Spectral Representation of Green Functions

The spectral representation writes a Green function or resolvent in the basis that diagonalizes the Hamiltonian. It makes the central message visible:

Green functions know the spectrum because they are built from factors such as $(z-E)^{-1}$.

For a Hamiltonian $H$, the resolvent is

$$R(z)=(zI-H)^{-1}.$$

The spectral theorem turns this into a sum or integral over spectral values. Poles, residues, branch cuts, and imaginary parts then encode bound states, continuum states, and densities of states.

Discrete Spectral Representation

Suppose the Hamiltonian has a discrete spectral decomposition

$$H=\sum_n E_nP_n,$$

where $P_n$ projects onto the eigenspace with energy $E_n$. Then

$$R(z) = \sum_n \frac{P_n}{z-E_n}.$$

If each eigenvalue is nondegenerate,

$$P_n=|n\rangle\langle n|,$$

so

$$R(z) = \sum_n \frac{|n\rangle\langle n|}{z-E_n}.$$

In coordinate representation,

$$G(x,x';z) = \langle x|R(z)|x'\rangle = \sum_n \frac{\psi_n(x)\psi_n^*(x')}{z-E_n}.$$

Here $\psi_n(x)=\langle x|n\rangle$. This formula is the energy-domain analogue of expanding a wavefunction in energy eigenstates.

Propagator Spectral Sum

The time-domain propagator kernel has a related but different spectral representation:

$$K(x_f,t;x_i,0) = \sum_n \psi_n(x_f)\psi_n^*(x_i) e^{-iE_nt/\hbar}.$$

The resolvent and propagator are both diagonal in the energy basis, but they apply different scalar functions to each energy:

$$R(z):\quad E_n\mapsto \frac{1}{z-E_n},$$

while

$$U(t):\quad E_n\mapsto e^{-iE_nt/\hbar}.$$

This is the cleanest way to see why Green functions and propagator kernels are related but not identical.

Continuous Spectral Representation

For a purely absolutely continuous spectrum, energy-normalized generalized eigenstates give the following identity resolution and integral. If bound states are also present, these continuum formulas describe only the absolutely continuous subspace: replace $I$ by its projector $P_{\mathrm{ac}}$ and add the bound-state projectors and resolvent terms separately. The Robin half-line propagator gives a concrete mixed-spectrum example. Singular continuous parts require the full spectral measure written below. In the energy-normalized notation,

$$H|E,\alpha\rangle = E|E,\alpha\rangle,$$

where $\alpha$ labels degeneracy or channels. The identity resolution is written formally as

$$I = \int dE\, \sum_\alpha |E,\alpha\rangle\langle E,\alpha|.$$

Then

$$R(z) = \int dE\, \sum_\alpha \frac{|E,\alpha\rangle\langle E,\alpha|}{z-E}.$$

In coordinate representation,

$$G(x,x';z) = \int dE\, \sum_\alpha \frac{\psi_{E,\alpha}(x)\psi_{E,\alpha}^*(x')}{z-E}.$$

This notation is convenient but formal. The rigorous version uses the projection-valued measure:

$$R(z) = \int_{\mathbb R} \frac{1}{z-\lambda}\,dE_H(\lambda).$$

The generalized eigenfunctions, measures, and degeneracy labels depend on the problem and normalization convention.

Poles and Residues

For an isolated bound-state energy $E_k$,

$$R(z) = \frac{P_k}{z-E_k} +R_{\rm reg}(z),$$

near $z=E_k$. The pole identifies the energy, and the residue is the spectral projector:

$$\operatorname*{Res}_{z=E_k}R(z) = P_k.$$

For a coordinate-space Green function, the residue becomes

$$\operatorname*{Res}_{z=E_k}G(x,x';z) = \sum_{a=1}^{d_k} \psi_{k,a}(x)\psi_{k,a}^*(x'),$$

where $d_k$ is the degeneracy of the bound state. The residue is therefore not merely a number; it reconstructs the bound-state wavefunctions up to basis choices inside the degenerate subspace.

Spectral Density

For real $E$, the boundary value

$$R(E+i0) = \lim_{\epsilon\to0^+}(E+i\epsilon-H)^{-1}$$

has an imaginary part tied to the spectral measure. The distribution identity

$$\frac{1}{E+i0-\lambda} = \operatorname{PV}\frac{1}{E-\lambda} -i\pi\delta(E-\lambda)$$

implies the formal operator relation

$$-\frac{1}{\pi}\operatorname{Im}R(E+i0) = \delta(E-H).$$

Here $\operatorname{Im}R=(R-R^\dagger)/(2i)$, interpreted through matrix elements or spectral measures. For a positive energy-bin function $f$, $\int f(E)\delta(E-H)dE=f(H)$ is positive; this is the precise meaning of a positive operator spectral density. Different subfields define spectral functions with different factors, for example $A(E)=-2\,\operatorname{Im}G^R(E)$ in common many-body conventions.

