13. Mechanical Effects of the Quantum Vacuum¶
PDF pages 533–578
Chapter 13 Mechanical Effects of the Quantum Vacuum Here we will examine mechanical forces on atoms due to the quantum vacuum. Of course, there is no absorption from the vacuum, but scattering of virtual photons can still produce forces. Complementary to these forces are shifts of atomic energy levels, which we will also calculate. The main effect we will examine is the Casimir–Polder effect,1 where a ground-state atom near a conducting surface, with the field in the vacuum state, is attracted to the surface. We will also investigate the Lamb shift of atomic transitions, which we can interpret as an ac Stark shift due to the vacuum field. Since the introduction of a conducting boundary modifies the field modes, we can thus interpret the Casimir–Polder effect as a space-dependent Lamb shift due to the local modification of the vacuum field modes. (However, unlike the Lamb shift, which must be observed by probing transitions among atomic levels, the Casimir–Polder shift is observable for an atom in a single level due to the mechanical action of the shift.) 13.1 Setup We want to compute the effect of the vacuum field due to a plane conductor on an atom. We can write the Hamiltonian for the free atom and field as H0 = X j
X k,\zeta ¯h\omegak a\dagger
, (13.1)
|g\rangle −\rightarrow |ej\rangle atomic transition, the wave vector k labels the field modes of different frequency and orientation, the index \zeta labels the two independent polarizations, and ak,\zeta is the annihilation operator for the (k, \zeta) mode. We will write the eigenstates that we need of the free Hamiltonian in the form |\alpha, nk,\zeta\rangle , which means that the atom is in state |\alpha\rangle , while the field mode (k, \zeta) has n photons (other modes not explicitly labeled are in the vacuum state). 1The original, and very readable, reference is of course H. B. G. Casimir and D. Polder, ‘‘The Influence of Retardation on the London-van der Waals Forces,’’ Physical Review 73, 360 (1948) (doi: 10.1103/PhysRev.73.360). However, they only obtain asymptotic results. Our results here, which include the intermediate regime, are consistent with the results of the seminal calculation of G. Barton, ‘‘Frequency shifts near an interface: Inadequacy of two-level atomic models,’’ Journal of Physics B 7, 2134 (1974) (doi: 10.1088/0022-3700/7/16/012); as well as the more recent calculations by A. O. Barut and J. P. Dowling, ‘‘Quantum electrodynamics based on self-energy, without second quantization: The Lamb shift and long-range Casimir-Polder van der Waals forces near boundaries,’’ Phys. Rev. A, 36, 2550 (1987) (doi: 10.1103/PhysRevA.36.2550); and D. Meschede, W. Jhe, and E. A. Hinds, ‘‘Radiative properties of atoms near a conducting plane: An old problem in a new light,’’ Phys. Rev. A, 41, 1587 (1990) (doi: 10.1103/PhysRevA.41.1587). See also S. Haroche, ‘‘Cavity Quantum Electrodynamics,’’ in Fundamental Systems in Quantum Optics: Proceedings of the Les Houches Summer School, Session LIII, 1990, J. Dalibard, J. M. Raimond, and J. Zinn-Justin, Eds. (Elsevier, 1992), Course 13, p. 767.
Chapter 13. Mechanical Effects of the Quantum Vacuum The quantum electric field from Eq. (8.56) is
X k,\zeta r ¯h\omegak 2ϵ0
(13.2) where the ak,\zeta are the field annihilation operators, and the unit-normalized mode functions of frequency \omegak in half-space from Section 8.4.3 are
r V ˆ\epsilonk,\zeta,∥sin kzz −iˆ\epsilonk,\zeta,z cos kzz eik∥\cdotr, (13.3) where V = L3 is the quantization volume. Recall that we are applying perfectly conducting boundary conditions at z = 0 and z = L, and periodic boundary conditions at x = 0 and x = L, as well as y = 0 and y = L. In the above mode functions, ˆ\epsilonk,\zeta is the unit polarization vector of the mode, and the ˆx\alpha are the Cartesian unit vectors along the x\alpha-direction. Recall also that the wave vectors are given by
L ,
L ,
L , (13.4) where nx and ny are any integers, and nz is nonnegative. In the above quantization volume, we regard the atom as being located at (L/2, L/2, z), although as we will see the transverse location is unimportant. 13.2 Atom–Vacuum Interaction We now account for the coupling of the atom with the field. We use the usual dipole form of the atom–field interaction Hamiltonian
(13.5) where we neglect the contribution of the polarization term in Eq. (9.38) in this gauge. With the above expression (13.2) of the electric field, along with the usual form of the dipole operator, we can write the interaction explicitly without the rotating-wave approximation as HAF = − X j X k,\zeta r ¯h\omegak 2ϵ0
h
k,\zeta i , (13.6) where \sigmaj := |g\rangle \langle j|, and we have assumed the dipole matrix element \langle g|d|ej\rangle to be real. The idea here is to compute the energy shift of the ground state |g\rangle in perturbation theory, which will give the standard Casimir–Polder potential. Up to second order,
X j X k,\zeta
Eg,0 −Eej,1k,\zeta . (13.7) The first two terms vanish, and the only nonvanishing contribution in the second-order term from HAF will be from the \sigmajak,\zeta and \sigma\dagger ja\dagger k,\zeta terms, which are not energy-conserving and thus are usually dropped in the rotating-wave approximation. Then we insert the atom and field energies: VCP = − X j X k,\zeta
. (13.8) Using Eq. (13.6), we can write
X j X k,\zeta \omegak 2ϵ0
(13.9)
13.2 Atom–Vacuum Interaction if the atom is at position r and using Eq. (13.3) for the mode functions,
X j X k,\zeta \omegak ϵ0V
, (13.10) where we have dropped cross-terms involving the parallel and perpendicular components of the dipole op- erator [these vanish in the upcoming angular integrals—convince yourself of this after working up through Eqs. (13.25)]. For compactness, we will henceforth write d 2 j,∥≡d 2 j,x + d 2
d 2
(13.11) so that Eq. (13.10) becomes VCP = −1 ϵ0V X j X k,\zeta \omegak
2d 2
j,z cos2 kzz , (13.12) where d 2 j,x = d 2 j,y for a spherically symmetric atom (technically, we are not assuming a spherically symmetric atom, but because of the symmetry of the setup, the x and y dependence of the solution must be equivalent to that of a spherically symmetric atom), and we have again discarded vanishing cross-terms. All that remains is to evaluate the sums over atomic states and field modes. The polarization sum is easy to evaluate, using the result from Eq. (8.190) X \zeta
k2 , (13.13) which becomes X \zeta
\alpha k2 , (13.14)
we may assume they are real, and the polarization sum is independent of complex basis transformations). In particular, we may write the two required sums X \zeta
X \zeta
z k2 X \zeta
X \zeta
X \zeta
x k2 −k 2 y
z k2 . (13.15) Then we can perform the polarization sums in the level shift, with the result VCP = −1 ϵ0V X j X k \omegak
1 1 + k 2 z k2 d 2 j,∥sin2 kzz + 1 −k 2 z k2 d 2 j,z cos2 kzz . (13.16) Expanding out the sin2 and cos2 functions, VCP = − 2ϵ0V X j X k \omegak
d 2
j,z + k 2 z k2 d 2 j,∥/2 −d 2 j,z − d 2 j,∥/2 −d 2 j,z + k 2 z k2 d 2
j,z cos(2kzz) . (13.17) Before evaluating the sum, we must be careful to remove divergences, which will simplify this expression.
Chapter 13. Mechanical Effects of the Quantum Vacuum 13.3 Renormalization Note that to compute the force on the atom due to the wall, we are really computing the difference in energy between this situation and the limit of a large box, where the particle is arbitrarily far away from the surface (we will also take the limit as the box size goes to infinity). That is, the ‘‘renormalized’’ potential is lim L,z0−\rightarrow \inftyVCP L 2 , L 2 , z −VCP L 2 , L 2 , z0 . (13.18) Hence the z-independent terms in Eq. (13.17), which are divergent under the mode summation, do not contribute to the answer, so we can explicitly drop it. Thus, we have VCP = 2ϵ0V X j X k \omegak
d 2 j,∥/2 −d 2 j,z + k 2 z k2 d 2
j,z cos(2kzz), (13.19) and what remains is to evaluate the wave-vector summation. The terms that we have dropped, which are present in free space, are responsible for the Lamb shift that we will return to later. In fact, from the perturbation result (13.9), which contains contributions from the \sigma\daggera\dagger and \sigmaa terms of the interaction Hamiltonian, we see that the effect we are considering corresponds to the following ‘‘one-loop’’ graph. |gÒ |gÒ |eÒ This corresponds to a ground-state atom emitting a photon and becoming excited, and then reabsorbing the photon and returning to the ground state. Roughly speaking, if the virtual photon ‘‘bounces off the mirror,’’ then we count it as contributing to the Casimir–Polder potential. Otherwise we count it as part of the free-space Lamb shift. 13.4 Large-Box Limit When the box becomes large (L −\rightarrow \infty), the spacing between the modes becomes small. In this limit, an integral of a function is equivalent to a sum weighted by the mode spacings. As usual, 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) (13.20) for an arbitrary function f(k). Since
Lx ,
Ly ,
Lz , (13.21) we can thus make the formal replacement X k −\rightarrow V
Z \infty −\infty dkx Z \infty −\infty dky Z \infty dkz, (13.22) where V = L3. Thus, we are left with the expression VCP = 8\pi3ϵ0 X j Z \infty −\infty dkx Z \infty −\infty dky Z \infty dkz k (kj0 + k) d 2 j,∥/2 −d 2 j,z + k 2 z k2 d 2
j,z cos(2kzz) = 16\pi3ϵ0 X j Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz k (kj0 + k) d 2 j,∥/2 −d 2 j,z + k 2 z k2 d 2
j,z cos(2kzz), (13.23) where \omegak = ck, and \omegaj0 = ckj0. We now just need to evaluate the integrals here.
13.5 Spherical Coordinates 13.5 Spherical Coordinates In Eq. (13.23), we basically just need to evaluate the integrals I1 = Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz k (k + k0) cos(2kzz) I2 = Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz k 2 z k(k + k0) cos(2kzz). (13.24) Writing I1 in spherical coordinates, and carrying out the \phi (azimuthal) integral,
Z \infty dk Z \pi d\theta k3 sin \theta cos(2kz cos \theta) k + k0
Z \infty dk k3 (k + k0) sin(2kz) kz
2z \partial 2 z Z \infty dk sin(2kz) k + k0 . (13.25) It appears that we are cheating by pulling the \partial 2 z out of the integral, since the integral is, strictly speaking, divergent since the integrand asymptotically scales as k for large k. However, in writing down the interaction Hamltonian (13.6), what we neglected is the fact that the atom is not perfectly localized, and the Hamiltonian should in fact be averaged over a localized distribution that represents the atom’s extent. If this distribution is h(r), then each factor of sin kzz and cos kzz in Eq. (13.10) and the subsequent expressions is multiplied by the Fourier transform ˜h(k). Then factors of cos 2kzz in the subsequent expressions are multiplied by ˜h2(k), which cuts off the high-frequency ends of the above integral. For example, if h(r) is a Gaussian function, then ˜h2(k) is also Gaussian, and we have no problems with the convergence of the integrals. We will not explicitly include these cutoff functions, except to note that they ensure that all the mode integrals are convergent, and omitting them is equivalent to performing the integrals properly and then taking the limit as the size of the atom vanishes. To evaluate the above integral, we need to introduce the auxiliary functions to the sine and cosine integrals,2 which have the definitions
h\pi 2 −Si(z) i
h\pi 2 −Si(z) i . (13.26) Here, the sine integral Si(x) and cosine integral Ci(x) are defined by
Z z sin t t dt
Z \infty z cos t t
Z z cos t −1 t dt, (13.27) where \gamma is Euler’s constant. The auxiliary functions are plotted here. 2See Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions (Dover, 1965), pp. 231-3.
