Microscopic electrodynamics Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 1 / 46
Maxwell s equations for electric field E and magnetic field B in terms of sources ρ and j The homogeneous pair: B = 0 E + t B = 0; t B = B t The inhomogeneous pair (sources): Electric and magnetic constants: E = ρ/ε 0 B 1 c 2 te = µ 0 j µ 0 ε 0 = 1 c 2 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 1 / 46
Vector algebra Basic operations: s (gradient) v = 0 (divergence); v = 0 (curl) Basic results: Divergence theorem: ˆ Curl theorem: ( v) = 0; ( s) = 0 ˆ S V ˆ ( v) dv = S ( v) nds = (v n) ds C (v t) dl Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 2 / 46
Continuity equation B 1 c 2 te = µ 0 j ( B) 1 }{{} c 2 t ( E) = µ }{{} 0 ( j) =0! ρ/ε 0 On integral form ˆ ˆ ( j) dv = V j + t ρ = 0 S ˆ (j n) ds = t ρdv = Q encl V t (charge conservation) Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 3 / 46
Boundary conditions Are the electric field E and the magnetic field B uniquely determined by their divergence (...) and curl (...)? The answer is NO!!!! The two vectors F 1 = (0, 0, 0) F 2 = (yz, zx, xy) both have zero divergence and curl. Boundary conditions must be introduced: E and B go to zero at infinity Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 4 / 46
Maxwell s equations: solutions Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 5 / 46
Maxwell s equations: The stationary case The homogeneous pair: The inhomogeneous pair: Implies steady currents: A useful formula (exercise): B = 0 E = 0 E = ρ/ε 0 B = µ 0 j j = 0 = t ρ ( F) = ( F) 2 F Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 5 / 46
Maxwell s equations: The stationary case Electrostatics 2 E = ( E) ( E) = ρ/ε }{{}}{{} 0 =0 ρ/ε 0 Each component of the electric field fulfills the Poisson equation: 2 Ψ (r 1, t) = f (r 1, t) Siméon Denis Poisson (1781-1840) with solutions Ψ (r 1, t) = 1 4π ˆ f (r2, t) r 12 d 3 r 2 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 6 / 46
Maxwell s equations: The stationary case Electrostatics E(r 1 ) = 1 ˆ 2 ρ (r 2 ) d 3 r 2 = 1 4πε 0 r 12 4πε 0 Coulomb s law: ˆ r12 ρ (r 2 ) d 3 r 2 r 3 12 Charles-Augustin de Coulomb (1736-1806) F(r 1) = q 1E(r 1) = ˆ q 1 (r1 r ) ρ (r ) dτ 4πε 0 r 1 r 3 Point charge: ρ(r ) = q 2δ ( r r 2 ) Scalar form: F(r 1) = q1q2r12 4πε 0r 3 12 r 12 = r 12n 12 F = q1q2 4πε 0r 2 12 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 7 / 46
Maxwell s equations: The stationary case Magnetostatics 2 B = ( B) ( B) = µ }{{}}{{} 0 j 4π c 2 j =0 Jean-Baptiste Biot Biot-Savart law: B(r 1 ) = µ 0 4π ˆ r12 j(r 2 ) d 3 r 2 r 3 12 It would be tempting to insert the expression for a moving point charge j ( r ) = q 2v 2δ ( r r 2(t) ) but this is wrong, since a moving charge is not a steady current. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 8 / 46
Maxwell s equations: the general case Electric field 1) B = 0; 3) E = ρ/ε 0 2) E + t B = 0; 4) B 1 c 2 te = µ 0 j We take the curl on both sides of 2) and use 4) ( E) + t ( B) = ( E) + 1 c 2 2 E t 2 + µ 0 t j = 0 Next, we use the vector relation to obtain ( E) = ( E) 2 E }{{} 4πρ 2 E = 4π ( ρ/ε 0 + µ 0 t j) ; 2 = 2 1 c 2 2 t 2 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 9 / 46