Relation to Density of States

The total density of states is formally

$$\rho(E)=\operatorname{Tr}\delta(E-H).$$

Using the spectral-density relation,

$$\rho(E) = -\frac{1}{\pi} \operatorname{Im}\operatorname{Tr}R(E+i0).$$

For a finite discrete spectrum,

$$\rho(E) = \sum_n d_n\,\delta(E-E_n),$$

where $d_n=\operatorname{Tr}P_n$ is the degeneracy. For infinite systems, one must specify whether $\rho(E)$ is total, per unit volume, per unit cell, per spin species, or per other normalization. See Green Functions and Density of States for the Green-function derivation and Density of States for a compact formula reference.

QFT Spectral Representation Preview

In field theory and many-body theory, Green functions are often correlation functions rather than one-particle wavefunction kernels. Spectral representations still express analytic structure through spectral weights, poles, cuts, and thresholds. Green Functions in Many-Body QM develops the nonrelativistic $N\pm1$ Lehmann representation, fermionic matrix positivity, occupation sum rule, and Matsubara bridge. Spectral Functions explains the cross-channel line-shape, quasiparticle, linewidth, and measured-intensity dictionary.

A schematic relativistic example is the Källén–Lehmann form:

$$\langle0|T\phi(x)\phi(0)|0\rangle = \int_0^\infty d\mu^2\, \rho(\mu^2)\, \Delta_F(x;\mu^2).$$

Here $\Delta_F(x;\mu^2)$ denotes the free time-ordered two-point function with the same overall-$i$ convention as the left side. Assume a Poincaré-invariant vacuum, a scalar field with vanishing one-point function, and a positive-metric physical Hilbert space. Then $\rho(\mu^2)\ge0$ records field overlaps with mass eigenstates and multiparticle continua; it is not an unweighted count of states. This is only a preview; QFT conventions add time ordering, Lorentz invariance, renormalization, and field normalization issues. The bridge entry is Green Functions.

From Correlation Functions to QFT Observables explains how poles, residues, thresholds, and operator overlaps enter mass extraction, response, and scattering.

Common Mistakes

• Treating generalized eigenstates in a continuum as ordinary normalizable vectors.
• Forgetting degeneracy labels in spectral sums or integrals.
• Reading a continuum branch cut as if it were a single isolated pole.
• Dropping the $i0$ prescription before taking imaginary parts.
• Confusing the density of states with an occupation probability.
• Comparing spectral functions across subfields without checking factors of $\pi$, $2\pi$, and $i$.

References

• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.

Exercises

1. For a nondegenerate discrete spectrum, derive the coordinate-space expression for $G(x,x';z)$ from $R(z)=\sum_n |n\rangle\langle n|/(z-E_n)$.

Solution

Insert position bras and kets:

$$G(x,x';z) = \langle x|R(z)|x'\rangle = \sum_n \frac{\langle x|n\rangle\langle n|x'\rangle}{z-E_n}.$$

Using $\psi_n(x)=\langle x|n\rangle$ and $\langle n|x'\rangle=\psi_n^*(x')$ gives

$$G(x,x';z) = \sum_n \frac{\psi_n(x)\psi_n^*(x')}{z-E_n}.$$

2. Use the distribution identity for $1/(E+i0-\lambda)$ to show why $-\pi^{-1}\operatorname{Im}R(E+i0)$ is a spectral-density operator.

Solution

The spectral representation is

$$R(E+i0) = \int \frac{1}{E+i0-\lambda}\,dE_H(\lambda).$$

Using

$$\frac{1}{E+i0-\lambda} = \operatorname{PV}\frac{1}{E-\lambda} -i\pi\delta(E-\lambda),$$

the imaginary part is

$$\operatorname{Im}R(E+i0) = -\pi \int\delta(E-\lambda)\,dE_H(\lambda).$$

Thus

$$-\frac{1}{\pi}\operatorname{Im}R(E+i0) = \delta(E-H)$$

in the spectral-measure sense.

3. Suppose an eigenvalue $E_k$ has degeneracy $d_k$. What is the residue of $R(z)$ at $z=E_k$?

Solution

The contribution of the degenerate eigenspace is

$$\frac{P_k}{z-E_k},$$

where

$$P_k=\sum_{a=1}^{d_k}|k,a\rangle\langle k,a|.$$

Therefore

$$\operatorname*{Res}_{z=E_k}R(z)=P_k.$$

4. Why does the spectral representation of the propagator contain $e^{-iE_nt/\hbar}$ while the resolvent contains $(z-E_n)^{-1}$?

Solution

Both objects are functions of the Hamiltonian. The time-evolution operator is $U(t)=e^{-iHt/\hbar}$, so each energy component is multiplied by $e^{-iE_nt/\hbar}$. The resolvent is $R(z)=(z-H)^{-1}$, so each energy component is multiplied by $(z-E_n)^{-1}$.