Atomic physics lecture notes, Lecture notes of Physics

Download Atomic physics lecture notes and more Lecture notes Physics in PDF only on Docsity! 1 ATOMIC PHYSICS (PHYS4011) LECTURE NOTES Lecture notes based on a course given by Tom Kirchner. The emphasis of the course is on solving atomic systems, in particular the Hydrogen atom through perturbation theory. Some relativistic quantum mechanics is introduced at the end York University, 2011 Presented by: TOM KIRCHNER LATEXNotes by: JEFF ASAF DROR 2011 YORK UNIVERSITY 2 CONTENTS I. Introduction: the field-free Schrodinger hydrogen atom 3 A. Reduction to an effective one-body problem 3 B. Central-field problem for relative motion 3 C. Solutions of the Coloumb problem 3 D. Assorted Remarks 5 II. Atoms in electric fields: the Stark effect 7 A. Non-Degenerate perturbation theory (PT) 8 1. Comments 10 B. Degenerate Perturbation Theory 12 C. Effect on excited states: the linear Stark effect 14 III. Interaction of Atoms with Radiation 18 A. The semiclassical Hamiltonian 18 B. Time-Dependent Perturbation Theory 21 1. General Formulation 21 2. Comments 23 3. Discussion of 1st order result 23 4. Example: Slowly varying perturbation 24 5. Solution of TDSE up to 1st order 25 6. Comments 25 7. Example: Sudden Perturbation 26 8. Example:Periodic perturbation 26 C. Photoionization 28 1. Transitions into the continuum: Fermi’s golden rule (FGR) 28 2. Dipole Approximation 30 D. Outlook on Field Quantization 31 1. Construction of HF 32 2. Creation and Annihilation Operators 34 3. Interaction Between Photon Field and Electrons 35 4. The Transition Matrix Elements 36 5. Spontaneous Emission 38 IV. Brief Introduction to Relativistic Quantum Mechanics 43 A. Klien-Gordon Equation 43 1. Setting up a relativistic wave equation 43 B. Discussion of KG equation 43 C. Dirac Equation 44 1. Free Particles 44 2. Solutions of the free Dirac equation 47 3. Add Electromagnetic Potentials 49 4. The relativistic hydrogen problem 49 5. Nonrelativistic limit of the Dirac equations 51 Lecture 3 - Jan 09, 2011 5 then ρn,`,mdr 3 is the probability to find the electron in the interval [r, r + dr]. We define the radial probability density by ρn,`(r) = r 2R2n,`(r) |Y`,m(φ, θ)| 2 dΩ (I.9) = r2R2n,`(r) (I.10) Lecture 4 - Jan 11, 2012 Next we consider the momentum space representation of the Hydrogen atom. Recall ψn,`,m(r) = 〈r|n, `m〉 (I.11) with Hrel |n, `m〉 = En |n, `,m〉. Alternatively we can project our states onto momentum space ψn,`,m(p) = 〈p|n, `m〉 (I.12) = 〈p| I |n, `,m〉 (I.13) = ∫ < 〈p|r〉 〈r|n, `,m〉 d3r (I.14) = 1 (2π~)3/2 ∫ < eip·rψn,`,m(r)d 3r (I.15) Use k = p~ and e−ik·r = 4π ∞∑ L=0 L∑ M=−L (−i)L Bessel︷ ︸︸ ︷ jL (kr)YLM (Ωk)Y ∗ LM (Ωr) (I.16) where jL is the spherical Bessel function. Note that here the Ωr represents normal θ and φ while Ωk represents θ and φ in spherical momentum coordinates. Using these relations ψn,`,m(p) = 4π (2π~)3/2 ∑ L,M (−i)L ∫ ∞ 0 r2jL (kr)Rn,`(r)dr δL,`δM,m︷ ︸︸ ︷∫ Y ∗L,M (Ωr)Y`,m(Ωr)dΩr YL,M (Ωk) (I.17) = 4π (2π~)3/2 (−i)` ∫ ∞ 0 r2j`(kr)Rn,`(r)drY`,m(Ωk) (I.18) = Pn,`(p)Y`,m(Ωp) (I.19) The probability density is ρn,`,m(p) = |φn,`,m(p)|2 (I.20) While the radial probability density in momentum space is ρn,`(p) = p 2 |φn,`,m(p)|2 (I.21) Lecture 13th, 2012 D. Assorted Remarks 1. Review Griffith’s chapters 4.1-4.3 2. Hydrogen-like ions are also solved (for Z ≥ 2). Energy scales like En ∝ Z2. 6 3. One can also look at a number of exotic systems using the same results as well. e.g. positronium(e+, e−), muonium (µ+, e−), muonic atom (p, µ−). When considering these systems the energy change is due to En ∝ µ = m1m2m1+m2 . For more on exotic systems consider C.T. book, volume I 4. There are corrections to be looked at which we will consider in detail later 5. Thus far we have used SI units. In these units we have the Hamiltonian (for Hydrogen) HSI = − ~2 2m ∇2 − e 2 4π0r (I.22) To eliminate some constants we introduce atomic units. To get rid of the constants we • measure mass in me = 1a.u. (atomic unit). • measure charge in units of e = 1a.u.. • measure angular momentum in units of ~ = 1a.u. • measure permittivity of 4πo = 1a.u. In short atomic units, a.u. are defined by me = e = ~ = 4πo = 1. As a consequence of this the Hamiltonian in atomic units are given by Ha.u. = − 1 2 ∇2 − 1 r (I.23) • For length in atomic units we use Bohr’s radius, ao = 4πo~ 2 mee2 = 0.53Å = 1a.u. (as a consequence of our earlier definitions). • For energy in atomic units we consider the Hydrogen ground state. En=1 = − ~ 2 2mea2o = −13.6eV = − 12a.u. (as a consequence of earlier definitions). Other units of energy may also be used: 1a.u.(of energy) = 27.2eV = 1 Hartree = 2 Rydberg (I.24) • For time in atomic units we use dimensional analysis in SI: time = distance speed = distance×mass momentum = distance2 ×mass angular momentum (I.25) Hence we define a unit of time: to = a2ome ~ = 2.4× 10−17s = 1a.u. (I.26) The Bohr-like revolution time of the electron around the proton in the Hydrogen ground state is τ = 2πto (I.27) • We can now infer a velocity in atomic units: τ = 2πto = 2πao vo (I.28) ⇒vo = ~ meao = 2.2× 106m/s = 1 137 c = 1a.u. (I.29) • The fine-structure constant is α = e2 4πo~c = ~ meaoc = 1 137 = ( 1 c ) a.u. (I.30) Lecture 6: January 16th, 2012 7 II. ATOMS IN ELECTRIC FIELDS: THE STARK EFFECT From classical electromagnetism we know that a uniform electric field in the z direction with field strength F is E = F k̂ (II.1) The electrostatic potential is φ(r) = −E · r (II.2) = −Fz (II.3) (II.4) The potential energy of an electron is W (r) = −eφ(r) (II.5) a = Fez (II.6) We need to solve the stationary Schrodinger equation: H |φα〉 = E |φα〉 (II.7) for (in atomic units) H = −1 2 ∇2 − 1 r + Fz (II.8) = Ho + +W (II.9) To get a better intuition on the problem we sketch the potential energies for x = y = 0. This is shown in figure 3. In principle there are no stationary state since “bound” electrons