Helium and two electron atoms

Transcription

1 1 Helium and two electron atoms e 2 r 12 e 1 r 2 r 1 +Ze Autumn 2013 Version:

2 2 (1) Coordinate system, Schrödinger Equation 3 slides Evaluation of repulsion term 2 slides Radial Integral - details 3 slides Two electrons and spin 1 slide Figures -helium spectra 3 slides Symmetry and Antisymmetry of wavefunctions 2 slides in Para... Parahelium and Orthohelium 4 slides + 1 Parahelium and Orthohelium in literature 1 slide + 1 Approximations to describe helium atom 3 slides Helium atom - Exchange interaction 1 slide Table - binding energies (or with discussion ) 2+1 slides Variational method and proof φ H φ E g.s slides Variational method with correlated function 1 slide Configuration mixing 2 slides Doubly excited states of Helium 3 slides + 4 Spin - Spin interaction - Ferromagnetism Atomic Units and formulae 2 slides optional

3 1 Description of Helium - 2 electron atom 3 A helium atom consists of a nucleus of charge +2 surrounded by two electrons. The coordinates are shown in the drawing e 2 r 12 e 1 r 2 r 1 +Ze The total energy is kinetic energy of the electrons (2 terms), interaction of electrons with the nucleus (2 terms) and the repulsion between the 2 electrons (1 term)

4 4 H = T 1 (r 1 ) + V 1 (r 1 ) + T 2 (r 2 ) + V 2 (r 2 ) + V 12 (r 1, r 2 ) e 2 r 12 e 1 r 2 r 1 +Ze [ h2 r 2 2m 1 e Z e2 h2 r 2 r 1 2m 2 e Z e2 e 2 ] + Ψ (r 1, r 2 ) = E Ψ (r 1, r 2 ) r 2 r 1 r 2

5 The wavefunction is a function of 6 variables, r 1 and r 2, i.e. x 1, y 1, z 1, x 2, y 2, z 2. Ψ (r 1, r 2 ) The details of the operators: H = T 1 (r 1 ) + V 1 (r 1 ) + T 2 (r 2 ) + V 2 (r 2 ) + V 12 (r 1, r 2 ) T 1 (r 1 ) h2 r 2 2m 1 e T 2 (r 2 ) h2 r 2 2m 2 e V 1 (r 1 ) = Z e2 r 1 Z e2 r 1 V 2 (r 2 ) = Z e2 r 2 e2 V 12 (r 1, r 2 ) = + r 1 r 2 + e2 r 12 Last term in the potential is repulsion between the electrons. Because of this term potential is not spherically symmetric in any two of the r 1, r 2. There are 4 terms which work only on r 1 or r 2, but the repulsion term mixes the two sets of variables 5

6 2 Two electrons and spin 6 Spin is a new degree of freedom, discovered already in 1920 s (W. Pauli) Even if spin is not present in our Hamilton operator, it plays a very important role in the spectra. Parahelium and Orthohelium We will show: Parahelium - singlet - includes the ground state Orthohelium - triplet - excited states (transitions forbidden ) Addition of spin (or angular momentum): S = s 1 + s 2 L = l 1 + l 2 J = j 1 + j 2 J 2 = J J J 2 j 1 j 2 ; JM = J(J +1) j 1 j 2 ; JM j 1 j 2 J j 1 +j 2

7 contrast, in the lectrons at the p. This behav- For the triplet nal, where the us; for the sinsed probability wo electrons to p in the singlet e 1a 0 from the 7 of the singlet lation can drain an atom: in he singlet state, states have the al occupations), han for the sinons of the eleccharges of the Computational Chemistry Figure 2. The lowest helium energy levels, calculated from experimental IPs. DOI /jcc Figure 1:

8 Figure 2: Para- and Ortho - Helium 8

9 9 Consider independent particles (prod- Symmetry - Exclusion Principle uct functions) without symmetry (exclusion principle): Pauli principle: Including spin variables ξ 1, ξ 1, Ψ(1, 2) = φ a (1)φ b (2) (2) Ψ(1, 2) = [ φ a (1)φ b (2) φ b (1)φ a (2) ] / 2 Ψ(1, 2) Ψ(r 1, ξ 1, r 2, ξ 2 ) Thus the total (independent) two particle wave function should be: Ψ(r 1, ξ 1, r 2, ξ 2 ) = [ φ a (r 1 )χ α (ξ 1 )φ b (r 2 )χ β (ξ 2 ) φ b (r 1 )χ β (ξ 1 )φ a (r 2 )χ α (ξ 2 ) ] / 2 But this is not the case; since spin and space are independent, we must at the same time have Ψ(1, 2) = Φ space (1, 2) Ξ spin (1, 2) How to combine these two - make each antisymmetric?

10 10 The total function must be product of space and spin part and change sign on exchange 1 2 Ψ(1, 2) = Φ Sym (r 1, r 2 ) Ξ Asym (ξ 1, ξ 2 ) Ψ(1, 2) = Φ Asym (r 1, r 2 ) Ξ Sym (ξ 1, ξ 2 ) This means in detail for the two types: symmetric space, antisymmetric spin ( ) ( ) [ φ a (r 1 )φ b (r 2 ) + φ b (r 1 )φ a (r 2 )] 2 [χ α (ξ 1 )χ β (ξ 2 ) χ β (ξ 1 )χ α (ξ 2 )] or antisymmetric space, symmetric spin ( ) ( ) [ φ a (r 1 )φ b (r 2 ) φ b (r 1 )φ a (r 2 )] 2 [χ α (ξ 1 )χ β (ξ 2 ) + χ β (ξ 1 )χ α (ξ 2 )] The mixed antisymmetric function mentioned first Ψ(r 1, ξ 1, r 2, ξ 2 ) = [ φ a (r 1 )χ α (ξ 1 )φ b (r 2 )χ β (ξ 2 ) φ b (r 1 )χ β (ξ 1 )φ a (r 2 )χ α (ξ 2 ) ] / 2 describes space and spin coupled. That is spin-orbit coupling or j-j coupling; in heavy atoms, not in helium.

11 11 There are four product combinations: χ (ξ 1 )χ (ξ 2 ), χ (ξ 1 )χ (ξ 2 ), χ (ξ 1 )χ (ξ 2 ) and χ (ξ 1 )χ (ξ 2 ) They can be combined to three symmetric (triplet) and a single one antisymmetric (singlet) χ (ξ 1 )χ (ξ 2 ) χ (ξ 1 )χ (ξ 2 ) 1 2 [χ (ξ 1 )χ (ξ 2 ) + χ (ξ 1 )χ (ξ 2 )] 1 2 [χ (ξ 1 )χ (ξ 2 ) χ (ξ 1 )χ (ξ 2 )] We can also use a more compact notation (1) (2) (1) (2) 1 2 [ (1) (2)+ (1) (2)] 1 2 [ (1) (2) (1) (2)]

12 12 3 Why are the Ortho-helium states lower in energy than the Para-helium states?

13 13 The SPATIAL symmetric combination Ψ S (r 1, r 2 ) = 1 2 [ φ a (r 1 )φ b (r 2 ) + φ b (r 1 )φ a (r 2 ) ] e 2 r 1 r 2 when r 1 r 2 Ψ S (r, r) 1 2 [ φ a (r)φ b (r) + φ b (r)φ a (r) ] 2φ a (r)φ b (r) S = 0 i.e. close to maximum. The the repulsion as large as possible, in the S = 0 singlet The SPATIAL antisymmetric combination (spin must be thus symmetric; triplet) Ψ A (r 1, r 2 ) = 1 2 [ φ a (r 1 )φ b (r 2 ) φ b (r 1 )φ a (r 2 ) ] when r 1 r 2 Ψ A (r, r) 1 2 [ φ a (r)φ b (r) φ b (r)φ a (r) ] 0 S = 1 Repulsion term - exchange interaction Singlet, S = 0, space symmetric Repulsion large Triplet, S = 1, space antisymmetric Repulsion small Can also be related to the sign of Exchange term

