Physics and Chemistry of the Interstellar Medium

0. Physics and Chemistry of the Interstellar Medium Pierre Hily-Blant Master2 Lecture, LAOG pierre.hilyblant@obs.ujf-grenoble.fr, Office 53 2012-2013 Pierre Hily-Blant (Master2) The ISM 2012-2013 1 / 220

0. I- Production of light in the ISM 1. Outline 2. Structure of Hydrogenic Atoms 3. Atoms with several electrons Pierre Hily-Blant (Master2) The ISM 2012-2013 29 / 220

1.Outline 1. Outline Pierre Hily-Blant (Master2) The ISM 2012-2013 30 / 220

Motivation 1.Outline Classical theory of radiation is unable to treat interactions between matter and radiation which involve one/few photons Some processes in the ISM require to take into account the quantum nature of atoms: e.g. HI-21cm line Light/matter interaction: e.g. photoelectric effect Review of hydrogen-like atoms Fundamentals of molecular physics Introduction to Spectroscopy for ISM astronomers Pierre Hily-Blant (Master2) The ISM 2012-2013 31 / 220

2.Structure of Hydrogenic Atoms 2. Structure of Hydrogenic Atoms Quantum Mechanical Formulation General Solution Spin Complete wave function Hydrogenic atoms Electronic configurations High-order effects Fine structure Hyperfine structure Pierre Hily-Blant (Master2) The ISM 2012-2013 32 / 220

2.Structure of Hydrogenic Atoms Quantum Mechanical Formulation Quantum mechanical formulation Time-dependent Schrödinger equation: i Ψ t = HΨ Stationary solutions: separating time and space if H is independent of time Ψ( r, t) = ψ( r)e iet/ ψ is solution of the time-independent Schrödinger equation Hψ = Eψ Pierre Hily-Blant (Master2) The ISM 2012-2013 33 / 220

2.Structure of Hydrogenic Atoms Quantum Mechanical Formulation In the absence of external fields, the non-relativistic Hamiltonian of an H atom consisting of an electron and an proton is H = T + V = ˆp2 2 m + V = 2 2 m 2 + V momentum operator: ˆp = i reduced mass of the e p system, m = mpme m p+m e m e central potential V (r) = ke2 r, k is a constant, here k = 1 4πɛ 0 Atomic state: spatial wavefunction ψ, solution of the time-independent Schrödinger equation Hψ = Eψ Pierre Hily-Blant (Master2) The ISM 2012-2013 34 / 220

2.Structure of Hydrogenic Atoms Quantum Mechanical Formulation Spherical coordinates (r, θ, φ): [ H = 2 2 2 m r 2 + 2 r r + 1 r 2 sin θ Hence, decompose H into a radial part: and an angular part: [ 1 L 2 = sin θ ( sin θ ) + θ θ R = r 2 r 2 r + 2 m 2 r2 [E V (r)] ( sin θ ) + 1 θ θ sin 2 θ 1 r 2 sin 2 θ 2 ] φ 2 = L2 2 2 ] φ 2 + V (r) Finally: Rψ = L 2 ψ Pierre Hily-Blant (Master2) The ISM 2012-2013 35 / 220

2.Structure of Hydrogenic Atoms Quantum Mechanical Formulation Consequences: ψ are eigenfunctions of L 2 since r 2 R is ind. of θ and φ. ψ are eigenfunctions of r 2 R since L is ind. of r look for ψ in the form ψ(r, θ, φ) = 1 r f(r)g(θ, φ). First solve for the ground state, then search for the general solution General considerations: boundary conditions (BC) give rise to quantization there are as many quantum numbers as boundary conditions separation of variables 3 functions 3 BC 3 quantum numbers: n, l, m l Pierre Hily-Blant (Master2) The ISM 2012-2013 36 / 220

2.Structure of Hydrogenic Atoms The ground state: L 2 = 0 Quantum Mechanical Formulation V (r) spherically symmetric look for ψ f(r): ( d 2 dr 2 + 2 ) d f + 2 m ( ) ke 2 r dr 2 r + E f = 0 At large radii, r so to leading order (V 0 free particle) d 2 f dr 2 + 2 me 2 f = 0 For E < 0, f(r) Ce r/δ. Replace for f in the equation: 2 δ = mke 2 = 4πɛ 0 2 ( ) me 2 = 107 2 = a 0 = 5.29 10 11 m m ec E = 2 2 ma 2 = 1 0 2 α2 mc 2 = 13.6 ev = 1Ry e 2 α = 4πɛ 0 c 1/137 Pierre Hily-Blant (Master2) The ISM 2012-2013 37 / 220