can always tunnel out of the potential well. This is V (-1/|z|)Coloumb(-1/|z|)Coloumb z Fz(F>0) Eo Eexcited FIG. 3. The potential energies of the Stark effect called ionization. In practice we only have ‘weak’ electric fields: Flab ∼ 104V/cm F1s ∼ e2 4πoa2o = 5× 109V/cm 10 1. Comments (a) These equations are non-defined only when E (0) α 6= E(0)β ∀α, β. In other words non-degenerate systems. (b) Convergence is difficult to check in perturbation theory but consistency checks can be done. One can check that ∣∣∣∣∣∣ 〈 φ0α ∣∣W ∣∣∣φ0β〉 E (0) α − E(0)β ∣∣∣∣∣∣ << 1 (II.41) If this is not fulfilled then it indicates that perturbation theory will likely fail. (c) Full calculations for the energies, Eα, beyond first order are in general not possible (due to the infinite sum). (d) Griffith, 6.1; Liboff, B.1 2. Applications to H(1s) in an electric field φ01s(r) = 1√ π e−r (II.42) E01s = − 1 2 (II.43) W = Fz (II.44) We can calculate the first order energy correction: ∆E (1) 1s = 〈 φ01s ∣∣Fz ∣∣φ01s〉 (II.45) = F π ∫ e−2rzd3r (II.46) = F π ∫ r cos θ sin θe−2rr2drdθdφ (II.47) = F π ∫ ∞ 0 r3e−2r   :0 ∫ π 0 1 2 sin 2θdθ ∫ dφ (II.48) = 0 (II.49) However this doesn’t tell us whether or not the rest of the correction orders are zero or non-zero. To get an idea for the second order correction we check the consistency criterion. Consider the overlap of the contribution of the 2p and 1s states. Note that this gives the smallest possible denominator (greatest correction).∣∣∣∣∣∣ 〈 φ (0) 1s ∣∣∣Fz |φ2p〉 E (0) 1s − E (0) n=2 ∣∣∣∣∣∣ = ∣∣∣∣∣∣ 〈 φ (0) 2p0 ∣∣∣ z ∣∣φ01s〉 − 12 + 1 8 ∣∣∣∣∣∣ (II.50) ≈ 2F (II.51) For the Stark effect the F ≈ 10−5 in atomic units. Hence we expect the second order effect to be small but non zero. Next we estimate the full second order correction ∣∣∣∆E(2)1s ∣∣∣ = ∣∣∣∣∣∣∣ ∑ β 6=1s ∣∣∣〈φ0β∣∣∣Fz ∣∣φ01s〉∣∣∣2 E (0) 1s − E (0) β ∣∣∣∣∣∣∣ (II.52) To get an upper bound for this estimate we can replace the denominator by it’s minimum value which corresponds to n = 2. ∣∣∣E(0)1s − E(0)β ∣∣∣ ≥ ∣∣∣E(0)1s − E(0)n=2∣∣∣ = 38 (II.53) 11 Thus we can write ∣∣∣∆E(2)1s ∣∣∣ = 83F 2 ∑ β 6=1s ∣∣〈φ0β∣∣ z ∣∣φ01s〉∣∣2 (II.54) = 8 3 F 2 ∑ β 6=1s ∣∣〈φ01s∣∣ z ∣∣φ0β〉 〈φ0β∣∣ z ∣∣φ01s〉∣∣ (II.55) = 8 3 F 2 ∣∣∣∣∣∣〈φ01s∣∣ z ∑ β 6=1s ∣∣φ0β〉 〈φ0β∣∣ z ∣∣φ01s〉 ∣∣∣∣∣∣ (II.56) but ∑ β 6=2s ∣∣φβ0〉 〈φ0β∣∣ = 1− ∣∣φ01s〉 〈φ01s∣∣ (II.57) Hence ∣∣∣∆E(2)1s ∣∣∣ = 83F 2 ∣∣〈φ01s∣∣ z (1− ∣∣φ01s〉 〈φ01s∣∣) z ∣∣φ01s〉∣∣ (II.58) = 8 3 F 2 {〈 φ01s ∣∣ z2 ∣∣φ01s〉− 〈φ01s∣∣ z ∣∣φ01s〉:0〈φ01s∣∣ z ∣∣φ01s〉 } (II.59) = 8F 2 3 〈 φ01s ∣∣ z2 ∣∣φ01s〉 (II.60) It turns out that the integral is easy to do ∣∣∣∆E(2)1s ∣∣∣ ≤ 8F 23 (II.61) We also know that the energy correction is negative, ∆E (2) 1s < 0. One can also calculate the exact result: ∆E (2) 1s = − 9 4 F 2 (II.62) This is called the “quadratic Stark effect”. Since F is small this is a very small correction. 3. Interpretation Consider a classical charge distribution in an electric field. The classical energy of a charge distribution is given by U = ∫ ρ(r)φ(r)d3r (II.63) = −F ∫ ρ(r)zd3r (II.64) = −Fpz (II.65) where pz ≡ z component of the dipole moment. Connection to QM: ρ(r) = − |φ1s(r)|2 (II.66) pz = − ∫ |φ1s(r)|2 zd3 (II.67) = −〈φ1s| z |φ1s〉 (II.68) = − 〈 φ (0) 1s + λφ (1) 2s + ... ∣∣∣ z ∣∣∣φ(0)1s + φ(1)1s + ...〉 (II.69) = − 〈 φ (0) 1s ∣∣∣ z ∣∣∣φ(0)1s 〉− λ〈φ(1)1s ∣∣∣ z ∣∣∣φ(0)1s 〉− λ〈φ(0)1s ∣∣∣ z ∣∣∣φ(1)1s 〉+O(λ2) + ... (II.70) 12 The first term is zero. By choosing real eigenstates we can pz = −2 ∑ β 6=1s 〈 φ (0) 1s ∣∣∣ z ∣∣∣φ(0)β 〉〈φ(0)β ∣∣∣Fz ∣∣∣φ(0)1s 〉 E (0) 1s − E (0) β (II.71) = −2F ∑ β 6=1s ∣∣∣〈φ1s| z ∣∣∣φ(0)β 〉∣∣∣2 E (0) 1s − Eβ(0) +O(λ2) (II.72) = − 2 F ∆E (2) 1s +O(λ2) (II.73) = 9 2 F +O(F 2) (II.74) In summary (a) ∆E (1) 1s = 0 ⇐⇒ p(0)z = − 〈 φ (0) 1s ∣∣∣ z ∣∣∣φ(0)1s 〉 = 0 (II.75) This is expected since a spherical charge distribution has no static dipole moment (and equivalently for a spherical probability distribution) (b) ∆E (2) 1s 6= 0 reflects that we have a nonzero induced dipole moment. This means that we have a nonzero polarizability αD = 1 F p(1)z = 9 2 (II.76) B. Degenerate Perturbation Theory We have a Hamiltonian given by H = Ho + λw (II.77) where Ho ∣∣φoα,j〉 = E(0)α ∣∣φoα,j〉 , j = 1, 2, ..., gα (II.78) H |φα,j〉 = Eα,j |φα,j〉 (II.79) Here we have assumed that after the perturbation the energy levels are non-degenerate (this is not always the case but it is the case for the Stark effect). The states{∣∣φoα,j〉 , j = 1, 2, ...gα} (II.80) span the subspace (of Hilbert space) associated with E (0) α . Any linear combination is an eigenstate of Ho for E (0) α . Effect of perturbation theory is shown in figure 4 Note that as the perturbation is increased we encounter a change in state to some linear combination of states. |φαj〉 → ∣∣φ0α,j〉 (II.81) We make the ansatz: Eα,j(λ) = E (0) α + λE (1) α,j + λ 2E (2) α,j + ... (II.82) |φα,j(λ)〉 = ∣∣φ0α,j〉+ λ ∣∣φ1α,j〉+ ... (II.83) Inserting this into the Schrodinger equation: (Ho + λw) {∣∣φ0α,j〉+ λ ∣∣φ1α,j〉+ ...} = {E(0)α + λE(1)α,j}(∣∣φ0α,j〉+ λ ∣∣∣φ(1)α,j〉+ ...) (II.84) Ho ∣∣φ0α,j〉+ λ{w ∣∣φ0α,j〉+Ho ∣∣∣φ(1)α,j〉}+O (λ2) = E(0)α ∣∣φ0α,j〉+ λ{E(1)α ∣∣φ0α,j〉+ E(0)α ∣∣∣φ(1)α,j〉}+O (λ2) (II.85) 15 Lecture 9 - January 25th, 2012 The selection rules can be written in terms of the Wigner 3j symbols. ( j1 j2 j3 m1 m2 m3 ) 6=⇐⇒  m1 +m2 +m3 = 0 and |j1 − j2| ≤ j3 ≤ j1 + j2(triangular condition) (II.106) ( j1 j2 j3 0 0 0 ) 6=⇐⇒  triangular condition and j1 + j2 + j3 = even (II.107) Applying these relations to our situation our integral is non zero only if m = m′ (II.108) ∆` = `− `′ = ±1 (II.109) These are sometimes called the electric dipole selection rules (E1) (special case). 