INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS

Transcription

1 4 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS 4. Resonant Faraday rotation When linearly polarized light propagates through a medium immersed in a magnetic field, the plane of light polarization at the output is rotated (Fig. 4.); this effect was observed by Michael Faraday almost one hundred and fifty years ago (Faraday 855). In 898, the Italian physicists D. Macaluso and O. M. Corbino discovered that Faraday rotation was resonantly enhanced near atomic absorption lines (Macaluso and Corbino 898). Circular Components Medium Linear Polarization Magnetic Field B FIG. 4. Conceptual setup for observing Faraday rotation. Linearly polarized light enters a medium subjected to a longitudinal magnetic field B. Left- and right-circularly polarized (σ + and σ, respectively) components of the light field acquire different phase shifts as they propagate through the medium, leading to rotation of the axis of light polarization by angle ϕ. (In general, there is also different absorption of the two circular components of the light field which gives rise to elliptical polarization at the output.) This phenomenon is known as resonant linear Faraday rotation or the Macaluso-Corbino effect. The Macaluso-Corbino effect is referred to as a linear effect because at sufficiently low light powers the rotation is light-power-independent. [For a detailed discussion of Faraday rotation and other closely related phenomena, see the recent review of linear and nonlinear magneto-optical effects by Budker et al. (00).] atomicphysicsproof 003/0/7 page 87 #97

2 88 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS } g B F B z m =- z m = 0 z m =+ z FIG. 4. Energy level diagram for an F = F = 0 atomic transition. Zeeman sublevels are shifted in the presence of a magnetic field, changing the resonance frequencies for left- and right-circularly polarized light. Consider an F = F = 0 atomic transition (Fig. 4.); F and F are the total angular momenta of the upper and lower state, respectively. Assume that the width of the transition is given by the spontaneous decay rate from the upper state γ 0 (i.e., there is no Doppler or other kinds of broadening), and the atomic vapor is of length l. A magnetic field is applied along the direction of light propagation. Derive the dependence of the Faraday rotation angle ϕ on the magnetic field and the detuning of the light frequency ω from the atomic resonance ω 0. Hint For this problem it is convenient to choose the axis of quantization along the magnetic field, i.e., the direction of light propagation. Solution Linearly polarized light incident on an atomic sample can be decomposed into left- (σ + ) and right- (σ ) circularly polarized components. When a magnetic field B z is applied to the sample along the direction of light propagation (the longitudinal direction, ẑ), the Zeeman shifts between adjacent magnetic sublevels (= g F µ 0 B z, where g F is the Landé factor and µ 0 is the Bohr magneton) cause the refractive indices for σ + and σ light to differ (circular birefringence). This, in turn, causes the circular components of the linearly polarized light to change their relative phase as they propagate through the medium leading to optical rotation. For the Doppler-free case and narrow-band light, in the absence of the magnetic field, the complex refractive index n(ω) can be described by a Lorentzian lineshape function [see, for example, Griffiths (999) and Problems 3., 3.3] n(ω) + πχ 0 γ 0 (ω ω 0 ) + iγ 0, (4.) atomicphysicsproof 003/0/7 page 88 #98

3 RESONANT FARADAY ROTATION 89 where χ 0 is the amplitude of the linear susceptibility. The magnetic field shifts the resonance frequencies for the two circular components, so the indices of refraction n ± (ω) for left- and right-circular polarizations in the presence of the longitudinal magnetic field become n ± (ω) + πχ 0 γ 0 (ω ω 0 g F µ 0 B z ) + iγ 0, (4.) According to Eq. (4.), the difference between the refractive indices for σ + and σ light is: 4g F µ 0 B z /γ 0 n + (ω) n (ω) = πχ 0 ( ). (4.3) (g F µ 0 B z /γ 0 ) + i( ω ω0 γ 0 ) The plane of light polarization is defined by the relative phase of the two circular components. For example, ˆɛ x = (ˆɛ ˆɛ + ) ˆɛ y = i (ˆɛ + + ˆɛ ), (4.4) where ˆɛ x and ˆɛ y represent light polarized along x and y, respectively. Suppose the light is initially linearly polarized along the x-axis. Then the optical electric field can be described by E = E0 ˆɛ x cos(kz ωt) (4.5) = E [ ] 0 ˆɛ e i(k z ωt) ˆɛ+ e i(k +z ωt) + c.c., (4.6) where E 0 is the amplitude of the optical electric field, c.c. denotes the complex conjugate, and the magnitude of the wavevectors k ± are given by k ± = n ±ω c. (4.7) The imaginary part of the wavevector causes absorption, and the real part leads to refraction. A difference in absorption of the two circular components causes the Note that there are a number of useful characterizations of the refractive index in terms of various physical quantities such as the linear susceptibility χ, the microscopic atomic polarizability α (see Problems. and.), and the dielectric constant ɛ, namely: where N is the atomic concentration. n = ɛ = + 4πχ + πχ = + πnα, atomicphysicsproof 003/0/7 page 89 #99

4 90 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS light to acquire elliptical polarization (Fig. 4.). The difference in the real parts of the indices of refraction causes optical rotation. As the light propagates through the atomic vapor, the two circular components acquire a relative phase shift of φ = ωl c Re(n + n ), (4.8) where l is the path length in the vapor. If φ = π, then the initially x-polarized light has become y-polarized light, i.e., ϕ = φ. (4.9) Rotation angle (rad) Normalized magnetic Field (b = g F 0 B z / 0 ) 0.6 (a) Rotation angle (rad) (b) Normalized detuning ( / ) FIG. 4.3 Magneto-optical rotation for one absorption length (l = l 0). (a) Magneto-optical rotation angle ϕ as a function of longitudinal magnetic field for the case of zero detuning (ω = ω 0). (b) Rotation angle as a function of detuning for b = [i.e., B z = γ 0 /(g F µ 0 )]. atomicphysicsproof 003/0/7 page 90 #00

