longer diagonalize the polarization operator already because the vectors ( ˜ Fk )$_{µ}$ and ( Fk )$_{µ}$ stop being mutually orthogonal, since their scalar product − k ˜ F Fk = G k 2 is now nonzero. When G = 0, the first eigenvector is expressed in terms of the fields by the same formula as in (7): and the first eigenvalue is given by the formula where the scalar function of the fields and momentum Λ$_{1}$ here, as well as other Λ’s below, is a linear superposition of the polarization tensor components Π$_{µν}$. The other two eigenvectors are the linear combinations (where the square root is understood algebraically: √$_{Z}$ 2 = Z , and not | Z | ) of two orthonor- malized vectors : thereby of the former basic vectors ( ˜ Fk )$_{µ}$ and ( Fk )$_{µ}$ , too. The corresponding two eigenvalues are The scalar coefficients in the linear combination (10) cannot be expressed in a universal way in terms of the field and momentum, but are irrational functions of the polarization tensor components. The reason is that the polarization operator is a linear combination of four independent matrices with four scalar coefficients, whereas there may be only three eigenvalues in accordance with three polarization degrees of freedom of a vector field. (When 10 κ 2 , 3 = 1 2 [ − (Λ$_{2}$ + Λ$_{4}$) ± √ (Λ$_{2}$ − Λ$_{4}$) 2 + 4Λ 2 3 ] . (12) c − µ = B ( Fk )$_{µ}$ + E ( ˜ Fk )$_{µ}$ ( B 2 + E $^{2}$)1 / $^{2}$( k $^{2}$E2 − kF $^{2}$k ) 1 / 2 ⇔ B ( Fk )$_{µ}$ + E ( ˜ Fk )$_{µ}$ ( B 2 + E $^{2}$) | k$_{⊥}$ | , c + µ = i E ( Fk )$_{µ}$ − B ( ˜ Fk )$_{µ}$ ( B 2 + $^{2}$)1 / $^{2}$( k $^{2}$B2 + kF $^{2}$k ) 1 / 2 ⇔ E ( Fk )$_{µ}$ − B ( ˜ Fk )$_{µ}$ ( B 2 + E $^{2}$)( k 2 0 − k 2 $_{3}$) 1 / $_{2}$, ( c $^{+}$c $^{−}$) = ( c $^{±}$♭ $^{(1)}$) = ( c $^{±}$k ) = 0 , ( c $^{±}$) 2 = 1 , (11) ♭ (2 , 3) µ = − 2Λ$_{3}$ c − µ + [ Λ$_{2}$ − Λ$_{4}$ ± √ (Λ$_{2}$ − Λ$_{4}$) 2 + 4Λ 2 3 ] c + µ (10) κ 1 = k $^{2}$( B 2 + E $^{2}$) k $^{2}$B2 + kF $^{2}$k Λ$_{1}$ ⇔ k 2 k 2 3 − k 2 0 Λ$_{1}$ , (9) ♭ (1) µ = ( F $^{2}$k )$_{µ}$ k 2 − k$_{µ}$ ( kF $^{2}$k ) , ( ♭ (1) ˜ Fk ) = ( ♭ $^{(1)}$Fk ) = 0 , ( ♭ $^{(1)}$)2 = k $^{2}$( k $^{2}$E2 − kF $^{2}$k )( k $^{2}$B2 + kF $^{2}$k ) ⇔ k $^{2}$( B 2 + E $^{2}$)$^{2}$k2 $_{⊥}$( k 2 3 − k 2 $_{0}$) (8)