Chapter 13. Mechanical Effects of the Quantum Vacuum foo(x) go(x) x foo(x), go(x) We note that even though they are composed of oscillatory functions, they are not themselves oscillatory. Now using the integral representation of the auxiliary function f(z) from Problem 13.1,3 Z \infty sin(ax)
(13.28) we can evaluate the above integral to obtain
2z \partial 2 z f(2k0z). (13.29) We can also write I2 in spherical coordinates, and after carrying out the \phi (azimuthal) integral,
Z \infty dk Z \pi d\theta k3 sin \theta cos2 \theta cos(2kz cos \theta) k + k0
2 \partial 2 z Z \infty dk k k + k0 Z \pi d\theta sin \theta cos(2kz cos \theta)
\partial 2 z z Z \infty dk sin(2kz) k + k0
\partial 2 z z f(2k0z). (13.30) In evaluating the derivatives, it is useful to note the equivalence of the operators 1 2z \partial 2 z −1
z3 ≡1 \partial 2 z z , (13.31) where both are assumed to operate on some function of z. Notice that I1 and I2 differ only by the ordering of \partial 2 z and (1/z). We can then write the Casimir–Polder potential as VCP = 16\pi3ϵ0 X j h d 2 j,∥/2 −d 2 j,z I1j + d 2
j,z I2j i , (13.32) where I1j is the same as I1 and I2j is the same as I2 but with k0 −\rightarrow kj0. Putting in the above values of I1 and I2, we find VCP = −
X j d 2 j,∥/2 −d 2 j,z 1 z \partial 2 z + d 2
j,z \partial 2 z z f(2kj0z). (13.33) 3See Abramowitz and Stegun, op. cit., or I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, English translation 7th ed., A. Jeffrey and D. Zwillinger, Eds. (Academic Press, 2007), Formula 3.722.1.
13.5.1 Thomas-Reiche-Kuhn Sum Rule¶
13.5 Spherical Coordinates This is not quite our final form. To simplify the antisymmetric part (involving the difference of d 2 j,∥and d 2 j,z), we must reintroduce the Thomas–Reiche–Kuhn sum rule. 13.5.1 Thomas–Reiche–Kuhn Sum Rule In the classical analysis of the Lorentz model in Section 1.2.1, we wrote down the Thomas–Reiche–Kuhn (TRK) sum rule X j f0j = 1. (13.34) For a fine-structure transition J −\rightarrow J′, we have the expression \GammaJ′J = e2\omega2 J′J 2\piϵ0mec3 2J + 1 2J′ + 1fJJ′ (13.35) that relates the oscillator strength to the decay rate. We further have the relation \GammaJ′J = \omega3 J′J 3\piϵ0¯hc3 2J + 1
(13.36) relating the decay rate to the lifetime, in which case the TRK sum rule becomes X J′ \omegaJJ′|dJJ′|2 = 3¯he2 2m . (13.37) This is closer to the usual quantum-mechanical form for the TRK sum rule, which we now derive directly. Recall that we derived the relation
(13.38) between the position and momentum matrix elements in Section 9.3.2, Eq. (9.52), where
¯h (13.39) is again the signed frequency of the transition between the energy eigenstates. The TRK sum rule follows from the following: X j
X j
= − i 2m X j
= − i
= −¯h 2m. (13.40) Thus, we have established the TRK sum rule (note the subscript ordering on \omegaj0): X j
2m. (13.41) (Thomas–Reiche–Kuhn sum rule)
13.5.2 Simplification¶
Chapter 13. Mechanical Effects of the Quantum Vacuum Using Eq. (13.38), we can also establish the alternate form of the TRK sum rule: X j
\omegaj0 = m¯h 2 . (13.42) In the Casimir–Polder potential, two combinations of dipole sums are particularly useful, as a corollary to the TRK sum rule: X j \omegaj0 d 2 j,∥+ 2d 2 j,z = 2¯he2 m X j \omegaj0 d 2 j,∥−2d 2 j,z = 0. (13.43) In writing these relations down, we have used ˆr\alpha \cdot d0j = e\langle 0|r\alpha|j\rangle . Of course, the label 0 can refer to any energy eigenstate, not just the ground state. As we indicated before in Section 1.2.1, the TRK sum rule can converge rather slowly, involving many bound and unbound states for a convergence of matrix elements in an atom.4 We can see in Eq. (13.41) a bit of why this is: In the ‘‘E gauge’’ (Section 9.3.2), calculations typically involve such sums over squared position matrix elements (where the corresponding sums in the ‘‘A gauge’’ involve sums over the squared momentum matrix elements as in Eq. (13.42). The E-gauge summations typically converge more quickly, since the dipole matrix elements to high-energy states typically drop off quickly (high-energy states tend to be far away from the nucleus, unlike low-energy states). However, the sum (13.41) is also weighted by the transition frequency, which again gives weight to high-energy states, and slowing the convergence of the sum. 13.5.2 Simplification In Eq. (13.33), we have the derivative of f(z). The derivatives of the auxiliary functions are given by
\partial 2
z −f(z), (13.44) so that the second derivative generates a term of the form X j d 2 j,∥/2 −d 2 j,z 2kj0 z2 , (13.45) which vanishes due to the TRK sum rule. Thus, we can write Eq. (13.33) as VCP =
X j " d 2 j,∥/2 −d 2 j,z 4k 2 j0 z − d 2
j,z \partial 2 z z # f(2kj0z). (Casimir–Polder potential for ground-state atom near perfect mirror) (13.46) Now defining the scaled coordinates z′ j := 2kj0z, (13.47) we can write VCP =
X j \omega 3 j0 " d 2 j,∥/2 −d 2 j,z 1 z′ j − d 2
j,z \partial 2 z′ j z′ j !# f(z′ j) (Casimir–Polder potential, scaled coordinates) (13.48) as the potential shift of the ground state. 4See Peter W. Milonni and Joseph H. Eberly, Lasers (Wiley, 1988), p. 239, where the terms due to ionized states in hydrogen make up a significant part of the sum.
13.5.3 Spherical Symmetry¶
13.6 Asymptotic Behavior 13.5.3 Spherical Symmetry For an atom with a spherically symmetric ground state (S-orbital), d 2 j,∥/2 −d 2 j,z = 1
(13.49) while d 2
j,z = 1
j,z. (13.50) Thus, the potential simplifies, and becomes VCP = −
X j d 2 j,z\partial 2 z z f(2kj0z). (Casimir–Polder potential, spherically symmetric atom) (13.51) Still, the f(z) function is unusual, and we will get some better feeling for it by examining its asymptotics and by carrying out the derivatives. 13.6 Asymptotic Behavior
VCP = − (4\piϵ0) 4z3 X j d 2 j,z. (near-field van der Waals potential, spherically symmetric atom) (13.52) We can see that this expression agrees with the classical result for the interaction of an induced, static dipole with its image. The interaction energy of two dipoles d1 and d2 is
(4\piϵ0)r 3 , (13.53) where r12 is the vector for the displacement between the two dipoles. For a dipole a distance z from the conducting surface interacting with its image, we have r12 = 2z, so that the dipole interaction energy reduces to Vdip = − (4\piϵ0) 16z3 d2 ∥+ 2d2 ⊥ , (13.54) where d∥and d⊥are the parallel and perpendicular components of the dipole, respectively. We have also added an extra factor of 1/2, since don’t want the direct interaction energy, but rather the energy required to bring the dipole from distance +\inftyto z from the surface (equivalently, the energy we derived amounts to a field integral over all space, but we only want the integral over the dipole fields over half space). This agrees with Eq. (13.52) if we interpret the dipole moments as ground-state expectation values d 2 ∥,⊥≡\langle g|d 2 ∥,⊥|g\rangle , (13.55) we find the quantum instantaneous dipole energy Vdip = − (4\piϵ0) 16z3 \langle g| d2 ∥+ 2d2 ⊥ |g\rangle . (13.56) If we further make the identifications d2
d2 ⊥= e2z2, (13.57) so that assuming an isotropic atom,
(13.58)
Chapter 13. Mechanical Effects of the Quantum Vacuum we can write the static-dipole interaction as Vdip = − (4\piϵ0)
(near-field van der Waals potential, spherically symmetric atom) (13.59) This is equivalent to Eq. (13.52) after inserting the identity summation over atomic states in the matrix element. This is the usual van der Waals atom–surface interaction. We can see that in the short range, this effect is largely a classical dipole–image interaction, except for the interpretation of the dipole as an expectation value. The force here is thus due to zero-point fluctuations of the atomic dipole, which induce a mean-square dipole that interacts with its image to produce the force. Asymptotically, f(z) ∼1/z and g(z) ∼1/z2 for |arg z| < \pi, so that for the spherically symmetric atom at large z, VCP ∼− 3c
X j d 2 j,z \omegaj0 = − 3¯hc\alpha0
(long-range Casimir–Polder potential, spherically symmetric atom) (13.60) where
X j 2d 2 j,z ¯h\omegaj0 (13.61) is the (classical or small-signal) static polarizability from Eq. (1.32), with \omega, \gammaj −\rightarrow 0, and Eq. (5.213) to convert the oscillator strength to the dipole matrix element. Thus, we recover the standard Casimir–Polder result in the large-z limit.5 Again, the effect is due to fluctuations of the atomic dipole, but now the dipole interacts with its retarded image, due to the long distance to the mirror and back. The retardation means that the dipole is no longer completely correlated with its image, and this is why the potential falls off more quickly (like z−4 instead of z−3) in the far-field regime. Note, however, that in this regime, the Casimir– Polder effect is regarded in this regime as a true quantum effect of the field, since a semiclassical argument with retardation does not reproduce the correct potential without an extra choice of field-operator ordering.6 13.7 Excited-Level Shift Now let us consider the corresponding shift of some excited atomic level |\alpha\rangle . Then we must modify the expression (13.9) to read
X
X k,\zeta \omegak 2ϵ0
, (13.62) When summing over a state |j\rangle of higher energy than |\alpha\rangle , then the term has the same form as before, since |\alpha\rangle acts effectively as a ground state. However, if |j\rangle is of lower energy than |\alpha\rangle , then |\alpha\rangle takes on the role of an excited state for the transition, and \omegaj\alpha < 0. In the interaction Hamiltonian, these modified terms are due to the usual energy-conserving terms \sigmaja\dagger
jak,\zeta, as opposed to the energy-nonconserving terms
ja\dagger k,\zeta terms that are responsible for the ground-state shift. To avoid problems with the pole in the k integration, we note that in perturbation theory, the integral is always taken to be the Cauchy principle value (Section 14.1.4.2), so that the singularity causes no difficulty in principle. The derivation then carries through as for the ground state, with possibly negative wave numbers in the solution (13.33):
X j d 2 j,∥/2 −d 2 j,z 1 z \partial 2 z + d 2
j,z \partial 2 z z f(2kj\alphaz), (13.63) 5Peter W. Milonni, The Quantum Vacuum (Academic Press, San Diego, 1994), p. 107. 6Stephen M. Barnett, Alain Aspect, and Peter W. Milonni, ‘‘On the quantum nature of the Casimir–Polder interaction,’’ Journal of Physics B: Atomic, Molecular, and Optical Physics 33, L143 (2000) (doi: 10.1088/0953-4075/33/4/106).
