Normal Modes. the True Story

Normal modes. The true story.

Emil Zak∗ Department of Physics and Astronomy, University College London, London, WC1E 6BT, UK (Dated: January 20, 2016)

I. INTRODUCTION KEO in arbitrary curvlinear coordinates was introduced by Yachmenev and Yurchenko [3] and successfully ap- The aim of this article is a comprehensive description plied to four-atomic systems by Chubb, Yachmenev and of normal modes of molecular vibrations. The starting Yurchenko [4]. Parallel analytic progress in the field has point is chosen to be a general molecular system with been made by Szalay [5], [6]. Several authors: Tennyson separated center of mass and an arbitrary embedding of et al. [7], [8], Jensen [9], Csaszar [10] and others also body-fixed axes. This allows to focus on internal de- succeeded to evaluate practical representations for many- grees of freedom only, leaving the problem of rotational body Hamiltonians in an arbitrary embedding of the motion behind. Nevertheless, for the sake of complete- body-fixed frame. In each of these cases one is theoreti- ness we first introduce a general quantum-mechanical cally able to linearize the internal coordinates and subse- Hamiltonian in Eckart-Watson representation, to make quently transfer into normal modes level. Nonetheless we a quick leap into a simplified harmonic description for shall follow the historically first attempt due to its strong internal motion and rigid rotor for rotational degrees of association with normal coordinates. After separating freedom. This regime constitutes a basis for more sophis- the nuclear center of mass, and moving into the Eckart ticated calculations. The first section introduces normal frame which guarantees minimal rotation-vibration cou- modes with emphasis on mathematical precision. Along- pling, some analytic manipulations yield the well known side we stress the points where approximations are made, Watson Hamiltonian: for the reader to be aware of limitations of the current model. Second section exemplifies presented methodol- ogy on a simple system of two masses connected with harmonic springs in 1D. Section V has supplementary 1 X 1 X character and underlines the formal route needed be Hˆ = Mˆ µ Mˆ + Pˆ2 + Uˆ + Vˆ (1) 2 i ij j 2 i taken to derive the Watson Hamiltonian, which in turn i,j i is immersed in the framework of normal vibrations. The last section (VI) is a detailed discussion of rigid rotor model. Current review is largely based on classic books where Uˆ is mass dependent contribution to the potential by Bunker& Jensen [1] and Wilson,Decius&Cross [2] sup- Vˆ : plied with original papers, and a few author’s own com- ments, derivations and theorems. 2 X 2 Uˆ = −~ µ ≡ −~ T rµ (2) 8 ii 8 II. GENERAL THEORY i

Before proceeding to the main topic, a few words of explanation are needed. Section V presents a detailed and µij is the inverse of the dynamical moment of inertia insight into the formal procedure of derivation of the tensor[11]. Mˆ i = Jî − pî − Lî stands for the i-th compo- arXiv:1601.04735v1 [quant-ph] 18 Jan 2016 Eckart-Watson form of quantum Hamiltonian for an ar- nent of the total angular momentum of the system and is bitrary many-body system. This is a popular but rather defined by the rovibronic angular momentum, vibrational specific approach, as is constrained to normal coordinates angular momentum and electronic angular momentum, only. Any set of coordinates which is not linearly related respectively. The potential operator V contains interac- to cartesian displacements brings significant complica- tions between all particles in the system, while Pˆ is the tions. Many efforts were made to evaluate an exact rep- linear momentum operator in normal coordinates repre- resentation of the kinetic energy operator (KEO) in curv- sentation. By imposing Born-Oppenheimer separation of linear coordinates and Eckart embedding. Most recently nuclear and electronic degrees of freedom the molecular an effective numerical procedure for expansion of nuclear Hamiltonian becomes a sum of an electronic and a rovibrational part:

∗ [email protected]; http://www.homepages.ucl.ac.uk/˜ucapejz/ Hˆ = Hˆel + Hˆrv (3) 2

The rovibrational Hamiltonian can be readily expanded later transfer to the rigid rotor approximation we are and arranged in the following fashion: left with internal nuclear motion problem. For a nonlinear molecule, or non-linear system of atoms this leaves 1 X 1 X Hˆ = µe Jˆ2 + Pˆ2+ 3N − 6 internal degrees of freedom. Lets start from a rv 2 αα α 2 i α i fully classical picture. After formulating a suitable rep- 1 X resentation of the Hamilton function, quantization will + (µ − µ ) Jˆ − pˆ Jˆ − Jˆ 2 αβ αβ α α β β be performed. From eq. 5 one can infer that the total αβ energy of the non-rotating system is a sum of kinetic and X e ˆ X e 2 ˆ ˆ potential energy: − µααJαpˆα + + µααpˆα + U + Vnn α α E = T (x˙ ) + V (x) (8) (4)

Mixing terms appearing in above expression are re- where x = x1, y1, z1, ..., x3N−6, y3N−6, z3N−6 and x˙ = sponsible for centrifugal distorsion and Coriolis cou- x˙ 1, y˙1, z˙1, ..., x˙ 3N−6, y˙3N−6, z˙3N−6 are position and veloc- plings. For suﬃciently rigid molecules they can be ne- ity vectors of all 3N − 6 atoms, respectively. No net glected. What remains is a purely rotational part added external force is exerted on the system and all internal to internal motion energy operators decoupled from ro- forces are conservative, hence the total energy is equal to tations: the Hamilton function. We should be able to write ad hoc both components in cartesian coordinates:

3N−6 1 X e 2 1 X 2 1 Hˆrv = µ Jˆ + Pˆ + Vˆnn (5) X X 2 2 αα α 2 i Tcart = miα˙i (9) α i 2 i=1 α=x,y,z The task now is to find normal coordinates for a general where α stands for cartesian coordinate in the molecule- many-body system, so that Watson approach may be re- i fixed (body-fixed) frame. It is very instructive to refor- alized. Hence it becomes clear that the choice of this par- mulate the above expression in a matrix form: ticular type of Hamiltonian representation was strongly dictated by its association to normal coordinates. How-     3N−6 m 0 0 x˙ ever from practical point of view, one usually prefer start 1 X i i T = (x ˙ , y˙ , z˙ ) 0 m 0 y˙ (10) with a set of geometrically defined internal coordinates or cart 2 i i i  i   i  i=1 0 0 m z˙ cartesian coordinates. Thus, below we present a detailed i i instructions to obtain normal coordinates from arbitrar- and further: ily chosen internal coordinates. Assuming the knowledge of normal coordinates of the system we may expand inter- 1 T = x˙ T Tx˙ (11) nuclear interaction potential energy as: cart 2 where x is a 3N − 6 dimensional position vector, while T 3N−6 3N−6 1 X 1 X stands for a (3N − 6) × (3N − 6) matrix representation V = λ Q2 + Φ Q Q Q + ... (6) nn 2 i i 6 ijk i j k of a diagonal metric tensor build from N − 2 diagonal i=1 i,j,k=1 blocks each containing three equal masses mi · 13. It and by truncating this expansion on quadratic terms we is clearly visible that the kinetic energy is a quadratic stay in harmonic approximation, in which normal coor- form in cartesian velocities. This observation will allow dinates were introduced. Now pick up the first two sums us to utilize some tools from quadratic forms theory in in eq. 5 and combine the second one with the truncated section IV. The internuclear potential energy is a func- potential energy. By neglecting all other terms we get: tion of 3N cartesian coordinates of all nuclei. However the functional form of potential energy is usually avail- able in a geometrically defined internal coordinates, i.e. 1 X 1 X Hˆ = µe Jˆ2 + Pˆ2 + λ Q2 (7) Vnn = Vnn(R1,R2, ..., R3N−6), where Ri is the i-th inter- rv 2 αα α 2 i i i nal coordinate. The reason for that is because the poten- α i tial energy function is in general a complicated function The first term in above Hamiltonian describes the energy of cartesian coordinates. Internal coordinates chosen in- of a rigid rotor - this model is solved and discussed in tuitively by geometrical considerations provide a more detail in section VI. The second term is the target of transparent perspective of types of internal motions and the present article. It is already shown in the final form, associated potential energy. Therefore internal coordi- however no recipe for finding λi and Qi was yet given. nate representation is more suitable at this stage. Nev- Until then this expression should be considered a black- ertheless, the potential energy function is typically also a box. fairly complicated function (but less complicated than in Having separated both center of mass motion and rota- cartesian representation) of internal coordinates. Thus, a tions via embedding in the Eckart body-fixed frame with for non-high-accuracy calculation purposes lowest terms 3 in the Taylor expansion are sufficient: where M is the total mass of the system, and

1 3N−6 1 3N−6 N X X eq X eq eq Vnn = fijRiRj + fijkRiRjRk + ... (12) R = µ m (y ∆z − z ∆y ) (17) 2 6 x xx i i i i i i,j=1 i,j,k=1 i=1 Note that the minimum of V is set to 0 and it corre- while y and z rotational coordinates are obtained by sponds to equilibrium values of all internal coordinates: cyclic permutation of cartesian coordinates in above def- ~ ~ −1 R = (0, 0, ..., 0) = Req. First derivatives vanish at this eq hPN eq 2 eq 2 i inition. µxx = i=1 mi((yi ) + (zi ) ) is the x, x point. fij, fijk, ... are appropriate derivatives of V with respect to internal coordinates taken in equilibrium component of the inverse moment of the inertia tensor. positions. After adding these relations, the B matrix is automati- cally reshaped into square invertable form. Further re- As it is exceptionally difficult to obtain internal coor- sults will show however, that this step is redundant if dinates representation of the kinetic energy (quantum- one requires only knowledge of normal coordinates. It mechanical operator) [10],[12] we are pushed to look for is rather intuitive, as neither translations nor rotations alternative approximate ways[13]. One such approach can affect the internal degrees of freedom in a decoupled is based on the assumption that in vicinity of the equi- model. As a matter of fact the rectangular form is suffi- librium the potential energy function can be well repre- cient to transform the KEO into normal modes form. sented only with the quadratic terms in the expansion 12. Note that time derivative of a cartesian displacement In such case the potential energy may be also written in is exactly equal to time derivative of cartesian absolute a matrix form: coordinate as equilibrium position doesn’t change over time. Thus, in the expression for kinetic energy in eq. 11 1 V = RT FR (13) we may responsibly put cartesian displacements: harm 2 1 T with F as force constants matrix, i.e. the matrix of sec- Tcart = ∆x˙ T∆x˙ (18) ond derivatives of potential energy with respect to appro- 2 priate internal coordinates taken in equilibrium. If one and readily transfer into our linearized coordinates: would like to include anharmonicities, then for example the cubic terms should be understood as a contraction of 1 T T 1 T −1 Tcart = S˙ B TBS˙ ≡ S˙ G S˙ (19) rank 3 tensor with components fijk with a rank 3 tensor 2 2 constructed from internal coordinate vectors R~. −1 T Still, internal coordinates are nonlinear functions of Here we introduced a new matrix: G = BT B . This cartesian displacements from equilibrium geometry, what definition of the G matrix indicates its dependence on makes it technically very difficult to transform the kinetic atomic masses and the equilibrium geometry of the sys- energy operator into internal coordinate representation. tem. Despite its simple form, evaluation of G matrix is a Thus, we are pushed to make another crude approxima- tedious job, especially for polyatomic systems. Elements tion, that is to expand each internal coordinate in a series of this matrix for several example systems are presented e in the appendix in ref.[2]. In contrast, the harmonic po- of cartesian displacements ∆αk = αk − α : k tential energy function transforms trivially:

N 1 T X X αk 2 Vharm = S FS (20) Ri = Bi ∆αk + o (∆αk) . (14) 2 k=1 α=x,y,z Hence we managed to represent the total classical en- αk ∂Ri ergy of a closed system of interacting point masses in Here Bi = ∂∆α depends only on equilibrium po- k eq the harmonic approximation and in terms of linearized sitions of atoms. Again in the matrix form we have: coordinates as:

R = B∆x (15) 1 T 1 H = S˙ G−1S˙ + ST FS (21) Because we started from general considerations for 3N 2 2 degrees of freedom, B matrix links the 3N cartesian dis- Unfortunately such form of the Hamilton function would placements with the 3N −6 linearized coordinates, hence result in coupled equations of motion due to non-diagonal is a rectangular matrix. In order to make it invertible we forms of both G and F. From this reason it is highly shall add 6 rows which give transformation laws for 3 desirable to transform this quadratic form into a diagonal translations and 3 rotations of the whole system. These representation. This task is equivalent to simultaneous are deﬁned as follows: diagonalization of two matrices. In general our wanted N transformation may be written as: 1 X T = m ∆α (16) α M 1/2 i i i=1 S = LQ (22) 4 where L is a square 3N − 6 dimensional transformation matrix and Q is a transformed 3N −6 dimensional vector assuring diagonal form of the total energy:

1 T 1 H = Q˙ Q˙ + QT ΛQ (23) 2 2 with Λ being a diagonal matrix. Above conditions uniquely define L matrix for a given G and F. The first one indicates that L is not in general an orthogonal matrix, by requiring: Figure 1. Visualization of the normal modes for water T −1 molecule. Arrows correspond to vectors having its beginning L = L G (24) in equilibrium positions of respective atoms and endings in points with cartesian coordinates corresponding to basis vec- whereas the second one refers to the force constant matrix tors of normal coordinates space from eq.34 transformation: LT FL = Λ (25) We shall make extensive use of the l matrix in section V. These can be unified into a single relation: What remains, is to recall the potential energy function defined in the beginning of this section and relate λi and fijk etc. to the L matrix: L−1GFL = Λ (26)

Hence L gains the interpretation of the matrix that di- 3N−6 3N−6 1 X 1 X agonalizes the product GF with normalization condition Vnn = fijRiRj + fijkRiRjRk + ... (32) T 2 6 L L = G. In general GF is not a symmetric matrix, i,j=1 i,j,k=1 even though both F and G are symmetric, since they nor- mally do not commute. As a consequence we don’t have Note that if the initial internal coordinates were chosen a guarantee that this matrix is diagonalizable. There- to be cartesian displacements then B matrix is a unit fore, it is practical to perform such a transformation of matrix, and a more familiar relation is retrieved: the original linearized coordinates so that G is a unit matrix. This problem is carefully discussed in section IV. Having this done, it is straightforward to relate cartesian L−1FL = Λ (33) displacements to normal coordinates and vice versa: − 1 and l = M 2 L. −1 ∆x = B LQ (27) 3N − 6 dimensional column vectors of type where it is immediate to show via Kronecker theorem  0  for determinants that B−1L is invertible. The inverse of  0  B can be expressed by other matrices, thereby avoiding  .   .  undefined operation (probably taking pseudoinverse of B  .  would give congruent result). In addition, from various  0  Q~ =   s (34) reasons it is convenient to express left hand side of above s  1    equation as mass weighted cartesian displacements:  0     .  1 1 T −1 . ∆q = M 2 ∆x = M 2 B G LQ (28)  .  0 which can be obtained by simple matrix manipulation. The last equation defines well known l matrix: form an orthonormal basis for normal coordinates space. Thus, by plotting cartesian displacements cor- 1 T −1 l := M 2 B G L (29) responding to each of the basis vector one may visualize geometry distortion caused by a particular normal vibra- −1 T which is orthogonal l = l so that the effective formu- tion. This is indicated with arrows in fig. II las for cartesian displacements and normal coordinates are respectively:

