Lecture notes for FYS KJM 4480

Transcription

1 Lecture notes for FYS KJM 4480 Quantum mechancs for many-partcle systems Smen Kvaal October 9, 2015

2 Contents 1 Fundamental formalsm Many-partcle systems Hlbert space and Hamltonan More on the space X, and fnte dmensonal spaces The manybody Hamltonan Separaton of varables Permutatons Partcle statstcs Slater determnants Second quantzaton The creaton and annhlaton operators Antcommutator relatons Occupaton number representaton Spn orbtals and orbtal dagrams Foc space Truncated bases Representaton of operators What we wll prove One-body operators Two-body operators Wc s Theorem A sort of summary and motvaton Vacuum expectaton values Normal orderng and contractons Statement of Wc s Theorem Vacuum expectaton values usng Wc s Theorem Proof of Wc s Theorem Usng Wc s Theorem Partcle-hole formalsm Operators on normal-order form (Not yet lectured) The number operator One-body operators Two-body operators Normal-ordered two-body Hamltonan Full expressons for the normal-ordered Hamltonan

3 2 The Standard Methods of approxmaton Introducton The varatonal prncple The Cauchy nterlace theorem and lnear models The Confguraton-nteracton method (CI) General descrpton Matrx elements of the CI method Computer mplementaton of CI methods Nave CI Drect CI Recpe for bt pattern representaton Hartree Foc theory (HF) The Hartree Foc equatons The Hartree Foc equatons n a gven bass: the Roothan Hall equatons Self-consstent feld teraton Bass expansons n HF sngle-partcle functons Restrcted Hartree Foc for electronc systems (RHF) Unrestrcted Hartree Foc for electronc systems (UHF) The symmetry dlemma Hartree Foc for the electron gas Normal-ordered Hamltonan n HF bass (Not yet lectured) Perturbaton theory for the ground-state (PT) Raylegh Schrödnger perturbaton theory (RSPT) Low-order RSPT A two-state example Møller Plesset perturbaton theory (MPPT) Feynman dagrams for Raylegh Schrödnger perturbaton theory 75 4 Coupled-cluster theory (CC) 76 A Mathematcal supplement 77 A.1 Calculus of varatons A.1.1 Functonals A.1.2 Functons of one real varable A.1.3 Functons of two real varables A.1.4 Extremalzaton of a functonal

4 Chapter 1 Fundamental formalsm Suggested readng for ths chapter: Rames [1], sectons , and Gross/Runge/Henonen [2], secton I.1. Chapter 1 of Szabo/Ostlund [3] contans a nce refresher on mathematcal topcs, ncludng lnear algebra. Dsclamer: Ths course s not a mathematcs course, so some mathematcal detals are glossed over. Examples nclude the fact that the Hamltonan s rarely a bounded operator, such that there exsts Ψ such that ĤΨ does not mae sense n Hlbert space. Often, operators have complcated spectra, ncludng contnuous parts. For ponts n the contnuous spectrum, no square-ntegrable egenfuncton exsts. As s usual, we gnore ths complcaton, and bascally calculate as f everythng were matrces of fnte dmenson. 1.1 Many-partcle systems Hlbert space and Hamltonan We dscuss the non-relatvstc quantum mechancal descrpton of a system of many partcles. For smplcty, we consder N dentcal partcles. Whereas the classcal state of such a system s a pont n phase space, the quantum state s a wavefuncton dependng on all the coordnates: Ψ = Ψ(x 1, x 2,, x N ), (1.1) where x s a pont n the confguraton space X, the space where each partcle lves. The confguraton space 1 for all N partcles s thus X N, and Ψ X N C. (1.2) Example: The confguraton space for an electron s R 3 { 1 2, + 1 }. A sngle electron s confguraton s 2 x = ( r, s z ), where s z s the proecton of the electron spn along some axs. The one-electron wavefuncton can thus be consdered a two-component wavefuncton. The N-electron wavefuncton s thus a functon of N coordnates r and N spn varables s z,, n total 2 N components. Example: A nucleon has two dscrete degrees of freedom: spn and sospn. Thus, X = R 3 { 1 2, } { 1 2, }, x = ( r, s z, z ). A sngle nonrelatvstc nucleon thus has a four-component wavefuncton, and N nucleons 4 N components. 1 Snce the partcles are dentcal, the confguraton space s actually the quotent space X N /S N, where S N s the permutaton group of N obects. Ths means that we dentfy ponts n X N that dffer only by a permutaton. Suppose X = R 3. Then X N s a flat space. But X N /S N s actually a curved space! For low-dmensonal systems, X = R 1 or X = R 2, one can show that partcle statstcs s not confned to only bosons or fermons. See [4]. 3

5 Remar: Mathematcally, X s a measure space, whch means that a functon ψ X C can be ntegrated over subsets of X. For subsets of R n, the standard measure s Lebesgue measure, whch gves an ntegral slghtly more general than the Remann ntegral encountered n ntroductory analyss courses. For dscrete sets, the standard measure s countng measure, where the ntegral s smply a sum. See also the small secton on fnte dmensonal spaces further down. Ths remar s for orentaton only. For us, we smply state that we ntegrate over contnuous degrees of freedom and sum over dscrete degrees of freedom. For X = R d S wth S = {s 1, s 2,, s n } a dscrete set, we defne f (x)dx = f ( r, s)d d r. X s S R d The wavefuncton has a probablstc nterpretaton: P(x 1,, x N ) = Ψ(x 1, x 2,, x N ) 2 s the probablty densty for locatng all partcles at the pont (x 1,, x N ) X N. Therefore, Ψ must be square ntegrable, and be n the Hlbert space L 2 (X N ), Ψ L 2 (X N ). (1.3) All physcs can be obtaned from the state Ψ. The governng equaton n non-relatvstc quantum mechancs s the tme-dependent Schrödnger equaton (TDSE): ĤΨ(x 1, x 2,, x N, t) = ħ t Ψ(x 1, x 2,, x N, t). (1.4) The system Hamltonan Ĥ s obtaned from ts classcal counterpart (f such exsts) by a procedure called Weyl quantzaton NB: Add reference. If Ĥ does not explctly depend on tme, the TDSE can be solved by nstead consderng the tme-ndependent Schrödnger equaton (TISE), ĤΨ(x 1, x 2,, x N ) = EΨ(x 1, x 2,, x N ). (1.5) The reason s well-nown: the evoluton operator s dagonal n the egenbass. The tme-ndependent Schödnger equaton s the man focus n ths course, and we wll only scratch the surface. Ψ s a very, very, very complcated functon. Intutvely, one mght thn that solvng for Ψ s N tmes as hard as solvng for an N = 1 wavefuncton. However, Ψ s a functon of all N coordnates. Resolvng each coordnate on a grd wth, say, K ponts requres K N ponts n total. For K = 2 (whch s rather coarse) and N = 40 (e.g., a 40 Ca nucleus), we need data ponts! Descrbng the correlated moton of N quantum partcles s harder than the poneers of quantum mechancs thought! Lterally thousands of researchers worldwde are mae a lvng out of devsng more or less clever schemes for fndng approxmate solutons More on the space X, and fnte dmensonal spaces The space L 2 (X) need not be nfnte dmensonal. Recall that ψ L 2 (X) f and only of ψ X C and Let ψ 2 = ψ(x) 2 dx < +. X = {1, 1, 2,, L} be a dscrete set, such that the ntegral s a sum. For example, we could dscretze space usng grd ponts, or a fnte collecton of bass functons such as Hermte functons, fnte element functons, etc. Then L 2 (X) conssts of functons ψ X C,.e., functons from the ntegers 1,..., L to C. But ths s ust an ordnary vector n C L! The ntegral becomes ψ 2 = L =1 ψ 2, whch s always fnte. Thus, usng countng measure on ths partcular X, L 2 (X) C L. 4