Maxwell s equations: the general case Magnetic field 1) B = 0; 3) E = ρ/ε 0 2) E + t B = 0; 4) B 1 c 2 te = µ 0 j We take the curl on both sides of 4) and use 2) ( B) 1 c 2 t ( E) = ( B)+ 1 2 B c 2 t 2 = µ 0 ( j) Next, we use the vector relation to obtain ( B) = ( B) 2 B }{{} =0 2 B = µ 0 ( j) ; 2 = 2 1 c 2 2 t 2 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 10 / 46
Electromagnetic waves Maxwell s equations with no sources The equations for the fields reduce to 2 E = 0; 2 B = 0 Each component of the fields is given by the homogeneous wave equation [ 2 Ψ (r, t) = 2 1 2 ] c 2 t 2 Ψ (r, t) = 0 with solutions given in terms of amplitude Ψ 0, wave vector k, angular frequency ω and phase constant δ Ψ (r, t) = Ψ 0 exp [i (k r ωt + δ)] Inserting the solution into the wave equation gives ) 2 Ψ (r, t) = ( k 2 + ω2 c 2 Ψ (r, t) = 0 : k = ± ω c Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 11 / 46
Electromagnetic plane waves Real part of Ψ (r, t) = Ψ 0 exp i (k r ωt + δ) }{{} phase The wave has its maximum value when the phase is zero (or a multiple of 2π). We split the position vector r in two parts: r = r + r, where r and r are parallel and perpendicular to the wave vector k, respectively. Fixing r fixes the value of k r since k r = 0: k r = k r = kr The perpendicular component r can be chosen freely, which means that a fixed value of k r defines a plane perpendicular to k. At time t = 0 and phase zero such a plane is situated at r = δ/k. At later time t the plane is situated at r = ωt k δ k = ±ct δ/k The wave function Ψ accordingly describes a plane wave propagating at the speed of light c. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 12 / 46
Electromagnetic plane waves The electric and magnetic components of the electromagnetic wave are written as E (r, t) = E 0 exp [i (k r ωt + δ)] ; B (r, t) = B 0 exp [i (k r ωt + δ)] where it is to be understood that the physical fields are the real parts of E and B. From Maxwell s equations it follows that the electric and magnetic fields are purely transversal and mutually orthogonal k E = 0; k B = 0; k E = ωb; E 0 = cb 0 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 13 / 46
Linear polarization Consider an electromagnetic wave travelling in the z-direction: k = ke z The most general form of the electric field is then ) E (r, t) = (E 0xe iδx e x + E 0y e iδy e y exp [i (kz ωt)] The phase constants δ x and δ y may be incorporated into complex amplitudes Ẽ 0x = E 0xe iδx ; Ẽ 0y = E 0y e iδy Identical phases δ x = δ y = δ represents a linearly polarized wave E (r, t) = Ẽ0 (cos θex + sin θey ) exp [i (kz ωt)] with complex amplitude Ẽ0 = e iδ E 2 0x + E 2 0y an angle θ = tan 1 ( E0y E 0x ) with the x axis. and making Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 14 / 46
Elliptical and circular polarization Different phase constants δ x and δ y leads to elliptically polarized waves. Circulary polarized waves (δ y = δ x ± π 2 ): E (r, t) = E 0 e iδx (e x ± ie y ) exp [i (kz ωt)] Setting δ x = 0 we see that the real parts of the electric field (the actual field) are E (r, t) = (E 0 cos (kz ωt), E 0 sin (kz ωt), 0) e ± = e x ± ie y : left/right circularly polarized light; At a fixed point in space E rotates counter-clockwise (L) or clockwise (R) for an observer facing the oncoming wave. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 15 / 46