2.Structure of Hydrogenic Atoms General Solution General Solution Eigenfunctions and eigenvalues of angular momentum: [L 2, L α ] = 0, α = x, y, z: e-f of L α are also e-f of L 2 L 2 ψ = l(l + 1)ψ, l = 0, 1/2, 1, 3/2,... L z ψ = m l ψ, m l = l, l + 1, l + 2,..., l 2, l 1, l g(θ, φ) are spherical harmonics noted Y l,ml (θ, φ) Y l,ml form an orthogonal set of functions L 2 Y l,ml = l 2 Y l,ml = l(l + 1) 2 Y l,ml L z Y l,ml = l z Y l,ml = m l Y l,ml Y l,m Y l,m = δ l,l δ m,m Pierre Hily-Blant (Master2) The ISM 2012-2013 38 / 220

2.Structure of Hydrogenic Atoms General Solution Eigenfunctions and eigenvalues of the radial part Rψ = L 2 ψ: [ ] Rψ = L 2 ψ d2 l(l+1) 2 em ke 2 dr 2 r 2 2 r f(r) = 2 em Ef(r) 2 BCs are f(0) = 0 and f( ) = 0 quantum number n Rewrite as R nl f nl = E n f nl General solution: Laguerre polynomials, noted P nl (r) Eigenvalues are: E n = 1 2 α2 mc 2 = Ry n 2, n N l can take n values: 0 l < n Degeneracy: l=0,n 1 (2l + 1) = n2 RP nl (r) = E n P nl (r) Pierre Hily-Blant (Master2) The ISM 2012-2013 39 / 220

2.Structure of Hydrogenic Atoms Spin Electron Spin: intrinsic oam Stern-Gerlach experiment (1921), silver atoms in spatially variable magnetic field: evidence of the intrinsic angular momentum associated to single valence electron, called spin Explained by Dirac theory (Relativistic effects) Generalization: the e possesses an intrinsic angular momentum, called spin, noted s (associated to operator S) Quantization: intrinsic angular momentum QN: s = 1/2 s 2 = s(s + 1) 2 = 3/4 2 s z = m s = ±1/2 Spin wave function: σ ms (s z ) = δ ms s z σ ms (s z ) σ m s (s z ) = +1/2 s σ z= 1/2 m (s s z)σ ms (s z ) = δ msm s S z σ ms (s z ) = m s σ ms (s z ) S 2 σ ms (s z ) = s(s + 1) 2 σ ms (s z ) Pierre Hily-Blant (Master2) The ISM 2012-2013 40 / 220

2.Structure of Hydrogenic Atoms Complete wave function Complete wave function Ψ(r, θ, φ, s z ) nlsm l m s nlsm l m s = 1 r P nl(r) Y lm (θ, φ) σ ms (s z ) HΨ = E n Ψ Total degeneracy: 2n 2 In terms of n, l, s, m l, m s : nlsm l m s = n lm l sm s L 2 lm l = l(l + 1) 2 lm l L z lm l = m l lm l S 2 sm s = s(s + 1) 2 sm s S z sm s = m s sm s In terms of the total angular momentum ( J = L + S): n, l, s, j, m j : Wave function jm = C(lsm l m s ; lsjm) lsm l m s m l m s J 2 jm j = j(j + 1) 2 jm j J z jm j = m j jm j C(lsm l m s ; lsjm): Clebsch-Gordon coefficients Pierre Hily-Blant (Master2) The ISM 2012-2013 41 / 220

2.Structure of Hydrogenic Atoms Complete wave function General solution for the H atom Ψ(r, θ, φ, s z ) = 1 r P nl(r)y lml (θ, φ)σ ms (s z ) Principal QN, n: energy level quantization, 2n 2 degeneracy E n = α2 mc 2 2n 2 = 13.6 n 2 ev Quantization of the Orbital Angular Momentum (oam): QN l < n l = l(l + 1) Quantization of the z-projection of oam: QN m l l (note: l z < l) l z = m l Pierre Hily-Blant (Master2) The ISM 2012-2013 42 / 220

2.Structure of Hydrogenic Atoms Complete wave function Quantization of the oam Pierre Hily-Blant (Master2) The ISM 2012-2013 43 / 220

