8. Quantization of the Electromagnetic Field¶
PDF pages 401–456
8.1 Classical Electromagnetic Field¶
Chapter 8 Quantization of the Electromagnetic Field Now we will switch to exclusively quantum-mechanical models. Before, we had treated the atom quantum mechanically, and introduced the idea of the photon, but now we will give a proper description of the quantum electromagnetic field. 8.1 Classical Electromagnetic Field Recall that we can write the source-free Maxwell equations in free space as
c2 \partial tE. (8.1) (Maxwell’s equations) In addition to the fields, we can consider the potentials A and \phi. For nonrelativistic calculations in vacuum, it is convenient to choose the Coulomb gauge,1 where \nabla \cdot A = 0. In this gauge, \phi = 0 in the absence of charges (we will reintroduce the sources later when we treat the atom–field interaction), so the fields are given by
(8.2) (Coulomb gauge) The last of the Maxwell equations thus implies the wave equation for the vector potential [using the gauge
\nabla 2A −1 c2 \partial 2 t A = 0. (8.3) (vector-potential wave equation) This equation is essentially the entire content of the Maxwell equations in this gauge, since the other three equations are implied by the Coulomb-gauge condition and the relations between the fields and the potentials. The first equation follows from
(8.4) 1The choice of Coulomb gauge is common in quantum optics, where the calculations are typically nonrelativistic, though relativistic field theorists prefer the Lorenz gauge. See L. Lorenz, ‘‘On the Identity of the Vibrations of Light with Electrical Currents,’’ Philosophical Magazine 34, 287 (1867). Incidentally, the Lorenz gauge is still commonly misattributed to Hendrik A. Lorentz; see J. van Bladel, ‘‘Lorenz or Lorentz?’’ IEEE Antennas and Propagation Magazine 33, No. 2 (April 1991); and Robert Nevels and Chang-Seok Shin, ‘‘Lorenz, Lorentz, and the Gauge,’’ IEEE Antennas and Propagation Magazine 43, No. 3 (June 2001) (doi: 10.1109/74.934904).
8.2.1 Variational Calculus¶
Chapter 8. Quantization of the Electromagnetic Field the second follows from
(8.5) and the third follows from
(8.6) The vector potential A thus compactly represents the fields. It also turns out to be fundamentally important in preserving the locality of quantum-mechanical particle-field interactions, so we will often work with it. 8.2 Hamiltonian Structure of the Classical Electromagnetic Field In order to quantize the field, we need the Lagrangian (and action principle), to identify the canonical coordinates and the Hamiltonian for the field, and then promote the canonical coordinates to operators.2 Before doing this, though, we will review briefly the ideas that underlie the Hamiltonian. 8.2.1 Variational Calculus First off, a (real-valued) functional3 is a function F : F −\rightarrow R, where F is a space of functions. Usually a functional will involve an integral to reduce functions to scalars, as in the action functional that we consider below. Let x(t) \in F; then F[x] is a real number, and the first variation of the functional F[x] is given by
ϵ\rightarrow 0
ϵ = d
ϵ=0 , (8.7) (first variation) where ϵ and \deltax are subject to the constraint that x + ϵ\deltax \in F. The first variation is essentially the ‘‘linear response’’ of F[x] to a small perturbation x(t) −\rightarrow x(t) + ϵ\deltax(t). We will then define the functional derivative \deltaF/\deltax such that \deltaF
:= Z t2 t1 \deltaF
(8.8) (functional derivative) Note that the brackets here denote an inner product of two vectors (here, functions), which is defined by the integral over the two vectors (functions). Thus, the functional derivative is the part of the first variation after dropping the integral and the variation \deltax. We can generalize this derivative in a couple of ways. A functional may depend on derivatives of the function; suppose that F[x, xt, xtt, . . . ; t] = Z t2 t1 f(t, x, xt, xtt, . . .) dt, (8.9) where xt ≡\partial x/\partial t. Then the first variation becomes
Z t2 t1 \partial f
\partial xt
\partial xtt
dt. (8.10) Integrating by parts,
Z t2 t1 \partial f \partial x −d dt \partial f \partial xt + d2 dt2 \partial f \partial xtt
\partial f \partial xt
t2 t1 , (8.11) 2Dirac discusses this basic ‘‘quantization recipe’’ and its advantages compared to alternatives in: Paul A. M. Dirac, Lectures on Quantum Mechanics (Belfer Graduate School of Science, 1964), Lecture 1. 3For more details on variational calculus and action principles, see P. J. Morrison, ‘‘Hamiltonian description of the ideal fluid,’’ Reviews of Modern Physics 70, 467 (1998) (doi: 10.1103/RevModPhys.70.467).
8.2.2 Action Principles¶
8.2 Hamiltonian Structure of the Classical Electromagnetic Field and typically the variation is arranged such that the surface terms vanish. This is usually enforced via
functional derivative, \deltaF
\partial x −d dt \partial f \partial xt + d2 dt2 \partial f \partial xtt
(8.12) being just the remaining bracketed expression in the integrand. The other generalization to the functional derivative is to many functions and dimensions, which occurs in the natural way, \deltaF
:= Z D \deltaF
(8.13) for functions y(x), where D is the domain of integration, and recall that we are implicitly summing over the repeated index \alpha. 8.2.2 Action Principles The calculus of variations is important in physics in setting up action principles, where equations of motion follow from the stationary points of some functional. For example, in a least-action principle, functions that minimize the functional correspond to physical solutions. As a concrete example, we can state an ‘‘action principle’’ for straight lines, in the sense of constructing an action principle that says that the shortest path between two points p1 and p2 is a straight line. That is, consider the length functional ℓfor the curve y(x), with x1,2 marking the points p1,2: ℓ[y] = Z x2 x1 s 1 + dy dx 2 dx. (8.14)
this from
Z x2 x1 " −d dx yx p 1 + y 2 x # dx = 0. (8.15) Setting the integrand to zero, after a bit of algebra we see that the only way the integrand can vanish is for y = \alphax + \beta, where \alpha and \beta are constants. 8.2.2.1 Lagrangian In general, we will define a scalar Lagrangian function L(q, ˙q; t) to describe our system. Naturally, this may be generalized to higher time derivatives. The action functional is defined by the integral S[L] := Z t2 t1 L(q, ˙q; t) dt, (8.16)
principle
(8.17) (Hamilton’s principle) implies the Euler–Lagrange equation \partial L
dt \partial L
(8.18) (Euler–Lagrange equation)
of Eq. (8.11). This proceeds along the lines of Eq. (8.13), noting that each variation \deltaq\alpha(t) is independent.
8.2.3 Electromagnetic Lagrangian and Hamiltonian¶
Chapter 8. Quantization of the Electromagnetic Field For a particle Lagrangian of the form L = 1 2m ˙q2 −V (q), (8.19) the Euler–Lagrange equation implies
(8.20) which is Newton’s Second Law. 8.2.2.2 Hamiltonian The Hamiltonian is defined by a Legendre transformation of the Lagrangian via
(8.21) (Hamiltonian) where the conjugate momentum to the generalized coordinate q\alpha is
(8.22) (conjugate momentum) The conjugate momentum is used to eliminate dependence on ˙q in the Hamiltonian introduced by the Lagrangian. Then the phase-space action S[q, p] := Z t2 t1
dt (8.23) (phase-space action) (note the bracketed quantity is basically the Lagrangian) along with the action principle \deltaS
\deltaS
= 0, (8.24) (phase-space action principle) imply Hamilton’s equations,
, (8.25) (Hamilton’s equations)
the Hamiltonian becomes H = p2 2m + V (q). (8.26) Then Hamilton’s equations become
m. (8.27) The first is again Newton’s Second Law, while the second is just the definition of the momentum. 8.2.3 Electromagnetic Lagrangian and Hamiltonian We now identify the Lagrangian for the electromagnetic field, which in terms of the vector potential is L = ϵ0 Z d3r
. (8.28) (electromagnetic Lagrangian)
8.2 Hamiltonian Structure of the Classical Electromagnetic Field We can see that this is the appropriate Lagrangian by employing the Euler–Lagrange equation (generalized for functional derivatives, since the generalized coordinates are fields) \deltaL
\deltaL
(8.29) where we take the generalized coordinate to be the vector potential A. Computing the functional derivatives (see below), we find
(8.30) which we see is equivalent to the wave equation (8.3) for the vector potential. Since A is the ‘‘position’’ coordinate, which we can see from the ‘‘kinetic energy’’ term (\partial tA)2 term in the Lagrangian, the conjugate momentum is given by \Pi := \deltaL
(8.31) (conjugate momentum to A) Then the Hamiltonian is given by the Legendre transform of the Lagrangian: H := Z
= ϵ0 Z d3r (\partial tA)2 −L = ϵ0 Z d3r
= ϵ0 Z d3r
. (8.32) In terms of the conjugate variables, the Hamiltonian is H = Z d3r \Pi2 2ϵ0 + 1
, (8.33) (electromagnetic Hamiltonian) while in terms of the fields, H = ϵ0 Z d3r E2 + c2B2 , (8.34) (electromagnetic Hamiltonian) it is clear that the Hamiltonian is just the total energy of the electromagnetic field. Hamilton’s equations then recover the Maxwell equations. The first Hamilton equation is
\deltaA, (8.35) which gives
(8.36) and thus yields the last Maxwell equation of Eqs. (8.1). The other Hamilton equation,
\delta\Pi, (8.37) contains essentially no information, as it implies
−\deltaE. (8.38) The other three Maxwell equations, as we indicated before, follow simply from the fact that the fields derive from the vector potential. Thus, we see how the Hamiltonian structure of the electromagnetic field arises within the Coulomb gauge, which will now allow us to quantize the field.
8.3 Quantization of a Single Field Mode¶
Chapter 8. Quantization of the Electromagnetic Field 8.2.3.1 Electromagnetic Functional Derivatives Here we compute the functional derivatives that arise in Eqs. (8.29), (8.35), and (8.37) as follows. Recall that a functional is a function that maps functions to scalars. The Lagrangian and Hamiltonian here satisfy this definition, due to the spatial integration, in the same way as the action integral above. The first variation of the Lagrangian is
ϵ=0 (8.39) for variations \deltaA and \delta(\partial tA) of the potential. This is easy to evaluate, as this is just a linearization in the variations, giving
Z d3r
. (8.40) Using the vector identity
(8.41) we can use the divergence theorem and integrate by parts to obtain
Z d3r
- ϵ0c2 Z surface
(8.42) We will assume a fixed-boundary variation, so that \deltaA = 0 on the surface of the integration volume, so that the surface term vanishes. We defined the functional derivatives in terms of the first variation via inner products with the varia- tions
\deltaL
- \deltaL
. (8.43) Interpreting the spatial integral with the dot product as the inner product here, we can write down the functional derivatives: \deltaL
\deltaL
(8.44) Note that from the form of the Hamiltonian, the same functional derivatives of the Hamiltonian have the same forms, except that the functional derivative with respect to A changes sign. 8.3 Quantization of a Single Field Mode Now we can proceed to quantize the field, considering only a single field mode.4 The idea is to take advantage of the fact that we have a linear field theory (because the QED Hamiltonian is quadratic in \Pi and A), so we can perform separation of variables and decompose the field operators into noninteracting normal modes. These normal modes are much simpler to deal with than the full fields. Taking an implicit Fourier transform, we can assume a monochromatic solution of frequency \omega:
(8.45) 4For further reading, see Peter W. Milonni, The Quantum Vacuum (Academic Press, 1993), Section 2.4, p. 40. The first quantum treatment of the electromagnetic field was M. Born, W. Heisenberg, and P. Jordan, ‘‘Zur Quantenmechanik II,’’ Zeitschrift für Physik 35, 557 (1926) (doi: 10.1007/BF01379806). Other important early papers on field quantization include P. A. M. Dirac, ‘‘The Quantum Theory of the Emission and Absorption of Radiation,’’ Proceedings of the Royal Society of London. Series A 114, 243 (1927) (doi: 10.1098/rspa.1927.0039); W. Heisenberg and W. Pauli, ‘‘Zur Quantendynamik der Wellenfelder,’’ Zeitschrift für Physik 56, 1 (1929) (doi: 10.1007/BF01340129); and Enrico Fermi, ‘‘Sopra l’etettrodinamica quantistica,’’ Atti della Reale Accademia Nazionale dei Lincei, 12, 431 (1930).
8.3 Quantization of a Single Field Mode The space and time dependences are now explicitly separated—the separation of variables is allowed by the form of the wave equation for A. The function f(r) is the mode function, which contains all the spatial dependence of the field. (In general, there are many possible mode functions for a given frequency, so we will simply choose one.) We assume them to be normalized such that Z
(8.46) The wave equation (8.3) then implies that the mode function satisfies the Helmholtz equation
(8.47) (Helmholtz equation for mode function)
We can now simplify the field Hamiltonian in the case of a single mode. We will need the relation Z
Z
= − Z
= k2 Z d3r A2, (8.48) where we again integrated by parts and discarded the surface term, then used the fact that the vector potential (8.45) satisfies the Helmholtz equation. Then the Hamiltonian becomes H = Z d3r \Pi2 2ϵ0 + 1 2ϵ0\omega2A2 . (8.49) This form suggests the Hamiltonian for a harmonic oscillator of frequency \omega and mass ϵ0, again with mo- mentum \Pi = −ϵ0E and position A. Of course, the spatial integral does not appear in the usual harmonic-oscillator Hamiltonian. However, the spatial dependence of the mode is fixed, so we can go ahead and carry out the integral to complete the analogy. Noting that the mode electric field is given by
(8.50) we can define a momentum coordinate to be the temporal part of \Pi, but with a different phase choice for \alpha(0),
(8.51) and a position coordinate to be the temporal part of A, with the same phase choice,
(8.52) so that p = ϵ0\partial tq. Then we can rewrite the Hamiltonian as H = ϵ0 Z d3r
= ϵ0 Z d3r
= ϵ0
= p2 2m + 1 2m\omega2q2, (8.53) with m = ϵ0. Here, the connection to the harmonic oscillator is more obvious, and the variables p and q are still clearly canonically conjugate.