3N−6 III. SIMPLE EXAMPLE 1 X − 2 ∆αi = mj lαj,iQj (30) j=1 In this section the just described procedure is exem- 3N−6 pliﬁed on an arguably simplest non-trivial model: two X 1 T 2 masses + three springs in one dimension. This is a classic Qi = mj (lαj,i) ∆αj (31) j=1 system presented in textbooks for pedagogical purposes. 5

so that the B matrix is a unit matrix:

1 0 B= = 1 (41) 0 1 2

Here one comment: we agreed that all degrees of freedom are vibrational. Therefore B matrix is already invertable and there’s no need to extend it by rotational and trans- Figure 2. Graphical representation of the system of two lational coordinates. However if one tries to calculate the masses on three springs. x-component of the translational coordinate: √ T = µ (∆x + ∆x ) (42) A. Potential energy x 1 2 one may argue that we should have one vibrational degree We have a system of two masses interconnected by a of freedom and one translational center of mass motion spring with force constant k. Additionally these masses described by above expression. Since we assumed infinite are connected to infinite mass, rigid walls by the same mass of walls the center of mass of the system as a whole type of springs. This gives 2 degrees of freedom repre- is always frozen. In such case Tx coordinate will become sented by catesian coordinates of respective masses. By a vibrational coordinate, which doesn’t affect the center investigating fig. 2 we may write down an expression for of mass. Such seemingly ambiguous situation is an arti- the potential energy of the system in harmonic approxi- fact of confinement of the system in 1D between static mation: walls. Hence we agree to call all our degrees of freedom 1 e 2 1 e 2 1 e e 2 vibrational. Moving on, the G matrix reads: V (x1, x2) = 2 k (x1 − x1) + 2 k (x2 − x2) + 2 k (x2 − x2 − x1 + x1) (35) 1 1 This is the cartesian representation. All two degrees of 1 0 m 0 1 0 m 0 G = 1 = 1 (43) freedom are identified as vibrational, as in 1D case there 0 1 0 m 0 1 0 m are no rotations, and our system is being held still by infinitely heavy walls, hence no center of mass motion. and Now we choose appropriate internal coordinates. In such case they can be simply displacements from equilibrium: −1 m 0 e G = (44) R1 := ∆x1 = x1 − x1, 0 m e (36) R2 := ∆x2 = x2 − x2 Finally from eq. 37 it is straightforward to read the ele- In this new representation the potential energy reads: ments of the force constant matrix: 1 1 2 2 V (R1,R2) = kR1 + kR2 − kR1R2 − kR2R1 (37) 2k −k 2 2 F = (45) −k 2k Newtonian equations of motion at this stage are coupled second order differential equations: and the Hamilton function takes the following form: R¨ − 2k k R 1 = m m 1 (38) ¨ k 2k R 1 T −1 1 T R2 m − m 2 H = S˙ G S˙ + S FS (46) 2 2 hence quite problematic to solve.

C. Normal coordinates B. Linearized coordinates We are looking for a matrix L that transforms nor- Second step in the procedure is evaluation of linearized mal to linearized coordinates S = LQ, diagonalizes GF coordinates: matrix and preserve the correct norm: 2 X ∂Ri Si = q∆xk (39) L−1GFL = Λ ∂∆xk k=1 e (47) LT G−1L = 1 In this particular case the transformation is trivial: 2 ∂R1 ∂R1 First lets calculate the GF matrix: S1 = ∆x1 + ∆x2 = ∆x1 ∂∆x1 eq ∂∆x2 eq 1 2k k (40) m 0 2k −k m − m 2 −1 GF = 1 = := ω0 ∂R2 ∂R2 0 −k 2k − k 2k −1 2 S2 = ∆x1 + ∆x2 = ∆x2 m m m ∂∆x1 eq ∂∆x2 eq (48) 6

k with ω0 = m . The Kernel of the operator GF − 1Λ is use the derived separable form of Hamilton function and spanned by two orthonormal vectors: apply the Podolsky trick. An excellent paper on this freedom of choice is given in ref.[14]. We choose the former 1 1 1 1 √ , √ (49) method as more straightforward in this case: 2 1 2 −1 ∂2 1 ∂2 ∂2 ∂2 ∂2 = + + 2 which constitute the prospective L matrix: ∂x2 2 ∂Q2 ∂Q2 ∂Q2 ∂Q2 1 1 2 1 2 (59) 2 2 2 2 2 1 1 1 ∂ 1 ∂ ∂ ∂ ∂ L = √ (50) 2 = 2 + 2 − 2 2 2 2 1 −1 ∂x2 2 ∂Q1 ∂Q2 ∂Q1 ∂Q2 and the eigenvalues matrix read: Note that only the sum of above two provides form invariance under linear transformation of cartesian coordi- ω 0 nates: Λ = 0 (51) 0 3ω0 2 2 ˆ 1 ∂ ∂ TQ = 2 + 2 (60) One easily checks that L is orthogonal, hence: 2m ∂Q1 ∂Q2 By setting Q = 1 and Q = 0 and reversely we may Q1 1 1 1 ∆x1 1 2 = √ (52) depict cartesian displacements along a chosen normal co- Q2 2 1 −1 ∆x2 ordinate. giving two non-degenerate normal modes: symmetric stretching and asymmetric stretching respectively. We may also want to evaluate the l matrix: IV. MATHEMATICAL BACKGROUND − 1 T −1 √ 1 1 l = M 2 B G L = µ (53) In order to understand the problem of finding of nor- 1 −1 mal vibrations at the deepest level, it is necessary to recall several mathematical definitions and theorems, com- to give: monly used in linear operators theory [15]: √ √ m∆x1 = µ (Q1 + Q2) √ √ (54) Definition 1 If an operator A is diagonalizable, and β m∆x2 = µ (Q1 − Q2) diagonalizes it by the similarity transformation: The Hamilton function is now a sum of two independent −1 components: β Aβ = D,Dij = λiδij (61)

H(Q1,Q2) = H1(Q1) + H2(Q2) (55) then for any real number γ we may define the power of the operator as follows: 1 ˙ 2 1 2 where Hi(Qi) = 2 Qi + 2 ΛiiQi . Having the classical expression for Hamilton function derived as well as having invertible transformation between cartesian and normal Aγ := βDγ β−1 (62) representation we are ready to turn into quantum mechanics. The kinetic energy operator in cartesian coordi- If operator A contains any zero’s in its spectrum, then nates reads: γ > 0. 2 2 ˆ 1 1 ∂ ∂ Theorem 1 Any matrix A defined above and any pair Tcart = (∆1 + ∆2) = 2 + 2 (56) 2m 2m ∂x1 ∂x2 real numbers compatible with def. 1 satisfies:

γ β (γ+β) Our ’curvlinear’ coordinates R~ = (R1, ..., R3N−6) role is A A = A (63) now taken by the derived normal coordinates: Proof 1

Q 1 1 1 ∆x Aγ Aβ = βDγ β−1βDββ−1 = βDγ Dββ−1 = ... (64) R~ (~x): 1 = √ 1 (57) Q2 2 1 −1 ∆x2 As long as D are diagonal, exponentiation converts into simple number operations: ~ ∆x1 1 1 1 Q1 ~x R : = √ (58) Dγ Dβ = (λγ λβI(γ+β)) = (λ(γ+β)I) (65) ∆x2 2 1 −1 Q2 kl k l kl k kl At this stage we could step forward in twofold way. We could either start from cartesian representation and ap- (γ+β) −1 (γ+β) ply chain rule to get internal coordinate representation or ... = βD β = A (66) 7

In classical and even more in quantum mechanics, special role is assigned to self-adjoint operators, in par- ∂2V ticular to those which are additionally positive definite ∆αi∆αj > 0 (72) (∀x (x, Ax) ≥ 0). It follows from the fact that positive ∂αi∂αj x=0 definite operators posses only non-negative eigenvalues, which is a physical requirement for observables. Non- where αi is a general cartesian coordinate of the the i-th negative spectrum is an immediate consequence of posi- atom, as defined in previous section. Thereby we may tive definitiness: assume that in the range of applicability of the normal coordinates the potential energy classical operator is real, (x, Ax) Ax = λx ⇒ λ = (67) symmetric and positive-definite: (x, x)

γ Lemma 1 If an operator A is self-adjoint, then A is (x, Vx) ≥ 0 (73) also self-adjoint. Proof 2 Matrix that diagonalizes a self-adjoint operator Lagrange function in this approximation reads: is unitary. It follows from the equality: 1 1 L = (x˙ , Tx˙ ) − (x, Vx) (74) 2 2 † (Aγ )† = UDγ U † = (U †)†Dγ U † = Aγ (68) And Euler-Lagrange equations because D is diagonal and have all its elements positive( as the eigenvalues of A are real). ∂L d ∂L − = 0 i = 1, ..., n (75) ∂α dt ∂α˙ Lets turn into application in classical mechanics. Ki- i i netic and potential energy functions may be expressed yield in equations of motion in the form: as expectation values of some linear operators (cf. 11), namely: n n X X Tijα¨j + Vijαj = 0 ı = 1, ..., n (76) n 1 X j=1 j=1 hT i = Tijα˙iα˙j (69) 2 i,j=1 whereas in operator notation:

where: Tij are time independent matrix elements of kinetic energy operator (classical!) in a basis constructed Tx¨ + Vx = 0 (77) from time derivatives of consecutive coordinates; n is the number of degrees of freedom. By utilizing natural scalar Here one faces numerous possible ways of solving of the product of Euclidean space (e , e ) = δ , i, j = 1, 2, 3 we i j ij above set of coupled differential equations. Of course we get: would like to see them decoupled. By and large it means that we want to find such basis in the euclidean space that 1 hT i = (x˙ , Tx˙ ) (70) both kinetic and potential energies are diagonal. Below 2 we present two alternative ways to achieve it: Round brackets are just another way to denote the scalar 1. A formal approach is to postulate the form of solu- product - here to distinguish it from Dirac brackets rep- tion in analogy to the simple harmonic oscillator: resenting scalar product in L2(R3, d3r) space. As far as the kinetic energy is required to be non-negative (and in quantum case positive) the operator T is positive- ±ωt 2 definite, real and symmetric, hence also self-adjoint. Po- αi(t) = ξe ⇒ (V − ω T )ξ = 0 (78) tential energy, as a complicated function of cartesian coordinates is usually expanded in Taylor series and, in our Solutions to this equation are given by a set be- approach, truncated at quadratic terms (see, eq. 6 and ing the Kernel of the V − ω2T operator, i.e. ξ ∈ eq. 20 ): Ker(V − ω2T ). Next we aim in reformulation of above equation into an eigenvalue problem for some n operator. 1 X 1 hV i = Vijxixj = (x, Vx) (71) To do that, lets first act on both sides with the 2 2 − 1 i,j=1 operator T 2 : Near the minimum, the second derivative is always positive, hence for a small displacement from equilibrium − 1 2 1 following inequality holds: T 2 V ξ = ω T 2 ξ (79) 8

− 1 1 Using the resolution of identity T 2 T 2 = id : because T commutes with V . After diagonalization of V − ω2T the Lagrangian takes separable form: − 1 − 1 1 2 1 T 2 VT 2 T 2 ξ = ω T 2 ξ (80)

1 1 1 1 h i 2 − 2 2 L = (Q˙ , Q˙ ) − (Q, ΛQ) (88) and denoting: η = T ξ, W = T VT we ﬁnd: 2

where: Λ is a diagonal matrix of normal frequen- W η = ω2η (81) cies, while Q is a superposition of original coordinates according to the transformation: Q = T 1 of course by deﬁnition of T and V , the W oper- O T 2 x , whereas O is transition matrix into the ator is symmetric and real, which guarantees the eigenbasis of the W operator. In consequence, one existence of an orthonormal eigenbasis: can easily transform into Hamilton function to get: 1 h i H = (Q˙ , Q˙ ) + (Q, ΛQ) (89) (i) (j) 2 (η , η ) = δij (82) Reader may compare this result to see that it represents exactly the same Hamilton function as in This orthonormality condition translates onto or- eq.23. thonormality of initial eigenstates: n (i) (j) 1 (i) 1 (j) (i) (j) X − 1 (η , η ) = (T 2 ξ ,T 2 ξ ) = (ξ , T ξ ) (83) αi = (T 2 O)ikQk (90) k=1 Therefore the special solution to the system of To avoid confusion we shall point out that V matrix equations 77 is given by: is identical with the force constant matrix F and was intentionally written to show the link with the (i) (i) ±ωit α± (t) = ξ e (84) V operator. The general solution is a linear combination of all 2. Lagrangian is simply a sum of two quadratic forms, special solutions: which stay in 1:1 correspondence with metric matrices. The task is to simultaneously diagonalize those two quadratic forms. Beforehand lets prove n the following statement: X (k) ±ωkt −±ωkt α(t) = ξ Cke + Dke (85) k=1 Theorem 2 Let V be euclidean space. In such case there is 1:1 correspondence between self-adjoint operators on V If provided initial conditions: α(0) = κ, α˙ (0) = β and symmetric metrics g˜ on V : and restricting only to real part of solutions, we ﬁnally obtain: g˜(x, y) = (x, Ay) (91) n (k) X (k) (k) (ξ , T β) and following relation holds: α(t) = ξ (ξ , T κ) cos ωkt + sin ωkt ωk k=1 (86) Kerg˜ = KerA (92)

Lets notice key features of the above solution: If however g, g˜,A are matrices in an arbitrary basis of two metrics on V and A, then: • a net motion is a superposition of vibrations of a precisely determined frequency • Relative vibrational amplitudes are deter- g˜ = gA (93) mined by components ξ(j) Proof of this theorem is not instructive, hence will On the margin note that be omitted. By using above theorem we may con- clude, that for any real number λ similar equality holds: 2 − 1 − 1 2 (V − ω T )ξ = 0 ⇒ (T 2 VT 2 − λI)ξ = 0, λ = ω (87) (g˜ − λg)(x, y) = (x, (A − λid)y) (94) 9

