\rom{(iii)}" $(d\mu_{\alpha})_{\text{\rom ac}}$ is supported on $L$, $(d\mu_{\alpha})_{\text{\rom sc}}$ is supported on $S_\alpha$. \item"\rom{(iv)}" For $\alpha\neq\beta$, $(d\mu_{\alpha})_{\text{\rom sing}}$ and $(d\mu_{\beta})_{\text{\rom sing}}$ are mutually singular. \endroster \endproclaim \remark{Remark} We say that a set $S$ supports'' a measure $d\nu$ if $\nu(\Bbb R\backslash S)=0$. \endremark \demo{Proof} (i) is trivial and (iv) follows from (i), given (ii) and (iii). That $L$ supports $(d\mu_{\alpha})_{\text{\rom ac}}$ is proven in the last section. By Theorem I.6(iv), $(d\mu_{\alpha})_{\text{\rom sing}}$ is supported by $\{E\mid\lim\limits_{\epsilon\downarrow 0} |F_{\alpha}(E+i\epsilon)| =\infty\}$. But $$F_{\alpha}(z)=\frac{F(z)}{1+\alpha F(z)}$$ says that $|F_{\alpha} (E+i\epsilon)|\to\infty$ if and only if $F(E+i\epsilon)\to -1/\alpha$. (ii), (iii) thus follow by Theorem I.6(iii) and the claim that when $F(E+i\epsilon)\to -1/\alpha$ (even if $G(E)=\infty$): \alignat2 \lim_{\epsilon\downarrow 0}\,\epsilon \text{ Im } F_{\alpha}(E+i\epsilon) &= 1/[\alpha^{2}G(E)] &&\qquad (0<\alpha<\infty) \tag II.2a \\ \lim_{\epsilon\downarrow 0}\, \epsilon \text { Im } (-F(E+i\epsilon))^{-1} &= 1/G(E) &&\qquad (\alpha=\infty). \tag II.2b \endalignat If $G(E)<\infty$, (II.2) follows from Theorem I.2(v) which implies that $F(E+i\epsilon)=-1/\alpha +i\epsilon G(x)+o(\epsilon)$ and the formula for $F_\alpha$. If $G(E)=\infty$ and $\alpha=\infty$, Theorem I.2(iv) implies $\epsilon^{-1}|F(x+i\epsilon)|\to\infty$ so $\left|\frac {\epsilon}{F(x+i\epsilon)}\right|\to 0$ proving II.2b in that case. If $G(E)=\infty$ and $0<\alpha<\infty$, by Theorem I.2(iv) again, $\epsilon^{-1} \text{ Im}(1+\alpha F)\to\infty$ so $\epsilon|1+\alpha F|^{- 1}\to 0$. Thus, since $F\to -1/\alpha$, $\epsilon|F/1+\alpha F|\to 0$. \qed \enddemo This theorem has interesting consequences for one-dimensional Schr\"odinger operators and Jacobi matrices: \proclaim{Theorem II.3} Let $A$ be a Schr\"odinger operator on $[0, \infty)$ with Neumann boundary conditions at $x=0$ as in {\rom{\S I.6}}. Then $G(E)<\infty$ if and only if both of the following hold: \roster \item"\rom{(i)}" $E$ is not an eigenvalue of $A$; \item"\rom{(ii)}" {\rom{(I.38)}} \rom(with $E=z$\rom) has a solution which is $L^2$ at $+\infty$. \endroster \endproclaim \demo{Proof} Since $G(E)$ finite implies that $F(E+i0)$ exists and is real and finite, Theorem II.2 implies that $G(E)<\infty$ if and only if $E$ is an eigenvalue of $A+\alpha\delta$ for some $\alpha\neq 0$ in $\Bbb R\cup\{\infty\}$. But this is equivalent to (i), (ii). \qed \enddemo \proclaim{Theorem II.4} Let $A$ be a Jacobi matrix on $\ell^{2}(\Bbb Z)$ and $\varphi=\delta_{0}$ as in {\rom{\S I.7}}. Then $G(E)<\infty$ if and only if \roster \item"\rom{(i)}" $E$ is not an eigenvalue of $A$; \item"\rom{(ii)}" One of the following holds: \item"\rom{(a)}" The equation $$u(n+1)+u(n-1)+V(n)u(n)=Eu(n) \tag II.3$$ has an $\ell^{2}$ solution on $(0, \infty)$ with $u(0)=0$; \item"\rom{(b)}" {\rom{(II.3)}} has an $\ell^{2}$ solution on $(- \infty, 0)$ with $u(0)=0$; \item"\rom{(c)}" {\rom{(II.3)}} has $\ell^{2}$ solutions $u_{\pm}$ on both $(0, \infty)$ and $(-\infty, 0)$ with both $u_{+}(0)\neq 0$ and $u_{-}(0)\neq 0$. \endroster \endproclaim \remark{Remarks} 1. To say $u$ is a solution on $(0, \infty)$ means $u$ is defined on $[0, \infty)$ but (II.3) holds for $n=1, 2,\dots$. 2. In particular, if (II.3) has $\ell^{2}$ solutions on both $(0, \infty)$ and $(-\infty, 0)$, then either $E$ is an eigenvalue of $A$ or $G(E)<\infty$. \endremark \demo{Proof} (a), (b) of (ii) hold if and only if $E$ is an eigenvalue of $A_\infty$. (c) holds if and only if $E$ is an eigenvalue of some $A_\alpha$ with $|\alpha|<\infty$. Given these, the proof is the same as for the last two theorems. \qed \enddemo \vskip 0.3in \head II.3 The Simon-Wolff Criterion \endhead \rightheadtext{The Simon-Wolff Criterion} \bigpagebreak The following is useful in discussing localization for random Jacobi matrices: \proclaim{Theorem II.5 ([47])} Fix an interval $[a, b]$. Then the following are equivalent: \roster \item"\rom{(a)}" $G(E)<\infty$ for a.e.~$E$ in $[a, b]$ \rom(w.r.t.~Lebesgue measure\rom). \item"\rom{(b)}" For a.e.~$\alpha$ \rom(w.r.t.~Lebesgue measure\rom), $A_\alpha$ has only pure point spectrum in $[a, b]$. \endroster \endproclaim \demo{Proof} Let $S=\{E\in [a, b]\mid G(E)=\infty\}$. Suppose (a) holds. Then $|S|=0$ by hypothesis. By Theorem I.8 that means $\int \mu_{\alpha}(S)\, d\alpha=0$ so $\mu_{\alpha}(S)=0$ for a.e.~$\alpha$. But if $\mu_{\alpha}(S)=0$, then in the language of Theorem II.2, $\mu_{\alpha}(L\cap [a, b])=\mu_{\alpha}(S_{\alpha}\cap [a, b])=0$. It follows by that theorem that $\mu_\alpha$ has only point spectrum, so (b) holds. Conversely, if $A_\alpha$ has only point spectrum on $[a, b]$, then $\mu_{\alpha}(S)=0$. If (b) holds, $\int \mu_{\alpha}(S)\, d\alpha =0$, so by Theorem I.8, $|S|=0$ and (a) holds. \qed \enddemo As an example of how to apply this theorem, we want to prove localization for the Anderson model in one dimension following [45]. We'll use various facts about that model proven elsewhere. \proclaim{Theorem II.6} Let $\{v_{\omega}(x)\}_{n\in\Bbb Z}$ be bounded i.i.d.s.~with distribution $p(\lambda)\,d\lambda$. Let $h_{\omega}=h_{0}+v_{\omega}$ as in {\rom{\S I.7}}. Then for a.e.