6 1.1.3 The manybody Hamltonan Havng ntroduced the wavefuncton, we now consder the Hamltonan. In ths course, we shall consder only Hamltonans on the followng generc form: Ĥ = N =1 ĥ() = Ĥ 0 + Ŵ. N, =1 ŵ(, ) (1.6) where ĥ() denotes a sngle-partcle operator actng only on the degrees of freedom of partcle, and ŵ(, ) denotes a two-body operator that acts only on the degrees of freedom of the par (, ),. Of course, one could consder three-body forces as well, and even hgher. Such occur n nuclear physcs. Let us tae the Hamltonan of an atom n the Born Oppenhemer approxmaton as an example. The Hamltonan for a free electron s ust ts netc energy, ˆt = 1 2m e p 2 = 1 2m e ( ħ ) 2 = ħ2 2m e 2. (1.7) If t s movng n an external feld, such as the Coulomb feld set up by an atomc nucleus of charge +Ze at the locaton R, we obtan the total sngle-partcle Hamltonan ĥ = ˆt + ˆv = ħ2 2 Ze2 2m e R r. (1.8) The Hamltonan for a system of N electrons, neglectng nter-electronc nteractons, becomes Ĥ 0 = N =1 ĥ() = N =1 [ ħ2 2m e 2 The electron par (, ) nteracts va the Coulomb force: w(, ) = Ze2 r R ]. (1.9) e 2 r r. (1.10) Thus, Ŵ = 1 2 N, =1 w(, ) = 1 2 N, =1 e 2 r r. (1.11) Separaton of varables If we neglect the two-body part Ŵ of the Hamltonan, we may solve the TISE by separaton of varables. We do ths now as a prelmnary step, before we dscuss the consequences of the partcles beng ndstngushable. We see an egenfuncton Ψ L 2 (X N ) to the non-nteractng Hamltonan Ĥ 0. Wrte Plug n to the TISE and dvde by Ψ to get Ψ(x 1,, x N ) = ψ 1 (x 1 )ψ 2 (x 2 ) ψ N (x N ). (1.12) ψ 1 [h()ψ ] = E. (1.13) 5

7 The rght hand sde s a constant. The left hand sde s a sum of functons f 1 + f 2 + f N, f = f (x ). Ths can only sum to a constant f f (x ) s a constant, ĥψ (x) = є ψ (x), (1.14) whch s ust the TISE for a sngle partcle! Thus, for any collecton of N egenvalues of the sngle-partcle problem, we get a soluton of the N partcle problem. We obtan that the total egenfuncton s wth egenvalue Ψ(x 1, x 2,, x N ) = ψ 1 (x 1 )ψ 2 (x 2 ) ψ N (x N ) (1.15) E = є є N. (1.16) One can also show that the converse s true: any egenfuncton Ψ can be taen on the above form Permutatons Are you unfamlar wth permutatons? As me, and I wll add some paragraphhs here Partcle statstcs Our partcles are dentcal, or ndstngushable. There s abundant evdence that all elementary partcles must be treated as such. That means that our probablty densty must be permutaton nvarant n the followng sense: let σ S N be a permutaton of N ndces, and let (x 1,, x N ) X N be a confguraton of the N partcles. Then we must have Ths s equvalent to Ψ(x 1, x 2,, x N ) 2 = Ψ(x σ(1), x σ(2),, x σ(n) ) 2. (1.17) Ψ(x 1,, x N ) = e α Ψ(x σ(1), x σ(2),, x σ(n) ) (1.18) for some real α, that may depend on σ. (Clearly, our separaton of varables egenfunctons do not satsfy ths!) Defne a lnear operator ˆP σ va ( ˆP σ Ψ)(x 1,, x N ) = Ψ(x σ(1), x σ(2),, x σ(n) ), (1.19) that s, the operator that evaluates Ψ at permuted coordnates. We have reformulated partcle ndstngushablty as: Ψ s an egenfuncton of ˆP σ for every σ S N, wth egenvalue possbly dependng on σ. One can show (see the exercses), that ether P σ Ψ = Ψ for every σ S N, or P σ Ψ = ( 1) σ Ψ for every σ S N, where σ s the number of transpostons n σ, and thus ( 1) σ s the sgn of the permutaton. In the former case, Ψ s totally symmetrc wth respect to permutatons, and n the latter case, totally ant-symmetrc. It s a postulate that partcles occurng n quantum theory (n three-dmensonal space) are of one of two types: bosons or fermons. Bosons have totally symmetrc wavefunctons only, and fermons have totally ant-symmetrc wavefunctons only. To cte Lenaas and Myrhem [4], The physcal consequences of ths postulate seem to be n good agreement wth expermental data. Wolfgang Paul proved (usng relatvstc consderatons) that wavefunctons of half-ntegral spn must be ant-symmetrc, and wavefunctons of partcles wth ntegral spn must be symmetrc, connectng the postulate wth the ntrnsc spn of partcles. To ths day, no partcles wth other spn values have been found. In ths course, we focus on fermons. See, e.g., [2] for the general case. 6

8 Exercse 1.1. In ths exercse, we prove that f Ψ L 2 (X N ) s an egenfuncton for all ˆP σ, then the egenvalue s ether 1 or ( 1) σ. We ntrodce transpostons: τ S N s transposton f t exhanges only a sngle par (, ),. Wrte ˆP ˆP τ. Assume that Ψ L 2 (X N ) s such that, for all σ S N, ˆP σ Ψ = s σ Ψ, s σ = e α(σ). Show that ˆP 2 = 1, and fnd all the possble egenvalues of ˆP. Under the assumpton on Ψ, show that f s s the egenvalue of ˆP, ˆP Ψ = s Ψ, then, for any other par (, ), the egenvalue s s = s. You wll probably need to use the group theoretcal propertes of permutatons. We have establshed that the egenvalue of a transposton s a characterstc of Ψ, let s = s. Compute the egenvalue of P σ for arbtrary σ n terms of s. Exercse 1.2. Let Ĥ = N =1 ĥ() + ŵ(, ). (, ) Show that Ĥ commutes wth P σ for any permutaton σ S N,.e., show that for any wavefuncton Ψ L 2 (X N ), ĤP σ Ψ = P σ ĤΨ. (1.20) Exercse 1.3. In ths exercse, we consder X = R 3,.e., no spn. Consder each of the below functons. 1. Ψ( r 1, r 2 ) = e α r1 r2. 2. Ψ( r 1, r 2 ) = sn( e z ( r 1 r 2 )), where e z s the unt vector n the z-drecton. 3. Ψ( r 1, r 2, r 3 ) = sn[ r 1 ( r 2 r 3 )]e r1 2 e r2 2 e r3 2 Answer the followng questons, per functon: Is the functon totally symmetrc wth respect to partcle permutatons? Is the functon totally antsymmetrc wth respect to partcle permutatons? Is the functon square ntegrable? 7