As a consequence, the subspace Vλ ≡ Ker(g˜ − λg) The most effective and practical turns out to be the sec- is non-trivial if and only if λ is an eigenvalue of ond approach, just described. Therefore we will try to the operator A, and this subspace is an eigensub- sketch steps needed to be taken in order to find nor- space of A. Note that if y ∈ Ker(g˜ − λg) then mal vibrations via simultaneous diagonalization of two for any vector x we have g˜(x, y) = λg(x, y). Thus, quadratic forms: subspaces Vλ orthogonal with respect to g are at the same time orthogonal with respect to g˜, while 1. Having found the Hessian matrix F (cf. 12, po- for metrics contracted to these subspaces there is a tential energy representation) as well as the kinetic energy matrix G (from geometric considerations), proportionality: g˜Vλ = λgVλ. It simply means that we want to find such basis transformation so that determination of the subspace Vλ brings us a basis where both g˜ and g are diagonal. This statement both mentioned matrices are diagonal. If our ini- is a subject of the next theorem: tial coordinates are cartesian, then G = T. First, solve the following equation: Theorem 3 Let g and g˜ be two symmetric forms on a real, finite dimensional space V , and assume that g 2 is positive-definite (it can be checked e.g. via Sylvester det(F − ω G) = 0 (100) criterion [15]). Let λ1, ..., λs be all possible real nu- mer for which the kernel g˜ − λig is non-trivial, i.e. to determine the allowed frequencies of vibrations Ker(g˜ − λig) 6= 0. Then V decomposed into a direct in the system. sum: 2 2. Find bases of the Kernel of the operator: F − ωi G where i = 1, .., n: s M V = Ker(g˜ − λig), (95) i=1 2 (F − ωi G)Q = 0 i = 1, ..., n (101) and this sum is orthogonal with respect to both metrics. In a basis orthonormal with respect to g, consistent with 3. If any of the modes is degenerate, then perform the space decomposition, the metric matrices take diago- orthonormalization with respect to T . The kinetic nal form: energy operator defines a quadratic form, which set a metric on euclidean space. The orthogonalization is performed with respect to this metric, what we   write as: (ei, Tej) = δij. λ11n1 0 ... 0 0 λ 1 ... 0 g’=1, g˜0 =  2 n2  (96) 4. As a result the Lagrangian takes separable form:  ......  0 0 ... λs1n n s 1 h i 1 X L = (Q˙ , βT GβQ˙ ) − (Q, βT FβQ) = Q˙ 2 − ω2Q2 2 2 i i i In practice, determination of Kernel Ker(g˜ − λig) i=1 is reduced to solution of a system of homogeneous (102) equations, together with a solvability condition: Transition matrix to the common eigenbasis is build of normalized vectors Q. The relation between original (cartesian) coordinates and new co- det(g˜ − λg) = 0 (97) ordinates (normal) is given as: Q = βx, where T T Fdiag = β Fβ , Gdiag = β Gβ. Gdiag has unit In the next step, for any found λ we solve the sys- values on its diagonal, while diagonal elements of tem of equations: Fdiag represent eigen-frequencies of the system. 5. The separated Lagrangian guarantees decoupled (g˜ − λg)x = 0 (98) equations of motion of harmonic oscillator type, which has a general solution: Linearly independent vectors corresponding to the same characteristic value are further orthogonal- Q = A cos ω t + B sin ω t i = 1, ..., n (103) ized with respect to g metric using Gram-Schmidt i i i i i method. After subsequent normalization we finally get: 6. Utilizing the relation between new and old coordinates (components respective vectors) we have: 0 0 0 T 0 T (e1, ..., en) = (e1, ..., en)β, g = β gβ, g˜ = β g˜β T − 1 (99) x = β Q = G 2 OQ (104) 10

7. Hence the explicit form for components reads: Normal coordinates are special case of curvlinear coordinates, same as straight line is a special case of a curve. n X T In what follows we introduce several frames of reference αi = βik (Ak cos ωkt + Bk sin ωkt) (105) that are commonly used by authors. After careful inves- k=1 tigation of the procedure for separation of the center of mass we shall focus on a special case of Eckart frame, to 8. From initial conditions x(0) = κ, x˙(0) = β we can yield so called Watson form of the rovibronic Hamilto- determine the values of coefficients Ak and Bk: nian. (k) Ak = (α , Gκ) (α(k), Gβ) (106) B. Rovibronic Coordinates Bk = ωk This section is intended to provide comprehensive, to Alternative and yet interesting method of finding of a reasonable extent, review of coordinate system choices normal coordinates may be the application of the Fourier and their implications on the form of molecular Hamilton transform to both sides of the equation for eigenvector operator. of both quadratic forms. By that means the equation is transformed into frequency domain, yielding readily separated problem. This can be done on account of first 1. Laboratory Frame powers of spatial coordinates appearing in the potential energy function. As a result we get a vector equation: Lets consider a system with N point particles. As a model for physical space we shall consider pair (Ξ,V ) constituting affine space over three dimensional vector 2 (V − ω T )xˆ(ω) = 0 (107) space V with metric gij = δij associated with set of points Ξ (Euclidean space). At this early stage we don’t where we used integration by parts twice and applied distinguish between electrons and nuclei, leaving the ex- the following trick: before we can act with the Fourier pressions more general. Such an approach may occur use- transform our functions must behave ’well’ enough, hence ful when investigating exotic systems like muon atoms we must assume that our system was under an action (cold fusion)[18, 19], Excitons [20], etc. The primary, of some dumping force parametrized by so that x(t) however poorly descriptive choice of coordinate system is vanishes in infinity. After that we take the limit → 0 space-fixed cartesian frame of reference depicted in fig.3 to retrieve the form of eq. 107. with origin marked as ∅0.

V. GENERAL THEORY OF NUCLEAR MOTION

A. Introduction

In this section we aim in giving a more detailed insight into the basics of nuclear motion theory. Unlike electronic structure, the nuclear motion calculations require 3 key factors: a) preferably exact representation of the nuclear kinetic energy operator in terms of chosen curvlinear coordinates, b) accurate potential energy surface, c) effective basis set, which allows for evaluation of matrix elements at the lowest possible cost. Having these provided a method of solution of the Schrödinger equation is also needed. Here we treat one aspect of point a). The last section, in turn, catches a bit of point Figure 3. An example of a space-fixed laboratory frame of c), serving rigorous theory that underlies commonly reference for 4-particle system. used rotational basis sets. Similar treatment of two most popular vibrational basis sets may be found in Given a set of physical quantities ( mass: mi, charge: ref. [16] and references therein. Methods of solution of Cie, electron g-factor: g, electron spin: si, nuclear g- SE are discussed in wide set of textbooks and papers; factor: gα, nuclear spin: Iα, nulcear charge quadrupole α α here we refer to author’s unpublished work on discrete moment: Qab, nuclear polarizability: χij, etc. ) one variable representation [17], as one of possible techniques. can try to extract information about a system by solving stationary Schrödingerequation. Resulting eigenval- 11 ues and eigenfunctions provide complete description of a system in quantum-mechanical sense. Hilbert space of such system divides into two orthogonal subspaces for (X1,Y1,Z1, ..., XN ,YN ,ZN ) → (XCM ,YCM ,ZCM , x2, y2, z2, ...xN , yN , zN ) (112) fermions and bosons, but this is yet another story [21]. where we have erased ’particle 1’ coordinates. Now Skipping less relevant digressions, a general Hamiltonian we need to check how momentum operators transform for a molecular system may be written in the following under the given coordinate change. These are 1-forms on form: Hilbert space. In cartesian laboratory frame (denoted as Hˆ = TˆT otal + VˆCoulomb + Hôther (108) L) the kinetic energy operator reads: N Hˆ stands here for other forms of energy not in- 2 X ∆L other Tˆ = −~ i (113) cluded in present discussion, e.g. mutual electron mag- 2 mi netic moment interactions, spin-orbit coupling, hyper- i=1 fine couplings, etc. These will be the subject of an- or in more general form can be regarded as the contrac- other article. As the system consist of N species, the tion of a metric tensor with a 2-vector formed from differ- quantum energy operator depend on 3N position opera- entiation operators. Lets transform the first derivatives tors: (X1,Y1,Z1, ...XN ,YN ,ZN ) and 3N momentum op- according to the chain rule: erators: (P1x,P1y,P1z, ...PNx,PNy,PNz). Our aim is to N ∂ X ∂xi ∂ ∂XCM ∂ change the coordinates in a way allowing us firstly to = = + ∂X ∂X ∂x ∂X ∂X perform easier separation of electronic motion from vi- 1 i=1 1 i 1 CM brations of nuclei and rotational motion entire molecule, N N X ∂xi ∂ m1 ∂ X m1 ∂ secondly to visualize the correspondence between expres- + = + (δ − ) ∂X ∂x M ∂X i1 M ∂x sions appearing in Hamiltonian and type of motion and i=2 1 i CM i=2 i finally to make some physically justified approximations (114) leading to some exactly solvable situations. These solutions may be later utilized as a basis for more accurate and calculations. N ∂ mk ∂ X mk ∂ a. The first step in changing to rovibronic coordi- = + (δ − ) (115) ∂X M ∂X ik M ∂x nates is separation of the center of mass. We therefore k 0 i=2 i ~ define isometric affine transformation (RCM ,MCM )∅LAB Due to the convention excluding first particle’s coordi- [22] as follows: nates from the coordinate system, the cartesian kinetic energy operator of the first particle transforms differently M : <3N → <3N : R~ = R~ + ~r (109) CM i CM i from the rest: N N ! where the translation of the origin is given by center of 2 2 2 2 2 ∂ m1 ∂ X ∂ X ∂ mass vector in laboratory frame of reference: 2 = 2 − 2 + ∂X M ∂X ∂XCM ∂xi ∂xi∂xj 1 CM i=2 i,j=2 N (116) 1 X R~ := m R~ (110) consequently, CM M j j j=1 N N ! 2 2 2 2 2 ∂ mk ∂ X ∂ X ∂ 2 = 2 − 2 + + Hence, the origin of the new coordinates system is cho- ∂X M ∂X ∂X0∂xi ∂xi∂xj k CM i=2 i,j=2 sen (by the above definition) in the center of mass of a 2 N 2 ! 2 mk ∂ X ∂ ∂ molecule, i.e. CM = LAB + R~ CM (cf. Fig.??). This ∅ ∅ + − 2 + 2 (117) M 2∂XCM ∂xk ∂xi∂xj ∂x implies: i,j=2 k

N One could tempt to draw conclusions about couplings X mj~rj = 0 (111) between particle’s motions at this stage. However, one j=1 remarkable feature of many-body systems should be pointed here: although some intermediate ’one-particle’ Thus, instead dealing with N position vectors of all par- terms may seem complicated, it is often the case, that ticles we can incorporate center of mass vector as one their collective treatment reveals numerous cancellations, of the coordinates and eliminate the coordinates of one entailing a simple final picture. Therefore one should not arbitrary particle, since there will always be exactly one rush for interpretation without analyzing the system as vector linearly dependent on others (cf. 111). This is a whole. Accordingly, after substituting derived Laplace formally another affine transformation with no transla- operators into kinetic energy terms we get: tional part, and linear part equal to identity except first   three relations replacing coordinates of the first particle N N 1 L m1 X X by center of mass coordinates, which are 0 in our sys- ∆ = ∆CM − 2 ∇~ CM · ∇~ j + ∇~ j · ∇~ i m 1 M 2   tem. Following this line of thought, we’ve passed from 1 j=2 j,i=2 laboratory to center of mass coordinates as shown below: (118) 12

 N N  1 mk X X ∆L = ∆ − 2 ∇~ · ∇~ + ∇~ · ∇~ + m 1 M 2  CM CM j j i k j=2 j,i=2

 N  2 X 1 + ∇~ · ∇~ − ∇~ · ∇~ + ∆ (119) M  CM k j j m k j=2 i Taking sums with respect to all particles the total kinetic energy operator reads:

2 N N ! ~ 1 X ∆i 1 X 0 Tˆ = − ∆CM + − ∇~ j · ∇~ i ≡ TˆCM +Tˆ0+Tˆ 2 M mi M i=2 j,i=2 (120) where we can identify the kinetic energy of the center of mass TˆCM , total kinetic energy of individual, inde- ˆ pendent particles T0 and kinetic energy correction term Figure 4. Three space-fixed frames of reference: Laboratory, resulting from coupling of correlated particles motion, total center of mass and nuclear center of mass. called often mass polarization or cross-terms Tˆ0. This term can be treated as a perturbation to uncoupled Hamiltonian, and in first order of perturbation theory the correction appears as so called mass polarization pa- Φ = Φ (X ,Y ,Z )Φ (x , y , z , ..., x , y , z ) PN D~ ~ E CM 0 0 0 int 2 2 2 N N N rameter: K = j,i=2 ∇j · ∇i [23]. (122) Lets notice at this point emerging coupling between where Φint is the wavefunction of internal motion of the the motions of particles. Because a change in position of system. any of the particles affects the position of center of mass, which we chose as dynamical variable of the system, and because the origin of coordinate system is located in the 2. Nuclear center of mass coordinate frame center of mass, numerical values of other particles positions undergo a change either. It is not a physical effect, It will be very useful to begin our discussion with the rather an artifact of particular coordinate system. It’s following theorem: the cost one must pay in order to separate the center of mass of a system and thereafter to reduce the dimension Theorem 4 (Center of mass coupling) If S denotes of the problem. This cost is however in most cases vi- the entire system and A is a system of particles that con- able due to the fact couplings are linear in the inverse of stitute a subset of S, then whenever transforming via total mass of the system, which in case of electrons cor- affine map MA from a space-fixed cartesian coordinate respond to non-Born-Oppenheimer (NBO) terms. These system with a frozen origin ∅0 to a cartesian coordinate are usually neglected on account of electron’s mass being system with origin ∅1 located in the center of mass of sub- three orders of magnitude smaller than that for light- set A, the corresponding kinetic energy operator of the est nucleus. In other words, all except first particle’s entire system S transforms into separable form of simple ˆ ˆ (which is excluded from the system) kinetic energy oper- sum of two kinetic energy operators TA ⊕ TS/A written in ators transform trivial when restricting to BO regime, i.e. explicit as follows: when neglecting linear terms of electron mass and total 2 2 ~ X ∆i ~ X (in practice nuclear) mass quotient. This is quite reason- TÂ = − + ∇~ j · ∇~ i (123) 2 mi 2MA able approximation, as the motion of light electrons very i∈A j,i∈A minutely affects the position of center of mass, therefore very weakly couples to motion of other particles. Culomb 2 2 ˆ ~ X ∆i ~ X ~ ~ potential energy depends only on distances between par- TS/A = − − ∇j · ∇i (124) 2 mi 2MA ticles, hence transforms identically to the center of mass i∈S/A j,i∈S/A coordinate system: where differentiation is taken with respect to coordinates N in ∅1 orthogonal coordinate system. 1 X q1q2 Vˆ = (121) 2 |~ri − ~rj| Proof 3 Linear part of the affine transformation relates i,j=2 coordinates of considered systems as: Because the Hamilton operator in 120 is separable, we 1 X can separate the wavefunction of center of mass as fol- ~ρi = ~ri − mj~rj (125) MA lows: j∈A 13 where i ∈ A. Using the chain rule we can write the ex- By rearranging summations one obtain separable opera- pression for the first derivative: tor:

3 2 A 2 2 X ∆i X ∆i X ∂ X X ∂ρjs ∂ Tˆ = − ~ − ~ − ~ ∇~ · ∇~ + = (126) ∅1 2 m 2 m M i j ∂xi ∂xi ∂ρjs i i A j∈S s=1 i=1 i∈S/A i,j∈S/A 2 ~ X ~ ~ ˆ ˆ Please note that the transformation does not mix dif- + ∇i · ∇j ≡ TA + TS/A (133) 2MA ferent components (can be decomposed into three trans- i,j∈A formations of orthogonal subspaces), hence summation over s vanishes immediately: Now, on the base of relation 120 we could claim that it is possible to pass from space-ﬁxed frozen-origin laboratory frame to A frame, on account of transitive relation ∂ X mi ∂ = δij − δiA (127) between these frames. To prove that and to reveal some ∂xi MA ∂ρjs j∈S formalism we write that: The symbol δ is deﬁned by: iA R~ CM ,MCM (134) ∅0 ( 1 when i ∈ A X δ = ≡ δ (128) We consider three transformations: iA 0 when i∈ / A il

l∈A 1. From the laboratory frame to the total center of mass frame, with associated affine transformation: Consequently, R~ CM ,MCM 2 2 ∅ ∂ X mi mi ∂ 0 2 = δij − δiA δik − δiA = ∂x MA MA ∂ρj1∂ρk1 i j,k∈S 2. From the total center of mass frame to the nuclear 2 2 2 2 center of mass frame: (~rNCM ,MNCM )∅ ∂ mi X ∂ mi X ∂ 1 = 2 − 2 δiA + δiA ∂ρi1 MA ∂ρi1∂ρj1 MA ∂ρj1∂ρk1 3. From the laboratory frame to the nuclear center of j∈S j,k∈S (129) mass frame: R~ NCM ,MLNCM ∅0 where ρk1 stands for 1-st component of pointing vector of If we are be able to show that combining the first two particle k in A center of mass system. Mixed terms can transformations yields in the third function, the proof is be evaluated in similar way to yield: be complete. In other words we ask if: 2 2 2 ∂ ∂ mi X ∂ ~ ~ = − δ − (~rNCM ,MNCM )∅ ◦ RCM ,MCM = RNCM ,MLNCM iA 1 ∅ ∅ ∂xi∂xj ∂ρi1∂ρj1 MA ∂ρj1∂ρk1 0 0 k∈S (135) 2 2 holds. Utilizing the properties of affine transforma- mj X ∂ mimj X ∂ − δ + δ δ jA 2 iA jA tion: (x, A)∅ ◦ (y, B)∅ = (x + Ay, AB)∅ and (x, A)∅ = MA ∂ρi1∂ρl1 M ∂ρk1∂ρl1 l∈S A k,l∈S (x + (A − id)a, A)∅+a we can write that: After extending above formulas to three dimensions, (~rNCM ,MNCM ) ◦ R~ CM ,MCM = (136) ∅1 the intermediate result appear as: ∅0 ~ ~ ! = ~rNCM − (MNCM − id)RCM ,MNCM ◦ RCM ,MCM 0 ∅ ∅ X ∆i X ∆i 1 X X ~ ~ 1 X ~ ~ 0 0 = − 2 ∇i · ∇j + ∇i · ∇j mi mi MA MA ~ ~ i∈S i∈S i∈A j∈S i,j∈S = ~rNCM + RCM ,MNCM MCM = RNCM ,MLNCM ∅0 ∅0 (130) In the present case the subsystem A stands for all nuclei And unlike in the approach of ref.[1], we obtain simple of the system (molecule), hence S/A gathers all electrons. cross terms transformation: Lets label by 1, 2, ..., Nnuc nuclei of the system and by Nnuc + 1,Nnuc + 2, ..., Nnuc + Nel, where Nnuc + Nel = X 0 0 ∇~ i · ∇~ j = 0 (131) N. Linear affine map into the new coordinates system is i,j∈S defined as:

3N−3 3N−3 Finally the total kinetic energy operator in ∅1 and the MNCM : < → < : ~ri = ~rNCM + ~ρi new coordinates reads : (137) where: 2 2 ~ X ∆i 2 1 X X ~ X ˆ ~ ~ ~ ~ N T∅1 = − +~ ∇i·∇j− ∇i·∇j nuc 2 mi MA 2MA 1 X i∈S i∈A j∈S i,j∈S ~r := m ~r (138) NCM M j j (132) n j=1 14

as in eq.133.

3. Euler Angles

Euler angles define the position of the rotated frame of reference with respect to the space fixed frame. There are several conventions yielding equivalent results, but for our purposes so called zyz’ convention will be most suitable [2]. It means that in order to transform one frame into the other we apply in proper order: φ angle rotation around z axis, then θ angle around y axis, and finally χ angle around z axis of already partially transformed frame. It is also convenient to introduce node line which represents positive sense of rotation from OZ to Oz and lays in the intersection of xy and XY planes. Full coverage of space is achieved when φ, χ ∈ [0, 2π] and Figure 5. Euler angles in zyz’ convention θ ∈ [0, π]. Explicit transformation relations for coordinates of a vector in rotating frame can be obtained by multiplying appropriate rotation matrices [24]. is nuclear center of mass in space fixed total center of     mass coordinates system. Therefore explicit transforma- 1 0 0 cos α 0 sin α tion may be written as: Rx(α) =  0 cos α − sin α  Ry(α) =  0 1 0  0 sin α cos α − sin α 0 cos α N me X (140) ~ρi = ~ri + ~rj (139)   Mn cos α − sin α 0 j=Nnuc+1 Rz(α) =  sin α cos α 0  The above theorem guarantees that we can make 0 0 1 straightforward transformation from a space-fixed coordinate system into a nuclear center of mass coordinate Therefore in zyz0 scheme the general rotation matrix in system obtaining the kinetic energy operator expressed physical space reads:

 cos θ cos φ cos χ − sin φ sin χ cos θ sin φ cos χ + cos φ sin χ − sin θ cos χ  Rzyz0 (φ, χ, θ) = Rz0 (θ)Ry(χ)Rz(φ) =  − cos θ cos φ sin χ − sin φ cos χ − cos θ sin φ sin χ + cos φ cos χ sin θ sin χ  sin θ cos φ sin θ sin φ cos θ (141) and this acts on the column representation of a vector in 4. Molecule-fixed frame the space fixed frame, yielding the column representation of a vector in the rotating frame: a. Classical approach Main advantage of the body-     fixed frames of reference together with rovibronic coor- x X dinates is intuitive form of associated Hamilton operator  y  = Rzyz0 (φ, χ, θ)  Y  (142) of a system, allowing to apply approximations leading z Z to separation of rotational and vibrational motion, etc. Terms responsible for particular physical effects may be then identified, providing convenient way of controlling Of course Rzyz0 (φ, χ, θ) ∈ SO(3), hence: approximation procedures. They also have considerable computational advantages such as use of 3−j symbols in T T Rzyz0 (φ, χ, θ)Rzyz0 (φ, χ, θ) = Rzyz0 (φ, χ, θ) Rzyz0 (φ, χ, θ) = 1 angular potential integrations instead of 6 − j in space- (143) fixed case [25]. Lets introduce the body-fixed frame, de- Rotations in Hilbert space associated with rotations in termine rotational coordinates. Mutual orientation of ~n body-fixed and space fixed frames is commonly given by physical space are given by unitary operator: U Rφ = set of three Euler angles (see fig.141). Origins of both co- exp − i J~ · ~nφ as for rotation about φ around ~n vector. ~ ordinate systems are located in molecular center of mass. 15

Previous assumptions lead to the conclusion that three The only thing needed to be specified is the embedding rotating axes must be embedded to the molecular system of the rotating frame, which will provide another three according to some arbitrary rule. So far we’ve got a set of relations. The natural choice and historical one is the 3N − 3 coordinates in molecular center of mass cartesian condition of vanishing angular momentum of a collection coordinate system: (x2, y2, z2, ..., xN , yN , zN ). We want a of all particles in rotating frame: transformation into rotating frame of reference described by set of new coordinates: (θ, φ, χ, Q1, ..., Q3N−6). These N ~ X ˙ are called rovibronic coordinates [1]. Qi stand for func- J = mα~r × ~rα = 0 (149) tions of cartesian coordinates. Herein, these functions α=1 are linear giving rise to normal coordinates. However one may choose any curvlinear internal coordinates (not Now if ~ρα is small for all α’s then we can within a good discussed here). Every molecule has its equilibrium ge- approximation write ~rα ≈ ~aα and ometry defined as a set of coordinates values minimizing globally internal potential energy of the molecule. Lets denote equilibrium position vectors of all particles in a N ~ X ˙ system as ~aα defining displacement from equilibrium po- J ≈ mα~a × ~rα = 0 (150) sition vectors as: α=1

~ρα = ~rα − ~aα (144) This approximation hold well for most of fairly rigid molecular systems, as the amplitudes of vibrations of According to ﬁg.4 we can write total velocity of α − th semi-rigid molecules generally don’t exceed 5% of bond particle of a system in laboratory cartesian coordinates length [2]. For more general point of view see, for ex- system. ample later paper by Schmiedt, et al.[26]. Keeping this intuition C. Eckart [27] postulated following general con- ˙ V~α = R~ α + ~ω × ~rα + ~vα (145) ditions:

The components of above equation correspond to center N X ˙ ˙ mα~a × ~ρα = 0 (151) of mass velocity R~ α, velocity due to rotation of body- α=1 fixed frame ~ω × ~rα and motion of particle in the rotating coordinate system ~v , respectively. Now lets write a α whereas eq.150 stating approximate angular momentum classical formula for the kinetic energy of our system of conservation law occurs as it’s direct consequence. The N particles in LAB frame: specification of the instantaneous position of the moving axes requires six numbers, which may be taken to be the N N three center of mass coordinates and the Eulerian angles X ~ 2 ~˙ 2 X 2T = mαVα = MRα + mα (~ω × ~rα) · (~ω × ~rα) + [27]. Hence we should have only 3N − 6 independent α=1 α=1 internal coordinates Qi. Utilizing these conditions we N N may write the kinetic energy in a new form X 2 X ~˙ + mα~vα + 2mαRα + (~ω × ~rα) + α=1 α=1 N N N ~˙ X 2 X X + 2Rα mα~vα + 2~ω · mα (~rα × ~vα) (146) 2T = mα (~ω × ~rα) · (~ω × ~rα) α=1 α=1 α=1 N where for the last term we’ve used cyclic invariance of X 2 + mα~vα+ mixed vector product. After separating centre of mass α=1 motion the expression simplifies a bit - this is equivalent N X to passing into center of mass frame of reference. This + 2~ω · m (~ρ × ~v ) yield in three relations confining the coordinates of all α α α α=1 particles of a system: where we have rotational energy, vibrational kinetic N energy and Coriolos coupling energy respectively. This X mα~rα = 0 (147) Eckart-derived form allows to separate rotations from vi- α=1 brations with the least cost, letting for most effective perturbation treatment of the Coriolis coupling. Note which is followed by (see 145) that Eckart equations have to be solved for euler an-

N gles. After introducing moment of inertia tensor Iij = X PN m (~r · ~rI − ~r ⊗ ~r ); and some algebraic manipu- mα~vα = 0 (148) α=1 α α α α=1 lations the kinetic energy reads 16

into the quantum mechanical formalism. Lets start from the definition of angular momentum: 2 2 2 2T = Ixxωx + Iyyωy + Izzωz − 2Ixyωxωy − 2Ixzωxωz− N N N X X 2 X J~ := m ~r × ~r˙ (158) − 2Iyzωyωz + mα~vα + +2~ω · mα (~ρα × ~vα) α α α=1 α=1 α=1 (152) Introducing angular velocity, moment of inertia and nor- Deriving the above relation would be a good simple ex- mal coordinates we get ercise, left for the reader. One of the possible ways of dealing with internal motion of molecule (relative mo- N N ~ X X tion of nuclei) is normal coordinates approach. They are J := mα~r × (~ω × ~rα) + mα~rα × ~vα = defined by linear relation to cartesian coordinates: α=1 α=1 N N 3N−6 X 2 X = mα r ~ω − (~ω · ~rα) ~rα + mα (~rα − ~aα) × ~vα = X α α ραi = ηikQk (153) α=1 α=1 k=1 =x ˆ (Ixxωx − Ixyωy − Ixzωz) +x ˆ (Iyxωx − Iyyωy − Iyzωz) + for i = 1, 2, 3 - index of displacement vector compo- 3N−6 α X ˙ nents, ηik stands for linear transformation matrix (3 × +z ˆ(Izxωx − Izyωy − Izzωz) + ~τsQt (159) N × 3N − 6) diagonalizing simultaneously kinetic energy s,t=1 and quadratic terms in potential energy of a system ; Qk is k-th normal coordinate. Of course α enumerates par- It is desired to relate angular and generalized momenta, ticles of the system, in practice those are the nuclei. We therefore we write the momentum conjugated to Q as don’t transfer the notation from section II, to save the s consistency with original derivations. Within the normal 3 ∂T X coordinates framework the Coriolis term reads P = = Q˙ + τ iω (160) s ˙ s s i N 3 N 3N−6 ∂Qs i=1 X X X X i α α ˙ ~ω · mα (~ρα × ~vα) = ω mαijkηjsηktQsQt = α=1 i,j,k=1 α=1 s,t=1 now we can construct vibrational angular momentum. 3 3N−6 Its presence can be easily observed when considering X X i ˙ = ω τisQt (154) vibrations of acetylene, where some motions of atoms i=1 s,t=1 that destroy linearity may contribute to some internal where we introduced Levi-Civita totally antisymmetric angular momentum of the molecule. Another example tensor ijk and defined τ matrix as follows: may be quasi-free rotations of methyl groups in hydro- carbons. These motions are in fact oscillations, which 3 N ’look’ like rotations, carrying also some internal angular X X α α τis := mαijkηjsηktQs (155) momentum. j,k=1 α=1 3N−6 3N−6 3N−6 X i X i ˙ X i Here for the first time so called Coriolis coupling con- ji = τkPk = τkQk + τk (~τ · ~ω) (161) stants ξ appeared, defined by the relation k=1 k=1 k=1

3 N It is clear that X X α α ξis := ijkηjsηkt (156) 3N−6 j,k=1 α=1 X 2T = J~ · ~ω + PkQ˙ k (162) The kinetic energy transforms now to the form k=1

and 2 2 2 2T = Ixxωx + Iyyωy + Izzωz − 2Ixyωxωy − 2Ixzωxωz− 3N−6 3 3N−6 Q˙ k = Pk − ~τs · ~ω (163) X X i ˙ X ˙ 2 − 2Iyzωyωz + ω τisQt + Qt (157) s,t=1 i=1 k=1 hence, Our aim is to make transition into quantum- 3N−6 mechanical expression for the kinetic energy, thus Hamil- ~ ~ X 2 2T = J − j · ~ω + Pk (164) ton picture is required, which, in turn involves general- k=1 ized momenta and coordinates. Historical and perhaps more intuitive route utilizes angular momentum repre- The only task now is to get rid of angular velocity. Mak- sentation, what de facto makes somewhat around way ing use of relation between angular momentum and an- 17 gular velocity what follows from 161

3N−6 X x ˙ Jx = Ixxωx − Ixyωy − Ixzωz + τk Qk k=1 3N−6 X y ˙ Jy = −Iyxωx + Iyyωy − Iyzωz + τk Qk (165) k=1 3N−6 X J = −I ω − I ω + I ω + τ zQ˙ z zx x zy y zz z 3N−6 k !k 3N−6 ! 3N−6 ! Xk=1 x 2 X x y X x z Jx − jx = Ixx − (τk ) ωx − Ixy + τk τk ωy − Ixz + τk τk ωz k=1 k=1 k=1 3N−6 ! 3N−6 ! 3N−6 ! X y x X y 2 X y z Jy − jy = − Iyx + τk τk ωx + Iyy − (τk ) ωy − Iyz + τk τk ωz (166) k=1 k=1 k=1 3N−6 ! 3N−6 ! 3N−6 ! X z x X z y X z 2 Jz − jz = − Izx + τk τk ωx − Izy + τk τk ωy + Izz − (τk ) ωz k=1 k=1 k=1