8.4 Quantization of Many Modes¶
Chapter 8. Quantization of the Electromagnetic Field The usual quantum relations for the coordinate operators in terms of the creation and annihilation operators are q = r ¯h 2m\omega
p = r m¯h\omega a −a\dagger i , (8.54)
relations, we can identify \alpha(t) −\rightarrow i r ¯h 2\omegaϵ0 a(t), (quantization replacement, single field mode) (8.55) which will be our ‘‘recipe’’ for quantization: after a rescaling, we replace the function \alpha(t) with the annihi- lation operator a(t) (and a scaling factor). We can thus write the quantum fields as
r ¯h 2\omegaϵ0 f(r) a(t) + H.c.
r ¯h\omega 2ϵ0 f(r) a(t) + H.c.
r ¯h 2\omegaϵ0
(8.56) (quantized fields) Note that the mode functions in the quantum fields are entirely classical; the quantum part of the field modes only enters in the ‘‘time dependence.’’ Also, with the relations (8.54), the Hamiltonian (8.53) becomes
. (8.57) (Hamiltonian for single field mode) Of course, this is the usual quantum Hamiltonian for the harmonic oscillator. Thus, we have explicitly shown that a single field mode behaves both classically and quantum-mechanically as an ordinary harmonic oscillator, and we have defined the annihilation operator for this oscillator. Of course, the energy level |n\rangle is colloquially called the ‘‘number of photons,’’ and a(t) is the annihilation operator that removes a photon from the field. 8.4 Quantization of Many Modes Classically, there are orthogonal modes corresponding to different wave vectors k: different frequencies correspond to different magnitudes |k|, and for the same frequency, different directions correspond to different spatial mode profiles. Also, there are two distinct polarizations for each possible wave vector (due the the
we have the orthonormal mode functions fk,\zeta(r), satisfying Z V
(8.58) (mode orthnormality) where V is the volume of the cavity enclosing the mode functions (the quantization volume). Then each mode is completely independent of the others, and by extending the above analysis, the Hamiltonian
8.4.2 Example: Quantization in Free Space¶
8.4 Quantization of Many Modes becomes H = X k,\zeta ¯h\omegak a\dagger
, (8.59) (many-mode Hamiltonian) where \omegak = c|k|. In this case, we also have h
k′,\zeta′ i = \delta3
(8.60) (bosonic commutation relation) for the bosonic commutation relations for the field operators. We can now write the field operators as a sum over the field modes, including the operator parts:
X k,\zeta i r ¯h 2\omegakϵ0
X k,\zeta − r ¯h\omegak 2ϵ0
X k,\zeta i r ¯h 2\omegakϵ0
(8.61) (quantized fields) Since we have quantized the classical field theory while preserving its Hamiltonian structure, we have per- formed canonical quantization or second quantization of the electromagnetic field. 8.4.1 Example: Quantization in a Perfectly Conducting Box For a perfectly conducting box of lengths Lx, Ly, and Lz (with one corner at the origin), the transverse components of the electric fields must vanish at the boundaries, and thus the mode functions become
r V ˆx(ˆ\epsilonk,\zeta \cdot ˆx) cos(kxx) sin(kyy) sin(kzz) + ˆy(ˆ\epsilonk,\zeta \cdot ˆy) sin(kxx) cos(kyy) sin(kzz) + ˆz(ˆ\epsilonk,\zeta \cdot ˆz) sin(kxx) sin(kyy) cos(kzz) , (mode functions, perfectly conducting box) (8.62) where V := LxLyLz, ˆ\epsilonk,\zeta is the unit polarization vector of the mode, and the ˆxj are the Cartesian unit vectors along the xj-direction. The wave vectors are given by
Lx ,
Ly ,
Lz , (8.63)
nxEx Lx + nyEy Ly + nzEz Lz = 0, (8.64) which constrains the number of independent polarizations per (nx, ny, nz) triple to at most 2. 8.4.2 Example: Quantization in Free Space Quantization in free space5 is similar to the case of the box cavity, and in fact free-space results can be obtained with some care as the limit of a box where V −\rightarrow \infty.6 However, it is aesthetically better to have 5For further reading, see Peter W. Milonni, The Quantum Vacuum (Academic Press, 1993), Section 2.5, p. 43. 6Note that in principle, we should always quantize in free space, if we treat all matter quantum mechanically. However, it should be a good approximation to treat macroscopic ‘‘boundaries’’ of matter in terms of classical boundary conditions, which justifies our quantization inside a cavity and in half space. See P. W. Milonni, ‘‘Casimir forces without the vacuum radiation field,’’ Physical Review A, 25, 1315 (1982) (doi: 10.1103/PhysRevA.25.1315); and J. D. Cresser, ‘‘Unequal Time EM Field Commutators in Quantum Optics,’’ Physica Scripta T21, 52 (1988) (doi: 10.1088/0031-8949/1988/T21/010).
8.4.3 Example: Quantization in Half Space¶
Chapter 8. Quantization of the Electromagnetic Field mode functions that have amplitudes that are independent of r. To avoid problems with the normalization, we will impose a fictitious array of boxes filling free space, each of volume V = L3, with periodic boundary conditions on the vector potential
(8.65) which is satisfied by the mode functions
\sqrt V
(8.66) (mode functions, free space) where the components of the wave vector k is given by kx,y,z = 2\pinx,y,z L
3\sqrt V , (8.67) and the nxj are any integers. We can write out the potential explicitly here as
r ¯h
= i r ¯h
(8.68) and the electric field similarly becomes
r ¯h\omega
= r ¯h\omega
(8.69) Strictly speaking, we must let V −\rightarrow \inftyin any calculation in free space, unless the problem obeys periodic boundary conditions. This limit is straightforward, where, for example, \delta3 k,k′ −\rightarrow \delta3(k −k′), and the sum over modes changes to an integral. 8.4.3 Example: Quantization in Half Space A case intermediate to the above two is the case of a perfectly conducting plane defined by z = 0, where we quantize the half-space z > 0. The parallel components of the field must vanish on the plane, and so we choose
r V ˆ\epsilonk,\zeta,∥sin kzz −iˆ\epsilonk,\zeta,z cos kzz eik∥\cdotr, (half-space mode functions) (8.70)
quantized inside a cube of length L = V 1/3, imposing periodic boundary conditions in the x- and y-directions and conducting boundary conditions in the z-direction. The components of the wave vector are then given by
L ,
L ,
L , (8.71) where nx and ny are any integers, and nz is nonnegative. The transverse condition \nabla \cdot E = 0 then implies
(8.72) again restricting the number of distinct polarizations to two.
8.4.4 Example: Quantization in a Spherical Cavity¶
8.4 Quantization of Many Modes Note that we may also write Eq. (8.70) in the form
i \sqrt 2V
, (8.73) where k−= kxˆx + kyˆy −kzˆz (8.74) is the wave vector reflected through the conducting plane, and
(8.75) is the unit polarization vector with the same reflection. In this form, it is clear that each mode consists of a plane wave propagating along k, along with its reflection off the mirror, which travels along k−. For the two polarizations, it is conventional to choose one polarization to be oriented parallel to the mirror. This is the transverse electric (TE) polarization—or S-polarization, for senkrecht or perpendicular to the plane of incidence of the incident wave—and is given by f (TE) k
r V ˆk∥\times ˆz sin kzz eik∥\cdotr. (8.76) (TE polarization) Here ˆk∥is the projection of the k into the plane of the conductor, renormalized to unit length. The other polarization, the transverse magnetic (TM)—or P-polarization, for parallel to the plane of incidence— is orthogonal to both the TE polarization vector and k, and is given by f (TM) k
r V ˆk∥ kz k sin kzz + iˆz k∥ k cos kzz eik∥\cdotr. (8.77) (TM polarization) It thus always suffices to assume that the unit polarization vectors ˆ\epsilonk,\zeta are real. 8.4.4 Example: Quantization in a Spherical Cavity An important but much more complicated cavity than the rectangular one is the spherical cavity. In spherical coordinates, the Laplacian in the Helmholtz equation (8.47) is
\phi = 1 r \partial 2 r r +
\phi , (8.78) where the derivative operators are understood to operate on everything to the right, including an arbitrary test function. 8.4.4.1 Scalar Field
equations r2\partial 2
c 2 1 − c 2 sin2 \theta \Theta = 0 \partial 2
2 \Phi = 0, (8.79)
Chapter 8. Quantization of the Electromagnetic Field
(8.80)
(8.81)
and regarding \Theta to be function of x, 1 −x2 \partial 2
c 2 1 − m2 1 −x2 \Theta = 0. (8.82) Taking c 2
1 −x2 \partial 2
l(l + 1) − m2 1 −x2 \Theta = 0. (8.83) which has nondivergent solutions on the domain [−1, 1] only if l is a nonnegative integer and |m| \le l. These solutions are the associated Legendre functions, denoted by P m l (x). They are given explicitly by P m
2ll!
x (x2 −1)l
P m
(l + m)!P m l (x). (8.84) Clearly, P m l (x) is a polynomial if m is even, and P 0 l (x) is an ordinary Legendre polynomial. The P m l (x) obey the orthogonality condition Z 1 −1 P m l (x) P m
2(l + m)!
(8.85) The full solution to the Helmholtz equation is also orthogonal for different values of m, due to the form of \Phi(\phi) above. The angular solutions are generally combined, and thus the solution \Theta(\theta)\Phi(\phi) is given by the spherical harmonics Y m
s (2l + 1)(l −m)!
P m l (cos \theta) eim\phi, (8.86) (spherical harmonic) which are more conveniently normalized such that Y m∗ l
l
Z dΩY m
(8.87) (orthonormality relations) (See also Section 7.3.2 for a more quantum-mechanical introduction.) They also obey the sum rule l X m=−l |Y m
4\pi , (8.88) (sum rule)
8.4 Quantization of Many Modes showing that the m ‘‘quantum number’’ determines the orientation of the modes; summing over it results in an isotropic angular distribution. Some examples of the lowest few (monopole and dipole) spherical harmonics are Y 0
\sqrt 4\pi , Y 0
r
Y \pm1
r
(8.89) The spherical harmonics form a complete set for the angular dependence of the scalar-field solutions. As for the radial dependence, the equation for the radial function R(r) becomes r2\partial 2
(8.90)
\sqrt kr R(r) leads to r2\partial 2
" k2r2 − l + 1 2#
(8.91) This is Bessel’s equation (with independent variable kr), and the solutions are ordinary Bessel functions of the first kind, Jl+1/2(kr), of order l + 1/2, as well as the ordinary Bessel functions of the second kind, Yl+1/2(kr), of the same order. The solutions R(r) are thus generally written as spherical Bessel functions of the first and second kind, defined by
r \pi
r \pi
(8.92) (spherical Bessel functions) respectively. Near the origin, these functions have the asymptotic forms7
rl (2l + 1)!!
rl+1 , (8.93)
Technically, the yl(r) are not even square-normalizable over the cavity for l > 0, but y0(r) can be normalized, so we can’t necessarily discard it based on normalizability or finite-energy arguments. However, when we go over to the vector-field case, the derivatives involved will also make it non-normalizable. In any case,
representation8
\infty X l=0 (2l + 1)iljl(kr)Pl(ˆk \cdot ˆr), (8.94)
l (x) is a Legendre polynomial. Since an arbitrary plane wave may be decomposed into spherical Bessel functions of the first kind, and plane waves are complete, so are the j1(r).
radius of the spherical cavity. But jl(r) is an oscillatory function, and so there is a countable infinity of k values where the boundary condition is satisfied. Thus, we will define these k values by
(8.95) (transcendental equation for knl) 7For this and other properties see Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions (Dover, 1965), pp. 437-41. 8Eugen Merzbacher, Quantum Mechanics, 3rd ed. (Wiley, 1998), p. 261.
Chapter 8. Quantization of the Electromagnetic Field where the solution is commonly written knl = anl R , (8.96) (allowed wave numbers) where anl is the nth positive zero of jl(r). We may thus write the scalar-field, spherical cavity modes as
(8.97) (scalar modes of spherical cavity) where the radial normalization factor is given by Nnl := sZ R dr r2 j 2 l (knlr) (8.98) (radial normalization factor) The integral in the normalization factor can be performed analytically, with the result (Problem 8.15) N −2 nl = Z R dr r2 j 2
2 [j′
2 j 2 l+1(knlR), (radial normalization factor) (8.99) We see that the modes are parameterized by three indices (quantum numbers, in the case of a quantum particle in a spherical cavity), as we expect for three dimensions: a radial number n, and two angular numbers l and m. 8.4.4.2 Vector Field
vector solutions are (Problem 8.14)
N = 1
(8.100)
for the cavity modes; as we will see later, only transverse fields transport energy. Alternately, starting from
M = 1
(8.101) We can therefore see that M and N are proportional to each others’ curl, and thus are obvious candidates to represent E and H. In general, we will only use the field modes to represent transverse waves, as is consistent with the above use of plane and standing waves in the free space and the rectangular cavity, so we need not consider the L field. Furthermore, L is orthogonal to M, since
(8.102) and evidently M is also orthogonal to r. Writing out the first field, using the form (8.97) for the scalar solution (though relaxing for the moment the boundary conditions, which we will apply directly to the vector solution),
(8.103) 9Julius Adams Stratton, Electromagnetic Theory (McGraw–Hill, 1941).
