5.1TheBorn-OppenheimerapproximationSofar we have tacitly assumed that themotion ofthe electrons can bedescribed separatelyfrom the motion of the nuclei. This assumption is also implicit in general chemistry, when wepicture a molecule as a nuclear framework bound together by molecular orbitals. Itstheoretical justification is the Born-Oppenheimer approximation, which is based on the factthat nuclei are much heavier than electrons (m,/me = 1836), and therefore much slower.Unless we need very high accuracy,themotion ofthe nuclei can be ignored completely whenwe are interested in the electrons. That is, we treat the nuclei as 'clamped' in position while weworkouttheelectronicwavefunction.Thisgivesusthe'clamped-nucleus'electronicHamiltonian,Helc = T, +V(Q,1)andtheelectronic Schrodingerequation,HeleVele (q,Q) = Eele (Q)Vele (q,Q)whichisthefirsthalfof theBorn-Oppenheimerapproximation
5.1 The Born-Oppenheimer approximation So far we have tacitly assumed that the motion of the electrons can be described separately from the motion of the nuclei. This assumption is also implicit in general chemistry, when we picture a molecule as a nuclear framework bound together by molecular orbitals. Its theoretical justification is the Born-Oppenheimer approximation, which is based on the fact that nuclei are much heavier than electrons (mp /me = 1836), and therefore much slower. Unless we need very high accuracy, the motion of the nuclei can be ignored completely when we are interested in the electrons. That is, we treat the nuclei as 'clamped' in position while we work out the electronic wavefunction. This gives us the 'clamped-nucleus' electronic Hamiltonian, and the electronic Schrodinger equation, which is the first half of the Born-Oppenheimer approximation. elec ˆ ˆ , H T V e Q q H E elec elec elec elec q Q Q q Q ; ;
Helee elee (q,Q) = Eelee (Q)V ele (q,0)(1)This equation includes the electronic kinetic energy T。 and the total potential energyV(Q, q), which depends on the positions of the electrons q, with the nuclei clamped atposition Q.If we are just interested in the electronic energy levels and orbitals at one particularnuclear geometry (e.g. the equilibrium geometry), then we solve eq. (1) once, with Q setequal to the nuclear positions at this geometry.However, if we also want to treat the nuclear motion, we must solve this eq at manydifferent nuclear positions Q, in order to obtain Eelec(Q) as a function of Q. Eelec(Q) isthen used as the potential energy in the nuclear hamiltonian,Hrue = TN + Vele (Q)where T is the nuclear kinetic energy operator. The nuclear wave function nuc(Q) iscalculated by solving the nuclear dynamics Schrodinger equation,HnucV nue (Q) = Ey nue (Q)where E is the total energy
This equation includes the electronic kinetic energy 𝑇 𝑒 and the total potential energy 𝑉(𝑸, 𝒒), which depends on the positions of the electrons 𝒒, with the nuclei clamped at position 𝑸. If we are just interested in the electronic energy levels and orbitals at one particular nuclear geometry (e.g. the equilibrium geometry), then we solve eq. (1) once, with 𝑸 set equal to the nuclear positions at this geometry. However, if we also want to treat the nuclear motion, we must solve this eq at many different nuclear positions 𝑸, in order to obtain 𝐸elec(𝑸) as a function of 𝑸. 𝐸elec(𝑸) is then used as the potential energy in the nuclear hamiltonian, where 𝑇 𝑁 is the nuclear kinetic energy operator. The nuclear wave function 𝜓nuc(𝑸) is calculated by solving the nuclear dynamics Schrodinger equation, where E is the total energy. nuc elec ˆ H T V N Q H E nuc nuc nuc Q Q H E elec elec elec elec q Q Q q Q ; ; (1)
E/cm: Note that each electronic energy level gives rise to itsBr2own potential energy function Eelec(Q). In general,30,000thesefunctionsarecompletelydifferent,becausethebonding described by each electronic energy level isBr(2Pa/2)+Br(2P1/2)20.000different.370Br(2Pg/2)+B(2Pa/2301. The diagram shows the potential energy curves Eelec(r)10.000for the three lowest electronic energy levels of the Br2molecule.Thesymbolsarestandardlabelsfortheelectronic energy levels which will be explained.234r/A
