《无机化学 Chemical Theory》课程教学资源（讲稿）Part II Molecular Spectroscopy Chapter 4 Vanishing Integrals 零积分

mPart IIISymmetry and BondingChapter 4 Vanishing Integrals (零积分)Prof.Dr.XinLu（吕鑫）Email: xinlu@xmu.edu.cnhttp://pcoss.xmu.edu.cn/xlv/index.htmlhttp://pcoss.xmu.edu.cn/xlv/courses/theochem/index.html

Part III Symmetry and Bonding Chapter 4 Vanishing Integrals（零积分） Prof. Dr. Xin Lu （吕鑫） Email: xinlu@xmu.edu.cn http://pcoss.xmu.edu.cn/xlv/index.html http://pcoss.xmu.edu.cn/xlv/courses/theochem/index.html

4. Vanishing integrals.One ofthemost powerful applications of GroupTheory is the ability todecidewhether a particular integral is zero or not without actually evaluating the integral. We will learn how this approach can be used to great advantage in constructingmolecular orbitals and in understanding spectroscopic selection rules.f(x)even·For example, it is clear that the integral ofthe evenfunction ispositive but that of the odd function must be zero. Using Group Theory we can generalise this property of odd andg(x)oddeven functions into a powerful method for deciding whether or notparticularintegralswill bezero

4. Vanishing integrals • One of the most powerful applications of Group Theory is the ability to decide whether a particular integral is zero or not without actually evaluating the integral. • We will learn how this approach can be used to great advantage in constructing molecular orbitals and in understanding spectroscopic selection rules. • For example, it is clear that the integral of the even function is positive but that of the odd function must be zero. • Using Group Theory we can generalise this property of odd and even functions into a powerful method for deciding whether or not particular integrals will be zero

4.1 Symmetry criteria for vanishing integrals. Now consider the integral of a general function y over all space: I= JydtI= Jj+ y(x, y,z)dxdydz (in cartesian coordinates):The integrandmusttransform asthetotallysymmetricIRto maketheintegral non-zero.Ifthe integrand transforms as some otherIR, theintegral is necessarily zeroThe value of the integral I =f ydt is necessarilyzero if ytransforms as anything otherthan the totally symmetric irreducible representation..In other words, if y transforms as a sum of IRs that contain the totally symmetric IR, theintegralisnotnecessarilyzero

4.1 Symmetry criteria for vanishing integrals • Now consider the integral of a general function ψ over all space: I = d • The integrand must transform as the totally symmetric IR to make the integral non-zero. • If the integrand transforms as some other IR, the integral is necessarily zero. I = −∞ +∞  𝑥, 𝑦, 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧 (𝑖𝑛 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠) The value of the integral 𝑰 = ψ𝒅𝝉 is necessarily zero if ψ transforms as anything other than the totally symmetric irreducible representation. • In other words, if ψ transforms as a sum of IRs that contain the totally symmetric IR, the integral is not necessarily zero

4.2 Overlap integralsA commonly encountered integral in quantum mechanics, and especially whenconstructing MOs, is the overlap integral between two wavefunctions ; and jJ yrwjdtSij =1.0(H)S0.80.6e0.40.2·InH,theoverlapoftwolsAOs0.02A356centred on differentatoms (ls(A)R/aoand 1s(B)):1s(A) M1s(B)1s(A)×1s(B)

4.2 Overlap integrals • A commonly encountered integral in quantum mechanics, and especially when constructing MOs, is the overlap integral between two wavefunctions ψi and ψj , • In H2 , the overlap of two 1s AOs centred on different atoms (1s(A) and 1s(B)): 𝑺𝒊𝒋 = ψ𝒊 ∗ψ𝒋𝒅𝝉

4.2Overlap integrals· Suppose that y; transforms as the IR [() and y; transforms as [0). Thus y* y; transforms asnro.. For the overlap integral to be non-zero the product r) must be (or at least contains)thetotally symmetricIR.. The only way for the product r to contain the totally symmetric IR is for (i) and 0)to be the same. This leads to a very important conclusion about the overlap integral:The overlap integral Si, = J y'y,dt is non-zero only if w; and yj,transform as thesame IR. (i.e., ‘symmetry compatible'!)

4.2 Overlap integrals • Suppose that ψi transforms as the IR Γ(i) and ψj transforms as Γ(j). Thus ψ𝒊 ∗ ψj transforms as Γ(i)⊗ Γ(j) . • For the overlap integral to be non-zero the product Γ(i)⊗ Γ(j) must be (or at least contains) the totally symmetric IR. • The only way for the product Γ(i)⊗Γ(j) to contain the totally symmetric IR is for Γ(i) and Γ(j) to be the same. This leads to a very important conclusion about the overlap integral: The overlap integral 𝑺𝒊𝒋 = ψ𝒊 ∗ψ𝒋 dτ is non-zero only if ψi and ψj transform as the same IR. (i.e., ‘symmetry compatible’!)

