research35 min studyLesson 553 in Physics

Green-function spectral weight and sum rules

Connect addition and removal spectra to canonical anticommutators and use normalization as a stringent numerical diagnostic.

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What you’ll understand

  • Define an energy-resolved retarded spectral convention.
  • Derive normalization and occupied-weight sum rules.
  • Distinguish lost measurement weight from a violation of the exact sum rule.

Adding one electron to an interacting material does not usually create a single energy eigenstate. It overlaps with many allowed excitations, distributing its spectral weight across energy. The one-electron Green function organizes those overlaps. Exact sum rules then constrain any proposed spectrum, even when its peaks are broad or no quasiparticle exists. These constraints are especially useful because a plausible-looking plotted peak can conceal missing weight, double counting or an inconsistent normalization.

Define the spectral convention explicitly

Use energy E measured relative to the chemical potential, rather than angular frequency. For one canonical spin-orbital define the retarded Green function as a function of complex energy in the upper half-plane. Its spectral representation is an integral over a nonnegative spectral function. The spectral function has inverse-energy units and the Green function has the same units. Choosing energy avoids silently mixing a frequency integration measure with an omitted reduced Planck constant.

The spectral peaks come from matrix elements connecting many-body states that differ by one electron. Electron addition and removal both contribute to the retarded anticommutator spectrum. At zero temperature the chemical potential separates occupied removal weight from unoccupied addition weight, apart from states exactly at zero energy requiring a limiting prescription. These weights are probabilities of overlaps for a specified operator; they are not a count of independent little electrons moving on each plotted spectral branch.

GR(z)=∫−∞∞A(E)z−E dE,A(E)=−1πIm⁡GR(E+i0),Im⁡z>0G^R(z)=\int_{-\infty}^{\infty}\frac{A(E)}{z-E}\,dE,\qquad A(E)=-\frac1\pi\operatorname{Im}G^R(E+i0),\qquad \operatorname{Im}z>0
E and z are energies relative to the chemical potential. For a canonical fermion orbital A is nonnegative and has inverse-energy units. This convention contains no extra 2π factor in the normalization integral.↔ Scroll wide equations horizontally.

Use completeness to fix total weight

Insert complete many-body energy eigenstates into the two terms of the anticommutator. Integrating over energy removes their delta functions and leaves the expectation of the equal-time anticommutator. For a normalized canonical orbital that operator is the identity, so the total spectral area equals one for each spin-orbital. Summing both spin spectra gives two, not one. Interactions redistribute the area but cannot change this algebraic normalization of the retained canonical operator.

Thermal occupation weights distinguish removal from addition. Multiplying the spectrum by the Fermi function and integrating gives the orbital occupation; the complementary weight is one minus that occupation. These are independent checks: a spectrum can have total area one yet put too much weight below the chemical potential. Momentum-resolved spectra obey corresponding rules for each momentum and spin. Projected noncanonical low-energy operators can have different anticommutators, so their reduced sum rules must be derived rather than copied unchanged.

∫A(E) dE=⟨{c,c†}⟩=1,n=∫fF(E)A(E) dE,∫[1−fF(E)]A(E) dE=1−n\int A(E)\,dE=\langle\{c,c^\dagger\}\rangle=1,\qquad n=\int f_F(E)A(E)\,dE,\qquad \int[1-f_F(E)]A(E)\,dE=1-n
fF is the Fermi occupation at the chosen temperature and chemical-potential reference; n is the occupation of one spin-orbital. The unit sum rule assumes canonical anticommutation and an integral over the complete energy axis.↔ Scroll wide equations horizontally.

Audit broadening and finite windows

An interacting atomic orbital provides two separated charge transitions, while hopping and lifetime effects can turn them into bands or broad peaks. A broadening procedure should conserve their integrated areas. Numerical smoothing that moves weight outside a finite plotting window can make the visible area too small without violating the exact full-axis sum rule. Renormalizing only the displayed window to unit area can then exaggerate the low-energy spectral residue and corrupt the inferred occupation.

