frame. Referring to the fact that in the special frame the momentum-containing invariants become with the two-dimensional vector k$_{⊥}$ being the momentum projection onto the plane orthogo- nal to the common direction 3 of the electric and magnetic field we shall use the equivalence relations throughout the paper. Polarization operator Π$_{µν}$( x, y ) is responsible for small perturbations above the constant- field background. It follows from the translation- Lorentz-, gauge-, PT- and charge- invariance [4, 5] that its Fourier transform can be presented in a diagonal form where ♭ ( a ) τ are its eigenvectors while the eigenvalues κ a are scalar functions of F , G , kF $^{2}$k and k ˜ F $^{2}$k. In the special case, where the second field invariant disappears, G = 0 , the three mean- ingful eigenvectors ♭ (1 , 2 , 3) µ are known [4, 5, 14] in the universal final form: 9 k ˜ F $^{2}$k = B $^{2}$( k 2 3 − k 2 $_{0}$) − E $^{2}$k2 $_{⊥}$, kF $^{2}$k = − B $^{2}$k2 ⊥ + E $^{2}$( k 2 3 − k 2 $_{0}$) , (3) k $^{2}$B2 + kF $^{2}$k B 2 + E 2 ⇔ k 2 3 − k 2 $_{0}$, k $^{2}$E2 − kF $^{2}$k B 2 + E 2 ⇔ k 2 ⊥ (4) Π$_{µτ}$( k, p ) = δ ( k − p )Π$_{µτ}$( k ) , Π$_{µτ}$( k ) = 3 ∑ a =1 κ a ♭ ( a ) µ ♭ ( a ) τ ( ♭ ( a $^{)}$)2 , (5) Π$_{µτ}$ ♭ ( a ) τ = κ a ♭ ( a ) µ , a = 1 , 2 , 3 , 4 , (6) The fourth eigenvector is trivial, ♭ (4) µ = k$_{µ}$ , so the fourth eigenvalue vanishes, κ 4 = 0 , as a consequence of the 4-transverseness of the polarization operator, Π$_{µτ}$ k$_{τ}$ = 0. All eigenvectors are mutually orthogonal, ♭ ( a ) µ ♭ ( b ) µ ∼ δ$_{ab}$ , this means that the first three ones are 4-transversal, ♭ ( a ) µ k$_{µ}$ = 0.