Fine structure in hydrogen - relativistic effects

Transcription

1 LNPhysiqueAtomique016 Fine structure in hydrogen - relativistic effects Electron spin ; relativistic effects In a spectrum from H (or from an alkali), one finds that spectral lines appears in pairs. take a Na spectrum as example: Moreover, the rays are slightly shifted in comparison with the non-relativistic theory The origins of these new effects: Electron spin Relativity Relativistic Hamiltonian Recall the one-electron Hmiltonian: H = H kin + V = p m Ze 4 " 0 r 1

2 LNPhysiqueAtomique016 So far, we have treated the term p m classically For a more exact solution, this has to be replaced with a relativistic version This gives the Dirac equation An analytical solution is possible, but very complex Instead, we treat the problem with perturbation theory

3 LNPhysiqueAtomique016 Perturbative treatment As a zero-order Hamiltonian, we take the nonrelativistic version We the treat the relativistic corrections as perturbations H = H 0 + H 0 = ~ m r + V (~r)+h 0 The corrections to the energy levels: E = h 0 H 0 0 i It turns out that the relativistic corrections can be divided into three parts: H 0 = H SO + H rel + H Darwin H SO : Spin-orbit interaction H rel : Relativistic treatment of the kinetic energy H Darwin : the Darwin term This is consistent with the exact treatment 3

4 LNPhysiqueAtomique016 Relativistic treatment of the kinetic energy Classical kinetic energy : H rel E 0 kin = p m Relativistic kinetic energy : E kin = p p c + m c 4 mc = p m p 4 8m 3 c +... First order correction (ignoring terms of order or higher) : H rel = p 4 8m 3 c = ~ 4 8m 3 c r4 v c 4 This does not depend on spin It is diagonal in n and l E rel,nl = E n0 (Z ) where = e 4 " 0 c ~ n 3 4 is the fine-structure constant E rel is more important for small n (small n small orbital radius high velocity) 4 l + 1 4

5 LNPhysiqueAtomique016 The Darwin term H Darwin Very difficult to explain Related to the singularity at r =0 ) ( H Darwin = ~ m c Ze 4 " (r) (Z ) E Darwin,nl = E n,0 n ; l =0 E Darwin,nl = 0 ; l 6= 0 5

6 Spin-orbit interaction LNPhysiqueAtomique016 H SO Interaction between the orbital angular momentum and the spin Spin An electron has a magnetic moment, which can be associated with a spin Wave function : s,m s Vectorial representation : sm s i The sm s i are eigenvectors to the operators: S sm s i = s(s + 1)~ sm s i S z sm s i = m s ~ sm s i For a single electron, we have (always): - s = 1 ) S = s(s + 1)~ = 3~ 4 = p 3~! - m s = ± 1 ) hs z i = ± ~ spin-up and spin-down 6

7 LNPhysiqueAtomique016 The total wave-function : nlml m s nlm l m s i Interaction between ~ l and ~s The electron has a magnetic moment : ~µ / ~s Consider a system of reference centered at the electron an orbiting proton (positive charge) An orbiting charge induced magnetic field : ~ B / ~ l 7

8 LNPhysiqueAtomique016 Interaction between ~ B and ~µ H SO = ~µ ~B / ~ l {z ~s } / L ~ {z ~S } vectors operators H SO is separable in radial and angular coordinates H SO = (r) ~ L ~S where (r) = 1 1 m c r dv dr For example, the hydrogenic potential: V = Ze 4 " 0 r ) (r) = 1 m c Ze 4 " 0 1 r 3 The energy shift due to the interaction is the expectation value of H SO h H SO i = h R nl (r) (r) R nl (r) i D lm l m s L ~ ~S E lm l m s h (r)i = 1 m c Ze 4 " 0 1 r 3 = 1 Ze Z 3 m c 4 " 0 a 3 0 n3 l(l + 1 )(l + 1) 8