Maxwell s equations: general solution E(r 1 ) = 1 ˆ { r12 ρ (r 2, t r ) 4π ε 0 r12 3 B(r 1 ) = µ ˆ { 0 r12 j (r 2, t r ) 4π r 3 12 + tρ (r 2, t r ) r 12 ε 0 r12 2 µ 0 t j (r 2, t r ) c 2 r 12 + r } 12 t j (r 2, t r ) d 3 r 2 cr ²12 } d 3 r 2 Note that we can always add the solutions of the homogeneous equations, that is the electromagnetic waves. Jefimenko s equations provide the general solution to Maxwell s equations, but typically lead to very complicated integrals. It is usually easier to work with electromagnetic potentials, rather than fields. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 16 / 46
Electromagnetic potentials Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 17 / 46
Helmholtz decomposition 2 F = ( F) + ( F) }{{}}{{} scalar vector Any vector function F (differentiable) who goes to zero faster than 1 r when r can be expressed as the sum of the gradient of a scalar and the curl of a vector F(r) = s(r) + v(r) The two parts of the vector function may be characterized according to the behaviour of the corresponding functions in reciprocal space ˆ F(r) = F(k)e ik r dk Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 17 / 46
Helmholtz decomposition Longitudinal component ( parallel ): F = s(r); F = 0 In reciprocal space: k F (k) = 0 Transversal component ( perpendicular ): F = v(r); F = 0 In reciprocal space: k F (k) = 0 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 18 / 46
Potentials Maxwell s equations: homogeneous pair B = 0 implies B = B = A(r) and B = 0 E + B t = 0 then becomes (E + ta) = 0 and one may write E + t A = φ(r) Longitudinal component: E = φ t A Transversal component: E = t A With the introduction of the scalar potential φ and the vector potential A, the homogeneous pair of Maxwell s equations is automatically satisfied. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 19 / 46
Potentials Maxwell s equations: inhomogeneous pair B 1 c 2 te = µ 0 j becomes [ 2 1 2 ] [ c 2 t 2 A ( A) + 1 ] φ c 2 = µ 0 j t E = ρ/ε 0 becomes 2 φ + t ( A) = ρ/ε 0 or [ 2 1 2 ] [ c 2 t 2 φ + t ( A) + 1 ] φ c 2 = ρ/ε 0 t Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 20 / 46
Special relativity There was an old lady called Wright who could travel much faster than light. She departed one day in a relative way and returned on the previous night. The principle of relativity: The invariance of physical laws in the universe. The invariance of the speed c of light in all inertial frames. The importance of relativistic effects is provided by the Lorentz factor: 1 γ = 1 v 2 c 2 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 21 / 46
Relativity and 4-vectors A simple way to see if a physical law is relativistic or not is to see if it can be expressed in terms of 4-vectors: 4-position: r µ = (x, y, z, ict) = (r, ict); r µ r µ = r 2 c 2 t 2 (the interval) 4-velocity: v µ = γ(v, ic); v µ v µ = c 2 4-gradient: ( µ =, i c t 4-potential: A µ = (A, ic ) φ 4-current: j µ = (j, icρ) Continuity equation: ( µ j µ =, i ) c t }{{} 4-gradient ) ; µ µ = 2 1 c 2 2 (j, icρ) }{{} 4-current t 2 = 2 (d Alembertian) = j + ρ t = 0 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 22 / 46
Maxwell s equations 4-vector notation We start from [ 2 1c 2 2 t 2 [ 2 1c 2 2 t 2 ] [ A ( A) + 1 φ c 2 t ] φ + t [ ( A) + 1 c 2 φ t This can be written more compactly as and finally compacted into 2 A ( νa ν) = j/ε 0c 2 2 φ + t ( νa ν) = ρ/ε 0 2 A µ µ ( νa ν) = 1 ε 0c 2 jµ ] ] = µ 0j = ρ/ε 0...demonstrating the relativistic nature of Maxwell s equations Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 23 / 46
