hydrogen atom: center-of-mass and relative

Transcription

1 hdrogen atom: center-of-mass and relative apple ~ m e e -particle problem (electron & proton) ~ m p p + V ( ~r e ~r p ) (~r e, ~r p )=E (~r e, ~r p ) separation in center-of-mass and relative coordinates R = m er e + m p r p m e + m p r = r e r p M = m e + m p µ = m em p m e + m p apple ~ ~ M R S(R) =E CM S(R) µ r + V ( ~r ) (~r) =E H (~r) (~r e, ~r p )=S( ~R) (~r) and E = E CM + E H

2 hdrogen atom: spherical separation apple ~ relative motion µ r + V ( ~r ) (~r) =E H (~r) ~ µ d dr + ~ µ spherical smmetr (~r) = u(r) r Y l,m (, ) l(l + ) r e 4 0 r u(r) =E H u(r) dimensionless units: ρ=κr with κ =m E /ħ and ρ0= me / (4πε0 ħ κ) d l(l + ) d + 0 u( ) =0

3 hdrogen atom: radial solution ansat (solve asmptotics) differential equation for L(s): recursion for coefficients u( ) = l+ w( ) e d w + (l + )dw d d +( 0 (l + ))w =0 X ansat: power series w( ) = a k k normaliabilit: recurrence must terminate at some finite k n l + k=0 a k+ = (k + l + ) 0 (k + )(k +l + ) a k

4 radial functions unl(r) = r Rnl(r) u, u,0 u, u 3,0 u 3, u 3, u 4,0 u 4, u 4, u 4,

5 radial functions Rnl(r) R, R,0 R, R 3,0 R 3, R 3, R 4,0 R 4, R 4, R 4,

6 periodic table s s s 3s 4s 5s 6s 7s 3d 4d 5d 6d p 3p 4p 5p 6p 4f 5f

7 atom in spherical mean-field approimation 00 Fe : [Ar] 3d 6 4s 4p 0 0 Z eff (r) r in Å

8 Atom- und Hbrid-Orbitale s sp sp p p p sp 3 sp 3 4 d d d d 3 - d -

9 numerical differentiation task: evaluate f (), onl knowing f() at some specified abscissae i idea: approimate f b a function that can be easil differentiated, e.g., a Talor epansion. Then combine the f(i) such that ecept for the desired derivative as man terms as possible are cancelled. eample: first derivative f ( 0 + h) = f ( 0 )+hf 0 ( 0 )+ h f 00 ( 0 )+ h3 6 f 000 ( 0 )+O(h 4 ) f ( 0 ) = f ( 0 ) f ( 0 h) = f ( 0 ) hf 0 ( 0 )+ h f 00 ( 0 ) h 3 6 f 000 ( 0 )+O(h 4 ) Then f ( 0 + h) f ( 0 h)=hf 0 ( 0 )+ h3 3 f 000 ( 0 )+O(h 4 ) or f 0 ( 0 )= f ( 0 + h) f ( 0 h) + O(h 3 ) h

10 numerical differentiation differences of similar numbers in numerator & small denominator 0 0 O(h n ) error from approimation abs(error) h ε/h error from finite machine accurac ε

11 method of undetermined coefficients idea: given a set of abscissae n, e.g., n=0+nh, n=-, 0,,, make an ansat with undetermined coefficients, e.g.,: f 0 ( 0 )= f ( )+ 0 f ( 0 )+ f ( )+ f ( ) h determine the coefficients αi b requiring that the formula differentiates polnomials of order, e.g., 0 to 3 eactl b solving the resulting sstem of linear equations O O O O Maple session: with linalg : n d : # formula for nth derivative mesh d 0Kh, 0, 0 C h, 0 C * h ; p d nops mesh : # abscissae i mesh := 0 K h, 0, 0 C h, 0 C h f_ d arra seq map / ^k, mesh, k = 0..pK ; # evaluate monomials on mesh f_ := 0 K h 0 0 C h 0 C h 0 K h 0 0 C h 0 C h 0 K h C h 3 0 C h 3 O O der_f d arra seq binomial k, n * n!* 0^ kkn, k = 0..pK ; # derivative of monomials der_f := coefficients d linsolve f_, der_f ; # coefficients alpha i coefficients := K 3 h K h h K 6 h

