946 C. GROUP THEORY IN SPECTROSCOPY a centre of inversion, the vibrations active in IR are inactive in Raman spectra and vice versa. When do the conclusions drawn from group theory fail? When deriving the selection rules the following assumptions have been made: • themoleculeisisolated, • elements are represented by the same isotope, • themoleculeisinastationary state, • the vibrations have small amplitudes, • the vibrations are harmonic, • the electromagnetic field interacts with the molecule only through the electric field-molecule interaction, • in the interaction of the molecule with the electromagnetic field only what are called the dipole transitions are involved. 47 However, in practice the molecule is never isolated. In particular, the interac- tions it undergoes in the liquid or solid state are sufficiently strong to deform the molecule. As a result, we have to do (especially in a liquid) with a population of molecules, each in a different geometry, usually devoid of any particular symmetry (for a single molecule this means a non-stationary state), although the molecule is not far away from perfect symmetry (“broken symmetry”). Suppose for a while that the molecule under consideration is indeed isolated. In a substance we usually have several isotopomers, with different distributions of isotopes in the molecules. In most cases this also means broken symmetry. Broken symmetry means that the selection rules are violated. In practice, broken symmetry means that the selection rules cause only a small intensity of forbidden transitions with respect to allowed transitions. When considering electronic transitions, we assumed that the molecule stays in its equilibrium geometry, often of high symmetry. This may be the most probable configuration, 48 but the vibrations and rotations deform it. An electronic excitation is fast, and usually takes place with a molecular geometry that differs slightly from the most probable and most symmetric one. This will cause a transition, forbidden for perfectly symmetric geometry, to have a non-negligible intensity. Deriving the selection rules for IR and Raman spectra, we assumed that the equivalent atoms can differ only by the sign of the deviation from the equilibrium position, but its absolute value is the same. This is how it would be for a harmonic 47 That is, the electric field of the electromagnetic wave within the molecule is assumed to be homoge- neous. Then the only term in the Hamiltonian related to the light-molecule interaction is −ˆμE,where ˆμ stands for the dipole moment operator of the molecule and E is the electric field intensity. 48 The maximum of the ground-state probability density for the harmonic oscillator indeed corresponds to the equilibrium geometry. This is why the selection rules work at all (although in an approximate way). 4 Integrals important in spectroscopy 947 oscillator. An anharmonicity therefore introduces another reason why a (harmon- ically) forbidden transition will have non-negligible intensity. The electromagnetic field has an electric and magnetic component. The selec- tion rules we have derived have not taken into account the presence of the mag- netic field. Taking into account the magnetic field introduces some additional selec- tion rules. Also, the wavelength of an electromagnetic wave in the UV-VIS region is of the order of thousands of Å, whereas the length of the molecule is usually of the order of a few Å. This means that the assumption that the electric field of the electromagnetic wave is homogeneous looks good, but in any case the field is not perfectly homogeneous. The deviations will be small, but non-zero. Taking this into account by including further terms besides −ˆμE, we obtain the interaction of the electric field gradient with the quadrupole moment of the molecule, as well as further terms. This also weakens the selection rules. Despite these complications, group theory allows for understanding the basic fea- tures of molecular spectra. It sometimes works even if the molecule being studied has no symmetry at all, because of a substituent that interferes. Some electronic or vibra- tional excitations are of a local spatial character and pertain to a portion of the mole- cule that is (nearly) symmetric. Due to this some optical transitions that are absolutely allowed, because the molecule as a whole does not have any symmetry, 49 will still have a very low intensity. 