Gauge freedom B = A implies that the longitudinal component A of the vector potential can be modified without changing B, that is A A = A + χ However E = φ A implies that a modification of A requires a t corresponding modification of the scalar potential φ φ = φ t χ The electric and magnetic fields are gauge invariant. The gauge transformations can be written with 4-vectors: A µ = (A, ic ) A + χ φ A µ = A µ + µ χ = φ t χ Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 24 / 46
Lorentz gauge α A α = (, i ) c t (A, ic ) φ = A + 1 c 2 tφ = 0 Maxwell s equations simplifies to: with solutions: φ(r 1, t) = 1 4πε 0 where apprears retarded time 2 A µ = 1 ε 0 c 2 j µ ˆ ρ(r2, t r ) r 1 r 2 d 3 r 2 ; A(r 1, t) = µ ˆ 0 j(r2, t r ) 4π r 1 r 2 d 3 r 2 t r = t r 12 c Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 25 / 46
Coulomb gauge Maxwell s equations reduces to: A = 0 with solutions: φ(r 1, t) = 1 4πε 0 2 φ = ρ/ε 0 2 A 1 c 2 tφ = µ 0 j ˆ ρ(r2, t) r 1 r 2 d 3 r 2 ; A (r 1, t) = µ ˆ 0 j (r 2, t r ) 4π r 1 r 2 d 3 r 2 The scalar potential describes the instantaneous Coulomb interaction,.....the complexities of relativistic interactions are hidden in the vector potential. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 26 / 46
Particles and fields Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 27 / 46
Particles and fields Complete Hamiltonian Fields specified: H = H p(articles) + H int(eraction) + H f(ields) (iγ µ µ mc) ψ = 0 Dirac equation Particles (sources) specified: Non-relativistic limit ( p 2 2m i ) ψ = 0 t Schrödinger equation Non-relativistic limit 2 A µ µ ( ν A ν ) = 1 ε 0 c 2 j µ??? Maxwell s equations Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 27 / 46
The non-relativistic limit of electrodynamics B = 0 B = 0 E + t B = 0 c E = 0 E = ρ/ε 0 E = ρ/ε 0 B 1 c 2 te = 1 ε 0 c 2 j B = 0 In the strict non-relativistic limit there are no magnetic fields and no effects of retardation! The Coulomb gauge bears its name because it singles out the instantaneous Coulomb interaction, which constitutes the proper non-relativistic limit of electrodynamics and which is the most important interaction in chemistry. All retardation effects as well as magnetic interactions are to be considered corrections of a perturbation series of the total interaction (in 1/c 2 ). Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 28 / 46
Maxwell s equations Lagrangian formalism The equation of motion for the electromagnetic field is given by Maxwell s equations ν F νµ = 2 A µ µ ( ν A ν ) = 1 ε 0 c 2 j µ; F νµ = ν A µ µ A µ which can be generated from the Lagrangian density L (A α, β A α ) = L int + L field = j α A α 1 4 ε 0c 2 F αβ F αβ by the Euler-Lagrance equations for continuous systems ( ) L L β = 0 A α β A α Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 29 / 46
Equations for particles or fields Complete Lagrangian: L = L p(articles) + L int(eraction) + L f(ields) Interaction term: ˆ ˆ [j L int = j µ A µ d 3 r ( = r, t ) A ( r, t ) ρ ( r, t ) φ ( r, t )] d 3 r Fixing the particles (as sources), the term L p drop out of the equations of motion, and we obtain Maxwell s equations. Let us now consider the equations of motion obtained for a single point charge in fixed fields Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 30 / 46
