Green Functions and Density of States

The density of states counts how many independent states live near an energy. Green functions reveal it because their singularities sit exactly at spectral values. In the most compact form,

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

The first expression is state counting. The second expression is the same state counting read from the boundary value of the resolvent. This page explains that relation and how it is used.

For compact formulas independent of Green-function derivations, see the reference entry Density of States.

Density of States

For a finite Hamiltonian with eigenvalues $E_n$, including degeneracies by repeating labels, the exact density of states is the distribution

$$\rho(E) = \sum_n\delta(E-E_n).$$

Equivalently, if $N(E)$ counts states with energies below $E$,

$$N(E) = \operatorname{Tr}\Theta(E-H),$$

then

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

For a finite-dimensional or purely discrete finite-volume problem, $\rho(E)$ is a sum of delta functions. In large systems, thermodynamic limits, coarse graining, or explicit broadening often turn it into a smooth density. The normalization must always be stated: total density of states, per volume, per unit cell, per spin species, per band, or per internal degeneracy.

Trace of the Resolvent

Let

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

In an eigenbasis,

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

Taking the trace gives

$$\operatorname{Tr}R(z) = \sum_n \frac{1}{z-E_n}.$$

The trace of the resolvent is therefore a meromorphic function for a finite-dimensional system. Its poles are the energy levels, and its residues count degeneracy. A finite spatial volume alone does not ensure that the resolvent is trace class: infinitely many high-energy levels can make the trace diverge. In that case use a spectral window, a suitable subtraction, or a regulator before taking a trace. This is separate from resolving individual peaks with a small $\eta$.

More generally, the spectral theorem gives

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

so a trace or matrix element of $R(z)$ is a transform of the spectral measure.

Imaginary Part of the Green Function

The retarded boundary value is

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

The distribution identity

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

implies

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

Taking the trace gives the density-of-states formula:

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

This is the same relation introduced in Spectral Representation of Green Functions, now emphasized as a practical state-counting tool.

Broadening and Numerical Spectra

In calculations one often keeps a small positive $\eta$ instead of taking the exact boundary value:

$$R_\eta(E) = \frac{1}{E+i\eta-H}.$$

Then each exact delta function becomes a Lorentzian:

$$-\frac{1}{\pi} \operatorname{Im} \frac{1}{E+i\eta-E_n} = \frac{1}{\pi} \frac{\eta} {(E-E_n)^2+\eta^2}.$$

Thus

$$\rho_\eta(E) = -\frac{1}{\pi} \operatorname{Im} \operatorname{Tr}R_\eta(E)$$

is a broadened density of states. This is the exact density of states convolved with a Lorentzian, not the exact spectral measure of the original closed Hamiltonian. Each broadened level still integrates to one and has full width $2\eta$. A physical linewidth requires a justified self-energy or dynamical model; choosing a plotting parameter does not establish a lifetime.

Counting inside an apparent gap

Broadening preserves the integral over the entire energy axis, but can move weight across the edges of a finite measurement or plotting window. Take the chapter’s two-level model, $H=\operatorname{diag}(0,\Delta)$ with $\Delta>0$. There are no levels in $[\Delta/4,3\Delta/4]$, yet the broadened count there is

$$\begin{aligned} N_\eta &=\int_{\Delta/4}^{3\Delta/4}\rho_\eta(E)\,dE\\ &=\frac{2}{\pi}\left[ \arctan\frac{3\Delta}{4\eta} -\arctan\frac{\Delta}{4\eta}\right]. \end{aligned}$$

At $\eta=\Delta/4$, this is approximately $0.2952$ states, despite the exactly empty window. For well-resolved levels,

$$N_\eta=\frac{16\eta}{3\pi\Delta} +O\!\left((\eta/\Delta)^3\right), \qquad \eta/\Delta\to0.$$

The positive weight is the tail of neighboring Lorentzians. Decreasing $\eta$ tests that explanation; the example predicts a linear decrease once $\eta\ll\Delta$. A numerical calculation must also resolve the narrower peaks and converge its basis or spatial discretization. Neither a single smooth plot nor the total-area sum rule alone distinguishes genuine in-gap states from broadening tails. The energy-window integral and its endpoint limits follow directly from the Stieltjes inversion formula discussed by Teschl; Exercise 2 makes those limits explicit.