~$\omega$, $h_\omega$ has only point spectrum. \endproclaim \remark{Remarks} 1. The general proof following the steps below requires a lot less than bounded or independent random variables; see [45]. 2. The proofs show that the eigenfunctions decay exponentially. 3. If $\text{supp}\,p=[a, b]$, then $\text{spec}(h_{\omega})=[a-2, b+2]$ with probability one, so this is point spectrum dense in an interval. \endremark \demo{Proof} 1. The transfer matrix, $A_{n}(E, \omega)$, is the two by two matrix defined by $\binom{u(n+1)}{u(n)}=A_{n}(E, \omega) \binom{u(1)}{u(0)}$ where $u$ obeys (II.3) for $v_{\omega}$. General principles (the subadditive ergodic theorem; see [45]) imply that for each fixed $E$, $$\lim_{|n|\to\infty} \frac{1}{|n|}\ln \|A_{n}(\omega, E)\|=\gamma(E) \tag II.4$$ for a.e.~$\omega$ and a fixed $\gamma$. Obviously, whether the limit in (II.4) holds only depends on $v_{\omega}$ in a neighborhood of infinity. 2. It can be proven that $\gamma(E)>0$ for almost all $E$. In the i.i.d.~ case under discussion, this follows from Furstenberg's theorem ([19]) for {\it{all}} $E$, but under much weaker conditions ($v$-non-deterministic), one gets the result for a.e.~$E$ from Kotani's theory [32,44]. 3. The theorem of Ruelle-Oseledec [39,37] implies that when (II.4) holds, with $\gamma(E)>0$, then there exist initial values $u_{\pm}(0),u_{\pm}(1)$ so that the corresponding solutions $u_{\pm}$ of (II.3) are $\ell^2$ at $\pm\infty$. (In fact, the solutions decay exponentially.) 4. Theorem II.4 then implies that if $\gamma(E)>0$ and (II.4) holds, then either $E$ is an eigenvalue of $h_{\omega}$ or $G(E, \omega)<\infty$ (with $G(E, \omega)$ associated to $h_{\omega}$ and $\delta_0$). 5. By Fubini's theorem, for a.e.~$\{v_{\omega}(n)\}_{n\neq 0}$, (II.4) holds for a.e.~$E$ with $\gamma>0$. Since $h_{\omega}$ has only countably many eigenvalues, we conclude that $G(E, \omega)<\infty$ for a.e.~$E$ (for this typical set of $\omega$). Thus, by Theorem II.5, $h_{\omega} + \alpha\delta_0$ has pure point spectrum for a.e.~$\alpha$. 6. By the hypothesis on the absolute continuity of the distribution of $v_{\omega}(0)$, $h_\omega$ has point spectrum for a.e.~$\omega$. \qed \enddemo Essentially, the identical argument proves \proclaim{Theorem II.7} Let $\{V_{\omega}(x)\}_{-\infty0$ for $E\in T$. \endroster \endproclaim \demo{Proof} By Theorem I.2(iv), the $G(E)$ associated to the Neumann operator with $\varphi=\delta(x)$ has $G(E)=\infty$ on $T_0$, a dense $G_\delta$ in $S$. Let $T$ be $T_{0}$ minus any eigenvalues for $\theta=\pi/2$. (i), (ii) follow by the results earlier in this chapter. \qed \enddemo \example{Example} If $V$ is random with $\text{spec}(H)=[0, \infty)$, this implies non-Lyapunov behavior for a dense $G_\delta$ on $[0, \infty)$, even though we know there is such behavior for a.e.~$E$. Dense $G_\delta$ sets are sometimes called Baire generic while complements of sets of measure zero are called Lebesgue generic. Both kinds of sets are locally uncountable. The theorem says the sets where one does/does not have Lyapunov behavior with $\gamma >0$ are intertwined, locally uncountable sets! \endexample We want to turn the $G(E)=\infty$ result into one about $\alpha$ values. \proclaim{Theorem II.10 ([12])} Let $d\mu$ be any measure on $\Bbb R$ obeying (I.4). Then the set of values $\{F(E+i0)\mid E\in\text{\rom{supp}}(d\mu)\text{ and } G(E)<\infty\}$ is the complement of a dense $G_\delta$. \endproclaim \demo{Proof} There is a general argument that the complement is a $G_\delta$ [46], so we focus on the fact that the set, $S$, is a countable union of nowhere dense sets. We sketch the idea behind the proof in [12], leaving the details for that paper. Write $\{E\mid\in\text{supp}(d\mu)\text{ and }G(E)<\infty\}$ as a countable union of sets $A_n$ so that \roster \item"\rom{(i)}" $G(E)$ is close to the constant, $\gamma_n$, on $A_n$. \item"\rom{(ii)}" For some $\delta_n$ and all $x\in A_n$, $\int\limits_{|x-y|<\delta_{n}}|x-y|^{-2}\,d\mu(y)$ is small compared to the variation of $G$ over $A_n$. \item"\rom{(iii)}" For each $x\in A_n$, consider the equilateral triangle, $T_x$, in $\Bbb C$ with one vertex at $x$ and base on the line $\text{Im }z=a\delta_n$ with a fixed number, $a$, (fixed in the proof in [12]). Then $\operatornamewithlimits{\cup}\limits_{x\in A_{n}}T_x$ is connected. \endroster By (i), (ii) $F'(E)$ is very close to the constant $\gamma_n$ on each $T_x$ so $|F(z)-F(y)-\gamma_{n}(z-y)|\leq\frac13\gamma_{n}|y-z|$ on $T_x$. By a simple geometric argument using (iii), $|F(z)-F(y)- \gamma_{n}(z-y)|\leq\frac23\gamma_{n}|y-z|$ for $y, z\in \operatornamewithlimits{\cup}\limits_{x\in A_{n}}T_x$ and so in $A_n$. Since $A_n$ is nowhere dense and this inequality says that $F$ on $A_n$ is the restriction of a homeomorphism from $\Bbb R$ to $\Bbb R$, we see that $F[A_{n}]$ is nowhere dense. \qed \enddemo Combining this with Theorem II.2, we immediately get \proclaim{Theorem II.11} In the general context of rank one perturbations, suppose that $[a, b]\subset\text{\rom{spec}}(A)$ and $\text{\rom{spec}}_{\text{\rom ac}}(A)\cap(a, b)=\emptyset$. Then for a dense $G_\delta$ of $\alpha$'s, $A_\alpha$ has purely singular continuous spectrum on $(a, b)$. \endproclaim We emphasize that there is a distinction between singular and point spectrum. If $\text{spec}_{\text{\rom ac}}(A)\cap (a, b)=\emptyset$ and $(a, b)\subset\text{spec}(A)$, there are {\it{always}} many $\alpha$'s for which $A_\alpha$ has singular continuous spectrum. There may or may not be $\alpha$'s when there is point spectrum there. The Simon-Wolff criterion gives cases where there is point spectrum for many $\alpha$'s, but the criterion may or may not hold. \remark{Remark} This result on singular continuous spectrum for dense $G_\delta$'s is one of many recently discovered [27,46,11,29]. For example, in [46], it is proven that in $C_{\infty}(\Bbb R^{n})$, the continuous functions vanishing at infinity, there is a dense $G_\delta$, $S$, so that if $V\in S$, then $-\Delta + V$ has purely singular continuous spectrum on $[0, \infty)$. \endremark \proclaim{Corollary II.12} Let $V_{\omega}(x)$ be a non-deterministic ergodic random process for $x\in\Bbb R$ so that $\text{\rom{spec}}\bigl(-\frac{d^2}{dx^2}+V\bigr)=[a, \infty)$ for some $a$. Let $H_{\theta}$ be $-\frac{d^2}{dx^2}+V$ on $L^{2}(0, \infty)$ with $\theta$ boundary condition at $x=0$. Then for a dense $G_\delta$ of $\theta$'s, $H_\theta$ has purely singular continuous spectrum on $[0, \infty)$ \rom(while for Lebesgue almost all, it has only point spectrum there\rom). \endproclaim \proclaim{Corollary II.13} Let $\{V_{\omega}(n)\}_{n\in\Bbb Z^{\nu}}$ be independent, identically distributed random variables with uniform distribution in $[a, b]$ and let $H_\omega$ be the corresponding Jacobi matrix on $\ell^{2}(\Bbb Z^{\nu})$. Let $|b-a|$ be so large that $H_\omega$ has only point spectrum for a.e.~$\omega$ \rom(see Chapter III\rom). Then for a.e.~$\omega$, there is a dense $G_\delta$, $S$ in $[a, b]$, so that for $\alpha\in S$ \alignat2 W(n) &= V_{\omega}(n) &&\qquad n\neq 0 \\ &= \alpha &&\qquad n=0 \endalignat yields a Jacobi matrix with only singular continuous spectrum. \endproclaim Both corollaries follow immediately from the theorem and earlier discussions of the examples. \vskip 0.3in \head II.5 Examples \endhead \rightheadtext{Examples} \bigpagebreak The astute reader may have noticed that $d\mu_{0}$ completely determines the spectral properties of the family $A_\alpha$ (assuming cyclicity, of course). Moreover, given $\mu_{0}$, let $A$ be multiplication by $x$ on $L^{2}(\Bbb R, d\mu_{0})$ and let $\varphi(x)\equiv 1$. Then $F(z)$ is just the Borel transform of $d\mu_{0}$. So we can describe examples by giving the measure $d\mu_{0}$, which we'll denote by $d\mu$. \example{Example 1 (The Friedrichs model)} Let $d\mu=dx\restriction [0, 1]+\delta_{1/2}$ be the sum of Lebesgue measure on $[0, 1]$ and a point mass at $1/2$. Then $G(x)=\infty$ on $[0, 1]$, and \alignat2 \frac{1}{\pi}\text{ Im }F(x+i0) &= 1 &&\qquad 00). The embedded eigenvalue dissolves. \endexample \example{Example 2} Let \{q_{n}\}^{\infty}_{n=1} be a counting of the rationals on [0, 1]. Let d\mu=dx\restriction [0, 1] + \sum^{\infty}_{n=1} 2^{-n}\delta_{q_n}. $$As above, all eigenvalues disappear for \alpha\neq 0. The interesting case is to take d\mu=d\mu_{-1}. The new A has no eigenvalues on [0, 1] but A+(\varphi, \cdot)\varphi suddenly has dense point spectrum embedded in the a.c.~spectrum. \endexample \example{Example 3} Let d\mu be conventional Cantor measure. We claim that G(x)=\infty on C, the Cantor set. In fact,$$ \text{Im }F(x+i\epsilon)\to\infty \tag II.5 $$for x\in C, which implies that G(x)=\infty. This fact about \text{Im }F implies that A_\alpha has neither point nor singular spectrum on C. In fact, each A_\alpha has one eigenvalue e_{j}(\alpha) in each interval, I_j in [0, 1]\backslash C. The corresponding eigenvectors, together with one, e_{0}(\alpha), in I_{0}\equiv\Bbb R\backslash [0, 1], are complete. \text{Spec} (A_{\alpha})=\operatornamewithlimits{\cup}\limits^{\infty}_{j=0}\{e_{j} (\alpha)\}\cup C since the limit points of the e's are precisely C. To prove (II.5), we claim first that$$ \text{Im }F(x_{0}+i\epsilon)\geq\frac{1}{2\epsilon}\,\mu (\{y\mid |x-y|\leq \epsilon\}) \tag II.6 $$for any measure \mu. Moreover, if x\in C, then$$ \mu(\{y\mid |x-y|\leq 2(3^{-n})\})\geq 2^{-n} $$so (II.6) implies (II.5). \endexample \example{Example 4 ([47])} Let d\mu_{n}=\frac{1}{2^n} \sum\limits^{2^n}_{j=1}\delta_{j/2^{n}} and \mu=\sum a_{n}\,d\mu_n where \sum a_{n}<\infty. If x\in [0, 1], then clearly, since there is a j with |x-(j/2^{n})|\leq 2^{-n}, we have$$ \int |x-y|^{-2}\, d\mu_{n}(y)\geq(2^{-n})^{-2} 2^{-n}=2^n $$so if \sum 2^{n}a_{n}=\infty, then G(x)=\infty on [0, 1] and thus A_\alpha has no point spectrum [0, 1]. So A has dense point spectrum on [0, 1] which turns into purely singular continuous spectrum on [0, 1] for all \alpha\neq 0. \endexample \example{Example 5} Let d\nu be Cantor measures and let$$ d\mu(x)=\chi_{[0, 1]}\sum^{\infty}_{n=1} n^{-2} \biggl[\,\sum^{2^n}_{j=1} \,d\nu\bigl(x-(j/2^{n})\bigr)\biggr]. $$Then \mu\{y\mid |y-x|\leq 2^{-n+1}\}\geq 3^{-n}n^{-2} for any \chi\in [0, 1] so G(x)\geq\frac14(2^{-n})^{-2}3^{-n}\mathbreak n^{-2}=\frac14 (\frac43)^{n}n^{-2} for any n and so G=\infty on [0, 1]. This yields an example with \sigma(A_{\alpha}) purely singular continuous on [0, 1] for all \alpha. As noted, this cannot happen for point spectrum. \endexample \example{Example 6 ([47])} Let x_n be arbitrary points in [0, 1]. Let a_n be a sequence with 0\leq a_{n}\leq C_{0}\alpha^{n} for some \alpha<1. Let$$ d\mu=\sum a_{n}\delta_{x_n}. $$We claim G(x)<\infty for a.e.