The Particle-Field Hamiltonian

Transcription

1 The Particle-Field Hamiltonian For a fundamental understanding of the interaction of a particle with the electromagnetic field we need to know the total energy of the system consisting of particle and field. This energy remains conserved; the particle can borrow energy from the field (absorption) or it can donate energy to it (emission). The total energy corresponds to the classical Hamiltonian H, which constitutes the Hamiltonian operator Ĥ encountered in quantum mechanics. For particles consisting of many charges, the Hamiltonian soon becomes a very complex function: it depends on the mutual interaction between the charges, their kinetic energies and the exchange of energy with the external field. To understand the interaction between a particle and an electromagnetic field we first consider a single point-like particle with mass m and charge q. Later we generalize the situation to systems consisting of multiple charges and with finite size. The Hamiltonian for a single charge in an electromagnetic field is found by first deriving a Lagrangian function L(r, ṙ) which satisfies the Lagrange Euler equation ( ) d L L = 0, q = x, y, z. (1) dt q q Here, q = (x, y, z) and q = (ẋ, ẏ, ż) denote the coordinates of the charge and the velocities, respectively. 1 To determine L, we first consider the (non-relativistic) equation of motion for the charge F = d [mṙ] = q (E + ṙ B), () dt and replace E and B by the vector potential A and scalar potential φ according to E(r, t) = A(r, t) φ(r, t), t (3) B(r, t) = A(r, t). (4) Now we consider the vector-components of Eq. () separately. For the x-component we obtain [ d φ [mẋ] = q dt x + A ] [ ( x Ay + q ẏ t x A ) ( x Ax ż y z A )] z x = x [ qφ + q (A xẋ + A y ẏ + A z ż)] [ Ax q + ẋ A x t x + ẏ A y y + ż A ] z. (5) z Identifying the last expression in brackets with da x /dt (total differential) and rearranging terms, the equation above can be written as d dt [mẋ + qa x] x [ qφ + q (A xẋ + A y ẏ + A z ż)] = 0. (6) 1 It is a convention of the Hamiltonian formalism to designate the generalized coordinates by the symbol q. Here, it should not be confused with the charge q. 1

2 This equation has almost the form of the Lagrange Euler equation (1). Therefore, we seek a Lagrangian of the form L = qφ + q (A x ẋ + A y ẏ + A z ż) + f(x, ẋ), (7) with f/ x = 0. With this choice, the first term in Eq. (1) leads to ( ) d L = d [ qa x + f ]. (8) dt q dt ẋ This expression has to be identical with the first term in Eq. (6), which leads to f/ ẋ = mẋ. The solution f(x, ẋ) = mẋ / can be substituted into Eq. (7) and, after generalizing to all degrees of freedom, we finally obtain L = qφ + q (A x ẋ + A y ẏ + A z ż) + 1 m ( ẋ + ẏ + ż ), (9) which can be written as L = q φ + q v A + m v v. (10) To determine the Hamiltonian H we first calculate the canonical momentum p = (p x, p y, p z ) conjugate to the coordinate q = (x, y, z) according to p i = L/ q i. The canonical momentum turns out to be p = mv + qa, (11) which is the sum of mechanical momentum mv and field momentum qa. According to Hamiltonian mechanics, the Hamiltonian is derived from the Lagrangian according to H(q, p) = i [p i q i L(q, q)], (1) in which all the velocities q i have to be expressed in terms of the coordinates q i and conjugate momenta p i. This is easily done by using Eq. (11) as q i = p i /m qa i /m. Using this substitution in Eqs. (9) and (1) we finally obtain H = 1 m [p qa] + qφ. (13) This is the Hamiltonian of a free charge q with mass m in an external electromagnetic field. The first term renders the kinetic mechanical energy and the second term the potential energy of the charge. Notice that the derivation of L and H is independent of gauge, i.e. we did not imply any condition on A. Using Hamilton s canonical equations q i = H/ p i and ṗ i = H/ q i it is straightforward to show that the Hamiltonian in Eq. (13) reproduces the equations of motion stated in Eq. (). The Hamiltonian of Eq. (13) is not yet the total Hamiltonian H tot of the system charge + field since we did not include the energy of the electromagnetic field. Furthermore, if the charge is interacting with other charges, as in the case of an atom or a molecule, we must take into account the interaction between the charges. In general, the total Hamiltonian for a system of charges can be written as H tot = H particle + H rad + H int. (14) Careful, we re using the same symbol for the dipole moment and the canonical momentum!

