Atomic Spectra in Astrophysics
Lida Oskinova, Helge Todt
Astrophysik Institut für Physik und Astronomie Universität Potsdam
WiSe 2016/2017
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 1 / 142 The Hydrogen Atom
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 2 / 142 Contents
importance of hydrogen, origin the hydrogen spectrum (brief) history of atom models quantum mechanics and solution of the central-force problem
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 3 / 142 Hydrogen
discovered 1766 by Cavendish (metal + acid), and found as constituent of water by de Lavoisir (1787) → hydrogen = generator of water simplest atom: proton & electron - mass 1.6738 × 10−27 kg Eion ≈ 13.6 eV
isotopes: deuterium (1 neutron) and tritium (2 neutrons) origin: Big Bang; deuterium from primordial nucleosynthesis (1 min after BB at 60 MK ˆ= 80 keV); recombination at 378 000 yr (z = 1100) → transparent universe fuel for stars (fusion) via proton-proton chain reaction or CNO cycle
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 4 / 142 The hydrogen spectrumI
Spectrum of a Balmer lamp:
→ low pressure gas-discharge tube (H. Geißler 1857) filled with hydrogen Ångström (1862): spectral lines of hydrogen in spectrum of sun
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 5 / 142 The hydrogen spectrumII
Balmer (1885): spectral lines of hydrogen given by
hm2 λ = (n = 2, m = 3, 4, 5,...) (1) m2 − n2
with h = 3645.6 × 10−10 m and 10−10 m = 1 Å, typical size of an atom predicted lines for m > were found in A stars Rydberg (1888): generalization to other series 1 1 1 7 −1 = RH 2 − 2 , RH = 1.096775854 × 10 m (2) λ n1 n2
generalization to H-like ions (e.g. He II, Li III): 1 2 1 1 = Z RX 2 − 2 (3) λ n1 n2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 6 / 142 Atom modelsI
Parmenides (500 v. Chr.): atoms (indivisible) as building blocks of the world
J. Dalton (1803): chemical elements consist of atoms of different mass
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 7 / 142 Atom modelsII
J.J. Thomson (1900): atoms contain negatively charged electrons in a positively charged continuum (cathode rays experiments)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 8 / 142 Atom modelsIII
Rutherford (1910): positively charged nucleus is smaller (10−15 m) than atom (scattering of helium nuclei on gold foil)
Problems of Rutherford model Why do electrons not fall into nucleus? → circular orbit Why don’t they emit like electric dipole, what about spectral lines? → Bohr model
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 9 / 142 Bohr modelI
Bohr’s postulates (1913) to explain observations: Motion of electrons in atoms obeys quantum rules 1 Electrons orbit in atoms the nucleus on so-called stationary orbits with discrete energies En. (The angular momentum is restricted to integer multiples of a fixed unit L = n~ (n = 1, 2, 3,...).) 2 Atoms can only gain or lose energy by the transition of an electron from one stationary orbit to another stationary orbit, this energy is discrete and given by ∆E = En − En+1 = hν for the involved photon of frequency ν.
Postulate (1) can be also written as phase space integral 1 I pdq = n (n = 1, 2, 3,...) (4) 2π ~
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 10 / 142 Bohr modelII Application to the hydrogen atom: Electron orbits in a Coulomb potential (polar coordinates) with Ze2 F = − 2 (5) 4πε0rn with stationary radius rn, balanced by centrifugal force 2 vn 2 Fz = mr = mrrnωn (6) rn with reduced mass mr so that 0 is in nucleus with mass mc
m2 me mr = = (7) 1 + m2 1 + me m1 mc Force balance yields
2 Ze 2 2 = mrrnωn (8) 4πε0rn
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 11 / 142 Bohr modelIII
As vn = rnωn the momentum and position are
pn = mrvn = mrrnωn, qn = rnφn (9) I Z 2π 1 1 2 2 pndqn = mrrn ωn dφ = mrrn ωn = n~ (10) 2π 2π 0 n~ → ωn = 2 (n = 1, 2, 3,...) (11) mrrn
Inserting force balance to elimnate ωn:
2 2 4πn ~ ε0 rn = 2 (n = 1, 2, 3,... ) (12) Zmre Z 2m e4 ω = r (13) n 2 3 3 2 16π n ~ ε0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 12 / 142 Bohr modelIV
This can be used to obtain the kinetic energy of the electron
m m Z 2m e4 E = r v 2 = r r 2ω2 = r (14) kin n n n 2 2 2 2 2 2 32π n ~ ε0
Its potential energy is given by qV (r) = − R Fdr:
Ze2 Z 2m e4 E = − = − r (15) pot 2 2 2 2 4πε0rn 16π n ~ ε0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 13 / 142 Bohr modelV both forces are conservative forces, → conservation of energy and the energy levels are therefore
Z 2m e4 E = E + E = − r (16) n kin pot 2 2 2 2 32π n ~ ε0 2 4 Z me e = − (n = 1, 2, 3,...) (17) 32π2n2 2ε2 1 + me ~ 0 mc from 2nd Bohr’s postulate (∆E = hν) and by ν = c/λ:
1 Z 2m e4 1 1 = e − (18) m 2 2 λ 32π2cε2 1 + e n1 n2 0 mc
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 14 / 142 Bohr modelVI
94 nm Lyman series 95 nm
97 nm
103 nm
122 nm 656 nm486 nm 434 nm n = 1 410 nm Balmer series 1875 nm n = 2 1282 nm
n = 3 1094 nm
n = 4 n = 5 n = 6 −1 1 1 λ = 911.8 Å 2 − 2 n1 n2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 15 / 142 Bohr modelVII
2 2 2 comparison to Rydberg’s formula 1/λ = Z RX 1/n1 − 1/n2 for hydrogen-like atoms suggests:
R R = ∞ (19) X 1 + me mc m e4 R = e = 1.097373177 × 107m−1 (20) ∞ 2 2 32π cε0
Application: Pickering lines of He II → blackboard note: R∞ sometimes used for atomic energies
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 16 / 142 Bohr modelVIII
Problems and limits of Bohr model Experiments: H-like spectra of alkali metals need more than one quantum number for discription (e.g. Na D line ≈ 5890 Å transition 3s − 3p) no explanation of other spectra Theory: Heisenberg’s uncertainty principle contradicts electron orbits Street light using → blackboard sodium lamp. Orange light from Na D line
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 17 / 142 The Schrödinger equationI wave equation, used for explanation of hydrogen spectrum Schrödinger (1926)
∂ ı |ψ(t)i = Hˆ |ψ(t)i (21) ~∂t ∂ 2 e.g. ı ψ(~r, t) = − ~ ∆ + V (~r, t) ψ(~r, t) (22) ~∂t 2m
derived from de Broglie and dispersion relation: p = ~k and E = ~ω describes unperturbed evolution in time of non-relativistic quantum systems, linear PDE of 2nd order with complex solutions ψ linearity: superposition principle of solutions probability of presence (find particle at position x): |ψ(~r, t)|2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 18 / 142 The Schrödinger equationII Stationary Schrödinger Eq. (∂/∂t = 0 ), separation: ψ(~r, t) = ϕ(~r) f (t) → blackboard
Hˆ ϕ(~r) = Eϕ(~r) (23) −ı E t ψ(x, t) = ϕ(x) e ~ (24) with eigenvalues E (energy) of Hamilton operator: the probability density |ψ(x, t)|2 = |ψ(x)|2 (Why?) does not depend on t, same holds for expectation values of dynamic variables p2 as H(x, p) = 2m + V (x) is classical Hamiltonian ≡ total energy → hHi = E (Why?) general solution is a linear combination of separable solutions:
∞ X −ıEnt/~ ψ(x, t) = cnψn(x)e (25) n=1
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 19 / 142 The Central-force problemI
3D: Potential is of form V (~r) = V (|~r|) The Laplace operator in 3D for spherical coordinates:
1 ∂ ∂ 1 ∂ ∂ 1 ∂2 ∆ = r 2 + sin θ + (26) r 2 ∂r ∂r r 2 sin θ ∂θ ∂θ r 2 sin2 θ ∂φ2
Use separation: ϕ(r, θ, φ) = R(r) · Y (θ, φ), thus
2 Y ∂ ∂R 2 1 ∂ ∂Y 2 R ∂2Y − ~ r 2 − ~ R sin θ − ~ 2m r 2 ∂r ∂r 2m r 2 sin θ ∂θ ∂θ 2m r 2 sin2 θ ∂φ2 + V (r) RY = ERY (27) with eigenvalue E
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 20 / 142 The Angular EquationI
2 1 Multiply ·r , and RY , again: obtain another separation constant
2 1 ∂ ∂R 2 1 ∂ ∂Y − ~ r 2 + r 2V (r) − Er 2 = ~ sin θ (28) 2m R ∂r ∂r 2m Y sin θ ∂θ ∂θ 2 1 ∂2Y + ~ 2m Y sin2 θ ∂φ2 2 = ~ [−`(` + 1)] (29) 2m Next separation: Y (θ, φ) = Θ(θ) · Φ(φ)
Φ ∂ ∂Θ Θ ∂2Φ sin θ + = −`(` + 1) (30) ΘΦ sin θ ∂θ ∂θ ΘΦ sin2 θ ∂φ2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 21 / 142 The Angular EquationII term Φ/Φ and Θ/Θ cancels out, multiply sin2 θ
1 ∂ ∂Θ 1 ∂2Φ sin θ sin θ + `(` + 1) sin2 θ = − = m2 (31) Θ ∂θ ∂θ Φ ∂φ2 with new separation constant m:
∂2Φ + m2Φ = 0 ⇒ solution: Φ(φ) = e±ımφ (32) ∂φ2 as rotation around φ = 2π in space means same as original state:
Φ(φ + 2π) =! Φ(φ) ⇒ m = 0, ±1, ±2,... (33)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 22 / 142 The Angular EquationIII The other equation is for Θ
