Now that we've developed our approximation methods, we can turn to solving the. 2m 4 r - 2e 4 r + e 2 0) 1

Transcription

1 11 Lctur 18 Now that w'v dvlopd our approximation mthods, w can turn to solving th hlium atom. As usual our Schrödingr quation is H ψ = E ψ, whr H = - ( + )- m 4 r - 4 r + 1 πε πε 4πε r 1 Lt s bgin by applying prturbation thory and s what w can larn from it. W ar intrstd in particular in th ground stat wavfunction. WHAT IS OUR UNPERTURBED HAMILTONIAN? [ () ˆ =- ( ˆ (1) ˆ 1+ )- - = HH + HH() m r1 r H 1, whr H H is th Hamiltonian for a hydrogn-lik atom.] WHAT IS OUR ψ ()? [ψ b g ( r, r ) = ψ ( 1 ) ψ ( ) s s Z 3/ a,whr ψ1s (1) = ( ) ]. Th nrgy of an lctron in a hydrogn-lik systm is π A givn by Zr 4 Z m E n = - (4πε ) n o WHAT IS THE ENERGY OF THE GROUND STATE OF OUR UNPERTURBED SYSTEM? -Z m (4 n + -Z m (4 n = -4m () E = E 1s(1)+ E 1s()= πε ) 1 πε ) (4 πε ) WHAT IS OUR PERTURBATION? [ ( 1) H = ] 4 πε r WHAT IS THE EQUATION WE USE TO CALCULATE THE CORRECTION TO OUR ENERGY? 1 (1) ()* () ()* () E= ψ Hˆ ψ d τ = ψ ψ dτ 4πε r 1

2 111 Whn w plug in th zro ordr wavfunction w just workd out w gt E = 5 4 Zm 8 (4πε ) So our total nrgy is E () + E = m (4πε ). Bcaus this is an approximation, w xpct that this nrgy will diffr from th tru nrgy. In fact w find that th xprimntal valu for this nrgy is EH(xp) = m (4πε ). Thus ignoring th scond ordr and highr trms lads to an rror of about 5%. Howvr, just including th scond ordr corrction improvs th nrgy to m (4πε ). Using svral ordrs of prturbation thory, w can improv th nrgy to within.1 % of th xprimntal valu. Thus for Hlium w can mak th following obsrvations. 1) Th first ordr corrction is larg, not small, sinc E (1) =.31 E (). ) Th rror rmaining aftr th first ordr corrction is about 5%. 3) If w go on to scond ordr prturbation thory, th rror is rducd to.%. What conclusions can w draw from ths obsrvations? Th main conclusion w can draw at this point is that th hlium wavfunction must b substantially diffrnt than th unprturbd wavfunctions w usd to calculat thir nrgis. To s this rmmbr that whn w drivd th formula for th first ordr corrction, w assumd that w could writ th nrgy and th wavfunction as E = E () + E (1), whr E (1) is small, and ψ = ψ () + ψ (1),

3 11 whr ψ (1) is small. Sinc E (1) was not small, w can conclud that ψ (1) was not small, and that thrfor th hlium atom ground stat wavfunction is substantially diffrnt than two lctrons in hydrogn-lik 1s orbitals. Now lt s turn to th variational mthod. First w writ down our Hamiltonian, H = - ( + )- m 4 r - 4 r + 1 πε πε 4πε r 1 Nxt w nd to choos a trial function. A logical choic is to us th product of two hydrogn-lik 1s orbitals, sinc this would corrspond to putting on lctron in ach of ths orbitals. If w writ this out this givs us 1 φ = 1 π ( Z 1 a ) 3/ 1 Z a ( π a ) Rmmbr that in ordr to us th variational mthod, w nd to choos a variational paramtr, a paramtr that w can vary in ordr to find th minimum nrgy for our trial function. For this trial function w'll lt th atomic numbr Z b our adjustabl paramtr. Essntially what w'r doing hr is asking, "What if th nuclar charg that ach lctron ss is lss than th actual charg? What will this "ffctiv nuclar charg" b?" Whn w plug this trial function into our formula for th nrgy Eφ of our trial function, w find that E = φ Zr * φ Hˆ φdτ * φ φdτ 3/ Zr a E φ = ( Z Z) 4 m (4πε )