14 4 Various formulations found in literature 14 The probability for small separations of the electrons is less for anti-symmetric than for symmetric space wave functions. (our formulation is better! ) If then the electrons are further apart on average, there will be less shielding of the nucleus by the ground state electron,and thus the excited state electron will be more exposed to the nucleus. This implies that the ortho-helium states will be more tightly bound and of lower energy. (our formulation is much better and much more correct! ) Helium energy levels (singly excited) 1) np state spin anti-parallel to the spin of 1s state: S=0, singlet state, Para-helium 2) np state spin parallel to the spin of 1s state: S=1, triplet state, Ortho-helium The S = 0 and S = 1 difference can also be related to the sign of Exchange term

15 5 Spin - Spin interaction - Ferromagnetism 15 The triplet - singlet difference in energy acts as a result of an effective spin-spin interaction. When spins are parallel (S = 1) repulsion weaker - i.e. effective attraction S = s 1 + s 2 S 2 = S S = (s 1 + s 2 ) (s 1 + s 2 ) = s s s 1 s 2 S S 2 S = S(S + 1) = S s 1 s 2 S S s 1 s 2 S = We see then that for the triplet 1 and singlet 0 S(S + 1) s 1 s 2 1 = s 1 s 2 0 = 3 4 We now attempt to write the two energies E(S) as function of S s 1 s 2 S E(S) = A B S s 1 s 2 S E(1) = A B 1 4 E(0) = A + B 3 4 It is easily seen that the energy constants A and B must be A = 3 4 E(1) + 1 E(0) B = E(0) E(1) (B > 0) 4 The energy difference E(0) - E (1) can be seen as a result of an effective interaction V (s 1, s 2 ) = A B (s 1 s 2 )

16 16 As one would expect, s 1 s 2 is positive for triplets S=1, the spins are parallel, and negative for S=0 singlet, the spins are antiparallel, as we have seen above in 1 s 1 s 2 1 = s 1 s 2 0 = 3 4 The energy difference is caused by the electrostatic repulsion term, being suppressed in the triplet (S=1) case due to the space-antisymmetry, or we can say Pauli principle. The nature of ferromagnetism is such spin-spin interaction, i.e. the correlation of spins is caused by electrostatic (atomic) interactions (that is why it is so strong). This correlates the spin magnetic moments, so that a magnetised material has a magnetic moment. The magnetic spin-spin interactions is much weaker and could not cause a permanent correlation.

17 6 Approximations to describe helium atom To obtain simplified solutions the Schrödinger we first disregard the repulsion term ( and then treat the repulsion using approximations ) [ h2 r 2 2m 1 e Z e2 h2 r 2 r 1 2m 2 e Z ] e2 + e2 Ψ (r 1, r 2 ) = E Ψ (r 1, r 2 ) r 2 r 1 r 2 Independent electron approach wave function is separable [ h2 r 2 2m 1 Z e2 h2 r 2 e r 1 2m 2 Z ] e2 Ψ 1 (r 1 ) Ψ 2 (r 2 ) = (E 1 +E 2 ) Ψ 1 (r 1 ) Ψ 2 (r 2 ) e r 2 [ h2 r 2 2m 1 e Z ] e2 Ψ 1 (r 1 ) = E 1 Ψ 1 (r 1 ) r 1 [ h2 r 2 2m 2 Z ] e2 Ψ 2 (r 2 ) = E 2 Ψ 2 (r 2 ) e r 2 Finally, the repulsion term can be included, - introducing various approximation methods (Perturbation theory, Variational method ) Go to 17

18 18 The above two eqs. Are identical to Schrödinger eq. for H-atom - but where Nuclear charge is +Ze instead of +e To obtain simplest approximation to the solutions of the Schrödinger we started by disregarding the repulsion term alltogether. The above Schrödinger equation is now separable into two equations - each for one electron (Independent electron approach: wave function is separable ) The energy of the ground state (see the table) E = E 1s (Z) + E 1s (Z) = 2Z E 0 Next approximation: perturbation theory Keep the same Independent electron approach wavefunctions, but evaluate expectation value of the repulsion integral e 2 = 5 r 1 r 2 8 Z E 0 ( = 17 ev for Z = 1 ) ( = 34 ev for Z = 2 ) Z Next improvement of independent (separable) wavefunction: Variational method Best results: functions including correlations; non-separable; not independent electrons Go to