9 Fine structure LNPhysiqueAtomique016 For the angular part of H SO : we have to look for eigenvectors to h ~ L ~Si What about the vector : lm l m s i? (eigenvector to the operators L, L z, and S z ) h i This will NOT do, since ~L S,Lz ~ 6= 0 and h i ~L S,Sz ~ 6= 0 We need some other operator (and quantum number) The total angular momentum We introduce : ~J = ~ L + ~ S J = L + ~ L ~S + S ) ~ L ~S = 1 J L S Consider the wave functions nljm j ( nljm j i ) that are eigenstates to H 0,L,J and J z : H 0 nljm j i = E 0 nljm j i L nljm j i = l(l + 1)~ nljm j i J nljm j i = j(j + 1)~ nljm j i J z nljm j i = m j ~ nljm j i 9

10 LNPhysiqueAtomique016 For hydrogen, we have one single electron ( s = 1 m s = ± 1 ( j = l ± 1, (l 6= 0) j = 1, (l = 0) j is a good quantum number it makes the total Hamiltonian diagonal m l and m s are NOT good quantum numbers due to the spin-orbit interaction, ~ L and ~ S will precess around each other thus, their projections are not constant The sum, ~ J, IS constant 10

11 LNPhysiqueAtomique016 The Fine-structure energy Expectation value of the angular part of the hamiltonian: D E ~L S ~ = 1 J L S = 1 = ~ ljmj J L S ljm j [j(j + 1) l(l + 1) s(s + 1)] with s = 1 : D ~L ~ S E = ~ apple j(j + 1) l(l + 1) 3 4 hh SO i = h (r)i D ~L ~ S E = 1 Ze Z 3 m c 4 " 0 = 1 apple j(j + 1) l(l + 1) ~ a 3 0 n3 l(l + 1 )(l + 1) 3 4 apple j(j + 1) l(l + 1) 3 4 The two possible values of m s ( ± 1 ) gives two possible values for j ( l + 1 and l 1 ) The energy level E 0 is split into a doublet E SO = H + SO HSO = l

12 LNPhysiqueAtomique016 Energy levels in hydrogen and spectroscopy Spectroscopic notation There is a convention for how to annote quantum numbers l-quantum numbers are described by a letter The s-quantum number does not need description The j-quantum number is given by its numerical value The combination of n and l (nl) is referred to as an orbital For a many-electron atom, the electron configuration is the list of all orbitals ( n 1 l 1,n l,... ) Coding for l : l =0! s l =1! p l =! d l =3! f l =4! g For a multi-electron atom, total angular momenta have to be defined. 1

13 LNPhysiqueAtomique016 These are given quantum numbers with capital letters: ~L = ~ l 1 + ~ l +... quantum numbers L and M L ~S = ~s 1 + ~s +... quantum numbers S and M S ~J = ~ l 1 + ~s 1 + ~ l + ~s +... quantum numbers J and M J Coding for L : L =0! S L =1! P L =! D L =3! F Coding for S (multiplicity) : S =0! (S + 1) = 1! singlet S =1/! (S + 1) =! doublet S =1! (S + 1) = 3! triplet S =3/! (S + 1) = 4! quartet J is given as its number (for a multi-electron atom, the definition of J is ambiguous) The L and S together gives the atomic term S+1 L (examples : 3 S, P, 4 D, 1 F ) (not always a good description) J is the fine structure level S+1 L J (examples : 1 S 0, P 1/ ) 13

14 LNPhysiqueAtomique016 Ground state Energy levels in hydrogen n =1 ) l =0 s =1/ ) j =1/ ) 1s S 1/ (1 level) Excites states - n =) l =0 l =1,s=1/ s ) j =1/ ) s S 1/ j =1/ ) p p ) P 1/ j =3/ ) p P 3/ 9 = ; (3 levels) - n =3) 8 < : l =0 l =1 l =,s=1/ 3s ) 3s S 1/ 3p ) 3p P 1/ and 3p P 3/ 3d ) 3d D 3/ and 3d D 5/ 9 = ; (5 levels) 14

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