8.4 Quantization of Many Modes Noting that¶
(8.104) Also noting that the following derivative of an angular function has the same form,
ˆ\theta1
(8.105) we can thus write
(8.106) The angular part here is often written in normalized form as a vector spherical harmonic Xm
i p l(l + 1)
(8.107) (vector spherical harmonic) which may also be written Xm
i p l(l + 1) ˆ\theta m
Y m
(8.108) (vector spherical harmonic) after writing out the gradient and cross product. These angular vector fields obey the orthonormality relations10 Z dΩXm′∗ l′
Z dΩXm′∗ l′
(8.109) (orthonormality relations) as well as the sum rule l X m=−l |Xm
4\pi , (8.110) (sum rule) which follows from the scalar sum rule (8.88). Thus, we finally write this solution as the transverse electric (TE) mode f (TE)
(8.111) (TE mode) so called because the polarization vector of f(r) (the same as for E) is parallel to the cavity surface and orthogonal to ˆr, as we can see from Eq. (8.104) or from Eq. (8.102). This is consistent with our previous notation in the half-space case of Section 8.4.3. Since the transverse component of the electric field vanishes at the cavity surface, the allowed wave numbers are identical to the scalar case,
(8.112) (allowed wave numbers, TE mode) 10See John David Jackson, Classical Electrodynamics, 3rd ed. (Wiley, 1999), p. 431.
Chapter 8. Quantization of the Electromagnetic Field and the normalization is likewise the same as for the scalar case: Nnl := sZ R dr r2 j 2 l (knlr) . (8.113) (radial normalization factor) Since the radial boundary condition is the same as for the scalar case, the integral in the normalization factor can again be performed analytically, with the result (Problem 8.15) N −2 nl = Z R dr r2 j 2
2 [j′
2 j 2 l+1(knlR), (radial normalization factor, TE mode) (8.114) However, the angular dependence is somewhat more complicated than for the scalar case, due to the vector nature of the field. Of course, we must deal with the other solutions N(r) to the vector Helmholtz equation. To do this, we essentially just compute the curls of the TE modes
knl
(8.115)
knl \nabla \times [jl(knlr) Xm
= Nnl knl [\nabla jl(knlr) \times Xm
= Nnl j′ l(knlr) ˆr \times Xm
knl jl(knlr) \nabla \times Xm
. (8.116) The first term is clearly transverse to the cavity surface, but the second isn’t necessarily. However, if this mode represents the electric field, then the orientation of the corresponding magnetic field is of the form
(8.117) which has no component along ˆr. Thus, the Nnlm(r) modes are called the transverse magnetic (TM) modes, f (TM)
knl \nabla \times [jl(knlr) Xm
(8.118) (TM mode) The TM mode function must satisfy the boundary condition that the ˆ\theta and the ˆ\phi components must vanish at the surface of the cavity. Noting that in spherical coordinates,
r sin \theta
+ ˆ\phi1 r
, (8.119) we see that the \partial r(rA\theta) and \partial r(rA\phi) terms vanish at the boundary provided \partial r r jl(knlr)
(8.120) (allowed wave numbers, TM mode) The other terms of the form \partial \phiAr and \partial \thetaAr vanish automatically, since ˆr \cdot Xm
8.4 Quantization of Many Modes factor in the integral form (8.113) is the same as for the TE mode, as we can verify by integrating by parts: Z d3r f (TM) nlm (r) 2 = k 2 nl Z d3r |\nabla \times Mnlm(r)|2 = k 2 nl Z d3r M∗
I da \cdot M∗
= −1 k 2 nl Z d3r M∗ nlm(r) \cdot \nabla 2Mnlm(r) = Z d3r M∗ nlm(r) \cdot Mnlm(r) = 1. (8.121) Here, we have used \nabla \cdot M = 0. We have also discarded the surface term, since it amounts to the surface integral of M∗\times N, or in other words the Poynting vector, and for cavity modes no energy is transported across the cavity boundary. However, the analytic solution for the integral has a somewhat different form, N −2 nl = Z R dr r2 j 2
1 −l(l + 1) k 2 nlR2 j 2 l (knlR), (radial normalization factor, TM mode) (8.122) because the boundary condition here is different from the TE case (Problem 8.15). In either the TE or the TM case, note that since the lowest scalar spherical harmonic Y 0
\sqrt 4\pi is constant, the corresponding vector spherical harmonic vanishes, X0
version. Thus, the lowest-order vector fields have l = 0, essentially because there are no monopolar vector
lrl−1/(2l+1)!!. If an atom is at the center of the spherical cavity, evidently it only has nonvanishing coupling to the l = 1 TM modes, since their mode functions involve j′ l(r). It is also useful to represent the vector solutions in terms of the scalar solution \psinlm(r), Eq. (8.97). Using the expression (8.100) for the TE mode M, as well as the expression (8.119) for the curl, we may write f (TE)
p l(l + 1) ˆ\theta
\psinlm(r), (8.123) (TE mode) where recall that we need the extra factor of p l(l + 1) to normalize the vector angular distribution. Writing out the scalar solution, f (TE)
N (TE) nl p l(l + 1) jl(knlr) ˆ\theta
Y m
(8.124) (TE mode) which is essentially what we have already written out in terms of the vector spherical harmonic. However, for the TM case, we can obtain a relatively simple explicit expression, compared to what we could otherwise get by expanding the above expressions. To write it, we use the expression (8.100) for N, again with the curl (8.119), to write p l(l + 1)f (TM)
ˆr kr sin \theta
+ ˆ\theta
ˆ\phi
(8.125) Now we use the fact that \psinlm satisfies Eqs. (8.79), and thus we may simplify the r component to the form f (TM)
p l(l + 1) kr
ˆ\theta p l(l + 1) kr
ˆ\phi p l(l + 1) kr sin \theta
(8.126) (TM mode)
Chapter 8. Quantization of the Electromagnetic Field so that we obtain the explicit form f (TM)
(TM) nl " ˆr p l(l + 1)jl(knlr) kr Y m
p l(l + 1) kr
sin \theta # . (TM mode) (8.127) after writing out the scalar solution. It is once again clear here that the TM solutions have a stronger presence near the origin than do the TE modes. 8.4.4.3 Asymptotics In general, the spherical Bessel functions make things somewhat difficult to work with. For example, we can’t write down analytic expressions for the allowed wave numbers knl. However, for very large spherical cavities, such that for a given wavelength many modes can be excited, it is useful to use the asymptotic forms of the spherical Bessel functions with large arguments. From Rayleigh’s formula
−1 z \partial z l sin z z , (8.128) we evidently have the asymptotic form jl(z) ∼(−1)l \partial l z sin z z
z sin z −l\pi + O(z−2). (8.129) This asymptotic form also follows from the asymptotic form for the ordinary (cylindrical) Bessel function, J\alpha(z) ∼ r \piz cos
2 −\pi , (8.130) along with the definition of jl(z) in Eq. (8.92). This form has zeroes whenever the argument of the sin is equal to n\pi for integer n, and thus for the TE modes leads from Eq. (8.112) to the asymptotic condition
2 l. (8.131)
obviously has n > 0. On the other hand, for the TM mode, the allowed modes from the condition (8.120) has a function of the form \partial z[zjl(z)] ∼\partial z sin z −l\pi
+ O(z−1), (8.132) and thus gives the condition.
2 (l + 1) (8.133)
The first positive zeros here occur for any n \ge 0, unlike the TE case. In this asymptotic regime, we can also analytically evaluate the normalization constant, since from Eq. (8.113), the radial normalization integral becomes N −2 nl = Z R dr r2 j 2
k 2 nl Z R dr sin2 knlr −l\pi = R 2k 2 nl , (8.134) and thus the normalization factor is Nnl \approx knl r R, (8.135) (radial normalization factor) with the appropriate value of knl for the TE or TM modes.
8.4 Quantization of Many Modes 8.4.4.4 Vector Multipole Modes Now that we have the mathematical apparatus, we may as well generalize the above spherical-cavity modes a bit. In the above treatment of the spherical cavity modes, we excluded the solutions yl(r) because they were divergent at the origin. However, if we consider the exterior modes of a spherical cavity, or we simply exclude from consideration a neighborhood around the origin, then these divergent modes are perfectly acceptable, because the divergence is removed. When they apply, it is common to define the spherical Hankel functions h(1) l
h(2) l
(8.136) Again from Rayleigh’s formulae
−1 z \partial z l sin z z
−1 z \partial z l cos z z , (8.137) we can write the corresponding formulae for the spherical Hankel functions h(1) l
−1 z \partial z l eiz z h(2) l
−1 z \partial z l e−iz z . (8.138) The phase dependence clearly indicates that h(1) l (kr) represents an outgoing wave, while h(2) l (kr) represents an ingoing wave. The jl(kr) and yl(kr) are thus spherical-coordinate analogues to the standing waves sin(kx) and cos(kx) in Cartesian coordinates, while the h(1) l (kr) and h(2) l (kr) are analogues to the traveling waves exp(ikx) and exp(−ikx). Then proceeding as above, but in free space, we may write the TE modes, but now separating them into ingoing and outgoing parts, as f (TE)\rightarrow klm
l (kr) Xm
f (TE)\leftarrow klm
l (kr) Xm
(8.139) (TE modes) where k is a positive, real number, while l and m are still integer indices, with l positive and m nonnegative. Of course, there may be further restrictions on k if there are boundary conditions, such as when treating the exterior modes of a spherical, conducting shell. Correspondingly, the outgoing and ingoing TM modes are f (TM)\rightarrow klm
h h(1) l (kr) Xm
i f (TM)\leftarrow klm
h h(2) l (kr) Xm
i . (8.140) (TM modes) If we work out the outgoing l = 1, m = 0 TE mode, we find for the radial part h(1)
eir r = −ieir r2 −eir r = −eir r 1 + i r , (8.141) and the angular part, X0
\sqrt
r
= −i r
r
(8.142)
Chapter 8. Quantization of the Electromagnetic Field so that the complete mode function is f (TE)\rightarrow k10
r 8\pi eikr k2 1 r2 −ik r sin \theta. (8.143)
derivatives
r
r 2\pi cos \theta r −1 r \partial rr −eikr kr 1 + i kr = −ieikr k2 1 r3 −i k r2 −k2 r , (8.144) so that we find the mode function f (TM)\rightarrow k10
r 8\pi eikr k2 ˆr 2 1 r3 −i k r2
1 r3 −i k r2 −k2 r sin \theta . (8.145) Recalling the electric-dipole fields from Eq. (1.42),
4\piϵ0
" d(+)(tr) r3 + ˙d(+)(tr) cr2 # + 4\piϵ0
¨d(+)(tr) c2r
" ˙d(+)(tr) cr2 + ¨d(+)(tr) c2r # , (8.146) which for monochromatic fields become,
4\piϵ0 eikr
1 r3 −i k r2
r
4\pi
k r2 −ik2 r . (8.147) Comparing to the above mode functions for ˆ\epsilon = ˆz, we see the dipole fields may be written as
\sqrt 6\pi
ϵ0c2 f (TM)\rightarrow k10 (r)
c \sqrt 6\pi f (TE)\rightarrow k10 (r). (8.148) Similarly, f (TE)\rightarrow k10 (r) represents the dimensionless electric-field mode profile due to a magnetic dipole, and f (TM)\rightarrow k10 (r) represents the dimensionless magnetic-field mode profile due to a magnetic dipole. In general, the f (TM)\rightarrow klm (r) represent the dimensionless electric-field mode profiles due to electric multi- poles of order l, and the f (TE)\rightarrow klm (r) represent the dimensionless magnetic-field mode profiles due to electric multipoles of order l. For magnetic multipoles, the profiles exchange identity. From the form of h(1) l (kr) in Eq. (8.138), we can see that in the far field, all the multipole fields decay as r−1, corresponding to radiation. Similarly, in the near field, for example, the electric field due to an electric multipole of order l goes as rl+2, while the magnetic field goes as rl+1.
8.5.1 Helmholtz Theorem¶
8.5 Transverse and Longitudinal Fields 8.5 Transverse and Longitudinal Fields 8.5.1 Helmholtz Theorem Before continuing, it is convenient to distinguish between transverse and longitudinal part of a vector field11. For an arbitrary field C(r), the Helmholtz theorem states that there is a unique decomposition
(8.149) (Helmholtz theorem) such that the transverse field is divergenceless,
(8.150) (transverse field condition) and the longitudinal field is irrotational,
(8.151) (longitudinal field condition) We can see this by starting with the delta-function identity
Z d3r′C(r′)\delta3(r −r′). (8.152) Thus using the delta-function identity \nabla 2
(8.153)
(8.154) we can write
4\pi Z d3r′C(r′)\nabla 2 |r −r′| = 1
Z d3r′ C(r′) |r −r′| −1
Z
|r −r′| = 1
Z d3r′ C(r′) |r −r′| + 1
Z
|r −r′| = 1
Z d3r′ C(r′) |r −r′| −1
Z
|r −r′|
(8.155) where we have assumed that the boundary terms vanish, and we have defined
Z d3r′ C(r′) |r −r′|
Z
|r −r′| (transverse and longitudinal fields) (8.156) 11For further reading, see Peter W. Milonni, The Quantum Vacuum (Academic Press, 1993), Appendix F, p. 501.
Chapter 8. Quantization of the Electromagnetic Field From their forms, the two components clearly have the desired properties. Now for uniqueness of the decomposition. Once the divergence and curl of a vector field are specified, along with its boundary conditions, the vector field itself is uniquely specified. That is, suppose that
(8.157)
(8.158) Then the difference field satisfies
(8.159) Because the curl of the difference vanishes, we can write it as the gradient of a scalar function,
(8.160) which, with the fact that the divergence vanishes, implies Laplace’s equation for h:
(8.161)
To establish this formally, consider the divergence theorem Z S
Z V
(8.162) Then letting K = h\nabla h, the surface integral becomes Z S
Z S
Z S
(8.163) if we assume that the normal component of C1 and C2 are equal on the boundary. (We need not specify the equality of the transverse components.) This implies that the volume integral vanishes, Z V
(8.164) so that we can use
(8.165) to write Z V
Z V
= − Z V
= 0, (8.166) since the divergences of the fields are equal. Finally, we have Z V
(8.167)
decomposition (8.156) is unique.