~x. For if |x-x_{n}|>D^{1/2} \alpha^{n/4} for {\it{all}} n, then$$ G(x)<\sum_{n}C_{0}\alpha^{n}D^{-1}\alpha^{-n/2}=(1-\alpha^{1/2})^{-1} C_{0}/D. $$Thus$$ \{x\mid G(x)>(1-\alpha^{1/2})^{-1}C_{0}/D\}\leq\sum 2D^{1/2}\alpha^{n/4}=2D^{1/2}(1-\alpha^{1/2})^{-1} $$goes to zero as D\to 0 proving that G(x)<\infty for a.e.~x. In this case, by the Simon-Wolff criterion, A_\alpha has only pure point spectrum for a.e.~\alpha, but for a dense G_\delta of \alpha, it has singular continuous spectrum, of course. Let x_n be the counting of the dyadic rationals. Let d\mu=\sum b_{n}\delta_{x_n}. Examples 4 and 6 say that depending on the b's, A_\alpha can have dense point or singular continuous spectrum for Lebesgue typical \alpha. Spectral properties depend heavily on \varphi as well as A! \endexample \example{Example 7 ([13])} Let \{q_{n}\}^{\infty}_{n=1} be a counting of the rationals in (0, 1). Define a_{n}=\min(3^{-n-1}, q_{n}, 1-q_{n}) and let S=\operatornamewithlimits{\cup}\limits_{n} (q_{n}-a_{n}, q_{n}+a_{n}) so |S|\leq 2\sum a_{n}\leq\frac13. Let d\mu_{0}=\chi_{S}\,dx so \text{spec}(A_{0})=[0, 1] since S is dense in [0, 1]. Since |[0, 1]\backslash S|\geq\frac23 and d\mu_{\alpha,\text{\rom ac}}([0, 1]\backslash S)=0, there must be a positive measure set of \alpha's with point and/or singular spectrum embedded in \text{spec}(A_{\alpha}). As in the last example, if |x-q_{n}|\geq a_{n}+\sqrt{\epsilon}\, 3^{-n/4} for all n, then G(x)\leq\epsilon^{-1}(1-\sqrt{1/3})^{-1}. Thus, \{x\in S\mid G(x)>\epsilon^{-1}(1-\sqrt{1/3})^{-1}\}=\sqrt{\epsilon}\, (1- \sqrt{1/3})^{-1} so G(x)<\infty for a.e.~x\in[0, 1]\backslash S. We conclude that for a.e.~\alpha, A_\alpha has no singular spectrum on S but it has some point spectrum there for a positive measure set of \alpha's. \endexample \example{Example 8 ([13])} Let S_{n}=\operatornamewithlimits{\cup} \limits^{n-1}_{j=1}\bigl(\frac{j}{2^n}-\frac{1}{4n^{2}2^{n}}, \frac{j}{2^n}+\frac{1}{4n^{2}2^{n}}\bigr) and S= \operatornamewithlimits{\cup}\limits_{n}S_{n}, so |S_{n}|\leq \frac{1}{2n^{2}} and |S|\leq\frac{\pi^{2}}{12}<1. Let d\mu_{0}= \chi_{S}\, dx. Then, \text{spec}(A_{0})=[0, 1] but since |[0, 1]\backslash S|>0, there must be a positive measure set of \alpha's with point and/or singular spectrum embedded in \text{spec}(A_{\alpha}). As in Example 4, G(x)=\infty on [0, 1] since for any n$$ G(x)\geq (2^{-n-1})^{-2}2^{-n-1}n^{-2}=2^{n}/8n^{2}\to\infty. $$Thus, for all \alpha we have no point spectrum in [0, 1] but we have singular continuous spectrum embedded in the a.c.~spectrum for a set of positive measure \alpha's. \endexample \newpage \title\chapter{3} Localization in the Anderson Model \\ Following Aizenman-Molchanov \endtitle \leftheadtext{Localization in the Anderson Model} \rightheadtext{Localization in the Anderson Model} \endtopmatter \vskip 0.5in In this chapter we'll present the approach of Aizenman-Molchanov [2] to the proof of localization for the large coupling Anderson model in arbitrary dimension. The proof is so simple, we won't even put it in separate sections. Results on exponential decay of Green's functions were first obtained by Fr\"ohlich-Spencer [18], with later extensions and simplifications by [17,15]. Even taking into account the fact that we have already developed some of the facts we need, this proof is something like a factor of ten shorter and has smaller constants. We will prove: \proclaim{Theorem III.1} Let \{V_{\omega}(n)\}_{n\in\Bbb Z^{\nu}} be independent identically distributed random variables with uniform density on [0, 1]. Then, there is a fixed constant, K, so that if |\lambda|>K\nu^{2}, then the Jacobi matrix, h_{0}+\lambda V_{\omega}, has only pure point spectrum \rom(and it is dense in [- 2\nu, 2\nu+\lambda]\rom). \endproclaim \remark{Remark} I believe that K=32 is what comes out of the minimization problem below. Suppose for a moment that we could prove that E_{\omega} \big(\sum\limits_{n}|G(n, 0; E+i0)|^{2}\bigr)\leq Ce^{-D|n|}. Then, by Fubini's theorem, we'd get that for a.e.~pair (E, \omega), \sum\limits_{n}|(\delta_{n}, (A-E-i0)^{-1}\delta_{0})|^{2}<\infty, so by Theorems II.8 and II.5, we'd have localization. The problem with this idea is that there is no chance that even E_{\omega}(|G(0,0; E+i0)|^{2}) is finite, since after all, G(0, 0; E+i0) is precisely F_{\alpha}(E+i0) for \alpha=V_{\omega}(0) and that is F/[1+\alpha F] and has a first order zero at \alpha=-1/F if there is to be an eigenvalue. But then, \int dv(0)(v(0)-a)^{-2} is infinite if a\in\text{supp}\,v(0). So the first key idea of [2] is to look at powers of (G)^s with 01, then for \{x_{n}\}^{N}_{n=1} all positive \rom(N=1, 2,\dots,\infty\rom):$$ \sum^{N}_{n=1}{x_{n}}^{s}\leq\biggl(\,\sum^{N}_{n=1} x_{n}\biggr)^s. $$If s<1, then$$ \sum^{N}_{n=1}{x_{n}}^{s}\geq \biggl(\,\sum^{N}_{n=1}x_{n}\biggr)^{s}. $$\endproclaim \demo{Proof} If s>1, {x_{n}}^{s-1}\leq\bigl(\sum\limits^{N}_{m=1} x_{m}\bigr)^{s-1}$$ {x_{n}}^{s}=x_{n}{x_{n}}^{s-1}\leq x_{n}\biggl(\,\sum^{N}_{m=1} x_{m}\biggr)^{s-1}. $$Now sum over n. If s<1, {x_{n}}^{s-1}\geq \bigl(\sum\limits^{N}_{m=1} x_{m}\bigr)^{s-1} and the proof is identical. \qed \enddemo In particular,$$ \biggl(\sum_{n} |G(n, 0; E+i\epsilon)|^{2}\biggr)^{1/4} \leq \sum_{n} |G(n, 0; E+i\epsilon)|^{1/2}. \tag III.1 $$So we need only control \sum\limits_{n} E_{\omega}(|G(n, 0; E+i\epsilon)|^{1/2}). A