2. Linear Stark effect for H(n = 2) Consider the degenerate stats { φo2s, φ o 2po , φ o 2p−1 , φ o 2p1 } (II.110) We consider the matrix eigenvalue problem:∑ `′,m′(n=2) 〈φo2`m|w − E (1) n=2 ∣∣φo2n,`′,m′〉 an=2`,m,`′,m′ = 0 (II.111) Consider wi,j = 〈φo2`m|w ∣∣φo2n,`′m′〉 (II.112) If i = j then wi,j = 0 because ∆` = 0. Further we have a symmetric matrix since the states are real. By using the selection rules we see that w =  0 w12 0 0w12 0 0 00 0 0 0 0 0 0 0  (II.113) Hence the only potentially nonzero element is the w12. Note that if the radial part is nonzero then the element may still be zero. It’s easy to calculate the element explicitly: w12 = 〈φo2s| z ∣∣φo2po〉 = −3a.u. (II.114) Hence we have the secular equation ∣∣∣∣∣∣∣∣ −E(1) w12 0 0 w12 −E(1) 0 0 0 0 −E(1) 0 0 0 0 −E(1) ∣∣∣∣∣∣∣∣ = 0 (II.115) This matrix is block diagonal and it’s easy to find the equation:( E(1) )2{( E(1) )2 − w212 } = 0 (II.116) E(1) = {0, 0, w12,−w12} (II.117) Hence the first order energy corrections are ∆E(1) = {0, 0,−3F, 3F} (II.118) We are left with three lines. Two lines stay degenerate as represented by the zeroes however the originally one line splits into three. Next we calculate the mixing coefficients of the expansion, a`,m,`′,m′ . We insert the eigenvalues into our equation. 16 (a) First consider E (1) 1 = E (1) 2 = 0  0 w12 0 0w12 0 0 00 0 0 0 0 0 0 0   a2sa2poa2p−1 a2p+1  =  000 0  (II.119) This is true only if a2s = a2po = 0. a2p−1 and a2p+1 are undetermined. The state that corresponds to this state is any linear combination of the 3rd and 4th states. We choose∣∣∣φ̃E1〉 = ∣∣∣φo2p−1〉 (II.120)∣∣∣φ̃E2〉 = ∣∣φ2p−1〉 (II.121) (b) Next we consider E (1) 3 = +w12: −w12 w12 0 0w12 −w12 0 00 0 −w12 0 0 0 0 −w12   a2sa2p0a2p−1 a2p+1  =  000 0  (II.122) This gives a2p−1 = a2p+1 = 0. The other two equations are − w12a2s + w12a2po = 0 (II.123) w12a2s − w12a2po = 0 (II.124) This requires that a2s = a2po . (c) The final eigenvectors for E4 = −w12 are given by a2s = −a2p0 , a2p−1 = a2p+1 = 0. We have fixed the components of the eigenvectors (the energy corrections). We should fix the normalization of the last two states (we include the previous states for concreteness):∣∣∣φ̃E(1)3 〉 = 1√2 (|φo2s〉+ ∣∣φo2po〉) (II.125)∣∣∣φ̃E(1)4 〉 = 1√2 (|φo2s〉 − ∣∣φo2po〉) (II.126)∣∣∣φ̃E(1)1 〉 = ∣∣∣φo2p−1〉 (II.127)∣∣∣φ̃E(1)2 〉 = ∣∣∣φo2p+1〉 (II.128) 3. Summary and interpretation (a) Splitting of energy levels is shown in figure (b) Note that 〈 φo2`,m ∣∣∣ z |φo2`m〉 = 0 hence the original states (φo2s, φ2p0 , φo2p±1) have no static dipole moment. However the Stark states ∣∣∣φ̃E(1)3 〉 , ∣∣∣φ̃E(1)4 〉 do have non zero static dipole moment p0z,3 = − 〈 φ E (1) 3 ∣∣∣ z ∣∣∣φE(1)3 〉) (II.129) = 3a.u. (II.130) pz,4 = −3a.u. (II.131) (c) The new states are shown in figure 6 17 F E m=+/- 1 m=0 m=0E4 E3 FIG. 5. The linear Stark splitting of energy levels z 2s x=y=0 2p0 z E x=y=0 3 z E x=y=0 4 FIG. 6. The Stark states (d) Note that we still have azimuthal symmetry. [`z, w] = 0 (II.132) Hence m is still a good quantum number (e) The last question to consider is the effect on the n = 3 shell states. We would need to consider the (3s, 3po, ..., 3d±2) (II.133) 20 • Next we perform quantization (p→ p̂ = ~i∇ ). Hence H = 1 2m (p̂ + eA) 2 − eφ+ V (III.14) Hψ(r, t) = 1 2m ( ~ i ∇+ eA(r, t) )( ~ i ∇+ eA(r, t) ) φ(r, t)− eφ(r)ψ (r, t) + V (r)φ (r, t) (III.15) = 1 2m ( −~2∇2φ+ ~e i ∇ (Aψ) + ~e i A∇ψ + e2A2ψ ) − eφψ + V ψ (III.16) = Ho︷ ︸︸ ︷( − ~ 2 2m ∇2 + V ) ψ + e~ mi A · ∇ψ + ( e~ 2mi ∇ ·A ) ψ + ( e2 2m A2 − eφ ) ψ (III.17) = Ho + W (t)︷ ︸︸ ︷ e m A · p + e 2m p ·A + e 2 2m A2 − eφ (III.18) where Ho is the Hydrogen Hamiltonian without electromagnetism and W (t) is the time dependent pertur- bation. • Assume for the following the electromagnetic is a free electromagnetic field which is characterized by no charges and no currents, ρ = 0; J = 0. Here its convenient to choose the Coloumb gauge, ∇ ·A = 0. This is useful because then we arrive at the equations ∇2A− 1 c2 ∂2A ∂t2 = 0 (III.19) and φ = 0 (III.20) Note this doesn’t mean that there is no electric field (since E = −∂A∂t ). Consider the monochromatic solution to the wave equation, a plane wave. A(r, t) = Π̂ |Ac| cos (k · r− ωt+ α) (III.21) One can perform two checks. Firstly ∇ ·A = −Π̂ · k |Ac| sin (k · r− ωt+ α) = 0 (III.22) Hence Π̂ ⊥ k. Thus the vector potential is a transverse wave. The second check is the wave equation ∇2A = 1 c2 ∂2A ∂t2 (III.23) −k2A = −ω 2 c2 A (III.24) Thus we see that this is a solution given that k = ωc . Going back to the perturbation we have (due to the Coloumb gauge p ·A one term disappears and another disappears because we don’t have a scalar potential) W (t) = e m A · p + e 2 2m A2 (III.25) Assume that A is weak. If this is the case we can neglect the second term: W (t) ≈ e m A · p (III.26) 21 Lecture 11 - February 1st, 2012 Test 1 is up to (and including) last lecture. In other words up to this point. We can write the Hamiltonian due to an external field by H = Ho +W (t) (III.27) where Ho = p2 2m + V (r); W (t) = e m A(r, t) · p + e 2 2m A2(r, t) (III.28) Note that since we have a time dependent Hamiltonian we cannot use the stationary Schrodinger equation. Thus we use time dependent perturbation theory. B. Time-Dependent Perturbation Theory 1. General Formulation Consider a perturbation W (t) = λw(t) on a Hamiltonian which is stationary. H(t) = Ho + λw(t) (III.29) The goal is to solve the time-dependent Schrodinger equation: i~ ∂ ∂t |ψ(t)〉 = H(t) |ψ(t)〉 (III.30) The SE is a initial value problem. We assume we know the state at some time. Assume for t ≥ to that W (t) = 0; |φ(to)〉 = |φo〉 (III.31) where Ho |ψj〉 = j |ψj〉 (j = 0, 1, ...). We assume that W is switched on at to and we look into how the system develops in time. Define |ψj(t)〉 = e−ijt/~ |φj〉 then inserting into the SE (before perturbation) by i~ ∂ ∂t |ψj(t)〉 = i~ ( ∂ ∂t e−ijt/~ |φj〉 ) (III.32) = je −ijt/~ |φj〉 (III.33) = Ho |ψj〉 (III.34) Hence if |φ〉 solves the SE then so does |ψ〉. We denote the |ψ(t)〉 by the solution after the perturbation has been turned on. We expand these states as |ψ(t)〉 = ∑ j cj |ψj(t)〉 (III.35) and insert into the time dependent SE i~ ∂ ∂t ∑ j cj(t) |ψj(t)〉  = (Ho +W (t))∑ j cj(t) |ψj(t)〉 (III.36) ∑ j (i~ċj(t) + j) e−ijt/~ |φj〉 = (Ho +W (t)) ∑ j cj(t)e −ijt/~ |φj(t)〉 (III.37) (III.38) 22 We project these states onto the state 〈ψk(t)| = 〈φk| eikt/~: ∑ j e i ~ (k−j)t δj,k︷ ︸︸ ︷ 〈φk|φj〉 (i~ċj(t) + cjj) = ∑ j e i ~ (k−j)tcj(t) 〈φk|Ho + λw(t) |φj〉 (III.39) i~ċk(t) +ck(t)k =ck(t)k + ∑ j e i ~ (k−j)tcj(t) 〈φk|λw(t) |φj〉 (III.40) i~ċk = ∑ j e i ~ (k−j)tcj(t)λwkj (III.41) where wkj ≡ 〈φk|w |φj〉. Note that thus far we have not made any approximations. These are sometimes called coupled-channel equations. If we assume that W (t > tf ) = 0 then for t > tf the matrix elements on the right side are 0 and hence ċk(t > tf ) = 0 (III.42) and the states are constant. The probabilities for transitions |φ0〉 → 〈φk〉 (equivalently |ψ0〉 → |ψk〉 ). The probability for transition to state k is pk = |ck|2t>tf = |〈ψk|ψ〉| 2 t>tf = |〈φk|ψ〉|2t>tf (III.43) To check the answer one can check that ∑ k pk = 1. This turns out to be true. We finally introduce perturbation theory. Notice that the only thing we can expand is the expansion coefficients. We use a power series expansion ck(t) = c (0) k (t) + λc (1) k (t) + λ 2c (2) k + ... (III.44) We insert this into the coupled channel equations i~ { ċ (1) k + λċ (1) k + ... } = λ ∑ j e i ~wkj(k−j)t { c (0) j + λc (1) j + ... } (III.45) consider zeroth order, λ0: i~ċ(0)k = 0 (III.46) Next consider first order i~ċ(1)k = ∑ j c (0) j e i ~ (k−j)twkj (III.47) Lastly consider second order i~ċ(2)k = ∑ j c (1) j e i ~ (k−j)twkj (III.48) Notice that you can in principle get a solution for all orders using this method. Once you have first order you can get second order and once you have second order you can get third order etc. In other words we solve these successively. For first order c (0) k (t) = const = δk,0 (III.49) Our initial condition was the ground state. Hence the coefficients are zero except the ground state coefficient. In zeroth order we just stay in the ground (initial) state. Input this into higher orders: i~ċ(1)k = ∑ j δj,0e i ~ (k−j)twk,j (III.50) = e i ~ (k−0)t 〈φ|w(t) |φ0〉 (III.51) c (1) k (t)− * 0 c (1) k (t0) = − i ~ ∫ t to e i ~ (k−0)t ′ 〈φk|w(t′) |φj〉 dt′ (III.52) c (1) k (t) = − i ~ ∫ t to e i ~ (k−0)t ′ 〈φk|w(t′) |φj〉 dt′ (III.53) 25 we have ck(t) = − i ~ ∫ t t0 eiωk0t ′ Wk0(t ′)dt′ (III.67) = − i ~  1iωk0 eiωk0t′Wk0(t′) ∣∣∣∣t t0 − 1 iωk0 ∫ t t0 eiωk0t ′ ≈0︷ ︸︸ ︷ W̄k0(t ′) dt′  (III.68) ≈ − 1 ~ωk0 Wk0(t)e iωk0t (III.69) = −〈φk|W (t) |φ0〉 k − 0 eiωk0t (III.70) p0→k = |ck|2 (III.71) = |〈φk|W (t) |φ0〉|2 (k − 0)2 (III.72) = 0(if for t > tf , W (tf ) = const) (III.73) This only works for a non-degenerate initial state (due to the energies in the denominator). 5. Solution of TDSE up to 1st order Consider the solution to the TDSE and insert in the 1st order coefficient |ψ(t)〉 = ∑ k ck(t)e − i~ kt |φk〉 (III.74) = c0(t)e − i~ 0t |φ0〉+ ∑ k 6=0 〈φk|W (t) |φ0〉 0 − k e− i ~ kt |φk〉 (III.75) Since we are using perturbation theory we require that c0 ≈ 1 in this case we have |ψ(t)〉 ≈  |φ̃0(t)〉︷ ︸︸ ︷ |φ0〉+ ∑ k 6=0 〈φk|W |φ0〉 0k |φk〉  e− i~ 0t (III.76) ∣∣∣φ̃0(t)〉 describes state of the system at time t up to 1st order corresponding to perturbed energy eigenvalue  (1) 0 (t) = 0 + 〈φ0|W (t) |φ0〉 (III.77) The system remains in the ground state of the total (instantaneous) Hamiltonian, H(t), at all times. This is often called an adiabatic situation. The approximation of neglecting the time derivative of W is called the adiabatic approximation. 6. Comments 1. This argument can be generalized to strong perturbations (all orders). If perturbation varies slowly with time the system is found in an eigenstate of the total Hamiltonian H(t) = H0 + W (t) at all times (“adiabatic approximation”). Ref: D.Bohm, Quantum Theory, Chapter 20 2. Realizations of this formulism can found in • Slow atomic collisions 26 • Stern-Gerlach experiment Lecture 13, Feb 13th, 2012 7. Example: Sudden Perturbation Consider the following perturbation: W (t) = { 0 t ≤ t0 W t > t0 (III.78) The first order amplitude is ck(t) = − i ~ ∫ t 0 ei(ω−ωk0)tWk0(t ′)dt′ (III.79) where ωk0 = k−0 ~ and Wk0 = 〈k|W (t) |0〉. Thus ck(t) = − i ~ Wk0 ∫ eiωk0dt′ (III.80) = −Wk0 ~ωk0 ( eiωk0t − 1 ) (III.81) P0→k(t) = |ck(t)|2 = |Wk0|2 ~2 ωk0 {2− 2 cosωk0t} (III.82) = 4 |Wk0|2 ~2 f (t, ωk0) (III.83) where f (t, ωk0) = ( sin2(ωk0t2 ) ω2k0 ) . The function f is independent of the particular perturbation. The perturbation is shown in figure 10. Note that noticeable transitions only occur within ∆Ω = 2πt . This corresponds to an energy range DΩ= 2 Πt -15-10 -5 5 10 15 Ωk0 0.05 0.10 0.15 0.20 0.25 f FIG. 10. The function f which determines the probability of transition ∆E = 2π~t . 8. Example:Periodic perturbation Consider the potential below: W (t) = { 0 t ≤ t0 = 0 Beiωt +B†e−iωt t > t0 = 0 (III.84) 27 Note that W = W †. We consider the first order amplitude for a transition from state |i〉 to |f〉. The amplitude is cfi(t) = − i ~ ∫ t 0 Wfi(t ′)eiωfit ′ dt′ (III.85) where ωfi = f−i ~ . cfi(t) = − i ~ { 〈f |B |i〉 ∫ t 0 ei(ωfi+ωt ′dt′) + 〈f |B† |i〉 ∫ t 0 ei(ωfi−ω)t ′ dt′ } (III.86) we now define Bfi ≡ 〈f |B |i〉 and note that 〈f |B† |i〉 = 〈i|B |f〉 = B∗if . We can now write cfi(t) = − { Bfi ~ (ωfi + ω) ( ei(ωfi+ω)t − 1 ) + B∗if ~ (ωfi − ω) ( ei(ωfi−ω)t − 1 )} (III.87) Pi→f (t) = |cfi(t)|2 (III.88) = |Bfi|2 ~2 (ωfi + ω)2 ∣∣∣ei(ωfi+ω)t − 1∣∣∣2 + |Bif |2~2 (ωfi − ω) ∣∣∣ei(ωfi−ω)t − 1∣∣∣2 + cross terms (III.89) The function is plotted in figure 11. fi = - fi = + FIG. 11. The probability of a transition with oscillatory perturbation We summarize the observations below • Consider the case of ωfi = −ω ⇐⇒ f = i−~ω. Hence we can only have a noticeable transition if the applied frequency corresponds to the difference between energy of the two energy levels is ~ω. This is called stimulated emission since this required a perturbation to emit energy. This is also called resonance de-excitation. • For the second peak we have ωfi = ω ⇐⇒ f = i − ~ω. Hence we can only have a noticeable