19 19!"!# $%& '#&& '()&* +(,&* -. / ), #D; <5=0= /,51< =051/.0)5?0 /0?5,? )?/5// / )51<< 0,5,<.//5,< /.=51< ),<5<<,>?51<!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. /@.A@#B5 "/=5/<< ".<>5>< "/,,5>< ",)05/< "1><5<< "?=?5/< &@0C>@- ".<5/<< "=,5>< ".?)5>< ")1=5/< "0?05<< ">==5/< D9E%97%349B "./5>01 "==5,1 ".?15,1 ")1?5>1 "0?=511 ">=?5>1!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. -#FF <51>>.51>> /51>> )51>>,51>> 051>> Figure 3: Table of approximation results (in electronvolts) E 1s (Z = 1) = 13.6eV = 0.5E 0 ( E 0 = 27.2eV is the atomic unit of energy ) E 1s (Z) = Z eV = 0.5Z 2 E 0 e 2 r 1 r 2 Z = 5 8 Z E 0 ( = 17 ev for Z = 1 ) ( = 34 ev for Z = 2 )

20 20 e 2 r 12 e 1 r 2 r 1 For general states in Helium [ h2 r 2 2m 1 Z e2 h2 r 2 e r 1 2m 2 e Z e2 r 2 + +Ze ] e 2 r 1 r 2 Ψ (r 1, r 2 ) = E Ψ (r 1, r 2 ) Φ HF a,b (r 1, r 2 ) = 1 2 ψ a (r 1 ) ψ b (r 1 ) ψ a (r 2 ) ψ b (r 2 ) and the energy becomes Φ HF Φ H HF = ψ a T Ze2 r + ψ a ψ b = 1 [ψ a (r 1 )ψ b (r 2 ) ψ b (r 1 )ψ a (r 2 )] 2 ψ a + ψ b T Ze2 r ψ b e 2 r r ψ a ψ b ψ a ψ b e 2 r r ψ b ψ a (3) The last term - exchange energy. THIS APPLIES to TRIPLET STATES. For SINGLET STATES - the exchange term sign is changed to +.

21 7 Repulsion term - multipole expansion 21 where 1 r 1 r 2 = LM 4π 2L + 1 r L < r L+1 > r < = r 1, r > = r 2 for r 1 < r 2 r < = r 2, r > = r 1 for r 1 > r 2 Evaluation of the matrix element in general case d 3 r 1 d 3 r 2 ψ n 1 l 1 m 1 (r 1 ) ψ n 2 l 2 m 2 (r 2 ) Y LM (ˆr 1 ) Y LM (ˆr 2 ) (4) 1 r 1 r 2 ψ n 1 l 1 m 1 (r 1 ) ψ n2 l 2 m 2 (r 2 ) (5) is performed separately over the radial and angular parts r1dr 2 1 dˆr 1 r2dr 2 2 dˆr 2 Rn 1 l 1 (r 1 )Yl 1 m 1 (ˆr 1 ) Rn 2 l 2 (r 2 )Yl 2 m 2 (ˆr 2 ) 1 r 1 r 2 where dˆr i means the integration over dω i = sin θ i dθ i dϕ i. Go to R n1 l 1 (r 1 )Y l1 m 1 (ˆr 1 ) R n2 l 2 (r 2 )Y l2 m 2 (ˆr 2 ) (6)