By inverting the above relations to express angular ve- from cartesian to rovibronic(curvlinear) coordinates: locities on left hand side of equations lets rewrite this −→ (x1, y1, z1, ..., xN , yN , zN ) = ξ (q1, ..., q3N ) where ξ is system in a matrix form. Formally we intoduce composed of transformation into molecular center of mass system and transformation molecule-fixed rotating frame ~ ~ ~ M = J − j (167) of reference by use of Euler angles 4. Both transformations are affine transformations. The first one is simply and ~ (RCM ,MCM )∅LAB and the latter has identity as trans-   lation part, while linear part is a combination of three ωx Euler rotations. Classical expression for total energy in ~ω =  ωy  (168) Lagrange formalism may be simply written as [14]: ωz

˙ ˙ 1 P √ ˙ ˙ relation 166 may be formulated in compact form E(~r1, ..., ~rN , ~r1, ..., ~rN ) = 2 ij mimjδij~ri · ~rj + V (~r1, ..., ~rN ) (171) M~ = I~ω (169) and corresponding Hamilton form: from which it’s straightforward to obtain angular veloc- 1 P δij ~pi·~pj H(~r1, ..., ~rN , ~p1, ..., ~pN ) = √ + V (~r1, ..., ~rN ) ities by inverting real symmetric matrix I. Then kinetic 2 ij mimj (172) energy takes more compact form, being a good starting Transformation into generalized coordinates aﬀects the point for the quantum Hamiltonian: expression for Lagrange-form energy in following way: 3N−6 1 1 X T = M~ T µM~ + P 2 (170) E(q , ..., q , q˙ , ..., q˙ ) = 1 P g q˙ q˙ + V (q , ..., q ) 2 2 k 1 3N 1 3N 2 ij ij i j 1 3N k=1 (173) where use of chain rule for coordinate change results in where M~ := J~ −~j and µ is the inverse matrix of I. This form of quadratic form metric tensor: exact expression plays a crucial role in classical mechanics of non-rigid rotating bodies [2]. X ∂rnα ∂rnα g = m (174) ij n ∂q ∂q n,α i j 5. Podolsky Trick We can use the deﬁnition of generalized momentum Pi = ∂(H−V ) When attempting to derive quantum-mechanical form ∂q˙ to transform above relations into Hamilton form: of the Hamilton operator for the general system discussed in previous paragraph, one might try to 3N 1 X apply Jordan rules straightforwardly to eq. 170. H(q , ..., q ,P , ..., P ) = gijP P +V (q , ..., q ) 1 3N 1 N 2 i j 1 3N Unfortunately such approach will result in wrong ij=1 answer, mainly because we’ve made transformation (175) 18

ij where g matrix is inverse of gij (we’ve simply raised two volume element transforms linearly with factor equal to 1 indices in metric tensor). Jacobian: dq = Jac[q(x)]dx = g 2 dx. Now demanding consistency: X ∂qi ∂qj gij = m−1 (176) Z Z Z n ∂r ∂r 2 2 2 1 n,α nα nα |ψ(x)| dx = |ψ(q)| dq = |ψ(q)| g 2 dx = 1 V V V For clarity of notation we will up from now use Einstein (185) summation convention, discerning upper and lower in- we find that 1 dices. Therefore classical Hamilton function will be writ- − 4 1 ij ψ(q) = g ψ(x) (186) ten as follows: H = 2 g PiPj + V (qi). In cartesian coordinates the metric tensor has unit matrix representation Derived expression for the kinetic energy 170 involves (up to a constant factor). However when transformed however angular momenta, while in Podolsky approach into, for example spherical or elliptical coordinates, it we utilize generalized momenta representation. Thus, the takes more complicated form. The proper form of the mapping from the classical expression into the quantum- quantum Hamiltonian for general coordinates system was mechanical form cannot be done directly. In order to given by B.Podolsky in 1928 [28]: apply the Podolsky procedure to our form of the Hamil- ton function, we must first find the form of the podolsky 1 − 1 − 1 ij − 1 Hˆ = g 4 pîg 2 g pˆjg 4 + Vˆ (177) Hamiltonian in the angular momentum representation, 2 then prove that this formula is consistent with the origi- where g is the determinant of the metric tensor. The nal one, i.e. gives the proper quantum-mechanical energy above expression can be derived in following way. First operator. Lets start from the assumption that general- we build Laplace-Beltrami operator [29] as ized momentum is linearly related to the rovibronic angular momentum. This is very strong statement, and will ∆ := div ∇~ (178) need a detailed proof. 3 with divergence of a vector field F defined as X 0 pi = aikMk (187) k=1 − 1 − 1 i divF := g 2 ∂ig 2 F (179) then classical expression for kinetic energy 170 takes the and gradient of a scalar field φ: form

i ij X ij 0 0 X 0 0 ∇~ φ := ∂ φ = g ∂jφ (180) 2T = aikajlg MkMl ≡ GklMkMl (188) i,j,k,l k,l Combining the two results in Ask then what conditions must be satisfied in order that − 1 − 1 ij ∆ := div ∇~ = g 2 ∂ g 2 g ∂ (181) i j 1 1 X 0 ij − 1 0 1 Hˆ = G 4 Mˆ G G 2 Mˆ G 4 + Vˆ (189) 2 i j Therefore if the transformation from cartesian to ij generalized coordinates is given by the relation i i while knowing G? First lets invert eq. 187 q = q (x1, y1, z1, ..., xN , yN , zN ) then the quantum- mechanical Hamilton operator transforms according to 3 3 0 X −1 X ik the expression: Mi = (a )ikpk ≡ a pk (190) k=1 k=1 1 − 1 − 1 ij Hˆ = g 2 pˆ g 2 g pˆ + Vˆ (182) 2 i j of course 3 and the corresponding Schrödingerequation reads X kj k aika = δi (191) 1 − 1 − 1 ij k=1 g 2 pˆ g 2 g pˆ ψ(q) + Vˆ − Eidˆ ψ(q) = 0 (183) 2 i j Inserting 190 into 189 yields in Note that wavefunction in cartesian representation was 1 1 1 X ik ij −1 − 1 jl 1 1 normalized according to the condition: Hˆ = a 2 g 4 a pˆ a a g a g 2 a p a 2 g 4 + Vˆ = 2 k ir tj l Z i,j,l,k,r,t 2 |ψ(x)| dx = 1 (184) 1 1 1 X ik ij −1 − 1 1 1 = a 2 g 4 a pˆ a g a g 2 p a 2 g 4 + Vˆ ≡ V 2 k ir l i,j,k,r,t where dx = dx1dy1dz1...dxN dyN dzN is the volume ele- 1 − 1 1 X − 1 ij 1 1 ment in space V . Now, after transformation our wave- ≡ s 2 g 4 pˆ g 2 g pˆ g 4 s 2 + Vˆ (192) 2 i j function is expressed by generalized coordinates. But i,j 19 only if Vector as a tensor object must be invariant to coordinate change, while its components transform according to gen- 1 X ik −1 − 1 eral tensor transformation. This implies vector equality a 2 a pˆkaira = a 2 pî (193) k,r ~ω0 = ~ω (195)

This result mean that if we’re able to find linear relation of angular velocities in both coordinate frames. General between angular (or any other quantity) and linear mo- expression for angular velocity in space-fixed frame reads mentum, and this relation would fulfill above condition, 0 0 ˆ0 0 ˆ0 0 ˆ0 then we can replace classical Hamilton function repre- ~ω = ωX i + ωY j + ωZ k (196) sented by 188 with 189. Knowing the coefficients aij and and in rotating molecule-fixed frame: the metric tensor gij is sufficient to find the quantum- mechanical expression for the Hamiltonian of a non-rigid ˆ ˆ ˆ body. This enables us to set up the next step, namely ~ω = ωxi + ωyj + ωzk (197) finding geometrical relation between rovibronic angular ˙ ˙ momentum and linear generalized momentum conjugated After that we need to express φ,~ θ,~ ~χ˙ components in a with euler angles. Consequently lets expand angular mo- space-fixed basis. Lets make use of geometrical relations: mentum in terms linear momenta via the chain rule: ˙ φ~ = φ˙kˆ0 ˆ0 ˆ ~˙ k × k ˙ 1 ˆ0 ˆ0 ˆ0 ˆ0 ˙ ˙ ˆ0 ˙ ˆ0 ˙ ∂φ˙ ∂χ˙ ˙ ∂φ˙ ∂χ˙ θ = θ = k × cos θ · k + sin θ cos φi + sin θ sin φj θ = cos φ · θ · j − sin φ · θ · i J~ = ∂T = ∂θ ∂T + ∂T + ∂T ≡ ∂θ p + p + p ~ 0 ~ ∂~ω ∂~ω ∂θ˙ ∂~ω ∂φ˙ ∂~ω ∂χ˙ ∂~ω θ ∂~ω φ ∂~ω χ ||k × k|| sin θ (194) (198) assuming that potential energy is independent of gener- ~χ˙ = cos θ · χ˙ · kˆ0 + sin θ cos φ · χ˙ · î0 + sin θ sin φ · χ˙ · ˆj0 alized velocities. Thereby, all we need is relation between ˙ ˙ (199) the components of φ,~ θ,~ ~χ˙ and components of angular velocity in molecule-fixed rotating frame ωx, ωy, ωz. Imag- On the other hand the components of angular velocities ine space and molecule-fixed frames of reference with an- in both frames are related by Euler angles transformation gular velocities ω drawn along the corresponding axes. matrix:

     0  ωx cos θ cos φ cos χ − sin φ sin χ cos θ sin φ cos χ − sin θ cos χ ωX 0  ωy  =  − cos θ cos φ sin χ − sin φ cos χ − cos θ sin φ sin χ + cos φ cos χ sin θ sin χ   ωY  (200) 0 ωz sin θ cos φ sin θ sin φ cos θ ωZ

Provided components of ~ω0 expressed by Euler angles time derivatives (cf.199) we can write ﬁnal transformation as

     ˙  ωx cos θ cos φ cos χ − sin φ sin χ cos θ sin φ cos χ − sin θ cos χ sin θ cos φχ˙ − sin φθ ˙  ωy  =  − cos θ cos φ sin χ − sin φ cos χ − cos θ sin φ sin χ + cos φ cos χ sin θ sin χ   sin θ cos φχ˙ − sin φθ  ωz sin θ cos φ sin θ sin φ cos θ φ˙ + cos θχ˙ (201) obtaining set of linear equations with respect to Euler tions): angles time derivatives. We can easily identify resulting transformation matrix and write inverse relations (it’s faster to inverse this linear system by simple substitu-  ˙      θ sin χ cos χ 0 ωx  φ˙  =  − csc θ cos χ sin χ csc θ 0   ωy  χ˙ cot θ cos χ − cot θ sin χ 1 ωz (202) 20

Now we are ready to write explicit relation from eq. 187:       Jx sin χ − csc θ cos χ cot θ cos χ pθ  Jy  =  cos χ sin χ csc θ − cot θ sin χ   pφ  Jz 0 0 1 pχ (203) After introducing vibrational angular momentum ~j to complete general momenta space, we are able to read coeﬃcients in eq.190:

 Mx   x x   pθ  sin χ − csc θ cos χ cot θ cos χ −τ1 ... −τ3N−6 My y y pφ    cos χ sin χ csc θ − cot θ sin χ −τ1 ... −τ3N−6     M  z z  p   z   0 0 1 −τ ... −τ   χ   P  =  1 3N−6   P  (204)  1     1   .     .   .   03N−6×3 13N−6   .  P3N−6 P3N−6

~ P ~ Note that we have 3 generalized momenta associated with j = k τk·Pk. Relation 190 requires also inverse a matrix Euler angles and 3N − 6 momenta incorporated in vibra- coeﬃcients, which can be obtained with little mainpula- tional angular momentum and vibrational kinetic energy tion on 205:

 p  y y  M  θ  sin χ cos χ 0 − sin χ · τ x cos χ · τ 0 ... − sin χ · τ x cos χ · τ 0  x p 1 1 3N−6 3N−6 M  φ  − sin χ cos χ sin χ sin θ cos θ − sin θ cos χτ x sin θ sin χτ y cos θτ z ... − sin θ cos χτ x sin θ sin χτ y cos θτ z  y   p   1 1 1 3N−6 3N−6 3N−6   M   χ   z z   z   P   0 0 1 0 0 τ1 ... 0 0 τ3N−6   P   1  =    1   .     .   .   0 1   .   .  3N−6×3 3N−6  .  P3N−6 P3N−6 (205)

Utilizing some trigonometric identities we ﬁnd that Now lets evaluate following a-matrix sums: −1 −1 1 a ≡ det a = sin θ ; of course from Cauchy matrix theorem a ≡ det a = sin θ. In the next step we should investigate if the condition

1 X ij −1 − 1 a 2 a pˆjaija = a 2 pˆi (206) i,j is satisﬁed. It may be rewritten to a simpler form:

X ji −1 a pîakja = 0 (207) i,j,k Consequently, X X ji −1 −1 X ji ak1 = sin χ − sin θ cos χ a akjpîa + a a pîakj = 0 (208) k i,j,k i,j,k X ak2 = cos χ − sin θ sin χ k X −1 −1 X ji X pˆka + a a pî akj = 0 (209) X a = cos θ + 1 k i,j k k3 k X x y ! akj = (sin χ − sin θ cos χ)τj + (cos χ − sin θ sin χ)τj + cos θ 1 X ij X k − + a pî akj = 0 (210) sin2 θ sin θ z i,j k + (cos θ + 1)τj + 1 j > 3 (211) 21

Acting with momenta operators on respective sums: two last terms give no contribution to the ﬁnal sum p1 ≡ pθ and eventually

X 2 pθ ak1 = − cos θ cos χ ... = − sin χ cos χ cos θ + cos χ cot θ + cos θ cos χ sin θ+ k 2 X + sin χ cos χ cos θ − cos θ sin χ cos χ + sin χ cot θ = cot θ pθ ak2 = cos θ sin χ (218) k X Coming back to 210 we find Podolsky condition fulfilled: pθ ak3 = − sin θ k cos θ cot θ X x y z − + = 0 (219) pθ akj = − cos θ cos χτj + cos θ sin χτj − sin θτj j > 3 sin2 θ sin θ k (212) This means that we can pass from classical quadratic p ≡ p form of the Hamilton function to the quantum- 2 φ mechanical operator form denoted below [30] X pφ ak1 = 0 1 1 X − 1 1 1 1 X − 1 1 k Hˆ = µ 4 Mˆ iµijµ 2 Mˆ jµ 4 + µ 4 Pˆkµ 2 Pˆkµ 4 +Vˆ X 2 2 p a = 0 i,j k φ k2 (220) k X where following relation must be satisfied: pφ ak3 = 0 k µ = aT · g · a (221) X pφ akj = 0 j > 3 (213) k C. Watson Simplification p3 ≡ pχ X pχ ak1 = cos χ + sin θ sin χ After almost 30 years after derivation of the Hamilto- k nian from eq.220 by Darling and Dennison [30] J.K.G. X Watson came up with a tricky way to simplify this op- p a = − sin χ + sin θ cos χ χ k2 erator using some sum rules and commutation relations k [31]. The final form of the Hamiltonian is very similar to X pχ ak3 = 0 the classical one k X 1 X 1 X 2 x Hˆ = Mˆ iµijMˆ j + Pˆ + Uˆ + Vˆ (222) pχ akj = (sin θ sin χ + cos χ)τj + 2 2 i k i,j i + (sin θ cos χ − sin χ)τ y j > 3 (214) j where Uˆ occurs to be mass dependent contribution to the pi ≡ pi, i > 3 potential

X 2 2 p3 ak1 = 0 X Uˆ = −~ µ ≡ −~ T rµ (223) k 8 ii 8 . i . To remind, in present case we are treating all particles of X p3N−6 ak3N−6 = 0 (215) a system as point masses, not distinguishing between nu- k clei and electrons. For the sake of convention lets denote This allows us to write explicit expression for second term instantaneous position of i − th particle in body fixed in 210: rotating frame ~ri = (rix, riy, riz). Reference configura- tion, not necessarily equilibrium, will be abbreviated as ! ! 0 X ij X X i1 X ~ri . We require now six constraints that allow to specify a pî akj = a pî ak1 (216) the position and orientation of moving axes at every in- i,j k i k stant (relative to set of particles). In fact it was already ! introduced as the Eckart frame 151.To summarize and X i1 X + a pî ak1 + gather all conditions within present notation: i k ! ! 1. X i3 X X ij X + a pî ak3 + a pî akj = ... N i k i,j>4 k X m ~r = 0 (224) (217) ı ı ı=1 22

setting rotating frame of reference of molecular cen- coordinates, say normal coordinates[32], related to carte- ter of mass. As long as there’s no external ﬁelds sian by orthogonal transformation Rˆ =∈ SO(3N). Of translation in space is the symmetry of the molecu- course the conditions for the body-ﬁxed frame may be lar Hamiltonian [1], due to space uniformity. Hence written in both types of coordinates, but normal coor- molecular center of mass motion can be separated dinates provide expressions with easily discernible rota- and corresponding constant energy taken as a ref- tional, vibrational and coupling parts, making the prob- erence. lem more transparent. In practise we have to decompose Rˆ into two: Rˆ = Lˆ · Wˆ . 2.