transition if we have absorption of ~ω. This is also called resonant excitation. Now we consider the connection to atom-radiation interaction. We found earlier that W (t) = e m A(r, t) · p (III.90) with A(r, t) = Π̂ |A0| cos (k · r− ωt+ α) (III.91) = Π̂ 2 { A0e i(k·r−ω)t +A∗0e −i(k·r−ωt) } (III.92) where A0 ≡ |A0| eiα. Hence we have W (t) = e 2m { A0e i(k·r−ωt)Π̂ · p +A∗0e−i(k·r−ωt) } (III.93) Recall that we had W (t) = B†e−ωt +Beiωt thus we an make the identification B = e 2m A∗0e −ik·rΠ̂ · p 30 To carry out the integral we define ω̃ ≡ ω′fi − ω and hence df = ~dω̃ P absi→f ≈ 4 ~ |Bfi|2 ρ(f ) ∫ sin2 ω̃t2 ω̃2 dω̃ (III.116) Technically the integral is from f−∆ to f+∆. However if we are centered on an absorption peak then contributions from other parts of the function are small. Thus we may as well just extend the limits of the integral to ±∞, which is a well known, analytic integal. With this we have P absi→f = 2πt ~ |Bfi|2 ρ (f ) t (III.117) with f = i + ~ω One can easily show that we get a similar equation for stimulated emission: PSEi→f = 2πt ~ |Bfi|2 ρ (f ) t (III.118) where f = i − ~ω. We define the transition rate by Wi→f = d dt Pi→f (III.119) This is given by Wi→f = 2π ~ |Bfi|2 ρ(f ) (III.120) where f = i ± ~ω. This is called Fermi’s Golden Rule (FGR). 2. Dipole Approximation Typical situation is that the wavelength of the light used is large compared to the characteristic distance of atoms. i.e. λ = 2πk  a0. In this case we use the dipole approximation that says eik·r = 1 + k · r + ... (III.121) ≈ 1 (III.122) Hence the field doesn’t change spatially across the atom. On Monday we said that Bfi = e 2m A∗0Π̂ 〈φf | e−ik·rp |φi〉 (III.123) Applying the dipole approximation we have Bfi ≈ e 2m A∗0Π̂ · 〈φf |p |φi〉 (III.124) We now use a commutator relation: p = im ~ [H0, r] (III.125) where Ho = p2 2m + V . With this relation 〈φf |p |φi〉 ≈ im ~ 〈φf |Hor− rH0 |φi〉 (III.126) If we have φf and φi as eigenstates then we have 〈φf |p |φi〉 = im ~ (f − i) Dipole Matrix Elements︷ ︸︸ ︷ 〈φf | r |φi〉 (III.127) 31 Thus we have (inserting back into previous equation) Bdipfi = ie 2~ A∗0 (f − i) Π̂ · 〈φf | r |φi〉 (III.128) For Π̂ = ẑ we have the standard selection rules, ∆m = 0,∆` = ±1. We can now figure out the transition rates using FGR. Using this one will find that W dip,ẑi→f ∝ cos 2 θ (III.129) for φi = s-states. This is plotted in figure 13 z FIG. 13. The transition probability as a function of θ for the 1s state Lecture 15th - February 27th, 2012 D. Outlook on Field Quantization We need to find a Hamiltonian for atom and electromagnetic field H = HA +HF +W (III.130) where HA is the Hamiltonian due to the atom, HF is the Hamiltonian due to the field, and W represents the interaction of the atom with the field. We can write H = Ho +W (III.131) where Ho = HA +HF . The steps toward a 1st-order PT treatment are 1. Determine H0 = HA + HF . We already know HA = p2 2m + e2 r . However we don’t know HF . One requirement we will use is that HF will be Hermitian, i.e. HF = H † F . 2. Solve the eigenvalue problem of H0. We already know the original eigenstates and energies: HA |φj〉 = j |φj〉 (III.132) but we don’t know the states and energies below HF |ρk〉 = ̃k |ρk〉 (III.133) 32 If we denote |ψ`〉 as the full unperturbed states then we have H0 |ψ`〉 = (HA +HF ) |φ`〉 |ρ`〉 (III.134) = (HA |φ`〉) |ρ`〉+ |φ`〉HF |ρ`〉 (III.135) = ` |ψ`〉+ ̃` |ψ`〉 = (` + ̃`) |ψ〉 (III.136) 3. Determine W . We use the ansatz W = e m A · p (III.137) 4. Obtain 1st - order transition rates (apply Fermi’s Golden Rule) • Need 〈ψf |W |ψi〉 (III.138) 1. Construction of HF The energy of a classical electromagnetic field in a vacuum of volume L3 is (denote this energy WEM ) WEM = o 2 ∫ ( E2 + c2B2 ) d3r (III.139) Recall for free electromagnetic waves (electric potential is zero) we have with the Coloumb gauge that ∇2A− 1 c2 ∂2A ∂t2 = 0 (III.140) Assume periodic boundary conditions for each side of cube A (x, y, z = 0) = A (x, y, z = L) A (x, y = 0, z) = A (x, y = L, z) A (x = 0, y, z) = A (x = L, y, z) This gives (kj = kx, ky, kz for j = 1, 2, 3) 1 = eikjL (III.141) and hence kz = 2π L nz (III.142) The total vector potential is given by A(r, t) = ∑ λ Aλ(r, t) (III.143) where Aλ(r, t) = Π̂ L3/2 { qλe i(kλ·r−ωλt) + q∗λe −i(kλ·r−ωλt) } λ is the mode index. Each mode is characterized by { kλ, Π̂λ, ωλ } . Π̂λ is the unit polarization vector, ωλ = ckλ, and kλ = 2π L ( nλx, n λ y , n λ z ) where nλx, n λ y , n λ z ∈ Z. We can now calculate E = −∂A∂t and B = ∇ ×A. By taking these derivatives and inserting into the equation for WEM above we get (after a lot of manipulations) WEM = 20 ∑ λ w2λq ∗ λqλ (III.144) 35 The occupation number obeys nλ |ψnλ〉 = nλ |ψnλ〉 (III.162) We use shorthand notation of |ψnλ〉 = |nλ〉. One must be careful with this notation. Since we may write the following equation n̂λ |nλ + 1〉 = (nλ + 1) |nλ + 1〉 (III.163) Aˆwas used to differentiate the operator n̂λ from the number. We now consider the state |nλ = 0〉 ≡ |0〉 (III.164) which we call the vacuum state. This gives nλ |0〉 = |∅〉 (III.165) This is not the same as |0〉! This is really the zero state. However since we have a ket on the left side we don’t want to write just 0. One can show that b†λ |nλ〉 = √ nλ + 1 |nλ + 1〉 bλ |nλ〉 = √ nλ |nλ − 1〉 In particular we have bλ |0〉 = |∅〉 (III.166) Lecture 17th - March 5th, 2012 Here we generalize these results. We use the rule that if the eigenvalue of your system is the sum of a set of eigenvalues then the resultant eigenvalues are the product of the eigenstates: HF |n1, n2, ...〉 = E |n1, n2, ...〉 (III.167) where E = ∑ λ ~ωλ ( nλ + 1 2 ) = ∑ λ ~ωλnλ+E0 and |n1, n2, ...〉 = |n1〉⊗|n2〉⊗ ... are the product states. The operator nλ counts the number of modes in mode λ: n̂λ |n1, n2, ..., nλ, ...〉 = nλ |n1, n2, ..., nλ, ...〉 b†λ |n1, n2, ..., nλ, ...〉 = √ nλ + 1 |n1, n2, ..., nλ + 1, ...〉 bλ |n1, n2, ..., nλ, ...〉 = √ nλ |n1, n2, ..., nλ, ...〉 