Solution Manual for Principles of Laser Spectroscopy and Quantum Optics by Berman and Malinovsky

1. complete document is available on https://solumanu.com/ *** contact me if site not loaded smtb98@gmail.com Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 smtb98@gmail.com Email: smtb98@gmail.com Telegram: https://t.me/solumanu Solutions Manual for Principles of Laser Spectroscopy and Quantum Optics Paul R. Berman Vladimir S. Malinovsky

2. Contents Preface v Chapter 1. Preliminaries 1 Chapter 2. Two-Level Quantum Systems 6 Chapter 3. Density Matrix for a Single Atom 25 Chapter 4. Applications of the Density Matrix Formalism 33 Chapter 5. Density Matrix Equations: Atomic Center-of-Mass Motion, Elementary Atom Optics and Laser Cooling 41 Chapter 6. Maxwell-Bloch Equations 61 Chapter 7. Two-level Atoms in Two or More Fields - Introduction to Saturation Spectroscopy 66 Chapter 8. Three-Level Atoms: Applications to Nonlinear Spectroscopy: Open Quantum Systems 81 Chapter 9. Three-Level Λ Atoms: Dark States, Adiabatic Following, and Slow Light 94 Chapter 10. Coherent Transients 105 Chapter 11. Atom Optics and Atom Interferometry 117 Chapter 12. The Quantized, Free Radiation Field 133 Chapter 13. Coherence Properties of the Electric Field 143 Chapter 14. Photon Counting and Interferometry 153 Chapter 15. Atom - Quantized Field Interactions 162 iii

3. iv CONTENTS Chapter 16. Spontaneous Decay 169 Chapter 17. Optical Pumping and Optical Lattices 179 Chapter 18. Sub-Doppler Laser Cooling 187 Chapter 19. Operator Approach to Atom-Field Interactions: Source-Field Equation 198 Chapter 20. Light Scattering 207 Chapter 21. Entanglement and Spin Squeezing 214

4. Preface We have prepared this solutions manual for the problems in Principles of Laser Spectroscopy and Quantum Optics. In the past, students, and even a few colleagues, have implied or said outright that some of the problems were “impossible”. In the movie Entrapment, Catherine Zeta-Jones proposes a theft to Sean Connery, who replies, “It’s impossible, but doable.” This manual is intended to provide evidence for the fact that all the problems are doable. This solution manual comes with a disclaimer. Although the solutions have been checked over, it is likely that there remain some errors. Moreover, the solutions that are presented are not necessarily the simplest or most elegant solutions of the problems. Therefore, this manual should be viewed as a guide for obtaining solutions to the problems, rather than a source of the definitive solutions to the problems. We welcome suggestions for alternative solutions and also would appreciate being notified of any errors in the solutions. Ann Arbor, MI, Hoboken, NJ, August, 2010 Paul R. Berman Vladimir S. Malinovsky

5. Email: smtb98@gmail.com Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 Telegram: https://t.me/solumanu Chapter 1. Preliminaries 1.1. Go online or to other sources to determine the fine and hyperfine separations in the 3S and 3P levels (as well as the 3S1/2-3P1/2,3/2separations) in23Na and the fine and hyperfine separations in the 5S and 5P levels (as well as the 5S1/2-5P1/2,3/2separations) in85Rb. Solution:Both23Na and85Rb are alkali earth metals with J = 1/2. 23Na has nuclear spin I = 3/2. Therefore its 3S ground state (L = 0) consists of two hyperfine states with F = 1,2. They are separated by 1.77 GHz. The 3P excited state (L = 1) consists of a fine structure doublet separated by 515.5 GHz. The J = 1/2 state has two hyperfine structure states with F = 1,2 separated by 189 MHz, while the J = 3/2 has four hyperfine structure states with F = 0,1,2,3 separated by 16 MHz, 34 MHz and 59 MHz The D1 line has a frequency of 5.0833×1014Hz and the D2 line 5.0885×1014Hz [see, for example, W. A. Wijngaarden and J. Li. Z., Phys. D, 32, 67 (1994) and W. Scherf et al, ibid., 36, 31 (1996)]. 85Rb has nuclear spin I = 5/2. Therefore its 3S ground state (L = 0) consists of two hyperfine states with F = 2,3. They are separated by 3.036 GHz. The 3P excited state (L = 1) consists of a fine structure doublet separated by 7128 GHz. The J = 1/2 state has two hyperfine structure states with F = 2,3 separated by 362 MHz, while the J = 3/2 has four hyperfine structure states with F = 1,2,3,4 separated by 29 MHz, 65 MHz and 121 MHz. The D1 line has a frequency of 3.77×1014 Hz and the D2 line 3.846×1014Hz [see, for example, D. Das and V. Natarajan, Eur. J. Phys. D, 37, 313 (2006) and http://www.phys.ksu.edu/personal/cocke/classes/phys506/sas.htm]. 1.2. Estimate the Doppler width and collision width on the 3S-3P transition in23Na. estimate the Doppler width, assume a temperature of 300K. To estimate the collision width per Torr of perturber gas, assume that the perturbers undergoing collisions with the sodium atoms are much more massive than sodium and that the collision cross section is 10˚ A2. Compare your answer with data on broadening of the sodium resonance line by rare gas perturbers. Solution: The full width at half maximum of a Gaussian To 1 [π2u]3/2e−v2/u2 W(v) = is 1.67u, where u is the most probable speed. For an ideal gas at temperature T, ? 2kBT M u = , where kBis Boltzmann’s constant. For sodium at 300K, u ? 4.6 × 102m/s such that the Doppler width is ΔfD≡ 1.67f(u/c) ≈ 1.3 GHz 1

