12 Putting everything together, the equation of motion takes the form This can be further decoupled into longitudinal and transverse components denoted by ˜ u$_{L}$ and ˜ u$_{T}$ respectively. The corresponding eigenvalues are and Of course, this is exactly what one would have obtained by assuming that the spectrum of the current-current response function consist of a single δ -function peak at ω$_{L}$ or ω$_{T}$ and requiring satisfaction of the first moment sum rule (see Ref. 35, Eq. 3.191 and Ref. 39). In Fig. (1) we plot the excitation spectrum of the ho- mogeneous electron gas calculated from Eqs. (94) and (95). We have used the static structure factor calculated in Ref. 36 by the quantum Monte Carlo method, and the kinetic energy has been computed from the parametrized correlation energy of Appendix B of Ref. 36, using the virial theorem. In the longitudinal channel, the exact spectrum is dominated at small q by the plasmon and at large q by free particle excitations (energy q $^{2}$/ 2 m ). There are also also electron-hole pair excitations at lower q and ω , as well as multiple electron-hole pair excitations all over the plane. The elastic approximation replaces this complex spectrum by a single branch of longitudinal excitations which has the correct spectral moment. In particular, we get the correct dispersion of the plasmon at small q and the correct free particle behavior at large q . In the transverse channel the plasmon and the high- q free particle excitations are absent. The exact transverse spectrum consists primarily of low-energy electron-hole pair excitations. The current vector, k + q / 2, where k and k + q are the momenta of the hole and the elec- tron, respectively, is essentially perpendicular to q when k and k + q both lie near the Fermi surface. On the other hand, high-energy excitations, with energy q $^{2}$/ 2 m are essentially longitudinal because the current vector is essentially parallel to q when q is much larger than the Fermi momentum. This is consistent with the fact that the frequency of our transverse collective mode ω$_{T}$ ( q ) grows linearly with q at large q . where the primes now denote derivatives with respect to r . Solving the eigenvalue problem with boundary con- dition n 1 / 2 0 ( r ) u$_{r}$ ( r ) → 0 for r → ∞ yields the correct A similar calculation can be done for hydrogen-like atoms of atomic number Z . Focusing for simplicity on excitations of spherical symmetry, we introduce a radial displacement field u$_{r}$ ( r ) which depends only on the radial coordinate r . Then u$_{r}$ satisfies the equation which are mutually orthogonal with respect to the scalar product (67). These are indeed proportional to the ma- trix elements of the current density operator between the ground-state and the n-th excited state. Solving the eigenvalue problem with the boundary con- dition of n 1 / 2 0 ( x ) u ( x ) → 0 as | x | → ∞ , we obtain the exact excitation spectra ω$_{n}$ = ± n Ω, where n = 1 , 2 , .. . The corresponding eigenfunctions are For a harmonic oscillator of natural frequency ω$_{0}$ , ex- ternal potential V$_{0}$ ( x ) = mω 2 $_{0}$x$^{2}$/ 2, and equilibrium den- sity n$_{0}$ ( x ) = e − x 2 /ℓ 2 √$_{πℓ}$ , where ℓ ≡ ( mω$_{0}$ ) − 1 / $^{2}$, the equation of motion (61) reduces to In order to demonstrate the exactness of the formu- lation for one-electron systems let us now consider the canonical examples of the one-dimensional harmonic os- cillator and the hydrogen atom. B. Linear harmonic oscillator and hydrogen atom mω $^{2}$˜ u = 2 3 t ( n ) [ 2 q ( q · ˜ u ) + q $^{2}$˜ u ] + q 2 4 m q ( q · ˜ u ) + mω 2 $_{p}$(ˆ q · ˜ u )ˆ q + mω 2 p n ∫ d q ′ (2 π ) $_{3}$[ S ( q − q ′ ) − S ( q ′ )]ˆ q ′ [ˆ q ′ · ˜ u ( q )] , (93) ω 2 $_{L}$( q ) = ω 2 p + 2 t ( n ) q 2 m + q 4 4 m 2 + ω 2 p n ∫ d q ′ (2 π ) 3 (ˆ q · ˆ q ′ ) $^{2}$[ S ( q − q ′ ) − S ( q ′ )] , (94) ω 2 $_{T}$( q ) = 2 3 t ( n ) q 2 m + ω 2 p 2 n ∫ d q ′ (2 π ) 3 (ˆ q × ˆ q ′ ) $^{2}$[ S ( q − q ′ ) − S ( q ′ )] . (95) 1 4 u ′′′′ − xu ′′′ + ( x 2 − 2) u ′′ + 3 xu ′ − ω 2 − ω 2 0 ω 2 0 u = 0 . (96) u$_{n}$ ( x ) ∝ $^{H$_{n}$}$− $_{1}$( x ) (97) 1 4 u ′′′′ r − ( 1 − 1 r ) u ′′′ r + ( 1 − 2 r − 1 r 2 ) u ′′ r + 3 r $_{2}$u ′ r − ( 2 r 3 + ω 2 Z 4 ) u$_{r}$ = 0 , (98)