5 KERR EFFECT IN AN ATOMIC MEDIUM 9 Finally, we can use our expression for the magnetic-field-induced difference in the refractive indices for σ ± light [Eq. (4.3)] in Eq. (4.9) to find the dependence of Faraday rotation on the magnetic field and the light detuning from resonance = ω ω 0 : ϕ = πχ 0ωl c b [ + b ( /γ 0 ) ] ( /γ 0 ) + [ + b ( /γ 0 ) ], (4.0) where b = g F µ 0 B z /γ 0. Equation (4.0) can also be expressed in terms of the unsaturated absorption length on resonance l 0 = (4πχ 0 ω/c) (defined for B z = 0, the absorption length is discussed in Problem 3.5). Figure 4.3 shows the magnetic field and light detuning dependence of the Faraday rotation. 3 When = 0, we have ϕ = πχ 0ωl c b + b = l l 0 b + b. (4.) It is interesting to note that the integral of rotation over light detuning is zero [Fig. 4.3(b)]. This is because Faraday rotation arises due to a relative shift in the refractive indices for left- and right-circularly polarized light, and the frequency integral over each of them is zero. 4. Kerr effect in an atomic medium If a transparent isotropic material is subject to an external electric field, the field induces a uniaxial anisotropy in the medium, which modifies its optical properties. In particular, light with linear polarization parallel to the direction of the field will experience a slightly different index of refraction compared to light linearly polarized perpendicular to the direction of the field (the Kerr effect). The difference in the indices of refraction can be detected, for example, by measuring the ellipticity induced in light which is initially linearly polarized at 45 to the field (Fig. 4.4): ε = KE πl λ, (4.) where K is the Kerr constant, E is the applied field, l is the length of the sample, and λ is the wavelength of light in vacuum. From Eq. (4.), we see that the Kerr 3 In the limit where b, the rotation angle is zero when ω ω 0 = γ 0/. In experiments where the homogeneous broadening exceeds the inhomogeneous broadening (e.g., measurements of pressure broadening), this property offers a convenient way to measure the homogeneous width γ of the transition. It is also significant that the separation between the zero-crossings is linearly dependent on γ even when γ is much smaller than the inhomogeneous width. atomicphysicsproof 003/0/7 page 9 #0

6 9 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS constant can be calculated from the difference of the indices of refraction for light with linear polarization parallel and perpendicular to the applied static electric field: KE = n n. (4.3) Estimate the Kerr constant K for the following systems, assuming the energy levels are as the s S 0 ( S ), sp P ( P ), and ss S 0 ( S ) in He, for near infrared and visible light (note that the frequency of such light is far from the S P resonance): (a) A two-level atom (the S and P states). (b) A three-level atom (all three of the above states). (c) Make an order of magnitude estimate of the Kerr constant for liquid helium. Hint The density of liquid helium is 0. g/cm 3, the refractive index of liquid helium is n.08, the lifetime of the P state for a free helium atom (which lies. ev above the ground state) is 0.56 ns. The electric dipole matrix elements connecting P, M = 0 with S (d ) and S (d ) states have the ratio d /d 6.9. FIG. 4.4 Simplified schematic of the Kerr effect measurement. PD and PD are photodetectors. Figure courtesy V. V. Yashchuk. atomicphysicsproof 003/0/7 page 9 #0

7 Solution KERR EFFECT IN AN ATOMIC MEDIUM 93 (a) We will estimate the Kerr constant in the following way. First, we will determine the light-dependent energy shift δ of the lower state in the presence of the static electric field. Then, we will connect this shift to the refractive index using the following relations: δ = αe, (4.4) where α is the polarizability (see Problems. and.) and αe is the magnitude of the dipole moment induced in an atom by the light electric field of magnitude E [the factor of / arises in Eq. (4.4) because the dipole moment itself is proportional to the applied field see, for example, Problem.] and n = ɛ + πnα, (4.5) where ɛ is the dielectric constant of the medium and N is the atomic concentration. [Eq. (4.5) follows from the fact that the electric induction is D = ɛe = E + 4πNα.] If we choose the direction of the applied static field E as the quantization axis, the effect of this field is to shift the ground state down by d E / ω P (where ω P is the frequency of the S P transition) and to shift the M = 0 Zeeman sublevel of the upper state up by the same amount [see Eq. (.) in Problem.]. The wavefunctions of the lower and the M = 0 component of the upper states mixed by the electric field [obtained by expanding the solutions of the two-level secular equation in the small mixing parameter d E/( ω P ), see Problem.4(b)] are: ( a d ) E ωp S d E P, M = 0, (4.6) ω P b d ( E S + d ) E ω P ωp P, M = 0. (4.7) (Note that the factors of / multiplying the d E /( ωp ) terms in the above equations come from normalization of the wavefunctions a and b, and we keep only terms up to second order in E in the expansion.) The dipole moment can be calculated from the lifetime τ using the following relation (see Problem 3.3): τ = 4ω3 P 3 c 3 J + S d P, (4.8) where J = is the total angular momentum of the upper state, and S d P is the reduced dipole moment [the Wigner-Eckart theorem (Appendix F) relates atomicphysicsproof 003/0/7 page 93 #03

8 94 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS S d P to d ]. Performing this calculation, one obtains d = S d P ea 0. (4.9) Consider now the case of the light field parallel to the static field. Using the wavefunctions (4.6) and (4.7) perturbed by the static field as the new basis, one finds for the dipole moment between the upper and the lower state: ) d ab = b d a = ( d E ωp d. (4.0) The light field couples a and b, leading to a light-induced shift described by d ab E δ (4.) ω P + d E ω P ) ) d ( E d ( E ω P ωp d E ωp (4.) ) d ( E 6d E ω P ωp, (4.3) where we used ω P ω E ω P. A similar calculation for the light field perpendicular to the static field yields: ) δ d ( E d E ω P ωp. (4.4) From Eqs. (4.4) and (4.5), we see that n is proportional to the lightinduced shift: We have n πnα = 4πN δ E. (4.5) n n 4πN E ( δ δ ) 4πN d ω P ( 4d E ) ω P (4.6) (4.7) = 4d E ωp (n ), (4.8) where we define n = 4πNd /( ω P ) to be the index of refraction in the absence of the static field. atomicphysicsproof 003/0/7 page 94 #04