22 22 The evaluation of general case - angular integrals of three Y lm s C L = Yl i m i (θ, ϕ)y LM (θ, ϕ)y li m i (θ, ϕ)dω (7) For the case of both s-states, l i = 0 m i = 0 only L = 0 M = 0 are nonzero; The sum reduces to one term. The angular factors give value one, since the (Y L=0M=0 ) 2 = (4π) 1 cancels the corresponding factor in the multipole expansion and due to the normalization. Thus the repulsion matrix element with the e 2 encluded d 3 r 1 d 3 r 2 ψ 100 (r 1 ) ψ 100 (r 2 ) is evaluated as the radial integral only r 2 1dr 1 e 2 r 1 r 2 ψ 100 (r 1 ) ψ 100 (r 2 ) (8) r 2 2dr 2 R 10(r 1 )R 10(r 2 ) e2 r > R 10 (r 1 )R 10 (r 2 ) (9)

23 8 Calculating the Radial Integral 23 Integral: r1dr 2 1 r2dr 2 2 R10(r 1 )R10(r 2 ) e2 R 10 (r 1 )R 10 (r 2 ) (10) r > ( ) 3 Z 2 Radial Part R 1,0 (r) = 2 e Z r a 0 = R1,0(r) a0 = = 2 4 ( Z a 0 ) 6 e 2 0 r 2 1 r 2 2 R 1,0 (r 1 ) 2 R 1,0 (r 2 ) 2 e2 r > dr 1 dr 2 ( ) Z e 2Z (r a 1 +r 2 ) 0 r 2 a 1 r 2 e r > dr 1 dr 2 e 2Z (r a 1 +r 2 ) 0 r 2 1 r2 2 1 dr 1 dr 2 r > With substitutions = 1 Ze 2 2 a 0 2Z a 0 r 1 r 1 and r 2 1r 2 2e r 1 e r 2 1 r > dr 1 dr 2 0 } 0 {{ } inta 2Z a 0 r 2 r 2 with e 2 a 0 = 1a.u. = E 0

24 24 We split the integrtion into two integrals. For each r 1 are we taking the integral over r 2 and than can we take the integrale over r 1 : ( r1 ) ( ) inta = e r 1 r 2 r 1 r2dr 2 2 dr 1 + e r 1 r 2 r1r 2 2 dr 2 dr 1 0 r 1 0 = 0 0 r 1 e r 1 r1 With partial integration one get: r2e 2 r 2 dr 2 dr } {{ } intb intb = 2 e r 1 (r r 1 + 2) intc = e r 1 (r 1 + 1) 0 r1e 2 r 1 e r 2 r 2 dr 2 dr 1 r } 1 {{} intc And with this you get by again merging the two split integrals: We use inta = 0 2r 1 e r 1 e 2r 1 (r r 1 )dr 1 0 x n e x dx = n!

25 25 If the exponent contains α, we make substitution x = 1 α y dx = 1 α dy so that We re-write inta as inta = 0 0 x n dx e αx = 1 α n+1 0 y n dy e y 2r 1 e r 1 dr 1 e 2r 1 r1dr 2 1 e 2r 1 2r 1 dr We see that the first integral has n = 1 and no constant in the exponential; thus we get 2. Second term contains n = 2 and α = 2. It thus gives ! = 1 4 The third term has n = 1 and α = 2. It gives ! = 1 2

26 26 The final expression for is thus A = 0 0 r 2 1r 2 2e r 1 e r 2 1 r > dr 1 dr 2 (11) A = = 5 4 And with this the whole integral becomes d 3 r 1 d 3 r 2 ψ 100 (r 1 ) ψ 100 (r 2 ) e 2 r 1 r 2 ψ 100 (r 1 ) ψ 100 (r 2 ) = 5 Ze 2 (12) 8 a 0 Acknowledgement: Alexander Sauter (PHYS261 fall 2006) has done lots of work on this part of the document

27 27 Variational methods for quantum mechanics are mainly based on this: The functional φ H φ E g.s. it means φ H φ has an absolute minimum for φ ϕ g.s. Suppose we know the exact solutions ϕ α (ground state ϕ g.s. ) H ϕ α = E α ϕ α E α E g.s. For any φ - i.e. also any approximation - we can use the expansion φ = c α ϕ α (13) φ H φ = c β cα ϕ β H ϕ α = c β cα E α ϕ β ϕ α Thus And thus φ H φ = c α 2 E α c α 2 E g.s. φ H φ E g.s. φ H φ ϕ g.s. H ϕ g.s. As the approximation φ improves, it approaches E g.s. from above.