4 113 Whn w tak th drivativ with rspct to Z, w find that th minimum nrgy w gt 4 m for this trial function is whn Z = 7/16, and is qual to (4πε ). This is an rror of about %, and is a prtty good rsult for a first try. Rmmbr that w said that whn w allow Z to vary, w ar trating it as an ffctiv nuclar charg. Th ffctiv charg w calculatd 7/16 = 1.69, is substantially smallr than th actual nuclar charg of two. What dos it man for th nuclus to hav an ffctiv charg smallr than its actual charg? It mans that th forc that th nuclus xrts on th lctron corrsponds to that smallr charg. To s why this maks sns lt s considr a pictur of a hlium atom. At any givn tim w hav two lctrons in th vicinity of th nuclus at two diffrnt radii, r1 and r. In most positions of th lctrons, thy will both fl th full charg of + from th hlium nuclus. Now suppos that th two lctrons ar lind up. Th lctron at r1 ss th full charg of th nuclus, but th lctron at r ss only a charg of + 1 bcaus th othr lctron ffctivly cancls part of th charg. W say that th first lctron partially scrns th scond lctron from th nuclus. Th rducd ffctiv charg rflcts th fact that thr ar a numbr of positions whr on lctron can scrn th nuclus for th othr.

5 114 Lctur 19 Now I'd lik to mov along in th priodic tabl and calculat th ground stat nrgy of lithium. Whn w startd to do this for hlium, w bgan with a trial function that put both lctrons in 1s orbitals, φh = φ1s(1)φ1s(). It would sm rasonabl to xtnd this to lithium and to say that its ground stat consists of thr lctrons in 1s orbitals, φli = φ1s(1)φ1s()φ1s(3). W all know that w can't hav thr 1s lctrons in th sam atom. Howvr, nothing w'v dvlopd at this point prcluds such a ground stat. To undrstand why such a ground stat cannot xist, w nd to introduc a nw physical proprty, th lctron spin, and a nw postulat of quantum mchanics, th Pauli Exclusion Principl. Th vidnc for this nw proprty, th socalld lctron spin, cam unxpctdly in an xprimnt prformd in 19 by Otto Strn and Waltr Grlach. In thir xprimnt, thy masurd th magntic stats of silvr atoms by passing a bam of silvr atoms through a magntic fild. What did thy xpct to s? First, raliz that th magntic fild will intract only with th magntic momnts of th atoms. Thus any ffct of th magntic fild on th bam of atoms will b as a rsult of this magnt-magnt intraction. What happns if w pass a moving magnt through a magntic fild? Suppos our magntic fild, B, is causd by a prmannt magnt. Such a magnt has two pols, a positiv pol and a ngativ pol. W'll put our magnt with th positiv pol facing

6 115 downward. Suppos that th travling magnt has its ngativ pol facing upward. Whn it intracts with th B fild from th prmannt magnt, it will b attractd, so its path will b dflctd upward. What if its positiv pol is facing upward? Bcaus idntical pols ar intracting, th travling magnt will b rplld and th bam will b dflctd downward. Th siz of th dflction dpnds on th orintation of th magnt passing through th fild. Th attraction of two magnts is maximizd whn th positiv and ngativ pols point dirctly at ach othr, and thus th upward dflction is maximizd for this cas. Th rpulsion is maximizd whn th two positiv pols point dirctly at ach othr, and thus th downward dflction is maximizd for this cas. If th pols of th two magnts ar at right angls to ach othr, thr is no rpulsion or attraction at all, and no dflction. If th travling magnt is pointing so its positiv pol is somwhr btwn prpndicular and pointing dirctly at th othr magnt, thn th attraction will b somwhr btwn and th maximum, and thr will b a dflction somwhat lss than th maximum. What do w xpct to s for a bam of silvr atoms? Th bam of atoms can b prpard so that all of th atoms ar charactrizd by th sam angular momntum L= ( ll ( + 1) 1/. This angular momntum stat dtrmins th strngth of th atom's magntic momnt. Th largr l is, th strongr th magnt is. For vry l thr ar l + 1 valus of m. Ths m valus tll us which way th magnt is pointing. An m valu of +l will hav th positiv pol of th magnt as clos as possibl to th positiv pol of th fild, so w will hav th maximum downward dflction of th bam of atoms. All othr positiv m valus will also rsult in a downward dflction, but to a lssr dgr. If th m valu is zro, th magnt will b prpndicular to th magntic fild, and thr will b

