where the sums are over the partcle labels. In general H = p2 2m + V s(r ) V j = V nt (jr, r j j) (5) where V s s the sngle-partcle potental and V nt
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1 Physcs 543 Quantum Mechancs II Fall 998 Hartree-Fock and the Self-consstent Feld Varatonal Methods In the dscusson of statonary perturbaton theory, I mentoned brey the dea of varatonal approxmaton schemes. The basc dea here s that the varatonal prncple: (h jhj,e h j ) = () j s equvalent to the Schrodnger equaton. In other words, the states j that satsfy ths equaton are egenstates of the Hamltonan. To be more explct, consder the quantty hh, E = d 3 r (r) H (r), E d 3 r (r) (r): (2) If we take the functonal dervatve of ths quantty wth respect to the functon (r) wehave hh, E = [H (r), E (r)] (r) d 3 r: (3) In order for ths to be statonary (the varaton s zero for all possble forms of (r)) we see that (r) must be a soluton to the Schrodnger equaton. We can also take the varaton wth respect to (r) and wll nd that hh,e s also statonary f (r) s a soluton to Schrodnger's equaton. The goal of these notes s to explore some of the mplcatons of usng a varatonal scheme to solve for the states of a many-partcle system. ote: I wll not go through the mathematcs of varatonal physcs and fucntonal dervatves. The methods used here are very smlar to those used n classcal physcs. Ths materal s avalable n most texts on Mathematcal Physcs. 2 Hartree Approxmaton Frst, consder a multpartcle system where the partcles are all dstngushable. The Hamltonan of the system has the form, H = H + 2 6=j V j (4)
2 where the sums are over the partcle labels. In general H = p2 2m + V s(r ) V j = V nt (jr, r j j) (5) where V s s the sngle-partcle potental and V nt s the nteracton potental. In the Hartee Approxmaton, one assumes that the egenstates of the total Hamltonan can be wrtten as a product of sngle partcle states. The varatonal approxmaton s then used to derve an equaton for these sngle partcle states. Explctly, assume the egenstates of H can be wrtten = (r) 2(r2) 3(r3) ::: (6) In ths case, the expectaton value of the total Hamltonan wll have the form d 3 r d 3 r2 d 3 r3::: (r) H + 2 6=j V j A (r) 2(r2):::: (7) Lookng at each of the terms, we see that the sums over H and V j reduce to one and two partcle expectaton values respectvely. For example, consder d 3 r d 3 r2 d 3 r3::: (r) 2(r2)::: H (r) 2(r2):::: (8) Because H only operates on the th partcle, the ntegrals over all the other partcle coordnates are equal to one. Therefore, ths term s smply, d 3 r (r ) H (r ): (9) Smlarly, the nteracton term reduces to d 3 r d 3 r j (r ) j (r j ) V j (r ) j (r j ): () Ths means that the quantty we wsh to be statonary s d 3 r (r ) H (r ) + 2 6=j, E Y d 3 r d 3 r j (r ) j (r j ) V j (r ) j (r j ) d 3 r (r ) (r ): () Takng the functonal dervatve of ths equaton wth respect to m(rm) for the m th partcle, we nd d 3 r Hm m (r m )+ 6=m d 3 r (r ) V m (r ) m (r m ), E m (r m ) A m (rm) =: (2) 2
3 The factor of two n front of the nteracton term drops out because there are two equal terms from the dervatve, m = and m = j. Ths means that each of the sngle partcle states n our guess at the total state must satsfy, H m m (r m 6=m d 3 r (r ) V m (r ) A m (r m )=E m (r m ) (3) Ths s usually called the \Hartree equaton". 3 Hartree Potental and Self-Consstency What we have shown above s that the Hartree approxmaton reduced our -partcle problem to a set of sngle partcle equatons that we know (n theory) how to solve. The nteracton between the partcles s reduced to a sngle potental term of the form V H (r m )= d 3 r j (r )j 2 V nt (jr, r m j) (4) 6=m Ths s the \Hartree potental". One usually thnks about ths potental as sayng that the m th partcle nteracts wth the probablty densty of all the other partcles, 6= m, (m) (r) = 6=m j (r)j 2 (5) through the potental V nt (jr, r2j). If we consder, for example, a coulomb nteracton between each of the partcles, the Hartree potental looks lke the nteracton wth the charge densty due to all the other partcles. V H (r m )=q d 3 r q(m) (r) jr m, rj : (6) The dculty wth solvng ths problem s that the Hartree potental connects the solutons of each of the sngle-partcle states. In other words, ths potental for partcle m depends on the states of all the other - partcles. For ths reason, t s sometmes referred to as the \Self-Consstent" potental. The dculty n solvng for the -partcle state s smply that of ndng a set of sngle partcle states (r) that together are solutons to the Hartree equaton. The approach to solvng ths problem numercally s essentally:. Guess a set of sngle partcle states,. 2. Compute the Hartree potental, V H, for each of the partcles usng the states guessed n (). 3. Solve the sngle-partcle (Hartee) equatons for the states usng V H. 4. If the states from (3) are the same as the states from (), you are done. 3