∂ ∂Θ sin θ sin θ + [`(` + 1) sin2 θ − m2] Θ = 0 (34) ∂θ ∂θ can be solved by so-called associated Legendre polynoms:
m Θ(θ) = P` (cos θ) (35) |m| m 2 |m| d = (−1) (1 − cos θ) 2 P (cos θ) (36) d(cos θ)|m| ` 1 d ` P (cos θ) = (cos2 θ − 1)` (37) ` 2``! d(cos θ)`
m Eq. (37) implies integer ` > 0. Eq. (36) says P` = 0 for |m| > ` and for ` are (2` + 1) values of m. Other solutions unphysical (e.g. not normalizable) → blackboard
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 23 / 142 The Angular EquationIV
R 2π R π 2 From normalization condition 0 0 |Y | sin θdθdφ follows Spherical harmonics
s (2` + 1) (` − |m|)! Y m(θ, φ) = eımφ Pm(cos θ) (38) ` 4π (` + |m|)! ` (−1)m for m ≥ 0 where = (39) 1 for m ≤ 0
0 0 E.g.: P0 = 1, P0 = 1 and P1 = cos θ, and therefore 0 1 1/2 0 3 1/2 ±1 3 1/2 ±ıφ Y0 = 4π and Y1 = 4π cos θ or Y1 = ∓ 8π sin θ e
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 24 / 142 Hydrogen atom - The Radial EquationI
The radial equation
d dR 2m r 2 r 2 − r [V (r) − E]R = `(` + 1)R (40) dr dr ~2
du 2 u dR r dr −u d 2 dR d u Substitution u(r) ≡ rR(r) → R = r , dr = r 2 , dr r dr = r dr 2 and so
2 2 2 ~ d u ~ `(` + 1) − 2 + V + 2 u = Eu (41) 2mr dr 2mr r → like 1D Schrödinger equation but with effective potential (with centrifugal term)
2 ~ `(` + 1) Veff = V + 2 (42) 2mr r
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 25 / 142 Hydrogen atom - The Radial EquationII Now for specific potential V (Coulomb potential)
e2 1 V (r) = − (43) 4π0 r so the radial equation becomes
2 2 2 2 ~ d u e 1 ~ `(` + 1) − 2 + − + 2 u = Eu (44) 2mr dr 4π0 r 2mr r √ Define κ = −2mrE and consider only bound states, where E < 0. Divide ~ Eq. (44) by E
2 2 1 d u mre 1 `(` + 1) 2 2 = 1 − 2 + 2 u (45) κ dr 2π0~ κ κr (κr)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 26 / 142 Hydrogen atom - The Radial EquationIII
2 mre Substitute ρ ≡ κr and ρ0 ≡ 2 : 2π0~ κ
d 2u ρ `(` + 1) = 1 − 0 + u (46) dρ2 ρ ρ2
d 2u for ρ → ∞ we obtain dρ2 = u and therefore
u(ρ) = Ae−ρ + Beρ (47)
For ρ → ∞ we get eρ → ∞, so we need B = 0 and hence u(ρ) ∼ Ae−ρ For ρ → 0 (and ` 6= 0):
d 2u `(` + 1) = u (48) dρ2 ρ2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 27 / 142 Hydrogen atom - The Radial EquationIV
Can be solved with
u(ρ) = Cρ`+1 + Dρ−` (49)
Again: For ρ → 0 the term ρ−` → ∞, thus D =! 0 and u(ρ) ∼ Cρ`+1 Let’s introduce v(ρ):
u(ρ) = ρ`+1 e−ρ v(ρ) (50)
Radial equation then reads
d 2v dv ρ + 2(` + 1 − ρ) + [ρ − 2(` + 1)]v = 0 (51) dρ2 dρ 0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 28 / 142 Hydrogen atom - The Radial EquationV
Assuming that v(ρ) can be written as a power series in ρ:
∞ X j v(ρ) = aj ρ (52) j=0 ∞ ∞ dv X j→ j+1 X → = ja ρj−1 =: (j + 1)a ρj (53) dρ j j+1 j=0 j=0 ∞ d 2v X → = j(j + 1)a ρj−1 (54) dρ2 j+1 j=0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 29 / 142 Hydrogen atom - The Radial EquationVI
So that radial equation now reads
∞ ∞ X j X j j(j + 1)aj+1ρ + 2(` + 1) (j + 1)aj+1ρ (55) j=0 j=0 ∞ ∞ X j X j −2 jaj ρ + [ρ0 − 2(` + 1)] aj ρ = 0 (56) j=0 j=0
The coefficients for any j must yield
j(j + 1)aj+1 + 2(` + 1)(j + 1)aj+1 − 2jaj + [ρ0 − 2(` + 1)]aj = 0 (57)
2(j + ` + 1) − ρ or a = 0 a (58) j+1 (j + 1)(j + 2` + 2) j
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 30 / 142 Hydrogen atom - The Radial EquationVII
Starting with a0 = A (A yet to be fixed) and check for large j:
2j 2 2j a ' a = a ⇒ a ' A (59) j+1 j(j + 1) j j + 1 j j j!
But if this were the exact solution:
∞ X 2j v(ρ) = A ρj = Ae2ρ ⇒ u(ρ) = Aρ`+1eρ (60) j! j=0
ρ where e → ∞ for ρ → ∞. So, series (60) must have maximum jmax, such that
ajmax+1 = 0 ⇒ 2(jmax + ` + 1) − ρ0 = 0 in Eq. (58) (61)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 31 / 142 Hydrogen atom - The Radial EquationVIII
Defining n ≡ jmax + ` + 1 → ρ0 = 2n, recalling that
2 √ mre −2mrE ρ0 = 2 and κ = (62) 2π0~ κ ~ 2κ2 mre4 → E = −~ = − (63) 2 2 2 2 2mr 8π 0~ ρ0
Bohr formula
2 m e 1 E1 En = − 2 2 = 2 n = 1, 2, 3,... (64) 2~ 4π0 n n → allowed energies of the Hydrogen atom
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 32 / 142 Hydrogen atom - The Radial EquationIX
Eq. (62) and ρ = 2n also yield Bohr radius:
2 mre 1 1 κ = 2 = (65) 4π0~ n an 2 4π0~ −10 where a = 2 = 0.529 × 10 m (66) mre r and ρ = (67) an Now, for normalization we remember that for the radial equation
u(r) ≡ rR(r) and our approach: (68) u(ρ) = ρ`+1e−ρv(ρ) (69) 1 → R = ρ`+1e−ρv(ρ) (70) n` r
Pjmax=n−`−1 j where v(ρ) = j=0 aj ρ
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 33 / 142 Hydrogen atom - The Radial EquationX Consider ground state, i.e. n = 1 Binding energy of hydrogen atom
" 2 2# mr e E1 = − 2 = −13.6 eV (71) 2~ 4π0
! So, 1 = n = jmax + ` + 1 → ` = 0 and therefore also m = 0 (Why?)
0 ϕ100(r, θ, φ) = R10Y0 (θ, φ) (72)
2(j+`+1)−ρ0 Our recursion formula aj+1 = (j+1)(j+2`+2) aj with j = 0 yields a1 = 0 (and 0 hence also for a2, a3,... → v(ρ) = a0ρ = const.
a R (r) = 0 e−r/a (73) 10 a L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 34 / 142 Hydrogen atom - The Radial EquationXI
R ∞ 2 2 ! Using Normalization condition 0 |R| r dr = 1 to determine a0:
Z ∞ 2 Z ∞ 2 2 |a0| −2r/a 2 2 a ! |R10| r dr = 2 e r dr = |a0| = 1 (74) 0 a 0 4 2 → a = √ (75) 0 a
With Y 0 = √1 : 0 4π
1 −r/a ϕ100(r, θ, φ) = √ e (76) πa3
independent of φ and θ → spherical symmetric with |ϕ|2 ∼ e−2r/a so-called s orbital
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 35 / 142 Hydrogen atom - The Radial EquationXII 2 For the first excited state n = 2 → E2 = E1/2 = −3.4 eV and
` = 0 ⇒ jmax = 1 jmax = n −` − 1 = (77) |{z} ` = 1 ⇒ jmax = 0 =2 → m = 0, +1, −1. Note: 4 different ϕn`m for one energy ( = eigenvalue of Hˆ ) → degeneracy
So for ` = 0 → Recursion formula gives a1 = −a0 (j = 0) and a2 = 0 (j = 1) → our polynomial v(ρ) = a0(1 − ρ):
a r ` = 0 R (r) = 0 1 − e−r/2a (78) 20 2a 2a a ` = 1 R (r) = 0 re−r/2a (79) 21 4a2 where a0 needs to be determined from normalization L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 36 / 142 Hydrogen atom - The Radial EquationXIII The number of different ϕ for any n, because of ` = 0, 1,..., n − 1 and for each ` exist (2` + 1) values of m, is:
n−1 X (2` + 1) = n2 (Proof!) (80) `=0
So, our polynomial v(ρ) can be written as
2`+1 v(ρ) = Ln−`−1(2ρ) (81) d p Lp (x) ≡ (−1)p L (x) (82) q−p dx q with the qth Laguerre polynomial: (83) d q L (x) ≡ ex (e−x xq) (84) q dx
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 37 / 142 Full solutionI
Now, the full solution reads:
s 2 3 (n − ` − 1)! 2r ` 2r ϕ (r, θ, φ) = e−r/na L2`+1 Y m(θ, φ) n`m na 2n[(n + `)!]3 na n−l−1 na `
ϕn`m e.g. with n = 2, ` = 1, m = 0:
1 1 3/2 r −r/(2a0) ϕ210(r, θ, φ) = √ e cos θ (85) 4π 2a a
independent of φ → rotationally symmetric w.r.t. the z-axis so-called p orbital → illustration of the probability of presence
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 38 / 142 Full solutionII
Summary: found stationary solutions for Coulomb potential (eigenfunctions for eigenvalues E ↔ Hˆ ) analytic solutions (real) → only certain functions for bound states due to normalization constraints, characterized by discrete quantum numbers quantum number n → from radial equation → associated Laguerre polynomials quantum numbers `, m → possible values depend on n and ` → spherical harmonics → associated Legendre polynomials eigenvalues E denpend only on n → degeneracy (only for pure Coulomb potential, central force) all solutions and quantum numbers verified by experiments but: one more quantum number, not deducible from Schrödinger equation (i.e. differential equation): spin quantum number
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 39 / 142 Full solutionIII
quantum number value / symbol formula 2 n 1, 2, 3,... En = −R∞/n K,L,M,. . . (shells) p ` 0, 1, 2, 3,..., n − 1 |~L| = `(` + 1)~ s, p, d, f ,... (orbitals)
m` 0, ±1, ±2,..., ±` Lz = m`~ 1 ms ± 2 Sz = ms ~
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 40 / 142 The angular momentum operatorI
In classical mechanics angular momentum ~L from cross product:
~L = ~r × ~p (86) Lx ypz − zpy Ly = zpx − xpz (87) Lz xpy − ypx
∂ In quantum mechanics: p → ~ ≡ ~∂ , so that x ı ∂x ı x
Lx y∂z − z∂y L = ~ z∂ − x∂ (88) y ı x z Lz x∂y − y∂x
Again, we are interested in the eigenvalues and eigenfunctions.