• Note that each electronic energy level gives rise to its own potential energy function 𝐸elec(𝑸). In general, these functions are completely different, because the bonding described by each electronic energy level is different. • The diagram shows the potential energy curves 𝐸elec(𝑟) for the three lowest electronic energy levels of the Br2 molecule. The symbols are standard labels for the electronic energy levels which will be explained
Each of the curves in the above diagram has a set of vibrational wave functions nuc(Q) andenergy levels E associated with it, which we could calculate by solvingHnuey nue (Q)= Ey/nuc (Q)separately for each curve. Note that E is the total (electronic + vibrational) energy, and that theenergy differences between successive E are the vibrational energy spacings. The zero pointenergy is the difference between the lowest E and the potential Eelec(Q) at equilibrium (i.e. thelowest energy point).In a practical calculation, one sometimes approximates each of the potential curves using aMorse potential, or, if only low-lying vibrations are of interest, using a harmonic oscillatorpotential.The total wave function tot(q, Q) describing the combined motion of all the electrons andnuclei in the molecule is the product,The Born-Oppenheimer approximation is usually very accurate, although it can break downwhen different potential surfaces get close together
Each of the curves in the above diagram has a set of vibrational wave functions 𝜓nuc(𝑸) and energy levels E associated with it, which we could calculate by solving separately for each curve. Note that E is the total (electronic + vibrational) energy, and that the energy differences between successive E are the vibrational energy spacings. The zero point energy is the difference between the lowest E and the potential 𝐸elec(𝑸) at equilibrium (i.e. the lowest energy point). In a practical calculation, one sometimes approximates each of the potential curves using a Morse potential, or, if only low-lying vibrations are of interest, using a harmonic oscillator potential. The total wave function 𝜓tot(𝒒, 𝑸) describing the combined motion of all the electrons and nuclei in the molecule is the product, The Born-Oppenheimer approximation is usually very accurate, although it can break down when different potential surfaces get close together. H E nuc nuc nuc Q Q
5.2ThehydrogenatomIn the hydrogen atom there is only one nucleus, and we take it to be clamped at the origin ofcoordinates.Weconsiderthegeneralhydrogen-like'one-electronatom,withnuclearchargeZeThepotential energyisZe?V:4元0randthekineticenergyisa2a?(a?h?h?_ p?Qx2ay2m2m。a2mIt is more convenient to express this in spherical polar coordinates, using the expression for v2 :a2aa1aa1h?1T=:sine2m.Tr2sin?0ap?r2 Orarrsino a0a0
5.2 The hydrogen atom In the hydrogen atom there is only one nucleus, and we take it to be clamped at the origin of coordinates. We consider the general 'hydrogen-like' one-electron atom, with nuclear charge Ze. The potential energy is and the kinetic energy is It is more convenient to express this in spherical polar coordinates, using the expression for 𝛻2 : 2 0 4 Ze V r ò 2 2 2 2 2 2 2 2 2 2 ˆ 2 2 2 e e e T m m x y z m p 2 2 2 2 2 2 2 2 1 1 1 ˆ sin 2 sin sin e T r m r r r r r
ThehydrogenatomHamiltonianFor the hydrogen-like atom, then,aZeh*aa11lddH=T+Vr? sin0 a00r? sin? ao?2m.4元0r- ararWe can recognise in the angular part of the kinetic energy the expression for the square of theangular momentum, but in the case of the hydrogen atom it is conventional to use the symbol ifor the orbital angular momentum of the electron, reserving J (or J) for the total angularmomentum including spin.ThecompleteHamiltonianisthen12h?a Ze?1aH2m,r22m。 r2 rOr4元Note that the radial part can be written in several equivalent ways:1 a2adrorry2ararQut
The hydrogen atom Hamiltonian For the hydrogen-like atom, then, We can recognise in the angular part of the kinetic energy the expression for the square of the angular momentum, but in the case of the hydrogen atom it is conventional to use the symbol መ𝒍 for the orbital angular momentum of the electron, reserving 𝑱 (or 𝒋 Ƹ) for the total angular momentum including spin. The complete Hamiltonian is then Note that the radial part can be written in several equivalent ways: 2 2 2 2 2 2 2 2 2 0 1 1 1 sin 2 sin sin 4 e Z e H T V r m r r r r r r ò 2 2 2 2 2 2 0 ˆ 1 2 2 4 e e Ze H r m r r r m r r ò l 2 2 2 2 2 2 1 1 2 r r r r r r r r r r