9 KERR EFFECT IN AN ATOMIC MEDIUM 95 S P E E S S S P E P E S, S S, S P P E E S S FIG. 4.5 Kerr effect as a four-wave-mixing process. The Kerr effect in the static field is obtained in the limit ω E 0; ω E ω E. The S state plays a significant role in the process as discussed in the text even though it does not couple to the S state directly, and there is no resonant enhancement of the effect. Consequently, from Eq. (4.3), we have K 4d ωp (n ). (4.9) (b) In the above estimate we have neglected the effect of the S state of He. This state is very close in energy to the P state, and, in fact, gives the dominant contribution to the electric polarizability of the latter. For this reason the S state plays an important role in a realistic treatment of the present problem. atomicphysicsproof 003/0/7 page 95 #05

10 96 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS The light-medium interaction responsible for the Kerr effect can be thought of as a nonlinear optical four-wave mixing (χ (3) ) process, 4 where three lowfrequency fields (E, E, and E) produce a new field (E ). Processes of this type may be represented by emission-absorption diagrams and by Feynman diagrams [see, for instance, Delone and Krainov (988) and Appendix H], examples of which are shown in Fig Let us compare the amplitude corresponding to a diagram with S as the intermediate state to that with S as the intermediate state (permutations of the static fields do not change anything, and we do not consider them). Using the Feynman diagram rules (Appendix H), we write the vertices and propagators for the case with S (ω E = ω E ): V S (4.30) d 4 (ω P ω E )( ω E )(ω P ω E ) + d 4 (ω P + ω E )(ω E )(ω P + ω E ) 4d4 ωp 3. For the case with S as the intermediate state, we have: V S (4.3) d d (ω P ω E )(ω S ω E )(ω P ω E ) + d d (ω P + ω E )(ω S + ω E )(ω P + ω E ) d d ωp 3. Comparing the two amplitudes, we see that since d 6.9d, the amplitude with S as the intermediate state is greater than that with S by a factor 4. We can therefore neglect the amplitude with S. (Note that although each individual diagram with S is resonantly enhanced due to the presence of a small quantity ω E in the denominator, the two diagrams different by the order of absorption and emission of light quanta nearly cancel. It is also important to note that the effect on the phase of light we are interested in is proportional to the forward scattering amplitude given by the Feynman diagrams, rather than its absolute value squared.) Finally, based on our calculation of V S and V S, for the three-level system, in place of Eq. (4.9), we can write K d ωp (n ) (4.3) by replacing the factor 4d in Eq. (4.9) with the factor +d. (c) For the purpose of the estimate, we model the liquid as a collection of free atoms (which works for helium, but is generally a poor approximation for 4 For an introduction to nonlinear optics, we recommend the book by Boyd (003). atomicphysicsproof 003/0/7 page 96 #06

11 THE HANLE EFFECT 97 molecular liquids). Substituting the values of d, ω P, n given in the hint yields K 0 4 (kv/cm). (4.33) As a numerical example, for the setup shown in Fig. 4.4, a 0-cm long sample, λ = µm, and E = 50 kv/cm, the induced ellipticity is ɛ 0 5. This should only be considered an order of magnitude estimate because the optical properties of the actual He system are not well-described by the three-level model that we adopted here. In fact, in this model the index of refraction n is [using Eqs. (4.4) and (4.5); S plays no role in this estimate]: n 4πNd ω P. (4.34) Substituting the numbers for liquid He, we find n 8 0 3, which is smaller than the experimental value for He gas scaled to the density of liquid He by a factor of The Hanle effect Consider the experimental arrangement shown in Fig. 4.6, where an ensemble of atoms with ground state angular momentum J = 0 is located in a small volume centered at the origin. At time t = 0, the atoms are exposed to a short pulse of circularly polarized photons propagating along ˆx, which drive a transition to an excited state with J =. The atoms are in a magnetic field B = B 0 ẑ. Because the incoming light is circularly polarized, the excited atoms initially have a projection of angular momentum along ˆx equal to (or ). However, the polarization vector then precesses with frequency Ω L (the Larmor frequency, proportional to B) because the excited atoms have magnetic moments (see, for example, Problem.6). The excited state has a finite lifetime τ = /γ. When an atom decays it may emit a photon into the solid angle subtended by the detector, which lies along the y-axis. The detector contains a circular polarization analyzer, so that only positive-helicity (σ + ) fluorescence photons are detected. What is the time dependence of the detector signal? How does the signal change with the sign of circular polarization of excitation photons and with the strength of the magnetic field B 0? Explain how one could use the described phenomenon, known as the Hanle effect, for excited state lifetime measurements. Solution In this problem it is convenient to use different quantization axes in different steps of the solution. To describe the population process, we choose the quantization atomicphysicsproof 003/0/7 page 97 #07

12 98 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS z y x B Detector Circularly polarized light beam Atomic vapor FIG. 4.6 Schematic setup for an experiment measuring the Hanle effect. An atomic vapor at the origin is illuminated by a pulse of circularly polarized light propagating in the ˆx-direction. A detector equipped with a circular polarization analyzer is positioned along the y-axis. The atomic vapor is immersed in a magnetic field B which points in the ẑ-direction. axis along ˆx. Then the J = states excited by left- and right-circularly polarized (σ + and σ, respectively) photons are just J =, m = = 0, (4.35) 0 J =, m = = 0 0, (4.36) respectively. Since the magnetic field is in the ẑ-direction, in order to describe the precession, it is convenient to choose the quantization axis along z. Note that an appropriate coordinate frame is obtained from the one we used to describe population by an Euler rotation (α = 0, β = π/, γ = 0) given by the following J = rotation matrix D(α, β, γ) (Appendix E): D(0, π/, 0) = 0. (4.37) Applying D(0, π/, 0) to the J =, m = ± states gives ψ(t = 0) = ±, (4.38) atomicphysicsproof 003/0/7 page 98 #08