2.Structure of Hydrogenic Atoms Hydrogenic atoms Hydrogenic atoms Generalization of the H-atom solutions to atoms with one electron, e.g. He +, Li ++, Be 3+,... Effective nuclear charge of the nucleus is Z > 1: potential V (r) = e2 Ze2 r r Correction to the Rydberg constant, due to the mass of the nucleus for H-atom, Ry = em m e R with R = 1 2 α2 m e c 2 = 109677 cm 1 nucleus of mass m A = A uma and m A = m e m A /(m e + m A ) Wavefunctions are unchanged Energy levels: Z 2 E n = R A n 2 ev R A = m A m e R 109677 (1 m e /m A ) cm 1 Pierre Hily-Blant (Master2) The ISM 2012-2013 44 / 220

Notation 2.Structure of Hydrogenic Atoms Electronic configurations For H-like atoms, the quantum state depends only on nl Set of n, l values for all the electrons electronic configuration principal quantum number: n identifies shells (K, L, M...) n shell l subshell oam: coded with letters l 0 1 2 3 4 5 code s p d f g h Pierre Hily-Blant (Master2) The ISM 2012-2013 45 / 220

2.Structure of Hydrogenic Atoms High-order effects High-order effects High precision atomic measurements require corrections to the Hamiltonian H 0 Due to relativistic effects (Dirac s equation) and angular momenta coupling electron spin (intrinsic): s oam of e : l nuclear spin (intrinsic): i = combination {spins of nucleons} (1/2 for p and n) Only total AM is conserved and its eigenvalue is good a quantum number: need to sum up all the AM How? In general, the strongest interactions are considered first. Each coupling leads to one AM. To the leading order (i.e. neglecting the weaker interactions). Zeeman effect discussed later (but not Stark effect and Lamb shift) Pierre Hily-Blant (Master2) The ISM 2012-2013 46 / 220

2.Structure of Hydrogenic Atoms High-order effects Angular momenta coupling Neglect the nuclear spin: the interaction of L with S leads to the fine structure. The total AM is j = l + s Take into account the total nuclear spin I: the interaction of total j with I leads to the hyperfine structure. The total AM is F = I + j In astrophysics, there is an intensive use of fine and hyperfine structure of atoms and molecules. Pierre Hily-Blant (Master2) The ISM 2012-2013 47 / 220

2.Structure of Hydrogenic Atoms High-order effects Addition of angular momenta (I) Note J 1 and J 2 2 a-m operators. Assume [J 1,α, J 2,β ] = 0, α = x, y, z, β = x, y, z Note ψ 1 (q 1 ) and ψ 2 (q 2 ) e-f of J 1 and J 2 resp. Therefore: J 2 k ψ k = j k (j k + 1) 2 ψ k and J k,z ψ k = m k ψ k Consider J = J 1 + J 2 : it is also an a-m operator (check the commutation properties). There must exist e-f ψ of both J 2 and J z, with e-v j(j + 1) 2 and m. Questions: How are j and m related to j 1, m 1, j 2, and m 2? How are ψ related to ψ 1 and ψ 2 (very difficult question)? Pierre Hily-Blant (Master2) The ISM 2012-2013 48 / 220

2.Structure of Hydrogenic Atoms High-order effects Answer to the first question: the eigenvalues J 2 ψ = j(j + 1) 2 ψ and J z ψ = m j ψ j = j 1 + j 2, j 1 + j 2 1,..., j 1 j 2 1, j 1 j 2 (min(2j 1 + 1, 2j 2 + 1)) values) For each value of j, 2j + 1 values of m = j, j + 1,..., j 1, j Restriction: m = m 1 + m 2 Total degeneracy: (2j 1 + 1)(2j 2 + 1) Pierre Hily-Blant (Master2) The ISM 2012-2013 49 / 220

2.Structure of Hydrogenic Atoms Fine structure: j= l+ s High-order effects Consider an electron orbiting a proton. e with oam l and spin s orbits the nucleus In the frame of the moving e, nucleus = current i creating a B-field interaction energy between B and intrinsic e magnetic dipole µ s Pierre Hily-Blant (Master2) The ISM 2012-2013 50 / 220