Local Density of States

In a position representation, the diagonal Green-function kernel gives the local density of states:

$$\rho(\mathbf x,E) = \langle\mathbf x|\delta(E-H)|\mathbf x\rangle.$$

Using the retarded Green function,

$$\rho(\mathbf x,E) = -\frac{1}{\pi} \operatorname{Im} G^R(\mathbf x,\mathbf x;E).$$

The total density of states is recovered by integrating over position:

$$\rho(E) = \int d^d x\, \rho(\mathbf x,E),$$

with the correct boundary, volume, and normalization conventions. In lattice models, the analogous statement is a site trace:

$$\rho(E) = \sum_i \rho_i(E), \qquad \rho_i(E) = -\frac{1}{\pi} \operatorname{Im}G^R_{ii}(E).$$

Local density of states is central in tunneling spectroscopy, impurity problems, mesoscopic systems, and numerical Green-function methods.

Bound and Continuous Spectra

For isolated bound states, the density of states contains delta-function peaks:

$$\rho_{\rm bound}(E) = \sum_b d_b\,\delta(E-E_b),$$

where $d_b$ is the degeneracy. In the resolvent, these appear as poles.

For familiar absolutely continuous spectra, a density is supported on allowed energy regions such as intervals or bands. Singular continuous spectral measures need the full measure formulation in Spectral Representation and need not admit an ordinary density. In the resolvent, this appears as a branch cut or as a limiting boundary value along the real axis. The imaginary part of the retarded Green function measures the discontinuity across that cut:

$$R^R(E)-R^A(E) = -2\pi i\,\delta(E-H).$$

This is why density of states, spectral functions, scattering thresholds, and band edges are all read from analytic structure. For the distinction between density of states, operator spectral functions, and measured intensity, see Spectral Functions.

Worked Example: Free Particle

For a spinless free particle in $d$ spatial dimensions,

$$E_{\mathbf k} = \frac{\hbar^2k^2}{2m}.$$

In a large box, the density of states per volume is

$$\frac{\rho_d(E)}{V} = \int\frac{d^d k}{(2\pi)^d}\, \delta\left( E-\frac{\hbar^2k^2}{2m} \right).$$

For $E\gt0$, using spherical coordinates in momentum space gives

$$\frac{\rho_d(E)}{V} = \frac{S_{d-1}}{2(2\pi)^d} \left( \frac{2m}{\hbar^2} \right)^{d/2} E^{d/2-1},$$

where

$$S_{d-1} = \frac{2\pi^{d/2}}{\Gamma(d/2)}$$

is the area of the unit sphere in $d$ dimensions. The dimension dependence is physically important:

Dimension	Spinless free-particle behavior
$d=1$	$\rho(E)/L\propto E^{-1/2}$
$d=2$	$\rho(E)/A$ is constant
$d=3$	$\rho(E)/V\propto E^{1/2}$

The Green-function formula gives the same answer:

$$\frac{\rho_d(E)}{V} = -\frac{1}{\pi} \operatorname{Im} \int\frac{d^d k}{(2\pi)^d} \frac{1}{E+i0-E_{\mathbf k}}.$$

The imaginary part turns the denominator into the delta function on the energy shell. The real part of the unregularized continuum integral can diverge, so do not assume its full coincident Green function exists merely because the spectral density is finite.