13 THE HANLE EFFECT 99 where the two signs correspond to the two possible circular polarizations of the incident photons. According to the time-dependent Schrödinger equation, the temporal evolution of this wavefunction is described by ψ(t) = e γt/ e iωlt ± e iω Lt. (4.39) Here Ω L = gµ 0 B 0 is the Larmor frequency (g is the appropriate Landé factor), and we have included the amplitude decay (hence γ/, not γ in the exponential factor) due to the natural lifetime of the excited state. Finally, detection is most conveniently described in a frame with quantization axis in the y-direction. This frame is obtained from the previously used one by an Euler rotation (α = π/, β = π/, γ = 0), so the wavefunction in the new frame takes on the form (see Appendix E): ψ (t) = 0 e γt/ ie iωlt ± ie iω Lt (4.40) ± + sin Ω Lt = sin Ω L t e γt/. (4.4) + sin Ω Lt The detector signal S(t) is proportional to population of the m = sublevel, and to the spontaneous decay rate: S(t) γ 4 ( ± sin Ω Lt) e γt. (4.4) This represents a decaying oscillatory signal: the detector sees maximum signal when the angular momentum precesses in such a way that it is pointing towards the detector. 5 If the temporal resolution of the detector is much better than /γ, the lifetime can be measured by fitting the observed time dependence to the expected dependence obtained above with γ = /τ being a free parameter. If the temporal resolution of the detector is poor (or the lifetime is short), the lifetime can be determined by measuring the dependence of the time-integrated 5 This statement is true when Ω L γ. If γ is large, the decay shifts the maxima in the detected intensity. atomicphysicsproof 003/0/7 page 99 #09

14 00 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS Signal Normalized Larmor Frequency ( L ) FIG. 4.7 Time-integrated Hanle signal as a function of Larmor frequency for left- and right-circularly polarized excitation light (solid and dashed curves, respectively). Fits to the curves allow one to extract the excited-state lifetime. signal on the magnetic field. The time-integrated signal is given by S(Ω L ) S(t)dt = γ ( γ ± Ω L γ + Ω ) γ L γ + 4Ω. (4.43) L The signal S(Ω L ) is plotted as a function of the Larmor frequency in Fig Note that the characteristic width of the dispersive part of the profile is γ. 4.4 Electric-field-induced decay of the hydrogen S / state In hydrogen, the S / state lies higher in energy than the P / state by the Lamb shift δ = 058 MHz. In the absence of external electric fields, the S / state has a very long natural lifetime ( /8 s) and decays by two-photon emission to the ground S / state. The P / state has a short lifetime (τ P s) since it can decay by single-photon electric dipole (E) emission to the ground state (Lyman α line). When an external electric field is applied, the S / state acquires an admixture of the P / state and its lifetime is shortened. (a) For weak electric fields E, show that if the S / state is populated at time t = 0, its population decays with inverse lifetime τ = 3γe a 0 E ( ω sp + γ /4 ), (4.44) atomicphysicsproof 003/0/7 page 00 #0

15 ELECTRIC-FIELD-INDUCED DECAY OF THE HYDROGEN S / STATE 0 where ω sp = πδ is the Lamb shift. Explain which electric fields can be considered weak. (b) Evaluate τ at E = 0 V/cm. Solution (a) Suppose the electric field is applied in the ẑ-direction. Consider a two-level system formed by the states S /, M J = / and P /, M J = / (to simplify notation, we will subsequently denote the S / state as S and the P / state as P ). The Hamiltonian for this system in the presence of an electric field is given by (where we set = ): ( ) ωsp de H =. (4.45) de iγ/ Here the energy of the unperturbed P / state is chosen to be zero (we neglect the width of the S state) and the dipole moment d is given by d = e S, M J = / z P, M J = / (4.46) ( ) = e, 0, 0 + z,,,, 0 + (4.47) 3 3 = e 3, 0, 0 z,, 0 = e 3, 0, 0 r cos θ,, 0, (4.48) where we have used the notation n, l, m l for the spatial wavefunctions and for the electron spin states we have employed our common notation in which m s = ±/ = ±. We can evaluate the expression for d using the appropriate hydrogen wavefunctions [Eqs. (.6) and (.63)] ψ 0 (r, θ, φ) = 4 π ψ 00 (r, θ, φ) = π and integrating. We find that a 3/ 0 a 3/ 0 r e r/a0 cos θ, a 0 ( r a 0 ) e r/a0, d = 3 ea 0. (4.49) Returning to the Hamiltonian matrix H, we see that since the electric field perturbation is nondiagonal, it does not cause first-order energy shifts. The wavefunctions do, however, acquire first-order corrections. The perturbed wavefunction atomicphysicsproof 003/0/7 page 0 #

16 0 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS corresponding to the S state, S, is found by substituting λ ω sp (since the energy is not shifted to first order) into the secular equation ( ) ( ωsp λ de a = 0, (4.50) de iγ/ λ b) where b ( b a ) is the small admixture amplitude of the P state into the S state. Equation (4.50) indicates that b a de ω sp + iγ/. (4.5) The decay rate of S is given by the relative probability of finding atoms in the admixed P state times the P decay rate: τ = γ d E ωsp + γ /4 = as advertised. The weak field condition is 3γe a 0 E ( ω sp + γ /4 ), (4.5) de ω sp. (4.53) In the opposite case of de ω sp, the lifetime does not depend on E, as the S and P states are fully mixed, and the lifetime of both eigenstates is τ. (b) We have the following values for the parameters in Eq. (4.5): which gives us ω sp = π s ; (4.54) γ = s ; (4.55) E = 0 V/cm = esu/cm; (4.56) 30 τ = 3.8 µs. (4.57) 4.5 Stark-induced transitions (T) Single-photon electric dipole (E) transitions between states n and m of the same parity are forbidden [neglecting the effects of parity nonconservation (PNC) which are usually very small, see below and Problem.3]. However, a nonzero atomicphysicsproof 003/0/7 page 0 #