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 41 / 142 The angular momentum operatorII
Unfortunately:
2 [L , L ]f = ~ (y∂ − x∂ ) f = ı L f (Proof!) (89) x y ı x y ~ z ⇒ [Lx , Ly ] = ı~Lz (90) and (91)
[Ly , Lz ] = ı~Lx , [Lz , Lx ] = ı~Ly (92)
As Lx , Ly , Lz do not commutate, from generalized uncertainty principle:
σ2 σ2 ≥ ~|hL i| (93) Lx Ly 2 z
→ incompatible observables, no common eigenfunctions of Lx and Lz , etc.
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 42 / 142 The angular momentum operatorIII Fortunately:
2 2 2 2 [L , Lx ] ≡ [Lx + Ly + Lz , Lx ] = 0 (Proof!) (94)
→ hopefully: eigenfunction f exists, so that
2 L f = λf and Lz f = µf (95)
Let’s introduce the ladder operator
L± ≡ Lx ± ıLy with: (96) [Lz , L±] = [Lz , Lx ] ± ı[Lz , Ly ] = ı~Ly ± ı(−ı~Lx ) (97) = ±~(Lx ± ıLy ) = ±~L± (98) 2 and [L , L±] = 0 (99)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 43 / 142 The angular momentum operatorIV
So, with [Lz , L±] = ±~L±:
2 2 L (L±f ) = L±(L f ) = L±(λf ) = λ(L±f ) (100)
Lz (L±f ) = (Lz L± − L±Lz )f + L±Lz f (101) = ±~L±f + L±(µf ) = (µ ± ~)(L±f ) (102)
2 → (L±f ) is an eigenfunction of L as well as f , with same eigenvalue λ → (L±f ) is also an eigenfunction of Lz , with new eigenvalue µ ± ~
Therefore: L+ raising operator → increases eigenvalue of Lz L− lowering operator → lowers eigenvalue by ~ 2 2 → Consecutive application of L+ until reaching ft with Lz = L , such that 2 L+ft = 0 with eigenvalue of ft , let’s call ~`, i.e. Lz ft = ~`ft and L ft = λft
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 44 / 142 The angular momentum operatorV
With help of the following relation
2 2 L±L∓ = (Lx ± ıLy )(Lx ∓ ıLy ) = Lx + Ly ∓ ı(Lx Ly − Ly Lx ) (103) 2 2 = L − Lz ∓ ~Lz (104) 2 2 → L = L±L∓ + Lz ∓ ~Lz (105) we find a relation for the eigenvalue λ of L2 in terms of the maxium eigenvalue of Lz :
2 2 L ft = (L−L+ + Lz + ~Lz )ft (106) 2 2 2 = (0 + ~ ` + ~`)ft = ~ `(` + 1)ft (107) 2 → λ = ~ `(` + 1) (108)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 45 / 142 The angular momentum operatorVI
Analogously, there is a minimum eigenvalue of Lz with eigenfunction fb:
2 L−fb = 0 with Lz fb = ~`bfb and L fb = λfb (109) 2 2 → L fb = (L+L− + Lz − ~Lz )fb (110) 2 2 2 = (0 + ~ `b − ~`)fb = ~ `b(`b − 1)fb (111) 2 → λ = ~ `b(`b − 1) (112)
Hence by combining both results for λ (e.v. of L2), we get
`(` + 1) = `b(`b − 1) → `b = −` (113)
→ eigenvalues of Lz are m~ with m = −`, . . . , 0, . . . , ` in N integer steps, i.e. ` = −` + N ⇒ ` = N/2 → ` is integer or half-integer
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 46 / 142 The angular momentum operatorVII
2 Eigenfunctions of Lz , L characterized by ` and m:
2 m 2 m m m L f` = ~ `(` + 1)f` Lz f` = ~mf` (114) 1 3 where ` = 0, , 1, ,... m = −`, . . . , 0, . . . , ` (115) 2 2
2 Note: by pure algebra we found the eigenvalues of Lz and L ` is also called the azimuthal quantum number and m the magnetic quantum number
→ now find eigenfunctions f
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 47 / 142 The angular momentum operatorVIII
~ ~ Remember that L = ı~r × ∇ and with the unit vectors ˆr etc. 1 1 ∇ = ˆr∂ + θˆ ∂ + φˆ ∂ (116) r r θ r sin θ φ 1 ⇒ ~L = ~ r(ˆr × ˆr)∂ + (ˆr × θˆ)∂ + (ˆr × φˆ) ∂ (117) ı r θ sin θ φ
As (ˆr × ˆr) = 0, (ˆr × θˆ) = φˆ, and (ˆr × φˆ) = −θˆ
1 ~L = ~ φ∂ˆ − θˆ ∂ (118) ı θ sin θ φ
Unit vectors θˆ, φˆ in Cartesian coordinates:
θˆ = (cos θ cos φ)xˆ + (cos θ sin φ)yˆ + −(sin θ)zˆ (119) φˆ = −(sin φ)xˆ + (cos φ)yˆ (120)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 48 / 142 The angular momentum operatorIX
~L = ~ (− sin φxˆ + cos φyˆ)∂ − (121) ı θ 1 (cos θ cos φxˆ + cos θ sin φyˆ − sin θzˆ) ∂ (122) sin θ φ ⇒ L = ~ (− sin φ∂ − cos φ cot θ∂ ) (123) x ı θ φ L = ~ (+ cos φ∂ − sin φ cot θ∂ ) (124) y ı θ φ L = ~∂ (125) z ı φ L = L + ıL = ~[(− sin φ ± ı cos φ)∂ − (126) ± x y ı θ (cos φ ± ı sin φ) cot θ∂φ] (127) ±ıφ ±ıφ = ±~e (∂θ ± ı cot θ∂φ) as: cos φ ± ı sin φ = e (128)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 49 / 142 The angular momentum operatorX m Now we can find eigenfunction f` (θ, φ) of Lz :
L f m(θ, φ) = ~∂ f m(θ, φ) = m f m(θ, φ) ⇒ f m(θ, φ) = g(θ)eımφ(129) z ` ı φ ` ~ ` ` m 2 Analogously, f` (θ, φ) is an eigenfunction of L with e.v. ~`(` + 1):
2 m 2 m 2 m L f` (θ, φ) = (L+L− + Lz − ~Lz ) f` (θ, φ) = ~ `(` + 1) f` (θ, φ) (130)
ımφ dg ımφ As ∂θf = e dθ and ∂φf = ıme g with our expression for L± and Lz we obtain (Proof!)