7 116 no dflction. Finally if th m valu is ngativ, th ngativ pol of th magnt will b facing th magntic fild, and th dflction will b upward. Th closr m is to - l, th strongr th dflction will b. Lt s prdict what will happn to th bam if th atoms ar prpard so that all of thm hav l =, l = 1, and l =. If l =, thr is only on m valu,, and whn th bam passs through th magntic fild it will b undflctd. Whn l = 1 w hav thr possibl m valus, m = -1,, 1. Thos atoms with m = 1 will b dflctd downward, thos atoms with m = will b undflctd, and thos atoms with m = -1 will b dflctd upward. Thus a bam of atoms with l = 1 will b split into thr bams whn it passs into a magntic fild. Lts considr l =. WHAT ARE THE POSSIBLE M VALUES? WHAT WILL HAPPEN TO A BEAM OF ATOMS WITH L = WHEN THEY PASS THROUGH THE MAGNETIC FIELD? So what did Strn and Grlach actually s? Whn thy passd thir bam of silvr atoms through th magntic fild it split into two componnts. From our analysis abov, w know that this can't b du to an l = stat which would lav th bam compltly undflctd, or to an l = 1 stat which would split th bam into thr componnts. Rmmbr that thr ar l + 1 m

8 117 stats for ach valu of l. W can us this to figur out th valu of th angular momntum quantum numbr that lads to this strang bhavior. Sinc w hav two componnts, w can writ = l + 1 and thrfor l = 1/. This has to b du to a nw proprty sinc w alrady know that th l quantum numbr, which dtrmins orbital angular momntum, is limitd to th valus l =, 1,,... Yt this nw proprty must b du to som kind of angular momntum, sinc a magntic momnt rsults. In 195, Wolfgang Pauli xplaind this bhavior by postulating that an lctron could xist in two stats. To labl ths stats, Pauli postulatd th xistnc of a fourth quantum numbr, ms, which w now call th spin quantum numbr. This spin quantum numbr can only tak on th valus ms = 1/ or -1/. Whn Pauli introducd this quantum numbr, h had not yt givn it an intrprtation. It was Gorg Uhlnbck and Samul Goudsmit of th Univrsity of Lidn who idntifid ths quantum numbrs with angular momntum stats of th lctron. Thy ar calld spin stats bcaus thy hav th sam magnitud no mattr how fast th lctron is travling, so this can't b du to orbital motion. If th lctron wr a classical objct th only sourc of angular momntum othr than orbital motion would b spinning, top lik motions, hnc th nam spin. W should b cautiond, howvr, against taking th nam too litrally. Th lctron is not actually spinning. Rathr w should viw ms as rfrring to th orintation of a magntic momnt intrinsic to th lctron. To clarify this point, Bohr has calculatd that for a spinning lctron to hav th xprimntally dtrmind magntic momnt, it would hav to spin at a spd many tims gratr than th

