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**Example text**

130) 1 Lint = − ∑ Q a 0 a ˙ [r A + s(ra − r A )] ds (ra − r A )A − [s(ra − r A ) × r˙ a ][∇ × A[r A + s(ra − r A )]] . 136) It is worth noting that the transverse electromagnetic ﬁeld appears in the new ˙ In parLagrangian in terms of the ﬁeld strengths B = ∇ × A and E⊥ = −A. ticular, the charges interact with the transverse ﬁeld via the coupling of the magnetization to the induction ﬁeld and the coupling of the polarization to the transverse electric ﬁeld, d3 r [P A (r)E⊥ (r) + M A (r)B(r)] Lint = = ∑ Qa a 1 0 ds (ra − r A )E⊥ [r A + s(ra − r A )] + [s(ra − r A ) × r˙ a ]B[r A + s(ra − r A )] .

170) for ρ = j = 0. Let us again restrict our attention to 23) Note that with respect to the integration measure ε (r)d3 r in the deﬁnition of scalar products, the differential operator ε−1 (r)∇ × ∇× is Hermitian. 24) Note that the photons deﬁned in this way do not only refer to the transverse part of the electromagnetic ﬁeld. 4 Dielectric background media nonmagnetic matter, so that Eq. 171) holds. 203) which obey the continuity equation ∇jN (r, ω ) = iωρN (r, ω ). 207) together with the boundary condition at inﬁnity, G (r, r , ω ) → 0 if |r − r | → 0 (for the properties of the Green tensor, see Appendix A).

83), with increasing V the modes become more and more dense in the k domain. 90) we see that in the limit as V → ∞ (i. 91) ˆ (r) accordingly. The commutation relations for the operators aˆ σ (k) and Π ˆ and a†σ (k) may be found from the original commutation relations as given 10) Nevertheless, the set of modes obtained from the boundary conditions that are realized by perfectly reﬂecting walls can be used to describe the electromagnetic ﬁeld inside the cavity correctly. Note that the number of modes which contribute to the ﬁeld can drastically increase with decreasing distance of the point of observation from the cavity walls, because of the wrong boundary conditions.