49 But they would be forbidden if the portion in question represented a separate molecule and were symmetric. D. ATWO-STATEMODEL The Schrödinger equation ˆ Hψ =Eψ is usually solved by expanding the unknown wave function ψ in a series 1 of complete basis set {φ i } N i=1 of states φ i ,whereN in principle equals ∞ (instead in practice we end up with a chosen large value of N). The expansion gives ˆ H j c j φ j = E j c j φ j ,or j c j ( ˆ Hφ j − Eφ j ) = 0 By multiplying this equation successively by φ ∗ i , i =12N, and integrating we obtain a set of N linear equations for the unknown coefficients 2 c i : j c j (H ij −ES ij ) =0 where the Hamiltonian matrix elements H ij ≡φ i | ˆ Hφ j , and the overlap integrals S ij ≡φ i |φ j The summation going to infinity makes impossible any simple insight into the physics of the problem. However, in many cases what matters most are only two states of comparable energies, while other states, being far away on the energy scale, do not count in practice (have negligible c j ). If indeed only two states were in the game (the two-state model), the situation could be analyzed in detail. The conclusions drawn are of great conceptual (and smaller numerical) importance. For the sake of simplicity, in further analysis the functions φ j will be assumed normalized and real. 3 Then, for N = 2wehaveH 12 =φ 1 | ˆ Hφ 2 = ˆ Hφ 1 |φ 2 = φ 2 | ˆ Hφ 1 =H 21 ,andallH ij are real numbers (in most practical applications H 12 H 11 H 22 0). The overlap integral will be denoted by S ≡φ 1 |φ 2 = φ 2 |φ 1 After introducing the abbreviation h ≡H 12 we have c 1 (H 11 −E) +c 2 (h −ES) = 0 c 1 (h −E) +c 2 (H 22 −ES) = 0 A non-trivial solution of these secular equations exists only if the secular deter- minant satisfies H 11 −Eh−ES h −ES H 22 −E =0 1 As a few examples just recall the CI, VB (Chapter 10), and MO (Chapter 8) methods. 2 The same set of equations (“secular equations”) is obtained after using the Ritz method (Chapter 5). 3 This pertains to almost all applications. For complex functions the equations are only slightly more complicated. 948 D. A TWO-STATE MODEL 949 After expanding the determinant, we obtain a quadratic equation for the un- known energy E: (H 11 −E)(H 22 −E) −(h −ES) 2 =0 with its two solutions 4 E ± = 1 1 −S 2 H 11 +H 22 2 −hS ∓ H 11 −H 22 2 2 + h −S H 11 H 22 2 +2hS H 11 H 22 − H 11 +H 22 2 After inserting the above energies into the secular equations we obtain the fol- lowing two sets of solutions c 1 and c 2 : c 1 c 2 ± = 1 (h −H 11 S) H 11 −H 22 2 ± H 11 −H 22 2 2 +(h −H 11 S)(h −H 22 S) Using the abbreviations = H 11 −H 22 2 and E ar =(H 11 +H 22 )/2 for the arithmetic mean, as well as E geom = √ H 11 H 22 for the geometric mean, we get a simpler formula for the energy E ± = 1 1 −S 2 E ar −hS ∓ 2 +(h −SE geom ) 2 +2hS(E geom −E ar ) Now, let us consider some important special cases. Case I. H 11 =H 22 and S =0(φ 1 and φ 2 correspond to the same energy and do not overlap). Then, =0, E ar =E geom =H 11 and we have E ± =H 11 ±h c 1 c 2 ± =±1 For h<0 this means that E + corresponds to stabilization (with respect to φ 1 or φ 2 states), while E − corresponds to destabilization (by the same value of |h|). The wave functions contain equal contributions of φ 1 and φ 2 and (after normalization) are 4 It is most practical to use Mathematica coding: Solve[(H11-EdS)*(H22-EdS)-(h-EdS*S)^2==0,EdS] Solve[(H11-EdS)*(H22-EdS)-(h-EdS*S)^2==0 &&c1*(H11-EdS)+c2*(h-EdS*S)==0 &&c1*(h-EdS*S)+c2*(H22-EdS)==0,{c1,c2},{EdS}] 950 D. A TWO-STATE MODEL ψ + = 1 √ 2 (φ 1 +φ 2 ) ψ − = 1 √ 2 (φ 1 −φ 2 ) Case II. H 11 = H 22 and S = 0(φ 1 and φ 2 correspond to the same energy, but their overlap integral is non-zero). Then, E ± = H 11 ±h 1 ±S c 1 c 2 ± =±1 Here also E + corresponds to stabilization, and E − to destabilization (because of the denominator, this time the destabilization is larger than the stabilization). The wave functions have the same contributions of φ 1 and φ 2 and (after normalization) are equal to ψ + = 1 √ 2(1 +S) (φ 1 +φ 2 ) ψ − = 1 √ 2(1 −S) (φ 1 −φ 2 ) Case III. H 11 =H 22 and S =0(φ 1 and φ 2 correspond to different energies and the overlap integral equals zero). This time E ± = E ar ∓ 2 +h 2 c 1 c 2 ± = 1 h ± 2 +h 2 (D.1) Here also the state of E + means stabilization, while E − corresponds to destabi- lization (both effects are equal). Let us consider a limiting case when the mean energy in state φ 1 is much lower than that in φ 2 (H 11 H 22 ), and in addition h 0. For the state with energy E + we have c 1 c 2 2 h , i.e. c 1 is very large, while c 2 is very small (this means that ψ + is very similar to φ 1 ). In state ψ − the same ratio of coefficients equals c 1 c 2 0 which means a domination of φ 2 Thus, if two states differ very much in their energies (or h