4 5. If the states from (3) are derent than the states from (), use these as a new guess at the states. 6. Repeat untl convergence (solutons = guesses). There are some subtletes n the process. The most mportant s that after solvng for the sngle partcle states n (3), the new guess at the states s a mxture of the old guess from () and the new solutons. If one uses the new states as the guesses, wld oscllatons n the numercs often occur. Of course, the soluton to the Hartree equaton s not exact. The approxmaton used s that the soluton to the problem can be wrtten as a product of sngle partcle states. Ths s the usual varatonal smplcatons where the regon of Hlbert space consdered s lmted by some guess as to the form of the states. \Exact" solutons to the -partcle problem nvolve lnear combnatons of these Hartee-lke bass states. Ths sort of soluton method has been able to produce exact dagonalzaton of Hamltonans of up to about dozen partcles, dependng on the physcal system. Of course, ths requres a lot of computer. 4 Hartee-Fock and Exchange The Hartree-Fock approxmaton follows drectly from what was done above for the Hatree approxmaton, only takng nto account the need for ant-symmetrc states for Fermons. The Hamltonan for the system s the same as for the Hartree approxmaton. The exchange requrement on the system of partcles wll add an extra nteracton term n the nal result. We start wth the Slater determnant state = p! (n ;n 2 ;:::) n n 2 n 3 ::: n (r) n2 (r2) n3 (r3) ::: (7) where the sum s over all possble permutatons of the ndces (n;n2;n3; :::). (ote that here I'm permutng the ndces labelng the sngle-partcle state functons rather than the ndces of the partcle coordnates as was done prevously. You mght want to convnce yourself that one gets the same states.) Frst, consder the matrx element of one of the sngle-partcle operators, H. Ths wll have the form, n n (n ;n 2 ;:::) (n ;:::) 2 n 3 ::: n n 2 n 3 ::: d 3 r d 3 r2 d 3 r3::: n (r ) n (r 2)::: H s n (r) n2 (r2):::: 2 ;n 2 (8) Because the operator only operates on the th state, ths can be rewrtten as (n ;n 2 ;:::) j6= (n ;n 2 ;:::) n n 2 n 3 ;::: n n 2 n 3 ;::: Y d 3 r j n j (r j ) nj (r j ) d 3 r s n (r ) H s n (r ): (9) 4
5 The sngle partcle states are orthogonal. Ths means that for there to be a non-zero matrx element, the sum on permutatons must have n j = n j for j 6= etc. But ths also means that n = n because that's the only state left. Ths gves only the dagonal terms (! of them) n the double sum over permutatons, and the product n n 2 n 3 ;::: n n 2 n 3 ;::: =. Therefore, the Hartree-Fock result for the sngle-partcle terms s the same as the Hartree result, d 3 r (r ) H (r ): (2) The nteracton potental term s a lttle more problematc. In ths case, we're lookng at a matrx element of the form, n n (n ;n 2 ;:::) (n ;:::) 2 n 3 ;::: n n 2 n 3 ;::: d 3 r d 3 r2 d 3 r3::: n (r ) n (r 2)::: V ;j n (r) n2 (r2):::: 2 ;n 2 (2) Agan, the ntegrals over the r k for k 6= ; j wll lmt the sum over the permutatons to those where n k = n k. For the and j states, there are two possbltes, ether n = n and n j = n j or n = n j and n j = n. Ths gves the matrx element, (n ;n 2 ;:::) d 3 r d 3 r2 V j [ n :::n :::n j ::: n :::n :::n j ::: n (r j ) n (r ) nj (r j ) + n :::n :::n j ::: n :::n j :::n ::: n (r j ) nj (r ) n (r j )]: (22) The product of the 's n the rst term wll equal because the order of the n k 's s the same n each. For the second term the product of the 's wll equal - because two ndces (n and n j )have been swtched. The sum over permutatons just gves a factor of!, so our nteracton potental term reduces to: d 3 r d 3 r j V j j n (r )j 2 j nj (r j )j 2, d 3 r d 3 r j V j n (r j ) nj (r ) n (r j ): (23) The rst term s the same as the Hartree analyss, the second term (sometmes called the Exchange or Fock term) arses from the Paul excluson prncple. The quantty that we need to mnmze n ths case s d 3 r (r ) H (r ) + 2, 2, E 6=j Y 6=j d 3 r d 3 r j n (r j ) V j n (r ) nj (r j ) d 3 r d 3 r j n (r j ) V j nj (r ) n (r j ) d 3 r (r ) (r ): (24) Performng the same mnmzaton process as above, we'll take the functonal dervatve of the terms above wth respect to some state m(r m ). The results, smlar to the Hartree case, are the sngle partcle equatons H m m (r m ) + 6=m d 3 r (r ) V m (r ) m (r m ) 5
6 , 6=m d 3 r (r ) V m (r m ) m (r )=E m (r m ) (25) The nal term s sometmes wrtten, d 3 6=m (r ) V m (r m ) m (r ) A (26) Ths s a \non-local nteracton" term for m. It depends on the value of m everywhere. Usually, nteracton terms are local, of the form V (r) (r). A non-local potental term has the form V (r) (r)! d 3 r U(r; r ) (r ) (27) Because of the ant-symmetrzaton requrement on Fermons, that the states \avod" each other, ths exchange energy depends on the value of the state everywhere. 6
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