Resolvent Methods

Resolvents connect three aspects of scattering: spectral densities on the real axis, perturbation equations away from it, and analytically continued poles. The first two require no resonance continuation. This page derives the spectral jump formula and a rank-one perturbation formula, then explains which conclusions follow from compact or trace-class resolvent information.

Required background. Resolvent Operators establishes the resolvent identities and the convention $R(z)=(H-z)^{-1}$. Stationary Scattering Theory specifies the weighted boundary spaces.

The resolvent jump recovers a spectral projection

Let $H$ be self-adjoint, with spectral measure $E_H$. For real $a<b$,

$$\frac{1}{2\pi i}\int_a^b [R(E+i\eta)-R(E-i\eta)]\,dE$$

converges strongly as $\eta\downarrow0$ to

$$E_H((a,b))+\tfrac12E_H(\{a\})+\tfrac12E_H(\{b\}).$$

In particular, if the endpoints are not eigenvalues, the limit is $E_H((a,b))$. This is Stone’s resolvent formula, distinct from Stone’s generator theorem.

For the proof, apply the spectral calculus to the scalar multiplier

$$\frac1\pi\int_a^b \frac{\eta}{(\lambda-E)^2+\eta^2}\,dE.$$

It lies between zero and one, tends to one inside $(a,b)$, to zero outside $[a,b]$, and to one half at each endpoint. Dominated convergence in each vector’s spectral measure proves the strong limit. The half weights are therefore part of the theorem, not a convention that can be discarded at atoms.

Where weighted boundary values exist with sufficient local continuity, the corresponding weak spectral density is

$$\frac{dE_H(E)}{dE} =\frac{1}{2\pi i}[R(E+i0)-R(E-i0)].$$

This notation describes pairings with suitably weighted test vectors. It does not make the density a bounded operator on the original $L^2$ space.

A rank-one perturbation can be solved exactly

Let $H_0$ be self-adjoint, $v\in\mathcal H$, and $g\in\mathbb R$. Define the bounded self-adjoint perturbation

$$H=H_0+g|v\rangle\langle v|, \qquad D(H)=D(H_0).$$

For nonreal $z$, put $R_0(z)=(H_0-z)^{-1}$ and $F(z)=\langle v,R_0(z)v\rangle$. Solving $(H-z)u=f$ gives

$$u=R_0(z)f-gR_0(z)v\langle v,u\rangle.$$

Pairing with $v$ determines the scalar coefficient. Hence

$$R(z)=R_0(z) -\frac{g}{1+gF(z)} R_0(z)|v\rangle\langle v|R_0(z).$$

All products here are bounded. The bra on the right means the functional $f\mapsto\langle v,R_0(z)f\rangle$; it is not obtained by replacing both resolvents with the same adjoint.

For $v\ne0$ and $\operatorname{Im}z>0$,

$$\operatorname{Im}F(z) =(\operatorname{Im}z)\|R_0(z)v\|^2>0.$$

Thus $1+gF(z)$ cannot vanish off the real axis for real nonzero $g$. This agrees with self-adjointness. Complex roots of a continued denominator, when a continuation exists, are a different matter.

A bound level emerging below a continuum

Take $\mathcal H=L^2(0,1)$, $(H_0f)(x)=xf(x)$, and $v(x)=1$. The unperturbed spectrum is the absolutely continuous interval $[0,1]$. For $z=-\kappa<0$,

$$F(-\kappa)=\int_0^1\frac{dx}{x+\kappa} =\log\frac{1+\kappa}{\kappa}.$$

For an attractive coupling $g=-\alpha$ with $\alpha>0$, the denominator vanishes when

$$\kappa=\frac{1}{e^{1/\alpha}-1}.$$

The vector $(x+\kappa)^{-1}$ belongs to $L^2(0,1)$ and solves the eigenvalue equation at energy $-\kappa$. As $\alpha\downarrow0$, the level approaches the band edge at an exponentially small distance. This behavior follows from the logarithmic scalar resolvent, not from an ordinary power series in the coupling.

The example also shows why compact perturbations can add bound states while preserving the essential spectrum. It does not imply that every small attractive potential in every spatial dimension has the same binding law.

Compactness and trace class support different conclusions

A compact resolvent difference preserves essential spectrum for self-adjoint operators. That spectral statement alone does not establish wave-operator existence or completeness.

Trace-class scattering theorems give stronger information. For example, the Kato–Rosenblum theorem states that a bounded self-adjoint trace-class difference $H-H_0$ ensures existence and completeness of the wave operators on the absolutely continuous sectors. The rank-one example satisfies that hypothesis.

The theorem is deep; its proof is not the elementary denominator calculation. Its conclusion concerns the absolutely continuous parts and does not exclude new bound states or establish absence of singular continuous spectrum. Replacing “trace class” by “compact” or “Hilbert–Schmidt” is not justified.

Meromorphic continuation is yet another input. Resonances supplies a compact-support model where it can be proved, rather than treating every pole suggested by an algebraic manipulation as physical.

Exercises

1. Show directly that the negative eigenvector in the example is unique up to a scalar.

Solution

At energy $-\kappa$, the equation gives $u=\alpha(H_0+\kappa)^{-1}v\langle v,u\rangle$. If the scalar pairing vanished, then $u=0$. Every nonzero solution is therefore proportional to $(x+\kappa)^{-1}$. The scalar denominator condition selects the allowed $\kappa$.

2. If an endpoint $a$ is an eigenvalue, why does integrating the resolvent jump over $[a,b]$ not return its full projection?

Solution

The Lorentzian kernel centered at $a$ has half its mass on each side of the endpoint in the zero-width limit. The integral therefore returns half of that atom. One can choose non-atomic endpoints or explicitly restore the missing half projection.

References

• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapter 12, author’s text.
• D. Yafaev, “Lectures on scattering theory,” 2004, arXiv:math/0403213.