28 9 The variational method for Helium 28 We start with hydrogen-like (one electron) problem H = T 1 + V 1. We remember that the kinetic energy contains only second derivatives of the wavefunction, while V i = Ze2 r i. We know that the ground state energy is (and the wave function) E 1s (Z) = 1 2 ( ) 3 e2 1 Z 2 Z2 Ψ 1,0,0 (r) = e Z r a 0 a 0 π a 0 We avoid unnecessary evaluations by using virial theorem. It states: T = 1 2 V

29 29 Since and we can see that H = E 1s (Z) = 1 2 H = T + V T = 1 2 Z2 E 0 Z2 e2 a 0 = 1 2 Z2 E 0 and V = Z 2 E 0 We introduce a new variable, meaning an unknown effective charge number z, defining the wavefunctions. When z = Z, the kinetic energy T is 1 2 Z2 E 0. As we mentioned, the kinetic energy contains only second derivatives, no Z. That means that when z becomes different from Z, there can only be z in the kinetic energy T result, thus T (z) = 1 2 z2 E 0

30 30 On the other hand, the potential energy contains Z, as seen above. Thus V (z) = z Z E 0 We look now at the total energy for two electrons including the repulsion H = T 1 + T 2 + V 1 + V 2 + V 12. The repulsion term V 12 is known for the hydrogen like orbitals, or repulsion between two electrons where both are in 1s orbital. For atomic number Z we obtained V 12 = 5 Ze 2 = 5 8 a 0 8 Z E 0. Again, there is no Z in the repulsion energy operator, therefore for the orbitals with effective z. V 12 (z) = 5 8 z E 0

31 31 Thus or ( ) ( ) E(z) = E 0 2 z2 zz + E 0 2 z2 zz + E 0 8 z. E(z) = (z 2 2zZ + 5 ) 8 z E 0. The variational method says that for the ground state the energy functional E(z) = (z 2 2zZ + 5 ) 8 z E 0. must be extremal: d dz E(z) = 0 2z 2Z = 0 z = Z 5 16.

32 32!"!# $%& '#&& '()&* +(,&* -. / ), :; <5</=> <5? /5=> 0511?50,.,5,/ / <50 /,50 >./50.>!"#$%&'(!')* +,$-./0 +.$1 +/$.0 +23$44 +..$,5 +3.$5. /@.A@#B5 ".5< ",5< "?5<< ".15<< "/05<< ")15<< &@0C>@- "<5)=0 "/5=0< "=5.) ".)50< "/.5>> ")/5/0 D9E%97%349B "<5,=) "/5>,> "=5// ".)51< "/.5?= ")/5)0!"#$%&'(!')* +,$-.0 +.$1,5 +/$.0 +23$44 +..$,5 +3.$5. -#FF <51>>.51>> /51>> )51>>,51>> 051>> Figure 4: Experimental ionization potentials and results of the approximations. Energies given in atomic units!"!# $%& '#&& '()&* +(,&* -. / ), #D; <5=0= /,51< =051/.0)5?0 /0?5,? )?/5// / )51<< 0,5,<.//5,< /.=51< ),<5<<,>?51<!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. /@.A@#B5 "/=5/<< ".<>5>< "/,,5>< ",)05/< "1><5<< "?=?5/< &@0C>@- ".<5/<< "=,5>< ".?)5>< ")1=5/< "0?05<< ">==5/< D9E%97%349B "./5>01 "==5,1 ".?15,1 ")1?5>1 "0?=511 ">=?5>1!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. -#FF <51>>.51>> /51>> )51>>,51>> 051>>