6. 2 Chapter 1 To estimate the collision rate for heavy perturbers, one uses the fact that the relative velocity of a collision is almost equal to the sodium velocity since the heavy perturber moves more slowly. Then the collision rate is simply Γ = Nuσ , where N is the perturber density and σ is the cross section. At 300K 1 Torr of rare gas is approx- imately (273/300) × 2.7×1025/760 = 3.3×1022atoms/m3which gives Γ/2π ? 0.24 MHz/Torr for a cross section of 10˚ A2. Typical collision rates for Na - rare gas collisions are of order 10 MHz/Torr [see J. Kielkopf, J. Phys. B, 9, L547 (1976)]. Thus the actual cross sections are about a factor 40 larger, or 400˚ A2; this is not uncommon for alkali metal - heavy rare gas pressure broadening cross sections. 1.3. Estimate the recoil frequency in Na and Rb. Solution: The recoil frequency can be defined as 1 ?k2 2M. fr= 2π For85Rb λ is of order 780 nm giving fr(Rb) ≈ 3.8 kHz and for23Na λ is of order 590 nm giving fr(Rb) ≈ 25 kHz. 1.4. Look up the transition matrix elements for85Rb to estimate the Rabi frequency for a laser field having 10 mW of power focussed to a spot size of 50 μm2. Solution: The electric field is 27.5√S = 3.8×105V/m. Since Ω0/2π = μE/h, we need a value for μ. The transition matrix element is of order (this will vary for different hyperfine transitions) 2.4ea0= 2×10−29C·m as can be found in the book of Metcalf and van der Straten on Laser Cooling and Trapping. Thus one finds a Rabi frequency Ω0/2π ? 12 GHz which is to be compared with a natural width of 10 MHz, such that this field easily saturates the transition. 1.5. Consider the two-dimensional vector A = i + 2j. Take as your basis states ?1 where u1,2=i ± j ? ?0 ? u1= ; u2= , 0 1 √2. Express the unit vectors i and j in this basis and find the coordinates. Show explicitly that A1u1+ A2u2= Axi + Ayj. Solution: i =u1+u2 √2 j =u1−u2 √2 ; . Therefore A = i +2j =u1+u2 + 2u1−u2 √2 3 √2u1− √2 1 √2; A2=−1 3 = √2u2; A1= √2

7. 3 Preliminaries such that 3 i + j √2 1 i − j √2 A1u1+ A2u2= √2 √2 − = i + 2j =Axi + Ayj. 1.6. Derive equation (1.27c). Solution: You are asked to prove −M 2?∇R[ˆ v · ˆ v]? = Ne?(v · ∇R)A(R,t)? + Ne?v × B(R,t)? , where ?ˆP − NeA(R,t) ? ˆ v = /M . We use the vector identity ∇(F · F) = 2[(F · ∇)F + F×(∇ × F)] to write ?∇2 R[ˆ v · ˆ v]?= −M [(v · ∇R)v + v×(∇R×v)]. −M 2 But the ∇Roperator acts only on the A(R,t) part of v such that −1 2 = Ne[(v · ∇R)A(R,t) + v × B(R,t)]. 1.7. Prove that dt ∂t Solution: ? ? ?? (v · ∇R)[−NeA(R,t)] ∇R×[−NeA(R,t)] ?∇2 R[ˆ v · ˆ v]?= −M +v× M M + (v · ∇)A(R,t) for a vector function A(R,t) with v =˙R. dA(R,t) =∂A(R,t) =∂A(R,t) +∂A(R,t) ∂X ˙X +∂A(R,t) ˙Y +∂A(R,t) dA(R,t) dt ˙Z ∂t ∂Y ∂Z =∂A(R,t) + (v · ∇)A(R,t). ∂t 1.8. Derive equation (1.44) from equation (1.41). Solution: We start with =ˆU(t)ˆHˆU†(t)Ψ(rj,RCM,t) − i?ˆU(t)∂ˆU†(t) i?∂Ψ(rj,RCM,t) Ψ(rj,RCM,t) (P1.1) ∂t ∂t and using the fact that eABe−A= B + [A,B] +1 2![A,[A,B]] + ..., with ˆP2 2M N ? 1 ˆH = [ˆ pj+ eA(RCM,t)]2+ˆVC + CM 2m j=1