X 0 ˙ mi~ri × ~ρi = 0 (225) − 1 X Q1 = M 2 miρi1 i i There’s no angular momentum of generated by the − 1 X Q2 = M 2 miρi2 system with respect to the rotating frame(in fact i P 0 the conditions are more general i mi~ri × ~ρi = 0) − 1 X Q3 = M 2 miρi3 (226) In order to separate the center of mass motion and ro- i tation of the axis system we introduce a new set of 3N

X 1 X 1 1  Q = I0 2 m 2 r0 m 2 ρ 4 1β i βγδ iγ i iδ β,γ,δ i   1 1 1  X 0 2 X 2 0 2  1 X i Q5 = I m βγδriγ m ρiδ 0 2 αβ 0 2β i i I mi ~ri × ~ρi (227) i β,γ,δ  i 1 1 1  X 0 2 X 2 0 2  Q6 = I m βγδr m ρiδ 3β i iγ i  β,γ,δ i

0. L matrix in one of possible ways can be understood as 3N × 3N − 6 rectangular transformation to normal 1 X 2 coordinates unitary matrix: Qk = lαikmi ρiα k = 7, ..., 3N (228) i,α

In order to apply some rotating axes embedding we make use of Eckart conditions and write Q(1) = ... = Q(6) =

1  2 1  m1 m 2 N 1 0 0 ... 1 0 0  M 2 M 2   1 1  2 2  m1 mN   0 1 0 ... 0 1 0   M 2 M 2   1 1  2 2  m1 mN   0 0 1 ... 0 0 1   M 2 M 2   1 1 1 1   P I0 2 m 2 r0 ... P I0 2 m 2 r0   β,γ 1β 1 βγ1 1γ β,γ 1β N βγ1 Nγ   1 1 1 1   P I0 2 m 2 r0 ... P I0 2 m 2 r0   β,γ 2β 1 βγ1 1γ β,γ 2β N βγ1 Nγ   1 1 1 1   P I0 2 m 2 r0 ... P I0 2 m 2 r0  (229)  β,γ 3β 1 βγ1 1γ β,γ 3β N βγ1 Nγ     l111 l211 l311 . . . l1N1 l2N1 l3N1     .   .     .   .   .     .   .     .   .  l113N−6 l213N−6 l313N−6 . . . l1N3N−6 l2N3N−6 l3N3N−6 23

Imposing unitarity of the above matrix LT L = LLT = 1 • The last relation arising from the orthogonality of L provides useful relations: is useful a transformation of cartesian coordinates into normal modes: •

3N m 3N−6 X T 1 − 1 L Ls1 = + 0 2 X −1 X 1s M riα = riα + mi lαikQk + M mjρjα+ s=1 k=1 i 1 1 X 0 2 0 2 0 0 1 X 0 0 − 2 + I αβ m1βγαηνα I αη r1γ r1ν + (230) + αβγ riγ I βν Rα β,γ,α,η,ν βγν 3N−6 (237) X 2 + l11k = 1 (231) k=1 the last to terms vanish in Eckart frame, yielding • ’standard’ relations between cartesian and normal N N coordinates. X X lαiklαil = δkl , α = 1, 2, 3 k = 1, ..., 3N − 6 i=1 i=1 (232) There are two kinds of vibration-rotation interaction coeﬃcients •

3 N X X − 1 1 I0 2 m 2 r0 l = 0 (233) 1β i βγα iγ αik 1. Coriolis coupling coefficients in vibrational angular i=1 β,γ,δ=1 momentum: Left hand side of the last equation can be rear- ranged to: X α ˆ ˆ jα = CklQkPl (238) 3 3 N k,l − 1 1 X 0 2 X X 2 0 I 1β mi βγαriγ lαik = 0 (234) β=1 γ,δ=1 i=1 where Pˆ is momentum conjugate to Qˆ i.e. ∂L = ∂Q˙ l We can discover that moment of inertia tensor P . The coefficients are defined in following way stays symmetric under index transposition, there- l fore provided that the entire above expression is zero, the other factor in the equation must be anti- X Cα := l l = −Cα (239) symmetric under odd index permutation. This im- kl αβγ βik γil lk 1 β,γ,i 3 N 1 P P P 2 0 2 plies that γ,δ=1 i=1 i mi βγδriγ m must be antisymmetric, and because Levi-Civita tensor is antisymmetric the following relation must be ful- α Note that Cα = P ~l × ~l , hence Cα = 0. filled: kl i ik il kk

1 1 X 2 0 X 2 0 mi riαlβik = mi riβlαik (235) 2. Interaction coeﬃcients (dependence of moment of i i inertia on normal coordinates):

• Finally summing over rows we get αβ ∂Iαβ 1 1 X 1 − 1 1 − 1 a := (240) M −1m 2 m 2 δ + m 2 I0 2 r0 m 2 I0 2 r0 + k i j ξφ i βγξ αβ iγ j ηνφ αη iν ∂Qk 0 α,β,γ,η,ν 3N−6 X + lξiklφjk = δijδξφ (236) Because moment of inertia tensor can be expanded k=1 in normal modes basis in following way 24

X X Iαβ = αγδβηδ miriγ riδ = γ,δ,η i 3N−6 ! 3N−6 ! 1 1 X X 0 − 2 X 0 − 2 X = αγδβηδ mi riγ + mi lγinQn riη + mi lηisQs = γ,δ,η i n=1 s=1 3N−6 3N−6 3N−6 ! 1 1 X X 0 0 0 X − 2 0 X − 2 −1 X = αγδβηδ mi riγ riη + riγ mi lηisQs + riη mi lγinQn + mi lηislγinQnQs (241) γ,δ,η i s=1 n=1 s,n=1

thus finally: Proof 4 We’ve shown that angular momentum is related to generalized momenta by linear expressions (cf.203). Then left hand side for choice of x and z components is − 1 − 1 αβ X 0 2 0 2 equal to (on account of linearity of commutators): ak = riγ mi lηik + riηmi lγik + γ,δ,η,i h i h i h i h i 3N−6 ! cos χ Jˆx, Jˆz = sin χPˆθ, Pˆχ − Pˆφ, Pˆχ + cot θ cos χPˆχ, Pˆχ = ... −1 X sin θ +mi (lηislγikQs + lηiklγisQs) αγδβηδ = (247) s=1 here comes perfect moment to introduce following useful − 1 X 0 2 lemma: = 2 riγ mi lηikαγδβηδ (242) γ,δ,η,i d df [f(x), pˆ ] = f(x)ˆp − p f(x) = −i f(x) + i + As one could expect our interaction coefficient is x x x ~ dx ~dx symmetric with respect to indices permutation. d d + i~f(x) = i~ f(x) (248) Now making use of above definitions and orthogonality dx dx of L matrix we can find the following sum rules [31] In other words, commutator of any differentiable function 3N−6 N of variable x with x component of quantum-mechanical X X 1 −1 Cα Cβ = δ − l l − aαγ I0 aδβ momentum operator is proportional to derivative of func- kn ln αβ βik αil 4 k γδ l n=1 i=1 tion f with respect to x. This relation simplifies many (243) commutator operations.

ˆ ˆ h ˆ i ˆ 3N−6 N ... = i~ cos χPθ + i~ sin χ csc θPφ + cot θ cos χ, Pχ Pχ = X αβ γδ X 0 0 0 0 a a = 4 δαβδγδmiriri − δαβmiriγ riδ − k k ˆ ˆ ˆ ˆ k=1 i=1 = i~ cos χPθ + sin χ csc θPφ − cot θ sin χPχ = i~Jy 0 0 0 0 − δγδmiriαriβ + δαγ miriβriδ− (249)  X 0 0 0 0 0−1 and similarly for other permutations. − αηγξθmiriβrimjrjδrjξ I ηθ  (244) j,θ,η,,ξ Above relations are historically named anomalous com- 3N−6 mutation relations to be distinguished from commutation X 1 X X ξα aβγ = a − aγ − relations for space-ﬁxed angular momentum operators, kl l 2 αβγ k αβ k l=1 where there’s no ” − ” in front of right hand side. Now X −1 lets turn our attention into vibrational angular momen- − m r0 r0 I0 aξα (245) βδ i iδ iγ γ k tum operators: i,,ξ,δ

For future purposes it is essential to derive commutation Theorem 6 Angular momentum associated with vibra- relations for angular momentum operator. Thereby we tions of atoms in molecules with respect to body-fixed claim that: frame (Eckart frame)obeys following commutation rela- Theorem 5 Angular momentum associated with rotations tions of body-fixed coordinates frame with respect to hˆ ˆ i ˆ space-fixed frame obeys following commutation relations jα, jβ = i~αβγ jγ (250) h i Jˆ , Jˆ = −i Jˆ (246) α β ~ αβγ γ Proof 5 Lets expand left hand side from definition of vi- 25 brational angular momentum As introduced in 170 the µαβ is defined as reciprocal of 0 Iαβ h i 3N−6 h i ˆ ˆ X α β ˆ ˆ ˆ ˆ jα, jβ = CklCmn QkPl, QmPn = 0−1 µαβ = I αβ (255) k,l,m,n=1 3N−6 where X α β ˆ ˆ ˆ ˆ = i~ CklCmn −QkδlmPn + QmδknPl = k,l,m,n=1 0 X α β I = Iαβ − C C QkQl (256) αβ kl lm X α β α β ˆ ˆ klm = i~ CklCmk − CmkCkl QmPl = k,l,m here terms linear in normal coordinates are the same in I αβ X α β α β ˆ ˆ and I0, therefore interaction coefficients a are the same = i~ CkmClm − CkmClm QkPl (251) k k,l,m for both. Now we want to express our refined moment of inertia by normal coordinates where the last equality comes from relabelling summation X X X X 0 0 indices: m → k, k → m, l → l with subsequent transpo- Iαβ = αγηβδηmiriγ riδ = αγηβδηmiriγ riδ+ sition of m, l on Coriolis coefficients. Further use of the γ,δ,η i γ,δη i sum rules 243-245 yields: X αβ X X X 0 + ak Qk + αγηβδηlηislγinQnQs = Iαβ+  k γ,δ,η i s,n h i 1 ˆ ˆ X X αγ 0−1 δβ X αβ X X X jα, jβ = i δαβδkl − lβiklαil + a I a + a Q + (δ δ − δ δ ) l l Q Q = ~  4 k γδ l k k αβ γη αη βγ ηis γin n s k,l i,γ,δ k γ,δ,η i s,n   0 X αβ X X X X 1 −1 = Iαβ + ak Qk + δαβ lγislγimQnQs− − δ δ − l l + aβγ I0 aδα Qˆ Pˆ =  βα kl αik βil 4 k γδ l  k l k γ i s,n i,γ,δ X X X 2 − lαislβinQnQs − δαβQk+ 3N−6 N X X ˆ ˆ i s,n k = i~ (lαiklβil − lβiklαil) QkPl− X X 1 X X αγ 0−1 δβ k,l=1 i=1 + l l Q Q + a I a Q Q = αil βik l k 4 k γδ l k l i k,l k,l γδ i~ X αγ 0−1 δβ ˆ ˆ ˆ ˆ − ak I γδ al QkPl − QlPk = ... 4 0 X αβ 1 X X αγ 0−1 δβ k,l,γ,δ = I + a Q + a I a Q Q αβ k k 4 k γδ l k l (252) k k,l γδ (257) in last transformation we changed summation indices k → l, γ → δ. Now contracting first term in last line by By taking a closer look at above expression one may infer noticing totally antisymmetric term, and applying Cori- that there exist a simpler factorized form of the last sum: olis coefficient definition: 0 X 00 0−1 00 0 00 0−1 00 Iαβ = Iαγ I Iδβ i.e. I = I I I 3N−6 N γδ X X ˆ ˆ γδ i~ (lαiklβil − lβiklαil) QkPl = (258) k,l=1 i=1 00 0 1 P αβ 0 where Iαβ = Iαβ + 2 k ak Qk. Note that I as a func- 3N−6 N tion of normal coordinates is much simpler than I, al- X X ˆ ˆ = i~ γαβlαiklβilQkPl = though physical significance of I0 is harder to visualize. k,l=1 i=1 That’s the reason for introducing I0 and µ subsequently. X α ˆ ˆ ˆ In order to rearrange Podolsky Hamiltonian into simpler = CklQkPl = i~αβγ jγ (253) form it will be necessary to calculate two types of com- k,l h i h i mutators, namely ˆjα, µ , ˆjα, µαβ , hence lets start with matrix form of µ operator, for easier manipulation: i X −1 ... = i ˆj − ~ aαγ I0 aδβ Qˆ Pˆ − Qˆ Pˆ ~ αβγ γ k γδ l k l l k h i h i h i h i 4 ˆ ˆ 00−1 0 00−1 ˆ 00−1 0 00−1 00−1 0 ˆ 00−1 k,l,γ,δ jα, µ = jα,I I I = jα,I I I + I I jα,I (254) (259) using the fact that I00I00−1 = 1 results in equality: but last expression is a product of totally symmetric (in- h 00i 00−1 00 h 00−1i ˆjα,I I + I ˆjα,I = 0 which when inserted teraction coefficients are and moment of inertia are symmetric tensors) and totally antisymmetric factor, with into above expression yields in: summation over all indices, summing up eventually to h i h i h i ˆ 00−1 ˆ 00 ˆ 00 00−1 zero. jα, µ = −I jα,I µ − µ jα,I I (260) 26