9 1.1.7 Slater determnants The set of totally antsymmetrc wavefunctons L 2 (X N ) AS n L 2 (X N ) form a closed subspace of Hlbert space: t s a lnear space whch s complete. Thus L 2 (X N ) AS s a Hlbert space n ts own rght, and from our perspectve t s the true Hlbert space of N dentcal fermons. The antsymmetry of a wavefuncton of N coordnates s a qute complcated constrant. We are also used to orthonormal bases, and t may seem dauntng to come up wth such a bass whch s also antsymmetrc. Slater determnants are the soluton. Exercse 1.4. Prove that L 2 (X N ) AS s a lnear space. Addtonally, f you have the mathematcal bacground, prove that t s a closed subspace usng the Hlbert space metrc. The orgnal space has a tensor product representaton: L 2 (X N ) = L 2 (X) L 2 (X) L 2 (X) (N factors). (1.21) Here, L 2 (X) s the Hlbert space of a sngle fermon. Let us assume that we have an orthonormal bass (ONB) ϕ 1, ϕ 2,, for ths space, such that we can expand any ψ L 2 (X) as wth and ψ(x) = c µ ϕ µ (x), (1.22) µ ϕ µ ϕ ν = δ µ,ν (1.23) ψ 2 = c µ 2. (1.24) µ Thus, ψ(x) s represented by an (nfnte) vector [c µ ] = (c 1, c 2, ). Because of Eq. (1.21), we may construct a bass for L 2 (X N ) by tensor products, Any Ψ L 2 (X N ) can be wrtten wth Φ µ1,,µ N (x 1,, x N ) = ϕ µ1 (x 1 )ϕ µ2 (x 2 ) ϕ µn (x N ). (1.25) Ψ(x 1,, x N ) = µ 1 µ N c µ1,,µ N Φ µ1,,µ N (x 1,, x N ), (1.26) Φ µ1,,µ N Φ ν1,,ν N = δ µ1,ν 1 δ µn,ν N. (1.27) In the N = 2 case, we see that Ψ(x 1, x 2 ) can be represented by an nfnte matrx [c µ1 µ 2 ], and n the N = 3 case a 3D matrx, and so on. Remar: Compare ths wth the separaton-of-varables treatment. If the set of egenfunctons ψ L 2 (X) of ĥ s complete, our separaton of varables egenfunctons Ψ = ψ 1ψ 2 ψ N form a complete set too. Another remar: For arbtrary N, the tensor product bass descrbed can be counted. For arbtrary N, let us ntroduce a generc ndex, a multndex, = (µ 1,, µ N ). There s a one-to-one mappng between multndces and the natural numbers N = {0, 1, 2,...}. Thus, wrtng ξ = (x 1,, x N ) Ψ(ξ) = c Φ(ξ), Φ Φ l = δ,l (1.28) 8

10 all the varous N are represented wth the same formula. There s nothng specal about c beng a vector, a matrx, a 3D matrx, etc. They are all fundamentally equvalent, snce the bass set can be counted. Important message so far: a sngle-partcle bass set {ϕ µ } can be used to construct a bass for L 2 (X N ). What about or actual Hlbert space, L 2 (X N ) AS? Can we construct a bass for ths usng our snglepartcle bass? Yes, ths s the role of Slater determnants. What s the smplest totally antsymmetrc wavefuncton we can create, startng wth some snglepartcle functons? If we start wth N = 2, and consder the product ϕ 1 (x 1 )ϕ 2 (x 2 ), ths s not antsymmetrc. But f we consder the lnear combnaton Φ(x 1, x 2 ) = ϕ 1 (x 1 )ϕ 2 (x 2 ) ϕ 2 (x 1 )ϕ 1 (x 2 ), (1.29) ths s antsymmetrc f we exchange x 1 and x 2. Contnung wth N = 3, we qucly realze that n order to obtan somethng antsymmetrc out of ϕ 1 (x 1 )ϕ 2 (x 2 )ϕ 3 (x 3 ), we must tae the lnear combnaton Φ(x 1, x 2, x 3 ) = ϕ 1 (x 1 )ϕ 2 (x 2 )ϕ 3 (x 3 ) ϕ 2 (x 1 )ϕ 1 (x 2 )ϕ 3 (x 3 ) ϕ 1 (x 1 )ϕ 3 (x 2 )ϕ 2 (x 3 ) ϕ 3 (x 1 )ϕ 2 (x 2 )ϕ 1 (x 3 ) + ϕ 2 (x 1 )ϕ 3 (x 2 )ϕ 1 (x 3 ) + ϕ 3 (x 1 )ϕ 1 (x 2 )ϕ 2 (x 3 ), (1.30) each term representng a permutaton of the ndces (123). There s nothng specal about (123) of course, (µ 1 µ 2 µ 3 ) also wors. Note that f one of these ndces are equal, then the whole lnear combnaton s zero as well. The generalzaton to N ndces s n fact a determnant, and we mae a defnton: Defnton 1.1. Let ϕ 1, ϕ 2,..., ϕ N be arbtrary sngle-partcle functons n L 2 (X) (not necessarly orthonormal). The Slater determnant defned by these functons s denoted by [ϕ 1 ϕ 2 ϕ N ], and s defned va the formula [ϕ 1, ϕ 2,, ϕ N ](x 1,, x N ) = = = 1 N! 1 N! 1 N! ϕ 1 (x 1 ) ϕ 1 (x 2 ) ϕ 1 (x N ) ϕ 2 (x 1 ) ϕ 2 (x 2 ) ϕ 2 (x N ) ϕ N (x 1 ) ϕ N (x 2 ) ϕ N (x N ) σ S N ( 1) σ σ S N ( 1) σ N =1 N =1 ϕ σ() (x ) ϕ (x σ() ) Note: the 1/ N! s there for normalzaton purposes, see later. The second formula n the defnton follows from the theory of matrx determnants. Exercse 1.5. Show that the two last lnes n Eq. (1.31) are equvalent. Ths requres some manpulaton of permutatons. (1.31) Exercse 1.6. Let A be an N N matrx. Let ϕ, = 1,, N be gven sngle-partcle functons, and let ψ, = 1,, N be defned by ψ = ϕ A. (1.32) Prove that [ψ 1, ψ 2,, ψ N ] = det(a)[ϕ 1, ϕ 2,, ϕ N ]. (1.33) (Hnt: use antsymmetry of Slater determnants wth respect to permutatons of sngle-partcle functons, and the expresson det(a) = σ SN ( 1) σ A 1σ(1) A 2σ(2) A Nσ(N).) 9

11 Exercse 1.7. NB: Ths exercse has been updated snce t was gven as part of Problem set 1 (H2015). The assumpton that the ndces were sorted was added. Suppose that {ϕ µ }, µ = 1, 2, are orthonormal. Prove that Φ µ1,,µ N = [ϕ µ1 ϕ µ2,, ϕ µn ] s normalzed, Prove that Φ µ1 µ 2 µ N Φ µ1 µ 2 µ N = 1. Φ µ1 µ 2 µ N Φ ν1ν 2 ν N = δ µ1ν 1 δ µn ν N, under the assumpton that µ and ν are sorted n ncreasng order. What do you get for the nner product f the ndces are not sorted? Observaton: Determnant propertes mply that permutaton of partcle ndces gves sgn change. Permutaton of functon ndces gves sgn change: [ϕ 1,, ϕ,, ϕ,, ϕ N ] = [ϕ 1,, ϕ,, ϕ,, ϕ N ] (1.34) [ϕ 1,, ϕ N ](x 1,, x,, x,, x N ) = [ϕ 1,, ϕ N ](x 1,, x,, x,, x N ). (1.35) Moreover, two equal rows (.e., equal functon ndces) means that two of the sngle-partcle functons are dentcal, gvng a vanshng determnant. If two columns n Eq. (1.31) are dentcal, the determnant vanshes. Two columns equal mean that we evaluate at some x = x. Ths s the Paul excluson prncple. Theorem 1.1. Let {ϕ µ } be an orthonormal bass for L 2 (X). Then, any Ψ L 2 (X N ) AS can be expanded n the Slater determnants [ϕ µ1, ϕ µ2,, ϕ µn ]. (1.36) Moreover, f we choose an orderng of the ndces µ, the Slater determnants satsfyng µ 1 < µ 2 < < µ N form an orthonormal bass for L 2 (X N ) AS. Proof. Step 1: Expand Ψ n the tensor product bass. Ψ(x 1,, x N ) = c µ1,,µ N Φ µ1,,µ N (x 1,, x N ). (1.37) µ 1,,µ N Step 2: Show that the coeffcents c µ are antsymmetrc under permutaton. For smplcty, consder a transposton of wth, < : ˆP Ψ(x 1,, x N ) = c µ1,,µ N ˆP Φ µ1,,µ N (x 1,, x,, x,, x N ) µ 1,,µ N = µ 1,,µ N c µ1,,µ N Φ µ1,,µ N (x 1,, x, x,, x N ) = µ 1,,µ N c µ1,,µ N Φ µ1,,µ,,µ,,µ N (x 1,, x N ) = µ 1,,µ N c µ1,,µ,,µ,,µ N Φ µ1,,µ N (x 1,, x N ) = µ 1,,µ N c µ1,,µ N Φ µ1,,µ N (x 1,, x N ) (1.38) Proectng the two last nequaltes onto Φ ν1,,ν N gves c ν1,,ν,,ν,,ν N = c ν1,,ν,,ν,,ν N. (1.39) 10