12 Numerov method one-dimensional Schrödinger equation u 00 ()+k ()u() =0 wherek () = m ~ (E V ()) numerical derivative f 00 ( 0 )= f ( 0 + h) f ( 0 )+f ( 0 h) h h f (4) ( 0 )+O(h 4 ) two-point iteration of wave function u(j): u j+ =( h k j ) u j u j + O(h 4 ) Numerov trick: remove leading error in derivative formula b using Schrödinger equation h u(4) ( j )+O(h 4 )=+ h d d k ( j )u( j ) + O(h 4 )= k j+ u j+ kj u j + kj u j Numerov iteration: u j± = ( 5h k j /)u j ( + h k j /)u j +h k j± / + O(h 6 ) + O(h 4 )

13 Numerov iteration close to eigenvalue

14 (in)stabilit of Numerov iteration

15 indistinguishabilit and statistics N-particle sstems described b wave-function with N particle degrees of freedom (tensor space): Ψ(,..., N) introduces labeling of particles indistinguishable particles: no observable eists to distinguish them in particular no observable can depend on labeling of particles consider permutations P of particle labels P (, )= (, ) with (, ) = (, ) P (, )=e i (, ) when P = Id e iɸ = ± (Ψ (anti)smmetric under permuation) antismmetric: Ψ(, ) = 0 (Pauli principle)

16 spin-statistics connection bosons (integer spin): smmetric wave-function fermions (half-integer spin): anti-smmetric wave-function Fenman Lectures III, 4-: Wh is it that particles with half-integral spin are Fermi particles whose amplitudes add with the minus sign, whereas particles with integral spin are Bose particles whose amplitudes add with the positive sign? We apologie for the fact that we cannot give ou an elementar eplanation. An eplanation has been worked out b Pauli from complicated arguments of quantum field theor and relativit. He has shown that the two must necessaril go together, but we have not been able to find a wa of reproducing his arguments on an elementar level. It appears to be one of the few places in phsics where there is a rule which can be stated ver simpl, but for which no one has found a simple and eas eplanation. The eplanation is deep down in relativistic quantum mechanics. This probabl means that we do not have a complete understanding of the fundamental principle involved. For the moment, ou will just have to take it as one of the rules of the world.

17 permutations in lower dimensions M. Berr et al.: spin-statistics connection from geometric phase when permuting particles along paths? t dimensional dimensional P Id: braiding statistics: anons fermions cannot pass

18 -particle wave-function: distinguishable two particles in (different) ortho-normal single-particle states φa() and φb() (, )=' a ( )' b ( ) or (, )=' b ( )' a ( ) epectation value of particle distance: M = (-) ( ) = + normalied = R d ' a( ) R d ' b ( ) = a = R R d ' a ( ) d ' b( ) = b = R d ' a ( ) R d ' b ( ) = hi a hi b ( ) = a + b hi a hi b = ( ) = ( ) observable does not distinguish particles

19 -particle wave-function: indistinguishable hmi ± = smmetric / anti-smmetric wave-function ±(, )= p (, ) ± (, ) cross-terms between product wave-functions hmi ±h M i±h M i+hmi observable does not distinguish particles = hmi ±h M i particle permutation: echange-terms orthogonal R = d ' a( )' b ( ) R d ' b ( )' a ( )= ab 0 R = d ' a ( )' b ( ) R d ' b( )' a ( )= 0 ba R = d ' a ( )' b ( ) R d ' b ( )' a ( )= hi ab hi ba ( ) ± = a + b hi a hi b hi ab Bosons prefer compan Fermions keep their distance

20 probabilit densit for particles in a bo φb() φa() distinguishable smmetric anti-smmetric

21 How about electrons on the moon? in principle we need to antismmetrie the wave-function for all electrons in the universe reall? product states of states with ero overlap will not give an echange contribution Z h M i = d d ' a ( )' b ( ) M(, ) ' b ( )' a ( ) ero overlap makes electrons actuall distinguishable b their coordinate in practice: can eclude electrons with negligible overlap from antismmetriation more practical eample: spin need not antismmetrie electrons of different spin when we are onl interested in observables that do not change spin