By considering matrix elements of µ we can continue calculations in following fashion: 00 1 1 00− 3 h 00i ... = µ Mˆ Mˆ + µ I 2 ˆj , − I 2 ˆj ,I − ij i j ij i 2 j h i h i h i P ˆj , µ = − P I00−1 ˆj ,I00 µ − µ ˆj ,I00 I00−1 = 0 1 h i h i 1 h h ii α α αβ α,γ,δ αγ α γδ δβ αγ α γδ δβ µ I00−2 ˆj ,I00 ˆj ,I00 = µ Mˆ Mˆ − µ I00−1 ˆj , ˆj ,I00 − (261) 2 ij i j ij i j 2 ij i j This result is quite surprising, as the vibrational angu- 1 00 1 h 00− 3 i h 00i 1 00−2 h 00i h 00i − µ I 2 ˆj ,I 2 ˆj ,I − µ I ˆj ,I ˆj ,I = lar momentum is a diﬀerential operator, which usually 2 ij i j 2 ij i j do not commute with functions of coordinates. Hav- 1 h h ii 3 h i h i = µ Mˆ Mˆ − µ I00−1 ˆj , ˆj ,I00 + µ I00−2 ˆj ,I00 ˆj ,I00 − ing that done, attention can be turned into the com- ij i j 2 ij i j 4 ij i j mutator of the determinant µ. Lets note that since 1 00−2 h 00i h 00i 2 − µ I ˆj ,I ˆj ,I = µ Mˆ Mˆ − µ = det I00−1 det I0 is expressed by the determinant 2 ij i j ij i j h i 00 ˆ 00 1 00−1 h h 00ii 1 00−2 h 00i h 00i of I we can consider only jα,I . After some calcula- − µ I ˆj , ˆj ,I + µ I ˆj ,I ˆj ,I 2 ij i j 4 ij i j tions we obtain: (265) Now consider the vibrational kinetic energy term in 220:

h 00i X 00 0 0−1 00 00−1 1 1 1 1 1 ˆ 4 ˆ − 2 ˆ 4 4 ˆ − 4 ˆ jα,I = i~ I γηKηδ I ξ Iξα I γδ µ Pkµ Pkµ = µ Pkµ Pk+ γ,η,,δ,ξ 1 1 h 1 i 4 ˆ − 2 ˆ 4 ˆ2 (262) + µ Pkµ Pk, µ = Pk +

At present stage we can attempt to write down the Podol- 1 h 1 i 1 1 h 1 i 4 ˆ − 4 ˆ 4 ˆ − 2 ˆ 4 sky Hamiltonian in a new form. Firstly lets deal with + µ Pk, µ Pk + µ Pkµ Pk, µ = ... (266) the Hamiltonian in 220 investigating the two terms separately: Evaluating separately appropriate commutators:

00 1 h 1 i 1 h 1 i 1 2 − 0− 4 00 0− 4 ∂I Pˆ , µ 4 = I Pˆ ,I 2 = −i I 1 − 1 1 1 − 1 1 k k ~ µ 4 Mˆ iµijµ 2 Mˆ jµ 4 = µijµ 4 Mˆ iµ 2 Mˆ jµ 4 = ∂Qk 00− 1 1 − 1 1 − 1 h 1 i h 1 i 1 h 1 i 1 ∂I 2 h 1 i = µ µ 4 Mˆ µ 4 Mˆ + µ µ 4 Mˆ µ 2 Mˆ , µ 4 = ˆ 4 0 4 ˆ 00− 2 0 4 ˆ 4 ij i j ij i j Pk, µ = I Pk,I = −i~ I Pk, µ = ∂Qk 1 h − 1 i − 1 h 1 i = µ Mˆ Mˆ + µ µ 4 Mˆ , µ 4 Mˆ + µ µ 4 Mˆ Mˆ , µ 4 + 1 00 ij i j ij i j ij i j 0− 2 ∂I = −i~ I (267) 1 h − 1 i h 1 i 1 h − 1 i ∂Qk + µijµ 4 Mˆ i, µ 2 Mˆ j, µ 4 = µijMˆ iMˆ j + µijµ 4 Mˆ i, µ 4 Mˆ j+ hence, − 1 h h 1 ii 1 h − 1 i h 1 i µijµ 4 Mˆ i, Mˆ j, µ 4 + µijµ 4 Mˆ i, µ 2 Mˆ j, µ 4 =

00− 1 h 00 1 i 00 1 h 00− 1 i 00 1 = µijMˆ iMˆ j + µijI 2 ˆji,I 2 Mˆ j + µijI 2 ˆjj,I 2 Mˆ i+ 1 ∂I 2 ∂ 1 h 1 i ˆ2 00− 2 − 4 ˆ ˆ 4 ... = Pk − i~I −i~ + µ Pk Pk, µ + 1 h h 1 ii 1 h i h 1 i ∂Qk ∂Qk 00 2 ˆ ˆ 00− 2 00− 2 ˆ 00 ˆ 00− 2 + µijµijI jj, jj,I + µijµijI ji,I jj,I = ... 00 1 1 h 1 i h 1 i 1 ∂I 2 ∂ 4 ˆ − 2 ˆ 4 ˆ2 2 00− 2 (263) + µ Pk, µ Pk, µ = Pk − ~ I − ∂Qk ∂Qk 2 00− 1 00− 1 00 00− 1 1 ∂ I 2 1 ∂I 2 ∂ 1 ∂I ∂I 2 2 00 2 2 00 2 2 00− 2 − ~ I 2 − ~ I − ~ I = where we’ve made use of commutation relation for µ ten- ∂Qk ∂Qk ∂Qk ∂Qk ∂Qk sor elements derived earlier, as well as of the expression 2 1 ∂2I00 1 ∂I00 for determinant of µ matrix. Lets treat ˆj as diﬀerential ˆ2 2 00−1 2 00−2 i = Pk + ~ I 2 − ~ I (268) operator (as it is indeed), then: 2 ∂Qk 4 ∂Qk Following Watson we denote respective terms in the residual Hamiltonian as U1,U2,U3,U4. h 00 1 i 1 00− 1 h 00i ˆj ,I 2 = I 2 ˆj ,I (264) First lets evaluate U : i 2 i 1 27

1 X h i h i 1 X U = I00−2µ ˆj ,I00 ˆj ,I00 = I0−1 + 1 8 αβ α β 8 αβ αβ αβ 00 X 0 0−1 00 00−1 X 0 0−1 00 00−1 + i~I γηKηδ I ξ (I )ξα I γδ + i~ θφψKφω I ψν (I )νβ I θω = γ,η,,ξ,δ θ,φ,ψ,ν,ω 2 X X X − ~ I00−1 I0 I00−1 K0 K0 I0−1 I00 I00−1 I0−1 I00 I00−1 = ... 8 αλ λκ κβ γη θφψ ηδ φω φξ ξα γδ ψν νβ θω α,β,λ,κ γ,η,,ξ,δ θ,φ,ψ,ν,ω (269)

and making use of the Kronecker deltas δξλ, δκν , δνφ we ﬁnally obtain

2 X X ... = −~ K0 K0 I00−1 I0−1 I00−1 8 γη θφψ ηδ φω γδ ψφ θω γ,η,,δ θ,φ,ψ,ω (270)

To simplify second term U2 lets ﬁrst compute following inner expression:

h i X ˆ 00 X 00−1 0 00−1 X 00 0 0−1 00 00−1 µαβ jα,I = I αγ Iγδ I δβ · i~I ηνφKνψ I φω Iωβ I ηψ = β β,γ,δ η,ν,φ,ψ,ω 00 X 00−1 00−1 0 00 X 00−1 0 X 00−1 = i~I I αγ I ηψ Kνψηνγ = i~I I ηψ Kνψ I αγ ηνγ = ... (271) γ,δ,η,ν,ψ δ,η,ν,ψ γ

using identity: in the meantime lets expand above commutator

X 00−1 X 00 00−1 00−1 I αγ ηνγ = αθµ (I )ηθ I γµ I γ θ,µ  h i X 1 (272) ˆj ,I00 = i I00 − I00 + α νµ ~  αν µ 2 ανµ we ﬁnd: ,η,ξ X 00 00−1 0 00 00 0 0−1 00 .. = i~I I Kνψ (I )ηθ (I )νµ αθµ = ηψ + νηKηµ I ξξ Iξα (275) δ,η,ν,ψ,θ,µ X 0 00 = i~ Kνψ (I )νµ αψµ ν,ψ,µ hence, (273)

With such preparation we’re ready to evaluate U2 expres- 2 X X ... = ~ I00−1 K0 I00 − . sion: 4 αψµ αν ψν µ α,β,ν,ψ µ,φ,η,ξ 1 1 X 00−1 h h 00ii 0 00 0 0 0−1 00 U = − I ˆj , µ ˆj ,I = − αψµανµKψν I + αψµνηKψν Kηµ I Iξα 2 4 α αβ β 2 φξ α,β (276)   1 X X = − I00−1 I00−1 ˆj , i K0 (I00) = 4  α ~ νψ νµ αψµ α,β ν,ψ,µ after using the identity: i X h i − ~I00−1 K0 ˆj ,I00 = ... (274) 4 αψµ ψν α νµ α,β,ν,ψ,µ αβγ αηφ = δβηδγφ − δβφδγη (277) 28

inserting into original expression, and after short manipulations we get 2 ~ 00−1 X 0 00 0 00 1 0 00 ... = I Kνν I − KµIµ − Kνν I + 4 2 2 X ,µ,ν,η,ψ,ν,ξ U = ~ I00−1 I00 2K0 − K0 K0 I0−1 4 4 αλ αλ λβδ θξ β δξ αθ 1 α,β,θ,ξ,,λ K0 I00 + K0 K0 I0−1 I00 = 2 µµ αψµ νη ψν ηµ φξ ξα (285) 2 ~ 00−1 X 0 0 0 0−1 00 I −K + K K I I Upon adding U1 and U3 we notice that last term in U3 4 µ αψµ νη ψν ηµ µ µ ,µ,ν,η,ψ,ν cancels U1, hence

(278) 2 ~ X X X 00−1 00−1 U1 + U3 = − I I × In order to calculate the third term we will need a general 8 αβ γδ k α,β,γ,δ ,η,ξ,θ lemma regarding differentiation of determinants: 0 0 0 0 × δαβδγδK − δαβKγδ − δγδKαβ + δαγ Kβδ Theorem 7 For a given reversible matrix A containing (286) elements remaining functions of x the following equality holds: and remaining terms sum up to: ∂ det A −1 ∂A 2 X = det (A) T r A (279) U + U = −~ I00−1 K0 I00 − 2K0 I00 = ∂x ∂x 2 4 4 αλ αλ αλ αλ α,λ (287) Recalling: 2 ~ 00−1 X 0 00 I KαλIαλ 2 4 2 00 α,λ ~ 00−2 X ∂I U3 = − I (280) 8 ∂Qk k making use of relation we may use the lemma to find that: 1 X I00 = I00 I00−1I00−1 αδ 2 αβγ δξ β ξγ (288) 00 ∂I00 γ,,ξ,β ∂I 00 X 00−1 αβ 1 X 00−1 αβ = I I αβ = I αβ ak ∂Qk ∂Qk 2 αβ αβ we find (281) 2 X U + U = −~ K0 I00−1I00−1 due to the introduced definition of interaction coefficient 2 4 8 αλ αβγ λξ β ξγ (289) α,λ,γ,,ξ,β 240, and the fact that I00 is linear in normal coordinates, 00 00 ∂Iαβ ∂Iαβ thus ∂Q = ∂Q . Coming back to U3, k k 0 2 ~ X 00−1 00−1 0 0 U = − I I δαβδγδK − δαβKγδ− 2 X X 8 αβ γδ U = −~ I00−2I002 I00−1 I00−1 aαβaγδ = α,β,γ,δ, 3 8 αβ γδ k k k α,β,γ,δ  X 2 0 0 0 00−1 00−1 X X X −δγδKαβ + δαγ Kβδ − Kαλαβγ λξIβ Iξγ  −~ I00−1 I00−1 δ δ K0 − 8 αβ γδ αβ γδ λ,ξ k α,β,γ,δ ,η,ξ,θ (290) −δ K0 − δ K0 + δ K0 − K0 K0 I0−1 αβ γδ γδ αβ αγ βδ αη γξθ β δξ ηθ changing indices in last sum above → α, ξ → δ and (282) utilizing fact that I00 is symmetric matrix we have: where we’ve made use of sum rule for interaction coef- 2 ~ X 00−1 00−1 0 0 ficients 244. Mostly in similar way we can evaluate the U = − I I δαβδγδK − δαβK − 8 αβ γδ γδ last term U4 α,β,γ,δ,,λ 0 0 0 00−1 00−1 2 2 00 − δγδKαβ + δαγ KβδKλβδλαγ I I ~ 00−1 X ∂ I β ξγ U4 = − I (283) 4 ∂Q2 (291) k k remembering determinant definition I00 = Finally it is required to use conversion formula between 1 P 00 00 00 Levi-Civita tensor product and Kronecker Deltas: 6 α,β,γ,δ,,ξ αβγ δξIαδIβIγξ

2 00 δλ δα δγ ∂ I 1 X 00 αβ γδ = I a a βδλαγ = δβλ δβα δβγ (292) ∂Q2 4 ηαγ λβδ ηλ k (284) k α,β,γ,δ,λ,η δδλ δδα δδγ 29 by cancelling emerging identical terms we continue to This is called the asymmetric top Schr¨odingerequation,

2 since all rotational constants take different values. The ~ X 00−1 00−1 0 simplest molecular asymmetric top model may be applied U = − I I δαδδβγ K + 8 αβ γδ + α,β,γ,δ,,λ to H2D cation, or simply water. There are three general 0 0 0 cases of top systems: +δαγ Kβδ − δβγ Kαδ − δαδKγβ (293) • The symmetric top molecule, in which two rota- rearranging some terms tional constants are equal, dividing into two groups: 2 ~ X 00−1 00−1 0 0 Prolate symmetric top when Ae > U = − I I δαδ δβγ K − Kγβ 8 αβ δγ Be = Ce (e.g. CH3Cl, trans − α,β,γ,δ,,λ Ni(H2O)4Cl2,CO2)(linear 0 0 +Kβδδαγ − Kαδδβγ (294) subtype),propyne (cigar or rugby shaped molecules) Now lets recall the definition of equilibrium moment of 0 0 0 0 00 Oblate symmetric top when Ae = Be > Ce inertia Iαβ = δαβKγγ − Kαβ As both Kαβ and I are − symmetric tensors and we perform summation over all (e.g. SO3,BF3,NO3 ,C6H6,NH3) (disk indices, hence last two term in brackets give antisymmet- or doughnut shaped molecules) ric contribution while I00 give symmetric contribution to • The spherical top molecule when Ae = Be = Ce the product, yielding vanishing expression. (e.g. SF6,CH4, SiH4) 2 2 ~ X 00−1 0 00−1 ~ U = − I I I = − T rµ • The asymmetric top molecule when Ae 6= Be 6= 8 δβ βγ γδ 8 (295) β,γ,δ Ce 6= Ae (e.g. H2O,NO2) which proves the result obtained by Watson. Eckart- It should be noted that symmetric top molecules must Watson Hamiltonian approach have been successfully ap- contain three-fold or higher symmetry axis, as an element plied to potassium cyanide rovibrational problem by Ten- of the point group. Molecules possessing at most two-fold nyson and Sutcliffe [33]. However, there are numerous ex- axis are asymmetric tops (most of larger molecules are). amples in the literature where other forms of the molecular Hamiltonian provide better convergence of energies, by proper treatment of large amplitude motions. A. Symmetric top molecule