12 We decompose an arbtrary σ S N nto transpostons, and obtan c µσ(1),µ σ(2),,µ σ(n) = ( 1) σ c µ1,,µ N. (1.40) Step 3: Rearrange summaton so that we exhbt Ψ as a lnear combnaton of Slater determnants. Note that we can wrte f (µ 1,, µ N ) = f (µ σ(1),, µ σ(n) ), (1.41) µ 1, µ N µ 1<µ 2< <µ N σ S N splttng the summaton over ordered multndces and permutatons of these. We now get Ψ = ( 1) σ c µ1,,µ N Φ µσ(1),µ σ(2),,µ σ(n) µ 1< <µ N σ = ( 1 N!c µ1,,µ N ) ( 1) σ Φ µσ(1),µ σ(2),,µ σ(n) µ 1< <µ N N! σ = µ 1< <µ N ( N!c µ1,,µ N )[ϕ µ1,, ϕ µn ]. (1.42) Ths n fact proves that the Slater determnants, when we only use ordered ndces, are suffcent to expand any ΨL 2 (X N ) AS. Clearly, f we omt one such Slater determnant, not all Ψ can be expanded. (In partcular, ths omtted Slater determnant cannot be expanded n the rest!) Thus, the Slater determnants wth ordered ndces form a bass. Exercse 1.8. How many terms are there n [ϕ 1 ϕ 2 ϕ 3 ϕ 4 ](x 1, x 2, x 3, x 4 ), when expanded as a lnear combnaton of tensor products? Wrte down the expanson explctly. Exercse 1.9. In ths exercse, we defne the antsymmetrzaton operator A as Now, A = 1 N! σ S N ( 1) σ ˆP σ. (1.43) [ϕ 1,, ϕ N ] = N!Aϕ 1 (x 1 ) ϕ N (x N ). (1.44) An operator U s an orthogonal proector f and only f U 2 = U and U = U. Prove that A s an orthogonal proector from L 2 (X N ) onto L 2 (X N ) AS. 1.2 Second quantzaton The creaton and annhlaton operators In ths secton, we ntroduce the followng shorthand: L 2 N L 2 (X N ) AS (1.45) 11

13 snce the space X s understood from context, and snce we only deal wth fermon spaces. We also ntroduce the bra/et notaton for wavefunctons. Recall that a bass for L 2 N could be formed from an orthonormal bass {ϕ µ} of L 2 (X), by computng a set of Slater determnants Φ µ1,,µ N = [ϕ µ1,, ϕ µn ], where µ 1 < µ 2 < < µ N were ordered. (If we permute the ndex set, we get the same functon wth a possble sgn change, so t s not an addtonal bass functon.) So far we have emphaszed that [ϕ µ1,, ϕ µn ] were functons, but n quantum mechancs the bra/et notaton s useful. We therefore ntroduce the et notaton ψ 1,, ψ N = [ψ 1,, ψ N ] (1.46) for an arbtrary Slater determnant. When {ϕ µ } s a sngle-partcle bass, we may choose to suppress all the ϕ s everywhere, and wrte µ = µ 1 µ 2 µ N, [ϕ µ1,, ϕ µn ](x 1,, x N ) = x 1 x N µ 1 µ N (1.47) for a Slater determnant. If µ = µ then µ = 0 s the zero vector. We recall the antsymmetry propertes, and more generally For any Ψ L 2 N, we have the bass expanson ˆP µ 1 µ µ µ N = µ 1 µ µ µ N (1.48) ˆP σ µ 1 µ N = ( 1) σ µ σ(1) µ σ(n). (1.49) Ψ = µ µ µ Ψ (1.50) connectng wth the earler treatment. The means that we sum only over ordered sets of ndces. As we saw earler, the coeffcents µ Ψ are permutaton antsymmetrc. So far, we have used Gree letters µ, ν, etc., as sngle-partcle ndces. There s nothng specal about ths, of course. We wll later also use p, q, r, etc. Loong at the determnant (1.31), we see that by addng a row contanng the ndex ν, and a column wth coordnate x N+1, we obtan an N + 1 partcle Slater determnant (modulo a constant factor): x 1 x N+1 νµ 1 µ 2 µ N = 1 (N + 1)! ϕ ν (x 1 ) ϕ ν (x 2 ) ϕ ν (x N ) ϕ ν (x N+1 ) ϕ µ1 (x 1 ) ϕ µ1 (x 2 ) ϕ µ1 (x N ) ϕ µ1 (x N+1 ) ϕ µ2 (x 1 ) ϕ µ2 (x 2 ) ϕ µ2 (x N ) ϕ µ2 (x N+1 ) ϕ µn (x 1 ) ϕ µn (x 2 ) ϕ µn (x N ) ϕ µn (x N+1 ) Smlarly, we can remove a row and column, and obtan an N 1 partcle Slater determnant. Ths nspres the creaton and annhlaton operators, that map wavefunctons between dfferent partcle number spaces: (1.51) c ν L 2 N L 2 N+1 (1.52) c ν L 2 N L 2 N 1 (1.53) The operator c ν s called a creaton operator and s, roughly defned, by nsertng a row and column as descrbed. The operator c ν s the Hermtan adont of c ν, and t wll be shown that ts acton on a Slater determnant corresponds to the mentoned removal of a row and column. We defne the space L 2 0 the zero partcle space as a one-dmensonal space spanned by the specal et, the vacuum state. There s nothng mysterous about ths, t s ust a defnton that wll be useful later. Note that 0. 12

14 Recall that a lnear operator s fully defned when we specfy ts acton on a bass set. Ths s how we defne c µ and c µ. Defnton of the creaton operator: For every sngle-partcle ndex ν, we defne the creaton operator c ν actng on the vacuum state by c ν = ν. (1.54) Snce ths s a Slater determnant wth a sngle partcle, we have, of course, x ν = ϕ ν (x). For an arbtrary Slater determnant wth N > 0, we defne the acton by c ν µ 1 µ N νµ 1 µ N. (1.55) We observe already that f there s a such that ν = µ, then νµ 1 µ N 0: ˆP 1 ν µ = ν µ = ν µ = 0, ν = µ. (1.56) In terms of determnant coordnate expressons as n Eq. (1.31), c ν nserts a column on the far rght wth x N+1 and nserts a row on the top wth the ndex ν. Fnally, the whole expresson s renormalzed. [Recall that the bass Slater determnants were the determnants that had ordered ndces. Assume that µ s ordered. Clearly, c ν s ether zero or equal to ( 1) µ 1 µ 2 µ νµ +1 µ N, whch s a new bass determnant. Here, s chosen such that the augmented ndex set s ordered.] Let us now consder the annhlaton operator. There are no partcles to remove n the vacuum state, so we set c ν 0. (1.57) Let µ be a multndex. If ν = µ for some, we defne c ν µ ( 1) 1 µ 1 µ 1 µ +1 µ N. (1.58) In terms of the coordnate determnant expresson, ths amounts to movng the th row to the top wth 1 transpostons, gvng the sgn factor, and then crossng out the far rght column and the frst row, now contanng the ndex ν. Ths movng of the th row may seem le a complcaton compared to the creaton operator, but note that for c ν we defned ts acton by nsertng ν on the top. Movng ν to the ( + 1)th poston wll nduce a ( 1). But c ν removes a row at an n prncple arbtrary locaton. Exercse Prove that c α and c α are Hermtan adonts of each other, as the notaton suggests. Thus, for any µ wth N ndces, and ν wth N + 1 ndces, show that µ (c α ν ) = [ ν (c α µ )] (1.59) Antcommutator relatons Recall that the antcommutator of two operators s defned by {Â, ˆB} Â ˆB + ˆBÂ. (1.60) In ths secton, we prove three mportant antcommutaton relatons: {c ν 1, c ν 2 } = 0 (1.61a) {c ν1, c ν2 } = 0 (1.61b) {c ν1, c ν 2 } = δ ν1,ν 2. (1.61c) 13