key lemma that we will need is \proclaim{Lemma III.3} There is a universal constant, C_0 so that for all complex numbers, \alpha, \beta:$$ \int\limits^{1}_{0} |x-\alpha|^{1/2} |x-\beta|^{-1/2}\, dx \geq C_{0}\int\limits^{1}_{0} |x-\beta|^{-1/2} \,dx. \tag III.2 $$\endproclaim \remark{Remark} I believe \alpha=\beta=\frac12 is the extreme point, in which case C_{0}=1/\sqrt8. \endremark \demo{Proof} It is easy to see the integral on the left side of (III.2) only decreases if \alpha is replaced by \text{Re }\alpha or if \alpha\notin [0, 1] and is replaced by the nearest point in [0, 1]. So we can suppose that \alpha\in [0, 1]. By x\to \frac12 -x symmetry, we can suppose \alpha\in [0, \frac12]. But then$$ \int\limits^{1}_{0} |x-\alpha|^{1/2} |x-\beta|^{-1/2} \, dx \geq \sqrt{1/4} \int\limits^{1}_{3/4} |x-\beta|^{-1/2} \, dx $$so we need only show that$$ \inf_{\beta\in\Bbb C} \biggl[\int\limits^{1}_{3/4} |x-\beta|^{-1/2}\, dx \big/\int\limits^{1}_{0}|x-\beta|^{-1/2}\, dx\biggr] >0. $$Let h(\beta) be the function in [\,\cdot\,]. It is clearly continuous on \Bbb C, non-vanishing and \lim\limits_{\beta\to\infty} h(\beta)=\frac14 so the \inf is strictly positive. \qed \enddemo One last lemma we will need concerns the free Laplacian, h_0. \proclaim{Lemma III.4} \roster\runinitem"\rom{(a)}" Let f, g\in\ell^{\infty}(\Bbb Z^{\nu}) be non-negative functions and suppose that$$ (1-ah_{0})f\leq g $$for some 04C^{-2}\nu^{2})$$\align k(n) &\leq (C_{0}\lambda^{1/2})^{-1} (1-(C\lambda^{1/2})^{-1}h_{0})^{- 1}(0, n) \\ &\leq (C_{0}\lambda^{1/2})^{-1} [(2\nu)(C_{0}\lambda^{1/2})^{-1}]^{|n|} (1-2\nu(C_{0}\lambda^{1/2})^{-1})^{-1}. \endalign $$Thus, \sum\limits_{n}k(n) is bounded uniformly in z, so by (III.1):$$ \sup_{z} E_{\omega}\biggl(\bigl(\sum_{n} |G(n, 0; z)|^{2}\bigr)^{1/4} \biggr) $$is bounded uniformly in z. Using Theorem II.8, this implies that for A=h_{\omega}, \varphi=\delta_{0}, we have$$ E_{\omega}(G(E)^{1/4})<\infty $$for all E\in\Bbb R. Thus for a.e.~pairs, (V, E), G(E)<\infty. Now apply Theorem II.5 to get localization. \qed \enddemo Actually, the proof shows that E_{\omega} \bigl((e^{\delta |n|} \sum\limits_{n} |G(n, 0; z)|^{2})^{1/4}\bigr) <\infty for some \delta>0 and this implies exponential decay of eigenfunctions (essentially by (I.14)). Indeed, one easily sees that each eigenvector \varphi obeys |\varphi(n)|\leq C_{\varphi, \epsilon} \bigl[\frac{\nu^2}{k|\lambda|}\bigr]^{(1-\epsilon)|n|/2} for each \epsilon >0. For extensions of these ideas, see [2,1]. \newpage \title\chapter{4} The Xi Function \endtitle \leftheadtext{The Xi Function} \endtopmatter \vskip 0.3in \head IV\!.0 Introduction: The Trace Formula \endhead \rightheadtext{The Trace Formula} \bigpagebreak In this chapter, I'll discuss a program of F. Gesztesy and mine (in part with Holden and Zhao) that deals with inverse spectral theory for one-dimensional Schr\"odinger and Jacobi matrices [22]. In the study of the inverse problem for periodic potentials, an important role is played by the trace formula originally proven by Gelfand-Levitan and exploited/refined by many authors subsequently; see [24] for a history. Let me describe this formula: let V(x+L)=V(x). Then \text{spec}\bigl(-\frac{d^2}{dx^2}+V(x)\bigr)=[E_{0}, E_{1}]\cup [E_{2}, E_{3}], \dots  where E_{0}, E_{3}, E_{4},\dots, E_{4n-1}, E_{4n}, \dots  are periodic eigenvalues, and E_{1}, E_{2}, \dots, E_{4n+1}, E_{4n+2}, \dots  are antiperiodic eigenvalues. Let \{\mu_{j}(x)\}^{\infty}_{j=1} be the eigenvalues on L^{2}(x, x+L) with u(x)=u(x+L)=0 Dirichlet boundary conditions. It can be shown that E_{2j-1}\leq\mu_{j}(x)\leq E_{2j}. The trace formula says that, at least when V is smooth enough (C^1 will do)$$ V(x)=E_{0}+\sum^{\infty}_{j=1} (E_{2j}+E_{2j-1}-2\mu_{j}(x)). \tag IV\!.1 $$One of our goals in this chapter is to extend this formula to more general V's than periodic. One issue to consider immediately is a case like V(x)=x^{2}-1 where V(x)\to\infty as |x|\to\infty. Then the bands, [E_{2j}, E_{2j+1}] collapse to a point and we have eigenvalues E_{0}0) will read: Let \xi(x, \lambda) be the Krein spectral shift for infinite coupling with A=-\frac{d^2}{dx^2}+V(x) and \varphi=\delta(x). Then$$ V(x)=\lim_{\gamma\to\infty} \int\limits^{\infty}_{0} \frac{\gamma^2} {(\lambda +\gamma)^{2}}\, (1-2\xi(x, \lambda)) \, dx. \tag IV\!.3 $$Note that |1-2\xi(x, \lambda)|=1 in all discrete spectrum cases, clearly illustrating the need for a summability method. \vskip 0.3in \head IV\!.1 Abstract Trace Formula \endhead \rightheadtext{Abstract Trace Formula} \bigpagebreak We recall some facts from \S I.5,6. As usual, let A\geq 0, A_{\alpha}=A+\alpha(\varphi, \cdot)\varphi with A_\infty defined as the strong resolvent limit of A_\alpha as \alpha\to\infty. Define$$ F(z)=(\varphi, (A-z)^{-1}\varphi) \tag IV\!.4 $$and \xi_\infty by$$ \xi_{\infty}(\lambda)=\frac{1}{\pi}\text{ Arg }F(\lambda +i0). \tag IV\!.5 $$Then$$ \text{Tr}(f(A)-f(A_{\infty}))=-\int f'(\lambda)\xi_{\infty}(\lambda)\, dx \tag IV\!.6 $$and$$ \int\limits^{\infty}_{0} \frac{\xi_{\infty}(\lambda)\, d\lambda} {(\lambda-z)^{2}}\, dx = \frac{d}{dz} \, \ln F(z). \tag IV\!.7 $$We also will consider a second problem with A^{(0)}, \xi^{(0)}_{\infty} and F^{(0)}. The point of a comparison is that with great generality, Green's functions, G(0, 0; z), have the same behavior, viz (2(-z)^{1/2})^{-1} at z=-\infty. \proclaim{Theorem