2.Structure of Hydrogenic Atoms High-order effects The e feels the electrostatic potential V (r) created by the central positive charge. Intrinsic magnetic momentum of the e (Stern & Gerlach 1921): µ e = g e µ B s = 9.3( 24) J/T, µ B = e /2m e, g e 2 In the electron rest-frame, the electric field associated to the varying potential V (r) is moving and is equivalent to an internal magnetic fields B int = v E/c 2 Magnetic interaction energy: E fs = µ e B int = C s l, C > 0 The interaction energy (sign and amplitude) depends on l s: s-o interaction breaks the level degeneracy. Energy depends on m l and m s through l s Note: actually, the fine structure of H-like atoms is due to 3 terms including the l-s coupling: relativstic correction to the kinetic energy and Darwin s term). Pierre Hily-Blant (Master2) The ISM 2012-2013 51 / 220

2.Structure of Hydrogenic Atoms Orders of magnitude High-order effects Perturbed hamiltonian: H = H 0 + H so, H so H 0 Perturbation theory: Energy shift due to s-o is E fs = H so = ψ 0 H so ψ 0 E n α 2 Z 2 nl(2l+1) Energy difference between two FS levels (j = l ± 1/2): α 2 Z 2 E fs = E n nl(l + 1) = R α 2 Z 4 A n 3 l(l + 1) FS α 2 Z 2 E n 10 3 10 5 E n Internal magnetic fields: B E/µ e 0.1 10 T Sodium doublet (KL3s): two transitions 3p 3s, λ = 0.6 nm, E fs = 2.1( 3) ev Pierre Hily-Blant (Master2) The ISM 2012-2013 52 / 220

2.Structure of Hydrogenic Atoms High-order effects Application of angular momenta coupling j = j 1 + j 2, j 1 + j 2 1,..., j 1 j 2 1, j 1 j 2 For each value of j, 2j + 1 values of m = j, j + 1,..., j 1, j Restriction: m = m 1 + m 2 Total degeneracy: (2j 1 + 1)(2j 2 + 1) Application to the H-atom j 1 = l, j 2 = s = 1/2: j = l + 1/2, l 1/2 m 1 = m l, m 2 = m s : m = m l + m s total am for l > 0 has e-v j = l ± 1/2 and m = m l + m s Pierre Hily-Blant (Master2) The ISM 2012-2013 53 / 220

2.Structure of Hydrogenic Atoms High-order effects Hyperfine structure: f= i+ j Nucleus possesses electromagnetic multipole moments So far, only electric monopole (charge) Take first correction: magnetic dipole µ i associated with the nucleus spin i, such that µ i = g p µ N i g N = Landé factor, g p = 5.5883 µ N = e = µ B = 5.05( 27)J/T nuclear magneton 2m p 1836 Two contributions to the perturbative Hamiltonian: E hfs µ i with B int i l: µ i B int = g N µ N i B int = C i l, C > 0 µ e -µ i interaction s i: non-zero even for l = 0 Nuclear magnetic moment: µ N µ e /1836 E hfs 10 3 E fs Pierre Hily-Blant (Master2) The ISM 2012-2013 54 / 220

2.Structure of Hydrogenic Atoms High-order effects Application to the ISM: the 21-cm line Ground state of hydrogen atom has n = 1, l = 0 only magnetic dipoles interaction s = 1/2, i = 1/2 f = s + i = 0 1, 1 3 E s,i = 4 µ 0 g eµ B g pµ N 3 2π s i a 3 0 s i = 1/2[f(f + 1) s(s + 1) i(i + 1)] = +1/4, 3/4 E hfs = 9.43( 25) [1/4 ( 3/4)] J = 0.04738 cm 1 = 5.9µeV ν = 1421MHz, and A ij = 2.87 10 15 s 1 = ( 10 7 y) 1 H atoms emit when the e and p flip their relative spins from parallel to antiparallel Frequency is (still) protected against commercial uses Pierre Hily-Blant (Master2) The ISM 2012-2013 55 / 220

2.Structure of Hydrogenic Atoms High-order effects High-order effects Fine Structure Splitting non-zero electronic oam l (spin s = ±1/2): j = l + s energy splitting: E fs = α2 Z 4 R A n 3 l(l + 1) = E α 2 Z 2 n nl(l + 1) α2 E n 10 4 E n Hyperfine Structure Splitting nuclear spin i and total electronic AM j: f = j + i energy splitting: E hfs E fs 1840 10 7 E n Pierre Hily-Blant (Master2) The ISM 2012-2013 56 / 220