15 Equaton (1.65) s called the fundamental antcommutator. Let ν 1, ν 2 be a two sngle-partcle ndces, and let N 0 be arbtrary. By the propertes of determnants, t s easy to see, that for any µ L 2 N, c ν 1 c ν 2 µ = c ν 2 c ν 1 µ. (1.62) Why? The rght hand sde s obtaned by exchangng the two frst rows of the determnant on the left hand sde. Snce ths equaton holds for any bass vector, we have shown that the two creaton operators antcommute {c ν 1, c ν 2 } c ν 1 c ν 2 + c ν 2 c ν 1 = 0. (1.63) Smlarly, two annhlaton operators antcommute, We now prove that Case 1: ν 1 = ν 2 = ν. Consder the expresson Case 1a: ν = µ for some. We get {c ν1, c ν2 } c ν1 c ν2 + c ν2 c ν1 = 0. (1.64) {c ν1, c ν 2 } c ν1 c ν 2 + c ν 2 c ν1 = δ ν1,ν 2. (1.65) c νc ν µ. (1.66) c νc ν µ 1 µ N = c ν( 1) 1 µ 1 µ 1 µ +1 µ N = ( 1) 1 µ µ 1 µ 1 µ +1 µ N = µ 1 µ N. (1.67) We also get c ν c ν µ 1 µ N = c ν µ µ 1 µ µ N = 0. (1.68) Case 1b: ν µ, ν s dstnct from all the µ. In ths case, c ν µ = 0, so On the other hand, Case 1 can be summarzed as as desred. Case 2: ν 1 ν 2. Let µ be arbtrary, and consder the expresson Case 2a: If ether ν 1 µ or ν 2 µ, the expresson vanshes. Smlarly, Case 2b: ν 1 µ and ν 2 µ (ν 2 = µ ): c νc ν µ 1 µ N = 0. (1.69) c ν c ν µ = c ν ν µ = ( 1) 0 µ. (1.70) {c ν, c ν} = 1 (1.71) c ν 1 c ν2 µ. (1.72) c ν2 c ν 1 µ = 0. (1.73) c ν 1 c ν2 µ = ( 1) 1 c ν 1 µ 1 µ 1 µ +1 µ N = µ 1 µ 1 ν 1 µ +1 µ N, (1.74).e., µ = ν 2 s replaced by ν 1. On the other hand, c ν2 c ν 1 µ = c ν2 ν 1 µ 1 µ 1 µ µ +1 µ N = ( 1) ν 1 µ 1 µ 1 µ +1 µ N = ( 1) µ 1 µ 1 ν 1 µ +1 µ N. (1.75) Summng, we see that Eq. (1.65) s proven n general. 14

16 1.2.3 Occupaton number representaton Consder a gven sngle-partcle bass {ϕ µ } and the correspondng bass of Slater determnants µ. Gven µ, we have N! rearrangements of the ndces. All of the rearrangements gve rse to the same Slater determnant, up to the sgn of the permutaton. If σ S N s the permutaton that sorts µ nto ν = σ( µ), then µ 1,, µ N = ( 1) σ µ σ(1),, µ σ(n) = ( 1) σ ν 1,, ν N. (1.76) So, the bass of Slater determnants can be chosen as those ndexed by sorted ndces. A sorted set of N ndces µ s n 1-1 correspondence wth a subset of ntegers, or equvalently, by a pcture of flled/unflled crcles, or occuped and unoccuped stes. One may say that the sngle-partcle functon ϕ µ s occuped n µ, whle ν µ s unoccuped. A common name for sngle-partcle functon n chemstry s orbtal, or spn-orbtal. We sometmes use the word orbtal for sngle-partcle functon. One can also consder µ as a bnary number wth N bts set: bt number ν s set f ν µ,.e., ν = µ for some = 1,, N. Thus, the Slater determnant µ 1 µ 2 µ 3 = 0, 1, 4 can be represented by the subset {µ 1, µ 2, µ 3 } = {0, 1, 4}, the pcture or the bnary number B = 2 µ1 + 2 µ2 + 2 µ3 = = = 19. (1.77) The dfferent bts are called occupaton numbers. The vacuum has no occuped sngle-partcle functons, and s represented by the bnary number 0 or the empty set. We use the notaton n 0 n 1 n µ to denote the Slater determnant wth occupaton numbers n µ {0, 1}, and by defnton we choose the one determnant out of the N! possble that has µ sorted: µ 1 < µ 2 < < µ N. We have n µ = 1 f and only f µ µ. In the above example, 0, 1, 4 = (1.78) If no ambguty can arse, we smply wrte etc. Agan, we stress that occupaton numbers only represent 1 of the N! Slater determnants possble to construct wth µ 1 through µ N, namely the one where all are sorted. But they stll form a bass. In the example, = 0, 1, 4 = 1, 0, 4 = 4, 1, 0 = 0, 4, 1 = + 1, 4, 0 = + 4, 0, 1, (1.79) exhaustng all possbltes of N! = 3! = 6 permutatons. All these determnants are clearly lnearly dependent. The followng defnton can be useful: Let µ be an ndex set, and let ν be an arbtrary ndex. Then #ν s the number of µ that satsfes µ < ν. Thus #ν counts the occuped sngle-partcle functons before ν. 15

17 Exercse Let µ = {µ 1,, µ N } be a gven set of occuped orbtals, wth occupaton number representaton n 0 n 1 n 2 Show that: c 0 f ν s occuped ν n 0 n 1 n 2 = ( 1) #ν n 1 n 1 n 2 n ν 1 1 ν n ν+1 f ν s unoccuped (1.80) 0 f ν s unoccuped c ν n 0 n 1 n 2 = ( 1) #ν n 1 n 1 n 2 n ν 1 0 ν n ν+1 f ν s occuped (1.81) Exercse Let ϕ, = 1, 2, 3 be three orthonormal sngle-partcle functons. Consder the determnants 1, 2, 3, 1, 3, 2, 2, 1, 3, 3, 2, 1, 2, 3, 1 and 3, 1, 2. a) Are there further N = 3 Slater determnants that can be created usng the sngle-partcle orbtals ϕ, = 1, 2, 3 only? b) Wrte down a bass for the space spanned by the sx determnants,.e., a bass for all the vectors on the form Ψ = a 1 1, 2, 3 + a 2 1, 3, 2 + a 3 2, 1, 3 + a 4 3, 2, 1 + a 5 2, 3, 1 + a 6 3, 1, 2. (Here, a are complex numbers.) Spn orbtals and orbtal dagrams Consder a system of electrons. Confguraton space s X N for N electrons, and for a sngle electron X = R d {+1, 1}, so L 2 (X) = L 2 (R 3 ) C 2. Ths means that each ψ L 2 (X) s a two-component functon, one for spn-up and one for spn-down. The notaton for spn can vary. Here, we use +1 for spn up and 1 for spn down, along the z-axs. Ths s arbtrary, of course. In chemstry, one often uses α for spn up, and β for spn down, as symbols. (Ths s the notaton n Szabo and Ostlund, for nstance.) Sometmes one uses arrows and, or and 1 2. If {φ p ( r)} s an orthonormal bass for L 2 (R 3 ), the space part, and χ +1 (σ) and χ 1 (σ) are bass functons for C 2, σ {+1, 1}, the spn space, we have a bass for L 2 (X) va tensor products: Typcally, one chooses That s, Or, yet another formula, ϕ µ (x) = ϕ p,α ( r, σ) = φ p ( r)χ α (σ). χ +1 = ( 1 0 ), χ 1 = ( 0 1 ) χ +1 (+1) = 1, χ +1 ( 1) = 0, χ 1 (+1) = 0, χ 1 ( 1) = 1. χ α (σ) = δ α,σ. 16