9 118 spd of light, an impossibility. Elctron spin is a purly quantum phnomnon. It has no classical analog. Th way that w'v introducd th lctron spin quantum numbr may rmind you of th Bohr hydrogn atom. Rmmbr that Bohr postulatd th quantization of th orbital angular momntum rathr than hav it com as a solution to his quations. This is rathr unsatisfactory, bcaus it dosn't yild much in th way of physical insight, by rlating th nw phnomnon to fundamntal principls. And in fact, th Schrödingr and Hisnbrg formulations of quantum mchanics cannot prdict th xistnc of th lctron spin. Fortunatly for quantum mchanics, P.A.M. Dirac, in th 193's, constructd a wav quation that combind th idas of Schrödingr with Einstin's spcial rlativity. This Dirac quation was immnsly succssful. Not only did its solution includ th lctron spin quantum numbr, but it prdictd th xistnc of positrons bfor thy wr first obsrvd. How do w dal quantitativly with lctron spin in our non-rlativistic formulation of quantum mchanics? Th spin quantum numbr, ms, as th Strn- Grlach xprimnt showd, is an xprimntal obsrvabl. According to our postulats, any obsrvabl is th ignvalu of an ignvalu quation of th form A ψ = aψ, so w nd to dfin appropriat wavfunctions and oprators. W tak th angular momntum ignvalu quations for th hydrogn atom, and m L Y = l(l +1)Y l L Y z m l = m Y m l m l

10 119 as our modl, and dfin spin angular momntum oprators S and S z and spin wavfunctions α and β. Th wavfunction α is charactrizd by th quations ˆ 1 1 S α= ( +1) α and S = + 1 zα α = m sα. which giv th spin angular momntum squard and th spin angular momntum in th z dirction, so th wavfunction α corrsponds to an lctron spin of 1/. Th wavfunction β is charactrizd by th ignvalu quations S = 1 ( 1 +1) β β and S = - 1 zβ β = m sβ. so th wavfunction β corrsponds to an lctron spin of -1/. Not that it is only th S z oprator that distinguishs btwn ths two ignfunctions. Th angular momntum of both stats is th sam, and is qual to S = s(s+1) = 3/ 4 Th S z oprator, which distinguishs th two wavfunctions, corrsponds to th orintation of th spin angular momntum rlativ to th z axis. Not that thr is only on magnitud of th spin quantum numbr. Rmmbr that for problms lik th harmonic oscillator or th rigid rotator, th wavfunction taks on classical bhavior whn th quantum numbr gts larg. Sinc th spin quantum

11 1 numbr is limitd to 1/, it can nvr bcom larg and lctron spin can nvr bhav classically. To ritrat, lctron spin is a purly quantum phnomnon. Th final part of our dfinition of th spin wavfunctions is to stat that th spin ignfunctions α and β ar orthonormal, i.., that w can writ and * * ααd σ= β βd σ=1 * * αβd σ= βαd σ= whr dσ is calld th spin variabl, which again has no classical analog. W now can s how to combin ths spin wavfunctions with our spatial wavfunctions. Th wavfunction will b a function of four variabls, th usual spatial variabls, r, θ, φ, and our nw variabl, th spin variabl σ, so w writ ψ = ψ (r, θ, φ, σ ) nlmms Notic that th wavfunction is now labld by four quantum numbrs, th thr spatial quantum numbrs, n, l and m, and th nw spin quantum numbr, ms. This mans that to b compltly dscribd, ach wavfunction must contain som valu of ach of ths four quantum numbrs. W'v alrady notd that this nw proprty of lctron spin is a strang on, vry unlik th othr dynamical proprtis w'v studid, so it sms rasonabl that it would b indpndnt of othr variabls. This is anothr way of saying that it should b sparabl, so w'll postulat this and writ th complt wavfunction as a product of a spatial wavfunction, lik our hydrogn-lik orbitals, and a spin function. For a wavfunction with ms = 1/, w hav ψ 1 = ψ (r, θ, φ ) α( σ ), and for a wavfunction with ms = -1/, nlm nlm nlmms

12 11 w hav ψ nlm = ψ nlm(r, θ, φ ) β( σ ). Ths complt on lctron wavfunctions ar 1 calld spin orbitals. As an xampl, th 1s orbital is th ground stat spatial orbital of a hydrogn-lik atom. Th first two spin orbitals of th hydrogn-lik atom ar ψ a 3 = ( Z π Zr 1/ a ) α and 1-1 ψ 3 = ( Z 3 a π Ths spin orbitals ar normalizd and orthogonal. Zr 1/ a ) β.