The HI sky 2.Structure of Hydrogenic Atoms High-order effects Mebold et al 1997 Pierre Hily-Blant (Master2) The ISM 2012-2013 57 / 220

The HI sky 2.Structure of Hydrogenic Atoms High-order effects Kalberla et al 2005: HI λ21cm integrated intensity in [ 400 : 400] km s 1 LSR. Pierre Hily-Blant (Master2) The ISM 2012-2013 58 / 220

2.Structure of Hydrogenic Atoms High-order effects The HI sky Top: 400 : 100 km s 1, Bottom: 100 : 400 km s 1 Pierre Hily-Blant (Master2) The ISM 2012-2013 59 / 220

3.Atoms with several electrons 3. Atoms with several electrons Independent particle approximation One Valence Electron More Than One Valence Electron High-order effects in complex atoms Summary Application to the ISM Pierre Hily-Blant (Master2) The ISM 2012-2013 60 / 220

3.Atoms with several electrons Independent particle approximation Independent particle approximation Wave equation can not be solved exactly Approximation: each e moves in spherically symmetrical field produced by {nucleus+other e } 1 self-consistent field approximation: V = V nucl + V elec = Z eff e 2 4πɛ 0r, 2 central field approximation: V t V (r) i.e. spherical symmetry Solve for the wave function of each e, defined by a set of 4 QN (similar to H atom) The potentiel is bracketed by the two asymptotic expressions: r Z eff Z (N 1) V (r) (Z N+1)e2 4πɛ 0 r r 0 Z eff Z V (r) Ze2 4πɛ 0 r Hydrogenic atoms: Z = N so that V (r) = e2 4πɛ 0 r Pierre Hily-Blant (Master2) The ISM 2012-2013 61 / 220

3.Atoms with several electrons Independent particle approximation ith electron state is described by φ i (called orbital), solution of ) ( 2 2 m 2 + V i (r) φ i ( r i ) = E i φ i ( r i ) Difference with H-atom: E i = E(n i, l i ) noted E ni l i The energy of the global system (in the central field approximation) is thus E c = i E ni l i The energy level E ni l i of each individual electron is thus 2(2l i + 1) times degenerate (instead of 2n 2 for the H atom) Complete wavefunction = spin-orbital: ψ i = φ i σ i Pauli exclusion principle: no 2 e can have the same QNs antisymmetric global wavefunction (e.g. Slater determinant) Pierre Hily-Blant (Master2) The ISM 2012-2013 62 / 220

3.Atoms with several electrons One Valence Electron Atoms with One Valence Electron The valence e moves in the central potential created by nucleus and inner-e Energy depends strongly on n and weakly on l (penetration of orbitals) Alkali-type atoms (Li, Na, K...) and ionized alkali-like atoms (Be +, B ++, C +++,...) Generalization of results on H atom by introducing the effective nuclear charge: Z eff = 1 for alkali-type atoms, Z eff = 2, 3, 4,... for ions. Nucleus mass is A in uma. Rydberg constant: R A = em A m e Energy levels: E n = R A Z 2 eff n 2 R Fine-structure splitting: E fs = R A α 2 Z 4 eff n 3 l(l+1) Pierre Hily-Blant (Master2) The ISM 2012-2013 63 / 220

3.Atoms with several electrons More Than One Valence Electron More Than One Valence Electron Each e state characterized by n, l, m l and m s (s = 1/2): φ φ nlml m s Difference with H atom: energy depends on n, l and also on spin Corrections to energy levels are of three types: 1 orbit-orbit (Coulomb repulsion between e ) o-o coupling 2 spin-orbit (Magnetic interaction) s-o coupling 3 spin-spin (Pauli exclusion principle) s-s coupling Different orders of coupling according to relative magnitudes of the interactions Usually, s-s s-o, o-o, but s-o wrt o-o? Composition of oam only (usually) for non-filled shell e Pierre Hily-Blant (Master2) The ISM 2012-2013 64 / 220

3.Atoms with several electrons Coupling schemes More Than One Valence Electron J = N i=1 ( l i + s i ) is conserved Question: in wich order shall we couple am s? Answer: the strongest interactions are coupled first. L S coupling (Russel-Saunders) o-o: usually (e.g. in light atoms), e e Coulomb interaction in the atom dominate, and do not affect the spins add the l i to form L add the s i to form S add L and S to form J = L + S j j coupling s-o: in other instances (large atoms), L S not the best scheme construct a set of j i = l i + s i add the j i s to construct the total J = i j i Pierre Hily-Blant (Master2) The ISM 2012-2013 65 / 220