d dg sin θ sin θ + [`(` + 1) sin2 θ − m2]g = 0 (131) dθ dθ
m → this is Θ(θ) of the spherical harmonics Y` (θ, φ) and – of course – f (φ) = eımφ is Φ(φ)! L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 50 / 142 The angular momentum operatorXI 2 ⇒ The spherical harmonics are the eigenfunctions of L and Lz :
2 2 Hψ = Eψ L ψ = ~ `(` + 1)ψ Lz ψ = ~mψ (132)
L z
Note 1: Found from analytical solution of the central-force problem that ` and m → integer but from algebraic solution for L angular momentum operator → ` and y m also half-integer Note 2: Absolute value of ~L is L x discrete, but also its component Lz → vector cone (see right figure), as
Jx , Jy unknown
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 51 / 142 The spin operatorI
Classical mechanics: ~L = ~r × ~p (motion of center of mass) and spin ~S = I~ω (motion about center of mass) Quantum mechanics of hydrogen atom: ~L→ orbital angular momentum of electron orbiting nucleus and ~S spin → angular momentum of electron itself → not a function of position variables r, θ, φ
Spin as intrinsic angular momentum cannot be decomposed into orbital angular momenta of constituent parts (e.g. electron is point-like)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 52 / 142 The spin operatorII
However, analogously to algebraic theory of angular momentum we postulate:
[Sx , Sy ] = ı~Sz , [Sy , Sz ] = ı~Sx , [Sz , Sx ] = ı~Sy (133)
And with ket notation (as eigenstates of spin are not functions):
2 2 S |smi = ~ s(s + 1) |smi; Sz |smi = ~m |smi (134) p S±|smi = ~ s(s + 1) − m(m ± 1) |s(m ± 1)i (135)
These relations do not exclude half-integer values of s, m. Interestingly, every elementary particle has specific, fixed value of s, e.g. 1/2 for electrons
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 53 / 142 The spin operatorIII
Theory of spin 1/2 simplest nontrival quantum system → only two possible eigenstates: 1 1 | 2 2 i → spin up 1 1 | 2 (− 2 )i → spin down Thus, general state can be written as 2-element column matrix (spinor):
a χ = = aχ + bχ where (136) b + −
1 0 χ = spin up χ = spin down (137) + 0 − 1
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 54 / 142 The spin operatorIV 2 The spin operators Sz , S , etc. are 2 × 2 matrices. From their eigenvalues (s = 1/2):
3 3 1 1 S 2χ = 2χ ; S 2χ = 2χ ; S χ = 2χ ; S χ = − 2χ (138) + 4~ + − 4~ − z + 2~ + z − 2~ − S+χ− = ~χ+; S−χ+ = ~χ−; S+χ+ = S−χ− = 0(139)
together with S± ≡ Sx ± ıSy , so that
1 1 S = (S + S ) S = (S − S ) (140) x 2 + − y 2ı + − and therefore
S χ = ~χ ; S χ = ~χ ; S χ = − ~ χ ; S χ = ~ χ ;(141) x + 2 − x − 2 + y + 2ı − y − 2ı +
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 55 / 142 The spin operatorV we find the algebraic representation of the spin operators
3 1 0 0 1 0 0 S2 = 2 ; S = ; S = (142) 4~ 0 1 + ~ 0 0 − ~ 1 0 0 1 0 −ı 1 0 S = ~ ; S = ~ ; S = ~ (143) x 2 1 0 y 2 ı 0 z 2 0 −1
~ One defines the Pauli spin matrices by σx = Sx / 2 etc. 2 Note: Sx , Sy , Sz , S are Hermitian (i.e. self-adjoint: † T T Sx = Sx = Sx = Sx ) → observables, while S+, S− are not Hermitian → not observable
We already know the eigenspinors (eigenstates) and eigenvalues of Sz :
1 0 χ = eigenvalue: + ~; χ = eigenvalue: − ~(144) + 0 2 − 1 2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 56 / 142 The spin operatorVI
For a particle in the general state
a χ = = aχ + bχ (145) b + −
~ 2 the probability to get for Sz the value + 2 is of course |a| and to get the ~ 2 value − 2 is |b| , so
|a|2 + |b|2 = 1 (i.e. normalized spinor) (146)
Accurate language 2 The probability |a| to get +~/2 for a measurement of Sz doesn’t mean 1 that the particle is in the state . In fact, the particle is in the state χ. 0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 57 / 142 The parity operator
The action of the parity operator P is just ~r → −~r:
Pψ(x, y, z) = ψ(−x, −y, −z) (147)
Obviously P2 = PP = 1 → only two eigenvalues: +1 → eigenstate of positive parity (even parity): ψ(~r) = ψ(−~r) −1 → eigenstate of negative parity (odd parity): ψ(~r) = −ψ(−~r) For a potential V (x, y, z) = V (−x, −y, −z), e.g. Coulomb potential → parity is good quantum number (i.e. conserved quantity)
m ımφ Parity of eigenfunctions of hydrogen atom ψ(r, θ, φ) = R(r)P` (cos θ)e Parity transformation θ → π − θ, φ → φ + π, r → r ⇒ parity depends only on transformation of associated Legendre polynoms even ` → even parity and odd ` → odd parity (Proof!)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 58 / 142 Application: Electron in magnetic fieldI spin + charged particle → magnetic dipole:
~µ = γ~S where γ := gyromagnetic ratio (148)
q Note: in classical electrodynamics: γ = 2m q q q in QM: γ = gJ 2m = gS 2m ≈ 2.002 2m magnetic dipole in magnetic field B~ experiences torque M~ :
M~ = ~µ × B~ (149)
→ torque tries to line ~µ up parallel to field (“compass needle”) Energy (and therefore Hamiltonian):
H = −~µ · B~ = −γB~ · ~S (150)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 59 / 142 Application: Electron in magnetic fieldII Larmor precession 1 Spin- 2 particle in a homogeneous magnetic field along z-direction B~ = B0~ez Hamiltonian (matrix!):
γB 1 0 H = −γB S = − 0~ (151) 0 z 2 0 −1
Hamiltonian H and Spin operator Sz share same eigenstates:
γB0~ χ+, energy E+ = − 2 (152) γB0~ χ−, energy E− = + 2 (153) where energy is lowest for dipole moment parallel to magnetic field (otherwise: → torque tries to align)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 60 / 142 Application: Electron in magnetic fieldIII
→ time-dependent solution of ı~∂t χ = Hχ expressed with stationary states:
+ıγB0t/~ −ıE+t/~ −ıE−t/~ ae χ(t) = aχ+e + bχ−e = (154) be−ıγB0t/~
fix coefficients by initial conditions, e.g.
a χ(t = 0) = , where |a|2 + |b|2 = 1 (155) b
Then (for real a, b) we can write a = cos(α/2) and b = sin(α/2), with constant α, so
cos(α/2)e+ıγB0t/2 χ(t) = (156) sin(α/2)e−ıγB0t/2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 61 / 142 Application: Electron in magnetic fieldIV therefore the expectation value of the spin as a function of time:
† hSx i = χ(t) Sx χ 0 1 cos(α/2)e+ıγB0t/2 = cos(α/2)e−ıγB0t/2 sin(α/2)e+ıγB0t/2 ~ 2 1 0 sin(α/2)e−ıγB0t/2
= + ~ sin α cos(γB t) 2 0 hS i = − ~ sin α sin(γB t) y 2 0 hS i = + ~ cos α z 2
→ h~Si tilted at constant angle α to z-axis, precesses about the field at Larmor frequency of a spinning electron
11 −1 −1 −1 ω = γB0 = 1.7606 × 10 rad s T · B0 → f = 28 GHz T · B0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 62 / 142 Addition of angular momentaI
1 Now, two spin- 2 particles (e.g. electron and proton in ground state of hydrogen, so ` = 0), composite system is in a linear combination of
↑↑, ↑↓, ↓↑, ↓↓ (157)
What is the total angular momentum of the atom? We define
~S ≡ ~S(1) + ~S(2) (158)
The z-component: simply adds, each composite state is eigenstate of Sz
(1) (2) (1) (2) Sz χ1χ2 = (Sz + Sz )χ1χ2 = (Sz χ1)χ2 + χ1(Sz χ2) (159) = (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2 (160)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 63 / 142 Addition of angular momentaII
Thus, m = m1 + m2:
↑↑ m = 1 (161) ↑↓ m = 0 (162) ↓↑ m = 0 (163) ↓↓ m = −1 (164) problem: two states with m = 0 → apply lowering operator (1) (2) S− = S− + S− to state ↑↑ to obtain the correct state:
(1) (2) S−(↑↑) = (S− ↑) ↑ + ↑ (S− ↑) (165) = (~ ↓) ↑ + ↑ (~ ↓) = ~(↓↑ + ↑↓) (166)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 64 / 142 Addition of angular momentaIII
Therefore, the states with s = 1 in |s mi notation
|1 1i = ↑↑ |1 0i = √1 (↑↓ + ↓↑) 2 → triplet: s = 1 (167) |1 − 1i = ↓↓
And orthogonal state |s mi = |0 0i
1 |0 0i = √ (↑↓− ↓↑ ) → singlet: s = 0 (168) 2
1 → system of two spin- 2 particles has total spin 1 or 0. Let us proof: triplet 2 2 states are eigenvectors of S with eigenvalues 2~ and singlet state is eigenvector with eigenvalue 0:
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 65 / 142 Addition of angular momentaIV
S2 = (~S(1) + ~S(2)) · (~S(1) + ~S(2)) = (S(1))2 + (S(2))2 + 2~S(1) · ~S(2) (169)
~ ~ ~ 2 3 2 Remember: Sx χ+ = 2 χ−, Sy χ+ = − 2ı χ−, Sz χ+ = 2 χ+, S χ+ = 4 ~ χ+ etc.
(1) (2) (1) (2) (1) (2) (1) (2) ~S · ~S (↑↓) = (Sx ↑)(Sx ↓) + (Sy ↑)(Sy ↓) + (Sz ↑)(Sz ↓) ı ı = ~ ↓ ~ ↑ + ~ ↓ − ~ ↑ + ~ ↑ −~ ↓ 2 2 2 2 2 2 2 = ~ (2 ↓↑ − ↑↓) (170) 4 and (171) 2 ~S(1) · ~S(2)(↓↑) = ~ (2 ↑↓ − ↓↑) (172) 4
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 66 / 142 Addition of angular momentaV
So we get
1 2 ~S(1) · ~S(2)|1 0i = ~ √ (2 ↓↑ − ↑↓ +2 ↑↓− ↓↑ ) = −~ |1 0i (173) 4 2 4 1 3 2 ~S(1) · ~S(2)|0 0i = ~ √ (2 ↓↑ − ↑↓− 2 ↑↓ + ↓↑) = − ~ |0 0i (174) 4 2 4 and therefore
3 2 3 2 2 S2|1 0i = ~ + ~ +2~ |1 0i = 2 2|1 0i (175) 4 4 4 ~ 3 2 3 2 2 S2|0 0i = ~ + ~ −2~ |0 0i = 0 (176) 4 4 4
2 2 2 2 and of course S |1 − 1i = 2~ |1 − 1i and S |1 1i = 2~ |1 1i
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 67 / 142 Addition of angular momentaVI
Thus, answer to the general problem of combining any s1, s2 or angular momenta:
s = (s1 + s2), (s1 + s2 − 1),..., |s1 − s2| (177) and also, e.g. for hydrogen atom, net angular momentum of electron (spin + orbital) j
1 1 j = ` + or j = ` − (178) 2 2 and with proton, total angular momentum of hydrogen atom:
J = ` + 1 or J = ` − 1 (179)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 68 / 142 Addition of angular momentaVII
Therefore, state |s mi with total spin s and z-component m → linear combination of composite states |s1 m1i |s2 m2i:
X s1 s2 s |s mi = Cm1 m2 m|s1 m1i |s2 m2i (180) m1+m2=m
s1 s2 s with Clebsch-Gordan coefficients Cm1 m2 m. E.g.