8.5.2 Transverse and Longitudinal Delta Functions¶
8.5 Transverse and Longitudinal Fields 8.5.1.1 Coulomb Gauge In the Coulomb gauge, the transverse and longitudinal components of the fields are easy to identify. The vector field is completely transverse, since \nabla \cdot A = 0. The magnetic field is similarly transverse, as funda- mentally \nabla \cdot B = 0. The electric field has both components; the source-free part due to A is E⊥= −\partial tA, which is clearly transverse. The part due to a source charge is E∥= −\nabla \phi, which is clearly longitudinal. 8.5.2 Transverse and Longitudinal Delta Functions Just as we can use the Kronecker delta and the delta function as a projection operator for a component of the field,
Z
(8.168) (summation is implied here by repeated indices), we can also define projection operators for the transverse and longitudinal parts of the field. The transverse delta function, defined by C⊥
Z d3r′\delta⊥
(defining relation, transverse delta function) (8.169) projects out the transverse part of the field, while the longitudinal delta function, defined by C∥
Z d3r′\delta∥
(defining relation, longitudinal delta function) (8.170) projects out the longitudinal part of the field. 8.5.2.1 Momentum Representation We will use the following normalization convention for the Fourier transform and inverse transform of the field:
Z d3r C(r)e−ik\cdotr
Z d3k ˜C(k)eik\cdotr. (8.171) The vector identity
k2 k[k \cdot ˜C(k)]
(8.172) is the k-space version of Eq. (8.155). The transverse and longitudinal components are easy to identify here from the conditions
(8.173) which are the Fourier-space versions of Eqs. (8.150) and (8.151). Then we can compute the inverse transform of the longitudinal part:
Z d3k 1 k2 k[k \cdot ˜C(k)]eik\cdotr = (2\pi)3 Z d3r′ Z d3k 1 k2 k[k \cdot C(r′)]eik\cdot(r−r′). (8.174) We can write this relation in components to find C∥
Z d3r′ (2\pi)3 Z d3k 1
(8.175)
Chapter 8. Quantization of the Electromagnetic Field Comparing this to Eq. (8.170), we can write the longitudinal delta function as \delta∥
(2\pi)3 Z
k2 eik\cdotr. (longitudinal delta function, momentum representation) (8.176) Eq. (8.149) implies that
(8.177) so the transverse delta function simply becomes \delta⊥
(2\pi)3 Z d3k
k2 eik\cdotr, (transverse delta function, momentum representation) (8.178) where we used the representation
(2\pi)3 Z d3k eik\cdotr (8.179) of the delta function in three dimensions. 8.5.2.2 Position Representation We can also evaluate the integrals above to obtain direct expressions for the transverse and longitudinal delta functions. Starting with Eq. (8.178), we note that the integral is not strictly convergent, so we will insert a convergence factor e−k\lambda, letting \lambda −\rightarrow 0 after the integration. This procedure is effectively equivalent to assuming that the fields on which the projectors operate are reasonably smooth (i.e., bandwidth-limited). Then \delta⊥
(2\pi)3 Z d3k
k2
= (2\pi)3
Z d3k eik\cdotr−k\lambda k2 = (2\pi)2
dk Z \pi
= (2\pi)2
dk 2 sin kr kr e−k\lambda = 2\pi2
r . (8.180) Letting \lambda −\rightarrow 0, we find \delta⊥
4\pi
r . (8.181)
1 r
r\beta
r5 , (8.182)
straightforward differentiation). The result is \delta⊥
4\pir3
r2 . (transverse delta function) (8.183)
8.6.1 Free-Space Commutators¶
8.6 Field Commutators Using Eq. (8.177), we can also write the corresponding expression \delta∥
4\pir3
r2 (longitudinal delta function) (8.184) for the longitudinal delta function. 8.6 Field Commutators In view of the bosonic commutator (8.60), the field operators will not in general commute.12 As these commutation relations are useful, we will spend some time exploring these. We start with the relations (8.56) for the quantum fields. We can then write, for example,
X k,\zeta ¯h 2\omegakϵ0
= i¯h ϵ0 X k,\zeta \omegak Im h
. (vector-potential commutator) (8.185) In this form, the commutator is not easy to interpret, but it is clear that the commutator depends on the boundary conditions that determine the mode functions fk,\zeta(r). Other useful relations include
ϵ0 X k,\zeta \omegak Im h
ϵ0 X k,\zeta \omegak
h
\partial ′ \sigmaf ∗
e−i\omegak(t−t′)i
ϵ0 X k,\zeta Re h
ϵ0 X k,\zeta Re h
µf ∗
. (various field commutators) (8.186) Again, the interpretation here is not transparent, so we will consider their specific form in free space. 8.6.1 Free-Space Commutators
8.6.1.1 Direction Cosines To evaluate the summations over the polarization index \zeta in the above commutators, we will need to compute the sum X \zeta
(8.187) 12For further reading, see Peter W. Milonni, The Quantum Vacuum (Academic Press, 1993), Section 2.8, p. 59; Leonard Mandel and Emil Wolf, Optical Coherence and Quantum Optics (Cambridge, 1995), Section 10.8, p. 500; and P. W. Milonni, ‘‘Casimir forces without the vacuum radiation field,’’ Physical Review A, 25, 1315 (1982) (doi: 10.1103/PhysRevA.25.1315).
Chapter 8. Quantization of the Electromagnetic Field To evaluate this sum, recall the direction cosines for the vector r. This vector makes angles \theta\alpha with the respective r\alpha-axes. The direction cosines are defined as \gamma\alpha = cos \theta\alpha = r \cdot ˆr\alpha/r, and thus they satisfy X \alpha \gamma 2
X \alpha r 2 \alpha r2 = 1. (8.188) More generally, if we have two orthonormal cartesian bases ˆr\alpha and ˆr′ \alpha, then we can define direction cosines between the coordinate systems of \gamma\alpha\beta := ˆr\alpha \cdot ˆr′ \beta. Then it follows from the orthogonality of the basis vectors that X µ
X µ
µ
µ
(8.189)
with ˆr\alpha forming another, we can apply the result (8.189) to obtain X \zeta
k2 , (8.190) which will prove to be a very useful relation in mode-summation problems. 8.6.1.2 Evaluation From Eq. (8.185), we can write
i¯h cϵ0V X k,\zeta k Im h eik\cdot(r−r′)e−i\omegak(t−t′)i
= −i¯h cϵ0V X k k eik\cdot(r−r′) sin[\omegak(t −t′)]
k2 , (8.191) where we used the fact that R d3k exp(ik \cdot r) is real. In free space, we take the limit of large quantization volume (V −\rightarrow \infty), and the spacing between the modes becomes correspondingly small. In this limit, an integral of a function is equivalent to a sum weighted by the mode spacings. Thus we can write X k f(k) ∆kx ∆ky ∆kz −\rightarrow Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz f(k) (8.192) for an arbitrary function f(k). Since ∆kx,y,z = 2\pi V 1/3 , (8.193) we can thus make the formal replacement X k −\rightarrow V (2\pi)3 Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz. (8.194) Thus, we write
cϵ0 (2\pi)3 Z d3k 1 k eik\cdot(r−r′) sin[\omegak(t −t′)]
k2 (vector-potential commutator, free space) (8.195) for the vector-potential commutator. Similarly, for the electric field we can write
ϵ0 (2\pi)3 Z d3k keik\cdot(r−r′) sin[\omegak(t −t′)]
k2 . (8.196)
8.6 Field Commutators We can simplify this commutator somewhat by introducing the singular D function:
(2\pi)3 Z d3k 1 k eik\cdotr sin \omegakt = − (2\pi)2 Z \infty dk Z \pi d\theta sin \theta keikr cos \theta sin \omegakt = − (2\pi)2 Z \infty dk 2 sin kr r sin \omegakt = 8\pi2r Z \infty dk eikr −e−ikr eickt −e−ickt = 8\pi2r Z \infty −\infty dk eik(r+ct) −eik(r−ct) =
(8.197)
ϵ0
\beta D(r −r′, t −t′), (electric-field commutator, free space) (8.198) where \partial ′
\alpha. This and the two field commutators that follow are the Jordan–Pauli commutators.13 The interpretation of this commutator is the electric field can be measured at two spacetime points, so long as they are not on the same light cone.14 More intuitively, making an electric-field measurement at (r, t) disturbs the field, and in vacuum the disturbance propagates occupies any spacetime point (r′, t′) on the
disturbance that propagates to (r, t), disturbing a measurement there. A similar calculation shows that the magnetic field has a commutator of almost the same form,
ϵ0c
D(r −r′, t −t′), (magnetic-field commutator, free space) (8.199) and so the same interpretation applies to the magnetic field. The mixed commutator for the electric and magnetic fields is slightly different. Using the same procedure as above, we can write
ϵ0 (2\pi)3 X \zeta Z d3k Re h
= ¯h ϵ0 (2\pi)3 Z d3k Re
k2 kµeik\cdot(r−r′)e−i\omegak(t−t′) = ¯h ϵ0 (2\pi)3 Z d3k Re h
= ¯h ϵ0
(2\pi)3 Z d3k kµeik\cdot(r−r′) cos [\omegak(t −t′)] , (8.200) 13P. Jordan and W. Pauli, Jr., ‘‘Zur Quantenelektrodynamik ladungsfreier Felder,’’ Zeitschrift für Physik 47, 151 (1928) (doi: 10.1007/BF02055793) (doi: 10.1007/BF02055793). 14The seminal discussion of simultaneous measurability of the quantum electric and magnetic fields can be found in Niels Bohr and Léon Rosenfeld, ‘‘Zur Frage der Messbarkeit der elektromagnetischen Feldgrössen,’’ Mathematisk-Fysiske Meddelelser 12 (1933); translation reprinted in Quantum Theory and Measurement, John Archibld Wheeler and Wojciech Hubert Zurek, Eds. (Princeton, 1983) p. 479.
Chapter 8. Quantization of the Electromagnetic Field and thus
ϵ0
(electric-magnetic-field commutator, free space) (8.201)
Finally, we have the commutator between the conjugate fields A and E,
ϵ0 (2\pi)3 X \zeta Z d3k Re h
, (8.202) so that
ϵ0 (2\pi)3 Z d3k Re h eik\cdot(r−r′)e−i\omegak(t−t′)i
k2 . (potential–field commutator) (8.203) This relation does not simplify as the ones for the E and B fields, a point that we will return to below. 8.6.1.3 Equal-Time Commutators in Free Space From Eq. (8.196), we can see that at equal times,
(8.204) (equal times) Similarly, we can see that
(8.205) (equal times) Thus, the fields can be measured at two different locations at the same time. However, for the mixed commutator for the electric and magnetic fields,
ϵ0
(2\pi)3 Z d3k kµeik\cdot(r−r′) = −i¯h ϵ0
(8.206) (equal times) The electric and magnetic fields thus cannot be measured at the same location at equal times. For the electric field and vector potential, we arrive at the important commutation relation
ϵ0 (2\pi)3 Z d3k eik\cdot(r−r′)
k2 = −i¯h ϵ0 \delta⊥
(8.207) (equal times) which is nonzero even off of the light cone. The vector potential is evidently not a local field in the same sense as the electric and magnetic fields (we will see this again in the Aharonov–Bohm effect in Section 9.4). Finally, it turns out that
(8.208) (equal times) so that the potential is more ‘‘compatible’’ with the magnetic field.
8.6.2 Half-Space Commutators¶
8.6 Field Commutators 8.6.2 Half-Space Commutators As we noted above, the field commutators depend on the boundary conditions through the form of the mode profile functions fk,\zeta(r). We can see this by considering the half-space mode functions of Section 8.4.3. Using the mode functions in the form of Eq. (8.73), and putting them into Eqs. (8.186), the mixed commutator for the vector potential and electric field becomes
i¯h 2ϵ0V X k,\zeta Re
eik−\cdot(r−r′)
eik\cdot(r−r′−) −
(8.209) In the expression here r−:= xˆx+yˆy−zˆz, and r−−r′ = r−r′ −2zˆz. Carrying out the sum over polarizations as usual, using the result (8.190), while noting the appropriate sign changes when the reflected polarization vectors are involved, X \zeta
k\alphak− \beta k2 , (8.210) we can write
i¯h 2ϵ0V X k Re "
k2
k− \alpha k− \beta k2 ! eik−\cdot(r−r′) −
\delta−
k\alphak− \beta k2 ! eik\cdot(r−r′−) − \delta−
k2 eik\cdot(r−−r′) # e−i\omegak(t−t′). (8.211) Here \delta− \alpha\beta is the same as the usual Kronecker delta, except that \delta− zz = −1. We can simplify this expression by considering the commutator only at equal times. Owing to the form of the wave vector in half-space, we can make the formal replacement X k −\rightarrow V 4\pi3 Z d3k, (8.212) where the integration is over half of reciprocal space. We can then extend the integration over all space and add a factor of 1/2 to eliminate the real-part operator, with the result
i¯h 2(2\pi)3ϵ0 Z d3k
k2
k− \alpha k− \beta k2 ! eik−\cdot(r−r′) −
\delta−
k\alphak− \beta k2 ! eik−\cdot(r−−r′) − \delta−
k2 eik\cdot(r−−r′) . (8.213) If we define the ‘‘reflected’’ transverse delta function by \delta⊤
2(2\pi)3 Z d3k
\delta−
k\alphak− \beta k2 !