IV\!.1 ([24])} Suppose that \lim\limits_{\gamma\to\infty} F(-\gamma)/F^{(0)}(-\gamma)=1. Then for z\in\Bbb C\backslash [0, \infty)$$ F(z)=F^{(0)}(z)\exp\biggl(\lim_{\gamma\to\infty} \int \frac{\gamma}{\lambda +\gamma}\,\, \frac{1}{\lambda - z}\, (\xi_{\infty}- \xi^{(0)}_{\infty})(x)\, dx \biggr). \tag IV\!.8 $$\endproclaim \demo{Proof} (IV\!.7) for F, F^{(0)} integrated says that$$ \ln (F(z)/F^{(0)}(z))-\ln (F(-\gamma)/F^{(0)}(-\gamma))=\int \biggl( \frac{1}{\lambda-z} - \frac{1}{\lambda+\gamma}\biggr) (\xi_{\infty} -\xi^{(0)}_{\infty})(\lambda)\, d\lambda. \tag IV\!.9 $$Moreover, by dominated convergence, for z fixed$$ \lim_{\gamma\to\infty} \int\frac{|z|}{|\lambda - z|}\,\, \frac{1}{\lambda +\gamma}\, |(\xi_{\infty}-\xi^{(0)}_{\infty}(\lambda)| \, d\lambda = 0. \tag IV\!.10 $$(IV\!.9,10) and \lim\limits_{\gamma\to\infty}\ln (F(-\gamma)/F^{(0)}(- \gamma))=0 imply (IV\!.8). \qed \enddemo \proclaim{Theorem IV\!.2: The Abstract Trace Formula ([24])} Supose that for some \epsilon>0, \operatornamewithlimits{\lim}\limits\Sb |\gamma|\to\infty \\ |\text{\rom{Arg }}\gamma|<\epsilon \endSb \gamma\bigl(\frac{F(-\gamma)}{F^{(0)}(-\gamma)}-1\bigr)=a. Then$$ a=\lim\Sb \gamma\to\infty \\ \gamma\text{\rom{ real}} \endSb \int \frac{\gamma^2}{(\lambda+\gamma)^{2}} (\xi_{\infty}-\xi^{(0)} _{\infty})(\lambda)\, d\lambda. $$\endproclaim \demo{Proof} By hypothesis$$ \lim\Sb |\gamma|\to\infty \\ |\text{Arg }\gamma|<\epsilon \endSb \gamma\, \ln \left[\frac{F(-\gamma)}{F^{(0)}(-\gamma)}\right]=a. $$By a simple contour argument$$ \lim\Sb \gamma\to\infty \\ \gamma\text{ real}\endSb \gamma^{2} \left(\frac{d}{dz}\,\ln \left[\frac{F(z)}{F^{(0)}(z)} \right]\right)_{z=-\gamma} =a. $$This plus (IV\!.7) implies the result. \qed \enddemo \vskip 0.3in \head IV\!.2 The Trace Formula for Schr\"odinger Operators \endhead \rightheadtext{The Trace Formula for Schr\"odinger Operators} \bigpagebreak We'll postpone the proof of the following until later in this section: \proclaim{Lemma IV\!.3} Suppose that V is bounded from below and is continuous at x. Let G be the Green's function for -\frac{d^2} {dx^2}+V(x) on L^{2}(-\infty, \infty). Then as |\gamma|\to\infty with |\text{\rom{Arg }}\gamma| < \pi/2:$$ G(x, x; -\gamma)=\frac12 \, (\gamma)^{-1/2} \biggl(1-\frac12 \,\gamma^{-1} V(x)+0(\gamma)^{-1}\biggr) \tag IV\!.11 $$\endproclaim \example{Example} This will be the comparison F^{(0)} so it is a critical example: Let A^{(0)}=-\frac{d^2}{dx^2} on L^{2}(\Bbb R, dx). Let \varphi=\delta_x. Then F^{(0)}(z)=G^{(0)}(x, x; z)= \frac12(-z)^{-1/2}. Thus F^{(0)}(x+i0)=\frac12(x)^{-1/2}i and \xi^{(0)}_{\infty}(x)=\frac12, x>0. \endexample We now apply the abstract theory of the last section: \proclaim{Theorem IV\!.4} Suppose that V is non-negative and is continuous at x. Let \xi(x, \lambda) be \xi_{\infty}(x) for A=-\frac{d^2}{dx^2}+V; \varphi=\delta_x. Then$$ V(x)=\lim_{\gamma\to\infty} \int\limits^{\infty}_{0} \frac{\gamma^2} {(\lambda+\gamma)^{2}}\, [1-2\xi(x, \lambda)]\, d\lambda \tag IV\!.12 $$and$$ G(x, x; z)=\frac12 \,(-z)^{-1/2} \exp \biggl(\lim_{\gamma\to\infty} \int\limits^{\infty}_{0} \frac{\gamma}{\gamma+\lambda}\, \frac{1} {\lambda - z} \biggl[\xi(x, \lambda)-\frac12\biggr] dx\biggr) \tag IV\!.13 $$\endproclaim \demo{Proof of Lemma {\rom{IV\!.3}} when V is bounded} W.l.o.g. suppose x=0. Let \psi(x)=(H_{0}+\gamma)^{-1}\delta_{0}(x)=\frac12 \gamma^{-1/2} e^{-\gamma^{1/2}|x|}, so \|\psi_{\gamma}\|=0(\gamma^{- 3/4}). By the second resolvent equation, since V is bounded$$ (\delta_{0}, (H_{0}+V+\gamma)^{-1}\delta_{0})=(\delta_{0}, (H_{0}+ \gamma)^{-1}\delta_{0})-(\psi_{\gamma}, V\psi_{\gamma})+R \tag IV\!.14 $$where \|R\|\leq\|\psi_{\gamma}\|^{2}\gamma^{-1}=0(\gamma^{-5/2}). The first term in (IV\!.14) is \frac12\gamma^{-1/2} and the second is$$ -\frac14 \,\gamma^{-1}\int\limits^{\infty}_{-\infty} e^{-2\gamma^{1/2}|x|} V(x)\, dx =-\frac14 \, \gamma^{-3/2}[V(0)+o(1)] $$since \int\limits^{\infty}_{-\infty} e^{-2\alpha|x|}\, dx=\alpha^{-1}, \alpha >0. \qed \enddemo \demo{Proof of Lemma {\rom{IV\!.3}} in the general case} We will prove with H=H_{0}+V that$$ e^{-tH}(x, x)=(4\pi t)^{-1/2} (1-tV(x)+o(t)). \tag IV\!.15 $$(IV\!.11) then follows from$$ G(x, x; -\gamma)=\int\limits^{\infty}_{0} e^{-\gamma s} e^{-sH} (x, x)\, ds $$and$$ \int\limits^{\infty}_{0} t^{-1/2} e^{-t}\, dt =\sqrt{\pi} \quad\text{and}\quad \int\limits^{\infty}_{0} t^{1/2}e^{-t}\, dt =\frac12 \sqrt{\pi}. $$To prove (IV\!.15), let \{\omega(s)\}_{0\leq s\leq 1} be the Brownian bridge [43] so the Feynman-Kac formula has the form [43]$$ e^{-tH}(x, x)=(4\pi t)^{-1/2} E_{\omega}\biggl(\exp\biggl[-t \int\limits^{1}_{0} V(x+2\sqrt{2t}\, \omega(s))\,ds\biggr]\biggr). \tag IV\!.16 $$A Levy-type estimate implies that$$ E_{\omega}\biggl(\sup_{0\leq s\leq 1} \omega(s)\geq \alpha \biggr)\leq C_{1} e^{-C_{2}\alpha^{2}}. $$Changing V outside (x-\delta, x+\delta) changes the right side of (IV\!.16) by 0(e^{-C/t}), so in proving (IV\!.15), we can suppose that V is bounded. But if V is bounded, it is easy to see that the integral is 1-tV(x)+o(t). \qed \enddemo Since we controlled e^{-tH}(x, x), we can prove the slightly stronger version of (IV\!.12), viz [24]:$$ V(x)=\lim_{\gamma\downarrow\infty} \int\limits^{\infty}_{0} e^{-\gamma\lambda} (1-2\xi(x, \lambda))\, d\lambda. $$This is all with V positive. If it is only bounded below and E_{0}\leq\inf\text{ spec}(H), then$$ V(x)=E_{0}+\lim_{\gamma\downarrow 0} \int\limits^{\infty}_{E_{0}} e^{-\gamma\lambda} (1-2\xi(x, \lambda))\,d\lambda. $$See [23] for higher order asymptotic formula involving moments of \xi. \vskip 0.3in \head IV\!.3 Examples \endhead \rightheadtext{Examples} \bigpagebreak \example{Example 1 (Periodic Potentials)} We begin with the motivating case of periodic potentials, V(x)=V(x+L). The spectrum of H is [E_{0}, E_{1}]\cup [E_{2}, E_{3}]\cup\cdots. General principles imply that \xi(x, \lambda) obeys$$ \xi(x,\lambda)=\cases \frac{1}{2}, &E_{2n}<\lambda 0. $$We see that if \int |R_{\ell}(\lambda)|\, d\lambda <\infty, then \int |1-2\xi(x, \lambda)|\,dx<\infty and one can take the limit inside the integral. See [21] for other short range examples. \endexample As a preliminary to the next example, we need a formula about how Dirichlet eigenvalues \mu(x) move as x changes. \proclaim{Theorem IV\!.6} Let (a, b) be an interval in \Bbb R\backslash\text{\rom{spec}}(H) \rom(i.e., a spectral gap of H\rom) and suppose \mu(x) is an eigenvalue of H_x, the operator with a Dirichlet boundary condition at x with \mu(x)\in (a, b). Let g(x, \lambda)\equiv G(x, x; \lambda). \roster \item"\rom{(i)}" \left.\frac{\partial g}{\partial x} (x, \lambda)\right|_{\lambda=\mu(x)} = \pm 1 with the value equal +1 \rom(resp.~-1\rom) if the eigenvalue lies on (-\infty, 0] \rom(resp.~[0, \infty)\rom). \item"\rom{(ii)}" With the same sign as in {\rom{(i)}}$$ \mu'(x)=\pm\biggl(\frac{\partial g}{\partial\mu}\, (x, \mu(x)\biggr)^{- 1}. \tag IV\!.17 $$\endroster \endproclaim \demo{Proof} (i) By general principles with \varphi_{\pm}\,\, L^{2} at \pm\infty:$$ g(x, \lambda)\equiv \frac{\varphi_{-}(x)\varphi_{+}(x)} {\varphi_{+}(x)\varphi'_{-}(x)-\varphi_{-}(x)\varphi'_{+}(x)} $$so$$ \frac{\partial g}{\partial x}(x, \lambda) = \frac{\varphi_{+}(x) \varphi'_{-}(x)+\varphi_{-}(x)\varphi'_{+}(x)} {\varphi_{+}(x)\varphi'_{-}(x)-\varphi_{-}(x)\varphi'_{+}(x)}. $$At \mu(x), either \varphi_{+}(x)=0, in which case \frac{\partial g}{\partial x}=-1, or \varphi_{-}(x)=0, in which case \frac{\partial g}{\partial x}=1 as claimed. (ii) Immediate from (i), given the implicit function theorem that says that \mu'=\mathbreak-[\partial g/\partial x]\big/[\partial g/\partial\mu]. \qed \enddemo \example{Example 4 (The Soliton)} A {\it{reflectionless}} potential is one with \xi(x, \lambda)=\frac12 for {\it all} x and a.e.~\lambda in \text{spec}_{\text{\rom ac}}(H). Let us look for reflectionless potentials with a single eigenvalue at E=-1. Let \mu(x) be the Dirichlet eigenvalue in (-1, 0]. Then$$ \xi(x, \lambda) = \cases 0, &\lambda <-1 \text{ or } \mu(x)<\lambda<0 \\ 1, &-1<\lambda<\mu(x) \\ \frac{1}{2}, &\lambda >0. \endcases $$The formulae for g and V then equal$$\align &V(x) = 2(-1-\mu(x)) \\ &g(x, \lambda) = \frac12 \,(-\lambda)^{-1/2}\,\,\frac{\mu(x)-\lambda} {-1-\lambda}. \endalign $$(IV\!.17) reads$$ \mu'(x)=\pm2(-\mu)^{1/2} (1+\mu). \tag IV\!.18 $$Let c(x)=(-\mu)^{1/2}. Then (IV\!.18) becomes$$ c'=\pm (1-c^{2}) $$whose solution is c(x)=\pm\tanh(x) so we get$$\align V(x)&= -2(1-\tanh^{2}(x)) \\ &= \frac{-2}{\cosh^{2}(x)}, \endalign $$the familiar soliton! This arguments does {\it{not}} show this potential is reflectionless, only that it is the only candidate. However, a comparison with the formula for \xi(x, \lambda) in Example 3 indeed verifies that R_{\ell}(\lambda)=0, \lambda >0 in the present case. \endexample \vskip 0.3in \head IV\!.4 The Trace Formula for Jacobi Matrices \endhead \rightheadtext{The Trace Formula for Jacobi Matrices} \bigpagebreak \proclaim{Theorem IV\!.7} Let H=H_{0}+V be a Jacobi matrix on \Bbb Z^{\nu} with V bounded. Let \xi(n, \lambda) be \xi_\infty for A=H_{0}+V and \varphi=\delta_n. Let E_{\pm}=\,^{\text{\rom sup}} _{\text{\rom inf}}\,\text{spec}(H). Then$$ V(n)=\frac12\, (E_{+}+E_{-})+\int\limits^{E_+}_{E_-} \biggl(\xi(n, \lambda) -\frac12\biggr)\, d\lambda. \tag IV\!.19 $$\endproclaim \demo{Proof} (IV\!.19) is unchanged if we add a constant, c, to V (since then V, E_+, and E_- all increase by c), so w.l.o.g.~we can suppose that H\geq 0. Since \xi=0 (resp.~1) for \lambdaE_{+}), we need only prove (IV\!.19) for some E_{+}>\sup \, \text{spec}(H) and E_{-}<\inf\,\text{spec}(H). Next we note that (IV\!.19) can be rewritten$$\align V(n)&= E_{-}+\int\limits^{E_+}_{E_-} (\xi(n, \lambda) -1)\, d\lambda = E_{-}+ \int\limits^{\infty}_{E_-} (\xi(n, \lambda) -1)\, d\lambda \\ &= E_{+}+\int\limits^{E_+}_{-\infty}\xi(n, x)\, dx \endalign $$so it suffices to prove$$ V(n)=\int\limits^{\infty}_{0} (\xi(n, x) -1)\, dx $$for H\geq 0. But noting that$$ (\delta_{n}, (H+\gamma)^{-1}\delta_{n})=\gamma^{-1}-\gamma^{-2} V(n) +O(\gamma^{-3}), $$this follows from (I.30) (Theorem I.17). \qed \enddemo \vskip 0.3in \head IV\!.5 A Regularity Theorem for the A.C.~Spectrum \endhead \rightheadtext{A Regularity Theorem for the A.C.~Spectrum} \bigpagebreak In this section we'll prove the following: \proclaim{Theorem IV\!.8 ([24])} Let V_{m}(x) and V_{\infty}(x) be functions on \Bbb R \rom(resp.~\Bbb Z\rom) so that \roster \item"\rom{(i)}" \operatornamewithlimits{\inf}\limits\Sb x\in\Bbb R\\ m=1, 2,\dots,\infty \endSb V_{m}(x)>-\infty and for each R>0, \operatornamewithlimits{\sup}\limits\Sb |x|\in R\\ m=1,2,\dots,\infty\endSb |V_{m}(x)|<\infty \rom(resp. \operatornamewithlimits{\sup}\limits \Sb x\in\Bbb Z \\ m=1,2,\dots,\infty\endSb |V_{m}(x)|<\infty\rom). \item"\rom{(ii)}" V_{m}\to V_{\infty} uniformly on compacts \rom(resp.