17 STARK-INDUCED TRANSITIONS (T) 03 E transition amplitude between these states may be induced by application of a static electric field E, which mixes states of opposite parity to both states n and m. Such Stark-induced transitions (Bouchiat and Bouchiat 975) have been used in several atomic-parity-nonconservation experiments [see, for example, Conti et al. (979), Bouchiat et al. (98), Wood et al. (997), Nguyen et al. (997), Guéna et al. (003)]. The transition rate W between same-parity states has contributions from both the Stark-induced transition amplitude A s and the parity-violating transition amplitude A pnc the overall rate has an interference term: W As + Re [ As A pnc] + Apnc As + Re [ As A pnc]. (4.58) The experimental method in which the interference term Re [ A s A pnc] is measured in order to determine A pnc is known as the Stark-interference technique. This technique serves both to enhance the PNC signal (which is now proportional to Re [ A s A pnc] as opposed to Apnc ) and provide a method for distinguishing the PNC signal from background effects (since, for example, the PNC signal Re [ A s A pnc] reverses with the sign of the electric field). (a) Show that the Stark-induced transition amplitude between the same-parity states m and n can be represented as A s = T (0) m U (0) n + T () m U () n + T () m U () n, (4.59) where T (κ) are the rank κ irreducible parts of the reducible second-rank tensor formed out of the light electric field E and the static electric field E, T ij = E i E j, (4.60) and U (κ) are the rank κ irreducible parts of the appropriate tensor formed from atomic vectors. What is the form of the matrix element m U ij n? Solution Stark-induced transitions occur when the static electric field E mixes states m and n with opposite-parity states, and then the light field E drives transitions between the mixed states. Based on this physical picture, the transition amplitude can be written as A s = [ m d E p p d E p p E E n + m d E p p d E n ], (4.6) m E n E p atomicphysicsproof 003/0/7 page 03 #3

18 04 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS where d is the electric-dipole operator, the sum goes over the intermediate states p of parity opposite to that of states n and m, E n, E m, and E p are the respective unperturbed energies of the states n, m, and p, and the two terms in (4.6) correspond to mixing of states p to the final and the initial state, respectively. Equation (4.6) can be expressed in the following form (where the sum over the repeated indices i, j, p is assumed): A s = m d ie i p E p E m p d j E j n + m d j E j p p d ie i n E n E p. (4.6) Factoring out the electric fields (which obviously commute with the atomic wavefunctions and operators), we write ( m di p p d j n A s = (E i E j ) + m d ) j p p d i n. (4.63) E p E m E n E p Thus the Stark-induced amplitude is described by the contraction of the rank-two tensor T (where T ij = E i E j ) with the rank-two tensor U: m U ij n = m d i p p d j n E p E m + m d j p p d i n E n E p. (4.64) The contraction of two rank-two tensors can be evaluated by first expanding each of the tensors into irreducible components (of rank zero, one, and two), and then taking the sum of scalar products of the irreducible components of the same rank κ (since the scalar product can only be formed between tensors of the same rank, see Appendix F). Thus the scalar product of T and m U n can be expressed in the form (see, for example, Varshalovich et al. 988): A s = T (0) m U (0) n + T () m U () n + T () m U () n. (4.65) (b) Write out the explicit form of the irreducible rank κ = 0,, tensors built from the electric fields in the spherical basis. Hint The general procedure for finding the explicit decomposition of a reducible rank κ tensor built from two irreducible tensors of rank κ and κ in terms of irreducible spherical tensors is as follows. If one has an irreducible rank κ tensor A κ and an irreducible rank κ tensor B κ, one can form the irreducible tensor product of atomicphysicsproof 003/0/7 page 04 #4

19 rank κ (Varshalovich et al. 988): STARK-INDUCED TRANSITIONS (T) 05 (A κ B κ ) κ q = q,q κ, q, κ, q κ, q A κ q B κ q. (4.66) One can also carry out the inverse decomposition of the rank κ + κ reducible tensor A κ q B κ q according to A κ q B κ q = κ +κ κ= κ κ κ, q, κ, q κ, q (A κ B κ ) κ q, (4.67) where q = q + q. The reducible rank-two tensor T, which is formed from the two irreducible rank-one tensors E and E, can be decomposed into irreducible rank κ = 0,, tensors T (κ), whose components are given by T κ q = q,q, q,, q κ, q ( E E ) κ q. (4.68) Equation (4.68) shows that the tensor components Tq κ are proportional to the irreducible tensor products (4.66) formed from the electric field vectors. Solution One may expect that the rank-zero part of the reducible tensor T should be proportional to the scalar product E E. Let us show this result formally, using the mathematical techniques discussed in the Hint. The electric-field vectors E i and E j are irreducible rank-one tensors, which can be expressed in the spherical basis according to the formulae (F.3)-(F.5) given in Appendix F: E ± = (E x ± ie y ), E 0 = E z (4.69) E ± = (E x ± ie y ), E 0 = E z (4.70) According to Eq. (4.66), the irreducible rank-zero tensor product of E and E is ( E E ) 0 0 = 3 E E 3 E 0E E E = 3 (E x E x + E y E y + E z E z ) = 3 E E, (4.7) atomicphysicsproof 003/0/7 page 05 #5

20 06 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS which differs by a numerical factor from the scalar product of the vectors. Thus for T0 0 we have T 0 0 = a E E, (4.7) where a is a constant. As we know, the only rank-one ( object ) that can be formed out of two vectors is the vector product, so Tq E E. This can be seen formally by carrying out the same procedure we used above to find T 0 0 : ( E E ) = E E 0 E 0E q = (E ze x E x E z ) + i (E ze y E y E z ), (4.73) and so on. In Eq. (4.73), we made use of Eqs. (4.69), (4.70) and (4.66). In general, we find the relation ( E E ) q = so the rank-one irreducible tensor components are given by where b is a constant. For the rank-two components, we have i ( E E )q, (4.74) ( ) Tq = ib E E, (4.75) q T q = c ( E E ) q, (4.76) where according to Eq. (4.66), ( E E ) = E E, (4.77) ( E E ) = ( E E 0 + E 0E ), (4.78) ( E E ) 0 = 6 ( E E + E 0E 0 + E E ), (4.79) ( E E ) = ( E E 0 + E 0E ), (4.80) ( E E ) = E E. (4.8) atomicphysicsproof 003/0/7 page 06 #6