L-S Coupling 3.Atoms with several electrons More Than One Valence Electron Splitting of npn p in L S coupling. Spectroscopic Notation: 2S+1 L J Pierre Hily-Blant (Master2) The ISM 2012-2013 66 / 220

3.Atoms with several electrons High-order effects High-order effects in complex atoms Fine Structure Splitting: L S L 0, J = L + S α 2 Zeff 2 E fs = E n nl(l + 1) α2 Zeff 2 E n Hyperfine Structure Splitting: I J I 0, F = J + I E hfs E fs 1840 Pierre Hily-Blant (Master2) The ISM 2012-2013 67 / 220

3.Atoms with several electrons Electronic configurations High-order effects in complex atoms Vocabulary: n: shells l: sub-shell electronic configuration: Distribution of the e in the atom for a given state n, l to a given configuration, several states spectroscopic notation Question: how to fill (sub-)shells? Answer: in increasing energy as a function of QN n, l, m l, m s Influence of oam and spins: Hund s rules Pierre Hily-Blant (Master2) The ISM 2012-2013 68 / 220

3.Atoms with several electrons High-order effects in complex atoms Observable QN Symbol Energy level E n n number Orbital am L l letter: s, p, d, f, g,... Total am J j subscript number Notation: nl S Ground state of H: n = 1, l = 0, j = 1/2; 1s 1/2 electron in p-state H: n = 2, l = 1, j = 1/2, 3/2; 2p 1/2 and 2p 3/2 For a given n, 2n 2 e For a given n, l: 2l + 1 values for m l, 2 values for m s 2(2l + 1) distinct states Pauli exclusion principle 2(2l + 1) e s 2, p 6, d 10, f 14 Pierre Hily-Blant (Master2) The ISM 2012-2013 69 / 220

3.Atoms with several electrons High-order effects in complex atoms Worked-out example: Carbon 6 e : 1s 2 2s 2 2p 2. K and L shells full, only L shell is of interest l 1 = l 2 = 1 so L = 0, 1, 2; s 1 = s 2 = 1/2 so S = 0, 1. L-S coupling: Pauli exclusion principle implies (admitted here) L + S even singulet states S = 0: J = L = 0 or 2 (Pauli): and only two singlet states: 1 S 0, 1 D 2. 1 P 1 is excluded. triplet states with S = 1: L = 1 only (Pauli), thus J = 0, 1, 2 and we have three triplet states 3 P 0,1,2. Finally: states for the ground state of Carbon are 3 P 0,1,2, 1 S 0, and 1 D 2. Pierre Hily-Blant (Master2) The ISM 2012-2013 70 / 220

3.Atoms with several electrons Spectroscopic notation High-order effects in complex atoms Element Electronic config. Spectro. notation L-S coupling H 1s 2 S 1/2 He 1s 2 1 S 0, 3 S 1 Li 1s 2 2s 2 S 1/2 C 1s 2 2s 2 2p 2 3 P 0,1,2, 1 S 0, 1 D 2 N 1s 2 2s 2 2p 3 4 S 3/2,... O 1s 2 2s 2 2p 4 3 P 2... Si 1s 2 2s 2 2p 6 3s 2 3p 2 3 P 0... In the nlsjm representation: 2S+1 L J nlsjm L = i l i, J = L + S L = 0, 1, 2, 3, 4...: S, P, D, F, G... ; J = L ± 1/2 Exercise: Two e with l 1 = 2 and l 2 = 1 (i.e. pd configuration). Determine the 15 coupled orbital-angular-momentum states. Hint: a coupled state corresponding to m 1 = 2 and m 2 = 1 has m = 1 which may be associated to any of l = 1, 2, and 3. Similarly, give the list of the m 1, m 2 couples which correspond to any possible value of m = 3, 2, 1, 0, 1, 2, 3. The total should be 15... Pierre Hily-Blant (Master2) The ISM 2012-2013 71 / 220

3.Atoms with several electrons High-order effects in complex atoms Pierre Hily-Blant (Master2) The ISM 2012-2013 72 / 220