8. 4 Chapter 1 and ? ? −i ˆU(t) = exp ?ˆ μ · A(RCM,t) . Remember that A(RCM,t) is not an operator. Thus ˆU(t)ˆ pjˆU†(t) = ˆ pj−i ?[ˆ μ · A(RCM,t),ˆ pj] + 0 = ˆ pj+ie ? j? = ˆ pj− eA(RCM,t); ˆU†(t) = 2M ? [rj? · A(RCM,t),ˆ pj] ?ˆP2 ? ?ˆP2 ? ˆU(t) +ˆVC +ˆVC CM 2M CM , where we used the fact that the commutator of ˆ μ = −e?rjand ˆ pjis a c number. Moreover, i?ˆU(t)∂ˆU†(t) ∂t = −ˆ μ ·∂A(RCM,t) = ˆ μ · E(RCM,t). ∂t Therefore, combining all the results in equation (P1.1), ⎡ ⎤ ⎣ˆP2 N ˆ p2 2m+ˆVC− ˆ μ · E⊥(RCM,t) ? i?∂Ψ(rj,RCM,t) j ⎦Ψ(rj,RCM,t). = + CM 2M ∂t j=1 1.9. Show that the analogue of the wave equation (1.5) for the displacement vector D(R,t) is ∂2D ∂t2 = −∇×( ∇ × P). ∇2D − μ0ε0 Solution: We can start with Maxwell’s Equations in the form ∇ · D = 0 ∇ × (D − P) = −ε0∂B ∇ × B = μ0∂D ∇ · B = 0, ∂t ∂t and take the curl of the second equation ∇×(∇ × D) = ∇×(∇ × P) − ε0∂∇ × B ∇(∇ · D) − ∇2D = ∇×(∇ × P) − ε0μ0∂D ∇2D − μ0ε0∂2D ; ∂t ∂t; ∂t2= −∇×(∇ × P).

9. 5 Preliminaries 1.10. For an infinite square well potential, show that an arbitrary initial wave packet will return to its initial state at integral multiples of a revival time τ = particle in the well and a is the width of the well. Solution: The eigenvalues are En=n2?2π2 4ma2 π?, where m is the mass of the 2ma2 and the wave function at any time is ? an(0)|n?e−in2?2π2 2ma2t. |ψ(t)? = n Clearly if τ =4ma2 π?q, where q is an integer, then ? These are called quantum revivals. 1.11. The radiative reaction rate γ2 for a classical oscillator having charge e, mass m, and frequency ω is given by 1 4π?0 Show that γ2 ω and estimate this ratio for an electron oscillator having a frequency corresponding to an optical frequency. Solution: The result follows directly for αFS= eV. Since mc2≈ 0.5 MeV, the ratio is of order 3 × 10−8giving γ2≈ 109s−1. ? an(0)|n?e−2πiqn2= |ψ(t)? = an(0)|n? = |ψ(0)? . n n 2 3 e2ω2 c3. γ2= ?ω mc2 = αFS e2 4π?0?c= 137. For an optical transition ?ω ≈ 2 1