3. Interaction Between Photon Field and Electrons In the beginning we discussed that we are using the semiclassical interaction with a perturbing Hamiltonian: W = e m A · p (III.168) with A (r, t) = ∑ λ Π̂λ L3 { qλe i(kλ·r−ωλt) + q∗λe −i(kλ·r−ωλt) } (III.169) To quantize this operator we made the transformation given earlier: qλ = 1 2 √ 0 ( Qλ + i Pλ ωλ ) = √ ~ 20ωλ bλ (III.170) 36 Hence we have the vector potential in its new form given by A (r, t) = ∑ λ Π̂λ γ︷ ︸︸ ︷ 1 L3 √ ~ 20ωλ { bλe i(kλ·r−ωλt) + b†λe −i(kλ·r−ωλt) } (III.171) This operator is time dependent. However we want an operator thats time independent. To achieve this we define bHλ = bλe −iωλt (III.172) For this to make sense we need to show that i~ d dt bHλ = [ bHλ , H H λ ] (III.173) where we denote operators in the Heisenberg picture by superscript H. Proof of statement above is shown below. The Hamiltonian in the Schrodinger interpretation is the same as in the Heisenberg interpretation. Thus we have RHS = [ bHλ , H (λ),H F ] = e−iωλt [ bλ, ~ω ( nλ + 1 2 )] (III.174) = ~ωλe−iωλt [bλ, nλ] (III.175) = ~ωλe−iωλtbλ (III.176) LHS = i~ω(−iωλ)bλe−iωλt = ~ωλe−iωλtbλ (III.177) = RHS (III.178) Hence our construction gave us an operator in the Heisenberg picture. Thus removing the time dependence is easy and gives (where γ was defined above) A(r) = γ ∑ λ Π̂ { bλe ikλr + b†λe −ikλ·r } (III.179) This is our quantized vector potential in the Schrodinger picture where the interaction is W = e m A · p (III.180) = −i β︷ ︸︸ ︷√ e2~3 2ωλ0m2L3 Π̂ { eikλr∇bλ + e−ikλ·r∇b†λ } (III.181) 4. The Transition Matrix Elements 〈φf |W |i〉 = ∑ λ 〈φf |Wλ |φi〉 (III.182) = ∑ λ 〈ψf | ⊗ 〈 nf1n f 2 ... ∣∣∣ (Wλ) ∣∣ni1ni2...〉⊗ |ψi〉 (III.183) where |ψ〉 are the electron states while |n1n2, ...〉 are the photon states. We have two parts to the equation the product states are made up of a photon part and an electron part. 〈φf |W |φi〉 = −iβ ∑ λ { 〈ψf | eikλ·rΠ̂λ · ∇ |φi〉 ⊗ 〈 nf1n f 2 ... ∣∣∣ bλ ∣∣ni1ni2...〉+ 〈φf | e−ikλ·rΠ̂λ∇ |φi〉 ⊗ 〈nf1nf2 ...∣∣∣ b†λ ∣∣ni1ni2...〉} (III.184) 37 Note that we already know the electron matrix elements since we dealt with them earlier (we will get back to them in more detail shortly). We now consider the photon matrix elements.〈 nf1n f 2 ...n f λ... ∣∣∣ bλ ∣∣ni1ni2...niλ...〉 = (δnf1 ,ni1δnf2 ,ni2 ...δnfλ,niλ−1...)√niλ (III.185) This matrix element is highly selective. Furthermore we have〈 nf1n f 2 ...n f λ... ∣∣∣ b†λ ∣∣ni1ni2...niλ...〉 = (δnf1 ,ni1δnf2 ,ni2 ...δnfλ,niλ+1...)√niλ + 1 (III.186) Hence the condition to have a non zero transition elements are that the photon numbers don’t change by more then one. Furthermore there can only be a single photon interaction at a time. The annihilation of a photon corresponds to absorption of a photon, while the creation of a photon corresponds to an emission of a photon. Summary: The matrix elements 〈φf |W |φi〉 6= 0 (III.187) are non-zero if and only if 1. ∣∣ni1ni2...〉 and ∣∣∣nf1nf2 ...〉 (III.188) differ in exactly one mode (by one photon). 2. If we apply the dipole approximation then we have the dipole selection rules for the electronic part in the non-zero mode (eikλ·r ≈ 1). In this case we can use the typical selection rules of ∆` = ±1,∆m = 0 3. Recall that we found in the constant perturbation or sudden approximation that the transition is small unless energy is conserved. i.e. Ei = Ef (III.189) H0 |φi〉 = (HA +HF ) |ψi〉 ∣∣ni1ni2...〉 (III.190) = (HA |φi〉) + |φi〉HF ∣∣ni1ni2...〉 (III.191) = ( εi + ∑ λ′ ~ωλ′ ( niλ′ + 1 2 )) |φi〉 (III.192) This gives Ef = εf + ∑ λ′ ~ωλ′ ( nfλ′ + 1 2 ) (III.193) Energy conservation Ei = Ef implies that εf = εi + ∑ λ′ ~ωλ′ ( niλ′ − n f λ′ ) (III.194) εi ± ~ωλ (III.195) where the plus sign corresponds to absorption while the minus sign corresponds to emission. Lecture 18 - March 7th, 2012 Note: Look carefully on last question of new assignment, it may be a test question Recall we can write 〈φf |W |φi〉 = −iβ ×  〈ψf | e−ikλ·rΠ̂λ · ∇ |ψi〉 √ niλ + 1 〈ψf | eikλ·rΠ̂λ · ∇ |φi〉 √ niλ 0 (III.196) Note we can only have one of above. 40 1. As an example consider a Hydrogen 2p→ 1s transition H(2p) H(1s) The lifetime is given by T dip2p→1s = 1 WS.E2p→1s ≈ 1.6× 10−9s (III.224) In a classical picture it takes the electron 10−16s to circle around the nucleus. Thus this is a very long lifetime with respect to this value. 2. We can consider a different decay of Hydrogen 2s→ 1s It turns out that T 1storder2s→1s =∞ (III.225) However experimentally we have T experiment2s→1s = 0.1s (III.226) This state is metastable. We need second order perturbation theory (more then FGR) to do this. This corre- sponds to a two-photon process. 3. For an N -photon process one needs to consider N th order perturbation theory. The W operator is linear in b and b† so in order to contribute a N photon process we need to combine more of these operators. Lecture 19th - March 12th, 2012 Recall that the total transition rate in the dipole approximation is W s.e.i→f = e2ω3 3π0~c3 ( |xif |2 + |yif |2 + |zif |2 ) (III.227) The lifetime is simplify given by Ti→f = 1 W s.e.i→f (III.228) Consider if we have an electron in the 3p state: φi = H(3p). This is shown in figure 14 The total decay rate is the 2s 1s 3p FIG. 14. A decay from a Hydrogen 3p state sum of the two rates: W s.e. = e2 3π0~c3 ∑ f(εf<εi) ω3if |rif | 2 (III.229) 41 Here the rates are uncoupled (the rate of 3p→ 2s doesn’t effect the rate of 3p→ 1s). In practice of course this is not the case. Concluding remarks on photons: Here we defined photons as the quanta of an electromagnetic field. The properties of the photon are • Can be created or annihilated (hence they are not stable) • Carry energy ~ωλ • One can show that they carry momentum ~kλ. The reasoning is as follows. One could start from a classical expression for momentum of the electromagnetic field: pEM = 0 ∫ V=L3 (E×B) d3r (III.230) For our cube we know the vector potential (Aλ(r, t) = ∑ λ ...) and hence we can find the electromagnetic fields E = −∂A ∂t B = ∇×A we can then insert this result into the momentum and find the momentum of the fields: pEM = 20 ∑ λ kλωλq ∗ λqλ (III.231) Quantizing this operator gives pF = ∑ λ ~kλb†λbλ = ∑ ~kλnλ (III.232) Hence we have pF |n1, n2, ...〉 = ∑ λ ~kλnλ |n1, n2, ...〉 (III.233) From this it is clear that momentum of each mode is ~kλ. • Similarly one can show that photons also carry angular momentum (often called photon spin) given by ±~. This is reasoned as follows. The angular momentum of an electromagnetic field is LEM = 0 ∫ L3 r× (E×B) d3r (III.234) This gives a spin of ±~. • Since one can show that the angular momentum is an integer the photons are bosons. Note that we didn’t find any restriction for the number of photons that can be in a given mode. This is another way of showing that photons are bosons. • The photons move with the velocity of light v = c since that the velocity of electromagnetic waves in a vacuum. By Einstein’s theory of special relativity we know that the photons have zero mass • The photon treatment can be found in the following resources: – Friedrich, Theoretical Atomic Physics – Sakurai and Napolitano, Modern Quantum Mechanics – Schiff, Quantum Mechanics • The semiclassical atom-radiation interaction can be found in – Liboff (Chap 13.5 - 13.9) – Cohen-Tannoudiji (Chap 13) 42 As an aside we now go over the solution to the practice problem put online. Consider the harmonic oscillator Hamiltonian: H0 = p2 2m + m 2 ω20x 2 (III.235) with the perturbation W = 1 2 mω2x2 cosωt (III.236) c1storderk (t) = δk0 − i ~ ∫ t t0 eiωk0t ′ 〈k|W (t′) |0〉 dt (III.237) We first need to find the matrix elements. 〈k| W̃ |0〉 ≡ 〈k|x2 |0〉 (III.238) Here we use the following x = √ ~ 2mω0 ( a+ a† ) (III.239) where a and a† are the annihilation and creation operators respectively. x2 = ~ 2mω0 ( a2 + a† 2 + aa†a†a ) (III.240) 〈k|W |0〉 = ~ 2mω0 (  :0〈k| a2 |0〉+ 〈k| aa† |0〉+ :0〈k| a†a |0〉+ 〈k| a† 2 |0〉 ) (III.241) but a† |0〉 = |1〉 , a |1〉 = |0〉 , and a† |1〉 = √ 2 |2〉. Thus 〈k|W |0〉 = ~ 2mω0 ( 〈k|0〉+ √ 2 〈k|2〉 ) (III.242) = ~ 2mω0 ( δk0 + √ 2δk2 ) (III.243) We now summarize our results as 〈k|W (t) |0〉 ∣∣∣∣ k 6=0 = √ 2~ 2mω0 mω2 4 ( eiωt + e−iωt ) (III.244) Thus c2(t) = −i √ 2 8 ω2 ω0 ∫ t 0 ei2ω0t ′ ( eiωt ′ + e−iωt ) dt′ (III.245) where we have ω20 = 2ω0. c2(t) = −i √ 2 8 ω2 ω0 ( 1 i (2ω0 + ω) ( ei(2ω0+ω)t − 1 ) + 1 i (2ω0 − ω) ( ei(2ω0−ω)t − 1 )) (III.246) For the case of ω = 2ω0 (the resonance condition) we have (must go back to the integral to show this) c2(t) = −i √ 2 2 ω0t (III.247) The probability is given by P2(t) = ω20 2 t2 (III.248) Lecture 20th, March 19th, 2012 45 We have our energy momentum relation E2 = p2c2 +m2c4 (IV.22) If we had the equation (which is not true!) E = pc+mc2 (IV.23) then we are first order and space and time. Dirac’s idea was to write H = α · pc+ βmc2 . (IV.24) Dirac realized that this an satz can fulfill the energy momentum relation if α and β are matrices. Trying to fulfill the requirements listed above gives the restrictions on α and β. All we know so far is that we have three α matrices and 1 β matrix. We call these N ×N matrices. Since these are matrices we require wavefunctions with a number of components equal to the dimensionality of the matrices. These wavefunctions are called ‘spinor wavefunctions’. We can denote this spinor wavefunction as follows Ψ =  ψ1(r, t) ψ2(r, t) ... ψN (r, t)  (IV.25) where N is the dimensionality of the matrices. The first requirement says that each component ψi(r, t) must fulfill the KG equation. Lecture 21 - March 21st, 2012 We now go through Dirac’s wishlist in order to find conditions for α and β: First each component of ψi(r, t) must fulfill the Klien Gordon equation. Consider the Dirac equation:( i~ ∂ ∂t −H ) Ψ = 0 (IV.26) iterating the Dirac equation gives( i~ d dt −H )( i~ d dt −H ) Ψ = 0 (IV.27) −~2 d 2 dt2 +H2 − 2i~ ( d dt H ) Ψ = 0 (IV.28) −~2 d 2 dt2 +H2 − 2H2Ψ = 0 (IV.29) − ~2 d 2 dt2 = H2Ψ 0 (IV.30) −~2 ∂ 2 ∂t2 Ψ = ( cα · p + βmc2 ) ( cα · p + βmc2 ) Ψ = c~ i ∑ j αj d dxj + βmc2 (c~ i ∑ k αk d dxk + βmc2 ) Ψ = −~2c2∑ j,k αjαk ∂2 ∂xj∂xk + ~mc3 i ∑ j (αjβ − βαj) d dxj + β2m2c4 Ψ = −~2c2∑ j,k αjαk + αkαj 2 ∂2 ∂xj∂xk + ~mc3 i ∑ j (αjβ − βαj) d dxj + β2m2c4 Ψ (IV.31) 46 Recall the Klien Gordan equation says that − ~2 ∂ 2 ∂t2 ψ = ( −~2c2∇2 +m2c4 ) ψ (IV.32) The left hand sides are trivially equal to one another. Thus we require certain conditions. Clearly there are no first order derivatives in the KG equation. Thus we require αjβ + βαj ≡ {αj , β} = 0 . (IV.33) The KG equation only has second derivatives in the form of a Laplacian. Thus all the cross terms such as ∂ 2 ∂x1x2 must be zero. This can be done by introducing a Kronecker delta. However the term must be 1 for each second derivative of a given variable (e.g. ∂ 2 ∂x21 ): αjαk + αkαj = 2δjk (IV.34) Lastly we require β2 = 1 (IV.35) This conditions must be fulfilled for all j, k = 1, 2, 3. Further requirements(one that wasn’t mentioned in Dirac’s wishlist but important) was that the Dirac Hamiltonian should be Hermitian: H = H† (IV.36) This requires αj = α † ; β = β† . (IV.37) Hence they have real eigenvalues. Notice our requirements tell us that α2j = 1 and β 2 = 1 Thus the eigenvalues must be ±1 for all four of these matrices. Consider the equation αjβ + βαj = 0 . If we multiply this equation on the right with β we we have αjβ 2 = −βαjβ αj = −βαjβ Tr(αj) = −Tr(βαjβ) Tr(αj) = −Tr(αj) (IV.38) where in the last step we used β2 = 1. Hence the trace of αj must be zero. The same can be shown for β by multiplying by αj instead of β in the beginning. Thus we have another condition on αj and β, αj and β are traceless. Since the matrices are tracelss the sum of the eigenvalues is zero (easy to prove). Since the eigenvalues are ±1 and the only way to have a sum of 0 we need to have an even dimension! We need 4 anticommuting matrices. The simplest choice for the dimension would be N = 