13.7 Excited-Level Shift with
(13.64) Note that since we now have in principle a complex-valued function, we are explicitly taking the real part
more complicated: Ci(z) has a branch cut along the (−z)-axis, and it picks up an additional term when the sign of the argument changes:7
(0 < arg z < \pi)
(−\pi < arg z < 0). (13.65) Note then that strictly speaking, we should not have a negative real argument of f(z). However, we are implementing the Cauchy principle value, which consists of deforming the k integral by adding \pmi0 to k, and then averaging the results, so that the two possible extra terms cancel. Thus, for our purposes, we may
(13.66) also for z real. Thus,
(13.67) where \Theta(z) is the Heaviside step function. That is, if the frequency is negative (for a term in the sum corresponding to a lower-energy level), the sign of the f(z) function changes, and an extra term appears. Thus,
X j
d 2 j,∥/2 −d 2 j,z 1 z \partial 2 z + d 2
j,z \partial 2 z z h
i . (13.68) Thus, extra, oscillatory terms are present when the level |\alpha\rangle is an excited state—recall that f(z) is not oscillatory. Redefining the scaled coordinates z′
(13.69) we can write the potential shift as
X j
" d 2 j,∥/2 −d 2 j,z z′ j \partial 2 z′ j ! + d 2
j,z \partial 2 z′ j z′ j !#h f(z′
j i . (13.70) The TRK sum rule again applies as in Eq. (13.48), and we can thus evaluate the derivatives in the above expression to obtain
X j
" d 2 j,∥/2 −d 2 j,z 1 z′ j − d 2
j,z \partial 2 z′ j z′ j !#h f(z′
j i . (13.71) This is our final result for the level shift of any level due to the presence of the conducting plane. However, let’s condense the notation just a bit more and write
X j
" d 2 j,∥/2 −d 2 j,z 1 z′ j − d 2
j,z \partial 2 z′ j z′ j #h f(z′ j) −\Thetaj\pi cos z′ j i , (Casimir–Polder potential, excited level) (13.72)
unity in the same case and vanishing otherwise. 7See Abramowitz and Stegun, op. cit., p. 232, Formula 5.2.20, or Gradshteyn and Ryzhik, op. cit., Formula 8.233.2.
13.7.1 Classical Antenna Behavior¶
Chapter 13. Mechanical Effects of the Quantum Vacuum 13.7.1 Classical Antenna Behavior Suppose now that we focus only on the oscillatory terms: V (osc) \alpha = X j<\alpha
4\piϵ0c3 " d 2 j,∥/2 −d 2 j,z 1 z′ j − d 2
j,z \partial 2 z′ j z′ j !# cos z′ j. (13.73) We can regard the sum here as extending over all states (even degenerate ones), in which case the formula
3\piϵ0¯hc3 (13.74) applies for the |j′\rangle −\rightarrow |j\rangle decay path. Thus, we find V (osc) \alpha = X j<\alpha 4¯h\Gamma\alphaj h ˆ\epsilon 2
j,⊥ − ˆ\epsilon 2
j,⊥ \partial 2 z′ j i cos z′ j z′ j , (13.75) where now ˆ\epsilon∥and ˆ\epsilon⊥are the projections of the dipole unit vector onto the components parallel and perpen- dicular to the surface, respectively. By comparison to the classical expression of a dipole near a mirror, from Eq. (1.130),
4\gamma h ˆ\epsilon 2
⊥ − ˆ\epsilon 2
⊥ \partial 2 z′ i cos z′ z′ , (13.76)
for a single transition, for here we must sum over all lower energy levels). Thus, we can interpret the part of the potential unique to excited states as a classical dipole potential due to the spontaneously radiated field. However, what the classical model misses is the f(z) part, which is a manifestation of the quantum vacuum. As we noted above, the ground-state Casimir–Polder potential is an effect of reflecting virtual photons, and we now see that the excited-state potential also includes effects due to real photons bouncing from the mirror. To examine this more quantitatively, consider a two-level atom,8 in which case the excited-state shift corresponding to Eq. (13.72) is \deltaEe ¯h = 3\Gamma 4\pi h ˆ\epsilon 2
⊥ − ˆ\epsilon 2
⊥ \partial 2 z′ i \pi cos z′ −f(z′) z′ . (13.77)
\pi/2 + O(z′), and thus the quantum near-field (van der Waals) shift is half the classical value. If we choose to identify the classical and quantum mean-square dipoles differently to make them agree in the near-field, then the quantum shift will be double the classical shift in the far field.9 Thus, because of the vacuum contribution, there is no simple way to exactly identify the shift of the atomic excited state of the two-level atom with that of the classical dipole radiator. 8Note that the two-level atom has some artifacts due to different cancellations than in the full summation for a real atom. For example, if we calculate the transition frequency shift to compare to the classical case, we get the combination \pi cos z′ −2f(z′), which vanishes to zeroth order. Thus, the 1/z3 leading-order contribution vanishes, which is not the case for real atoms, because the coupling to other levels still generate this term. See G. Barton, op. cit. 9E. A. Hinds and V. Sandoghdar, ‘‘Cavity QED level shifts of simple atoms,’’ Physical Review A 43, 398 (1991) (doi: 10.1103/PhysRevA.43.398). See also S. Haroche, op. cit.
13.8 Power–Zienau Transformation in Half-Space 13.8 Power–Zienau Transformation in Half-Space One little detail that we have ignored is that the complete dipole interaction Hamiltonian that we derived before in Eq. (9.38) has the form HAF = 1 Z
2ϵ0 Z d3r P ⊥2(r) = 1 Z
2ϵ0 Z d3r P ⊥2(r) = ˜HCoulomb + ˜Hd\cdotE + ˜Hself, (13.78) where the last term is a polarization energy due to the dipole, with P ⊥
(13.79) Note that we are also now including the interaction of the electron charge density \rho(r) with the scalar potential, because now we have a coupling to a longitudinal electric field: an instantaneous dipole moment of the atom shows up as an instantaneous image dipole, to enforce the boundary conditions at the conducting plane. The r \cdot E term does not account for this, as it was derived from the coupling to the vector potential and thus includes only the coupling to the transverse field. We have neglected the first and last terms thus far, although we have gotten the correct interaction, which yields the instantaneous dipole-dipole coupling at short range and the retarded scaling at long range. We have already shown in Eq. (13.56) that the instantaneous dipole-image interaction due to HCoulomb is given by ∆ECoulomb = − (4\piϵ0)
d2 ∥+ 2d2 ⊥
(13.80) We have neglected the last term in Eq. (13.78), the dipole self-energy. This easy to justify, since the self- energy is integrated over all space and thus independent of r. The contribution is thus z-independent and disappears in the renormalization of the Casimir–Polder potential. Note that this term is important in computing the Lamb shift, as we will show later. Of course, the above interaction Hamiltonian was derived using a field commutator, specifically a free- space field commutator. However, for this calculation we have been living in half space, and we should verify that our interaction Hamiltonian is still appropriate. From our previous treatment of the Power–Zienau transformation, specifically Eq. (9.33), the transformed electric field operator is given by ˜E⊥
= E⊥
¯h re,\alpha h A\alpha(zˆz), E⊥ \beta (r) i , (13.81)
arbitrary). We now use the commutator [from Eq. (8.215)]
ϵ0 \delta⊥
, (13.82) where the transverse and ‘‘reflected transverse’’ delta functions are [from Eqs. (8.178) and (8.214), respec- tively] \delta⊥
(2\pi)3 Z d3k
k2 eik\cdotr \delta⊤
(2\pi)3 Z d3k \delta−
k2 eik\cdotr. (13.83)
Chapter 13. Mechanical Effects of the Quantum Vacuum (Recall that k−
transformed field in half-space as ˜E⊥
ϵ0 re,\alpha \delta⊥
. (13.84) We now see that the transformed field has an extra contribution at the location of the image dipole. The field energy thus transforms as U(re) Z z>0 d3r E⊥(r) 2
Z z>0 d3r U(re)E⊥(r)U \dagger(re) 2 = Z z>0 d3r E⊥(r) 2 + 2e ϵ0
ϵ0 re \cdot E⊤(−zˆz) + 1 ϵ 2 Z z>0 d3r P⊥(r −zˆz) 2 + 1 ϵ 2 Z z>0 d3r P⊤(r + zˆz) 2 + 2 ϵ 2 Z d3r P⊥(r −zˆz) \cdot P⊤(r + zˆz). (13.85) We have defined here the reflected field E⊤
Z z>0 d3r′ \delta⊤
E⊥ \beta (r′), (13.86) which for our purposes here is the same as the usual transverse field, but with the opposite sign for the z-component, and the reflected polarization P ⊤
(13.87) The interaction Hamiltonian then becomes ˜HAF = 1 Z
- 1 2ϵ0 Z z>0 d3r P⊥(r + zˆz) 2 + 1 2ϵ0 Z z>0 d3r P⊥(r −zˆz) 2 + 1 ϵ0 Z z>0 d3r P⊥(r −zˆz) \cdot P⊤(r + zˆz). (13.88) Thus we see that in half-space, there is an additional dipole-field interaction term and two additional self- energy terms. However, the field operator E⊤(−zˆz) vanishes, since it refers to the field amplitude behind the conductor, where it vanishes. The other polarization terms reduce very simply, using the properties
(13.89) and
(13.90) In this case, the interaction Hamiltonian becomes ˜HAF = 1 Z
2ϵ0 Z all z d3r P⊥(r) 2 + 1 2ϵ0 Z all z d3r P⊥(r −zˆz) \cdot P⊤(r + zˆz) = ˜HCoulomb + ˜Hd\cdotE + ˜Hself + ˜Hdipole-image. (13.91) Thus, we recover the usual dipole self-energy term, plus a second that is evidently due to the interaction of the dipole with its image in the mirror. The usual term is z-independent and does not contribute to the Casimir–Polder potential after renormalization. We will examine this term more closely in Section 13.12.2.1. However, the new term does, and we will now evaluate it.
13.8 Power–Zienau Transformation in Half-Space To evaluate the dipole-image part of the Hamiltonian, ˜Hdipole-image = 2ϵ0 Z all z d3r P⊥(r −zˆz) \cdot P⊥(r + zˆz), (13.92) we first evaluate the integral
Z d3r \delta⊥
= (2\pi)6 Z d3r Z d3k Z d3k′
k′ \alphak′ \beta k′2 \delta−
k2 eik\cdot(r−zˆz)eik′\cdot(r+zˆz) = (2\pi)3 Z d3k Z d3k′
k′ \alphak′ \beta k′2 \delta−
k2 \delta(k + k′) e−i(kz−k′ z)z = (2\pi)3 Z d3k
k2 \delta−
k2 e−i2kzz = (2\pi)3 Z d3k \delta−
\gamma k2 e−i2kzz. (13.93) Note that the z coordinate here is the location of the atom, not a component of the integration variable r. Now we can evaluate the components on a case-by-case basis. Note that the integrand is axially symmetric, except for the tensor part, and thus the integral vanishes if \alpha̸ = \gamma, as in this case the dependence on the axial angle \phi will be sin \theta, cos \theta, or sin \theta cos \theta. We can also see that the \delta− \alpha\gamma never contributes, in view of the integral Z d3k e−i2kzz = Z d3k cos(2kz cos \theta)
Z \infty dk k2 Z \pi d\theta sin \theta cos(2kz cos \theta)
z Z \infty dk k sin(2kz) = lim \sigma\rightarrow 0 2\pi z Z \infty dk k sin(2kz)e−k\sigma = lim \sigma\rightarrow 0
= 0. (13.94) If \alpha = \gamma = z, then we have
(2\pi)3 Z d3k k 2 z k2 e−i2kzz = (2\pi)3 −1 4\partial 2 z Z d3k 1 k2 cos(2kzz) = (2\pi)2 −1 4\partial 2 z Z \infty dk Z \pi d\theta sin \theta cos(2kz cos \theta) = (2\pi)2 −1 4\partial 2 z Z \infty dk sin(2kz) kz = (2\pi)2 −1 4\partial 2 z \pi 2z = − 16\piz3 . (13.95) On the other hand, if \alpha = \gamma = x or y, then
(2\pi)3 Z d3k k 2 ∥ 2k2 e−i2kzz = − 32\piz3 , (13.96)
Chapter 13. Mechanical Effects of the Quantum Vacuum where we have used the fact that the x- and y-directions are equivalent in this axisymmetric problem, and k 2 ∥= k 2 x + k 2 y = k2 −k2 z. Thus,
(13.97) To second order in the atomic dipole moment (i.e., to order e2), it is sufficient to compute the shift due to the dipole-image Hamiltonian to first order in perturbation theory. Thus, for the shift of level \alpha,
= e2 2ϵ0
=
h d 2 j,∥+ 2d 2 j,z i
(13.98) This is the opposite of the static-dipole energy shift due to the Coulomb Hamiltonian. Note that this result—where contributions boundary terms from the Power–Zienau transformation cancel the static image energies—always holds, independent of the shape of the boundary. Thus, when using the dipole Hamiltonian, it is sufficient to use the free-space version without worrying about Coulomb interactions with images.10 13.9 Calculation in the Coulomb Gauge Now we will show that the same Casimir–Polder potential obtains if we use the A-gauge form H(A) AF = e me
2me A2 (13.99) of the interaction Hamiltonian from Eq. (9.49), including the sometimes-neglected A2 term. The ground-state shift of the first term follows from adapting Eq. (13.32) to read
16\pi3ϵ0 X j h d 2 j,∥/2 −d 2 j,z I′ 1j + d 2
j,z I′ 2j i , (13.100) where the above form is the same as before, but written in terms of new integrals I′ 1 = k 2 Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz k(k + k0) cos(2kzz) I′ 2 = k 2 Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz k 2 z k3(k + k0) cos(2kzz). (13.101) This is because the matrix elements of the r \cdot E and p \cdot A interaction Hamiltonians differ in magnitude by a factor of \omega/\omegaj0, as we showed in Section 9.3.2. Thus, the integrands are multiplied by factors of (kj0/k)2 compared to the previous calculation in the E gauge. Evaluating the I′ 1 integral, I′
Z \infty dk Z \pi d\theta k sin \theta cos(2kz cos \theta) (k + k0)
Z \infty dk k (k + k0) sin(2kz) kz
z Z \infty dk sin(2kz) (k + k0)
z f(2k0z). (13.102) 10E. A. Power and T. Thirunamachandran, ‘‘Quantum electrodynamics in a cavity,’’ Physical Review A 25, 2473 (1982) (doi: 10.1103/PhysRevA.25.2473).