2(2\pi)3 Z d3k \delta−
k2 eik\cdotr = (2\pi)3 Z d3k \delta−
k2 eik\cdotr, (8.214) this commutator simplifies to
ϵ0 \delta⊥
. (8.215) (equal times, half-space)
8.7 Unconfined Mode Functions¶
Chapter 8. Quantization of the Electromagnetic Field
contains contributions from two intervals: the first is the direct separation r −r′, which is the same as in the free-space case, while the second is the separation including one bounce from the conducting surface (mirror). The second contribution also contains a flipped orientation of the delta function, which accounts for the reversal of an image-charge distribution with respect to the source-charge distribution. This seems a physically reasonable modification to the free-space commutator. Similarly, we can write the electric-field commutator as
i¯hc 2ϵ0V X k,\zeta k Im
eik−\cdot(r−r′)
eik\cdot(r−r′−) −
= i¯hc ϵ0 (2\pi)3 Z d3k k "
k2
k− \alpha k− \beta k2 ! eik−\cdot(r−r′) −
\delta−
k\alphak− \beta k2 ! eik\cdot(r−r′−) − \delta−
k2 eik\cdot(r−−r′) #
(8.216) Carrying out the same procedure as before, we can write this commutator in terms of the singular D function as
ϵ0
\beta D(r −r′, t −t′) −i¯hc ϵ0 " \delta− \alpha\beta
\beta # D(r−−r′, t −t′), (half-space) (8.217) where again \partial ′
\alpha. Again, we see that we have the same form as before, but now the simultaneous measureability is excluded also by an additional term corresponding to the light cone that includes a bounce off of the mirror.15 The second term has precisely the same form as before, except for sign differences in the z-related components. Recall that the orientation of a dipole image has similar sign modifications compared to the original. 8.7 Unconfined Mode Functions Recall from Section 8.4 that we quantized electromagnetic field modes inside a quantization volume. In sit- uations without a cavity, such as in the Weisskopf–Wigner calculation for the rate of spontaneous emission in free space (Chapter 11), the quantization volume corresponds to a fictitious cavity. Generally, the quanti- zation volume cancels in the relevant physical quantities for these calculations. For example, the free-space mode functions from Section 8.4.2 are
\sqrt V
(8.218) where again k takes on discrete values due to the periodic boundary conditions, and the mode functions are normalized according to Z V
(8.219) 15P. W. Milonni, ‘‘Casimir forces without the vacuum radiation field,’’ Physical Review A 25, 1315 (1982) (doi: 10.1103/Phys- RevA.25.1315).
8.8 Hamiltonian Viewpoint of Electromagnetic Gauge Freedom¶
8.8 Hamiltonian Viewpoint of Electromagnetic Gauge Freedom The sum over modes amounts to something of the form [c.f. Eq. (11.23)] X k −\rightarrow V (2\pi)3 Z d3k, (8.220) where the integration extends over all possible orientations of k. Generally, physical quantities involve an integrand quadratic in the mode function, and thus the factors of V cancel. The subsequent limit V −\rightarrow \infty for true free space is then trivial. Thus, for problems where no physical cavity is involved, it is convenient to define mode functions where the limit V −\rightarrow \inftyis already taken. We can take the large-volume limit of Eq (8.219) by noting that in this limit the spacings between adjacent values of kx, ky, and kz become small (scaling as 2\pi/ 3\sqrt V ), and so we can make the replacement \delta3
V \delta3(k −k′). (8.221) This is again so that the sum of the left-hand behaves as the integral over the right-hand side. Then we can write Z V
V
(8.222) in the large-volume limit. Then absorbing a factor of \sqrt V into the definition of the mode function by the rescaling fk,\zeta(r) −\rightarrow \sqrt V fk,\zeta(r), (8.223) the normalization for unbounded mode functions becomes Z
(8.224) The sum over modes, assuming a summand quadratic in the mode functions (as is usually the case), then is given by the correspondence X k −\rightarrow (2\pi)3 Z d3k. (8.225) The free-space mode functions are then
(8.226) but now a continuous vector index k. In general, the mode functions for any situation (e.g., half-space) will be independent of V , and can be obtained by setting V −\rightarrow 1 in the expressions for the functions quantized in a finite volume. 8.8 Hamiltonian Viewpoint of Electromagnetic Gauge Freedom Finally, we will return to the Hamiltonian structure of the electromagnetic field, and examine more closely the gauge freedom that we swept under the rug by choosing a particular gauge.16 We start with the Lagrangian for the mass-free electromagnetic field, with sources, including both the vector potential A and the scalar potential \phi: L = Lfree + Lsource Lfree = ϵ0 Z d3r
Lsource = − Z
(electromagnetic Lagrangian) (8.227) 16This is essentially Dirac’s treatment of Hamiltonian dynamics with constraints, P. A. M. Dirac, ‘‘Generalized Hamiltonian dynamics,’’ Canadian Journal of Mathematics 2, 129 (1950) (doi: 10.4153/CJM-1950-012-1); but applied to the electromagnetic field, as in P. A. M. Dirac, ‘‘The Hamiltonian form of field dynamics,’’ Canadian Journal of Mathematics 3, 1 (1951) (doi: 10.4153/CJM-1951-001-2). I learned the following treatment from Tanmoy Bhattacharya (unpublished).
8.8.1 Hamiltonian¶
Chapter 8. Quantization of the Electromagnetic Field Here, we have broken the Lagrangian into the free part and a source part that represents the coupling to the source fields \rho (charge) and j (current density). We will take this Lagrangian to be the fundamental starting point. However, we can also note that the form of the Lagrangian motivates the definitions
(8.228) (electromagnetic fields) for the electric and magnetic fields. Then in terms of these fields, the simple quadratic nature of the Lagrangian is more apparent: Lfree = ϵ0 Z d3r E2 −c2B2 . (free electromagnetic Lagrangian) (8.229) We will also soon see that the electric field is again essentially the momentum field conjugate to the coordinate field A. The minus sign in the definition of the electric field simply makes the gradient of the potential agree
8.8.1 Hamiltonian To obtain the Hamiltonian for the electromagnetic field, we functionally differentiate the Lagrangian:
= 0
\delta ˙A
(8.230) (conjugate momenta) The fact that the momentum \Pi0 conjugate to \phi vanishes indicates some funniness, and will ultimately lead to constraints and the gauge freedom on the field. Now the Hamiltonian is given by H = Z d3r h
i −L. (8.231) We can drop the vanishing \Pi0 piece, and simplifying, we can write the Hamiltonian as H = Hfree + Hsource Hfree = Z d3r \Pi2 2ϵ0
Hsource = Z
(electromagnetic Hamiltonian) (8.232) Note again that while \phi appears here, the conjugate momentum does not. In fact, since we are constrained to \Pi0 = 0, the Hamiltonian is arbitrary up to a term proportional to \Pi0 anyway. Thus we should generalize the Hamiltonian so that H = Hfree + Hsource + Hgauge Hgauge = Z d3r \Pi0g, (8.233) (electromagnetic Hamiltonian) where g(\Pi0, \Pi, \phi, A; r, t) is an arbitrary function, and will represent part of the gauge freedom of the field.
8.8.2 Hamilton Equations and Gauge Freedom¶
8.8 Hamiltonian Viewpoint of Electromagnetic Gauge Freedom 8.8.2 Hamilton Equations and Gauge Freedom Now we can work out the dynamics of the field in terms of Hamilton’s equations. First we start with the equation of motion for the scalar momentum:
(8.234) However, \Pi0 = 0, so the last expression here must vanish, leading to one of the Maxwell equations:
ϵ0 . (8.235) (Gauss’ law) Note that there is no time derivative here: this is a constraint equation, not an evolution equation. We ostensibly started with an evolution equation, but constraining \Pi0 = 0 turned this into a constraint. Next, the evolution equation for the scalar potential is
\delta\Pi0
\deltag \delta\Pi0 = g. (8.236) The last equality follows by assuming the derivative of g to be finite—then the last term vanishes, since it includes a factor of \Pi0 = 0. The time derivative of \phi is thus given by a completely arbitrary function, and so \phi itself may be arbitrarily chosen. This is our first gauge freedom. A common choice is to use this freedom to set \phi = 0, which implies the choice g = 0. Note that we can already introduce a second gauge freedom,
modify the gauge Hamiltonian according to Hgauge = Z d3r h
i , (8.237) (gauge Hamiltonian) where h(\Pi0, \Pi, \phi, A; r, t) is another arbitrary function. Note that introducing the h term changes nothing that we have done so far, since we have not yet differentiated with respect to \Pi. Working out the equation of motion for \Pi,
(8.238) Converting to the usual fields, we find the next Maxwell equation
c2 ˙E + µ0j, (8.239) (Ampère’s law) noting that c2 = 1/µ0ϵ0. The other two Maxwell equations follow from the definition of the fields: \nabla \cdot B =
Faraday’s law. The equation of motion for the vector potential A is
ϵ0
(8.240) The interpretation here is that ˙A is arbitrary up to a a gradient of a function, which has the same form as
above, \phi can be modified arbitrarily, (8.241) (first gauge freedom) this second gauge freedom amounts to freedom to modify the vector potential under transformations of the form
(8.242) (second gauge freedom)
8.9.1 Fresnel Coefficients: Modes of a Planar, Dielectric Interface¶
Chapter 8. Quantization of the Electromagnetic Field In starting with four coordinate fields (\phi, A) and four momentum fields (\Pi0, \Pi), we eliminate two momenta via \Pi0 = 0 and \nabla \cdot\Pi = −\rho, and by finding that we can introduce two arbitrary fields, \phi and f, we essentially eliminate two of the coordinate fields. Thus, the remaining independent canonical coordinates—two momenta and two configuration coordinates—correspond to two independent degrees of freedom, corresponding to the two independent polarizations in the transverse fields. The longitudinal fields are counted separately, since we have implicitly introduced them via the source charge \rho. To connect this gauge freedom with the more usual treatment, we can solve (8.240) for the momentum to obtain
(8.243) This expression is the same as in Eq. (8.230), except for the presence of the gradient of h. In the Hamil- tonian formulation, this is not an inconsistency: after deriving Eq. (8.230) for the momentum field and the Hamiltonian, we have modified the Hamiltonian by introducing Hgauge, and so in this case the momentum changes to reflect this. However, to compare more directly to the Lagrangian picture, we can require that the momentum field not change. In this case, we should use the arbitrariness in the scalar potential to modify it to cancel the extra h term, which would mean setting \phi −\rightarrow \phi −h, to correspond to ˙A −\rightarrow ˙A + \nabla h. Thus, we have a combined gauge invariance under the combined transformation
(8.244) (Lagrangian gauge freedom) However, in the Hamiltonian picture, this link between the transformations is not explicitly enforced; if \phi is set to zero, then for example to model the effects of a static charge, we must still pick an appropriate form of A. 8.8.3 Continuity Constraint One last detail comes from again considering Eq. (8.234), where the right-hand-side expression \nabla \cdot \Pi + \rho vanishes, and is thus a constant of the motion. We can then differentiate it,
(8.245) where we have used Eq. (8.238) for ˙\Pi. This quantity vanishes, and we are left with
(8.246) (continuity constraint) which is the usual continuity condition for the source fields. 8.9 Quantization with Dielectric Media We have so far restricted quantization to free space, possibly with boundary conditions that modify the spatial mode functions. Here we will generalize field quantization to handle dielectrics modeled by permittivities ϵ(r). We will not handle dispersion here, though, which causes considerably more complication for general quantization. We will handle dispersion only in the linear-response regime in Chapter 14, where we will not need to explicitly quantize the field. 8.9.1 Fresnel Coefficients: Modes of a Planar, Dielectric Interface As another example of unconfined mode functions in the sense of the last section, and as a starting point for field quantization with dielectric media, we will consider the field modes in the presence of an interface between two dielectric media. We will take Maxwell’s equations for the electromagnetic fields in a medium,
8.9 Quantization with Dielectric Media but without sources, to be our starting point:
(8.247) (Maxwell’s equations) These equations are covered in more detail later (see Section 14.1, so we will use them without much discussion. Here, D is the electric flux density or electric displacement, B is the magnetic flux density, and E and H are the usual electric and magnetic fields, respectively. We will ignore magnetic effects, so that B = µ0H, (8.248) and the electric fields are related by D = ϵE, (8.249)
functions of space for an inhomogeneous optical medium. Now we will decouple the equations to derive a wave equation for the electric field. The curl of the third Maxwell equation is
(8.250) and with B = µ0H and the fourth Maxwell equation, we arrive at the wave equation
t E, (8.251) (electric-field wave equation) after decoupling the electric displacement with D = ϵE. For a homogeneous medium of constant ϵ, this
t E. (electric-field wave equation, homogeneous medium) (8.252) The homogeneous wave equation has plane-wave solutions of the form
(8.253) (plane-wave solution) provided \omega k = \sqrtµ0ϵ. (8.254) (dispersion relation/phase velocity) This is also the phase velocity of the wave, denoted c, which is commonly written in terms of the refractive index n := p ϵ/ϵ0. The last Maxwell equation implies an accompanying magnetic field of the same form, with the same phase, but orthogonal to E. More generally, the transverse nature of the plane wave solutions is apparent from the last two Maxwell equations, from which it follows that E, B, and k form a mutually orthogonal vector basis, provided the medium is homogeneous and isotropic. Further, E \times B points in the direction of k. 8.9.1.1 Boundary Conditions To analyze the electric field at the interface, we will need to consider the boundary conditions on the field at the material discontinuity at the interface. Because the interface produces only a finite discontinuity in the
Chapter 8. Quantization of the Electromagnetic Field wave equation, we will use this to enforce continuity conditions across the interface. Since we are considering the inhomogeneous part of space, we can write out Eq. (8.251) which for the monochromatic mode becomes
(8.255) It is convenient to separate this equation into normal and transverse components. The normal component is
(8.256) or since Ez = 0
(8.257) and canceling the z derivatives,
(8.258) where \nabla 2
y is the transverse Laplacian. Examining the behavior of this equation in the z direction, since ϵ only makes a finite step, we can integrate this equation from z = −\delta to \delta, letting \delta −\rightarrow 0. Only the
\nabla T \cdot E is continuous across the boundary. To obtain a second boundary condition, we return to Eq. (8.255), considering now the components parallel to the interface:
(8.259) Separating out the z derivatives, we then have
TE∥−\partial 2
(8.260) Again, we consider this to be a differential equation in z, where the only z derivatives appear in the first and fourth terms. Since the coefficients of the equation are bounded, we can assume all the terms of this equation are also bounded in the neighborhood of the interface, which holds so long as the field itself is bounded in the region near the boundary, which is a reasonable physical assumption in the absence of localized sources there. Then either \partial zEz is finitely discontinuous or it is continuous. In either case, we can integrate over a vanishingly small interval in z across the boundary, in which case we obtain
(8.261) where the \delta symbols here indicate reference the change in the following quantity over the interface. This means that the change in the derivative of E∥is bounded, which means that E∥itself is continuous across the boundary, which is our second condition. Actually, this turns out to be equivalent to the first condition, since \nabla ⊥will be equivalent to k∥at the interface, and we will show below that all parallel components of the wave vectors are the same at the boundary. Thus, we will allow the second condition to supersede the first one, keeping the simpler requirement of continuity of E∥. To obtain the other boundary condition we will need, we can return to the wave equation (8.251), and integrate around a rectangular contour in a plane normal to the (local) interface that encompasses the boundary as shown below. d l c
8.9 Quantization with Dielectric Media The width \delta of the loop will contract to zero, while we assume the length ℓof the loop is short compared to any length scale of the interface (in the case of a nonplanar interface) or field. Then integrating \nabla \times E around this contour, we can then use Stokes’ theorem and then Eq. (8.251) to find I
Z
Z (ϵ\partial 2 t E) \cdot da, (8.262) where a is the area-normal vector, and the integrals run over the area enclosed by the contour. As \delta −\rightarrow 0, the last integral vanishes, and only the segments of length ℓcontribute to the contour integral on the left- hand side, such that the contributions of the two segments cancel. Since this argument can be repeated in any plane normal to the interface it implies that the component of \nabla \times E parallel to the interface is continuous across the boundary. Thus, our second boundary condition is that (\nabla \times E)∥is continuous across the interface. It is possible to derive another boundary condition, though it is one we will not use here. Computing the divergence of Eq. (8.251) for a monochromatic field, we have
(8.263) This is a statement that according to the wave equation, ϵE must be a transverse field, or that E must be ‘‘ϵ-transverse’’ in this sense. We note that the only term involving a z derivative is \partial zϵEz, so integrating this equation over a short interval across the interface in the z direction shows that ϵEz must be continuous across the boundary. In summary, we have E∥is continuous, (\nabla \times E)∥is continuous (8.264) (boundary conditions) for the boundary conditions at the interface for the solution E of the vector wave equation. Note that in this form, the boundary conditions only refer to the electric field itself, and is thus more ‘‘self-contained’’ than boundary conditions that are typically written for both E and H. (Though note that B satisfies the same boundary conditions, since it satisfies the same wave equation as does E.) 8.9.1.2 TE Modes Now we will consider the field modes in the presence of a dielectric interface, where the dielectric covers z < 0 and z > 0 is vacuum. Since the permeability is constant in each region, the solutions are plane waves (or superpositions of plane waves) in each region. First, we will make the ansatz of a solution that consists of three parts: an incident plane wave and a reflected plane wave on the vacuum side, and a transmitted wave on the dielectric side. We will take all of the propagation vectors to lie in the same plane. The transverse electric field also has two independent components, which we will treat separately. The two polarizations that respect the symmetry of the surface are TE (transverse electric, meaning that the electric field is parallel to the interface) and TM (transverse magnetic, with the magnetic field parallel to the interface) polarizations. First, we will consider TE polarization, and sketch our ansatz below.