Point charge in fixed fields Lagrangian Point charge with trajectory r(t): ρ(r ) = qδ (r r (t)) ; j(r ) = qv δ (r r (t)) Interaction term: L int = ˆ [j (r, t) A (r, t) ρ (r, t) φ (r, t)] d 3 r = qv(t) A(r, t) qφ(r, t) Euler-Lagrange equation F = d ( ) Lp = L int dt v r + d dt ( Lint v ) = L int + q da dt = qe + q (v B) (Lorentz force) Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 31 / 46
Point charge in fixed fields Hamiltonian Momentum p = L v = π + qa; Hamiltonian π = L p v (mechanical momentum) H (r, p, t) = p v L (r, v, t) = (π v L p ) + qφ(r, t) = H p (r, π, t) + qφ(r, t) Principle of minimal electromagnetic coupling p µ p µ qa µ This relativistic coupling of particles and fields is also used in the non-relativistic domain. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 32 / 46
Non-relativistic Hamiltonian in external fields We may start from Minimal substitution ĥ p = ˆp2 2m p π = p + ea; E E + eφ then gives h = π2 p2 eφ = 2m 2m + e e2 {A p + p A} + 2m 2m A2 = p2 2m + e m A p + e2 2m A2 eφ;..but we miss all spin interaction. (Coulomb gauge) Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 33 / 46
Non-relativistic Hamiltonian in external fields Alternatively we use the Dirac identity (σ A) (σ B) = A B + iσ (A B) to write ĥ p = Minimal substitution now gives (σ p)2 2m ĥ = (σ π)2 2m Further manipulation gives π2 iσ eφ = eφ + (π π) 2m 2m ĥ = p2 2m + e m A p + e2 2m A2 eφ + e 2m σ B Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 34 / 46
Energy flow Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 35 / 46
The Lorentz force and work Lorentz force: force on the charge Work on the charge F = qe + q(v B) W = ˆ r(tb ) r(t a) F dr = Insertion of the Lorentz force gives W = ˆ tb t a (qe v) dt + ˆ tb Magnetic forces do no work! ˆ tb t a q(v B) vdt = t a (F v) dt; v = dr dt ˆ tb t a (E qv) dt + ˆ tb q (v v) Bdt t a }{{} The substitution p π = p + ea corresponds to the removal of the magnetic contribution from the former. =0 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 35 / 46
Energy flow: the Poynting vector Replacing a point charge by a charge distribution, the work is given by ˆ tb ˆ W = (E (r) j (r)) d 3 rdt Instantanous power: rate of mechanical work ˆ dw = (E (r) j (r)) d 3 r dt t a Using Mawell s equations and vector algebra we rearrange to dw dt + d {ˆ 1 dt V 2 ε ( 0 E 2 + c 2 B 2) } ˆ d 3 r = ε 0 c 2 (E B) nds S }{{}}{{} Poynting vector S Energy stored in electromagnetic field Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 36 / 46
Poynting vector and intensity For an electromagnetic wave the Poynting vector is given by S = ε 0 c 2 (E B) = ε 0c 2 ω (E (k E)) = 1 ω ε 0c 2 E 2 k = ε 0 ce 2 e k The corresponding intensity is the energy flux averaged over a period of the wave I = S T = 1 T ˆ T 2 T 2 ε 0 ce 2 dt = 1 2 ε 0cE 2 0...where we used that for θ = (k r ωt + δ) cos 2 θ + sin 2 θ = 1 cos 2 θ T + sin 2 θ T = 1 2 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 37 / 46
Multipolar gauge Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 38 / 46
Multipolar gauge The expectation value of the interaction Hamiltonian is given in terms of the electromagnetic potentials ˆ H int = [ρ(r)φ(r) j(r) A(r)] d 3 r Is it possible to express the interaction Hamiltonian directly in terms of electromagnetic fields? The answer is: Yes, using multipolar gauge. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 38 / 46