is small, which means the overlap integral is also small), they do not change in practice (do not mix together). This is why at the beginning of this appendix, we admitted only φ 1 and φ 2 of comparable energies. E. DIRAC DELTA FUNCTION Paul Dirac introduced some useful formal tools (for example his notation for inte- grals and operators, p. 19) including an object then unknown to mathematicians, which turned out to be very useful in physics. This is called the Dirac delta function δ(x). We may think of it as of a function 1 • which is non-zero only very close to x =0, where its value is +∞; • the surface under its plot is equal to 1, which is highlighted by a symbolic equa- tion ∞ −∞ δ(x) dx =1 When we look at a straight thin reed protruding from a lake (the water level = 0), then we have to do with something similar to the Dirac delta function. The only importance of the Dirac delta function lies in its specific behaviour, when integrat- ing the product f (x)δ(x) and the integration includes the point x =0, namely: b a f (x)δ(x) dx =f(0) (E.1) This result is well understandable: the integral means the surface under the curve f(x)δ(x),butsinceδ(x) is so concentrated at x = 0, then it pays to take seriously only those f(x) that are “extremely close” to x = 0. Over there f (x) is equal to f(0). The constant f(0) can be taken out of the integral, which itself therefore has the form ∞ −∞ δ(x) dx =1. This is why we get the right hand side of the previous equation. Of course, δ(x −c) represents the same narrow peak, but at x =c, therefore, for a c b we have b a f(x)δ(x−c)dx =f(c) (E.2) 1 APPROXIMATIONS TO δ(x) The Dirac delta function δ(x) can be approximated by many functions, which de- pend on a certain parameter and have the following properties: 1 More precisely this is not a function, but what is called a distribution. The theory of distributions was developed by mathematicians only after Dirac. 951 952 E. DIRAC DELTA FUNCTION • when the parameter tends to a limit, the values of the functions for x distant from 0 become smaller and smaller, while for x close to zero they get larger and larger (a peak close to x =0); • the integral of the function tends to (or is close to) 1 when the parameter ap- proaches its limit value. Here are several functions that approximate the Dirac delta function: • a rectangular function centred at x = 0 with the surface of the rectangle equal to 1 (a →0): f 1 (x;a) = ⎧ ⎨ ⎩ 1 a for − a 2 x a 2 0 for other; • a (normalized to 1) Gaussian function 2 (a →∞): f 2 (x;a) = a π e −ax 2 ; • a function: f 3 (x;a) = 1 π lim sinax x when a →∞; • the last function is (we will use this when considering the interaction of matter with radiation): 3 f 4 (x;a) = 1 πa lim sin 2 (ax) x 2 when a →∞ 2 Let us see how an approximation f 2 = a π e −ax 2 does the job of the Dirac delta function when a →∞. Let us take a function f(x)=(x −5) 2 and consider the integral ∞ −∞ f(x)f 2 (x) dx = a π ∞ −∞ (x −5) 2 e −ax 2 dx = a π 1 4a π a +0 +25 π a = 1 4a +25 When a →∞, the value of the integral tends to 25 =f(0), as it has to be when the Dirac delta function is used instead of f 2 . 3 The function under the limit symbol may be treated as A[sin(ax)] 2 with amplitude A decaying as A = 1/x 2 ,when|x|→∞. For small values of x,thesin(ax) changes as ax (as seen from its Taylor expansion), hence for small x the function changes as a 2 . This means that when a →∞,therewillbea dominant peak close to x =0, although there will be some smaller side-peaks clustering around x =0. The surface of the dominating peak may be approximated by a triangle of base 2π/a and height a 2 ,and we obtain its surface equal to πa, hence the “approximate normalization factor” 1/(πa) in f 4 . 2 Properties of δ(x) 953 2 PROPERTIES OF δ(x) Function δ(cx) Let us see what δ(cx) is equal to: δ(cx) = lim a→∞ a π exp −ac 2 x 2 = lim a→∞ ac 2 πc 2 exp −ac 2 x 2 = 1 |c| lim ac 2 →∞ ac 2 π exp −ac 2 x 2 = 1 |c| δ(x) Therefore, δ(cx) = 1 | c | δ(x) (E.3) Dirac δ in 3D The 3D Dirac delta function is defined in the Cartesian coordinate system as δ(r) =δ(x)δ(y)δ(z) where r = (xyz). Then, δ(r) denotes a peak of infinite height at r = 0,and δ(r −A) denotes an identical peak at the position shown by the vector A from the origin. Each of the peaks is normalized to 1, i.e. the integral over the whole 3D space is equal to 1. This means that the formula (E.1) is satisfied, but this time x ∈R 3 . 