Chapter 8. Quantization of the Electromagnetic Field c z = 0 qi qr qt E0i (+) H0i (+) E0r (+) H0r (+) E0t (+) H0t (+) Although we will refer only to the electric field in the following, we have sketched in the proper relative orien- tations of the magnetic fields relative to the (electric-field) polarization vectors. Writing out our assumptions mathematically, we have the monochromatic TE field mode E(+)
0i ˆ\epsilonk,TE eik\cdotr\Theta(z) + rTEeikr\cdotr\Theta(z) + tTEeikt\cdotr\Theta(−z) , (8.265) where k, kr, and kt are the incident, reflected, and transmitted wave vectors, E0i, E0r, and E0t are the respective field amplitudes, and we have defined rTE := E0r/E0i and tTE := E0t/E0i as the field reflection and transmission coefficients, respectively. A common time dependence of exp(−i\omegat) is implied for all the terms. But while \omega is the same on both sides of the boundary, in view of the dispersion relation (8.254), k should be viewed as a function of ϵ (k ∝\sqrtϵ ∝n). The first boundary condition (8.264), continuity of the surface-parallel component E∥, gives simply for the mode (8.265) eik∥\cdotr∥+ rTEeikr∥\cdotr∥= tTEeikt∥\cdotr∥, (8.266) since the polarizations are all already transverse. Here, the notation r⊥refers to only the transverse part of
every point on the interfaces is for the transverse parts of all the wave vectors themselves to be separately equal,
(8.267) The first condition here leads to kr = k−, (8.268) where k−= kxˆx + kyˆy −kzˆz, since we assumed the incident and reflected waves to propagate toward and away from the interface, respectively. This in turn leads to the usual reflection law \thetai = \thetar. The other equality in the phase condition, k∥= kt∥, (8.269) can be rewritten as ki sin \thetai = kt sin \thetat, (8.270) or more familiarly as Snell’s law, ni sin \thetai = nt sin \thetat, (8.271) if we write the relation in terms of the refractive index. The first boundary condition (8.266) then becomes
(8.272) The second of the boundary conditions (8.264), i.e., enforcing the continuity of (\nabla \times E)∥, becomes
(8.273)
8.9 Quantization with Dielectric Media or working out the cross products,¶
(8.274) Eqs. (8.272) and (8.274) have the solutions rTE = kz −kt,z kz + kt,z , tTE = 2kz kz + kt,z . (8.275) (TE Fresnel coefficients) Another useful representation of these coefficients comes from noting that kz = k cos \thetai and kt,z = kt cos \thetat = k p
q 1 −sin2 \thetat = k p
s 1 −
= k q
= k p
(8.276) so that
p
p
tTE = 2\xi
p
(8.277) (TE Fresnel coefficients) where \xi = cos \thetai. Still another form comes from writing out the angles more explicitly, and using ki = n1k0 and kt = n2k0, where k0 is the vacuum wave number, and n1 and n2 are the refractive indices on the left- and right-and sides of the interface, respectively, so that rTE = n1 cos \thetai −n2 cos \thetat n1 cos \thetai + n2 cos \thetat , tTE = 2n1 cos \thetai n1 cos \thetai + n2 cos \thetat . (TE Fresnel coefficients) (8.278) (While we are nominally assuming a vacuum on one side of the interface, the k-vector treatment is actually more general than this.) Fixing these coefficients also fixes the form of the mode ansatz (8.265). We can then define a normalized mode function fk,TE = ˆ\epsilonk,TE eik\cdotr\Theta(z) + rTEeikr\cdotr\Theta(z) + tTEeikt\cdotr\Theta(−z) . (normalized mode function) (8.279) To work out the normalization of this mode function, first we write out Z d3r ϵ(r) ϵ0 f∗ k,TE(r) \cdot fk′,TE(r) = Z d3r ei(k′−k)\cdotr\Theta(z) + |rTE|2ei(k′
h rTEei(k′−kr)\cdotr + r∗ TEei(k−k′ r)\cdotri \Theta(z) + ϵ ϵ0 |tTE|2ei(k′ t−kt)\cdotr\Theta(−z) , (8.280) assuming the two waves are incident from the vacuum side. If we consider the real part of this equation, the conjugate terms are equivalent to the same terms already present, but with \Theta(z) −\rightarrow \Theta(−z). Thus, we can
Chapter 8. Quantization of the Electromagnetic Field carry out the integrals to obtain \delta functions, with the result 2Re Z d3r ϵ(r) ϵ0 f∗ k,TE(r) \cdot fk′,TE(r)
rTE\delta3(k′ −kr) + r∗ TE\delta3(k −k′ r) + ϵ ϵ0
t −kt)
r −kr) + ϵ ϵ0
t −kt), (8.281) where we have removed the \delta functions that are always zero due to the relative orientations of incident and reflected waves, under the assumption that both modes are incident from the same direction. To simplify this expression further, we should write the wave vectors in the last term in terms of the incident wave vector. This procedure is slightly involved, and goes as follows.17 First, recalling that k 2 t = ϵ ϵ0 k2 (8.282) and from Snell’s law, kt∥= k∥, (8.283) then we will now relate the z components of the incident and transmitted wave vectors. The incident version satisfies k 2 z = k2 −k2 ∥, (8.284) while the transmitted version satisfies k 2 t,z = k 2 t −kt ∥= ϵ ϵ0 k2 −k2 ∥. (8.285) Rearranging and dividing these last two equations gives k 2 t,z + k2 ∥ k 2 z + k2 ∥ = ϵ ϵ0 . (8.286) If we instead introduce a subtraction with these same two relations in terms of k′ before division, we obtain (k′ 2 t,z −k 2 t,z) + (k′2 ∥−k2 ∥) (k′ 2 z −k 2 z ) + (k′2 ∥−k2 ∥) = ϵ ϵ0 . (8.287) Then using this relation, we have the relation \delta3(k′
t∥−kt∥) \delta(k′ t,z −kt,z)
t∥−kt∥) \delta[(k′ t,z −kt,z)(k′ t,z + kt,z)]
t∥−kt∥) \delta(k′ 2 t,z −k 2 t,z) = 2ϵ0 ϵ |kt,z|\delta2(k′ t∥−kt∥) \delta(k′ 2 z −k 2 z ) = ϵ0 ϵ
kt,z kz \delta2(k′ t∥−kt∥) \delta(k′ z −kz) = ϵ0 ϵ
kt,z kz \delta2(k′ ∥−k∥) \delta(k′ z −kz) = ϵ0 ϵ kt,z kz \delta3(k′ −k), (8.288) 17C. K. Carniglia and L. Mandel, ‘‘Quantization of Evanescent Electromagnetic Waves,’’ Physical Review D 3, 280 (1971) (doi: 10.1103/PhysRevD.3.280).
8.9 Quantization with Dielectric Media where in the last step we noted that kt,z and kz have the same sign by construction. Eq. (8.281) then becomes 2Re Z d3r ϵ(r) ϵ0 f∗
r −kr) + kt,z kz
(8.289) To simplify this even further, we can multiply Eq. (8.272) with Eq. (8.274) to give 1 = r 2 TE + kt,z kz t 2 TE =: RTE + TTE. (8.290) (reflectance and transmittance) The interpretation here regarding complex-valued Fresnel coefficients is a bit subtle. Briefly, rTE is real unless its modulus is unity. In the former case, tTE is real. In the latter case, the transmitted wave is evanescent (decaying with z), TTE is defined to vanish, and the tTE wave is normalizable, so it does not contribute to the orthonormality relation. which we can interpret as conservation of energy: R and T here are the reflection and transmission coefficients of the intensity, respectively called the reflectance and transmittance. In this case, Eq. (8.289) reduces to Re Z d3r ϵ(r) ϵ0 f∗
(8.291) Repeating the above argument for the imaginary part is simple: while the real part gave integrals over cosines leading to \delta functions, the imaginary part gives corresponding integrals over sines, which simply vanish. Therefore, our orthonormality relation is Z d3r ϵ(r) ϵ0 f∗
(8.292) (orthonormality relation) Note that for ϵ = ϵ0, this reduces to the relation we expect for free-space mode functions. Technically, we need to verify this relation in a couple more situations. First, we should consider the case where both waves are incident from the dielectric side. This goes through in the same way as we have shown. (Though note that evanescent waves are possible in this case, in which case kt becomes complex.) Also, we should verify the case where one wave is incident from each side. In this case, Eq. (8.280) becomes Z d3r ϵ(r) ϵ0 f∗ k,TE(r) \cdot fk′,TE(r) = Z d3r tTEei(k′
TEtTEei(k′ t−kr)\cdotr\Theta(z) + ϵ ϵ0 t∗ TEei(k′−kt)\cdotr\Theta(−z) + ϵ ϵ0 rTEt∗ TEei(k′ r−kt)\cdotr\Theta(−z)
tTE\delta3(k′ t −k) + r∗ TEtTE\delta3(k′ t −kr) + ϵ ϵ0 t∗
ϵ0 rTEt∗ TE\delta3(k′ r −kt)
r∗ TEtTE\delta3(k′ t −kr) + ϵ ϵ0 rTEt∗ TE\delta3(k′ r −kt) , (8.293) where we have assumed k to be incident from the vacuum side, and k′ to be incident from the dielectric side. We have also dropped the interaction terms between incident and transmitted waves that always vanish. Now we must also be careful with the interpretation of the reflection and transmitted coefficients. The conjugated coefficients are incident from the vacuum side, whereas the normal ones are incident from the dielectric side. From Eq. (8.275), this means in particular that r∗ TE = −rTE and t∗
terms here cancel by applying Eq. (8.288) to the second \delta function. Thus, oppositely-propagating modes are always orthogonal.