1 1 1 |21 i = √ |22 i|1 −1i + √ |21 i|10 i − √ |20 i|11 i (181) 3 6 2
Note that the z-components have to add to m = 1 → For two particles of spin 2 and spin 1 with total spin 2 and total (1) 1 1 z-component 1: measure Sz and get 2~ (probability 3 ), ~ (probability 6 ), 1 or 0 (probability 2 )
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 69 / 142 The helium atomI
Hamiltonian of helium (Z = 2), neglecting spin
2 2 2 2 2 ~ 2 1 2e ~ 2 1 2e 1 e H = − ∇1 − + − ∇2 − + 2m 4π0 r1 2m 4π0 r2 4π0 |r~1 − r~2| → without repulsion term :
ψ(~r1,~r2) = ψn`m(~r1)ψn0`0m0 (~r2) (182)
aH with a = 2 and E = 4EH,
E = 4(En + En0 ) → E0 = 8 · (−13.6 eV) = −109 eV (183)
(But measured: −79 eV = −24.6 eV − 54.4 eV, because we neglected e−e−-interaction)
8 with ψ (~r ,~r ) = ψ (~r )ψ (~r ) = e−2(r1+r2)/a (184) 0 1 2 100 1 100 2 πa3
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 70 / 142 The helium atomII Electrons are identical particles, therefore
ψ(~r1,~r2) 6= ψa(~r1)ψb(~r2) but (185)
ψ−(~r1,~r2) = A[ψa(~r1)ψb(~r2) − ψb(~r1)ψa(~r2)] (fermions) (186)
two electrons cannot occupy same state → Pauli exclusion principle (Proof!) ψ for fermions must be antisymmetric under exchange:
ψ(~r1,~r2) = −ψ(~r2,~r1) (187)
with ψ → ψ(~r1,~r2)χ(~r1,~r2) r +r as the ground state ψ0 ∼ e 1 2 of helium is symmetric → spin state must be antisymmetric → singlet, electrons oppositely aligned 1 √ (↑↓− ↓↑ ) (188) 2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 71 / 142 The helium atomIII
Excited helium atom:
combination ψn`mψ100 1 S 1 Po 1 D 3 S 3 Po 3 D 0 0 0 13 17 16 12 14 15 (What about ψn`mψn ` m ?) 7 11 10 6 8 9 5 3 4 symmetric ψn`mψ100 2 → antisymmetric spin configuration (singlet) or antisymmetric ψn`mψ100
singlet E triplet → symmetric spin configuration (triplet) thus, two different kinds of excited states: singlet and triplet
singlet states have sligthly higher 1 E than corresponding triplets 0 1 2 0 1 2
(because of symmetric ψ L L − − → closer e e → larger Erepuls)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 72 / 142 The helium atomIV
Intercombination Note that total wavefunction must be antisymmetric by ψa(~r1,~r2)χs(~r1,~r2) ψ(~r1,~r2) = (189) ψs(~r1,~r2)χa(~r1,~r2)
ψ χ the interchange operator can be written as P12 = P12 P12 ψ Hamiltonian does not depend on spin → [H, P12] = 0 − ı Ht time evolution operator U(t) = e ~ → [U, P12] = 0
if for any t0: P12|ψ(t0)i = −|ψ(t0), then for all times:
P12|ψ(t)i = P12U(t)|ψ(t0)i = U(t)P12|ψ(t0)i (190)
= −U(t)|ψ(t0)i = −|ψ(t)i (191)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 73 / 142 The helium atomV
→ symmetry of ψ(~r1,~r2) does not change with time → symmetry of χ(~r1,~r2) does not change with time Selection rule for multiplicity Transitions between different multiplicities are forbidden:
∆s = 0 (192)
(exact Hamiltonian depends on spin → small modification of selection rule) Parity m 0 ψ composed of ψn`mψ100 = Rn`Y` R10Y0 → parity good quantum number, odd for odd `, even for even ` odd parity indicated by small lettero right from captial letter for net orbital angular momentum → e.g. 1Po
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 74 / 142 Multi-electron atomsI
Analogously, ground-state electron configurations of other atoms described by
N 2 2 2 X ~ 2 1 Ze X 1 e H = − ∇i − + (193) 2m 4π0 ri 4π0 |~ri − r~j | i=1 i6=j
neglect repulsion term → each electron occupies a one-particle hydrogenic state (n, `, m) → orbital due to Pauli exclusion principle → only two e− per orbital: n2 wavefunction for each shell n, e.g. n = 1 → two electrons, n = 2 → eight electrons, etc. Periodic system: horizontal rows ↔ filling out each shell How to fill n = 2 (` = 0 or ` = 1) with single e−? → screening by inner electrons, favor lowest `, e.g. `=0 `=1 `=0 `=2 En=2 < En=2, En=4 < En=3 (larger ` → larger hri → stronger screening of nucleus → less binding energy)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 75 / 142 Multi-electron atomsII
Total angular momentum state of electron represented by pair n`, where ` is a letter, e.g. s for ` = 0 – m not listed, but exponent for number of electrons in `, e.g. He → 1s2 (i.e. two electrons in n = 1, ` = 0) total orbital angular momentum indicated by capital L (letter), total spin by captial S (multiplicity, number), and total angular momentum by capital J (number) total configuration of atom listed as
2S+1 LJ (194)
2 e.g. groundstate of hydrogen S = 1/2, L = S, J = 1/2 → S1/2
LS-coupling of multi-electrons ~ P ~ ~ P ~ ~ ~ If spin-orbit interaction negligible: L = `i , S = ~si and J = L + S
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 76 / 142 Multi-electron atomsIII Spectroscopic nomenclature for atomic states
Table : Angular momenta of electrons
` 0 1 2 3 . . . letter s p d f ... name sharp principal diffuse fundamental ?
Z El. Configuration 2 1 H 1s S1/2 2 1 Ground-state 2 He 1s S0 2 electron 3 Li (He)2s S1/2 2 1 configurations of 4 Be (He)2s S0 first elements 2 2 o 5 B (He)2s 2p P1/2 ... 2 2 21 Sc (Ar)4s 3d D3/2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 77 / 142 Multi-electron atomsIV Note: Screening effect becomes larger for larger ` → overlap of shells
1s 2s 3s 4s 5s 6s 7s 8s 9s 10s 2p 3p 4p 5p 6p 7p 8p 9p 10p 3d 4d 5d 6d 7d 8d 9d 10d 4f 5f 6f 7f 8f 9f 10f 5g 6g 7g 8g 9g 10g 6h 7h 8h 9h 10h 7i 8i 9i 10i 8k 9k 10k 9l 10l 10m
→ for determining the energy order of terms of a one-electron system, just go along the diagonals, e.g. 1s 2s 2p 3s 3p 4s 3d 4p 5s.
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 78 / 142 Multi-electron atomsV
How to distribute electrons to shells? → find configuration of minium energy → Hund’s rules (minimum energy principle)
1 The total angular momentum J of completely filled shells (n) or subshells (`) is 0. E.g. He: 1s2, Be: (He)2s2, Ne: (He)2s2 2p6 2 Spins of electrons are pereferably parallel, i.e. electron are distributed on subshells with m`, such that multiplicity 2S + 1 is maximal. E.g. N → (He) 2s2 2p3: in 2p shell: ↑↑↑ → 4S. (sometimes referred to as the 1st rule or S rule) 3 For states of same S, electrons are distributed such that largest
angular momentum L is achieved (larger ` → larger hri → larger ∆xe−e− 1 1 → less repulsion), e.g. E( D2) < E( S0). (sometimes referred to as the 2nd rule or L rule)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 79 / 142 Multi-electron atomsVI
4 If outermost subshell half-filled or less: configuration with lowest total angular momentum J = |L − S| (i.e. with smallest J) is preferred. If outermost subshell more than half-filled: configuration with highest J = L + S is preferred. → reason: electron-electron interaction → only valid for LS coupling (no spin-orbit interaction → light atoms, Z < 10) Example: groundstate of carbon 1s2 2s2 2p2 two e− in 2p2 → 1S, 1D, or 3P? ↑↓ , ↑↓ , ↑ ↑ Using S rule → largest mulitplicity S preferred, i.e. 3P 3 3 3 J = L + S → L = 1 and S = −1, 0, +1 → P0, P1, or P2 Using J rule → smallest J, i.e. J = 0 3 ⇒ P0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 80 / 142 Spin-orbit interactionI
When solved Schrödinger equation for hydrogen, used simple Hamiltonian
2 e2 1 H = − ~ ∇2 − (195) 2m 4π0 r But: electron orbiting around nucleus → from electrons point of view: proton orbits electron → setting up magnetic field B~ , exerting a torque on spining electron to align its magnetic moment ~µe along B~
e Hso = −~µ · B~ = − − ~S B~ (196) e m
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 81 / 142 Spin-orbit interactionII From Biot-Savart law (→ blackboard):
µ I B = 0 (197) 2r where current I = e/T with L = rmv = 2πmr 2/T and B~ ||~L, so
~ 1 e ~ B = 2 3 L (198) 4π0 mc r √ used c = 1/ 0µ0 to eliminate µ0. So the Spin-Orbit Interaction is described by
2 so 1 e 1 ~ ~ H = 2 2 3 S · L (199) 2 4π0 m c r where factor 1/2 is from Thomas precession: back transformation into rest frame of proton (electron: not an inertial system, accelerated) L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 82 / 142 Spin-orbit interactionIII
Now: [H,~L] 6= 0 and [H, ~S] 6= 0 → not longer separatly conserved quantities, instead:
~J ≡ ~L + ~S (200)
2 2 commutes with H, as well as L , S ,Jz and