3 AN APPLICATION OF THE DIRAC DELTA FUNCTION When may such a concept as the Dirac delta function be useful? Here is an exam- ple. Let us imagine that we have (in 3D space) two molecular charge distributions: ρ A (r) and ρ B (r). Each of the distributions consists of an electronic part and an nuclear part. How can such charge distributions be represented mathematically? There is no problem in mathematical representation of the electronic parts, they are simply some functions of the position r in space: −ρ elA (r) and −ρ elB (r) for each mole- cule. The integrals of the corresponding electronic distributions yield, of course, −N A and −N B (in a.u.), or minus the number of the electrons (minus, because the electrons carry negative charge). How do we write the nuclear charge distribution as a function of r? There is no way to do it without the Dirac delta function. With the function our task is simple: ρ nuclA (r) = a∈A Z Aa δ(r −r a ) 954 E. DIRAC DELTA FUNCTION ρ nuclB (r) = b∈B Z Bb δ(r −r b ) We put delta functions with the “intensities” equal to the nuclear charges in the nuclear positions . For neutral molecules ρ nuclA (r) d 3 r and ρ nuclB (r) d 3 r have to give +N A and +N B , respectively. Indeed, we have ρ nuclA (r) d 3 r = a∈A Z Aa δ(r −r a ) d 3 r = a∈A Z Aa =N A ρ nuclB (r) d 3 r = b∈B Z Bb δ(r −r b ) d 3 r = b∈B Z Bb =N B Thus the Dirac delta function enables us to write the total charge distributions and their interactions in an elegant way: ρ A (r) =−ρ elA (r) +ρ nuclA (r) ρ B (r) =−ρ elB (r) +ρ nuclB (r) To demonstrate the difference, let us write the electrostatic interaction of the two charge distributions with and without the Dirac delta functions. The first gives the following expression E inter = a∈A b∈B Z Aa Z Bb |r a −r b | − a∈A ρ elB (r) Z Aa |r −r a | d 3 r − b∈B ρ elA (r) Z Bb |r −r b | d 3 r + ρ elA (r)ρ elB (r ) |r −r | d 3 r d 3 r The four terms mean the following interactions respectively: nuclei of A – nuclei of B, nuclei of A – electrons of B, electrons of A – nuclei of B, electrons of A – electrons of B. With the Dirac delta function the same expression reads: E inter = ρ A (r)ρ B (r ) |r −r | d 3 r d 3 r The last expression comes from the definition of the Coulomb interaction and the definition of the integral. 4 No matter how the charge distributions looks, whether they are diffuse (the electronic ones) or point-like (those of the nuclei), the formula is always the same. 4 Of course, the two notations are equivalent, because inserting the total charge distributions into the last integral as well as using the properties of the Dirac delta function, gives the first expression for E inter . F. TRANSLATION VS MOMENTUM AND ROTATION VS ANGULAR MOMENTUM It was shown in Chapter 2 that the Hamiltonian ˆ H commutes with any translation (p. 61) or rotation (p. 63) operator, denoted as ˆ U: ˆ H ˆ U =0 (F.1) 1 THE FORM OF THE ˆ U OPERATOR Below it will be demonstrated for κ, meaning first a translation vector, and then a rotation angle about an axis in 3D space, that operator ˆ U is of the form ˆ U =exp − i ¯ h κ · ˆ K (F.2) where ˆ K stands for a Hermitian operator (with xy z components) acting on func- tions of points in 3D Cartesian space. Translation and momentum operators Translation of a function by a vector r just represents function f in the coordinate system translated in the opposite direction, i.e. f(r −r) see Fig. 2.3 and p. 62. If vector r is infinitesimally small, then, in order to establish the relation between f(r − r) and f(r), it is of course sufficient to know the gradient of f (neglecting, obviously, the quadratic and higher terms in the Taylor expansion): f(r −r) =f(r) −r ·∇f = (1 −r ·∇)f (r) (F.3) We will compose a large translation of a function (by vector T)fromanumber of small increments r = 1 N T,whereN is a veeery large natural number. Such a tiny translation will be repeated N times, thus recovering the translation of the function by T. In order for the gradient formula to be exact, we have to ensure N tending to infinity. Recalling the definition exp(ax) =lim N→∞ (1 + a x ) N ,wehave: ˆ U(T)f (r) = f(r −T) = lim N→∞ 1 − T N ∇ N f(r) = exp(−T·∇)f =exp − i ¯ h T · ˆ p f(r) 955 . following interactions respectively: nuclei of A – nuclei of B, nuclei of A – electrons of B, electrons of A – nuclei of B, electrons of A – electrons of B. With the Dirac delta function the same. region is of the order of thousands of Å, whereas the length of the molecule is usually of the order of a few Å. This means that the assumption that the electric field of the electromagnetic wave. intensity of forbidden transitions with respect to allowed transitions. When considering electronic transitions, we assumed that the molecule stays in its equilibrium geometry, often of high symmetry.