13.9 Calculation in the Coulomb Gauge Recalling that¶
1 z \partial 2 z
z f(2k0z) −\pik0 z2 , (13.103) we showed above that the second term vanished in the sum over levels by the TRK sum rule (13.43). Thus, the two integrals give equivalent results for the asymmetric-dipole part of the Casimir–Polder potential. To evaluate the second integral I′ 2, the procedure is similar to the case for I2: I′
Z \infty dk Z \pi d\theta k sin \theta cos2 \theta cos(2kz cos \theta) (k + k0)
Z \infty dk k (k + k0) sin(2kz) 2kz + 2cos(2kz) (2kz)2 −2sin(2kz) (2kz)3
1 2z \partial 2 z −1
z3 Z \infty dk sin(2kz) k2(k + k0)
2 k 2 \partial 2 z z Z \infty dk sin(2kz) k2(k + k0)
2 k 2 \partial 2 z z −4\partial −2 z Z \infty dk sin(2kz) (k + k0)
2 k 2 \partial 2 z z −4\partial −2 z f(2k0z)
\partial 2 z z
\partial 2 z z + \pik0 \partial 2 z z [z log(2k0z) −z]
\partial 2 z z
2z3 −\pik0 z2 , (13.104) where we used the antiderivative formula \partial −2
(13.105) The \pi/2 is one of the constants of integration from evaluating the antiderivatives, while the other gives a term of the form cz, which vanishes under the derivative. These constants are set by noting that the integral
\partial 2 z z f(2k0z), (13.106) we see that there are two terms in I′ 2 that do not appear in I2 that we must explain. The second term in I′ leads to an energy shift of the form 32\piϵ0z3 X j d 2
j,z = 4\piϵ0 16z3 \langle g| d 2 ∥+ d 2 z |g\rangle . (13.107) This has the same form, except for the opposite sign, as the static dipole energy (13.54), if we interpret the classical squared dipole moments as quantum expectation values. Thus, this term cancels the static Coulomb energy of the instantaneous dipole moment interacting with the boundary, which we have so far neglected to include, given by the Hamiltonian HCoulomb = 1 Z
(13.108) where \rho(r) is the charge density corresponding to the atomic dipole, and \phi(r) is the scalar potential, which gives the energy contribution of the longitudinal field. This energy corresponds to an unretarded energy, and clearly must be canceled to produce the correct quantum (retarded) shift.
Chapter 13. Mechanical Effects of the Quantum Vacuum The third term in I′ 2, which also does not appear in I1, also scales as k0 and therefore, by the TRK sum rule (13.43), becomes independent of the level: kj0 16\pi2ϵ0z2 X j d 2
j,z = − e2¯h 16\pi2ϵ0mecz2 . (13.109) We will show below that this term exactly cancels the contribution from the A2 term of the A-gauge interaction Hamiltonian. Thus, we see that the A-gauge interaction Hamiltonian gives the same result as the total E-gauge dipole interaction Hamiltonian. The same is true of the excited-level shifts, because the extra cos(2k0z) terms that appear there do not generate extra terms when differentiated or integrated twice, and the extra terms we generated turned out to be level-independent. Thus, the same branch-cut argument above produces the same terms in the A gauge. However we see explicitly here the importance of summing over all the excited levels; had we made a two-level atom approximation, we would have gotten different results in the two gauges.11 Then what about the rest of the A-gauge Hamiltonian? We still have left the field self-energy part HAF,2 = e2 2me A2 (13.110) of the interaction. Since our calculation is valid to lowest (second) nonvanishing order in the dipole matrix element, we can compute the shift due to this Hamiltonian to order e2, and thus it suffices to compute the shift to first order in perturbation theory:
= e2 2me
(13.111) Using the expression
X k,\zeta i r ¯h 2\omegakϵ0
(13.112) for the quantum vector potential in terms of the mode functions from Eq. (8.61), we find that only the terms of the form ak,\zetaa\dagger k,\zeta, where both operators correspond to the same mode, contribute to the vacuum expectation value above. Thus, ∆EA2 = e2¯h 4meϵ0 X k,\zeta \omegak |fk,\zeta(r)|2. (13.113) Using the half-space mode functions in Eq. (13.3), the shift then becomes ∆EA2 = e2¯h 2meϵ0V X k,\zeta \omegak
= e2¯h 2meϵ0V X k \omegak 1 + k 2 z k2 sin2 kzz + 1 −k 2 z k2 cos2 kzz = e2¯h 2meϵ0V X k \omegak 1 + k 2 z k2 sin2 kzz −cos2 kzz = e2¯h 2meϵ0V X k \omegak 1 −k 2 z k2 cos(2kzz) . (13.114) As usual, we can change this to an integral over all reciprocal space in the large-volume limit: ∆EA2 = e2¯h 8\pi2meϵ0c Z \pi
Z \infty dk k 1 −k 2 z k2 cos(2kzz) . (13.115) 11This point was made eloquently by G. Barton, op. cit.
13.10 Evaluation The first term in the integral is z-independent, and gives a correction to the Lamb shift: ∆EA2,Lamb = e2¯h 4\pi2meϵ0c Z \infty dk k. (13.116) Renormalization thus removes this part from the Casimir–Polder shift. The remainder of the energy shift is, using the same integration procedures as before, ∆EA2 = − e2¯h 8\pi2meϵ0c Z \infty dk Z \pi
z k cos(2kzz) = − e2¯h 8\pi2meϵ0c Z \infty dk k Z \pi d\theta sin \theta cos2 \theta cos(2kz cos \theta) = − e2¯h 8\pi2meϵ0c −1 4\partial 2 z z Z \infty dk sin(2kz) k2 = − e2¯h 8\pi2meϵ0c \partial 2 z z \partial −2 z Z \infty dk sin(2kz) = − e2¯h 16\pi2meϵ0c \partial 2 z z \partial −2 z 1 z = − e2¯h 16\pi2meϵ0c \partial 2 z z (z log z −z) = − e2¯h
z log z = − e2¯h
4\pimez2 , (13.117) where \alpha0 := e2/4\piϵ0¯hc is the fine-structure constant. There is here no constant of integration from evaluating
the other constant of integration contributes a linear term that vanishes subsequently under the derivative. Note that this is exactly the opposite of the extra term that we found in Eq. (13.109), and thus this contribution cancels that one. Again, this term is level-independent, and thus this cancellation occurs for any atomic level, not just the ground state. 13.10 Evaluation If we are to evaluate the Casimir–Polder potential, it is useful to expand out the rather compact form of Eq. (13.72). We can do this by noting \partial 2 z′ j z′ j h f(z′ j) −\Thetaj\pi cos z′ j i = 1 z′2 j +
z′3 j −1 z′ j ! h f(z′ j) −\Thetaj\pi cos z′ j i + 2 z′2 j h g(z′ j) −\Thetaj\pi sin z′ j i , (13.118) and thus
X j
d 2 j,∥/2 −d 2 j,z 1 z′ j h f(z′ j) −\Thetaj\pi cos z′ j i − X j
d 2
j,z " z′2 j +
z′3 j −1 z′ j ! h f(z′ j) −\Thetaj\pi cos z′ j i + 2 z′2 j h g(z′ j) −\Thetaj\pi sin z′ j i# , (general Casimir–Polder potential, expanded derivatives) (13.119) we can then evaluate this expression computationally by summing over all the states, evaluating the auxiliary functions in terms of the sine and cosine integrals.
Chapter 13. Mechanical Effects of the Quantum Vacuum 13.11 Numerical Evaluation: 87Rb The dipole moments defined above must be written in terms of the appropriate dipole matrix element connecting two hyperfine levels |F, mF\rangle and |F ′, m′ F\rangle (primes referring to the excited level). The hyperfine dipole moment can be factored in terms of a Wigner 3-j symbol, \langle F mF|erq|F ′ m′
2F + 1 F ′ F m′ F q −mF , (13.120) where I is the nuclear spin and J is the composite electron spin. This can be further factored in terms of a 6-j symbol and a reduced matrix element for the fine-structure transition:
J J′ F ′ F I . (13.121) This form is particularly convenient, as this dipole matrix element may be written in terms of the partial lifetime for the decay path J′ −\rightarrow J: \tauJ′J = \omega3 3\piϵ0¯hc3 2J + 1
(13.122) Experimental measurements usually consider only the total lifetime of a level, given by summing over all possible decay paths
X J \tauJ′J . (13.123) However, using the partial lifetimes avoids confusion with myriad normalization conventions for dipole matrix elements, oscillator strengths, etc. We are only considering broadband ‘‘light,’’ so to good approximation we do not need to explicitly consider hyperfine splittings. Thus, it is convenient to simply sum over the dipole matrix elements originating from a particular hyperfine state |F, mF\rangle . For the ground and D1 (5P1/2) states, we can evaluate this sum with the above formulae, and it turns out to be particularly simple: X F ′
(13.124) For the D2 excited states, the sum is somewhat more complicated but can be written in terms of a number
possibilities depending on the type of transition. If we consider coupling to the ground (5S1/2) state, then J = 1/2, J′ = 3/2, and for the perpendicular dipole moment, X F ′
F ′ = 3 1/6, F ′ = 2 (1 + 6m2 F)/30, F ′ = 1 1/6, F ′ = 0 . (13.125) For the parallel dipole moment, X F ′
=
(12 + m2
1/2, F ′ = 2 (13 −6m2
1/6m2 F, F ′ = 0 . (13.126)
13.11 Numerical Evaluation: 87Rb If we consider coupling to higher-lying excited states, then J = 3/2, J′ = 1/2, 3/2, 5/2, and for the perpen- dicular dipole moment, X F ′
1/3,
(1 + 6m2 F)/15,
1/3,
(9 + 4m2 F)/75,
1/3,
(17 + 24m2 F)/75,
1/3,
(29 −m2 F)/75,
1/3,
(7 + 2m2 F)/25,
1/3,
. (13.127) For the parallel dipole moment, X F ′
=
(6 + m2 F)/15,
2/3,
(14 −6m2
2/3,
(66 −45m2
2/3,
(41 −24m2
2/3,
(46 + m2 F)/75,
2/3,
(18 −2m2
2/3,
. (13.128) These formulae are sufficient to determine the dipole moments in the Casimir–Polder energy shift for the excited state. 13.11.1 Tabulated Data Now we tabulate the lines involved in the trap-depth calculations. For the ground-state shift, we need the series of nP3/2 and nP1/2 transitions, as given here. Sources here are listed as Steck12, NIST13, Morton14, Gomez15, and Safronova16. 12Daniel A. Steck, ‘‘Rubidium 87 D Line Data,’’ available online at http://steck.us/alkalidata. 13NIST Atomic Spectra Database (version 3.0), Available online at http://physics.nist.gov/PhysRefData /ASD/index.html. 14Donald C. Morton, ‘‘Atomic Data for Resonance Absorption Lines. II. Wavelengths Longward of the Lyman Limit for Heavy Elements,’’ Astrophys. J. Supp. Ser. 130, 403 (2000) (doi: 10.1086/317349). 15E. Gomez, F. Baumer, A. D. Lange, G. D. Sprouse, and L. A. Orozco, ‘‘Lifetime measurement of the 6s level of rubidium,’’ Phys. Rev. A 72, 012502 (2005) (doi: 10.1103/PhysRevA.72.012502). 16M. S. Safronova, Carl J. Williams, and Charles W. Clark, ‘‘Relativistic many-body calculations of electric-dipole matrix elements, lifetimes, and polarizabilities in rubidium,’’ Phys. Rev. A 69, 022509 (2004) (doi: 10.1103/PhysRevA.69.022509).