18 The operators S x, S y and S z act on spn degrees of freedom only, and ther matrces are gven by the Paul matrces: σ S χ α = 1 2 ħσ,ασ. Thus, The N-body spn operator s σ S x χ α = ħ 2 ( ) (1.82) σ S y χ α = ħ 2 (0 0 ) (1.83) σ S z χ α = ħ 2 ( ) = ħαδ ασ (1.84) Ŝ = N =1 S () (1.85) where S () acts only on the spn of partcle. Suppose the Hamltonan of the electronc system s ndependent of spn,.e., the Hamltonan acts only on the degrees of freedom r, and not σ for each partcle. Then, [Ĥ, Ŝ z ] = 0 and we can fnd a common set of egenvectors for Ĥ and Ŝ z. Consder the one-body part Ĥ 0 of the Hamltonan, whch now s a purely spatal operator: Ĥ 0 = ĥ( r ). (1.86) Let ĥ, an operator on the space L2 (R 3 ), have a complete set of egenfunctons, φ p ( r), ĥφ p ( r) = є p φ p ( r). Then, as operator on L 2 (X), we have the complete set ϕ µ ( r, σ) = φ p ( r)χ α (σ), and we see that the snglepartcle functons are doubly degenerate: ĥϕ p,α (x) = є p ϕ p,α (x), α {+1, 1}. Here, µ = (p, α) s the combned space/spn quantum numbers. In chemstry parlance, φ p ( r) s an orbtal, whle ϕ µ (x) s a spn-orbtal. Only the spn-orbtal s a sngle-partcle functon n the sense that we use n ths text,.e., a bona fde element n L 2 (X), the snglepartcle Hlbert space. The orbtal s an element n L 2 (R 3 ) and must be adoned wth a spn bass functon to become a sngle-partcle functon, a spn-orbtal. Each spn-orbtal can be occuped by only one electron, but each orbtal has room for two one spn up and one spn down. One typcally llustrates the egenfunctons and the occupatons of Slater determnants va a dagram le Fgure 1.1. In the fgure, sx spn-orbtals are occuped, and three orbtals are doubly occuped. The llustrated state s Ths s the N = 6-electron ground-state wavefuncton of Ĥ 0. c 0,+1 c 0, 1 c 1,+1 c 1, 1 c 2,+1 c 2, 1. (1.87) 17

19 ε 4 ε 3 ε 2 ε 1 ε 0 Fgure 1.1: Illustraton of spn-orbtals and a Slater determnant of 6 electrons Foc space The space L 2 N has the bass consstng of the Slater determnants n 0n 1 n 2 n 3 wth n total N occuped orbtals, or N bts set n the bnary representaton. The creaton operator c µ nserts a bt n poston µ f t s zero, and gves the zero vector f t was already 1. Smlarly, the operator c µ turns a bt off, see Exercse 1.11 It s natural to consder Foc space, the drect sum of all L 2 N : F = L 2 N. (1.88) N=0 By defnton, µ ν = 0 f ν and µ have dfferent number of partcles,.e., dfferent number of occuped sngle-partcle functons. Thus = 0 (1.89) for example, snce the number of partcles dffer n the two functons. Now, c µ F F maps entrely nsde F, and smlarly wth c µ. A bass for F s the set of all n 0 n 1 wth arbtrary number of orbtal occuped. The bnary number representaton s qute useful for computer programs nvolvng Slater determnants, as easly can be magned. A specal operator, the number operator: Let ν be arbtrary. We have that and furthermore that Thus, Therefore, we defne ν c ν n 0 n 1 = ( 1) #ν n ν n 0 n 1 0 ν, (1.90) c νc ν n 0 n 1 = ( 1) 2#ν n ν n 0 n 1. (1.91) c νc ν n 0 n 1 = n ν n 0 n 1 = N n 0 n 1. (1.92) ν ˆN c νc n u. (1.93) ν Ths operator extracts the number of fermons n a state Ψ n the sense that for any Ψ F, ˆN Ψ = N Ψ f and only f Ψ L 2 N. 18

20 1.2.6 Truncated bases For physcal partcles, the Hlbert space s nfnte dmensonal. But, as we have seen n exercses, especally Exercse 1.13, we can select a few sngle-partcle functons ϕ µ, and construct Slater determnants out of these. These wll be fnte n number. From a mathematcal perspectve, we can consder these fnte sngle-partcle functons to defne a sngle-partcle space on ther own: Thus, ψ V 1 means V 1 = span{ϕ 1,, ϕ L } L 2 (X). (1.94) ψ(x) = L µ=1 ψ µ ϕ µ (x). Havng selected the fnte bass, we obtan for dfferent N a Slater determnant bass, spannng V N L 2 (X N ) AS. Clearly, as we have only L sngle-partcle functons avalable, we cannot create more than N partcles from vacuum wthout gettng at least one repeated creaton operator,.e., we must have L N to have nonzero dmenson. The general dmenson s dm(v N ) = ( L N ). In computatonal settngs, the truncaton of the nfte bass nto a fnte one s almost unversally done. Of, course, we can only numercally dagonalze a fnte matrx! But we would stll le the bass to be as large as possble to acheve the greatest accuracy. At least ntutvely, we expect that as we nclude more and more sngle-partcle functons, the numercal results wll approach the exact result. Under mld assumptons on the bass set and the Hamltonan under consderaton, ths s n fact true. Sometmes, the fnte truncaton s done after a detaled consderaton of the physcs of the system. Ths can gve consderable physcal nsght, gvng great explanatory power to the second quantzed pcture. As an example, tae the physcal explanaton of the prncples of a laser. (See for example wpeda.org/w/populaton_nverson.) Another example s the Hubbard model from soldstate physcs, see for example Exercse [Note: Ths exercse has been updated snce t was gven as a weely exercse.] Let ϕ µ, µ = 1, 2,, 6 be gven orthonormal sngle-partcle functons. a) Usng the µ 1,, µ N notaton, wrte down a bass for the fnte dmensonal subspace of L 2 (X N ) AS for N = 2, N = 3 and N = 4, that you can construct usng the gven sngle-partcle functons. (Mae sure you nclude only lnearly ndependent Slater determnants.) b) Can you construct a Slater determnant for N = 10 partcles usng the gven ϕ µ? c) Usng the occupaton number notaton n 1 n 2 n 6 notaton, wrte down a bass for the same spaces as n exercse a). c) What s the dmenson of the subspace of Foc space you can create wth the 6 sngle-partcle functons? e) Assume that you have L orbtals nstead of ust 6. What s the dmenson of the N-partcle spaces you can buld? What s the dmenson of the Foc space you can buld? 19

21 Exercse 1.14 (Note: Ths exercse has been updated snce t was gven as a weely exercse.). Consder the followng pcture: ϕ 3 ϕ 2 ϕ 1 ϕ 0 We have four horzontal lnes, each representng a sngle-partcle functon ϕ µ. The crcle represents an occuped sngle-partcle functon,.e., the Slater determnant 0. a) In a smlar fashon as the the above pcture, draw a pctures of all the dstnct Slater determnants you can create usng the four sngle-partcle functons. Mae sure you consder all possble partcle numbers. Capton each pcture wth the correspondng µ 1 µ 2 µ N. We now consder electrons. Consder 4 spn-orbtals φ p ( r),.e., 8 spn-orbtals ϕ µ ( r, σ). The correspondng dagram for the Slater determnant 0, 0 s: φ 3 φ 2 φ 1 φ 0 Each level now can hold 2 electrons, spn up and spn down. b) Draw all possble 2-electron Slater determnants. Mar those that have total spn proecton 0. c) Consder the one-body operator gven by Ĥ 0 = є p (c p c p + c p c p ). p Here, є p are numbers such that є 1 < є 2 <. In frst quantzaton, Ĥ 0 = N =1 ĥ(). Wrte down the matrx of the (sngle-electron) operator ĥ n the spn-orbtal bass {ϕ pσ} and fnd ts egenfunctons. Interpret the spn-orbtal dagram n terms of your results. Fnd the N = 4 ground state of Ĥ 0, and draw a pcture of t. 20