2. However we already know that the Pauli matrices make up three independent anticommuting matrices and there cannot be a fourth. Hence we need to go up to a larger dimension. We choose the next lowest dimension, namely N = 4. Matrices are obey all the conditions we have above are αx =  0 0 0 10 0 1 00 1 0 0 1 0 0 0  ; αy =  0 0 0 −i0 0 i 00 −i 0 0 i 0 0 0  ; αz =  0 0 1 00 0 0 −11 0 0 0 0 −1 0 0  ; β =  1 0 0 00 1 0 00 0 −1 0 0 0 0 −1  (IV.39) 47 Note that we can rewrite these as αx = ( 0 σx σx 0 ) ; αy = ( 0 σy σy 0 ) ; αz = ( σz 0 0 σz ) ; β = ( 1 0 0 −1 ) (IV.40) We now take a look at the Dirac equation more explicitly: i~ ∂ ∂t  ψ1ψ2ψ3 ψ4  = {c (αxpx + αypy + αzpz) + βmc2}  ψ1ψ2ψ3 ψ4  (IV.41) These are four coupled equations because α have all off-diagonal terms. Working out these equations gives (when you multiply by α and β matrices) i~ ∂ ∂t ψ1 = ~c i (( ∂ ∂x − i ∂ ∂y ) ψ4 + ∂ ∂z ψ3 ) +mc2ψ1 (IV.42) i~ ∂ ∂t ψ2 = ~c i (( ∂ ∂x + i ∂ ∂y ) ψ3 − ∂ ∂z ψ4 ) +mc2ψ2 (IV.43) i~ ∂ ∂t ψ3 = ~c i (( ∂ ∂x − i ∂ ∂y ) ψ2 + ∂ ∂z ψ1 ) −mc2ψ3 (IV.44) i~ ∂ ∂t ψ4 = ~c i (( ∂ ∂x + i ∂ ∂y ) ψ1 − ∂ ∂z ψ2 ) −mc2ψ4 (IV.45) 2. Solutions of the free Dirac equation To solve them we use the ansatz ψj(r, t) = uje i(k·r−ωt) (IV.46) with E = ~ω and p = ~k. This gives four linearly independent solutions. The two solutions u(1) =  10χ1 χ2  ; u(2) =  01χ′1 χ′2  (IV.47) correspond to an energy of E = + √ p2c2 +m2c4 (IV.48) The factors χ1, χ2, χ ′ 1, χ ′ 2 are kinematic factors. In the limit of v  c these factors are zero. The two other solutions, u(3) =  φ1φ21 0  ; u(4) =  φ ′ 1 φ′2 0 1  (IV.49) correspond to negative energy, E = − √ p2c2 +m2c4 with φ1, φ2, φ ′ 1, φ ′ 2 vc−−−→ 0. These negative solutions are difficult to interpret. Dirac’s interpretation is shown in figure 15. Dirac says that the negative energy levels are all occupied by negative energy electrons. Since electrons are fermions they cannot all pile in these negative energy states. These particles make up the vacuum. Thus we use a photon with high enough energy we can excited these particles and create a hole in the Dirac sea. Dirac inferred that these holes are observable as antiparticles called the positrons. These are holes in the negative energy spectrum and hence have positive energy. This interpretation says that the vacuum is a a fully occupied Dirac sea with no electrons with energy E > mc2. 50 (i) δj ≥ 0 for Zα ≤ 1 (ii) En,j 6= En,j′ this accounts for fine-structure. (iii) One can expand En,j in powers of the small parameters (Zα) 2 . This gives En,j = mc 2 ( 1− (Zα) 2 2n2 − (Zα) 4 2n3 ( 1 j + 12 − 3 4n )) (IV.62) The first term (0th order) is the rest energy mc2. The second term (1st order) we have the non relativistic binding energy since mc2α2 = ~ ma20 (IV.63) and E(2)n = − ~ 2ma20 Z2 n2 (IV.64) The third term (2nd order) gives a relavistic correction which is j dependent (fine structure). The fine structure is shown in figure ??: Continuum E=mc2 n=1 1s 1.8 x 10^-4 eV1s1/2 n=2 DiracSchrodinger 2s,2p 2s1/2 2p3/2 , 2p1/2 10^-4 eV n=3 3s,3p,3d 3s1/2 3p3/2 , 2p1/2 , 3d1/2 , 3d 3/2 3d5/2 Notice that in Schrodinger’s theory our reference frame was zero. On the other hand with Dirac we have relativity and we use E = mc2 as our new reference frame. Further corrections of bound state hydrogen problem: (i) Hyperfine structure coupling of magnetic moments of the electron to the nuclear magnetic moment. e.g. Continuum E=mc2 n=1 1s1/2 'trilet' 'singlet' 6 x 10 eV-6 (ii) QED effects (the Lamb shift) e.g. 51 Continuum E=mc2 n=2 2s1/2 4.4 x 10 eV-6,2p1/2 1/22p 2s 1/2 5. Nonrelativistic limit of the Dirac equations The starting point is the stationary Dirac equation:( cα · p + βmc2 + V (r) ) Φ = EΦ(r) (IV.65) we write the 4-spinor as Φ = ( φ χ ) =  φ1φ2χ1 χ2  (IV.66) We can write αj = ( 0 σj σj 0 ) with σj as the Pauli matrices. Further we have β = ( 1 0 0 −1 ) . Inserting into the Dirac equation gives c ( 0 σ σ 0 ) · p ( φ χ ) = ( E − V (r)−mc2 ( 1 0 0 −1 )) Φ(r) (IV.67) This gives two coupled matrices for φ and χ: cσ · pχ = ( E − V (r)−mc2 ) φ cσ · pφ = ( E − V (r) +mc2 ) χ Isolating the second equation for χ gives χ = c E − V (r) +mc2 σ · pφ (IV.68) Inserting this equation into the top equation gives σ · p { c2 E − V (r) +mc2 σ · p } φ = ( E − V (r)−mc2 ) φ (IV.69) Lecture 24th, 2012 Isolating for φ we have φ = 1 (E − V (r)−mc2) σ · p { c2 E − V (r) +mc2 σ · p } φ (IV.70) We now make a weak relativistic approximation. We define ε = E −mc2  mc2 Further we assume V (r) mc2 52 We now expand consider the term in the left hand side of equation IV.69:{ c2 ε+ 2mc2 − V (r) } = 1 2m ( 1 + ε−V2mc2 ) ≈ 1 2m ( 1− ε− V 2mc2 ) (IV.71) This gives 1 2m [ σ · p ( 1− ε− V 2mc2 ) σ · p ] φ = (ε− V )φ (IV.72) 1 2m [( 1− ε− V 2mc2 ) (σ · p)2 + ~ i σ · ∇V 2mc2 σ · p ] φ = (ε− V )φ (IV.73) Now for Pauli matrices we can write (σ ·A) (σ ·B) = ∇ (A ·B) + iσ · (A×B) (IV.74) In our case we have A×B = p×p = 0. Further we have that the gradient of a function is (given no radial dependence ∇V = 1 r dV dr dr (IV.75) Inserting in these relations we have 1 2m [( 1− ε− V 2mc2 ) p2 + ~ i 1 r dV dr (σ · rσ · p 2mc2 )] φ = (ε− V )φ (IV.76) p 2m [( 1− ε− V 2mc2 ) + ~ i 1 4m2c2r dV dr (r · p) + ~ 4m2c2r dV dr σ · L ] φ = (ε− V )φ (IV.77) where we have used equation IV.74 We define these terms T1, T2 and T3 respectively. Interpretation of Terms As preparation consider a term acting on a non-relativistic wavefunction ( ε−mc2 − V (r) ) ψ = p2 2m ψ This is not the term we have. In our case we have T1φ = ( 1− ε− V 2mc2 ) p2 2m ≈ p 2 2m − 1 2mc2 ( p2 2m )2 (IV.78) = p2 2m − p 4 8m3c2 (IV.79) Here we see the relativistic correction of kinetic energy. The second term T2 is not Hermitian. However it can easily be made Hermitian by considering Hermitian average T̄2 ≡ 1 2 ( T2 + T † 2 ) (IV.80) = 1 2 ~i 1r dVdr r · p− ~i 1r dVdr r︸ ︷︷ ︸ F(r) ·p  † (IV.81) Now (F · p)† = p†F†(r) (IV.82) = p† · F† (IV.83) = p · F (IV.84)