Sum Rules and Completeness Tricks

Perturbation theory often produces sums over intermediate states. Sum rules and completeness tricks simplify those sums, check calculations, and sometimes avoid explicit summation altogether. The general many-body hierarchy of spectral moments, nested commutators, high-energy response, and density $f$-sums is developed on Sum Rules; here the goal is to simplify and test stationary perturbation sums.

The central idea is that a complete set of exact or unperturbed states resolves the identity:

$$\sum_m \lvert m\rangle\langle m\rvert=I,$$

with integrals over continuum states included when needed.

Required background. Use spectral decomposition to resolve the identity and commutators to evaluate weighted sums.

Completeness and spectral strength

If $\{\lvert m\rangle\}$ is a complete orthonormal basis, then for a state in the domain of $A$,

$$\sum_m |\langle m|A|n\rangle|^2 = \langle n|A^\dagger A|n\rangle.$$

If the state $\lvert n\rangle$ should be excluded from the sum, then

$$\sum_{m\ne n} |\langle m|A|n\rangle|^2 = \langle n|A^\dagger A|n\rangle - |\langle n|A|n\rangle|^2.$$

This identity is exact when the basis is complete. In truncated numerical bases, it becomes a convergence test.

Second-Order Sums

A typical second-order energy shift is

$$E_n^{(2)} = \sum_{m\ne n} \frac{|V_{mn}|^2}{E_n^{(0)}-E_m^{(0)}}.$$

The derivation, signs, polarizability interpretation, and continuum form are given in Second-Order Energy Corrections. This page focuses on methods for reorganizing or checking the resulting spectral sums.

Completeness alone does not remove the energy denominator. Sum rules become useful when the numerator and denominator can be related by commutators, when the denominator is approximated by an average energy, or when an auxiliary equation is solved instead of summing explicitly.

Closure Approximation

If a denominator varies slowly over the important intermediate states, one may replace it by an effective average $\Delta$:

$$\frac{1}{E_n-E_m} \approx \frac{1}{\Delta}.$$

Then

$$\sum_{m\ne n} \frac{|\langle m|A|n\rangle|^2}{E_n-E_m} \approx \frac{1}{\Delta} \left( \langle n|A^\dagger A|n\rangle - |\langle n|A|n\rangle|^2 \right).$$

This is an approximation, not an identity. Here $\Delta$ is a signed energy denominator, negative for a ground-state target. For nonnegative strengths $w_m=|A_{mn}|^2$, write $E_n-E_m=\Delta+\delta_m$. If all contributing terms satisfy $|\delta_m|\le\epsilon|\Delta|$ with $\epsilon<1$, then

$$\left|\sum_m\frac{w_m}{\Delta+\delta_m}-\frac{\sum_m w_m}{\Delta}\right| \le\frac{\epsilon}{1-\epsilon}\frac{\sum_mw_m}{|\Delta|}.$$

The estimate follows from subtracting the two reciprocal denominators. It gives a concrete condition for closure; a spectral tail outside this interval must be bounded separately.

Commutator Sum Rules

Let $H\lvert n\rangle=E_n\lvert n\rangle$ and let $A$ be Hermitian. Then

$$\frac12 \langle n|[A,[H,A]]|n\rangle = \sum_m (E_m-E_n) |\langle m|A|n\rangle|^2.$$

This identity is exact under the usual domain and completeness assumptions. It turns an energy-weighted sum over states into an expectation value of a double commutator.

For a one-dimensional Hamiltonian

$$H=\frac{P^2}{2m}+V(X),$$

using $A=X$ gives

$$[X,[H,X]] = \frac{\hbar^2}{m}.$$

Therefore

$$\sum_m (E_m-E_n) |\langle m|X|n\rangle|^2 = \frac{\hbar^2}{2m}.$$

This is the basic energy-weighted position sum rule for one particle in one dimension.

Oscillator Strength Preview

In atomic physics, dipole matrix elements and excitation energies are often packaged into oscillator strengths. The Thomas–Reiche–Kuhn sum rule is the many-state statement that the total oscillator strength is fixed by commutation relations and particle number. Oscillator Strengths develops the spectroscopic definitions, degeneracy conventions, continuum budget, and relation to integrated absorption.

For perturbation calculations, the useful consequence is that not every sum over unknown excited states is unconstrained. Commutators can enforce exact global checks.

Dalgarno–Lewis inhomogeneous equation

Some second-order sums can be avoided by solving an inhomogeneous equation. Suppose one needs

$$\sum_{m\ne n} \frac{\langle n|A|m\rangle\langle m|A|n\rangle} {E_n-E_m}.$$

Assume $H_0|n\rangle=E_n|n\rangle$ with a simple isolated eigenvalue and Hermitian $A$. Define an auxiliary state $\lvert F_n\rangle$ by

$$(H_0-E_n) \lvert F_n\rangle = - (A-\langle A\rangle_n) \lvert n\rangle,$$

with $\langle n|F_n\rangle=0$. This orthogonality is essential: adding $c|n\rangle$ would change the extracted answer by $c\langle A\rangle_n$. Then the desired sum can be extracted from

$$\langle n|A|F_n\rangle.$$

This method is especially useful for polarizabilities, where explicit sums over infinitely many excited and continuum states are inconvenient.

Numerical Use

In a truncated basis, sum rules are powerful diagnostics. For example, compute

$$S_N = \sum_{m=0}^{N} (E_m-E_n) |\langle m|X|n\rangle|^2$$

and compare it with $\hbar^2/(2m)$. Failure to approach the sum rule can indicate that the basis is too small, the continuum is missing, or matrix elements are inconsistent.

No finite matrix pair obeys $[X,P]=i\hbar I$ exactly, since the trace of a commutator is zero. A spectral sum can agree perfectly with a double commutator computed from the same truncated matrices while both violate the continuum sum rule. Compare against the independently derived $\hbar^2/(2m)$ and examine low states as the basis grows.

The same warning applies to perturbative sums: a partial sum may look converged for low states but still violate an exact sum rule sensitive to high-energy tails.

Exercises

1. Prove the simple completeness identity

$$\sum_m |\langle m|A|n\rangle|^2 = \langle n|A^\dagger A|n\rangle.$$

Solution

Insert the identity between $A^\dagger$ and $A$:

$$\langle n|A^\dagger A|n\rangle = \langle n|A^\dagger \left( \sum_m\lvert m\rangle\langle m\rvert \right) A|n\rangle.$$

This becomes

$$\sum_m \langle n|A^\dagger|m\rangle \langle m|A|n\rangle = \sum_m |\langle m|A|n\rangle|^2.$$

2. For $H=P^2/(2m)+V(X)$, show that $[X,[H,X]]=\hbar^2/m$.

Solution

Since $X$ commutes with $V(X)$,

$$[H,X] = \frac{1}{2m}[P^2,X].$$

Using $[P,X]=-i\hbar$,

$$[P^2,X] = P[P,X]+[P,X]P = -2i\hbar P.$$

Thus

$$[H,X] = - \frac{i\hbar}{m}P.$$

Then

$$[X,[H,X]] = - \frac{i\hbar}{m}[X,P] = \frac{\hbar^2}{m}.$$

References

• H. A. Bethe and E. E. Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, Springer, 1957.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• A. Dalgarno and J. T. Lewis, “The exact calculation of long-range forces between atoms by perturbation theory,” Proceedings of the Royal Society A 233, 70–74, 1955.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• W. Thomas, “Über die Zahl der Dispersionselektronen, die einem stationären Zustande zugeordnet sind,” Naturwissenschaften 13, 627, 1925.