22 1.3 Representaton of operators What we wll prove In ths secton, we shall demonstrate the followng representaton of one-body operators: Ĥ 0 = N =1 ĥ() = µ ĥ ν c µc ν. (1.95) µν Note that the last expresson does not contan N explctly. Here, note that µ s a sngle-partcle functon t s the Slater determnant ϕ µ (x 1 ). The number µ ĥ ν s the matrx element of the sngle-partcle operator ĥ n the gven one-partcle bass, µ ĥ ν = dxϕ µ (x) ĥϕ ν (x). (1.96) Eq. (??) gves a nce mage of how ˆ H 0 acts on a bass functon: each term n the sum manpulates the Slater determnant s occuped orbtals and weghs t wth a matrx element. Smple, and not at all obvous from the sngle quantzed form. We shall also prove the followng formula for the two-body operator: Ŵ = N (, ) ŵ(, ) = 1 2 where the orderng of the annhlaton operators should be noted. Here, wµν αβ c µc νc β c α, (1.97) µναβ w µν αβ = dx 1 dx 2 ϕ µ (x 1 ) ϕ ν (x 2 ) w(x 1, x 2 )ϕ α (x 1 )ϕ β (x 2 ) (1.98) s a matrx element usng tensor product two-body functons, not Slater determnants. Usng Slater determnant matrx elements we n fact have a smlar expanson, Ŵ = 1 4 µν ŵ αβ c µc νc β c α, (1.99) µναβ where thus the matrx elements are antsymmetrc, computed as a matrx element usng two-body Slater determnants. A word of warnng: notaton for two-body matrx elements s notorouly varyng between sources. Some authors use the notaton ϕ α ϕ β ŵ ϕ µ ϕ ν for the matrx element wµν αβ, whch s not antsymmetrc. In our case, the notaton clashes wth the Slater determnant matrx element, but we wll stll sn n ths respect. Some authors wrte ϕ α ϕ β ŵ ϕ µ ϕ ν AS for the ant-symmetrc Slater-determnant matrx element (and sometmes we wll too), whch s equal to: ϕ α ϕ β ŵ ϕ µ ϕ ν AS = αβ ŵ µν = w αβ µν w αβ νµ. (1.100) Ths can cause some confuson, as the expansons usng tensor products and Slater determnants dffer by a factor 2... The proofs gven n ths secton borrow heavly from [5]. Lemma 1.1. Let µ 1 µ 2 µ N be a Slater determnant bult from orthonormal sngle-partcle functons ϕ µ, no partcular orderng assumed. The operator c νc α replaces ϕ α wth ϕ ν (or gves zero of α s not present n µ), wth no sgn change. Smlarly, c ν 1 c ν 2 c α2 c α1 replaces α 1 wth ν 1, and α 2 wth ν 2, or gves zero f one of α 1 or α 2 s not present n µ. 21

23 Exercse Prove the lemma One-body operators We prove Eq. (1.95) by showng that the actons of the left- and rght-hand sdes on an arbtrary Slater determnant agree. Let therefore {ϕ µ } be a sngle-partcle bass as usual. Consder the acton of Ĥ 0 = ĥ() on an arbtrary Slater determnant: 1 Ĥ 0 ϕ µ1, ϕ µ2,, ϕ µn = ( ĥ()) ( 1) σ ˆP σ ϕ ν1 (x 1 ) ϕ νn (x N ) N! σ 1 = ( 1) σ ˆP σ ( ĥ()) ϕ ν1 (x 1 ) ϕ νn (x N ) N! σ = (ĥϕ ν 1 ), ϕ ν2,, ϕ νn + ϕ ν1, (ĥϕ ν 2 ),, ϕ νn + + ϕ ν1, ϕ ν2,, (ĥϕ ν N ) (1.101) Here, we used that ˆP σ commutes wth Ĥ 0. Consder the operator ĥ actng on a sngle-partcle functon ϕ µ. The result, ψ, can be expanded n the bass: ψ(x) = ĥϕ µ(x) = ϕ ν (x) ν ĥ µ. (1.102) ν We nsert ths expanson: Ĥ 0 ϕ µ1, ϕ µ2,, ϕ µn = (ĥϕ ν 1 ), ϕ ν2,, ϕ νn + ϕ ν1, (ĥϕ ν 2 ),, ϕ νn + + ϕ ν1, ϕ ν2,, (ĥϕ ν N ) = ν ν ĥ µ 1 νµ 2,, µ N + ν ν ĥ µ 2 µ 1 νµ 3,, µ N + + ν ĥ µ N µ 1 µ 2,, ν ν (1.103) Now, we note that whch we plug n: µ 1,, µ 1 νµ +1 µ N = c νc µ µ 1,, µ N, (1.104) Ĥ 0 ϕ µ1, ϕ µ2,, ϕ µn = ν = [ ν ν ĥ µ 1 νµ 2,, µ N + ν ν ĥ µ 1 c νc µ1 + ν ν ĥ µ 2 µ 1 νµ 3,, µ N + + ν ĥ µ N µ 1 µ 2,, ν ν ν ĥ µ 2 c νc µ2 + + ν ĥ µ N c νc µn ] µ 1,, µ N. ν (1.105) Fnally, we note that c µ µ 1 µ N = 0 whenever µ µ, so we may extend the summaton over µ to all of µ, resultng n: Ĥ 0 ϕ µ1, ϕ µ2,, ϕ µn = [ ν ν ĥ µ 1 c νc µ1 + ν = ν ĥ µ c νc µ µ 1,, µ N. µν ν ĥ µ 2 c νc µ2 + + ν ĥ µ N c νc µn ] µ 1,, µ N ν Snce µ 1,, µ N was an arbtrary Slater determnant, we have proven Eq. (1.95). (1.106) 22

24 1.3.3 Two-body operators The operator Ŵ = < ŵ(, ) s a two-body operator. The operator ŵ(1, 2) s thus an operator on L 2 (X 2 ) that s completely characterzed by ts acton on a bass: the tensor products ϕ µ1 (x 1 )ϕ µ2 (x 2 ). Thus, ŵ(1, 2)ϕ µ1 (x 1 )ϕ µ2 (x 2 ) = ν 1ν 2 w ν1ν2 µ 1 µ 2 ϕ ν1 (x 1 )ϕ ν2 (x 2 ), (1.107) where the matrx elements w ν1ν2 µ 1 µ 2 are gven by the formula (1.98). There s nothng specal about the ndces (1, 2), t may ust as well be (, ). Note also the symmetry property w ν1ν2 µ 1 µ 2 = w ν2ν1 µ 2 µ 1. As for the one-body case, Ŵ commutes wth ˆP σ, and we get, usng Eq. (1.107), 1 Ŵ ϕ µ1 ϕ µn = ( 1) σ ˆP σ ŵ(, )ϕ µ1 (x 1 ) ϕ µn (x N ) N! σ < 1 = ( 1) σ ˆP σ ŵ(, )ϕ µ1 (x 1 ) ϕ µn (x N ) N! σ < 1 = ( 1) σ ˆP σ wµ N! σ ν1ν2 µ ϕ µ1 ϕ ν1 (x ) ϕ ν2 (x ) ϕ µn (x N ) < ν 1ν 2 = µ µ ϕ µ1 ϕ ν1 ϕ ν2 ϕ µn w ν1ν2 < ν 1ν 2 = w ν1ν2 < ν 1ν 2 µ µ c ν 1 c ν 2 c µ c µ ϕ µ1 ϕ µn. (1.108) Here, we used Lemma 1.1 about replacement behavour of the c c cc product. We are currently summng over µ and µ, such that <. Includng = gves zero contrbuton (why?), and ncludng > gves equal contrbuton: w ν1ν2 < ν 1ν 2 = µ µ c ν 1 c ν 2 c µ c µ ϕ µ1 ϕ µn = w ν1ν2 < ν 1ν 2 = w ν2ν1 < ν 2ν 1 w ν1ν2 < ν 1ν 2 µ µ c ν 2 c ν 1 c µ c µ ϕ µ1 ϕ µn = w ν1ν2 < ν 1ν 2 µ µ c ν 1 c ν 2 c µ c µ ϕ µ1 ϕ µn = w ν1ν2 < ν 2ν 1 µ µ c ν 1 c ν 2 c µ c µ ϕ µ1 ϕ µn µ µ c ν 2 c ν 1 c µ c µ ϕ µ1 ϕ µn µ µ c ν 1 c ν 2 c µ c µ ϕ µ1 ϕ µn (1.109) Here, we used the antcommutators and symmetry of the matrx elements. Assemblng the two contrbutons, Ŵ ϕ µ1 ϕ µn = 1 2 ν 2ν 1 w ν1ν2 µ µ c ν 1 c ν 2 c µ c µ ϕ µ1 ϕ µn. (1.110) We note that the sum over s really a sum over two occuped orbtals µ and µ. We can therefore extend the sum to all unoccuped orbtals as well, snce c α µ gves zero contrbutons for such orbtals. Thus, Eq. (1.97) s proven. We leave the proof of the antsymmetrzed verson as an exercse. Exercse Prove Eq. (1.99). Start wth showng Eq. (1.100). 23