A peak becoming narrow need not mean its area grows; height, width and integrated weight answer different questions. Experimental photoemission also multiplies occupied spectral weight by matrix elements and the Fermi function, then convolves with resolution and adds background. Its raw intensity is therefore not automatically a unit-normalized spectral function. The probability lab illustrates exact normalized weights in a two-outcome reference distribution, but it contains no Green function, energy axis, many-body transition matrix elements or photoemission response.

Aatom(E)=(1−nσˉ)δ(E+μ)+nσˉδ(E+μ−U)A_{\rm atom}(E)=(1-n_{\bar\sigma})\delta(E+\mu)+n_{\bar\sigma}\delta(E+\mu-U)
This exact isolated-site spectrum is for one spin σ in the Hubbard atomic model, with the opposite-spin occupation nσ̄ fixed by its equilibrium state. Each delta function has inverse-energy units and their weights sum to one.↔ Scroll wide equations horizontally.

Read occupancy and a spectral moment

At zero temperature an illustrative canonical-orbital spectrum is A(E)=0.30δ(E+2 eV)+0.70δ(E−1 eV)A(E)=0.30\delta(E+2\,\mathrm{eV})+0.70\delta(E-1\,\mathrm{eV}). Check its total weight, find its occupation and compute its first energy moment.

  1. Each delta function integrates to one, so total spectral weight is 0.30+0.70=10.30+0.70=1. This is the canonical single-spin-orbital normalization, not a spin-summed spectrum.

  2. The negative-energy peak lies below the chemical potential and is occupied at zero temperature; the positive-energy peak is empty. Therefore the orbital occupation is n=0.30n=0.30, while addition weight is 0.700.70.

  3. Weight each energy by its spectral area: M1=0.30(−2)+0.70(1)=0.10 eVM_1=0.30(-2)+0.70(1)=0.10\,\mathrm{eV}. The moment has energy units even though the total area and occupation are dimensionless.

The result

The spectrum is normalized, its occupation is 0.30 and its first energy moment is +0.10 eV.

Explore the model

Interactive lab

Interactive lab · probability

Normalized weights without an energy spectrum

Reset and set BOTH Success probability p = 0.30 and Trials n = 1; predict outcome weights 0.70 and 0.30 with total one. Reset and set p = 0.70 and n = 1; predict exchanged weights. These exact binomial probabilities illustrate weight bookkeeping only. They do not encode spectral energy positions, addition/removal matrix elements or an interacting Green function.

Normalized weights without an energy spectrumThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNormalization constrains a distribution without determining where its weight lies.

Check your understanding

For one normalized canonical fermionic spin-orbital, what fixes the integrated retarded spectral weight?

Try it yourself

A normalized Lorentzian is A(E)=Γ/[π(E2+Γ2)]A(E)=\Gamma/[\pi(E^2+\Gamma^2)] with positive energy width Γ. Find the fraction of its total area visible in −Γ≤E≤Γ-\Gamma\le E\le\Gamma. What would happen if that window alone were renormalized to unit area?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use arctan(E/Γ)/π as an antiderivative and compare the finite-window result with the full-axis area.

Compare with the solution

The window integral is [arctan⁡(1)−arctan⁡(−1)]/π=1/2[\arctan(1)-\arctan(-1)]/\pi=1/2. Half the normalized spectral weight lies outside that chosen window. Rescaling the window to unit area would double its apparent intensity and would no longer represent the original canonical spectral function on the full axis. A smaller visible area can therefore arise from finite coverage rather than lost particles or broken anticommutation. The width controls how the weight is distributed, while the exact infinite-window normalization remains one.

Take it with you

  • Canonical anticommutation fixes total one-orbital spectral area.
  • Occupied spectral area and full spectral area are different sum rules.
  • Finite windows, resolution and matrix elements complicate an experimental normalization test.

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