Chapter 8. Quantization of the Electromagnetic Field 8.9.1.3 TM Modes The other (TM) polarization follows in essentially the same wave as for the TE polarization. We will again sketch our ansatz and sign convention for the fields below, again recalling that the dielectric covers the region z < 0. c z = 0 qi qr qt E0i (+) H0i (+) H0r (+) E0r (+) E0t (+) H0t (+) Again, we will refer only to the electric field in the following, but we have sketched in the proper relative orientations of the magnetic fields relative to the (electric-field) polarization vectors. Mathematically, we have the monochromatic TM field mode E(+)
0i h (ˆ\epsilonk,TE \times ˆk) eik\cdotr\Theta(z) + (ˆ\epsilonk,TE \times ˆkr) rTMeikr\cdotr\Theta(z) + (ˆ\epsilonk,TE \times ˆkt) tTMeikt\cdotr\Theta(−z) i . (8.294) Clearly, this mode is orthogonal to any TM mode. Applying the first of the boundary conditions (8.264) again implies the phase conditions (8.267)-(8.271). For the vector part of the mode, continuity of E∥implies
(8.295) [Note the similarity to Eq. (8.273) here.] Working out the cross products, we have kz
kt tTM. (8.296) The second boundary condition, continuity of (\nabla \times E)∥gives
(8.297) so that working out the cross products and dividing through by k gives 1 −rTM = kt k tTM. (8.298) Eqs. (8.296) and (8.298) have the solutions rTM = k2kt,z −k 2 t kz k2kt,z + k 2 t kz , tTM = 2kktkz k2kt,z + k 2 t kz . (8.299) (TM Fresnel coefficients) which is the general form for a dielectric–dielectric interface, or using kt/k = \sqrtϵ for a vacuum-dielectric interface, rTM = kt,z −ϵkz kt,z −ϵkz , tTM = 2\sqrtϵkz kt,z + ϵkz . (8.300) (TM Fresnel coefficients)
8.9.2 Quantum Fields at a Planar Dielectric Interface¶
8.9 Quantization with Dielectric Media¶
rTM = p
p
, tTM =
p
. (TM Fresnel coefficients) (8.301) Additionally, we can write rTM = n1 cos \thetat −n2 cos \thetai n1 cos \thetat + n2 cos \thetai , tTM = 2n1 cos \thetai n1 cos \thetat + n2 cos \thetai (TM Fresnel coefficients) (8.302) for a dielectric–dielectric interface with refractive indices n1 and n2 on the incident and transmitted side, respectively. Now if we multiply together Eqs. (8.296) and (8.298), we obtain 1 = r 2 TM + kt,z kz t 2 TM =: RTM + TTM, (8.303) (reflectance and transmittance)
reflectance and transmittance for intensities has the same definition in terms of the respective reflection and transmission coefficients as for TE polarization. Then defining the mode function
(normalized mode function) (8.304) and noting that the phase behavior is the same as for the TE modes, we see that the orthonormality argument for the TE case goes through here as well. Furthermore, by construction the TE modes are orthogonal to the TM modes, so we have the general orthonormality relation Z d3r ϵ(r) ϵ0 f∗
(8.305) (orthonormality relation) where \zeta is either TE or TM. 8.9.2 Quantum Fields at a Planar Dielectric Interface With the mode functions in hand, we can establish the form for the quantized fields, just as we did in free space. The derivation proceeds mostly in the same way as before, but with a few modifications.18 8.9.2.1 Hamiltonian Structure The free-space Lagrangian (8.28) had the form L = 1 Z d3r ϵ0E2 −1 µ0 B2 (8.306) in terms of the electromagnetic fields, and has the obvious generalization L = 1 Z d3r ϵ(r)E2 −1 µ0 B2 (dielectric electromagnetic Lagrangian) (8.307) 18Here we are following the lucid treatment of Roy J. Glauber and M. Lewenstein, ‘‘Quantum optics of dielectric media,’’ Physical Review A 43, 467 (1991), Eq. (2.16a) (doi: 10.1103/PhysRevA.43.467).
Chapter 8. Quantization of the Electromagnetic Field in the presence of a dielectric. Again taking A to be the generalized coordinate, the conjugate momentum is := \deltaL
(8.308) (conjugate momentum to A) This leads to the Hamiltonian H = Z d3r \Pi2 2ϵ(r) + 2µ0
= 1 Z d3r ϵ(r)E2 + 1 µ0 B2 , (dielectric electromagnetic Hamiltonian) (8.309)
vector potential, this gives the wave equation
ϵ0c2 \partial 2 t A = 0. (8.310) (dielectric wave equation)
(8.311) (generalized Coulomb gauge) although the Maxwell equation is gauge-invariant. More fundamentally, this transverse-field condition arises in the form
(8.312) (dielectric wave equation) by introducing the scalar potential \phi, and considering its Hamilton equation, as in Eq. (8.234). The second
follows as before from differentiating this definition and the definition E = −\partial tA. 8.9.2.2 Mode Functions As before, we can separate variables in the wave equation by assuming a harmonic time dependence
(8.313) such that the mode functions satisfy the wave equation (8.310) in the form
ϵ0c2 \omega2f(r), (8.314) (generalized Helmholtz equation) and the mode functions are constrained by Eq. (8.312) such that
(8.315) (mode-function transverse constraint) We will assume the mode functions to be normalized such that Z d3r ϵ(r) ϵ0
(8.316) This is consistent with the orthonormality relation (8.305) for a planar dielectric interface. In the multimode case, note that if we choose
p ϵ(r) g(r), (8.317)
8.9 Quantization with Dielectric Media the wave equation (8.314) becomes p ϵ(r)¶
p ϵ(r)
ϵ0c2 g(r), (8.318) in which case g(r) is an eigenfunction of a Hermitian operator. Then we can assume the full set of solutions gk,\zeta(r) to be orthonormal, Z d3r gk,\zeta(r) g∗
kk′, (8.319) in which case we have the orthonormality relation Z d3r ϵ(r) ϵ0 fk,\zeta(r) f∗
kk′, (orthonormality of mode functions) (8.320) which again is consistent with the special case (8.305) when the k sum is compared to an integral. 8.9.2.3 Quantized Fields In the more general separation of variables, we can write the (classical) multimode field as
\sqrtϵ0 X k,\zeta
(8.321)
second term in the Hamiltonian (8.309) then reduces to 2µ0 Z
2µ0 Z
= 2µ0ϵ0 X k,\zeta X k′,\zeta′ qk,\zetaq∗ k′,\zeta′ Z
k′,\zeta′ = 2µ0ϵ0 X k,\zeta X k′,\zeta′ qk,\zetaq∗ k′,\zeta′ Z d3r ϵ ϵ0c2 \omega2 k′fk,\zetaf∗ k′,\zeta′ = 1 X k,\zeta X k′,\zeta′ \omega2 k′qk,\zetaq∗ k′,\zeta′ Z d3r ϵ ϵ0 fk,\zetaf∗ k′,\zeta′ = 1 2\omega2 kq 2 k,\zeta, (8.322) where we used Eqs. (8.314) and (8.320). Similarly, we can write the conjugate field as a superposition of mode functions,
\sqrtϵ0 X k,\zeta
(8.323)
ϵ(r) here is necessary to satisfy the constraint (8.312). The first term in the Hamiltonian (8.309) then reduces to Z d3r \Pi2
X k,\zeta X k′,\zeta′ pk,\zetap∗ k′,\zeta′ Z d3r ϵ ϵ0 fk,\zetaf∗ k′,\zeta′ = 1 2p 2 k,\zeta. (8.324) Thus, the total electromagnetic Hamiltonian becomes H = X k,\zeta
p 2 k,\zeta + 1 2\omega2 kq 2 k,\zeta ! , (8.325)
8.9.3 Transverse and Longitudinal Fields: Dielectric¶
Chapter 8. Quantization of the Electromagnetic Field which has the form of a set of uncoupled (complex) harmonic oscillators of unit mass and frequencies \omegak. Then quantization of the field amounts to promoting the canonical normal coordinates to operators, with the usual relations (for unit mass):
r ¯h 2\omegak
k,\zeta
r ¯h\omegak
k,\zeta i ! . (8.326) With these operator forms, the conjugate fields (8.321) and (8.323) become
X k,\zeta r ¯h 2\omegakϵ0 fk,\zeta(r)
k,\zeta
X k,\zeta r ¯h\omegak 2ϵ0 fk,\zeta(r)
k,\zeta , (8.327) Since the fields are real, but the mode functions in general are not, any complex mode must be accompanied by its complex conjugate as another mode. Thus, by rearranging terms among the modes, we can alternately write our fields as
X k,\zeta i r ¯h 2\omegakϵ0
X k,\zeta r ¯h\omegak 2ϵ0
(8.328) We have also shifted our phase convention by letting ak,\zeta −\rightarrow iak,\zeta. With these changes, the A field has the same form as in free space, Eq. (8.61), and the momentum field is consistent with the electric field from before via Eq. (8.308). Thus, the quantized fields in the presence of an inhomogeneous dielectric, have the same form as before,
X k,\zeta i r ¯h 2\omegakϵ0
X k,\zeta − r ¯h\omegak 2ϵ0
X k,\zeta i r ¯h 2\omegakϵ0
(quantized fields, with dielectric) (8.329) except that the mode functions are modified by the presence of dielectric, as we saw in the dielectric-interface example of Section 8.9.1. 8.9.3 Transverse and Longitudinal Fields: Dielectric In the free-space case, it was useful to introduce the concept of transverse and longitudinal fields, for example, to represent commutators of the quantum fields. It is still useful to do this with an inhomogeneous dielectric, but the definitions must be modified somewhat. 8.9.3.1 Scalar Green Function First, a small digression to recall a tool we will need to generalize the notation for the transverse and longitudinal projection operators. From the Maxwell equation \nabla \cdot E = \rho/ϵ0 and the potential relation
8.9 Quantization with Dielectric Media E = −\partial tA −\nabla \phi = −\nabla \phi for a static field, we have the Poisson equation
ϵ0 . (8.330) For a point charge q at r0,
(8.331) the well-known potential in free space is
q 4\piϵ0 Z d3r′ |r −r0|. (8.332) The scalar Green function G0(r, r′) is the solution to the Poisson equation (8.330) when the charge density \rho(r)/ϵ0 is replaced by a unit point charge at r′:
(scalar Green function, defining equation) (8.333) By comparing this expression to the solution (8.332), we can identify the free-space scalar Green function
4\pi|r −r′|. (8.334) (scalar Green function, free space) Multiplying through by \rho(r′) and integrating in Eq. (8.333), and then comparing to Eq. (8.330) shows that the solution for a general charge density is
ϵ0 Z d3r′ G0(r, r′) \rho(r′). (solution in terms of Green function) (8.335) Note in particular that we recover the solution (8.332) in the case of a point charge (8.331). 8.9.3.2 Transverse and Longitudinal Fields: Free Space Recall [Section 8.5] that according to the Helmholtz theorem, any vector field can be decomposed into transverse and longitudinal parts,
(8.336)
according to these properties,
(8.337) since the transverse part vanishes under the divergence operator. We can formally invert this equation as
(8.338) (longitudinal projection operator) where the vector integral operator \nabla −1 is defined such that this reproduces Eq. (8.337) under the divergence operator. Hence, we see that \nabla −1\nabla is a representation of the longitudinal projection operator. To give meaning to the \nabla −1 operator, note that we can similarly rewrite Eq. (8.333) as
(8.339) If we multiply through by a scalar field f(r′) and integrate with respect to r′,
Z d3r′ G0(r, r′) f(r′), (8.340) (integral representation of \nabla −1)
Chapter 8. Quantization of the Electromagnetic Field which serves to explicitly define the action of \nabla −1. Now suppose instead that we multiply through Eq. (8.339) by \nabla \cdot C(r′), and then integrate over r′:
Z
Z d3r′ G0(r, r′)\nabla ′ \cdot C(r′). (8.341) Using Eq. (8.338) on the left-hand side and Eq. (8.334) on the right-hand side, we obtain
Z
4\pi|r −r′| . (8.342) This is the integral representation of the longitudinal field that we derived before in Eq. (8.156). Because we have the decomposition (8.336), we can then write the transverse field as
(8.343) or
\cdot C(r), (8.344) (transverse projection operator) where here the ‘‘1‘‘ is a dyadic identity operator. Now since we have the defining relations [Eqs. (8.169) and (8.170)] C⊥
Z d3r′\delta⊥
C∥
Z d3r′\delta∥
(8.345) for the transverse and longitudinal \delta functions, we can put Eq. (8.341) into index notation to obtain C∥
Z d3r′ G0(r, r′)\partial ′
= Z
(8.346) and compare the result to the defining relation to yield \delta∥
\betaG0(r, r′). (8.347) (longitudinal \delta function) The derivatives can then be expanded to give the longitudinal \delta function in the position representation [cf. Eqs. (8.182) and (8.184). Similarly, for the transverse projector, we have \delta⊥
\betaG0(r, r′) (8.348) (transverse \delta function) in terms of the free-space scalar Green function. 8.9.3.3 Scalar Green Function: Dielectric Case
leads to the generalization
(8.349) of the Poisson equation. In analogy with the free-space case, we can then define the inhomogeneous scalar Green function by replacing the charge with a \delta function:
(scalar Green function, defining equation) (8.350) where we have maintained the same dimensions as in the free-space case in this replacement. For simple dielectric geometries, the solution can be evaluated for example by the method of images. However, this equation cannot be solved analytically for arbitrary ϵ(r). In any case, the solution for a general charge distribution is still given by Eq. (8.335).
