B = 0. E = 1 c. E = 4πρ

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1 Photons In this section, we will treat the electromagnetic field quantum mechanically. We start by recording the Maxwell equations. As usual, we expect these equations to hold both classically and quantum mechanically. In the latter case, the various fields and sources become operators. The Maxwell equations are No Magnetic Monopoles Faraday s Law Gauss s Law B = 0 E = c E = 4πρ B t Ampere s Law B = 4π c J + c From elementary physics, the energy density is in Gaussian units) given by 8π E + B ) Once we introduce operators and commutation relations, this expression will be part of the full Hamiltonian of our system. To quantize the system, we introduce potentials; the scalar potential ϕ, and the vector potential A. These are related to the field strengths in the usual way by E = A c t ϕ, B = A Although the field strengths E, B are normally thought of as fundamental, there is no way known to quantize the electromagnetic field without potentials. This makes the potentials more significant in the quantum case than they are in the classical case. In particular, the presence of a vector potential can be detected in quantum mechanics. This is the Bohm-Aharanov effect. We will not pause to study this here, but go on to the quantization of the electromagnetic field. Periodic Boundary Conditions It is very convenient to imbed our system in a spacial box with periodic boundary conditions. The box volume is = L x L y L z. We expand functions of x in the basis functions, e i k x =< x k >, E t

2 where the components of k are Two useful identities are: k x = πn x, n x = 0, ±, ±,..., etc. L x k < x k >< k x >=< x x >= δ 3 x x ) and d 3 x < k x >< x k >= δ k, k Action, Lagrangian, Hamiltonian In particle quantum mechanics, one starts with a classical Lagrangian, constructs a classical Hamiltonian, and them quantizes the system by turning the coordinates and momenta in the Hamiltonian into operators, and imposing equal time commutation rules. This is essentially the same procedure used in a field system. However, in the case of fields, the Lagrangian and Hamiltonian are integrals over space of corresponding densities, denoted as L and H, respectively. The most fundamental starting point is the classical action, defined as in integral over all space along with an integral over time between two fixed times, t and t, of the Lagrange density, L. We will start with the action for the electromagnetic field, coupled to an external charge density ρx, t), and current density, Jx, t). The action for this situation is then t { S = dt d 3 x E 8π B ) d 3 xρϕ } J c A) t This is still a completely classical expression. When E and B are expressed in terms of the potentials, ϕ and A, the first two Maxwell equations, B = 0, and Faraday s law are automatic. The remaining two equations, i.e. Gauss s law and Ampere s law follow from the action principle. This states that the action is stationary when ϕ and A are subject to small variations which vanish at the time endpoints, t, and t. Stationarity of S means that when the classical field equations are satisfied, the first order change in S vanishes. Exercise Subject ϕ and A to small variations ϕ ϕ+δϕ, A A+δ A, with both variations vanishing at t and t. Demand that the first order variation in the action vanish for arbitrary δϕx, t) and δ Ax, t) and show that Gauss s law and Ampere s law are the result.

3 Coulomb gauge Electrodynamics is an example of a gauge theory. Different choices of ϕ, and A lead to the same equations of motion and the same physical results. One s first inclination is to formulate electrodynamics, either classical or quantum, in such a way that every step is manifestly gauge invariant. However, this is not possible. In our study of the motion of a charge in a magnetic field, it was already clear that a gauge choice was needed, since the Schrödinger equation involves the vector and scalar potentials. There we investigated the symmetric and Landau gauges. Both gauges gave the same energy levels. This was a case there the electromagnetic field was treated classically, and only the motion of the charge was treated with quantum mechanics. The situation is similar in the present case where we wish to quantize the electromagnetic field. A gauge choice must be made to proceed. However, no physical result can depend on the choice of gauge. arious choices are possible, but the most appropriate gauge for describing the interactions of photons with atoms is the Coulomb gauge. In the Coulomb gauge, the vector potential has no divergence, i.e. A = 0 In the Coulomb gauge, the part of the action involving the electric field simplifies. Writing out the spacial integral involving the electric field, we have d 3 x E) 8π = d 3 x ϕ) 8π + A c t ) + ϕ ) A c t The cross term between ϕ and A turns out to vanish. This is shown by integration by parts. The boundary term vanishes due to periodic boundary conditions, and the gets transferred to A. But this term also vanishes, since A = 0. The action now takes the form, S = t t dt d 3 x { 8π [ c A } t ) + ϕ) A) ] ρϕ + J c A. We note that the action in addition to the fact that ϕ and A are decoupled, does not contain any time derivatives of ϕ. This means that ϕ is not really a dynamical quantity, i.e. there is no momentum density variable associated with ϕ. Minimizing the action with respect to ϕ gives Gauss s law as mentioned before, i.e. ϕ = 4πρ This just states that the scalar potential is determined by the charge density, which is equivalent to ϕx, t) = d 3 x x x ρ x, t), which is nothing other than Coulomb s law. When ϕ is determined in this way by Coulomb s law, we will call the result ϕ c. The action now breaks into two separate terms, S = SA) + Sϕ c ). 3