Atomic unitsTheHatomHamiltonianisratherclutteredwithfundamental constants.Togetridoftheclutter, we use atomic units:UnitDefinitionSI valueSymbolNamebohrLength4元e.h2/m.e252.917721pma.Mass0.910938×10-30kgm.electronmassChargee1.6021765×10-19Cproton chargeEhEnergyHartreee?/4E.a.4.359744×10-18Angularh1.0545717×10-34JsmomentumWhen we work in atomic units, me, e, h and 4πe.all have a numerical value of 1, as do thequantities ao and E, derived from themThissimplifiestheequations,buthasthedisadvantagethatitbecomesdifficulttocheckthedimensions ofanexpression
The H atom Hamiltonian is rather cluttered with fundamental constants. To get rid of the clutter, we use atomic units: When we work in atomic units, me , e, ℏ and 4𝜋𝜖o all have a numerical value of 1, as do the quantities a0 and Eh derived from them. This simplifies the equations, but has the disadvantage that it becomes difficult to check the dimensions of an expression. Unit SymbolName Definition SI value Length a0 bohr 4𝜋𝜖oℏ 2/𝑚𝑒𝑒 2 52.917721 pm Mass me electron mass 0.910938 × 10-30 kg Charge e proton charge 1.6021765 × 10-19 C Energy Eh Hartree 𝑒 2/4𝜋𝜖o𝑎o 4.359744 × 10-18 J Angular momentum ℏ 1.0545717 × 10-34 Js Atomic units
We write r' = r /ao, and find thatZe?ZeV :4元蝌4元a0Similarly, writing I' = i/n, the angular kinetic energy becomesi2112112h?h?m,e?m,a 4元o,h2 2r'22m.l1122ma.In this way we find1a114a7H/E2r'aror, dropping the primes, and remembering that energies will be in Hartree,12a1aZH:2r22r2arOr
We write 𝑟 ′ = 𝑟/a0 , and find that Similarly, writing መ𝒍′ = መ𝒍/ℏ, the angular kinetic energy becomes In this way we find or, dropping the primes, and remembering that energies will be in Hartree, 2 2 0 0 0 4 4 h Ze Ze Z V E r a r' r' 蝌 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 ˆ ˆ ˆ ˆ 2 2 4 2 2 e h e e e ' ' ' m e E m r m a r' m a r' r' ò l l l l 2 2 2 2 ˆ 1 / 2 2 h ' Z H E r' r' r' r' r' r' l 2 2 2 2 ˆ 1 2 2 Z H r r r r r r l
5.3WavefunctionsforthehydrogenatomThe Hamiltonian contains the angular variables only as the operator i2; the potential energydepends only on r: Knowing that the eigenfunctions of 12 are the spherical harmonics, we lookfor solutions of the form (r, , ) = R(r)Y (m(, ). We get12a1RYm -ZRYm = ERYmHy:12lr? orarand since 12Yim =l(l +1)Ylm we can cancel out Yim to get the radial equation1 ,2ORI(I+1)R-ZR= ER2r? rOr2rFor each value of I (i.e., O, 1, 2, ...) there are infinitely many solutions of this equation. Theyare conventionally labelled by theprincipal guantum number n, which runs from l +1 to ooThe quantum number m has dropped out of this eq, so the radial wavefunctions don't dependon m. We label them Rnl, and the complete wavefunction is nlm = Rnl(r)Y m(, )
5.3 Wavefunctions for the hydrogen atom The Hamiltonian contains the angular variables only as the operator መ𝒍 2 ; the potential energy depends only on r. Knowing that the eigenfunctions of መ𝒍 2 are the spherical harmonics, we look for solutions of the form 𝜓(𝑟, 𝜃,𝜑) = 𝑅(𝑟)𝑌𝑙𝑚(𝜃,𝜑). We get and since መ𝒍 2𝑌𝑙𝑚 = 𝑙(𝑙 + 1)𝑌𝑙𝑚 we can cancel out 𝑌𝑙𝑚 to get the radial equation For each value of l (i.e., 0, 1, 2, .) there are infinitely many solutions of this equation. They are conventionally labelled by the principal quantum number n, which runs from 𝑙 + 1 to ∞. The quantum number m has dropped out of this eq, so the radial wavefunctions don't depend on m. We label them Rnl, and the complete wavefunction is 𝜓𝑛𝑙𝑚 = 𝑅𝑛𝑙(𝑟)𝑌𝑙𝑚(𝜃,𝜑). 2 2 2 2 ˆ 1 1 2 lm lm lm Z H r RY RY ERY r r r r r l 2 2 2 1 1 2 2 R Z l l r R R ER r r r r r
Solution ofRequation1(1+1)1aaR-R=ER2r2Or2r.2araaInthiseq,writep=Zr.Thenand we getarap>Z1(1+1) ,aaRR_Z2R=ERap2pao0ora1(1+1)aR1E22p? opap2p7pSo the radial equation is the same for all Z, except that the energies scale with Z77E=RiscalledRydbergconstantwiththevalueof 13.6eVn8元manecessarycondition:n≥I+1n:Principaln = 1, 2, 3,..quantum number = 0, 1, 2,
Solution of R equation 2 2 2 1 1 2 2 R Z l l r R R ER r r r r r In this eq, write 𝜌 = 𝑍𝑟. Then 𝜕 𝜕𝑟 = 𝑍 𝜕 𝜕𝜌 and we get or So the radial equation is the same for all Z, except that the energies scale with Z 2 . 2 2 2 2 2 2 1 2 2 Z R Z Z l l R R ER 2 2 2 2 1 1 1 2 2 R E l l R R R Z 2 2 2 2 2 2 2 0 ( ) 8 h Z Z E R ma n n R is called Rydberg constant with the value of 13.6 eV necessary condition: n l+1 n = 1, 2, 3,. l = 0, 1, 2, . n: Principal quantum number