13 1 Lctur Sinc th lctron spin was first discovrd bcaus atoms showd an unxpctd rspons to a magntic fild, it is logical that th nrgy of spin wavfunctions in a magntic fild will dpnd on th valu of th lctron spin quantum numbr. Lt's work out this nrgy diffrnc. Th magntic momnt of th lctron is proportional to th spin angular momntum and is givn by µ s = - g m S This is idntical to th quation for th orbital magntic momnt, with th xcption of th constant g, calld th lctron g factor and qual to.3. Bcaus by convntion any xtrnal fild is along th z-axis, th important part of th magntic momnt is th z componnt, givn by µ = - g z z m S. Sz is th componnt of th spin angular momntum in th z dirction and is qual to ms. This mans that w can rwrit th z componnt of th magntic momnt as µ m m = -g µ m z = - g s B s Th classical potntial nrgy whn w plac this lctron in a magntic fild of strngth B is V = g µ m B B s If w add this trm to th Hamiltonian for a hydrogn atom in a magntic fild, our nw Hamiltonian including lctron spin bcoms

14 13 H = H + B (L z + gs z ). m whr H is th Hamiltonian for a hydrogn atom in th absnc of an xtrnal fild. Th nrgy ignvalus now dpnd on thr quantum numbrs, n, m, and ms, and ar now givn by 4 m Z E = - nlmms s (4 n + B (m+ g m ). πε ) m Not that th rsult of this nw magntic momnt is to furthr split th hydrogn atom nrgy stats. Considr a s orbital first. Thr is no ffct on th nrgy du to th orbital angular momntum, but th prsnc of th spin angular momntum splits th stat into two distinct nrgy lvls. Now considr a p orbital. In th absnc of an xtrnal fild th six p spin orbitals ar dgnrat. Adding th ffct of th orbital magntic momnt has th ffct of splitting ths into thr nrgy lvls for m = 1, and -1. Finally adding th ffct of th lctron spin splits ach of ths m valus into two mor distinct nrgis. Not that although th stat with m = 1 and ms = -1/ appars to hav th sam nrgy as th stat with m = -1 and ms = +1/, bcaus th g factor is not xactly, ths stats hav nrgis that ar clos to idntical, but not xactly idntical. Such a situation is dscribd as nar dgnracy. Nar dgnracy is important in crystal fild thory, whr wak ligand filds will split d- orbitals in two groups of two and thr orbitals with diffrnt nrgis. Howvr, th

15 14 occupancy of ths orbitals will still b th sam as for normal d-orbitals bcaus th splitting is small compard to th thrmal nrgis of th lctrons. Whn w introducd this nw proprty of lctrons, th spin, w claimd that it would allow us to liminat a ground stat for th lithium atom in which all thr lctrons wr in th 1s orbital. To accomplish this, th lctron spin by itslf is not sufficint - w nd to add a nw postulat, th Pauli Exclusion Principl. In frshman chmistry, this is usually statd as No two lctrons in an atom can hav th sam valus of all four quantum numbrs, n, l, m, and ms.. This statmnt immdiatly hlps us find ground stat lctron configurations. For xampl, th lithium atom has thr lctrons. Th first lctron will go into th orbital ψ 1 1 = 1sα. Th scond lctron will go into th orbital ψ 1 1 = 1sβ. Th third lctron can t go into th 1s orbital or on of th quantum numbrs would hav to b rpatd. Thus for th third lctron, w nd to go to th nxt lowst orbital, ψ 1 = sα. So you s that invoking this vrsion of th Pauli xclusion principl rsults in a limitation of two lctrons in a givn orbital, and givs ris immdiatly to th bulk of th form of th priodic tabl. Unfortunatly, th ground stat wavfunction for lithium which this vrsion of th xclusion principl prdicts, ψli = 1sα 1sβ sα, has implications which ar not supportd by xprimnt. To obtain a mor corrct ground stat wavfunction w nd to turn to a mor fundamntal form of th xclusion principl, which hlps us to dtrmin th form of multilctron wavfunctions as wll. W will prsnt this form of th Pauli xclusion principl as a sixth postulat of quantum mchanics, but first w nd to introduc a nw ida, that of an antisymmtric wavfunction.