3.Atoms with several electrons High-order effects in complex atoms Addition of angular momenta (II) Recall Note J 1 and J 2 2 a-m operators. Assume [J 1,α, J 2,β ] = 0, α = x, y, z, β = x, y, z Note ψ 1 (q 1 ) and ψ 2 (q 2 ) e-f of J 1 and J 2 resp. Answer to the second question: the eigenfunctions Uncoupled e-f: j 1 m 1 j 2 m 2 = ψ j1 m 1 j 2 m 2 (q 1, q 2 ) ψ j1,m 1 (q 1 )ψ j2,m 2 (q 2 ) Coupled e-f: j 1 j 2 jm ψ j1 j 2 jm(q 1, q 2 ) Clebsh-Gordon coef.: j 1 j 2 jm = m 1 m 2 C(j 1 j 2 m 1 m 2 ; jm) j 1 m 1 j 2 m 2 Question: C =? Pierre Hily-Blant (Master2) The ISM 2012-2013 73 / 220

3.Atoms with several electrons Physical insight High-order effects in complex atoms J1, J 2, and J = J 1 + J 2 General, internal torques between 1 and 2: J 1 and J 2 not conserved If no external torque, J is conserved J 1 and J 2 precess about J and J > J z z-components of J 1 and J 2 vary: m 1 and m 2 are not good quantum numbers, but m = m 1 + m 2 is because J z is constant. Pierre Hily-Blant (Master2) The ISM 2012-2013 74 / 220

3.Atoms with several electrons High-order effects in complex atoms Weak 1 2 interaction: J 1 and J 2 precess slowly about J magnitude still well-defined (not their direction) j 1, j 2, j, m are good QN. Quantum state = combination of j 1 m 1 j 2 m 2 or j 1 j 2 jm (with prescriptions on m 1 and m 2 ) Strong 1 2 interaction: precession is fast, magnitude and direction are lost only j, m are good QN. Quantum state = mixture of all j 1 j 2 jm with correct j 1 and j 2. Presence of an external torque: e.g. magnetic fields along z-axis J precesses around z axis If strong fields, precession is fast: magnitude of J is lost Only good quantum number: m Quantum state = combination of all j 1 j 2 jm with correct j 1, j 2, and j m. Pierre Hily-Blant (Master2) The ISM 2012-2013 75 / 220

Summary (1) 3.Atoms with several electrons Summary Cowan 1981 Pierre Hily-Blant (Master2) The ISM 2012-2013 76 / 220

Summary (2) 3.Atoms with several electrons Summary Hydrogenic Complex Value Energy levels R A Z 2 /n 2 N i=1 E n i,l i ev Degeneracy 2n 2 2j + 1 States nlm l m s nlsjm J Notation nl 2S+1 L (p) J Fine structure j = l + s J = L + S Efs α 2 Z 2 E n Hyperfine structure f = i + j F = I + J Ehfs E fs /m p Pierre Hily-Blant (Master2) The ISM 2012-2013 77 / 220

Summary (3) 3.Atoms with several electrons Summary Quantity Symbol Expression Value Unit FS constant α e 2 4πɛ 0 c 1/137 α 2 5 10 5 Bohr radius a 0 αm ec 5.29 10 11 m 1 Rydberg R 2 α2 m e c 2 109677 cm 1 1 Ry 2 α2 m H c 2 13.6 ev 1 R A 2 α2 m A c 2 electronic levels E n R AZeff 2 n 2 10 100 ev FS-splitting E fs R Aα 2 Zeff 4 n 3 l(l+1) 10 3 10 1 ev HFS-splitting E hfs E fs 1840 10 7 10 4 ev Pierre Hily-Blant (Master2) The ISM 2012-2013 78 / 220

3.Atoms with several electrons Application to the ISM Application to the ISM: atomic FS 1 D2 1 D2 6392 9850 O 0 6363 6300 A o C 0 9823 9809 A o C + 3 P 0 3 P 1 146 µm 63 µm 3 P 2 3 P 0 3 P 1 3 P 2 370 µm 609 µm 2 P 3/2 2 P 1/2 158 µm 1s 2 2s 2 2p 4 1s 2 2s 2 2p 2 1s 2 2s 2 2p 1 3 P 0,1,2 3 P 0,1,2 2 P 1/2,3/2 Magnetic dipole transitions: S = L = 0, J = ±1 Pierre Hily-Blant (Master2) The ISM 2012-2013 79 / 220

3.Atoms with several electrons Application to the ISM FS lines of CII and NII by COBE Pierre Hily-Blant (Master2) The ISM 2012-2013 80 / 220