Chapter 13. Mechanical Effects of the Quantum Vacuum transition \lambda (nm) source \tauJ′J (ns) type source
780.241209686(13) Steck 26.236(11) expt Steck
794.9788509(8) Steck 27.679(27) expt Steck
420.2989(10) NIST 565(28) expt Morton
421.6726(10) NIST 667(33) expt Morton
358.8073(10) NIST 2530(130) expt Morton
359.2597(10) NIST 3460(170) expt Morton
334.9658(10) NIST 7300(360) expt Morton
335.1775(10) NIST 1.122(56) \times 104 expt Morton
322.8911(10) NIST 1.563(78) \times 104 expt Morton
323.0088(10) NIST 2.60(13) \times 104 expt Morton
315.8444(10) NIST 2.96(15) \times 104 expt Morton
315.9173(10) NIST 4.98(25) \times 104 expt Morton
311.3468(10) NIST 3.98(20) \times 104 expt Morton
311.3950(10) NIST 7.87(39) \times 104 expt Morton
308.2893(10) NIST 6.71(34) \times 104 expt Morton
308.3229(10) NIST 1.43(72) \times 105 expt Morton
306.1131(10) NIST 9.52(48) \times 104 expt Morton
306.1375(10) NIST 2.22(11) \times 105 expt Morton The additional data for the treatment of the D2 excited (5P3/2) state are as follows: transition \lambda (nm) source \tauJ′J (ns) type source
1366.875(10) NIST 68.35(26) expt Gomez
741.02136(10) NIST 219.7(88) theory Safronova
616.13310(10) NIST 458(18) theory Safronova
1529.261(10) NIST 563(23) theory Safronova
1529.366(10) NIST 93.7(37) theory Safronova
776.15716(10) NIST 1490(340) theory Safronova
775.97855(10) NIST 254(56) theory Safronova
630.09666(10) NIST 1586(20) theory Safronova
630.00670(10) NIST 269(20) theory Safronova The additional data for the treatment of the D1 excited (5P1/2) state are as follows: transition \lambda (nm) source \tauJ′J (ns) type source
1323.879(10) NIST 136.71(51) expt Gomez
728.20028(10) NIST 419(17) theory Safronova
607.24355(10) NIST 870(35) theory Safronova
1475.644(10) NIST 103(41) theory Safronova
762.10304(10) NIST 335(82) theory Safronova
620.80263(10) NIST 339(29) theory Safronova 13.11.2 Results The ground-level shift is shown here with the long-distance Casimir–Polder result (∼1/z4), where the de- viation is clear at distances much smaller than the main transition wavelengths around 800 nm. The shifts in the long-distance approximation computed by summing the dipole moments agree with those computed
17Daniel A. Steck, ‘‘Rubidium 87 D Line Data’’ unpublished, available online at http://steck.us/alkalidata.
13.11 Numerical Evaluation: 87Rb general result large-distance approximation z (nm) level shift (MHz) -10 -5 The shift is much smaller at large distances. Here, the general and large-distance results are visually indis- tinguishable. general result large-distance approximation z (nm) 10000 level shift (Hz) -100 -50 Comparing the general result instead to the small-distance (static-dipole) expression, we also see deviations in the intermediate regime. It appears that the small-distance result is only very accurate in the regime of such short distances that the accuracy of this calculation is questionable.
Chapter 13. Mechanical Effects of the Quantum Vacuum general result static-dipole approximation z (nm) level shift (MHz) -50 The shift of the D1 excited manifold is much stronger and oscillates due to the standing-wave pattern formed by the atomic radiation. The shift here is scalar, so that all hyperfine levels in the manifold are shifted equally. z (nm) 10000 level shift (MHz) -10 -5 The shifts of the levels in the D2 excited manifold depend substantially on the level, but the same general oscillatory behavior is apparent.
13.12 Lamb Shift¶
z (nm) 10000 level shift (MHz) 0.5 -1 -0.5 13.12 Lamb Shift The effect that we ignored when renormalizing the divergence in Section 13.3 is the Lamb shift,18 which is the shift of any atomic transition frequency due to the quantum vacuum, even in free space. 13.12.1 Coulomb Gauge We will now first examine a conventional nonrelativistic calculation of the Lamb shift using the usual Coulomb-gauge interaction Hamiltonian H(A) AF = e me
2me A2. (13.129) Again using second-order perturbation theory to compute the shift of level \alpha, we can write
X j X k,\zeta
, (13.130) which we will again evaluate to order e2 (second order in the atomic dipole moment). We will start by evaluating the first-order perturbation, to which only the A2 term contributes. Thus, we will write the first-order shift as ∆E(1) \alpha = e2 2me
(13.131) Using the expression
r ¯h
(13.132) 18Willis E. Lamb, Jr. and Robert C. Retherford, ‘‘Fine Structure of the Hydrogen Atom by a Microwave Method,’’ Physical Review 72, 241 (1947) (doi: 10.1103/PhysRev.72.241); H. A. Bethe, ‘‘The Electromagnetic Shift of Energy Levels,’’ Physical Review 72, 339 (1947) (doi: 10.1103/PhysRev.72.339); H. A. Bethe, L. M. Brown, and J. R. Stehn, ‘‘Numerical Value of the Lamb Shift,’’ Physical Review 77, 370 (1950) (doi: 10.1103/PhysRev.77.370); Edwin A. Power, ‘‘Zero-Point Energy and the Lamb Shift,’’ American Journal of Physics 34, 516 (1966) (doi: 10.1119/1.1973082).
Chapter 13. Mechanical Effects of the Quantum Vacuum from Eq. (8.68) for the vector potential quantized in free space, we can proceed in the usual way so that the first-order shift becomes ∆E(1) \alpha = e2¯h 4meϵ0V X k,\zeta \omegak = e2¯h 2meϵ0V X k \omegak = e2¯h 2(2\pi)3meϵ0c Z d3k 1 k = e2¯h 4\pi2meϵ0c Z \infty dk k. (13.133) This result is independent of the state |\alpha\rangle , and therefore only produces an overall shift of the atomic energy level. Since it does not contribute to the observable shifts of the atomic transition energies, we can neglect it. In the second-order shift, only the pe \cdot A term contributes, and thus
\alpha = e2 m 2e X j X k,\zeta
= e2 m 2e X j X k,\zeta
= −e2 m 2e ¯h 2ϵ0V X j X k,\zeta \omegak
. (13.134) Recall here that ¯h\omegaj\alpha := Ej −E\alpha and thus can be positive if level j is an excited state or negative if |j\rangle is a ground state with respect to |\alpha\rangle . In the continuum limit, the angular part of mode sum amounts an average over the relative orientation of the dipole and the field direction. We can thus take a uniform average over all dipole orientations and replace the squared dot product by a factor of 1/3. Continuing with the calculation,
\alpha = − e2 3m 2e ϵ0V X j X k
= − e2 3(2\pi)3m 2e ϵ0c2 X j
Z d3k
= − e2 6\pi2m 2e ϵ0c2 X j
Z \infty dk k
(13.135) Clearly this expression is divergent, and the divergence is asymptotically linear in k. The basic problem here is that high energies are only treated correctly in relativistic theory, and this is a nonrelativistic calculation. Nevertheless, Bethe19 used this nonrelativistic theory in a clever way to produce a finite prediction for the energy-level shift. The first step in Bethe’s argument is to realize that to get a finite energy shift, we will need to cut off the integral. Supposing that we cut off contributions from energies larger than some large energy \Lambda, we can write
\alpha = − e2 6\pi2m 2e ϵ0c2 X j
Z \Lambda/¯hc dk k
(13.136) The next step is Bethe’s mass renormalization. The idea is that the above energy shift contains the energy of the free electron, which is unobservable; we can only observe the shifts in transition frequencies, which 19H. A. Bethe, ‘‘The Electromagnetic Shift of Energy Levels,’’ Physical Review 72, 339 (1947) (doi: 10.1103/PhysRev.72.339).
13.12 Lamb Shift are related to the electron energies only in bound states. We can find the free-electron energy by taking the limit kj\alpha −\rightarrow 0 in Eq. (13.136), corresponding to a vanishing binding potential, so that ∆Efree \alpha = − e2 6\pi2m 2e ϵ0c2 X j
Z \Lambda/¯hc dk. (13.137) Subtracting this from Eq. (13.136), we use the integrand subtraction k
kj\alpha
(13.138) and thus we find the renormalized energy ∆E\alpha −∆Efree \alpha = e2 6\pi2m 2e ϵ0c2 X j
Z \Lambda/¯hc dk
(13.139) This renormalization is important in that it has reduced the divergence from linear to logarithmic, and thus the result is now relatively insensitive to the value of the cutoff \Lambda. The interpretation of the renormalization is as follows: in relativistic theory, the free electron energy due to the field coupling shifts the atomic rest mass, but by using the observed (or renormalized) mass me, we have already included this contribution, and should not double-count it. This is the rationale for subtracting it after computing the atom-field coupling energy.20 Now we can carry out the integration in Eq. (13.139), using Z t dx
Z t d(x + a) x + a
t = log |t + a| −log |a| = log
t a + 1
\approx log t |a|, (13.140) where the last equality holds if t > 0 and t ≫|a|. Thus, we obtain the Bethe logarithm ∆E\alpha −∆Efree \alpha = e2 6\pi2m 2e ϵ0c2 X j
\Lambda
(13.141) The final step is to choose the cutoff \Lambda. Bethe chose \Lambda = mec2 as a reasonable energy beyond which the nonrelativistic theory should fail. Thus, ∆E\alpha −∆Efree \alpha = e2 6\pi2m 2e ϵ0c2 X j
(13.142) (Lamb shift) This is the basic, nonrelativistic result for the Lamb shift of level |\alpha\rangle . Notice that for the ground state, kjg > 0, and thus the Lamb shift is positive. Viewed as a Stark shift, evidently the ultraviolet modes far above the dominant transition frequencies are the most important in determining the shift. 20For a much more detailed discussion of this point, as well as a nice historical account and many viewpoints of the Lamb shift, see Peter W. Milonni, op. cit., Sections 3.4-3.9, pp. 82-96.