21 STARK-INDUCED TRANSITIONS (T) 07 Similarly to how we have expressed the tensors T (κ) in the spherical basis [Eqs. (4.7), (4.75), and (4.76)], the tensors U (κ) can also be expressed in the spherical basis. The Stark-induced transition amplitude can then be expressed in terms of irreducible spherical tensors [using Eq. (4.65) and Eq. (F.3) from Appendix F]: ( ) A s = a E E m U0 0 n + ib( ) q( ) E E m U q n q q + q c( ) q( E E ) q m U q n. (4.8) (c) Use the Wigner-Eckart Theorem (Appendix F) to write the Stark-induced transition amplitude between states m, F, M and n, F, M in terms of Clebsch- Gordan coefficients. Solution From Eq. (4.8) and the Wigner-Eckart theorem (F.), we find for the Starkinduced amplitude: ( ) A s = a E m, F U 0 n, F E F, M, 0, 0 F, M F + + ib( ) M M ( ) E m, F U n, F E F, M,, M M F, M q=m M F + + c( ) M M ( E E ) κ= m, F U n, F F, M,, M M F, M. q=m M F + (4.83) In the literature [see, for example, Bouchiat and Bouchiat (975); Drell and Commins (985); Bowers et al. (999); Bennett and Wieman (999)], the Starkinduced amplitude is defined in terms of the real parameters α, β, and γ, known as the scalar, vector, and tensor transition polarizabilities, respectively: A s =α E E ( ) δ F,F δ M,M + iβ E E F, M,, M M F, M q=m M + γ ( E E ) q=m M F, M,, M M F, M. (4.84) The fact that the transition polarizabilities α, β, and γ are real follows from timereversal invariance (Bouchiat and Bouchiat 975). (d) Discuss the angular momentum selection rules for transitions described by these polarizabilities. atomicphysicsproof 003/0/7 page 07 #7

22 08 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS TABLE 4. Various contributions to the transition polarizability and the corresponding selection rules for the total angular momentum F. Rank Selection rules α 0 F = 0 β F = 0, ±; 0 0 γ F = 0, ±, ±; 0 0; ; 0 ; 0 Solution Angular momentum selection rules for various terms in Eq. (4.84) are the usual ones for tensors of appropriate rank; see Table 4.. These selection rules follow from the properties of the Clebsch-Gordan coefficients. Depending on the transition, different combinations of α, β, and γ may contribute. For example, for F = / F = /, the electric-field-induced transition amplitude has a scalar and a vector component (Bouchiat and Bouchiat 975), while for F = 0 F =, only the vector amplitude contributes. (e) Discuss the limits of the approximations employed in the above analysis. Solution In this tutorial, we have used first-order perturbation theory, which correspondingly limits the applicability of the result. In particular, the present results are only valid when the Stark shifts of the levels involved are much smaller than separations between these levels. 4.6 Magnetic deflection of light Magneto-optical effects are usually observed by measuring how the Stokes parameters (Appendix D) change when a light field traverses a medium exposed to a magnetic field [see Problems 4., 4.3, and 4.7, as well as the review by Budker et al. (00)]. Here we consider another magneto-optical effect, observed by Schlesser and Weis (99): the deflection of a light beam as it passes through a medium immersed in a magnetic field. Consider an isotropic medium to which a homogeneous magnetic field B is applied. atomicphysicsproof 003/0/7 page 08 #8

23 MAGNETIC DEFLECTION OF LIGHT 09. k H B D FIG. 4.8 Geometrical relationship between the wavevector k, the induction vector D and magnetic field H of the light, and the static magnetic field B applied to the medium. (a) Recall that the components of the induction vector D are related to the components of the electric field E via D i = ε ij E j, (4.85) where ε ij is the dielectric permeability (permittivity) tensor. From symmetry considerations, show that ε ij is given by: ε ij ( B) = εδ ij + i γɛ ijk B k, (4.86) where ε and γ are frequency-dependent complex scalars, and ɛ ijk is the Levi-Civita totally antisymmetric tensor. (b) Show that if a linearly polarized wave with wavevector k enters such a medium perpendicular to its boundary, the Poynting vector S = c 4π E H, (4.87) inside the medium (i.e., the direction of energy flow in the light beam) has a timeaveraged value given by: S ce ( 0 [ˆk + Im( γ) sin ϕ cos ϕ B 8π + sin ϕ (ˆk B) )], (4.88) where E and H are the electric and magnetic fields of the light, respectively, ϕ is the angle between the incident light polarization and the magnetic field B, and B is assumed to lie in the plane perpendicular to ˆk (Fig. 4.8). Assume a nonmagnetic, weakly absorbing medium: Im( ε), Im( γb) Re( ε). The quantities E and B are the magnitudes of the optical electric field and the applied static magnetic field, respectively, and E 0 is the amplitude of the optical field [i.e., E = E 0 cos(ωt), where ω is the light frequency]. atomicphysicsproof 003/0/7 page 09 #9

24 0 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS S E S S E. B vapor cell FIG. 4.9 The change in the direction of E, and hence the Poynting vector S, inside a medium (e.g., an atomic vapor) leads to deflection of the light beam. The geometry shown in the figure is optimal for observation of deflection and corresponds to ϕ = π/ (see text). (c) For the most favorable geometry, estimate the magnitude of the magnetic-fieldinduced deflection of a laser beam upon traversal of a cell of length l containing resonant atomic vapor (Fig. 4.9). Hint In part (c), make use of the fact that the component of the dielectric tensor ε ij responsible for magneto-optical phenomena such as magnetic-field-induced circular birefringence and dichroism (Problem 4.) is also responsible for magnetic deflection. Since the complex index of refraction is related to ε via n = ε, (4.89) one can use the formulae from Problem 4. to determine the magnitude of the magnetic-field-induced change of the refractive index. Solution (a) The first term in Eq. (4.86) is the usual form of the permittivity tensor [see, for example, Griffiths (999) and the text on Electrodynamics of Continuous Media by Landau et al. (995)] for isotropic media. For such media, there are no preferred directions, and the induction vector D has to be collinear with E: D = ε E. (4.90) When the magnetic field is applied, there appears another possibility of building a vector quantity out of the vectors of the problem (or more precisely, a vector of the light electric field E and the pseudovector B): B E. This is represented by the second term in (4.86). The factor i is written explicitly in this term because the quantities ε and γ must be real in the case of transparent media. The relation between D and E should be invariant with respect to reversal of time. Both D atomicphysicsproof 003/0/7 page 0 #0