16 15 To dfin what an antisymmtric wavfunction is lt's go back to th hlium atom. Using th argumnts w just usd for lithium, w can writ th ground stat wavfunction of hlium as ψ(1,) = 1sα(1) 1sβ(). In othr words, w put lctron 1 in a 1sα orbital, and w put lctron in a 1sβ orbital. Howvr, thr's a problm with this bcaus it implis that our two lctrons ar distinguishabl, i.., that thr's som way to tll thm apart. In othr words, w could writ anothr ground stat wavfunction, ψ(, 1) = 1sα() 1sβ(1) Again, this implis that lctrons 1 and ar distinguishabl, bcaus w ar saying that w can tll which lctron is in which spin stat. This is not th cas. Rmmbr, that nothing in quantum mchanics is maningful unlss it is ultimatly linkd to an obsrvabl, and vn if w had a microscop with sufficint rsolution, w wouldn't s any labl on th lctrons tlling us which was lctron 1 and which was lctron. Unfortunatly, this is not just a philosophical problm, bcaus if w us our ground stat wavfunction, as writtn abov, as a variational trial function, th nrgis com out all wrong. What w nd to do is find a way to writ th wavfunction so that it placs an lctron ach in th 1sα and 1sβ orbitals, and not spcific lctrons in ach orbital. W do this by dfining two nw ground stat wavfunctions that ar linar combinations of th two abov, ψ1(1,) = ψ(1,) + ψ(,1) = 1sα(1) 1sβ() + 1sα() 1sβ(1) and ψ(1,) = ψ(1,) - ψ(,1) = 1sα(1) 1sβ() - 1sα() 1sβ(1).

17 16 Th lctrons ar indistinguishabl in ths two wavfunctions bcaus thy both contain both possibl ground stat configurations. Bcaus both configurations ar includd, it s sort of lik saying that w will always find on lctron in th 1sα spin orbital and on in th 1sβ spin orbital without spcifying which is which. Ths two orbitals ar not idntical. To s th diffrnc lts intrchang lctrons 1 and for ach of thm. For ψ1, ψ1(, 1) = 1sα() 1sβ(1) + 1sα(1) 1sβ() = 1sα(1) 1sβ() + 1sα() 1sβ(1) = ψ1(1,). For this orbital, whn w chang th ordr of th lctrons in th orbitals, th wavfunction rmains compltly unchangd. W say that this wavfunction is symmtric to th intrchang of lctrons. In contrast considr ψ. For this wavfunction, w hav ψ(,1) = 1sα() 1sβ(1) - 1sα(1) 1sβ() = - (1sα(1) 1sβ() - 1sα() 1sβ(1)) = -ψ(1,). Whn w rvrs th ordr of th lctrons, th sign of th wavfunction changs. W call such a wavfunction antisymmtric with rspct to intrchang of lctrons. Now that w hav dfind ths nw trms, w ar rady to stat our sixth postulat, th most complt form of th Pauli Exclusion Principl. Postulat Six All lctronic wavfunctions must b antisymmtric undr th xchang of two lctrons. According to this postulat, of th two wavfunctions that w'v constructd for Hlium, only th scond occurs in natur. So th ground stat of hlium is ψ(h) = 1sα(1) 1sβ() - 1sα() 1sβ(1). Not that whn w writ 1s, w ar not hr rfrring to hydrogn-lik 1s orbitals, but th tru 1s orbitals of H, calculatd by th appropriat approximation mthods.