8.9 Quantization with Dielectric Media 8.9.3.4 Dielectric Transverse and Longitudinal Fields: Green-Function Form Even in the presence of an inhomogeneous dielectric ϵ(r), we can decompose a field into transverse and longitudinal parts as in Section 8.9.3.2, since in general the field has nothing to do with the dielectric. However, there is also another sense in which we can decompose fields when an inhomogeneous measure ϵ(r) is available. For example, in the absence of charge, Maxwell’s equations state that \nabla \cdotD = 0, so that D = D⊥ is purely transverse. But in this case, E = D/ϵ is not purely transverse. Then what is the decomposition of E that gives rise to the transverse and longitudinal parts of D? Another example is in the choice of a generalized Coulomb gauge \nabla \cdot ϵA = 0. In this case, \nabla \cdot A̸ = 0 in general. Then what is the decomposition of A into components, such that one of them vanishes in the sense of \nabla \cdot ϵA = 0? Sticking to the gauge example, suppose that we make the (unique) decomposition
(8.351) (ϵ-Helmholtz decomposition) such that \nabla \cdot ϵAϵ⊥= 0 and \nabla \times ϵAϵ∥= 0. Our goal here is to find the projection operators corresponding to these ‘‘ϵ-transverse’’ and ‘‘ϵ-longitudinal’’ components. We can begin by noting that to make this de- composition, we can simply decompose ϵA in the usual way, so that our constraints are clearly satisfied by choosing Aϵ⊥= ϵ−1[ϵA]⊥, Aϵ∥= ϵ−1[ϵA]∥. (8.352) Taking the longitudinal case first, we can use Eq. (8.338) to obtain
Z d3r′ G0(r, r′) \nabla ′ \cdot [ϵA(r′)], (8.353) where we used Eq. (8.340) for \nabla −1. Now multiply through by ϵ(r) and compute the divergence. Note that we lose no information in doing so, since the vector fields are longitudinal on both sides of the equality:
Z d3r′ G0(r, r′) \nabla ′ \cdot [ϵA(r′)]. (8.354) Then we can combine Eqs. (8.333) and (8.350) to obtain
(8.355) so that
Z d3r′ G(r, r′) \nabla ′ \cdot [ϵA(r′)]. (8.356) Undoing the overall divergence operator (as we justified above), and dividing through by ϵ(r) gives
0 \nabla Z d3r′ G(r, r′) \nabla ′ \cdot [ϵA(r′)]. (8.357) Writing this out in components, Aϵ∥
Z d3r′ G(r, r′) \partial ′
= ϵ−1 Z
(8.358) Therefore, we can define the ϵ-longitudinal projection Aϵ∥
Z d3r′ ϵ(r′) ϵ0
(8.359) (ϵ-longitudinal projection)
Chapter 8. Quantization of the Electromagnetic Field where \deltaϵ∥
\betaG(r, r′). (8.360) (ϵ-longitudinal \delta function)
\deltaϵ∥(r −r′). Note also that the projection and the \delta function reduce to their free-space counterparts when ϵ −\rightarrow ϵ0. Similarly, the ϵ-transverse projector is given by subtracting the ϵ-longitudinal projector from the iden- tity, \deltaϵ⊥
\betaG(r, r′). (8.361) (ϵ-transverse \delta function) including the correct measure on the identity such that in the projection integral, we have19 Aϵ⊥
Z d3r′ ϵ(r′) ϵ0 \deltaϵ⊥
(8.362) (ϵ-transverse projection) We may verify the proper action of the ϵ-transverse projector simply as Z d3r ϵ(r′) ϵ0 \deltaϵ⊥
Z d3r′ ϵ(r′) ϵ0 ϵ0
\betaG(r, r′) A\beta(r′)
\alpha (r) = Aϵ⊥ \alpha (r), (8.363) using the ϵ-decomposition (8.351). 8.9.3.5 Completeness The other route to representing the projection operators20 is to recall that the functions g(r) are eigenfunc- tions of a Hermitian operator in Eq. (8.318). We can then choose a set of orthonormal functions via (8.319), and in doing so, we can sum over elements of the form gk,\zeta,\alpha(r) g∗ k,\zeta,\beta to represent the identity operator. Normally, for a set of complete, vector-valued functions on R3, the identity operator would be \delta\alpha\beta\delta3(r −r′).
identity on this restricted subspace. Transforming back to the f(r) functions, we can therefore write the identity as \deltaϵ⊥
X k,\zeta
(8.364) (completeness relation)
this projection operator, via the decomposition (8.351). 19For an in-depth discussion of the projector in for a half-space dielectric, see Robert Zietal, Quantum Electrodynamics Near Material Boundaries, Ph.D. thesis, University of Sussex (2010), Chapter 4. 20Roy J. Glauber and M. Lewenstein, op. cit.
8.10 Exercises 8.10 Exercises Problem 8.1 Geometric optics, or ray optics, can be formulated in terms of the action principle (Fermat’s principle): \delta Z
(8.365)
system. Take the coordinate z to be the ‘‘time’’ variable and the coordinate y to be the position coordinate. Let’s simplify things and consider only the two-dimensional case, so x is an ignorable coordinate, and note that z is also ignorable in the sense of being completely determined by x, y, and s. (a) Draw an analogy to classical mechanics, and write down the ray-optics Lagrangian. Then show that the conjugate momentum p for the generalized coordinate y is n dy/ds. Finally, write down the ray-optics Hamiltonian, which you should write in terms of the canonical coordinates p and y, but not y′. (b) Make the paraxial approximation (small p, small y), and assume that the refractive index may be written as a small variation on a large baseline, n = n0 + \deltan, where \deltan/n0 ≪1. Keep only lowest- order terms in p, y, and \deltan (dropping higher-order cross-terms in these variables), and show that the Hamiltonian takes the form of a classical particle Hamiltonian, with effective mass n0 and potential −\deltay. Problem 8.2 The usual Euler–Lagrange equation, \partial L \partial q −d dt \partial L
(8.366) applies to Lagrangians of the form L(q, ˙q; t). (a) Generalize the Euler–Lagrange equation to handle Lagrangians of the form L(q, ˙q, ¨q, ...q ; t). Indicate any conditions you impose on the endpoints of the variation. (b) One might hope to write down a Lagrangian for the Abraham–Lorentz force (F ∝...x) by considering Lagrangians of the form L = 1
(8.367) Use your result from part (a) to write down the equation of motion for this Lagrangian. Problem 8.3 Consider a thin string of linear mass density µ, stretched under tension T0 nearly along the x-axis between positions x1 and x2. Consider only small (i.e., linear) vibrations of this string, so that the wave function y(x, t) is always much smaller than the string’s length. (a) Derive the following Lagrangian for the string:
Z x2 x1 µy 2 t −T0y 2 x dx. (8.368) To do this, take a small segment of string of length dℓ, and compute its kinetic energy, integrating the result to get the total kinetic energy. Then compute the potential energy by computing the length of the string, and then considering what this means in terms of the energy. Assume that both y and yx are small, and ignore any longitudinal motion of the segment. (b) Noting that the Lagrangian has the form
Z x2 x1 f(y, yt, yx) dx, (8.369)
Chapter 8. Quantization of the Electromagnetic Field derive a suitably generalized Euler–Lagrange equation for the action principle \deltaS = 0 in terms of the integrand f. Then use your result to derive the wave equation for the string. Problem 8.4 Compute the functional derivative of S = Z d3r Z dt
, (8.370) with respect to the fields \phi∗(r, t) and \phi(r, t), ignoring surface terms. What are the equations of motion obtained from the action principles \deltaS/\delta\phi = 0 and \deltaS/\delta\phi∗= 0? What is the canonically conjugate momentum field to \phi? To \phi∗? Problem 8.5 In each of the following, you may ignore surface terms. (a) Compute the functional derivative of S = Z dt f(t) f (201)(t), (8.371) where f (n)(t) is the nth derivative of f(t). (b) Compute the functional derivative of S = Z dt f (199)(t) f (201)(t). (8.372) Problem 8.6 Consider the functional F[a, b] = Z 1 −1 dx f(a, b), (8.373) where
ab2 a2 + b2 otherwise (8.374)
this functional?21 Problem 8.7 (a) Show that \delta⊥
(b) Show that \delta⊥
Problem 8.8 The electric field due to an oscillating dipole at frequency \omega = ck has the somewhat messy form
4\piϵ0
1 r3 −i k r2
r
(8.375) where ˆr is a unit vector in the r direction, and ˆ\epsilon is a unit vector marking the dipole orientation. (The plus superscripts here indicate an implied time dependence of e−i\omegat.) 21This problem stolen from P. J. Morrison, ‘‘Hamiltonian description of the ideal fluid,’’ Reviews of Modern Physics 70, 467 (1998) (doi: 10.1103/RevModPhys.70.467).
8.10 Exercises A naïve guess for the radiation field is to adapt the scalar spherical wave eikr/r, tacking on the orientation and magnitude of the dipole moment to make a vector field (we can also tack on a factor of 4\piϵ0 for good measure): E(+)
4\piϵ0
r . (8.376) Obviously this is wrong. However, show that the correct dipole field arises by using the transverse
E(+) \alpha
guess,\beta. (8.377) Note that we associate this operator with transverse projection by examining the transverse delta function: \delta⊥
(2\pi)3 Z d3k
k2
(2\pi)3 Z d3k 1 k2
eik\cdotr. (8.378) Thus, up to a factor of 1/k2 the transverse delta function is a Fourier transform of the projection
Problem 8.9 Following the steps in the notes, show that the transverse and longitudinal delta functions, \delta⊥
(2\pi)3 Z d3k
k2 eik\cdotr \delta∥
(2\pi)3 Z
k2 eik\cdotr, (8.379) can be expressed as \delta⊥
4\pir3
r2 \delta∥
4\pir3
r2 (8.380) in the position representation. Problem 8.10 Starting with the general relation
ϵ0 X k,\zeta Re h
, (8.381) follow the notes and derive the following commutator in half space:
ϵ0 \delta⊥
. (8.382) Again, the interpretation is that the vector potential and electric field are ‘‘connected’’ at two spacetime points if they lie on the same light cone with respect to paths that are either direct or have one bounce off the mirror. Problem 8.11 Show by using the momentum-space representation of the transverse delta function that Z d3r \delta⊥
(8.383)
Chapter 8. Quantization of the Electromagnetic Field Problem 8.12 The quantum coherence functions g(n)(\tau) are defined in terms of the quantum fields in the same way we defined them for the classical counterparts, except that the classical time average is replaced by an expectation value with respect to the state of the field. (a) Derive expressions for g(1)(\tau) and g(2)(\tau) for a single mode of the electromagnetic field. (b) Evaluate your expressions from part (a) for a field in a number state (Fock state) |n\rangle . Show that the first-order coherence can be interpreted classically, but that this state is nonclassical at the second order of coherence. Problem 8.13 Consider the squeezed vacuum state
(8.384) of the harmonic oscillator (or a single field mode), where |0\rangle is the vacuum state, and S(\zeta) is the squeezing operator22
1
. (8.385) Note that the squeezing operator reduces to the identity for \zeta = 0. Like the vacuum state, the squeezed vacuum is a minimum-uncertainty Gaussian state, but with a different set of variances (the vacuum state is literally stretched or ‘‘squeezed’’ into a different Gaussian, keeping the uncertainty product constant in some basis). Compute the initial value of the second-order correlation function
(8.386) analogous to the one we studied to find antibunching in the resonance fluorescence in the two-level atom. From this result, what can you conclude about the classicality of the squeezed vacuum? Is there anything physically funny about your solution in the limit \zeta −\rightarrow 0? You may use without proof the transformation rules
(8.387) for the ladder operators, where \zeta = rei\theta. Problem 8.14 Given the solution \psi(r) to the scalar Helmholtz equation
(8.388) show that the vector fields
N = 1
(8.389) satisfy the vector Helmholtz equation
(8.390) 22See David Stoler, ‘‘Equivalence Classes of Minimum Uncertainty Packets,’’ Physical Review D 1, 3217 (1970) (doi: 10.1103/PhysRevD.1.3217); Carlton M. Caves, ‘‘Quantum-mechanical noise in an interferometer,’’ Physical Review D 23, 1693 (1981) (doi: 10.1103/PhysRevD.23.1693).
8.10 Exercises Problem 8.15 In this problem you will work out the normalization factors for the spherical-cavity modes. (a) Noting that the ordinary Bessel function Jn(x) satisfies Bessel’s equation x\partial x [xJ′ n(x)] + 1 −n2 x2
(8.391) show that Z R dr rJ2
[J′ n(kR)]2 −Jn(kR)J′′ n(kR) −1 kRJn(kR)J′ n(kR) . (8.392) Do this by multiplying Bessel’s equation by rJn(k′r), then switching k′ \leftarrow \rightarrow k and subtracting the resulting equations. Then integrate by parts and let k′ −\rightarrow k, being careful to keep the lowest-order nonvanishing terms in k′ −k. (b) Use the result of part (a) to show that the normalization integral from Eq. (8.98) or (8.113) is N −2 nl = Z R dr r2 j 2
2 [j′
2 j 2 l+1(knlR), (8.393) for modes subject to the radial boundary condition (8.112)
(8.394) as is appropriate for a perfectly conducting spherical cavity for scalar or TE vector waves. (c) Show that the same normalization integral, subject to the radial boundary condition (8.120) \partial r r jl(knlr)
(8.395) as is appropriate for a perfectly conducting spherical cavity for TM vector waves, becomes N −2 nl = Z R dr r2 j 2
1 −l(l + 1) k 2 nlR2 j 2 l (kR). (8.396) To start, it will help to use Bessel’s equation again to eliminate the J′′ n(x) in the result from part (a). Problem 8.16 Work out the Hamiltonian structure of the massive electromagnetic field (Proca field), paralleling the massless treatment of Section 8.8. Use as your Lagrangian L = Lfree + Lsource + Lmass Lfree = ϵ0 Z d3r
Lsource = − Z
Lmass = −ϵ0 m2c4 ¯h2 Z d3r A2 −\phi2 c2 , (8.397) where Lfree and Lsource are defined as before, and notice that the particle mass m enters with the proper dimensions as the Compton length ¯h/mc. In your treatment you should cover the following: (a) Derive the canonical momenta and the Hamiltonian. (b) Write down Hamilton’s equations and the generalized Maxwell equations, treating the explicitly any gauge freedom for this field theory that arises from the structure of the Hamiltonian.
Chapter 8. Quantization of the Electromagnetic Field (c) Show that charge is only conserved in the Lorenz gauge, and thus that this field theory is not truly gauge-invariant. (d) Write down the wave equations for the potentials in the Lorenz gauge.
are real amplitudes. Derive the dispersion relation between \omega and k. By considering (A, \phi) to be the components of a four-dimensional vector, show that in the limit m −\rightarrow 0 limit, the longitudinal field becomes orthogonal to the source four-vector (you should work out the form of the source field). Thus, in this limit, the longitudinal field decouples from any sources, leaving only the two transverse fields in massless electromagnetism.