Chapter 13. Mechanical Effects of the Quantum Vacuum 13.12.1.1 Evaluation Using this energy shift, Bethe computed a numerical value for the Lamb shift of the 2S −\rightarrow 2P fine-structure transition in atomic hydrogen—a transition between degenerate states according to the solutions to the Dirac equation. As a rough approximation, Bethe noted that the arguments of the logarithms are large and relatively weakly dependent on the transition frequencies. As a rough approximation, he assumed it was constant in the level summation and replaced it by an average excitation energy. Thus, the level sum amounts to the sum X j
¯hc X j
= 1 ¯hc X j
= −i
= −i
= ¯h
= ¯h 2c Z
e V (re)
c
(13.143) where in the last step we used the Coulomb binding potential
re , (13.144) so that \nabla 2
(13.145)
nonvanishing at the nucleus. We can then write the observed Lamb shift from (13.142) as ∆Eobserved \alpha \approx 4\alpha0Z e¯h mec 2
mec2
(13.146) where again \alpha0 := e2/4\piϵ0¯hc is the fine-structure constant. Then for a hydrogenic S state in orbital n,
\pi Z na0 3 , (13.147) where a0 = ¯h/mec\alpha0 is the Bohr radius, so that ∆Eobserved n \approx 8\alpha 3 0 Z4 3\pin3 R\inftylog mec2
(13.148) (S-state Lamb shift)
here is much larger than R\inftysince evidently ionized states make a large contribution to the sum. Using these results, Bethe arrived at a shift of 1040 MHz for the 2S state (with negligible shift for the 2P state). This is in surprisingly good agreement with the modern value of about 1058 MHz, considering the ad hoc nature of the calculation.
13.12 Lamb Shift 13.12.2 Electric Dipole Interaction It is again interesting to see how the Lamb shift arises from the electric-dipole Hamiltonian, which we showed in Eq. (9.38) has the form in free space takes the form
2ϵ0 Z d3r P ⊥2(r), (13.149) where the last term is the dipole self-energy, with P ⊥
(13.150) Starting with the second-order shift due to the −d\cdotE⊥(0) part of the interaction, we have already computed as the position-independent part of the Casimir–Polder expression (13.17), which we can write as a shift for level |\alpha\rangle as ∆Ed\cdotE \alpha = − 2ϵ0V X j X k \omegak
d 2
j\alpha,z + k 2 z k2 d 2
j\alpha,z . (13.151) For a spherically symmetric atom, the asymmetric part of the dipole vanishes, and so ∆Ed\cdotE \alpha = − 3ϵ0V X j |dj\alpha|2 X k \omegak
= − 3(2\pi)3ϵ0 X j |dj\alpha|2 Z d3k k
= − 6\pi2ϵ0 X j |dj\alpha|2 Z \infty dk k3
(13.152) Again, the infinite upper limit here is understood to be an appropriate ultraviolet cutoff. Note, however, that the asymptotic scaling of the integrand is now k2, where it was k0 in the p \cdot A calculation. Obviously we are missing some contributions that will make the scaling correct. 13.12.2.1 Dipole Self-Energy Now to evaluate the dipole self-energy term HP⊥= 2ϵ0 Z d3r P ⊥2(r) (13.153) of the interaction Hamiltonian. Since \partial \alpha\delta⊥
transverse vector field for any particular value of \beta, and thus we may write (see also Problem 8.11) Z d3r \delta⊥
(13.154) Thus, HP⊥= 2ϵ0 Z d3r P ⊥2(r) = e2 2ϵ0 Z
= e2 2ϵ0
\alpha\beta(0) = e2
Z d3k
k2 , (13.155)
Chapter 13. Mechanical Effects of the Quantum Vacuum where we have used the momentum-space representation of the transverse delta function from Eq. (8.178). Again, since we are calculating the Lamb shift to order e2, it is sufficient to consider the level shift due to the dipole self-energy to first order in perturbation theory. In this case, the shift of level \alpha is ∆EP⊥ \alpha
= e2
Z d3k
k2 = e2 2ϵ0(2\pi)3 X j
Z d3k
k2 = 2ϵ0(2\pi)3 X j (d\alphaj)µ(d∗
Z d3k
k2 . (13.156) Notice that for a spherically symmetric atom, the dipole matrix elements are independent of direction, and thus (d\alphaj)µ is independent of µ, and we can carry out the sum to write ∆EP⊥ \alpha
= 2ϵ0(2\pi)3 X j |d\alphaj|2 Z d3k 3 −kµkµ k2 = 3ϵ0(2\pi)3 X j |d\alphaj|2 Z d3k = 6\pi2ϵ0 X j |d\alphaj|2 Z \infty dk k2. (13.157) We can then write the total shift as
\alpha + ∆EP⊥ \alpha = 6\pi2ϵ0 X j |dj\alpha|2 Z \infty dk − k3
= 6\pi2ϵ0 X j
Z \infty dk k2
(13.158) By accounting for the dipole self-energy, we have reduced the order of the divergence, and we are on track to obtain the correct Lamb shift.21 13.12.2.2 Mass Renormalization Again, we must subtract the free-electron energy, which we will calculate from the Coulomb-gauge Hamilto- nian (13.129). We computed the contribution from the A2 Hamiltonian in Eq. (13.133) and found ∆E(1), free \alpha = e2¯h 4\pi2meϵ0c Z \infty dk k. (13.159) Recall from Eq. (13.41) that the TRK sum rule is X j
¯h 2me , (13.160) 21The importance of the dipole self-energy and the following renormalization procedure were pointed out by E. A. Power and S. Zienau, ‘‘Coulomb Gauge in Non-Relativistic Quantum Electro-Dynamics and the Shape of Spectral Lines,’’ Philo- sophical Transactions of the Royal Society of London. Series A, Mathematical and Physical Sciences 251, 427 (1959) (doi: 10.1098/rsta.1959.0008).
13.13 Casimir–Polder Potential for a Rarefied Dielectric Surface which we can also write as X j
2me . (13.161) This gives ∆E(1), free \alpha = 6\pi2ϵ0 X j
Z \infty dk k. (13.162) Subtracting this part of the free-electron energy from the Lamb shift (13.158), we find ∆E\alpha −∆E(1), free \alpha = 6\pi2ϵ0 X j
Z \infty dk k2
= − 6\pi2ϵ0 X j k 2
Z \infty dk k
(13.163) This result is, in fact, equivalent to the result (13.135) in the Coulomb-gauge Hamiltonian before mass renormalization in that gauge. We can again perform the same renormalization with the p \cdot A part of the Hamiltonian, which we have already computed in Eq. (13.137): ∆E(2), free \alpha = − e2 6\pi2m 2e ϵ0c2 X j
Z \infty dk = − 6\pi2ϵ0 X j k 2
Z \infty dk. (13.164) Subtracting this part (i.e., the rest) of the electron free energy, we find ∆E\alpha −∆Efree \alpha = 6\pi2ϵ0 X j k 2
Z \infty dk − k
= 6\pi2ϵ0 X j k 3
Z \infty dk
= 6\pi2ϵ0 X j k 3 j\alpha|dj\alpha|2 log mec2
(13.165) This result is exactly equivalent to the Coulomb-gauge result (13.142), if we use the conversion between the momentum and dipole matrix elements. 13.13 Casimir–Polder Potential for a Rarefied Dielectric Surface At this point, we return to the atom–surface potential, but instead of a perfect conductor, as we have been considering, we will consider instead a dielectric planar surface, with the dielectric filling half of all space. This problem is better handled with the powerful formalism of electromagnetic Green tensors [Section 14.3.5.6, particularly Eqs. (14.211) and (14.212)]; however, this setup is a nice example of the mode-summation formalism applied to a different set of modes.22 We will also approach the problem slightly differently, computing the local (renormalized) energy density of the electromagnetic field due to the surface, and then convert this into a Casimir–Polder potential. To keep things simple, we will assume a ‘‘rarefied’’ dielectric, such that the dielectric susceptibility \chi [see Eqs. (14.16) and (14.18)] is small (i.e., we will work to first order in \chi). We will also ignore dispersion of the dielectric (i.e., \chi is independent of frequency). 22for a general approach to the Casimir–Polder potential of an atom near a planar dielectric via mode summation, see Shin-Tza Wu and Claudia Eberlein, ‘‘Quantum Electrodynamics of an Atom in Front of a Non-Dispersive Dielectric Half- Space,’’ Proceedings of the Royal Society of London A: Mathematical, Physical and Engineering Sciences 455, 2487 (1999) (doi: 10.1098/rspa.1999.0413). See also Nicola Bartolo and Roberto Passante, ‘‘Electromagnetic-field fluctuations near a dielectric-vacuum boundary and surface divergences in the ideal conductor limit,’’ Physical Review A 86, 012112 (2012) (doi: 10.1103/PhysRevA.86.012122).
Chapter 13. Mechanical Effects of the Quantum Vacuum z c For the geometry, we will assume as before that the atom is a distance z above the dielectric, with the dielectric occupying the half space z < 0. We want to calculate the electromagnetic energy density for the vacuum state of the field, which is the expectation value of the Hamiltonian (8.34),
= ϵ0
E2(r, t)
- ϵ0c2
B2(r, t)
= ϵ0
(13.166) where we use ϵ0 here instead of ϵ(r) since we are interested in the energy density outside the dielectric. Note that the two contributions here, from the electric and magnetic fields, are in fact equal in vacuum, but we will see that their changes due to a planar boundary are not the same. We will begin with the electric-field energy, which is the important part for a polarizable, nonmagnetic atom interacting with the dielectric, as it will not ‘‘see’’ the magnetic field (more generally, the interaction via the magnetic dipole moment will be much weaker than via the electric dipole moment). The quantum electric field from Eq. (8.56) is
X k,\zeta r ¯h\omegak 2ϵ0
(13.167) where the ak,\zeta are the field annihilation operators, and the fk,\zeta(r) are the unit-normalized mode functions. Noting that in the vacuum expectation value, the only nonvanishing terms have the form of aa\dagger for some mode, and so
X k,\zeta ¯h\omegak
X k,\zeta ¯h\omegak 2 |fk,\zeta(r)|2, (13.168) which is half of the zero-point energy of each mode, spatially distributed via the mode function, and then summed over all modes. We only have a half of the zero-point energy because we are ignoring the half due to the magnetic fields thus far. To set up the modes, we will need the Fresnel reflection coefficients23
p
p
= −
p
p
= 1 cos2 \theta −2
(13.169) which give the amplitude of the reflection of a plane wave from the dielectric surface, normalized to the incident amplitude. Here, \theta is the angle of incidence, measured from the normal, and the two polarizations are transverse-electric (TE, where the electric field is parallel to the surface) and transverse-magnetic (TM, where the magnetic field is parallel to the surface). We have already expanded these expressions to lowest order in \chi, in anticipation of our perturbative calculation. 23Daniel A. Steck, Classical and Modern Optics (2006), Chapter 9. Available at http://steck.us/teaching. Note that rTE ≡rS and rTM ≡rP in the notation there. See also Eq. (14.175) for further usage information.