4.2 Overlap integrals: Example: the overlap between an s orbital and a p, orbital (the internuclear axis is z)·Qualitativepicture:The overlap ofthetwopositivepartsofpositivethewavefunctions(positiveoverlap)iscancelledbythe2pxoverlapnegative overlap'betweenthe positive and negativepartsofthewavefunctions.·Symmetry argument:Thes orbital is symmetric with1srespect to a mirror plane coming out of the plane of thepaper whereas the pr, orbital is anti-symmetric. The twonegativeoverlapAOsdoNOThavethesamesymmetryandthereforedonotoverlap

4.2 Overlap integrals • Example: the overlap between an s orbital and a px orbital (the internuclear axis is z). • Qualitative picture: The overlap of the two positive parts of the wavefunctions (positive overlap) is cancelled by the ‘negative overlap’ between the positive and negative parts of the wavefunctions. • Symmetry argument: The s orbital is symmetric with respect to a mirror plane coming out of the plane of the paper whereas the px orbital is anti-symmetric. The two AOs do NOT have the same symmetry and therefore do not overlap

O--HZMore practical exampleH 1s vs. 0 2s, 2pz, 2px, 2pFigureoutthesymmetry-compatibleoverlapofAOsinOH:OH belongstoCoy point group.: Simple inspection:H1s ~2+(A)02s~ E+ (A)Symmetry compatibleA,@A, =AI02pz~(A)Symmetry compatibleSymmetryincompatible!A,E,=E,O(2px,2p)~II(E)CoyE2C-(α)...0Z+x2 +y2;2?111(A1)zR-Z(A2)11-1II0(xz,yz)(E)2(x,y)2cosα(Rr,Ry)0A(E2)22cos2α(r2 -y2,2xy)@20(E3)2cos3α

More practical example Figure out the symmetry-compatible overlap of AOs in OH: H 1s vs. O 2s, 2pz , 2px , 2py H 1s O 2s O-H z • Simple inspection： O 2pz O (2px , 2py ) ~  + (A1 ) ~  + (A1 ) ~ (E1 ) ~  + (A1 ) Symmetry compatible A1⊗ A1 =A1 Symmetry compatible Symmetry incompatible! • OH belongs to C? point group. v A1⊗ E1 = E1

4.3 Matrix elementsAnotherverycommonlyencounteredintegralinquantummechanicsisQij =Jy,Qy,dtQ is a quantum mechanical operator. Qij is often described as a matrix element. The size ofthe matrix is determined by the number of basis functions , we choose to use.: Apparently, Qi is of course just a number (scalar).To make the integral non-zero, the integrand (or part of it) must transform as the totallysymmetricrepresentation.. Accordingly we need to work out the triple direct product.i) @(@) @[0)where [, r and ( are the IRs of , , and Q, respectively

4.3 Matrix elements • Another very commonly encountered integral in quantum mechanics is: 𝑸𝒊𝒋 = ψ𝒊 ∗𝑸 ψ𝒋 dτ 𝑸 is a quantum mechanical operator. 𝑸𝒊𝒋 is often described as a matrix element. The size of the matrix is determined by the number of basis functions ψ𝒊 we choose to use. • Apparently, 𝑸𝒊𝒋 is of course just a number (scalar). • To make the integral non-zero, the integrand (or part of it) must transform as the totally symmetric representation. • Accordingly we need to work out the triple direct product: Γ(i) ⊗ Γ ( 𝑸)⊗ Γ(j) where Γ(i) , Γ(j) and Γ(Q) are the IRs of ψ𝒊 , ψ𝒋 , and 𝑸 , respectively

4.3.1 Interactions leading to the formation of MOs:Atomic orbitals (AOs)can overlap to form molecular orbitals (MOs): In the simplest molecule, H, the ls AOs of two H atoms can overlap to givei)anin-phasecombination-the bonding MOenergy=α-βantibondingi) an out-of-phase combination1s1sr- the antibonding MObondingenergy =α +βThebonding MOis lower inenergythan the AOs, and the antibondingnote:βis negativeenergy is higher

4.3.1 Interactions leading to the formation of MOs • Atomic orbitals (AOs) can overlap to form molecular orbitals (MOs). The bonding MO is lower in energy than the AOs, and the antibonding energy is higher. • In the simplest molecule, H2 , the 1s AOs of two H atoms can overlap to give i) an in-phase combination – the bonding MO ii) an out-of-phase combination – the antibonding MO

4.3.1 Interactions leading to the formation of MOs: The MOs were generally written as a linear combination of the ls AOs S. and S, on thedifferent atoms: y =caSa+ c,Sb(for H2, Ical = Icbl, why?):The resulting energies of the MOs were:E=a±β[+for the bonding MO, - for the antibonding MO]. The quantities a and β are both energies, and are defined via the integrals:α = J s,Hsadt = J s,Hs,dt ; β = J s,Hsidt = J s,Hsadt ; both a and β are negative.H is just the Hamiltonian for the electron and β is sometimes called the resonance integral

4.3.1 Interactions leading to the formation of MOs • The MOs were generally written as a linear combination of the 1s AOs sa and sb on the different atoms: ψ = ca sa+ cb sb • The resulting energies of the MOs were: E = α ± β [+ for the bonding MO, - for the antibonding MO]. • The quantities α and β are both energies, and are defined via the integrals: 𝜶 = sa𝑯 sadτ = s𝒃𝑯 sbdτ ; 𝜷 = sa𝑯 sbdτ = sb𝑯 sadτ ; both α and β are negative. 𝑯 is just the Hamiltonian for the electron and β is sometimes called the resonance integral. (for H2 , |ca | = |cb |, why?)