J2 = (~L + ~S) · (~L + ~S) = L2 + S2 + 2~L · ~S (201) 1 → ~L · ~S = (J2 − L2 − S2) (202) 2
1 with eigenvalues (where s = 2 in our case):
2 ~ [j(j + 1) − `(` + 1) − s(s + 1)] (203) 2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 83 / 142 Spin-orbit interactionIV
so 1 For H we also need r 3 :
1 1 = (204) r 3 1 3 3 `(` + 2 )(` + 1)n a Thus the energy of SO interaction is
2 3 e2 1 (~ /2[j(j + 1) − `(` + 1) − ] E so = hHsoi = 4 (205) 8π m2c2 1 3 3 0 `(` + 2 )(` + 1)n a E 2 n[j(j + 1) − `(` + 1) − 3 ] E 2 = n 4 ∼ n (206) mc2 1 mc2 `(` + 2 )(` + 1)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 84 / 142 Spin-orbit interactionV
However, only half of the truth, need correction for relativistic motion of electron, at least for larger Z, which is
" # E 2 4n E r = − n − 3 (207) 2mc2 1 ` + 2
so r So in total we have En + E + E : Fine-structure energy levels of hydrogen
" !# 13.6 eV α2 n 3 E = − 1 + − (208) nj n2 n2 1 4 j + 2
→ breaks degeneracy in ` (different eigenvalues of H for same n) → m` and ms are not longer “good” quantum numbers (stationary states now linear combinations of states with different m`, ms )
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 85 / 142 Spin-orbit interactionVI
→ n, `, s, j, mj “good” quantum numbers (write |j mj i as linear combination of |` m`i|s ms i with help of Clebsch-Gordan coefficents)
Fine structure splitting of hydrogen
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 86 / 142 LS couplingI
For light elements Z < 10, total angular momentum ~J = ~L + ~S, where ~L = P ~` and ~S = P~s of single electrons Example: LS-coupling of np4 electrons
equivalent p electrons: must differ in either m` or ms → for 2p: 3 m` states × 2 ms states = 6 different states, each electron in one of these states (Paul exclusion principle) 6 4 electrons in 6 states: combinations = 6! = 15 4 (6−4)!4! configurations (see below) However: degeneracy, only L and S are “good” quantum numbers Spins can couple to either S = 0 (e.g. ↑↓,↑,↓) or S = 1 (e.g. ↑↓,↑,↑) Four electrons with ` = 1 → angular momenta can couple to L = 0, 1, 2 (at maximum two electrons in m` = +1)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 87 / 142 LS couplingII
therefore: → possible states 2S+1L: 1S, 1P, 1D, 3S, 3P, 3D However, some states forbidden, because of Pauli exclusion principle (complicated to find out, which one), so rule for equivalent electrons (same n and `): L + S must be even → only 1S, 3P, 1D allowed (see below)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 88 / 142 LS couplingIII
m` = −1 m` = 0 m` = +1 ML MS Total/Term 1 ↑↓ ↑↓ 0 0 S0 ↑↓ ↓ ↓ -1 -1 ↑↓ ↑ ↓ -1 0 ↑↓ ↑ ↑ -1 1 ↓ ↑↓ ↓ 0 -1 LS 3 ↑ ↑↓ ↓ 0 0 P0,1,2 coupling ↑ ↑↓ ↑ 0 1 4 for np ↓ ↓ ↑↓ 1 -1 electrons ↑ ↓ ↑↓ 1 0 ↑ ↑ ↑↓ 1 1 ↑↓ ↑↓ -2 0 ↑↓ ↓ ↑ -1 0 1 ↓ ↑↓ ↑ 0 0 D2 ↓ ↑ ↑↓ 1 0 ↑↓ ↑↓ 2 0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 89 / 142 jj-couplingI
For heavy atoms (Z > 10, e.g. Pb Z = 82) → spin-orbit interactions ≈ spin-spin interactions → for each electron individual combination of ~` and ~s to individual ~j: X X ~J = ~j = (~` +~s) (209)
→ resulting angular momentum (S,P,. . . ) of electrons not longer useful But: pure jj coupling only in heaviest atoms
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 90 / 142 Zeeman effect
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 91 / 142 Summary: Spin and orbitI
Hydrogen
Bohr formula from Schrödinger Eq. → degeneracy of solutions φn`m` , i.e. same eigenvalue E for different ` and m` 13.6 eV E = − (210) n n2 take spin-orbit interaction (and relativistic corrections) into account: interaction of internal magnetic field caused by electron orbit with electron spin → fine structure splitting, breaks E degeneracy " !# 13.6 eV α2 n 3 E = − 1 + − (211) nj n2 n2 1 4 j + 2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 92 / 142 Summary: Spin and orbitII
5 4 +2 +1 3 0 5/2 -1 3/2 -2 1/2
+1 energy 2 0 -1 3/2
1/2
1 0
1/2
Bohr: solution of fine structure Schro¨dinger Eq. spin-orbit interaction w/o spin + relativistic correction
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 93 / 142 Zeeman effect in QMI
An atom in an external magnetic field B~ ex adds interaction term to hamiltonian, i.e. energy shift:
~ HZ = − (~µ` + ~µs ) · Bext (212) e e e = − − ~L − ~S · B~ = (~L + 2~S) · B~ (213) 2m m ext 2m ext
For Bext BSO: fine structure dominates → HZ treated as small perturbation For Bext BSO: HZ dominates → HSO treated as small perturbation For Bext ≈ BSO: degenerate perturbation theory
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 94 / 142 Zeeman effect in QMII
Estimation of internal magnetic field BSO:
~ 1 e ~ B = 2 3 L (214) 4π0 mc r e.g. for L = ~, r = a (215) 1.60 × 10−19 C · 1.05 × 10−34 J s B = 4π 8.9 × 10−12 C2/N m2 9.1 × 10−31 kg (3.0 × 108 m/s)2 (0.53 × 10−10 m)3 (216) = 12.3 J s C−1 m2 = 12.3 T (217)
→ magnetic field Bext 10 T is “strong” Zeeman field → magnetic field Bext 10 T is “weak” Zeeman field e.g. magnetic field of the Earth ≈ 1 G = 10−4 T is “weak” Zeeman field
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 95 / 142 Zeeman effect in QMIII Weak-field Zeeman effect
for Bext Bint → fine structure dominates ~ ~ good quantum numbers: n, `, j, mj (but not m` and ms as L and S are not separately conserved in spin-orbit coupling) from 1st order perturbation theory: e E = hn`jm |H |n`jm i = B~ · h~L + 2~Si (218) Z j Z j 2m ext
as ~J = ~L + ~S →~L + 2~S = ~J + ~S, with constant ~J = ~L + ~S
so time average of ~S is its projection along ~J:
(~S · ~J) S¯ = ~J (219) J2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 96 / 142 Zeeman effect in QMIV
and ~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S and therefore
1 2 ~S · ~J = (J2 + S2 − L2) = ~ [j(j + 1) + s(s + 1) − `(` + 1)] (220) 2 2 which implies
* ! + ~S · ~J h~L + 2~Si = 1 + ~J (221) J2 " # j(j + 1) − `(` + 1) + 3 = 1 + 4 h~Ji = g h~Ji (222) 2j(j + 1) J
with the Landé g-factor gJ . Note: For single-electron system, if only “orbit magnetism” (S = 0) → gJ = 1, if only “spin magnetism” (` = 0) → gJ ≈ 2
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 97 / 142 Zeeman effect in QMV
With z-axis aligned with B~ ext:
EZ = µBgJ Bextmj (223)
and the Bohr magneton
e µ = ~ = 5.788 × 10−5 eV T−1 (224) B 2m
→ total energy is Enj + EZ, e.g. ground state of hydrogen (n = 1, 1 ` = 0, j = 2 → gJ = 2) splits into 2 levels:
α2 E = −13.6 eV 1 + ± µ B (225) 4 B ext
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 98 / 142 Zeeman effect in QMVI
E µe Bext
≈ mj = 1/2 Weak-field splitting of the hydrogen ground state, slope -13.6 eV(1 + α2/4) dE/d(µBBext) = ±1
m = 1/2 j −
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 99 / 142 Zeeman effect in QMVII
Strong-field Zeeman effect (Paschen-Back effect)
for Bext Bint → Zeeman effect dominates
let B~ ext = ~ez Bext
good quantum numbers: n, `, m`, ms – but not j, mj , as total angular momentum ~J not conserved for external torque (but Lz and Sz are conserved) Zeeman Hamiltonian:
e H = B (L + 2S ) (226) Z 2m ext z z so “unperturbed” energies:
13.6 eV E = − + µ B (m + 2m ) (227) nm`ms n2 B ext ` s
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 100 / 142 Zeeman effect in QMVIII from 1st order perturbation theory: fine-structure correction
Efs = hn`m`ms |(Hrel. + HSO)|n`m`ms i (228)
2 h i En 4n where Erel. is same as before, i.e. Erel. = − 2mc2 `+1/2 − 3 for SO-interaction we need
~ ~ 2 hS · Li = hSx ihLx i + hSy ihLy i + hSz ihLz i = ~ m`ms (229)
(because hSx i = hSy i = hLx i = hLy i = 0 for eigenstates of Sz and Lz ) therefore
13.6 eV 3 `(` + 1) − m m E = α2 − ` s (230) fs n3 4n `(` + 1/2)(` + 1)
and total energy is Efs + Enm`ms Note: square bracketed term in Eq. 230 indeterminate for ` = 0, L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 101 / 142 Zeeman effect in QMIX
but then j = s, mj = ms , so e e E = B~ · h~L + 2~Si = B 2m = 2m µ B (231) Z 2m ext 2m ext s ~ s B ext and fine-structure energies (includingn relativistic correction)
13.6 eV α2 3 E = − 1 + n − (232) nj n2 n2 4 as j = 1/2, so total energy
13.6 eV α2 3 E = − 1 + n − + 2m µ B (233) n2 n2 4 s B ext and fine-structure is α2 term:
