(relativistic effects kinetic energy & spin-orbit coupling) 3. Hyperfine structure: ) (spin-spin coupling of e & p + magnetic moments) 4.

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1 4 Time-ind. Perturbation Theory II We said we solved the Hydrogen atom exactly, but we lied. There are a number of physical effects our solution of the Hamiltonian H = p /m e /r left out. We already said motion of proton can be accounted for by nearly trivial reduced mass correction not important, since spectra of original problem unchanged except for tiny overall energy scaling. In fact there are a hierarchy of other effects we left out, each of which can be treated in perturbation theory on top of the other. In the most complete theory, almost all degeneracies are split. Classify formally in powers of fine structure constant α e / hc /37.. Bohr levels: E n = R/n O(α mc ) ( n j+/ 3 4. Fine structure: R α O(α 4 mc ) n 4 (relativistic effects kinetic energy & spin-orbit coupling) 3. Hyperfine structure: ) µ 0 ge 3πm p ma 3 0 S p S e O(m/m p )(α 4 mc ) (spin-spin coupling of e & p + magnetic moments) 4. Lamb shift: 3π R n ( α n ) 3 ln mc ( E m E n ) avg O(α 5 mc ) (quantization of photon field really larger than hyperfine) 4. Fine Structure spin-orbit coupling st effect of interest is coupling of i) spin of electron to ii) its motion, things which we certainly treated independently before. Full derivation of proper ˆV is hairy, give only hand-waving derivation here, misses factor of. Force on electron is ee = e r/(4πɛ 0 r 3 ). Recall that observer moving in electric field E sees magnetic field

2 B = v c E v c () So electron with magnetic moment µ = g e S, m g () should have energy so with L r mv, get ˆV = µ B (3) = g e m S e 4πɛ 0 r 3r v (4) c ˆV = ge m c 4πɛ 0 r 3 L S (5) Off by factor of from full relativistic calculation (neglected Thomas precession!): Now define total angular momentum ˆV = ge 4m c 4πɛ 0 r3l S (6) (7) = f(r)l S (8) J = L + S (9) whose square is so Ĵ = ˆL + Ŝ + L S, (0) ˆV = f(r) (Ĵ ˆL Ŝ ) () Verify: Ĵ, ˆL, Ŝ, and Ĵz now commute with H = H 0 + ˆV. ˆLz and Ŝz no longer do. We choose therefore to label states as

3 Thus we aren t quite following prescription we set out, by looking at expectation value of ˆV in original eigenstates of H 0. Why? Because we can find the new exact eigenstates with no work. So energy shift calculation is exact except for higher order relativistic effects. In this basis we have (Ĵ ˆL Ŝ ) nlsjm j = h (j(j + ) l(l + ) 3 4 ) nlsjm j () so energy shift due to spin-orbit coupling is For f(r) we need δe so = nlsjm j ˆV nlsjm j = h (j(j + ) l(l + ) 3 ) f(r) (3) 4 r 3 = l(l + /)(l + )n 3 a 3 0 (l 0) (4) (see Griffiths Probs. 6.34, 6.35). Spin-orbit energy shift is therefore δe so = E n mc = E n mc n(j(j + ) l(l + ) 3 4 ) l(l + (5) )(l + ) n l + j + (check for j = l ± /!) (6) Note although (5) looks singular at l 0, proper relativistic treatment 3

4 by Dirac shows nonsingular result (6) is correct for all l not obvious though! Relativistic kinetic energy corrections. Up to now we have neglected relativistic effects in kinetic energy term ˆT. Properly we should probably write ˆT = ˆp c + m c 4 mc ˆp m ˆp4 8m 3 c +... (7) Then we need to find the energy shift in ψ to leading order in the perturbation ˆV = ˆp4 8m 3 c : δe rel = 8m 3 c ψ ˆp4 ψ = 8m 3 c ˆp ψ ˆp ψ (8) The rhs supposed to be evaluated using unperturbed wave fctns, so use (ˆp /m e /(4πɛ 0 r)) ψ = E n ψ to find δe rel = mc [E n + E n e + e 4 ] (9) 4πɛ 0 r (4πɛ 0 ) r The expectation values depend only on a 0, n and l as usual: = ; = r n a 0 r (l + /)n 3 a 0 (0) So full relativistic kinetic energy shift is δe rel = E n 4n mc l + / 3 () and full fine structure energy splitting is δe fine = δe so + δe rel () yielding an energy spectrum for hydrogen up to fine structure corrections of E = E n + δe fine = R + α n n n j + / 3/4 (3) 4

5 4. Hyperfine Interaction Figure : Hydrogen fine structure spin-orbit coupling Hyperfine splitting of atomic spectral lines arises from interaction of electron & nuclear dipole moments. Whereas ground state in hydrogen not split by fine structure, it is split by hyperfine interaction, allowing for transition between two levels, producing radiation of wavelength cm ( 0 7 Ryd, allows tracking of hydrogen by astronomers, measurements of proper motion of celestial objects (Doppler shifts of line), & construction of stable atomic clocks. Calculation is lengthy, but can be solved by elementary methods to high accuracy ( real quantum mechanics problem!). Magnetic field due to dipole moment µ: B = 3r( µ r) µ 4πɛ 0 r 5 r 3, (4) whereas the interaction of a magnetic moment with external field is U = µ B (5) 5