Common Mistakes

• Forgetting that $\rho(E)$ is a distribution before smoothing or taking a thermodynamic limit.
• Mixing total density of states with density per volume, per unit cell, or per spin.
• Taking the imaginary part of the wrong Green function or using the wrong $i0$ sign.
• Treating numerical broadening as a physical lifetime without justification.
• Assuming the local density of states is uniform in an inhomogeneous system.
• Applying a smooth energy-shell formula at a critical point where $\nabla_{\mathbf k}E=0$.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• E. N. Economou, Green’s Functions in Quantum Physics, 3rd ed., Springer, 2006.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.
• G. D. Mahan, Many-Particle Physics, 3rd ed., Kluwer Academic/Plenum, 2000.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• S. H. Simon, The Oxford Solid State Basics, Oxford University Press, 2013.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, American Mathematical Society, 2009, §3.4, Theorem 3.21, author text.

Exercises

1. Starting from a discrete spectral decomposition, derive $\rho(E)=-(1/\pi)\operatorname{Im}\operatorname{Tr}(E+i0-H)^{-1}$.

Solution

For a discrete spectrum,

$$\operatorname{Tr}(E+i0-H)^{-1} = \sum_n \frac{1}{E+i0-E_n}.$$

Using

$$\operatorname{Im} \frac{1}{x+i0} = -\pi\delta(x),$$

one finds

$$-\frac{1}{\pi} \operatorname{Im} \operatorname{Tr}(E+i0-H)^{-1} = \sum_n\delta(E-E_n) = \rho(E).$$

2. Show that finite $\eta$ broadens a delta function into a Lorentzian. Integrate a single broadened level over $[a,b]$, with $a<b$. What happens as $\eta\to0^+$ if the level lies strictly inside, strictly outside, or exactly at an endpoint?

Solution

For real $x$,

$$\frac{1}{x+i\eta} = \frac{x-i\eta}{x^2+\eta^2}.$$

Therefore

$$-\frac{1}{\pi} \operatorname{Im} \frac{1}{x+i\eta} = \frac{1}{\pi} \frac{\eta}{x^2+\eta^2}.$$

With $x=E-E_n$, this is the Lorentzian approximation to $\delta(E-E_n)$ as $\eta\to0^+$.

Its window weight is

$$w_\eta(E_n;[a,b]) =\frac{1}{\pi}\left[ \arctan\frac{b-E_n}{\eta} -\arctan\frac{a-E_n}{\eta}\right].$$

Taking the limits of the two arctangents gives $1$ for $a<E_n<b$, $0$ for $E_n<a$ or $E_n>b$, and $1/2$ for $E_n=a$ or $E_n=b$. Thus symmetric Lorentzian smoothing assigns half of an endpoint atom to the limiting window integral; it does not automatically implement either a strictly open or a closed spectral window. Move the endpoints away from levels or specify the desired endpoint convention before interpreting the result as an integer state count. Spectral Projections explains the corresponding projection-valued distinction.

3. Derive the spinless three-dimensional free-particle density of states per volume.

Solution

Start from

$$\frac{\rho_3(E)}{V} = \int\frac{d^3k}{(2\pi)^3} \delta\left( E-\frac{\hbar^2k^2}{2m} \right).$$

In spherical coordinates,

$$\frac{\rho_3(E)}{V} = \frac{4\pi}{(2\pi)^3} \int_0^\infty dk\,k^2 \delta(E-ak^2), \qquad a=\frac{\hbar^2}{2m}.$$

The root is $k_E=\sqrt{E/a}$, and

$$\delta(E-ak^2) = \frac{\delta(k-k_E)}{2ak_E}$$

for $k\ge0$. Hence

$$\frac{\rho_3(E)}{V} = \frac{1}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt{E},$$

for $E\gt0$, before spin degeneracy factors.

4. Show that the local density of states integrates to the total density of states.

Solution

By definition,

$$\rho(\mathbf x,E) = \langle\mathbf x|\delta(E-H)|\mathbf x\rangle.$$

Integrating over position and using the coordinate resolution of identity gives

$$\int d^d x\,\rho(\mathbf x,E) = \int d^d x\, \langle\mathbf x|\delta(E-H)|\mathbf x\rangle = \operatorname{Tr}\delta(E-H) = \rho(E).$$

The same argument becomes a sum over sites or basis states in a lattice model.