33 33!"!# $%& '#&& '()&* +(,&* -. / ), #D; <5=0= /,51< =051/.0)5?0 /0?5,? )?/5// / )51<< 0,5,<.//5,< /.=51< ),<5<<,>?51<!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. /@.A@#B5 "/=5/<< ".<>5>< "/,,5>< ",)05/< "1><5<< "?=?5/< &@0C>@- ".<5/<< "=,5>< ".?)5>< ")1=5/< "0?05<< ">==5/< D9E%97%349B "./5>01 "==5,1 ".?15,1 ")1?5>1 "0?=511 ">=?5>1!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. -#FF <51>>.51>> /51>> )51>>,51>> 051>> Figure 5: Experimental ionization potentials and results of the approximations. Energies given in electronvolts

34 34!"!# $%& '#&& '()&* +(,&* -. / ), #D; <5=0= /,51< =051/.0)5?0 /0?5,? )?/5// / )51<< 0,5,<.//5,< /.=51< ),<5<<,>?51<!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. /@.A@#B5 "/=5/<< ".<>5>< "/,,5>< ",)05/< "1><5<< "?=?5/< &@0C>@- ".<5/<< "=,5>< ".?)5>< ")1=5/< "0?05<< ">==5/< D9E%97%349B "./5>01 "==5,1 ".?15,1 ")1?5>1 "0?=511 ">=?5>1!"#$%&'(!')* +25$3-/ +/1$,, +210$,. +3/2$ $ $0. -#FF <51>>.51>> /51>> )51>>,51>> 051>> Figure 6: Table of approximation results (in electronvolts) E 1s (Z = 1) = 13.6eV = 0.5E 0 ( E 0 = 27.2eV is the atomic unit of energy ) E 1s (Z) = Z eV = 0.5Z 2 E 0 e 2 r 1 r 2 Z = 5 8 Z E 0 ( = 17 ev for Z = 1 = 34 ev for Z = 2 )

35 10 Hylleraas variational function for ground state of helium 35 Our variational method assumed the product function for two electrons. One can construct various functions where the variables can not be separated into a product ( ) 3 1 Z 2 Ψ(r 1, r 2 ) ψ(r 1 )ψ(r 2 ) ψ 1,0,0 (r i ) = Zr i e a 0 π a 0 Already in 1929 Norwegian physicist Egil Hylleraas worked with variational method for a correlated function of a special type - using transformed coordinates (r 1, r 2 ) s = r 1 + r 2 t = r 1 r 2 u = r 12 = r 1 r 2 s (0, ) t (, ) u (0, ) Ψ(r 1, r 2 ) ψ(s, t, u) = e zs N l,m,n=0 c l,2m,n s l t 2m u n The variational parameters here are the constant z and all the coefficients c l,2m,n Note that e zs = e zr 1 e zr 2 is up to a normalisation the product function with the effective charge number z used in our variational method. Hylleraas (1929) used 6 variational parameters.

36 11 Configuration mixing 36 Consider the usual: H x (x)ϕ α (x) = E α ϕ α (x) H y (y)χ β (y) = E β χ β (y) For any Φ(x) Φ(x) = c α ϕ α (x) For any Ξ(x) Ξ(y) = d β χ β (y) Take now a general Ψ(x, y). First look at y as a parameter, Ψ(x, y 0 ) Ψ(x, y 0 ) Φ(x) = c α (y 0 )ϕ α (x) for every y 0 ; Thus we get a new function of y; c α (y) = d β (α)χ β (y) Inserting back: Ψ(x, y) = d β (α)χ β (y)ϕ α (x) Or, with a simpler notation Ψ(x, y) = d βα χ β (y)ϕ α (x)

37 37 In the case of Helium, for example, the H(x) and H(y) are identical and so are the χ β (y) and ϕ α (x). This becomes configuration mixing. Ψ(x, y) = d βα ϕ β (y)ϕ α (x) The coefficients are found by diagonalization. For three coordinate sets - e.g. for Lithium : Ψ(x, y, z) = D γβα ϕ(z)ϕ β (y)ϕ α (x)

38 38 E(eV) n=4 n=3 n=2 + + He + e + e Doubly excited levels of helium (autoionizing) He + e n= He (2 S) 1 He (1 S) S=0 S=1 3 He (2 S) He Figure 7: Excited states of helium