18 17 I know that this sixth postulat sounds strang and that it sms unconnctd to physical rality, but in fact it givs ris to th Pauli xclusion principl as you larnd it in gnral chmistry. Sinc w know that th Pauli xclusion principl has bn xprimntally vrifid many tims ovr, with th structur of th priodic tabl bing only on xampl, w hav ampl proof that this odd postulat is valid. Thr is mor dirct vidnc in support of this rquirmnt for antisymmtric wavfunctions, and that is th natur of th first xcitd stats of hlium. Th lowst nrgy wavfunction would b on in which on lctron was in a 1s orbital and on in a s orbital. As was th cas for th ground stat, this can b writtn two diffrnt ways, ψa(1,) = 1s(1) s() and ψb(,1) = 1s() s(1). As in our discussion of th hlium ground stat, ths two wavfunctions imply that th lctrons ar distinguishabl, so w nd to tak linar combinations of th two, yilding ψ(1,) = 1s(1) s() + 1s() s(1) and ψ1(1,) = 1s(1) s() - 1s() s(1). Th wavfunction labld ψ1 with an nrgy of -59. V is lowr in nrgy than ψ, with nrgy V, and thus ψ1 is th first xcitd stat of hlium. According to th Pauli xclusion principl, th ovrall wavfunction, including th lctron spin, must b antisymmtric. A usful fact hr is that th symmtry of th product of two wavfunctions follows th ruls 1) Symmtric x symmtric = symmtric ) Antisymmtric x antisymmtric = symmtric and 3) Antisymmtric x symmtric = symmtric x antisymmtric = antisymmtric.

19 18 Sinc th first xcitd stat wavfunction without th spin is antisymmtric, th spin functions that w add must b symmtric. Thr ar thr possibl symmtric spin functions that w can considr, α(1) α(), in which both lctrons hav spin up, β(1) β(), whr both lctrons hav spin down, and (α(1) β() + β(1) α()) th symmtric stat whr both lctrons hav opposit spins. Thus thr ar thr diffrnt antisymmtric wavfunctions for this first xcitd stat, ψ1a(1,) = -1/ (1s(1) s() - 1s() s(1)) α(1) α() ψ1b(1,) = -1/ (1s(1) s() - 1s() s(1))(α(1) β() + β(1) α()) and ψ1c = -1/ (1s(1) s() - 1s() s(1)) β(1) β(). Thus th first xcitd stat of hlium in th absnc of xtrnal filds is triply dgnrat. Bcaus thr ar thr diffrnt stats with th sam spatial wavfunction and th sam nrgy, w call this a triplt stat. Th diffrnc btwn ths stats is th orintation of th angular momntum vctor, so w should b abl to obsrv a diffrnc in nrgis whn th hlium atom is placd in a magntic fild. As xpctd th first xcitd stat splits into thr stats with diffrnt nrgis, with ψ1a having th highst nrgy, ψ1b having th intrmdiat nrgy, and ψ1c having th lowst nrgy. What about ψ(1,), th scond xcitd stat of hlium? Rmmbr that its spatial wavfunction is symmtric. To gt an ovrall antisymmtric wavfunction including spin, w nd to multiply by an antisymmtric spin wavfunction. Th only antisymmtric spin wavfunction for two lctrons is (α(1)β() -β() α(1)). Thus thr is only on wavfunction, including spin, for th scond xcitd stat of H, ψ(1,) = (1s(1) s() + 1s() s(1)) (α(1)β() -β() α(1)).

20 19 Sinc thr is only on wavfunction that has th nrgy of th scond xcitd stat, th stat is singly dgnrat, and w call it a singlt stat. Thus in addition to th structur of th priodic tabl, our sixth postulat allows us to corrctly prdict th triplt and singlt naturs of th first two xcitd stats of hlium.