Common convention for symmetric tops is to choose VI. RIGID ROTOR APPROXIMATION space-fixed z axis as the one with the distinct rotational constant Ae. Prolate symmetric top yields in following In the last section of this paper we present solution SE: and a short discussion of the model of rigid rotor in 3D. −2 A Jˆ2 + B (Jˆ2 + Jˆ2) Φ (θ, φ, χ) = E Φ (θ, φ, χ) Resulting eigenstates are ubiquitous in nuclear motion ~ e z e b c rot rot rot theory, thereby this model should be deemed as impor- (299) tant. In section II we ended with an approximate form Angular momentum operators defined in the previous for rotational Hamiltonian: section, together with the use of chain rule results in: 3 1 X Hˆ = µe Jˆ2 (296) 1 ∂ ∂ 1 ∂2 rot 2 αα α sin θ + + α=1 sin θ ∂θ ∂θ sin2 θ ∂φ2 2 2 Hence the corresponding rigid rotor Schrödingerequation 2 Ae ∂ cos θ ∂ Erot + cot θ + 2 − 2 2 + Φrot(θ, φ, χ) = 0 reads: Be ∂χ sin θ ∂φ∂χ Be (300) 3 1 X µe Jˆ2Φ (θ, φ, χ) = E Φ (θ, φ, χ) (297) As Euler angles χ, φ occur only in derivatives, they are 2 αα α rot rot rot α=1 cyclic coordinates hence, we can make an anzatz:

imφ ikχ For spectroscopic purposes it will be convenient to Φrot(θ, φ, χ) = Θ(θ)e e (301) rewrite the above equation into wavenumber units, because energy scale of rotational transitions is of order Cyclic boundary conditions Φrot(θ, φ, χ + 2π) = of tens/hundreds wavenumbers. Introducing rotational Φrot(θ, φ + 2π, χ) = Φrot(θ, φ, χ) imply m, k ∈ Z. Af- 2 e ~ µaa ter some manipulations we obtain the following equation constants, e.g. Ae = , with convention that Ae ≥ 2hc for Θ(θ): Be ≥ Ce we find a new form of SE: h 2 2 i −2 ˆ2 ˆ2 ˆ2 1 d d m −2mk cos θ+k ~ AeJa + BeJb + CeJc Φrot(θ, φ, χ) = ErotΦrot(θ, φ, χ) sin θ dθ sin θ dθ + ∆ − sin2 θ Θ(θ) = 0 (298) (302) 30 where The a0 coefficient is chosen so that the rotational wavefunction is normalised. In order to Θ(θ) be a proper E − (A − B )k2 ∆ = rot E e (303) representation of wavevector it must be finite, thus the Be series expansion 310 must truncate at finite term labeled This is already one dimensional second-order linear dif- by nmax, then ferential equation, which can be solved analytically by writing anmax+1 = 0 (314) |k−m| |k+m| Θ(θ) = x 2 (1 − x) 2 F (x) (304) giving the condition, which provides eigenvalues of rigid (1−cos θ) rotor SE (cf.299) where x = 2 . Note that also |k−m| |k+m| θ θ 2 θ Θ(θ) = sin cos F (sin ) (305) βnmax + nmax(nmax − 1) − γ = 0 (315) 2 2 2 Substituting eqs. 303,308,309 and making the abbrevia- After some lengthy differentiation and algebra eq. 302 |k+m|+|k−m| reduces to tion J = nmax + 2 gives the eigenenergies as functions of J and k: d2F dF x(1 − x) + (α − βx) + γF = 0 (306) dx2 dx E = B J(J + 1) + (A − B )k2 (316) with rot e e e α = 1 + |k − m| (307) We can easily figure out that: J = 0, 1, 2, ... k = 0, ±1, ±2, ..., ±J and β = α + 1 + |k + m| (308) m = 0, ±1, ±2, ..., ±J (317) and It turns out that F (x) solution is hypergeometric func- β(β − 2) γ = ∆ − (309) tion [34], which enables us writing rotational wavefunc- 4 tion in compact form: Following Frobenius method we write expansion form of our solution: |k−m| |k+m| 2 2 1 1 imφ ikχ ΦJkm(θ, φ, χ) = NJkmx (1 − x) F ( 2 β − J − 1, 2 β + J; α, x)e e ∞ X (318) F (x) = a xn (310) n The normalization constant NJkm is defined by the con- n=0 dition: First and second derivative read: ∞ Z 2π Z 2π Z π dF X n−1 ∗ = annx ΦJkmΦJkm sin θdθdφdχ = 1 (319) dx 0 0 0 n=1 ∞ d2F X and a choice of phase factor. In fact symmetric top rota- = a n(n − 1)xn−2 (311) dx2 n tional wavefunction have many different representations. n=2 One most popular is given below: Substituting into 306: r 2J + 1 Φ (θ, φ, χ) = D(J)(θ, φ, χ) (320) γa0 + αa1 + [(γ − β) a1 + 2 (1 + α)) a2] x Jkm 8π2 km ∞ X + [(γ − n(n − 1) − βn) a + (n + 1)(n + α)a ] xn = 0 (J) n n+1 where Dkm(θ, φ, χ) is Wigner D-matrix[1]. Sometimes n=2 abbreviation of type ΦJkm(θ, φ, χ) ≡ |J, k, mi is made, (312) but it should be treated with caution, as ket vector rep- Due to linear independence of basis monomials we can resent state vector from Hilbert space and ΦJkm states equate consequent coefficients to 0 for spectral representation of this state vector (hence is representation-dependent) and belongs to L2(C3, d3) γ space of square-integrable functions, which in fact is iso- a1 = a0 2 morphic with the original Hilbert space. The explicit form β − γ of the wavefunction Φ (θ, φ, χ) is a = a Jkm 2 1 2 + 2α h (cos 1 θ)2J+k−m−2σ (− sin 1 θ)m−k+2σ i −γ + βn + n(n − 1) P σ 2 2 imφ ikχ N σ(−1) σ!(J−m−σ)!(m−k+σ)!(J+k−σ)! e e an+1 = an (313) (n + 1)(n + α) (321) 31

ˆ± ˆ ˆ± where, to JmJz|J, k, mi + Jm|J, k, mi. Now utilizing eigenequa- ˆ 1 tion for Jz we find (J + m)!(J − m)!(J + k)!(J − k)!(2J + 1) 2 N = ˆ ˆ± ˆ± 8π2 JzJm|J, k, mi = ~(k ∓ 1)Jm|J, k, mi (327) (322) laddered function to be also eigenfunction of Jˆ but to and σ runs from 0 or (k − m), whichever is the larger, z eigenvalue shifted by 1. That’s the general property of up to (J − m) or (J + k), whichever is smaller [1]. operators obeying 326 commutation relations. The in- Note that the symmetric top Hamiltonian commutes with verse in sign here is due to anomal molecule fixed com- Jˆ2, Jˆ , Jˆ , therefore they have common eigenfunctions. ρ3 z mutation relations 330. Therefore we presume that lad- Apparently different situation emerges in case of asym- dered function should be proportional to its ’neighbour’ metric top (all three rotational constants are different). ˆ2 ˆ with shifted k value: Here the Hamiltonian 296 commutes with J , Jρ3 but not ˆ ˆ± with Jz: Jm|J, k, mi = NJ,k|J, k ∓ 1, mi (328) h i h i Requiring orthonormality for both hand sides states it’s H,ˆ Jˆz = Jˆx, Jˆy (Be − Ae) (323) + straightforward to write down final relation: where molecule fixed angular momentum operators sat- Jˆ±|J, k, mi = pJ(J + 1) − k(k ∓ 1)|J, k∓1, mi (329) isfy commutation relations: m ~ where we made use of equality: Jˆ∓Jˆ± = Jˆ2 −Jˆ (Jˆ ∓ ). h ˆ ˆ i ˆ m m z z ~ Ji, Jj = −i~ijkJk (324) In order to obtain general symmetric top wavefunction from some known and easy to obtain generating function, Therefore Hˆ will have common eigenbasis with Jˆ2 and it’s needed to find a way to ladder m values (see ref. [16]), ˆ ˆ Jρ3 but not with Jz. In order to obtain eigenfunctions, i.e. space fixed z-axis projection of angular momentum a good starting point would be to express Hamilton ma- quantum number. This goal may be acheived by intro- trix in the symmetric top eigenbasis. Such approach pro- ducing space fixed angular momentum ladder operators vides final states representation as linear combination of in similar way to molecule fixed case. Now angular mo- Wigner functions with different k values: mentum operators satisfy normal commutation relations: h ˆ ˆ i ˆ     Jρi, Jρj = i~ijkJρk (330) Ψ1 |J, −J, mi . .  .  = C  .  . ˆ± ˆ  .  J,m  .  and analogical definition of ladder operators Js := Jρ1 ± Ψ2J+1 |J, J, mi iJˆρ2 may be applied to retrieve result similar to the previous case: where C diagonalizes Hˆ block for a given (J, m). At this ˆ ˆ± ˆ± stage it is convenient to write explicitly Hamilton matrix Jρ3Js |J, k, mi = ~(m ± 1)Js |J, k, mi (331) elements in symmetric top basis. To do this, lets rewrite and Hamiltonian as below ˆ± p Js |J, k, mi = ~ J(J + 1) − m(m ± 1)|J, k, m ± 1i −2 1 2 1 2 Hˆ = (Be + Ce) Jˆ + Ae − (Be + Ce) Jˆ + (332) ~ 2 2 z allowing to derive the following important relation: 1 + 2 − 2 + (Be − Ce) (Jˆ ) + (Jˆ ) (325) k m 4 m m ˆ∓ ˆ± |J, ±|k|, ±|m| >= N Jm Js |J, 0, 0i (333) where we introduced molecule fixed ladder operators: ˆ± ˆ ˆ Normalisation factor is given by Jm := Jx ± iJy. The purpose of abbreviating this new operators lays within their marvellous abilities to ladder s −(|k|+|m|) (J − |m|)!(J − |k|)! and lower k quantum number in our basis. And this fea- N = ~ (334) ture is due to special commutation relations fulfilled by (J + |m|)!(J + |k|)! ladder operators: Note the analogy to the second quantization procedures. h i Here we obtained a general state as a result of acting Jˆ , Jˆ± = ∓ Jˆ± (326) z m ~ m with creation operators on the ’vacuum’ state of rotations space. These operators were derived from position repre- ˆ ˆ Because |J, k, mi are eigenfunctions of Jz: Jz|J, k, mi = sentation differential operators, giving rare link between ˆ± ~k|J, k, mi the laddered function Jm|J, k, mi is also traditional quantum mechanics and quantum field the- ˆ eigenfunction of Jz to the ~(k ∓ 1) eigenvalue. Firstly ory formalism. Knowing the above symmetric top wave- lets act on standard basis element with a product of op- functions, it’s easy to write down matrix elements of a ˆ ˆ± erators: JzJm|J, k, mi. This is equal, on account of 326 particular angular momentum associated operators: 32

which is 5×5 lets build similar symmetry adapted states:

ˆ2 2 1 < J, k, m|J |J, k, m >= ~ J(J + 1) |2, 2, 0,E±i = √ (|2, 2, 0i ± |2, −2, 0i) 2 < J, k, m|Jˆz|J, k, m >= k ~ + ˆ |2, 0, 0,E i = |2, 0, 0i (339) < J, k, m|Jρ3|J, k, m >= m~ ± 1 ˆ± p |2, 1, 0,O i = √ (|2, 1, 0i ± |2, −1, 0i) (340) < J, k, m ± 1|Js |J, k, m >= ~ J(J + 1) − m(m ± 1) 2 < J, k ∓ 1, m|Jˆ±|J, k, m >= pJ(J + 1) − k(k ∓ 1) m ~ The block factorizes into subblocks E+,E−,O+,O−. For (335) + J+2 J even (as in our case) the E block has dimension 2 J − At this stage we’re ready to build the Hamilton matrix for while the other three 2 , and for J odd the E block 2 J−1 J+1 the asymmetric top, which a J(2J +1) dimensional sym- has dimension 2 while the other three 2 . Hence in metric matrix. Nevertheless on account of relations from our case E+ block is two dimensional and the remaining eq.335 all elements between states with different J van- three are one dimensional, providing eigenvalues: ish, making the matrix block-diagonal. Because eigenen- − − ergies of Hˆ don’t depend on m (see 325) each block with h2, 2, 0,E |Hˆ |2, 2, 0,E i = 4Ae + Be + Ce fixed m will yield with identical eigenvalues, hence we h2, 1, 0,O+|Hˆ |2, 1, 0,O+i = A + 4B + C (341) can restrict to only one of them, e.g. m = 0. Every e e e − ˆ − m replica is 2J + 1 dimensional with states labelled by h2, 1, 0,O |H|2, 1, 0,O i = Ae + Be + 4Ce k = −J, −J +1, ..., J −1,J. For a given J there are 2J +1 identical blocks because m = −J, −J +1, ..., J −1,J. The The 2 × 2 block whole matrix may be theoretically analytically diagonal- √ ized by diagonalising subsequent blocks with rising J. √3(Be + Ce) 3(Be − Ce) And so for J = 0, m = 0 and k = 0, hence it’s already di- 3(Be − Ce) 4Ae + Be + Ce agonalized with eigenvalue equal to 0, which means that can be easily diagonalized to give appropriate eigenvec- for asymmetric top we can find states with no rotational tors and eigenvalues, what we leave for the reader as an motion at all! The anticipated wavefunction is a constant exercise. The important fact is that basis states with q 1 |0, 0, 0i = 8π2 . For J = 1 we have three dimensional even/odd k couple only to states with even/odd k and block with basis functions |1, −1, 0i, |1, 0, 0i, |1, 1, 0i. In states with ± parity couple to states with ± parity. order to diagonalize this block we can take advantage of one of the symmetries of Hamiltonian, namely the parity. Due to Hamiltonian invariance under parity operation ACKNOWLEDGEMENTS the states must transform according to irreducible representation of parity group, which in fact is Ci with two I would like to thank the homeless stranger from representations: A and B. For rotations this means that King Cross station who kindly spoke to me in follow- the proper wavefunction must take into account clock- ing words: Before you reach for all those superb whiskies wise and anti-clockwise rotations of a systems. Equiva- and liqueurs for the bold and the beautiful, try to deeply lently states with k and −k must be allowed. Symmetry understand and appreciate the taste of simple beer, unless adapted states are therefore in-phase and out-of-phase you want to end up as a useless drunker like me. linear combinations of the mentioned two: 1 |1, 1, 0, ±i = √ (|1, 1, 0i ± |1, −1, 0i) (336) 2 By lucky coincidence this unitary transformation already diagonalizes the J = 1 block. The corresponding eigenvalues read

h1, 1, 0, +|Hˆ |1, 1, 0, +i = Ae + Be

h1, 1, 0, −|Hˆ |1, 1, 0, −i = Ae + Ce (337)

h1, 0, 0|Hˆ |1, 0, 0i = Be + Ce (338)

Results have proper dimension of cm−1. Note that in present case the |k| no longer labels the states (the states are not eigenstates of Jz), therefore cannot constitute a good quantum number. Instead new index called parity has been introduced. Heading now towards J = 2 block 33

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