13.13 Casimir–Polder Potential for a Rarefied Dielectric Surface 13.13.1 TE Energy Density To proceed with the computation of the energy density, we will begin with the TE modes. Modes incident from above must include incident, reflected, and transmitted plane-wave components, and the modes have the form f↓
h
p 1 −|rTE|2eikt\cdotr\Theta(−z) i , (13.170) where the arrow superscript indicates the direction of incidence, k is the incident wave vector (with kz < 0), k−= kxˆx + kyˆy −kzˆz is the reflected wave vector, and kt is the transmitted wave vector. Notice that the polarizations of all fields match, as required by continuity of the surface-transverse component of the electric field.24 (Although kt is determined by Snell’s Law, its precise form turns out to be irrelevent to the calculation here.) We have written the transmitted field in terms of the reflection coefficient so that the mode is explicitly normalized to a unit integral like the free-space mode functions, which is sufficient for our purposes here (technically, they should be normalized under an ϵ integral measure, but since we are staying outside the dielectric, this will not matter). Notice also that we have adopted unconfined mode functions (Section 8.7) here, so there is no explicit quantization volume. The contribution of these modes to the electric-field energy density (13.168) is EE ↓
¯hc 4(2\pi)3 Z kz<0 d3k k|f↓ k,TE(r)|2, (13.171) where we have written the sum as an integral and written \omegak = ck. Then for z > 0, EE ↓
¯hc 4(2\pi)3 Z kz<0 d3k k
= ¯hc 4(2\pi)3 Z kz<0 d3k k 1 + r 2 TE + 2rTE cos(2kzz) . (13.172) Since we only want the renormalized energy, we drop the z-independent terms: EE ↓
¯hc 2(2\pi)3 Z kz<0 d3k krTE(\theta) cos(2kzz). (13.173) Before proceeding, consider the TE modes incident from below, which are of the form f↑
h eik\cdotr\Theta(−z) −rTE(\theta)eik−\cdotr\Theta(−z) + p 1 −|rTE|2eikt\cdotr\Theta(z) i , (13.174) where the sign of the reflection is changed to respect the unitarity of the interface reflection, and kt is once again the transmitted wave vector, which we will leave unspecified. The associated energy density is EE ↑
¯hc 4(2\pi)3 Z kz>0 d3k k|f↑
¯hc 4(2\pi)3 Z kz>0 d3k k 1 −|rTE|2 , (13.175) which is z-independent, and thus should be entirely dropped upon renormalization. Note that the total we have discarded so far is equivalent to the total energy density of the free TE field, and the only contribution that we have kept is the interference term between the incident and reflected fields. Thus, the total TE energy density after renormalization is
¯hc 2(2\pi)3 Z kz<0 d3k krTE(\theta) cos(2kzz). (13.176) This integral turns out to be somewhat tricky to evaluate. Writing out the integral in spherical coordinates, we have
8\pi2 Z \infty dk Z \pi/2 d\theta sin \theta k3rTE(\theta) cos(2kz cos \theta) = ¯hc 8\pi2 Z \infty dk Z 1 d\xi k3rTE(\xi) cos(2kz\xi), (13.177) 24Daniel A. Steck, op. cit.
Chapter 13. Mechanical Effects of the Quantum Vacuum where we have chosen the range of \theta integration to be compatible with the definition for the reflection- coefficient angle, and
(13.178) The k integral can be performed most easily by inserting a convergence factor of exp(−ak) to cut off high frequencies, and the letting a −\rightarrow 0 afterwards (recall the discussion of convergence in Section 13.5, with the result
3¯hc 64\pi2z4 Z 1
\xi4 . (13.179) This integral unfortunately diverges at \xi = 0; inserting the reflection coefficient doesn’t help, since it is
result blows up! 13.13.1.1 Digression: Back to the Perfect Conductor For inspiration on how to proceed, let’s briefly go back to the familiar territory of the perfect conductor, where rTE = −1. We can then evaluate the integral as follows:
8\pi2 Z \infty dk Z \pi/2 d\theta sin \theta k3 cos(2kz cos \theta) = −¯hc 8\pi2 Z \infty dk k3 sin(2kz) 2kz = − ¯hc 8\pi2(2z)4 Z \infty dk k2 sin k = −lim a\rightarrow 0 ¯hc 8\pi2(2z)4 Z \infty dk k2e−ak sin k = − ¯hc 8\pi2(2z)4 (−2) = ¯hc 64\pi2z4 , (13.180) where we have first carried out the angular integral, and then carried out the k integral by inserting a convergence factor. Finally, the result is
3¯hc 32\pi2z4 1 . (TE energy density, perfect conductor) (13.181) This is the energy density in the perfect-conductor limit, and we will comment on this after we complete the TE and TM calculations for the rarefied dielectric. 13.13.1.2 Integral for the Rarefied Dielectric Evidently, the key to the success in the perfect-conductor case is in first carrying out the angular integration, and we will be more careful about the order of integration. Writing out the integral (13.176) again in spherical coordinates, we have
8\pi2 Z 1
Z \infty dk k3 cos(2kz\xi). (13.182) The problem here is that while rTE(\xi) is a well-behaved function, it is complex enough that it is difficult to perform the integration with the cos factor. And we can simplify the expression by using the lowest-order
13.13 Casimir–Polder Potential for a Rarefied Dielectric Surface first carrying out the k integral by parts to build up powers of \xi that will regularize the integral at \xi = 0. Performing the k integral by parts four times, we have
8\pi2 (2z)4 Z 1
Z \infty dk k7 cos(2kz\xi). (13.183) We have ignored boundary terms here. At k = 0, the integrand always involves a positive power of k, so this is obvious. At k = \infty, we are again forcing the integral to converge by cutting it off with an exponential convergence factor, or equivalently by regarding the cosine integration as being shifted in the complex plane off of the real axis as cos k = 1
, (13.184) so that the value at infinity is suppressed without affecting the value of the integral. We then proceed by
8\pi2 z4 210\chi Z \infty dk k7 Z 1
= − ¯hc
Z \infty dk k7 Z 1
= − ¯hc
Z \infty dk k7 2k cos k + (k2 −2) sin k k3 = − ¯hc
= 3¯hc 32\pi2z4 \chi , (13.185) where we changed variables k −\rightarrow k/2z. The final result is
3¯hc 32\pi2z4 \chi . (TE electric energy density, rarefied dielectric) (13.186) Note that this is equivalent to the energy density (13.181) for the perfect conductor, up to a factor 6\chi/40. 13.13.2 TM Energy Density Now we can repeat the calculation for the TM modes. The polarization of the modes here works out to be slightly more complicated, because the polarization is not parallel to the surface, but only the parallel component is continuous across the dielectric interface. The resulting mode is25 f↓
k,TM p 1 −|rTM|2eikt\cdotr\Theta(−z), (13.187) where the reflected polarization vector ˆ\epsilon− k,TM is the same as the incident vector ˆ\epsilonk,TM, but with the z compo- nent reversed, and ˆ\epsilont k,TM is the polarization vector of the transmitted wave, whose precise form we will not need. However, note that both polarization vectors are orthogonal to k and each other, so that
ˆ\epsilon−
ˆ\epsilont
(13.188) where ˆk−also has its z-component reversed compared to ˆk. Considering only the z > 0 region, the only z-dependent part of the mode envelope that we will need is the cross-term between the incident and reflected fields: |f↓
(13.189) 25Daniel A. Steck, op. cit.
Chapter 13. Mechanical Effects of the Quantum Vacuum where the angular factor comes from the inner product of the incident and reflected polarization vectors. As in the TE case, the modes incident from the dielectric side do not contribute any z-dependent terms. Thus, the TM contribution to the electric energy density (13.168) is
¯hc 2(2\pi)3 Z kz<0 d3k krTM(\theta) 2 cos2 \theta −1 cos(2kzz), (13.190) after renormalizing away the vacuum contribution, changing the mode sum to an integral, and using the double-angle formula cos 2\theta = 2 cos2 \theta −1. We will adopt the same method of evaluation as in the TE case. Writing out the integral in spherical coordinates,
8\pi2 Z 1
Z \infty dk k3 cos(2kz\xi), (13.191) where again \xi = cos \theta. Integrating by parts in k four times,
8\pi2 (2z)4 Z 1
Z \infty dk k7 cos(2kz\xi), (13.192) where we dropped boundary terms as before.
proceeding as in the perfect-conductor case,
8\pi2 z4 210\chi Z 1
1 \xi2 −2 Z \infty dk k7 cos(2kz\xi) = ¯hc 8\pi2 z4 210\chi Z \infty dk k7 Z 1 d\xi
cos(2kz\xi) = ¯hc
Z \infty dk k7 Z 1 d\xi
cos(k\xi), (13.193) where we have again changed variables k −\rightarrow k/2z. Carrying out the \xi integral,
¯hc
Z \infty dk
sin k −2k 1440 −192k2 + 5k4 cos k = ¯hc
= 43¯hc
(13.194) where we have again used a convergence factor to perform the k integration.
3¯hc 32\pi2z4 43\chi , (TM electric energy density, rarefied dielectric) (13.195) Note that this is equivalent to the TE energy density (13.186) up to an overall constant.
13.13 Casimir–Polder Potential for a Rarefied Dielectric Surface 13.13.2.1 Digression: TM Energy Density for the Perfect Conductor For completeness, we can also compute the TM energy for the perfect-conductor case. Putting rTM = −1 into Eq. (13.191), we have
8\pi2 Z \infty dk k3 Z 1
= − ¯hc 8\pi2(2z)4 Z \infty dk k3 Z 1
= − ¯hc 8\pi2(2z)4 Z \infty dk 4k cos k + (k2 −4) sin k = − ¯hc 8\pi2(2z)4 (−10), (13.196) with the same procedure as for the dielectric. Finally, the result is
3¯hc 32\pi2z4 5 . (TM electric energy density, perfect conductor) (13.197) This is the same as the TE energy density for the perfect conductor (13.181), but with an extra factor of 5. 13.13.3 Total Casimir–Polder Potential for a Polarizable Atom Summing the two polarization contributions (13.186) and (13.195) for a rarefied dielectric, the total electric- field energy density
3¯hc 32\pi2z4 23\chi . (electric energy density, rarefied dielectric) (13.198) Notice that the contribution of the TM mode to the energy density is much larger than the contribution of the TE mode (43/120 vs. 1/40). Similarly, summing the two contributions (13.181) and (13.197) for the perfect conductor gives
3¯hc 32\pi2z4 . (electric energy density, perfect conductor) (13.199) Note that we have written all of the energy densities as a fraction of this one, which is the largest of all the cases. Again, the contribution of the TM mode to the energy density is much larger than the contribution of the TE mode (5/6 vs. 1/6). This may be somewhat surprising, as in the dielectric case, |rTM| < |rTE| (recall that TM light is transmitted perfectly at Brewster’s angle, but not TE light), and in both cases, the interference is imperfect due to polarization mismatching in the TM case. To relate the electric energy density to the Casimir–Polder potential for an atom, recall [see Eq. (1.60)] that the dipole potential for an atom interacting via its induced dipole moment with an electric field has the
where only the low-frequency modes contribute to the potential (the equivalent of the limit of large atomic resonance frequencies). There is no dc field, but we have exactly computed the vacuum expectation value of E2 (with a factor of ϵ0/2), so we wish to consider Vdipole = −1 2\alpha0
E2 . (13.200) Combining this with Eq. (13.166), we can relate the atomic potential directly to the energy density via Vdipole = −\alpha0 ϵ0 EE(r). (13.201) (far-field Casimir–Polder potential)
Chapter 13. Mechanical Effects of the Quantum Vacuum Using this result, the perfect-conductor Casimir–Polder potential becomes Vdipole = 3¯hc\alpha0 32\pi2ϵ0z4 (Casimir–Polder potential, perfect conductor) (13.202) in the far field, which agrees with Eq. (13.60). For the rarified dielectric, the Casimir–Polder potential is Vdipole = 3¯hc\alpha0 32\pi2ϵ0z4 23\chi , (Casimir–Polder potential, rarefied dielectric) (13.203) or a factor of 23\chi/60 times the perfect-conductor result, valid for \chi ≪1. The results here agree with the limiting cases of the general treatment in Section 14.3.5.6. 13.13.4 Magnetic-Field Energies While the magnetic fields do not contribute to the Casimir–Polder potential of an electrically polarizable atom, it is nonetheless interesting to compute the magnetic-field energies, and sum these to find the total energies. While the electric and magnetic field energies are equivalent in vacuum, the changes in the energy densities due to the dielectric are not the same. The derivation will be approximately the same as for the electric-field case, but with two modifications, due to the magnetic fields are orthogonal to both the electric fields and wave vector (i.e., the magnetic-field modes involve \nabla \times f rather than f. The first modification is that the interference terms from the incident and reflected fields in both polarizations gain a relative minus sign compared to the electric-field case, because of this orthogonality condition (in particular, E \times B must point in the direction of k. So for example, for normal incidence, if the incident and reflected waves are assumed to have parallel (electric-field) polarizations, the magnetic-field vectors will be antiparallel. The second modification is that the geometry factor of cos 2\theta in the TM expression (13.189), which later becomes a factor of 2\xi2 −1, moves from the TM case to the TE case for the magnetic fields. These considerations are best visualized by comparing a diagram of the electric and magnetic fields at a dielectric interface for TE polarization, c z = 0 qi qr qt E0i (+) H0i (+) E0r (+) H0r (+) E0t (+) H0t (+) and for TM polarization,
13.13 Casimir–Polder Potential for a Rarefied Dielectric Surface c z = 0 qi qr qt E0i (+) H0i (+) H0r (+) E0r (+) E0t (+) H0t (+) noting the relative orientations of the incident and reflected vectors in each case. 13.13.4.1 Magnetic TE Energy Densities Starting with the expression (13.182) and making the sign and geometric-factor changes, we have
8\pi2 Z 1
Z \infty dk k3 cos(2kz\xi). (13.204) The procedure to evaluate this is to integrate by parts four times,
8\pi2 (2z)4 Z 1
Z \infty dk k7 cos(2kz\xi), (13.205)
as in the perfect-conductor case,
8\pi2 z4 210\chi Z 1
1 \xi2 Z \infty dk k7 cos(2kz\xi) = ¯hc 8\pi2 z4 210\chi Z \infty dk k7 Z 1
cos(2kz\xi) = ¯hc
Z \infty dk k7 Z 1
cos(k\xi) = ¯hc
Z \infty dk k2 6k(k2 −8) cos k + (48 −22k2 + k4) sin k = ¯hc
(13.206) or simplifying the result,
3¯hc 32\pi2z4 −7\chi . (TE magnetic energy density, rarefied dielectric) (13.207) For the perfect conductor case, we instead set rTE = −1. This is the same as the TM electric energy density for the perfect conductor, so Eq. (13.197) can be adapted here as
3¯hc 32\pi2z4 −5 , (TE magnetic energy density, perfect conductor) (13.208) where we have only had to change the overall minus sign.
Chapter 13. Mechanical Effects of the Quantum Vacuum 13.13.4.2 Magnetic TM Energy Densities For the TM polarization, we take the electric TE expression, remove the geometric factor, and put in an overall minus sign, so Eq. (13.191) becomes
8\pi2 Z 1
Z \infty dk k3 cos(2kz\xi), (13.209)
8\pi2 (2z)4 Z 1
Z \infty dk k7 cos(2kz\xi) = −¯hc 8\pi2 z4 210\chi Z 1 d\xi \xi4 1 \xi2 −2 Z \infty dk k7 cos(2kz\xi) = −¯hc 8\pi2 z4 210\chi Z \infty dk k7 Z 1
cos(2kz\xi) = − ¯hc
Z \infty dk k7 Z 1
cos(k\xi). (13.210)
small \chi. Thus, we have the same contribution as in Eq. (13.207).
3¯hc 32\pi2z4 −7\chi . (TM magnetic energy density, rarefied dielectric) (13.211) For the perfect conductor case, we again instead set rTM = −1. This is the same as the TE electric energy density for the perfect conductor, so Eq. (13.181) can be adapted here as
3¯hc 32\pi2z4 −1 , (TM magnetic energy density, perfect conductor) (13.212) where we have again only changed the overall minus sign. 13.13.4.3 Total Magnetic and Electromagnetic Energy Densities The total magnetic energy density for the rarefied dielectric is the sum of Eqs. (13.207) and (13.211) or
3¯hc 32\pi2z4 −7\chi . (magnetic energy density, rarefied dielectric) (13.213) Then in the case of the rarefied dielectric, the total electromagnetic energy is given by the sum of Eqs. (13.198) and (13.213), which is
3¯hc 32\pi2z4 4\chi . (total electromagnetic energy density, rarefied dielectric) (13.214) Thus, the magnetic energy density cancels some of the electric energy density (after renormalization against vacuum). For the perfectly conducting plane, the total magnetic energy density is the sum of Eqs. (13.208) and (13.211)
3¯hc 32\pi2z4 , (magnetic energy density, perfect conductor) (13.215)
13.13 Casimir–Polder Potential for a Rarefied Dielectric Surface which is exactly the opposite of the electric-field contribution (13.199), so that the total energy density is
(total electromagnetic energy density, perfect conductor) (13.216) Remarkably, the change in the total energy density vanishes for the perfect conductor. While the dielectric boundary increases the fluctuations (and thus the energy) of the electric field, it tends to suppress those of the magnetic field, at least outside the dielectric. This means that a hypothetical atom that was not electri- cally polarizable, but purely magnetic, would experience a repulsive Casimir–Polder force near a dielectric boundary, at least in the limits of a weak and strong dielectric.
Chapter 13. Mechanical Effects of the Quantum Vacuum 13.14 Exercises Problem 13.1 The sine integral is defined as
2 − Z \infty x sin t t dt, (13.217) and the cosine integral is defined as
Z \infty x cos t t dt. (13.218) The auxiliary functions are also defined as
h\pi 2 −Si(z) i
h\pi 2 −Si(z) i . (13.219) (a) Show that
Z \infty sin t t + z dt (|arg z| < \pi)
Z \infty cos t t + z dt (|arg z| < \pi). (13.220) (b) Show that
Z \infty e−zt 1 + t2 dt (Re[z] > 0)
Z \infty t e−zt 1 + t2 dt (Re[z] > 0). (13.221) You can do this by considering the combination g(z) + if(z), writing out its integral expression, and then changing to an integral along the positive imaginary axis. Note that simple changes of variable justify the more general integral formulae26 Z \infty sin(ax)
Z \infty cos(ax)
Z \infty e−µx
\beta (Re[µ] > 0, Re[\beta] > 0) Z \infty x e−µx
(Re[µ] > 0, Re[\beta] > 0) (13.222) that are useful for evaluating the Casimir–Polder potential near perfectly conducting planes. Problem 13.2 Prove the ‘‘branch-cut formulae’’ for the cosine integral,
(0 < arg z < \pi)
(−\pi < arg z < 0), (13.223) 26See I. S. Gradstein and I. M. Ryzhik, Table of Integrals, Series, and Products, English translation 6th ed., A. Jeffrey and D. Zwillinger, Eds. (Academic Press, 2000), Integrals 3.722.1, 3.722.3, 3.354.1, and 3.354.2.
13.14 Exercises where recall that we define the cosine integral by¶
Z \infty x cos t t dt. (13.224) You can do this as follows. (a) Defining the exponential integral
Z \infty x e−t t dt, (13.225) show that
\infty X j=1 (−1)jxj jj! , (13.226) where \gamma \approx 0.577 215 664 901 532 860 607 is the Euler–Mascheroni constant, defined to be the asymptotic difference between the partial sums of the harmonic series and the logarithmic function:
n\rightarrow \infty 1 + 1 2 + 1
n −log n . (13.227) Hint: you may find it helpful to use the integral relation
Z \infty e−t log t dt, (13.228) but if you use it, you should prove it. Whether or not you use it, it will help to get started by proving that 1 + 1 2 + 1
n = Z 1 1 −(1 −t)n t dt. (13.229) You should prove this by induction (if you don’t know what that means, then make sure you find out). Then recalling that the exponential function may be defined by ex = lim n\rightarrow \infty 1 + x n n , (13.230) you can, for example, establish Eq. (13.228). In general, you should be integrating by parts like crazy in this problem. (b) Let t −\rightarrow xt in Eq. (13.225) to remove x from the integration limit, and then show that
E1(ix) + E1(−ix) . (13.231) (c) Use the result of (a) to show that
Z x 1 −cos t t dt. (13.232) (d) Now recall that the log function has a branch cut along the negative real axis, and is otherwise
prove Eqs. (13.223).
Chapter 13. Mechanical Effects of the Quantum Vacuum Problem 13.3 Show that the integrals I1 = Z d3k k (k + k0) cos(2kzz) I2 = Z d3k k 2 z k(k + k0) cos(2kzz). (13.233) can also be performed in cylindrical coordinates, and agree with the results from spherical-coordinate integration in the region z > 0. Be careful to avoid infinities! Problem 13.4 Show that the integrals I′ 1 = k 2 Z d3k k(k + k0) cos(2kzz) I′ 2 = k 2 Z d3k k 2 z k3(k + k0) cos(2kzz). (13.234) can also be performed in cylindrical coordinates, and agree with the results from spherical-coordinate integration in the region z > 0. Problem 13.5 (a) Show that expression (13.46) for the Casimir–Polder potential VCP =
X j " d 2 j,∥/2 −d 2 j,z 4k 2 j0 z − d 2
j,z \partial 2 z z # f(2kj0z) (13.235) reduces in the long-distance limit to VCP = − X j c
d 2 j,∥+ d 2 j,z 1 z4 = − X j cd 2 j
(13.236) so that even for an anisotropic molecule, the far-field potential is independent of the molecular orien- tation and the potential still scales as z−4. (b) Defining the normal and parallel static polarizabilities by
X j 2d 2 j,z ¯h\omegaj0
X j d 2 j,∥ ¯h\omegaj0 , (13.237) show that the far-field Casimir–Polder potential can then be written VCP = − ¯hc
1 z4 . (13.238) Justify the above definitions of the static polarizability on the basis of the usual expression (13.61). Also show that for an isotropic atom, the Casimir–Polder potential here reduces to the usual far-field formula, Eq. (13.60). (c) Argue that the anisotropic correction—the term involving (d 2 j,∥/2−d 2 j,z)—to the near-field potential (van der Waals regime) should be negligible compared to the usual component for the spherically symmetric part of the atom.
13.14 Exercises Problem 13.6 Starting with the integral expression (13.23) for the Casimir–Polder potential, VCP = 16\pi3ϵ0 X j Z \infty −\infty dkx Z \infty −\infty dky Z \infty −\infty dkz k (kj0 + k) d 2 j,∥/2 −d 2 j,z + k 2 z k2 d 2
j,z cos(2kzz), (13.239) obtain directly the asymptotic form for large z for a spherically symmetric atom (that is, do not use the full expression in terms of the auxiliary function f(z) as an intermediate step). To do this, note that for large z, cos(2kzz) is a rapidly oscillating function of kz (and thus of k), and thus the integrand only gives a nonvanishing contribution where kz (and thus k) is close to zero. Problem 13.7 Work through the entire derivation of the Casimir–Polder potential near a planar conductor, but with the following simplifications, which you should implement as early as possible in the derivation: assume a spherically symmetric atom from the beginning, and work in the far field, where only low frequencies are important (see the comment in Problem 13.6). Also, start with the mode functions inside a rectangular cavity. Problem 13.8 Compute the Casimir–Polder potential for a spherically symmetric atom near an L-shaped conductor, where the conductor occupies the region x < 0 and z < 0, with the atom at (x > 0, y, z > 0). Use the setup of Problem 13.7, including the assumption of being in the far-field regime. z x In particular, you should find a potential of the form
CP + V (z) CP + V (xz) CP , (13.240) where V (x) CP and V (z) CP have the form for atom-plane Casimir–Polder energies, and V (xz) CP is a nonadditive correction for the atom interacting with both surfaces. You should be able to give a simple interpreta- tion for V (xz) CP once simplified (in fact it is the Casimir–Polder potential for an atom–surface distance \sqrt x2 + z2. Problem 13.9 Work out an integral expression for the Casimir–Polder potential, in analogy to Eq. (13.23), for the electromagnetic field in two dimensions (that is, two-dimensional space with a line conductor, which of course does not correspond to reality; instead, consider the physical problem of an atom constrained to be halfway between two planar, parallel conductors, spaced by the very small distance ℓ, and the atom is a distance z from a third planar conductor that intersects the other two conductors perpendicularly). For simplicity, just work out the case of a spherically symmetric atom. Show that the potential scales asymptotically as z−3 at large z. Problem 13.10 Recompute the Casimir–Polder force for a spherically symmetric atom near a perfectly conducting
Chapter 13. Mechanical Effects of the Quantum Vacuum plane, but using only the TE modes [Eq. (8.76)] f (TE) k
r V ˆk∥\times ˆz sin kzz eik∥\cdotr (13.241) (i.e., ignoring the contribution of the TM modes). Show that in the far-field regime, the TE modes contribute only 1/6 of the total potential. Problem 13.11 Derive an expression for the ground-state Casimir–Polder potential of an atom near a perfectly con- ducting, infinite surface, generalized to the case where the electromagnetic field is at temperature T.