25 MAGNETIC DEFLECTION OF LIGHT and E are time-reversal-invariant (T-even), but B is T-odd. Since under time reversal one should take complex conjugates of all operators (see Problem.3), the i ensures the proper time-reversal symmetry. (b) Since the light enters the medium at normal incidence, the direction of k does not change. The direction of the magnetic field of the light is also unchanged. It follows from Maxwell s equations that ˆk, D, and H are mutually perpendicular. Thus the induction D is directed along the electric field of the incident light, i.e., along k H. However in general, the electric field in the medium E is not collinear with D. From Equation (4.86), we write: D = ε E i γ B E. (4.9) Since H = ˆk E, in order to find the Poynting vector, one needs to evaluate: S = c E 4π ) (ˆk E. (4.9) Applying a well-known vector identity, ) E (ˆk E = ˆk ( E ) E ) (ˆk E, (4.93) we have S = ce 4π ) (ˆk E ˆk E E. (4.94) First, we evaluate ˆk E using (4.9) and the fact that ˆk D = 0 (Fig. 4.8): ˆk E = i γ ε ˆk ( ) B E i γ ε ˆk ( ) B D, (4.95) where we have neglected terms γ B. Since B D is along ˆk (Fig. 4.8), and the magnitude of the induction vector D εe, we have ˆk E i γbe ε Thus we have for the Poynting vector S ce 4π [ ˆk i γ where again we ignore terms γ B. sin ϕ. (4.96) ( ) ] B sin ϕd D, (4.97) atomicphysicsproof 003/0/7 page #

26 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS Since k B, we can resolve ( ) D B D = B B + Using Eq. (4.98) in (4.97), we obtain Eq. (4.88): [ )] ) D (ˆk B (ˆk B B. (4.98) S ce ( 0 [ˆk + Im( γ) sin ϕ cos ϕb 8π + sin ϕ(ˆk B) )]. Here we have taken into account (by taking the real part of the second term) that only the component of the light magnetic field in-phase with the electric field contributes to the average Poynting vector (4.87), and used the time average of E = E 0 /. (c) From (4.88) it is seen that the largest deflection occurs at ϕ = π/. In order to estimate the magnitude of the displacement for a near-resonant vapor, we follow the suggestion outlined in the hint to say that the magnitude δn of the magnetic-field-induced change of the complex refractive index of the medium can be estimated as (see Problem 4.): δn gµb (n ). (4.99) Γ Here gµb is the Zeeman shift, Γ is the width of the transition (e.g., Doppler width), n is the magnetic-field-independent complex refractive index, and a weak magnetic field is assumed ( gµb Γ). In the vicinity of a resonance, the maximal magnitudes of the real and imaginary parts of n are comparable, and can be estimated from 4πIm(n) l 0 /λ, (4.00) where l 0 is the absorption length (Problem 3.7) and λ is the wavelength of light. Using this, we can estimate the magnitude of the relevant magnetic-fielddependent term in the dielectric tensor: Im( γ)b gµb λ. (4.0) 4π Γ l 0 Using (4.0) and (4.88), we find that upon traversal of a length l in the medium, a light beam is deflected by: gµb l λ. (4.0) 4π Γ l 0 Note that the scale of the displacement is determined by the wavelength. atomicphysicsproof 003/0/7 page #

27 CLASSICAL MODEL OF AN OPTICAL-PUMPING MAGNETOMETER 3 In the work by Schlesser and Weis (99), a room temperature vapor cell with l/l 0 was used. Light was tuned to the D line (85 nm), and a magnetic field of 50 G corresponding to gµb/γ 0. was applied. The observed beam displacement was 30 nm. 4.7 Classical model of an optical-pumping magnetometer Figure 4.0 depicts a schematic diagram of an optical-pumping magnetometer operating in the so-called M x scheme (the origin of this terminology will become apparent later on in this problem). Circularly polarized resonant light Alkali metal vapor cell Photodetector Lock-in amplifier Reference Voltage-controlled oscillator Magnetic field H To coil Magnetic coil (rf) Phase rotator Frequency counter (output) FIG. 4.0 A simplified schematic of an optical pumping magnetometer of the M x type. The central element of the magnetometer is a cell containing the vapor of one of the alkali metals, usually Rb, Cs, or K. The vapor is illuminated with circularly polarized light resonant with either the D or D transition. The intensity of the transmitted light is detected with a photodetector connected to a phase-sensitive (lock-in) amplifier. The reference for the lock-in amplifier is provided by an oscillator that also drives radio-frequency current through a magnetic coil surrounding the vapor cell. The magnetic field produced by the coil is collinear with the light propagation direction. The frequency of the oscillator is controlled by the (nearly dc) voltage at its input provided by the output of the lock-in amplifier. In this problem, we will show that under proper operating conditions, the photodetector signal and the rf magnetic field oscillate at the Larmor frequency atomicphysicsproof 003/0/7 page 3 #3

28 4 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS corresponding to the field H. Measuring the frequency (for example, with a frequency counter) it is possible to determine the magnitude of H using the known value of the gyromagnetic ratio. 6 In order to provide the simplest model of the magnetometer while still retaining its salient features, we will make a number of simplifying assumptions. First, instead of considering the quantum mechanical problem involving the rather complicated hyperfine-structure energy levels of the alkalis, we will model atoms as classical spins with gyromagnetic ratio γ, such that the spin exposed to a static magnetic field H precesses around the direction of the field with Larmor frequency γh. The component of the magnetization along H relaxes at a rate Γ (longitudinal relaxation), and the components perpendicular to H relax at a rate Γ (transverse relaxation). 7 Second, we will assume that the amplitude of the rf magnetic field is small compared with H and will neglect the component of the rf magnetic field along ẑ, the direction of H. This component leads to a fast modulation of the Larmor frequency, which is not important for us here. Third, the remaining component of the rf magnetic field which oscillates in the direction perpendicular to ẑ can be resolved into two counter-rotating components, one of which rotates in the same direction as the magnetic moments, while the other rotates in the opposite direction. Since we will be considering near-resonant conditions (i.e., when the rf frequency is close to the Larmor frequency), we will neglect the latter component (the rotating wave approximation see Problem.7). Let us assume that the optical pumping rate and the relaxation rates are all much smaller than the Larmor frequency. Then, in the absence of the rf magnetic field, there is a steady-state magnetization (which we designate M 0 ) along ẑ. Indeed, while optical pumping produces magnetization at an angle to ẑ, Larmor precession spreads the magnetic moment vectors, so their tips are uniformly distributed on a circle in the xy-plane, so only the ẑ-component survives averaging over the atomic ensemble. 8 6 For precise magnetic field measurements in the Earth-field range where M x magnetometers are commonly used, it is necessary to take into account the nonlinearity in the Zeeman energy shifts due to the mixing of different hyperfine components of the alkali atom ground states caused by the magnetic field (Problem.4). 7 The difference between the transverse and longitudinal relaxation in an ensemble may arise when individual spins precessing in the magnetic field see slightly different magnetic fields (Problem.8). In this situation, if, for example, the spins are originally oriented in the same direction at an angle to the magnetic field, they will eventually spread around due to unequal precession rates, so the transverse magnetization would vanish, while longitudinal magnetization would persist. It is always true that Γ Γ. 8 No net magnetization is created in the special case of H perpendicular to the light propagation direction. This results in the appearance of a dark zone of the magnetometer an orientation of the device where it loses sensitivity to the magnetic field. Another such dark zone in the case of M x magnetometers is when the magnetic field H is along the light propagation direction, so the field applied by the rf magnetic coil is entirely longitudinal. atomicphysicsproof 003/0/7 page 4 #4