3 Here, H rad is the Hamiltonian of the radiation field in the absence of the charges and H particle is the Hamiltonian of the system of charges (particle) in the absence of the electromagnetic field. The interaction between the two systems is described by the interaction Hamiltonian H int. Let us determine the individual contributions. The particle Hamiltonian H particle is determined by a sum of the kinetic energies p n p n /(m n ) of the N charges and the potential energies V (r m, r n ) between the charges (intramolecular potential), i.e. H particle = [ ] pn p n + V (r m, r n ), (15) m n,m n where the nth particle is specified by its charge q n, mass m n, and coordinate r n. Notice that V (r m, r n ) is determined in the absence of the external radiation field. This term is solely due to the Coulomb interaction between the charges. H rad is defined by integrating the electromagnetic energy density W of the radiation field over all space as 3 H rad = 1 [εo E + µ 1 o B ] dv, (16) where E = E and B = B. It should be noted that the inclusion of H rad is essential for a rigorous quantum electrodynamical treatment of light matter interactions. This term ensures that the system consisting of particles and fields is conservative; it permits the interchange of energy between the atomic states and the states of the radiation field. Spontaneous emission is a direct consequence of the inclusion of H rad and cannot be derived by semiclassical calculations in which H rad is not included. Finally, to determine H int we first consider each charge separately. Each charge contributes to H int a term that can be derived from Eq. (13) as H p p m = q q [p A + A p] + A A + qφ. (17) m m Here, we subtracted the kinetic energy of the charge from the classical particle field Hamiltonian since this term is already included in H particle. Using p A = A p and then summing the contributions of all N charges in the system we can write H int as 4 H int = n [ q n m n A(r n, t) p n + ] q n A(r n, t) A(r n, t) + q n φ(r n, t). m n In the next section we will show that H int can be expanded into a multipole series similar to our previous results for V E and V M. (18) Multipole expansion of the interaction Hamiltonian The Hamiltonian expressed in terms of the vector potential A and scalar potential φ is not unique. This is caused by the freedom of gauge, i.e. if the potentials are replaced by new potentials Ã, 3 This integration leads necessarily to an infinite result which caused difficulties in the development of the quantum theory of light. 4 In quantum mechanics, the canonical momentum p is converted to an operator according to p i (Jordan rule), which also turns H int into an operator. p and A commute only if the Coulomb gauge ( A = 0) is adopted. 3

4 φ according to A Ã + χ and φ φ χ/ t, (19) with χ(r, t) being an arbitrary gauge function, Maxwell s equations remain unaffected. This is easily seen by introducing the above substitutions in the definitions of A and φ (Eqs. (3) and (4)). To remove the ambiguity caused by the freedom of gauge we need to express H int in terms of the original fields E and B. To do this, we first expand the electric and magnetic fields in a Taylor series with origin r = 0 E(r) = E(0) + [r ]E(0) + 1! [r ] E(0) +, (0) B(r) = B(0) + [r ]B(0) + 1! [r ] B(0) +, (1) and introduce these expansions in the definitions for A and φ (Eqs. (3) and (4)). The task is now to find an expansion of A and φ in terms of E and B such that the left- and right-hand sides of Eqs. (3) and (4) are identical. These expansions have been determined by Barron and Gray as φ(r) = φ(0) i=0 r [r ] i (i+1)! E(0), A(r) = i=0 [r ] i B(0) r. () (i+) i! Inserting into the expression for H int in Eq. (18) leads to the so-called multipolar interaction Hamiltonian H int = q tot φ(0, t) p E(0, t) m B(0, t) [ Q ] E(0, t).. (3) in which we used the following definitions q tot = n q n, p = n q n r n, m = n (q n /m n ) r n p n, Q = n (q n /)r n r n q tot is the total charge of the system, p denotes the total electric dipole moment, m the total magnetic dipole moment, and Q the total electric quadrupole moment. If the system of charges is charge neutral, the first term in H int vanishes and we are left with an expansion which looks very much like the former expansion of the potential energy V E + V M. However, the two expansions are not identical! First, the new magnetic dipole moment is defined in terms of the canonical momenta p n and not by the mechanical momenta m n ṙ n. 5 Second, the expansion of H int contains a term nonlinear in B(0, t), which is non-existent in the expansion of V E + V M. The nonlinear term arises from the term A A of the Hamiltonian and is referred to as the diamagnetic term. It reads as (qn/8m n ) [r n B(0, t)]. (5) n Our previous expressions for V E and V M have been derived by neglecting retardation and assuming weak fields. In this limit, the nonlinear term in Eq. (5) can be neglected and the canonical momentum can be approximated by the mechanical momentum. 5 A gauge transformation also transforms the canonical momenta. Therefore, the canonical momenta p n are different from the original canonical momenta p n. (4) 4

5 The multipolar interaction Hamiltonian can easily be converted to an operator by simply applying Jordan s rule p i and replacing the fields E and B by the corresponding electric and magnetic field operators. However, this is beyond the present scope. Notice that the Hamiltonian H int in Eq. (3) is gauge independent. The gauge affects H int only when the latter is expressed in terms of A and φ but not when it is represented by the original fields E and B. The first term in the multipolar Hamiltonian of a charge neutral system is the dipole interaction, which is identical to the corresponding term in V E. In most circumstances, it is sufficiently accurate to reject the higher terms in the multipolar expansion. This is especially true for far-field interactions where the magnetic dipole and electric quadrupole interactions are roughly two orders of magnitude weaker than the electric dipole interaction. Therefore, standard selection rules for optical transitions are based on the electric dipole interaction. However, in strongly confined optical fields, as encountered in near-field optics, higher-order terms in the expansion of H int can become important and the standard selection rules can be violated. Finally, it should be noted that the multipolar form of H int can also be derived from Eq. (18) by a unitary transformation. This transformation, commonly referred to as the Power Zienau Woolley transformation, plays an important role in quantum optics. 5