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43 43 Resonances as discussed in the topic Light and Atoms

44 12 Atomic Units 44 Unit of length is the Bohr radius: ( a 0 = h2 m e e 2 = 4πɛ 0 h 2 m e e 2 ) The first is in atomic units, second in SI-units. This quantity can be remembered by recalling the virial theorem, i.e. that in absolute value, half of the potential energy is equal to the kinetic energy. This gives us e 2 1 = 2 a 0 h2 2m e a 0 2 and if we accept this relation, we have the above value of a 0. The so called fine structure constant α = e2 hc expresses in general the weakness of electromagnetic interaction.

45 45 Some Constants and Quantities v 0 = αc = m s 1 Bohr velocity a 0 = m Bohr radius h = ev s Planck s constant k B = ev K 1 Boltzmann constant R = N A k B N A = Avogadro s number µ B = ev (Tesla) 1 Bohr magneton Plank s formula ρ(ω ba ) = hω3 ba π 2 c 3 1 e hω/kt 1 Useful formulae and informations P 0 (cosθ) = 1 P 1 (cosθ) = cosθ

46 46 Go to the Full Main Topics Coordinate system Atomic Units Approximations to describe helium atom Evaluation of repulsion term Radial Integral Figure -helium spectra Table - binding energies Variational method Doubly excited states of Helium Parahelium and Orthohelium

47 47 Calculations from other sources - future extension - must be modified Part 1 e ik r = 4π i L j L (Kr)YLM( ˆK)Y LM (ˆr) (14) LM Part 2 1 r j R(t) = 4π 2L + 1 LM r< L r> L+1 Y LM( ˆR)Y LM (ˆr) (15) Part 3 where then: V fi =< Φ f (r) 4π 2L + 1 LM r< L r> L+1 r < = r, r > = R for r < R r < = R, r > = r for r > R Y LM( ˆR)Y LM (ˆr) Φ i (r) > (16)

48 48 V fi =< R f (r)y l f m f (ˆr) LM V fi = LM 4π R 2L R 4π 2L + 1 rl r< L r> L+1 Y LM( ˆR)Y LM (ˆr) R i (r)y li m i (ˆr) > (17) r 2 dr R L+1 R f (r)[yl f m f (ˆr)Y LM( ˆR)Y LM (ˆr)]R i (r)y li m i (ˆr)+ r 2 dr RL r L+1 R f (r)[y l f m f (ˆr)Y LM (ˆr)Y LM( ˆR)]R i (r)y li m i (ˆr) (18) This can be described by a simple notation V fi = LM 4π 2L + 1 [GL fi(r(t))][y LM( ˆR)][C L ] (19) where G L fi (R(t)) is called The G-function and CL is composed of Clebsch-Gordan coefficients. The matrix element is different from zero only if: M = m i + m f and l i l f L l i + l f

49 49 on l f + L + l i is even. As can be seen above, we denote G L fi[r(t)] = Rf (r) rl < 0 r L+1 > R i(r)r 2 dr = 1 R R L+1 r L r 2 drrf (r)r i (r) + R L 1 0 R r L+1 r2 drrf (r)r i (r) (20) R i (r) and R f (r) are the radial wave functions for initial and final states.

50 50 The integration over the angular parts, which is the integral over three spherical harmonics, gives C L = Yl f m f (θ, ϕ)y LM (θ, ϕ)y li m i (θ, ϕ)dω (21) ( 1) m f [ (2l f + 1)(2L + 1)(2l i + 1) ] 1 2 4π ( ) ( ) lf L l i lf L l i m f M m i ( a b c This is known as Gaunts formula, and the numerical values d e f 3j symbols, up to a factor equal to Clebsch-Gordan coefficients. ) are Wigner-

51 Coordinate system Atomic Units Approximations to describe helium atom Evaluation of repulsion term Radial Integral Figure -helium spectra Table - binding energies Variational method Doubly excited states of Helium Parahelium and Orthohelium 51