~pointwise\rom). \item"\rom{(iii)}" Each V_m is periodic \rom(but the period can be m-dependent\rom). \endroster Let H_{m}, H_\infty be the Hamiltonians -\frac{d^2}{dx^2}+V_m \rom(resp.~the Jacobi matrices on \ell^{2}(\Bbb Z), \mathbreak h_{0}+V_{m}\rom). Then \rom(with |\, \cdot \, | = Lebesgue measure\rom):$$ |\sigma_{\text{\rom ac}}(H_{\infty})|\geq\varlimsup \, |\sigma_{\text{\rom ac}}(H_{m})|. $$\endproclaim The most interesting application of this result is to prove a slightly strengthened version of a striking result of Last [34]: \proclaim{Corollary IV\!.9} Let V(n)=\lambda\cos(\pi\alpha n+\theta) and let h be the corresponding Jacobi matrix. Then |\sigma_{\text{\rom ac}}(h)|\geq 4-2|\lambda|. \endproclaim \demo{Proof} If \alpha=p/q is rational, this is proven in [7]. If \alpha is arbitrary, let p_{m}/q_{m}\to\alpha and let V_{m}(n)=\lambda\cos(\pi_{m}\alpha n+\theta). Then V, V_{m} obey the hypotheses of Theorem IV\!.8. So, by the theorem, we can take limits. \qed \enddemo \remark{Remark} If \alpha is a Liouville number (irrational but well approximated by rationals), it is known for \lambda >2, that h has only singular continuous spectrum and before Last [34], some believed that was true also for |\lambda|<2. Last showed that if |\lambda|<2, |\sigma_{\text{\rom ac}}(h)|>0. \endremark To prove Theorem IV\!.8, we begin with \proclaim{Theorem IV\!.10} Under the hypotheses {\rom{(i), (ii)}} of Theorem {\rom{IV\!.8}},$$ \xi^{(m)}_{\infty}(x,\lambda)\, dx \to \xi^{(\infty)}_{\infty}(x, \lambda)\, dx $$weakly as measures where \xi^{(m)}_{\infty} is the spectral shift for -\frac{d^2}{dx^2}+V_m with \varphi=\delta(x). \endproclaim \demo{Proof} It suffices to prove that for all z\in\Bbb C\backslash \Bbb R$$ \int\frac{\xi^{(m)}_{\infty}(x, \lambda)\, d\lambda} {(\lambda-z)^{2}} \longrightarrow \int\frac{\xi^{(\infty)}_{\infty}(x, \lambda) \,d\lambda}{(\lambda - z)^{2}} $$by a standard density argument. By (I.29) and the Cauchy formula, it suffices to prove that F_{m}(z)\to F_{\infty}(z) where F_{m}(z)=G_{V_m}(0, 0; z). Writing$$ \bigl[(H_{n}-z)^{-1}-(H_{\infty}-z)^{-1}\bigr] (H_{\infty}-z)\psi =(H_{n}-z)^{-1}(V_{\infty}-V_{n})\psi $$for \psi\in C^{\infty}_{0}(\Bbb R), we see that$$ \text{s-lim }(H_{n}-z)^{-1} = (H_{\infty}-z)^{-1}. \tag IV\!.20 $$Let H_{0}=-\frac{d^2}{dx^2}. Then (H_{\infty}+1)^{+1/2} (H_{n}-z)^{-1} (H_{0}+1)^{1/2} is uniformly bounded, so (IV\!.20) implies that$$ \text{s-lim }(H_{\infty}+1)^{1/2} (H_{n}-z)^{-1} (H_{0}+1)^{1/2} = (H_{0}+1)^{1/2} (H_{\infty}-z)^{-1} (H_{0}+1)^{1/2}. $$Since (H_{0}+1)^{-1/2}\delta_{0} is in L^2, we can take matrix elements in this vector and conclude that F_{m}(z)\to F_{\infty}(z) is as required. \qed \enddemo \demo{Proof of Theorem {\rom{IV\!.8}}} Since V_m is periodic, \xi_{m}(x,\lambda)=0,1\text{ or } \frac12 for a.e.~\lambda. Fix [a,b]\subset\Bbb R bounded and let A_{m}=\{\lambda\mid\xi_{m}(x, \lambda)=0\}\cap [a, b] for m=1, 2,\dots, \infty. Then$$ 0=\int\limits_{A_\infty} \xi_{\infty}(x, \lambda)\, d\lambda = \lim \int\limits_{A_\infty}\xi_{m}(x, \lambda)\, d\lambda \geq \lim\frac12 \,|A_{\infty}\backslash A_{m}|\geq\frac12\, (|A_{\infty}|- \varliminf |A_{m}|). $$We conclude that$$ |A_{\infty}|\leq\varliminf |A_{m}|. $$A similar argument with \xi_{m}=1 shows that$$ |\{\lambda\mid 0<\xi_{\infty}(x, \lambda)<1\}\cap [a, b]| \geq \varlimsup\, \left|\biggl\{\lambda\mid\xi_{m}(x, \lambda)=\frac12\biggr\} \cap [a, b]\right|. $$But \xi_{\infty}(x, \lambda)=\frac{1}{\pi}\text { Arg } (F_{\infty}(\lambda +i0)) lies in (0, 1) if and only if \text{Im } F_{\infty}>0 (at points with F_{\infty}(\lambda+i0) finite). So$$ |\sigma_{\text{\rom ac}}(H_{m})\cap [a,b]| \geq |\{\lambda\mid 0 < \xi_{m}(x, \lambda)<1\}\cap [a, b]| $$with equality if m is finite. \qed \enddemo \vskip 0.3in \head IV\!.6 Inverse Problems \endhead \rightheadtext{Inverse Problems} \bigpagebreak These ideas play an important role in understanding inverse spectral theory. We'll give an example here; see [24] (and a paper in preparation) for further examples. We'll rely on the Gelfand-Levitan theory [20,35] which allows one to recover V(x) from any m_{\alpha}(z) and, in particular, from G_{\theta=\pi/2}(0, 0; z). In 1952 Borg [8] and Mar\v cenko [36] proved that: \proclaim{Theorem IV\!.11} Let V(x)\to\infty to infinity. Let \{E^{(\alpha)}_{n}\}^{\infty}_{n=1} be the eigenvalues for - \frac{d^2}{dx^2}+V(x) with \theta boundary conditions at x=0 where \alpha=-\cot(\theta). Then V can be recovered from \{E^{(\alpha)}_{n}\}^{\infty}_{n=1} for any two distinct values of \alpha. \endproclaim Actually, Mar\v cenko [36] proved in addition that the two values of \alpha can be recovered as well from the two sets \{E^{(\alpha)}_{n}\}^{\infty}_{n=1}. We'll prove a generalization of this. \proclaim{Theorem IV\!.12} Suppose that V is bounded below. Let \xi_{\alpha, \beta}(\lambda) (\alpha>\beta) be the spectral shift for going from A_\beta to A_\alpha. Then \alpha, \beta, and V can be recovered from \xi_{\alpha, \beta}(\lambda). \endproclaim \remark{Remark} This implies Theorem IV\!.11. For in that case E^{(\beta)}_{n}\leq E^{(\alpha)}_{n}\leq E^{(\beta)}_{n+1} and$$ \xi_{\alpha, \beta}(\lambda)=\cases 1 &E^{(\beta)}_{n}<\lambda< E^{(\alpha)}_{n} \\ 0 &E^{(\alpha)}_{n}<\lambda