13.6 eV 3 13.6 eV 3 E = − α2 n − = α2 − 1 (234) fs n4 4 n3 4n → square bracketed term is 1 for ` = 0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 102 / 142 Zeeman effect in QMX
Intermediate-field Zeeman effect
as neither HZ nor Hfs dominates, both are equal perturbations to Bohr 0 Hamiltonian H = Hfs + HZ
consider case n = 2 → ` = 0, 1, express basis |jmj i (makes matrix elements of Hfs easier) by linear combination of |`m`i|sms i
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 103 / 142 Zeeman effect in QMXI
2 . . . eigenvalues are with γ ≡ 13.6 eV(α/8) and β ≡ µBBext for n = 2:
1 = E2 − 5γ + β
2 = E2 − 5γ − β
3 = E2 − γ + 2β
4 = E2 − γ − 2β p 2 2 5 = E2 − 3γ + β/2 + 4γ + (2/3)γβ + β /4 p 2 2 6 = E2 − 3γ + β/2 − 4γ + (2/3)γβ + β /4 p 2 2 7 = E2 − 3γ − β/2 + 4γ + (2/3)γβ + β /4 p 2 2 8 = E2 − 3γ − β/2 − 4γ + (2/3)γβ + β /4
for β = 0 (zero-field limit) → fine-structure energies for β γ (weak-field) → as in weak-field treatment for β γ (strong-field) → as in strong-field case
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 104 / 142 Classical description of Zeeman effectI
Normal Zeeman effect classical description (w/o QM) by Lorentz:
e 2 − ~ 3 B
1 2 3
emission by bound electron (circular orbit) → projection into one direction → linear oscillator, split into 1 and 2 + 3 no force on 1 → δω = 0, linearly polarized light
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 105 / 142 Classical description of Zeeman effectII consider only orbital angular momentum of electron ~` e ~ ~ → magnetic moment ~µ = 2m ` = γ`:
~ EZ = −(~µ · Bext) = −µz Bext = −γ`z Bext (235)
→ “Larmor precession” (but with g = 1, because of orbit instead of spin) around z-axis with constant `z and frequency
e ω = γB = B (236) L 2m
→ perturbation of electron orbit (with Bohr frequency ω0), especially component perpendicular to B~ without external magnetic field → frequency ω0 for 2 , 3
2 2 Ze mω0~r = 3~r (237) 4π0r L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 106 / 142 Classical description of Zeeman effectIII now with additional external field → Lorentz force, e.g. in cartesian coordinates
2 ~ mx¨ + mω0x − ey˙ B0 = 0 (238) 2 ~ my¨ + mω0y − ex˙ B0 = 0 (239) 2 mx¨ + mω0z = 0 (240)
→ equations of motion for electron with u = x + ıy and v = x − ıy: e u = u exp[ı(ω − B)t] (241) 0 0 2m e v = v exp[ı(ω + B)t] (242) 0 0 2m e → left- and right-circular oscillation with ω0 ± 2m B → emission / absorption of circular polarized light of frequency e ω0± 2m B (Lorentz, Nobel prize 1902)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 107 / 142 Classical description of Zeeman effectIV
e frequency shift δω = 2m B of spectral lines does not depend on frequency ω0 circular polarization along B~ , linear polarization perpendicular to B~ → σ+ (right-handed circularly pol.∗, +δω) and σ− (left-handed circularly pol.∗, −δω) radiation ∗relative to B~ no emission of unshifted line π along B~ → only triple splitting of lines explained
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 108 / 142 L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 109 / 142 Application of Zeeman effectI
Astrophysics
consider only weak fields where gJ is good quantum number interaction energy (we already know), aligned with z-axis:
~ V = ~µ · B → V = −µz B = mJ gJ µB B (243) e~ mJ = −J, −J + 1,..., J and µB = (244) 2me
compare two Zeeman components with ∆mJ = ±1 and λ1, λ2:
4πmc −1 −1 ∆E = gJ µB B → B = (λ2 − λ1 ) (245) egJ 4πmc ∆λ → B ≈ 2 (246) egJ λ1
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 110 / 142 Application of Zeeman effectII
e.g. for single electron above filled shell (e.g. H, Li, . . . ), s = 1/2 → gs = gS = 2 and j = ` ± s:
g − 1 g = g = 1 ± S (247) J j 2` + 1
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 111 / 142 Application of Zeeman effectIII Magnetic field of a sunspot
16 Spectrum of a sun spot: right-hand 14 circularly polarized
(tell.) + 2 (σ , blue) and 12 Fe I O left-hand circularly Normalized flux − 10 polarized (σ , red) light of a Fe I 6302.3 6302.5 6302.7 6302.9 λ o / A line (gJ = 5/2)
∆λ B ≈ 0.02141 · 2 TÅ ≈ 0.2 T (248) gJ λ0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 112 / 142 Application of Zeeman effectIV Weak stellar magnetic fields line splitting (π and σ components) approximately:
eBλ2 ∆λ = g (249) B J 4πmc2
assume that Zeeman line splitting ∆λB ∆λDoppler, Stark → individual line shifts in subpixel regime measure so-called stokes parameter I , V :
2 2 I = P0◦ + P90◦ = hEx + Ey i → unpolarized (250)
V = Prc − Plc (251) V 1 f o − f e f o − f e = o e − o e (252) I 2 f + f α=−45◦ f + f α=+45◦ where f is flux measured from ordinary and extraordinary beam, α is angle of polarisator
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 113 / 142 Application of Zeeman effectV in weak field regime (Landstreet 1982), ensemble of spectral lines:
V g eλ2 1 dI = − eff hB i (253) I 4πmc2 I dλ z e.g. FORS2 of NGC 1514 → slope gives magnetic field of −250 G
0.005
(%) 0.000 I / V
-0.005
-1.00E-06 0.0 1.00E-06 4.67 10-13 λ2 1/I dI /dλ [10-6 G-1 ]
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 114 / 142 Selection rules for radiative line transitions
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 115 / 142 Selection rule for multiplicity
From helium atom we already learnt a selection rule: considering time evolution of the position-space wave function, then transitions between different multiplicities are forbidden:
∆S = 0 (254)
→ radiative line transitions only within singlet or triplet states small modification of the this rule are possible (why?)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 116 / 142 Fermi’s Golden Rule
So far, only stationary states (“pure” Hamiltonian H0). How to consider transitions? 0 → perturbation theory (H = H0 + H ), especially: → Fermi’s Golden Rule:
gives transition rate Ti→ f from one energy eigenstate |ii into another energy eigenstate |f i, due to a perturbation H0 (perturbing Hamiltonian) Fermi’s Golden Rule
2π 0 2 Ti→ f = |hf |H |ii| %(Ef ) (255) ~
where %(Ef ) is the space density of the final states hf |H0|ii is called matrix element (in bra–ket notation) of the perturbation H0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 117 / 142 Interaction of a two-level atom with coherent lightI
Let us consider the time-dependent Schrödinger equation
2 dψ(~r, t) − ~ ∆ + V + V ψ(~r, t) = ı (256) 2m S ~ dt with potential of the nucleus, V ; VS corresponds to potential of “light field” with electric field (plane wave)
F~ = F~0 cos(kx − ωt) (257)
Atom is placed at ~r = 0, consider only light with 2π/k = λ datom, so x = 0 in Eq. (257), i.e. spatially constant light field
F~ = F~0 cos(ωt) (258) with polarization in z-direction: F~0 = (0, 0, F0)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 118 / 142 Interaction of a two-level atom with coherent lightII
Then, force of electric field F~ on electron: −eF~ , therefore (why?)
VS = eF0z cos ωt (259)
Assumption: already solved unperturbed Schrödinger eq. (VS = 0), found ϕi , ϕj with energies Ei , Ej . Now: wave function ψ of perturbed system as superposition of ϕi , ϕj Hence matrix element for coherent light:
Z S ∗ Hij = ϕi (~r) eF0z ϕj (~r)dV cos ωt (260)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 119 / 142 Hamiltonian of EM fieldI
Hamiltonian for electromagnetic field with generalized momentum
(~p − eA~ )2 ~p 2 e H = + eΦ(~r) ≈ − A~ · ~p + eΦ(~r) (261) 2m 2m m with not too strong vector potential A~ and electrostatic potential Φ, i.e. e ~ time-dependent perturbation m A(~r, t) · ~p irradiated light as plane wave with polarization vector ~:
~ A~ (~r, t) ∼ ~ cos(~k ·~r − ωt) ∼ ~eık·~r (262)
hence the perturbation term
~ e~p · A~ ∼ e~p · ~eık·~r (263)
and the matrix element Z ∗ ı~k·~r e ϕn(~r)~p · ~e ϕm(~r)dV (264)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 120 / 142 Hamiltonian of EM fieldII
approximation for large λ, i.e. ~k ·~r = 2πr/λ 1 compared to extent r of ϕ, series expansion of e:
~ eık·~r ≈ 1 + ı~k ·~r + ... ≈ 1 (265)
matrix element, lowest order
Z ∗ ϕn(~r)e~p · ~ϕm(~r)dV (266)
~ ~p because of ı m = [H,~r ], matrix element (266) is (proof!)