4 The term Sϕ c ) is Sϕ c ) = t t dt { } d 3 x 8π [ ϕ c ) ] ρϕ c = t dt d 3 xρϕ c, t where the last term on the right hand side is generated by using Gauss s law and integrating by parts. This term will simply generate the usual Coulomb interactions between whatever charges are present in the problem. The term SA) is given by SA) = t t dt d 3 x { 8π [ c A } t ) + A) ] + J c A This term contains the dynamics of photons coupled to an external current. Note the strong resemblance to a simple forced harmonic oscillator, where the action is S = t t dt[ m Q m ω Q + F t)q] The coupling of A to J in SA), is analogous to the coupling of Q to the external force F t) in the forced oscillator problem. Classical Hamiltonian By definition, the quantity which is integrated over space and time in the action is the Lagrange density. For our problem the action is S A + Sϕ c ), so the Lagrange density is { L = 8π [ } A c t ) + A) ] + J c A ρϕ c This formula is closely analogous to the corresponding result for the string, L = ρ u t ) T u x ) + fx, t)u For the string, we defined π = ρ u t, and constructed the Hamiltonian density H as H = π ρ + T u x ) +fx, t)u Following similar steps for the electromagnetic case, we define Π = A 4πc t 4

5 and construct H by We finally have H = d 3 xh = H = Π A t L. { 4πc d 3 x Π Π + 8π A) J c A + } ρϕ c Quantization We are now ready to quantize the system. The generalized momentum density is Π, and the field variable is A. Quantization means both become operators, satisfying equal time commutation rules. For this case, we require [Π j x, 0), A l x, 0)] = i δt jlx x ) One might have expected simply δ jl δx x ) on the right hand side, but this is not compatible with A = 0. The so-called transverse delta function is defined as δ T jlx x ) = The terms in k j k l are necessary. To see this consider In k space, l A l will produce k e k x [δ jl k jk l k ] [Π j x, 0), l A l x, 0)] = [Π j x, 0), Ax, 0)] = 0 k l [δ jl k jk l k ] = k j k j k l k l k = 0 as it should. This shows that the extra terms in k j k l are necessary. Expansion of A in Creation and Destruction Operators We can deduce the correct expansion of A in terms of photon creation and destruction operators by comparing to the case of the string for periodic boundary conditions. For the string we had L ρ L = dx u t ) T ) u x ) 0 for the Lagrangian and ux, t) = ) a k e ikx iωkt + a k Lρω e ikx+iω kt k k 5

6 for the expansion in creation and destruction operators for periodic boundary conditions. Going from a one dimensional string with periodic boundary conditions to a three dimensional box with periodic boundary conditions is simple. We have where expikx) expi k x), L k = πn x L x, πn y, πn z ), L y L z and = L x L y L z. To obtain the correct square root factor, we compare the terms in time derivatives in the two Lagrangians. For the string, we have B ) ρ dx u t ), while for photons we have 0 dx 8πc ) A t ). We see that in going from strings to photons, Ax) = k ρ 4πc. We now can make an educated and correct) guess that the expansion of A is ) 4πc a k expi k x) + a k exp i k x), ω k where from the wave equation, ω k = k c. To insure that A = 0, we must have k ak = 0. This is done by using polarization vectors. We expand a k = λ a kλ ϵ λ k), where the ϵ λ k) are any two linearly independent vectors satisfying ϵ λ k) k = 0. A common and useful choice is to use so-called circular polarization vectors. For k along the +z axis, we would have ϵ k) = ê + iê ), ϵ k) = ê iê ) 6