10. Chapter 2. Two-Level Quantum Systems Unless noted otherwise (as in Problems 2.4-2.6), the rotating wave approximation can be used. 2.1. Express the Rabi frequency Ω0in terms of the time averaged power density of the light field. Assuming some reasonable values for the matrix elements, estimate Ω0for typical cw and pulsed laser fields. Solution: Recall that Ω0is in units of s−1and that Ω0/2π is in units of Hz. This problem has been solved effectively in the introductory remarks. Expressed in units of s−1, Ω0= μE/?, where E is the field amplitude and μ is a dipole moment matrix element. In atoms, the magnitude of μ is of order of the Bohr radius or μ ≈ 10−29C·m. To estimate the value of E, it is useful to express E in terms of the power density of the field. In mks units, the time-averaged power density is S = E2/(2μ0c), where μ0= 4π × 10−7. ?s−1?= μE/? =μ so that a power of 1 W/cm2corresponds to atom-field coupling strength on the order of Ω0≈ 3×108 s−1or a frequency of order Ω0/2π ≈ 5 × 107Hz=50 MHz. For a HeNe laser having 1 mW of power focussed in 1 mm2, S ≈ 103W/m2and for an Ar laser having 10 watts of cw power focussed in 1 mm2, S ≈ 107W/m2. Pulsed lasers provide much higher powers (but for short intervals of time so that the average power never exceeds a Joule or so). In 1965, NdYag lasers produced 1 mJ in 1 μ s - if focused to 1 mm2, S ≈ 109W/m2. Current state of the art pulsed lasers such as alexandrite lasers produce 1 J in 100 fs. If focused to 1 μm2, the power density is S ≈ 1025W/m2= 1021W/cm2. Current power output of the Hercules laser at the University of Michigan is on the order of 100 TW = 1014with power densities approaching 1022 W/cm2. 2.2. The appropriate Hamiltonian in the field interaction representation is ?−δ(t) |Ω0(t)| where Ω0(t) = |Ω0(t)|eiϕ(t)and δ(t) = δ+˙φ(t). For Ω0(t) = real and constant, explicitly diagonalize ˜ H to obtain its eigenvalues and eigenfunctions. Show that the amplitudes in this field interaction representation evolve as ˜ c(t) = T†eiΩtσz/2T˜ c(0), ? ? ? W/m2? 2μ0cS ? ≈ 2.6 × 106 Ω0 , S ? ˜ H(t) = (?/2) , |Ω0(t)| δ(t) 6

11. 7 Two-Level Quantum Systems where Ω2= |Ω0|2+ δ2and T˜ HT†is diagonal. Using the matrix T you find, prove that this result agrees with equation (2.116). Solution: Diagonalizing˜ H, one finds that for ?cosθ −sinθ and tan2θ =|Ω0| ? cosθ = 2 ?−Ω 0 Let ˜ c = T†b. Then the equation of motion i?d˜ c/dt =˜ H˜ c leads to i?T†db/dt =˜ HT†b i?db/dt = T˜ HT†b = − (?Ω/2)σzb b(t) = eiΩtσz/2b(0) T˜ c(t) = eiΩtσz/2T˜ c(0) ? T = , sinθ cosθ , δ ? ? ? ? 1 −δ 1 +δ sinθ = 1 2 , Ω ? 1 . Ω ?1 0 ? ? T˜ HT†=(?/2) = −(?Ω/2) = −(?Ω/2)σz. 0 Ω 0 −1 ?eiΩt/2 ? 0 ˜ c(t) = T†eiΩtσz/2T˜ c(0) = T† T˜ c(0). 0 e−iΩt/2 Carrying out the matrix multiplication, one reproduces equation (2.116) 2.3. With˜ H given as in Problem 2 calculate ˜ c(t/T) for Ω0T = 5, δT = 3. Compare your result with a direct evaluation of e−i˜ Ht/?obtained using MatrixExponential of Mathematica or some equivalent program and plot |˜ c2(t/T)|2as a function of (t/T) given ˜ c2(0) = 0. Solution: From equation (2.116) ?cos?ΩT ⎛ ⎜ i 2 T −i If ˜ c2(0) = 0, |˜ c2(t/T)|2=25 ?+ iδ ?√34 ?√34 Ωsin?ΩT t T ? Ωsin?ΩT ?√34 cos?√34t? ? ? iΩ0 t T Ωsin?ΩT cos √34cosθsin?ΩT 5 √34sin t T t T ˜ c(t/T) = ˜ c(0) ? cos?ΩT 5 √34sin ?− iδ 2 T Ωsin?ΩT ⎟ ⎠ ? 2 2 2 iΩ0 t T t T t T 2 2 2 ? ⎞ ⎟ ? t 2 ? i ⎜ ⎝ +i 3 t T ⎜ ⎟ = ˜ c(0). 2 ⎜ ⎟ ? ?√34 ? t √34sin 3 t T 2 ?√34 ? t T 34sin2 . 2 2.4. For the two-level problem including the counter-rotating terms [equation (2.55) or (2.179)], solve for a1(t) and a2(t) given that a1(0) = 1 and a2(0) = 0 in the following cases: