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 continuous spectrum, sums become spectral integrals. In a simplified generalized-eigenstate 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

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

More carefully, this statement is interpreted through matrix elements, traces, or spectral measures. 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 owns the nonrelativistic $N\pm1$ Lehmann representation, fermionic matrix positivity, occupation sum rule, and Matsubara bridge. Spectral Functions owns the cross-channel line-shape, quasiparticle, linewidth, and measured-intensity dictionary.

A schematic relativistic example is the Källén–Lehmann form:

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

The spectral density $\rho(\mu^2)$ records which mass values or multiparticle continua can propagate. 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 $i$.

Cross-Links

• Resolvent Operator
• Energy Green Function
• Green Functions and Response Preview
• What Is a Green Function?
• Spectral Theorem, Practical Version
• Propagator Kernel
• Spectral Decomposition of the Propagator
• Green Functions and Density of States
• Time-Dependent Correlations
• Green Functions in Many-Body QM
• Spectral Functions
• Density of States
• Many-Body Correlation Functions
• From Correlation Functions to QFT Observables
• QFT Bridge: Green Functions

References

• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• 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.

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 $(z-E_n)^{-1}$.