Z En − Em ∗ −~ım ϕn(~r)e~rϕm(~r)dV (267) ~ → contains term e~r → electric dipole moment
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 121 / 142 Matrix elements by symmetry considerationsI
In one dimension, e.g. for ~ k x-axis matrix element I = hψ|x|ψi |ψ(x )|2
Z +∞ I = ψ∗(x) x ψ(x)dx |x → −x −∞ Z −∞ = ψ∗(−x)(−x)ψ(−x)d(−x) + +∞ x Z +∞ = ψ∗(−x)(−x)ψ(−x)d(x) - −∞
because symmetry of wavefunction ψ(−x) = ±ψ(x), → ψ∗ψ invariant under P transformation, so without detailed calculation:
⇒ I = −I ⇒ I = 0 (268)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 122 / 142 Selection rule for the magnetic quantum numberI
Electric dipole matrix element between two different hydrogen wave functions
Z ∗ Iz = ψn0,`0,m0 (~r) z ψn,`,m(~r)dV (269) Z ∗ = dV ψn0,`0,m0 (r, θ, φ) r cos θψ n,`,m(r, θ, φ) (270)
ımφ because of ψ(r, φ) = e ψ(r, 0) (with integer m), for rotations φ0 around z-axis:
0 −ı(m−m )φ0 Iz = e Iz (271)
0 Therefore: either Iz = 0, or, if Iz 6= 0 → m = m
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 123 / 142 Selection rule for the magnetic quantum numberII
Now for x- and y-component:
Z ∗ Ix = dV ψn0,`0,m0 (r, θ, φ) x ψn,`,m(r, θ, φ) (272) Z ∗ Iy = dV ψn0,`0,m0 (r, θ, φ) y ψn,`,m(r, θ, φ) (273) multiply Iy with ı → x + ıy = r sin θ cos φ + r sin θı sin φ
Z ∗ ıφ Ix + ıIy = dV ψn,`,m(r, θ, φ) r sin θe ψn0,`0,m0 (r, θ, φ) (274) again, rotation around z-axis by angle φ0
0 −ı(m−1−m )φ0 Ix + ıIy = e (Ix + ıIy ) (275)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 124 / 142 Selection rule for the magnetic quantum numberIII Therefore: 0 Ix + ıIy = 0 if m 6= m + 1 (276) 0 Ix − ıIy = 0 if m 6= m − 1 (277) Or summarized: if m 6= m0 + 1 and m 6= m0 − 1 then
Ix = 0, Iy = 0 (278)
Selection rules for magnetic quantum number of one-electron systems For light polarized in z-direction:
∆m = 0 (279)
For light polarized in x- or y-direction:
∆m = ±1 (280)
→ corresponds to π (linearly pol.) and σ (circularly pol.) transitions L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 125 / 142 General selection rulesI
In general, use spherical components of dipole moment (for one e−)
1 r (±) = √ (x ± ıy) , r (0) = z (281) 2 hence the product of ~r and polarization vector ~
1 X ~r · ~ = r (+)(−) + r (−)(+) + r (0)(0) = r ν(−ν) (282) ν=−1 trick → write spherical components of ~r with help of spherical harmonics and r = px2 + y 2 + z2 (why?):
r4π r4π r (±) = rY (θ, φ) , r (0) = rY (θ) with (283) 3 1,±1 3 1,0 r 3 r 3 Y = ∓ sin θe±ıφ , Y = cos θ (284) 1,±1 8π 1,0 4π
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 126 / 142 General selection rulesII
For one-electron wave function (w/o spin)
ϕ` ϕ`0 ψ (~r) = Y (θ, φ) , ψ (~r) = Y 0 0 (θ, φ) (285) i r `,m f r ` ,m
(ν) matrix elements of spherical components rfi can be written as integral over three spherical harmonics, which be reduced to an integral over radial wavefunctions with help of Wigner-Eckart theorem:
Z ∞ r Z (ν) ∗ 4π ∗ rfi = ϕ`0 r ϕ`(r)dr dΩY`0,m0 Y1,νY`,m (286) 0 3 Z ∞ ∗ 0 0 0 = ϕ`0 r ϕ`(r)dr F (` , `)h` , m |1, ν, `, mi (287) 0 with Clebsch-Gordan coefficient (CG) h`0, m0|1, ν, `, mi for coupling of initial angular momentum `, m with angular momentum 1, ν of spherical component of ~r to the final state `0, m0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 127 / 142 General selection rulesIII quantum numbers `0, 1, ` must satisfy triangular condition for CG coefficients (selection rule for CG coefficients):
|` − 1| ≤ `0 ≤ ` + 1 (288)
→ implies ∆` ≤ 1, moreover from parity of spherical harmonics
` PY`,m(θ, φ) = (−1) Y`,m(θ, φ) (289) follows that ` + 1 + `0 must be even, otherwise parity of integrand ∗ R Y`0,m0 Y1,νY`,m would be negative → integral dΩ would vanish another selection rule for CG coefficients states that
m + ν = m0 (290)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 128 / 142 General selection rulesIV
Selection rules for angular momentum of one-electron systems
∆` = `f − `i = ±1 , ∆m = mf − mi = 0, ±1 (291) transitions ψi → ψi which do not satisfy (291) are forbidden (in dipole approximation)
Addition: factor F (`0, `) is in this case
p`0/(2`0 + 1) if `0 = ` + 1, F (`0, `) = (292) −p`/(2`0 + 1) if `0 = ` − 1
One-electron wave functions with spin:
(ν) (ν) 0 0 0 rfi = hΦf |r |Φi i = hj ||~r ||jihj , m |1, ν, j, mi (293)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 129 / 142 General selection rulesV where m corresponds to eigenvalues of z-component of total angular momentum ~J = ~L + ~S again: used Wigner-Eckart theorem (WE) to get reduced matrix element hj0||~r ||ji (independent of m and ν) → in general, WE theorem states that for matrix elements of operators in eigenstates of angular momentum operator m and ν dependence is only via CG coefficients of coupling between initial state and component of operator to final state → from selection rules for CG coefficients: Selection rules for total angular momentum of one-electron systems
f i ∆j = jf − ji = 0, ±1 , ∆mj = mj − mj = 0, ±1 (294) j = 0 = 0 , mj = 0 = 0 if ∆j = 0 (295)
→ Interpretation: EM field (photon) carries angular momentum (spin ±1) → angular momentum conservation
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 130 / 142 Selection rules of multi-electron systemsI
Multi-electron wave functions with spin: by applying WE theorem: Selections rules for total angular momentum of multi-electron systems ~ ~ ~ For quantum numbers J, MJ of total angular momentum J = L + S
f i ∆J = Jf − Ji = 0, ±1 , ∆MJ = MJ − MJ = 0, ±1 (296) J = 0 = 0 , MJ = 0 = 0 if ∆J = 0 (297)
for LS coupling: total orbital angular momentum ~L = P ~` and total spin ~ P S = ~s are “good” quantum numbers, selections rules for L, ML
f i ∆L = Lf − Li = 0, ±1 , ∆ML = ML − ML = 0, ±1 (298) L = 0 = 0 , ML = 0 = 0 if ∆L = 0 (299)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 131 / 142 Selection rules of multi-electron systemsII
Selection rules for spin, as the dipole operator e~r doesn’t act on spins: Selections rules for total spin of multi-electron systems
∆S = 0 , ∆MS = 0 (300) holds exact only for weak spin-orbit interaction → LS-coupling of light atoms → already derived for helium: no transition between singlet and triplet system However, for heavy atoms, e.g. with jj-coupling → intercombination possible Intercombination transition for strong spin-orbit interaction
∆S = ±1 , ∆L = 0, ±1, ±2 (301)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 132 / 142 Selection rules of multi-electron systemsIII
Selection rule for parity → remember example in one dimension, |ψ(x )|2 Z +∞ I = ψ∗(x) x ψ(x)dx |x → −x −∞ (302) Z −∞ + ∗ = ψ (−x)(−x)ψ(−x)d(−x) (303) x +∞ Z +∞ - = ψ∗(−x)(−x)ψ(−x)d(x) (304) −∞ we used symmetry of wavefunction ψ(−x) = ±ψ(x), especially equal parity of ψ∗ and ψ to show that ψ∗ψ invariant under P transformation and therefore ⇒ I = −I ⇒ I = 0
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 133 / 142 Selection rules of multi-electron systemsIV
Selection rule for parity of electric dipole transitions
Pi Pf = −1 → change of parity (305)
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 134 / 142 Selection rules of multi-electron systemsV
However, transitions which are forbidden for electric dipole operator (E1) may be allowed by electric 2k pole operator (e.g. E2 electric quadrupole) or Mk magnetic 2k pole operator (e.g. M1 magnetic dipole, M3 magnetic octupole) Selection rule for E2, E4, . . . and M1, M3, . . . radiation
Pi Pf = 1 → same parity of intial and final state (306) for even electric and odd magnetic multipole
Selection rule for E1, E3, . . . and M2, M4, . . . radiation
Pi Pf = −1 → change of parity (307) for odd electric and even magnetic multipole
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 135 / 142 Collisions
Collisional transitions not restricted by selection rules
States for which the radiative transition to any lower state is forbidden are called meta stable. 1 S 1 Po 1 D 3 S 3 Po 3 D 13 17 16 12 14 15 7 11 10 6 8 9 5 3 4 e.g. 1st excited state of He I: 2 triplet (configuration?)
transition to ground state singlet E triplet (singlet) forbidden (by which rule?) but: level is depopulated by 1 collisions 0 1 2 0 1 2
L L If collisional deexcitation negligible (low density) → forbidden lines can be observed, e.g. in planetary nebulae, solar corona
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 136 / 142 Oscillator strengths
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 137 / 142 Oscillator strengthsI
Characterization of electric dipole transitions between Ψi and Ψf by dimensionless quantity from Fermi’s Golden Rule
N (x) 2m X 2 ffi = 2 ~ω|hΨf| xj |Ψii| (308) ~ j=1 in cartesian coordinates for N-electron atom, where ω = Ef − Ei , by summing up over all three components
(x) (y) (z) 2m 2 ffi = f + f + f = ω|hΨf|~r|Ψii| (309) fi fi fi ~ → f is called oscillator strength of the corresponding electric dipole line transition
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 138 / 142 Oscillator strengthsII
Example: Cross section of line transition
Absorption σ(E) of x-polarized photons (~ = ~ex ) by Ψi → Ψf line transition (of energy E)
2 2 2 e ~ (x) σ = 4π f δ(Ef − Ei − E) (310) ~c 2m fi
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 139 / 142 Oscillator strengthsIII
From commutator relation of position and momentum follows sum rule for oscillator strengths over all final states n:
X (x) fni = N (311) n By summing up over x, y, z → Thomas-Reiche-Kuhn sum rule
X X (x) (y) (z) fni = fni + fni + fni = 3N (312) n n
→ typical values of allowed transitions 0.1 < f < 1 → f values of multi-electron systems may be larger than 1 → f values of E2, M1, etc. usually 1
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 140 / 142 Oscillator strengthsIV
f -values of Lyman lines Analytic expression (Menzel & Pekeris 1935):
29n05(n0 − 1)2n0−4 g f 0 = (313) 1 n 1 3(n0 + 1)2n0+4
What is value of g1? For Lyα (1 → 2): f = 0.4162 What is the detailed transition of Lyα (angular momentum)?
L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2016/2017 141 / 142 Literature
D. Griffiths: Introduction to Quantum Mechanics H. Hänsel, W. Neumann: Atome · Atomkerne · Elementarteilchen H. Haken, H. C. Wolf: Atom- und Quantenphysik H. Friedrich: Theoretische Atomphysik J. D. Landstreet, 1982, ApJ, 258, 639
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