7 For an arbitrary direction of k, we rotate the system composed of the z axis, ϵ and ϵ to the direction of k. The expansion of A can now be written, Ax) = 4πc ) a kλ ϵ λ ω k k) expi k x) + a kλ ϵ λ k) expi ) k x), ) kλ Πx) = i kλ ω k 4πc ) a kλ ϵ λ k) expi k x) a kλ ϵ λ k) expi ) k x), ) To gain insight into the indices on the polarization vectors, consider a photon traveling up the zaxis with polarization vector ϵ. Denoting the no photon state by 0 >, i.e. all the harmonic oscillators in their ground states, this one photon state would be a kẑ, 0 >. Calculating the matrix element < 0 Ax)a kẑ, 0 > we have < 0 Ax)a kẑ, 0 >= 4πc ω k expikz) ê + iê ) 3) This calculation is easy. Just substitute the expansion for A and use the commutation rules [a kλ, a k λ ] = δ k, k δ λ,λ ) Now looking at Eq.3) we see a plane wave headed up the z axis, multiplied by ê + iê. This expression is basically the wave function of the photon, with the polarization vector being the spin wave function. The + i character of ϵ implies that the spin projection along k is +, so for a general direction of k, ϵ ± k) corresponds to a photon spin projection along k of ± in units of.) This clarifies the meaning of the indices on polarization vectors and the meaning of the statement that a photon has spin. This is true, but it is also important to note that the spin projection of a photon along its wave vector can only be ±, never 0. Time Dependence of the ector Potential The Hamiltonian of the system of free photons is straightforward to write down. In terms of creation and destruction operators, we have H γ = ω k a k,λ a k,λ + ), 4) k,λ the expected form for a set of harmonic oscillators. It is also of interest to write H γ in terms of electric and magnetic fields. Recall the formula for the electric field; E = c A t ϕ. 7

8 The Coulomb gauge allows us to cleanly split E into a photon part and a Coulomb part. We have E = E γ + E coul, where E γ = A c t, and E coul = ϕ, and since we are not presently considering applied external magnetic fields, we also have B = B γ = A. In terms of E γ and B γ, the Hamiltonian is also given by H γ = dx E 8π γ ) + B ) γ ) 5) Showing that Eq.4) is the same as Eq.5) is done by using the expansion of A from Eq.) to calculate the electric and magnetic fields, and carrying out the integral dx in Eq.5). Both Eqs.4) and 5) represent the energy present in the form of photons. The rest of the electromagnetic energy is in the Coulomb interactions between charges. It is now easy to get the time dependence of A. By definition, Ax, t) = exp ih γt ) Ax, 0) exp ih γt ) Now since what we have is a set of independent harmonic oscillators, the time dependence of the creation and destruction operators is the usual form, exp ih γt )a k,λ exp ih γt ) = a k,λ exp iω k t), and exp ih γt )a k,λ exp ih γt ) = a k,λ expiω k t), Using these formulas along with Eq.) we finally have Ax, t) = kλ 4πc ω k a kλ ϵ λ k) expi k x ω k t) + a kλ ϵ λ k) expi ) k x + iω k t), 6) It is worthwhile to examine the coefficients of the creation and destruction operators in Eq.6). For the coefficient of a k,λ we have 4πc ω k expi k x iω k t)ϵ λ ) 8

9 In calculations, this factor will accompany a photon in the initial state of a process. Its structure is fairly easy to understand. The factor expi k x iω k t)ϵ λ ) is just a plane wave factor along with a spin wave function. In the prefactor square root ω k is familiar in every harmonic oscillator problem. Finally 4πc can be deduced from the /8πc accompanying A t ) in the Lagrangian. In a similar way, the factors accompanying a k,λ in Eq.6) are the complex conjugates of those accompanying a k,λ, and in calculations, these will accompany a photon in the final state of a process. 9