6 Figure : If µ p ẑ, r ˆx, B ẑ, then µ e µ p in lowest energy configuration. If µ p ẑ, r ẑ, B ẑ, then µ e µ p for lowest energy. Compare with (5). so interaction is where r = r e r p ˆV = 4πɛ 0 3( µ e r)( µ p r) + ( µ e µ p ) r 5 r 3 (6) Magnetic moment operators: µ e = g e e M e S e ; µ p = g p e M p S p (7) with gyromagnetic ratios g e.00, g p Perturbation theory Want to use ˆV in (5) as perturbation, write H = H 0 + H fine + ˆV, look for hyperfine energy shift δe hf in ground state ψ 0. Can write ground state as ψ 0 = ψ 00 (r) m e m p (8) where st factor is spatial part, old friend ψ 00 = e r/a 0/ πa 3 0, nd factor represents spin state of electron and now proton, too (before we ignored proton spin since it didn t contribute to energy, now we must include it as Note we will focus exclusively on the ground state of the H-atom, where the spin-orbit coupling has already been accounted for. Although in general states of definite m p, m e are not eigenstates because of the L S coupling, for the S / ground state, m j (e value of Ĵ z / h) is the same as m s (e value of Ŝz/ h) always since l = 0. We call m s of electron m e here, and m s for proton m p. 6

7 dynamical variable). Note there are four degenerate states before taking ˆV into account, since both m e and m p can be ±/. A priori we expect that since ˆV commutes with Ŝ = (S e + S p ), eigenstates of new H with hyperfine term should be eigenstates of Ŝ, i.e. either S = (triplet) or S = 0 (singlet). cm spectral line we re looking for will be difference in shifts of 3 degenerate triplet states [,, + )/ ], and singlet state )/. So we want to evaluate 3 = g eg p e 6πɛ 0 M e M p ψ 00 (r) SM δe hf = ψ 0 ˆV ψ 0 (9) ( 3(Se r)(s p r) (S e S p ) r 5 r 3 ) SM ψ 00 (r) = g eg p e 6πɛ 0 M e M p I αβ SM ŜeαŜpβ SM (30) where the SM are e states of total spin Ŝ and total Ŝz, and I αβ d 3 r ψ 00 (r) 3r α r β r 5 δαβ r 3 (3) is a tensor we have to evaluate using the wave function ψ 00. Tedious but straightforward to show that I αβ = Iδ αβ, with I = 8π ψ 00 (0) /3 (see below!). I diagonal = only spin matrix elements we need are SM S e S p SM = SM (Ŝ Ŝ e Ŝ p) SM = h = (+) (+ ) (+ ) S = 0 (+ ) (+ ) S = 0 h /4 S = 3 h /4 S = 0 (3) So energy shifts for singlet and triplet states are Check that Ŝ commutes with ˆV by first writing Ŝ = (S e + S p) = Ŝ e + Ŝ p + S e S p. First of all, Ŝ e commutes with ˆV since [Ŝ e, Ŝe,α] = 0, similarly for Ŝ p. Now S e S p obviously commutes with second term of ˆV (see (5)), harder part is to see [S e S p, (S e r)(s p r)] = 0. To see this, use relation σ i σ j = δ ij + iɛ ijk σ k. 3 Need to use states ψ 0 > in which ˆV is diagonal these are precisely the SM! 7

8 δe hf = g e g p e 6πɛ 0 M e M p = g eg p e 8 M e 4πɛ 0 M p h 4 8π ψ 00 (0) 3 8Me 3 e 6 3 h 6 S = 3 S = 0 h /4 S = 3 h /4 S = 0 (33) So splitting between triplet and singlet (energy of photon emitted in triplet decay) is and frequency of transition is hω = δe hf (S = ) δe hf (S = 0) = g eg p e 8 M e 6πɛ 0 M p h 4 (34) ν = ω/π = g eg p e 8 Me π ɛ 0 M p h 5 (35) = ± 0.08Hz (36) or λ = c/ν =. cm. Value given is current measurement! Pf. of I αβ = 8π ψ 00 (0) δ αβ /3 Recall definition I αβ d 3 r ψ 00 (r) ( 3rα r β r 5 δ ) αβ r 3 (37) Quantity I αβ is singular integral and must be handled carefully. Note 8

9 . Off-diagonal terms vanish because in l = 0 state answer must be invariant under rotation of coordinate system, so I αβ = Iδ αβ.. Diagonal terms funky, since angular part of, e.g. I zz gives dω (3z r ) = πr d cos θ(3 cos θ ) = 0 (38) but radial integral is proportional to dr e r/a 0 r = (39) So we have to find way to take limit 0 sensibly. Note singularity in (38) comes entirely from r = 0; if we were to exclude r = 0 from range of integration integral would vanish!. So consider very small sphere around r = 0, of radius r 0 so small we can approximate smoothly varying fctn. ψ 00 (r) by ψ 00 (0). Consider diagonal elt. I zz (others must be same by rotation symmetry), express in cylindrical coords: ( I zz = ψ 00 (0) r0 3z d 3 r 0 r ) 5 r 3 = ψ 00 (0) r0 [ r0 dz z dρ 3z ] π r 0 0 (ρ + z ) 5/ (ρ + z ) 3/ [ = ψ 00 (0) r0 z ] ρ dz π r 0 (ρ + z ) + =r0 z 3/ (ρ + z ) / ρ =0 ( ) = ψ 00 (0) r0 dz π z r 0 r 0 r 0 = 8π 3 ψ 00(0) (40) 9