Multipolar gauge Taylor expansion of electromagnetic potentials Scalar potential: φ (r, t) = φ [0] + r i φ [1] i + 1 2 r ir j φ [2] ij +... where we use the Einstein summation convention and the notation X [n] n X j 1 j 2...j n = r j1 r j2... r jn Likewise, we have A i (r, t) = A [0] i + r j A [1] i;j + 1 2 r jr k A [2] i;j,k +... 0 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 39 / 46
Multipolar gauge Replacing potentials by fields Using the relation E = φ A t we obtain φ(r) = φ [0] r i E [0] i 1 2 r ir j E [1] i;j... A [0] i r i t 1 2 r ir j A [0] i;j t which can be written as a gauge transformation... with the gauge function χ = n=0 φ(r) = φ (r) χ t 1 ( ) (n + 1)! r j 1 r j2... r jn r A [n] j 1 j 2...j n Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 40 / 46
Multipolar gauge Final expressions We next carry out the gauge transformation A = A χ After some further manipulations we arrive at the final expressions φ(r, t) = φ [0] A(r, t) = n=0 n=0 1 ( ) (n + 1)! r j 1 r j2... r jn r E [n] j 1 j 2...j n n + 1 ( ) (n + 2)! r j 1 r j2... r jn r B [n] j 1 j 2...j n What happended to gauge freedom? Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 41 / 46
Multipolar gauge Examples φ(r, t) = φ [0] n=0 1 ( ) (n + 1)! r j 1 r j2... r jn r E [n] j 1j 2...j n A(r, t) = Uniform electric field: n=0 n + 1 ( ) (n + 2)! r j 1 r j2... r jn r B [n] j 1j 2...j n φ(r, t) = r E [0] ; A(r, t) = 0 Uniform magnetic field: φ(r, t) = 0; A(r, t) = 1 (r B [0]) 2 For a time-dependent uniform magnetic field we get a non-uniform electric field as well E(r, t) = φ t A = 1 ( ) r B [0] t 2 Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 42 / 46
Multipole expansions In multipolar gauge the expectation value of the interaction Hamiltonian takes the form ˆ H int = [ρ(r)φ(r) j(r) A(r)] d 3 r = Q [0] φ [0] n=1 1 n! Q[n] j 1j 2...j n 1 E [n 1] j 1...j n 1 n=1 where appears electric multipoles ˆ Q [n] j 1...j n = r j1 r j2... r jn ρ(r)d 3 r 1 n! m[n] j 1j 2...j n 1 B [n 1] j 1...j n 1 and magnetic multipoles m [n] j 1...j n 1 = n n + 1 ˆ r j1 r j2... r jn 1 [r j(r)] d 3 r Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 43 / 46
Electric and magnetic multipoles Examples Electric dipole: ˆ µ i = Q [1] i = r i ρ(r)d 3 r Electric quadrupole: Magnetic dipole: ˆ Q [2] ij = m [1] i = 1 ˆ 2 r i r j ρ(r)d 3 r (r j(r)) i d 3 r Many molecular properties are defined in terms of (induced) electric and magnetic multipoles. Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 44 / 46
Truncated multipole expansions Consider the expectation value of the interaction Hamiltonian H int = Q [0] φ [0] n=1 1 n! Q[n] j 1j 2...j n 1 E [n 1] j 1...j n 1 n=1 1 n! m[n] j 1j 2...j n 1 B [n 1] j 1...j n 1 For electromagnetic waves it is convenient to develop the interaction in orders of the wave vector k. The electric and magnetic fields are given by E (r, t) = Ẽ 0 e iωt exp [i (k r)] ; B (r, t) = 1 k E (r, t) ω such that E [n] j 1...j n = (i) n k j1... k jn Ẽ 0 e iωt O (k n ) ; B [n] j 1...j n O ( k n+1) Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 45 / 46
Truncated multipole expansions a hierarchy of approximations Electric dipole approximation: k 0 : Q [1] E [0] Electric-quadrupole magnetic-dipole approximation: k 1 : 1 2 Q[2] j E [1] j m [1] B [0] Trond Saue (LCPQ, Toulouse) Microscopic electrodynamics Virginia Tech 2015 46 / 46