25 Exercse a) Let ˆF = N =1 ˆf () be a frst-quantzaton operator. Wrte down the second-quantzed form of ths operator. Let Ĝ = < ĝ(, ) be a general two-body operator, where ĝ(1, 2) = ĝ(2, 1). Wrte down the second-quantzed form. b) Usng the fundamental antcommutator relatons, compute the matrx element µ 1 µ 2 ˆF µ 1 µ 2 c) Usng the fundamental antcommutator relatons, compute the matrx element µ 1 µ 2 µ 3 ˆF µ 1 µ 2 µ 3 d) Usng the fundamental antcommutator relatons, compute the matrx element µ 1 µ 2 Ĝ µ 1 µ 2 e) Usng the fundamental antcommutator relatons, compute the matrx element f) Compute the matrx element g) Compute the matrx element µ 1 µ 2 µ 3 Ĝ µ 1 µ 2 µ 3 µ 1, µ 2,, µ N ˆF µ 1, µ 2,, µ N µ 1, µ 2,, µ N Ĝ µ 1, µ 2,, µ N Exercse (Tedous, but very nstructve.) In ths exercse, we prove the so-called Slater Condon rules: the explct expressons of matrx elements of one- and two-body operators n a Slater determnant bass. We do not assume any partcular orderng of the occuped sngle-partcle functons consdered. If you solved Exercse 1.17, you solved parts of ths exercse. a) Usng the fundamental antcommutator relatons, compute µ Ĥ 0 µ and µ Ŵ µ and prove that µ Ĥ 0 µ = µ Ŵ µ = N =1 N < b) Let ν be equal to µ, except for one occuped orbtal,.e., h µ µ, (1.111) µ µ ŵ µ µ AS = 1 2 µ µ ŵ µ µ AS. (1.112) ν = c ν c µ µ, ν µ. (1.113) Usng the fundamental antcommutator relatons, compute µ Ĥ 0 µ and µ Ŵ ν, and fnd µ Ĥ 0 ν = h µ ν, (1.114) µ Ŵ ν = µ µ ŵ µ ν AS. (1.115) 24

26 p = 3 p = 2 p = 1 Fgure 1.2: Schematc plot of the possble sngle-partcle levels wth double degeneracy. The flled crcles ndcate occuped partcle states. The spacng between each level p s constant n ths pcture. We show some possble two-partcle states. c) Let ν be equal to µ, except for two ndces,.e., ν = c ν c ν c µ c µ µ,. (1.116) Usng the fundamental antcommutator relatons, compute µ Ŵ ν and fnd µ Ĥ 0 ν = 0, (1.117) µ Ŵ ν = µ µ ŵ ν ν AS. (1.118) d) Explan that f ν dffers from µ n more than two occuped functons, then µ Ŵ ν = 0. Exercse (Ths exercse adapted from an exercse by Morten Horth-Jensen.) We wll now consder a smple three-level problem, depcted n Fgure 1.2. The sngle-partcle states are labelled by the quantum number p and can accomodate up to two sngle partcles, vz., every snglepartcle state s doubly degenerate (you could thn of ths as one state havng spn up and the other spn down). We let the spacng between the doubly degenerate sngle-partcle states be constant, wth value d. The frst state has energy d. There are only three avalable sngle-partcle states, p = 1, p = 2 and p = 3, as llustrated n the fgure. a) How many two-partcle Slater determnants can we construct n ths space? b) We lmt ourselves to a system wth only the two lowest sngle-partcle orbts and two partcles, p = 1 and p = 2. We assume that we can wrte the Hamltonan as Ĥ = Ĥ 0 + Ĥ I, and that the onebody part of the Hamltonan wth sngle-partcle operator ĥ0 has the property ĥ 0 ψ pσ = p dψ pσ, where we have added a spn quantum number σ. We assume also that the only two-partcle states that can exst are those where two partcles are n the same state p, as shown by the two possbltes to the left n the fgure. The two-partcle matrx elements of Ĥ I have all a constant value, g. Show then that the Hamltonan matrx can be wrtten as ( 2d g g g 4d g ), 25

27 and fnd the egenvalues and egenvectors. What s mxng of the state wth two partcles n p = 2 to the wave functon wth two-partcles n p = 1? Dscuss your results n terms of a lnear combnaton of Slater determnants. c) Add the possblty that the two partcles can be n the state wth p = 3 as well and fnd the Hamltonan matrx, the egenvalues and the egenvectors. We stll nsst that we only have two-partcle states composed of two partcles beng n the same level p. You can dagonalze numercally your 3 3 matrx. Ths smple model catches several brds wth a stone. It demonstrates how we can buld lnear combnatons of Slater determnants and nterpret these as dfferent admxtures to a gven state. It represents also the way we are gong to nterpret these contrbutons. The two-partcle states above p = 1 wll be nterpreted as exctatons from the ground state confguraton, p = 1 here. The relablty of ths ansatz for the ground state, wth two partcles n p = 1, depends on the strength of the nteracton g and the sngle-partcle spacng d. Fnally, ths model s a smple schematc ansatz for studes of parng correlatons and thereby superfludty/superconductvty n fermonc systems. 1.4 Wc s Theorem Supportng materal for ths secton s Shavtt/Bartlett Ch. 3, Gross/Runge/Henonen Ch A sort of summary and motvaton Let us tae a loo at what we have so far. In the preceedng sectons, we ntroduced a collecton of tools for descrbng (1) many-fermon states n many-partcle Hlbert space, and (2) second-quantzaton language for expressng these states and, mportantly, the quantum-mechancal Hamltonan. The quantum mechancal Hlbert space for N fermons s defned solely n terms of the confguraton space X of a sngle fermon. The Hlbert space of a sngle partcle s L 2 (X), the set of square ntegrable sngle-partcle functons ψ X C. Such a space always has an orthonormal bass, say {ϕ µ }. Formng Slater determnants ϕ µ1 ϕ µ2 ϕ µn, we obtan totally antsymmetrc bass functons for L 2 (X N ) AS. Furthermore, we defned Foc space F, F = L 2 (X N ) AS. N=0 Insde the Foc space, every possble wavefuncton of a system of some number of fermons exst. Gven a wavefuncton Ψ N F, t could be expanded n Slater determnants, Ψ N = µ µ 1 µ N µ 1 µ N Ψ N. (1.119) Here, the subscrpt N only ndcates that we now the number of partcles. The notaton µ 1 µ N ndcates that a certan sngle-partcle bass {ϕ µ } has been chosen, snce we only lst the ndces µ. Each determnant can be constructed from vacuum usng creaton operators c µ (these, of course, depend on the bass), µ 1 µ N = c µ 1 c µ 2 c µ N. (1.120) 26