29 CLASSICAL MODEL OF AN OPTICAL-PUMPING MAGNETOMETER 5 (a) Write down the differential equations describing the classical evolution of the Cartesian components of the magnetization, including the effects of both the static and rotating magnetic field and the longitudinal (Γ ) and transverse (Γ ) relaxation. (b) Find the steady state solution of these equations. Consider the cases of Γ = Γ and Γ Γ. (c) Referring to Fig. 4.0, explain the origin of the modulation of the light transmission signal. What is the purpose of the phase rotator? Hint It is convenient to write the equations in the rotating frame, i.e., the frame in which the rotating rf field is static. First write the equations neglecting relaxation, and then add the relaxation terms by hand. The resulting equations are known as the Bloch equations, as they were first derived in 946 by magnetic resonance pioneer Felix Bloch. Solution (a) The gyromagnetic ratio γ is the proportionality coefficient between the magnetic moment of an atom M and its angular momentum F : M = γ F. (4.03) The time derivative of the angular momentum due to the magnetic torque is: d F dt = M H t, (4.04) where H t is the total magnetic field (static plus rotating), so that [multiplying both sides of Eq. (4.04) by γ and using (4.04)] d M dt = γ M H t. (4.05) Next, in order to eliminate explicit time dependence of the magnetic field, we go to the frame co-rotating with the rf field at frequency ω. The time derivatives of the magnetic moment in the rotating frame and the lab frame are connected atomicphysicsproof 003/0/7 page 5 #5

30 6 INTERACTION OF LIGHT WITH ATOMS IN EXTERNAL FIELDS according to (see Problem.6) ( dm ) ( d = ) ( M ω M dt dt = γm H t + ω ). (4.06) γ rot lab We will conduct the calculation in the rotating frame [coordinates x, y, z], and will not write the subscript. Rewriting Eq. (4.06) in terms of Cartesian components, assuming that the rotating field is pointing in the ˆx -direction, and adding the relaxation terms, we obtain the sought-for Bloch equations: dm x dt dm y dt ( = γm y H ω ) Γ M x, (4.07) γ ( = γm x H ω ) + γm z H r Γ M y, (4.08) γ dm z dt = γm y H r Γ (M z M 0 ). (4.09) Here H r is the magnitude of the rotating field and M 0 is the equilibrium magnetization in the absence of the rf field. Note that we have chosen the direction of the rotation of the rf field so that the magnitude of the z-directed magnetic field is reduced in the rotating frame. (b) Setting to zero the derivatives on the left-hand-side of Eqs. (4.07,4.08,4.09) (which we are allowed to do because the problem is stationary in the rotating frame), we obtain an inhomogeneous linear system Γ 0 Γ ω r M x M y = 0 0. (4.0) 0 ω r Γ M z Γ M 0 Here = γh ω, and ω r magnetization, we obtain: = γh r. Solving for the components of the ω r M x = M 0 Γ + + Γ, (4.) Γ ωr ω r Γ M y = M 0 Γ +, (4.) + Γ Γ ωr M z = M 0 + Γ Γ + + Γ Γ ω r. (4.3) atomicphysicsproof 003/0/7 page 6 #6

31 SEARCHES FOR PERMANENT ELECTRIC DIPOLE MOMENTS (T) 7 On resonance ( = 0), we have: M x = 0, (4.4) M y = M 0 ω r Γ Γ + Γ Γ ω r M z = M 0 Γ Γ + Γ Γ ω r ω r Γ = M 0 Γ Γ + ωr, (4.5) Γ Γ = M 0 Γ Γ + ωr, (4.6) i.e., in the rotating frame, the average magnetization is in the y z-plane at an angle ( ) tan ωr (4.7) Γ to the z-axis. The magnitude of the M y component as a function of ω r reaches maximum at ω r = Γ Γ. For ω r equal to this value, and for Γ = Γ, M y = M z, so the magnetization is at π/4 to the z-direction. For Γ Γ, the magnetization is at a small angle of Γ /Γ to the z-axis. (c) Going back to the laboratory frame, the steady-state (i.e., time-independent in the rotating frame) magnetization precesses around the z-axis with frequency ω. This precession leads to a temporal variation (with frequency ω) of the projection of the magnetization on the light propagation direction. Since light is circularly polarized, this leads to a modulation in the transmission coefficient (see Problem 3.9). It is important to note that there is a phase shift of π/ in the modulation resulting from M x and M y. With an appropriate choice of phase between the rf field and the lock-in amplifier (facilitated by the phase rotator shown in Fig. 4.0), the detector may be made sensitive to the modulation due to M x, which, according to Eq. (4.), is an odd function of the mismatch between the Larmor frequency corresponding to the field to be measured, H, and the rf frequency. The phase is adjusted in such a way as to lock the frequency output by the voltage-controlled oscillator to the Larmor frequency. Thus, the magnetometer is a self-oscillating frequency generator, whose frequency is determined by the external magnetic field. 4.8 Searches for permanent electric dipole moments (T) Here we explore the possibility that an atom could possess a permanent electric dipole moment (EDM). We show that this can occur, for example, if the electron possesses an EDM. A long series of experiments have been performed to search for EDMs of many different particles: most prominently the neutron, various nuclei, atomicphysicsproof 003/0/7 page 7 #7