A Concise Method for Kinetic Energy Quantisation
aa r X i v : . [ phy s i c s . c h e m - ph ] O c t A Concise Method for Kinetic Energy Quantisation
Yonggang Yang ∗ and Oliver K¨uhn † Institut f¨ur Chemie und Biochemie, Freie Universit¨at Berlin,Takustr. 3, D-14195 Berlin, Germany (Dated: November 15, 2018)
Abstract
We present a straightforward method for obtaining exact classical and quantum molecularHamiltonians in terms of arbitrary coordinates. As compared to other approaches the resultingexpression are rather compact, the physical meaning of each quantity is quite transparent,and in some cases the calculation effort will be greatly reduced. We also investigate systemswith constrains to find the suggested method to be applicable in contrast to most conventionalapproaches to kinetic energy operators which cannot directly be applied to constrained systems.Two examples are discussed in detail, the monohydrated hydroxide anion and the protonatedammonia dimer.
Keywords: Vibrational Dynamics, Kinetic Energy Operator, Quantisation, Strong Hydrogen Bonds ∗ Electronic address: [email protected] † New Address: Institut f¨ur Physik, Universit¨at Rostock, Universit¨atsplatz 3, D-18051 Rostock, Germany,Tel. +493814986950, Fax: +493814986942; Electronic address: [email protected] . INTRODUCTION Theoretical modeling of moelcular reaction dynamics requires to have at hand a Hamil-tonian. For few-atom systems this may involve defining the potential energy surfaces(PES) explicitly in terms of properly chosen coordinates such as bond length and an-gles or distance vectors. In the latter case, for instance, it has been shown that PESin full-dimensionality including proper account of exchange symmetry can be generatedfor systems like the five-atomic H O − [1] or the seven-atomic H O +2 [2]. With increas-ing dimensionality this approach is no longer feasible due to the exponential scaling ofthe computational effort for explicit calculation of the PES on a numerical grid. Here,reaction surface concepts have been proven to be a versatile compromise, mapping outthat PES region which is relevant for the considered reaction by combining suitable largeamplitude reaction coordinates with orthogonal harmonic motions[3, 4, 5, 6, 7].Besides the PES the kinetic energy operator (KEO) needs to be known in terms of thechosen coordinate system. In principle, kinetic energy quantisation had been discussed byPodolsky in terms of intricate tensor analysis right after the foundation of the Schr¨odingerequation had been laid [8]. Later studies are to some extent based on the same mathemat-ical techniques [9, 10]. Approximate KEOs using normal mode coordinates arewidely adopted which go back to contributions by Eckart [11], Wilson [12], andWatson [13]. In particular, the Eckart equations [11] enable one to determinean orientation of molecule fixed axes suitable for using normal modes. Thedetails of the derivation of a rovibrational Hamiltonian can be found in Ref.[12]. A significant simplification of this rovibrational Hamiltonian in terms ofnormal coordinates has been achieved by Watson [13, 14]. In fact the WatsonHamiltonian served as a starting point for many later investigations and isimplemented, e.g., in the MULTIMODE program of Bowman, Carter, andcoworkers [15]. In principle normal mode coordinates provide a very compactrepresentation of the PES which is taylored, however, to the stationary pointfor which they have been defined. For large amplitude motions away fromthis stationary point the above mentioned reaction surface approach or theuse of general curvilinear coordinates are more suitable. A detailed discus-sion of advantages and disadvantages of various coordinates for rovibrationalHamiltonians has been presented, e.g., in Ref. [16].
In recent years Gatti and coworkers have developed a general scheme for obtaining a2EO using vector parameterization, i.e., the non-trivial coordinates are expressed with N − N atoms [17, 18, 19]. This scheme is quite successful andconvenient for a full dimensional description [20]. However, often one needs a reduceddimensional description leading to a problem with imposed constraints [21, 22]. Generallyspeaking in this case the coordinates can not be expressed as components of real vectors.Finally, we should point out that a numerical alternative to the analytical determinationof the KEO has been suggested in Ref. [23].In the present work we introduce a simple physical method of kinetic energy quanti-sation which directly quantises the classical Lagrange/Hamilton functions. The formaltheory is presented in Section II. Here it is also shown that our method can be directlyapplied to constrained systems. In Section III the approach is applied to two molecularsystems, that is, H O − which is treated in full dimensionality and N H +7 for which certainconstraints are applied. II. GENERAL THEORY FOR OBTAINING A KEOA. Formal Development
In the following we focus on the most common cases where the kinetic energy includespurely quadratic terms and the potential energy does not depend on the velocity, althougha generalisation would be straightforward. Further we adopt the laboratory referenceframe (LRF). The classical Lagrange/Hamiltonian in terms of arbitrary variables can bewritten as L ( Q , ˙ Q , t ) = T ( ˙ Q ) − V ( Q ) = 12 ˙ Q † M ˙ Q − V ( Q ) H ( Q , P ) = T ( P ) + V ( Q ) = 12 P † M − P + V ( Q ) , (1)where Q and P are single column vectors of the generalised coordinates and correspond-ing conjugate momenta, respectively. The generalised momentum vector is defined as P = ∂L/∂ ˙ Q = M ˙ Q . The generalised mass matrix, M , in general is a function of the co-ordinates. We assume that it is defined such as to be Hermitian and “ † ” means Hermitianconjugate which is equivalent to the transpose in classical mechanics. In order to obtainthe quantum mechanical Hamiltonian operator, the major task is to derive the quantumKEO in coordinate representation.Since the choice of coordinates is quite arbitrary in Eq. (1), we start with Cartesian ones { X , P X } ; in the following all the subscripts “ X ” are associated with Cartesian coordinates.3he operator for each component of the momentum ( ˆ P † X ) j = ( ˆ P X ) j = − i ~ ∂/∂X j isHermitian. For non-Cartesian coordinates we define the momentum operator in the sameway: ˆ P = − i ~ ∂∂ Q . (2)To obtain the quantum KEO in terms of non-Cartesian coordinates, we must exploit thecoordinate transform relations, Q = Q ( X ). The following relations can be derived easily˙ Q = ∂ Q ∂ X ˙ X ˆ P = (cid:18) ∂ X ∂ Q (cid:19) † ˆ P X ∂ Q ∂ X = (cid:18) ∂ X ∂ Q (cid:19) − , (3)where the elements of the transformation matrices are defined as ( ∂ Q /∂ X ) ij = ∂Q i /∂X j and ( ∂ X /∂ Q ) ij = ∂X i /∂Q j . The quantum KEO in terms of non-Cartesian coordinatescan be obtained by the simple coordinate transform from Cartesian coordinatesˆ T = 12 ˆ P † X M − ˆ P X = 12 ˆ P † ∂ Q ∂ X M − (cid:18) ∂ Q ∂ X (cid:19) † ˆ P , (4)where M X is the diagonal Cartesian mass matrix consisting of real mass associated witheach Cartesian coordinate.The momentum operator vector ˆ P can be replaced by the derivative operator ˆ P = − i ~ ∂/∂ Q . However, the expression for the Hermitian conjugate of momentum operator(HCMO) vector ˆ P † needs to be established yet. Since one has the relation between ˆ P and the Cartesian one ˆ P X it is not difficult to directly apply the definition of Hermitianconjugation to Eq. (3) (cid:16) ˆ P † (cid:17) i = (cid:18) ∂ X ∂ Q (cid:19) † ˆ P X ! † i = (cid:18) ˆ P † X ∂ X ∂ Q (cid:19) i = X j (cid:16) ˆ P X (cid:17) j (cid:18) ∂ X ∂ Q (cid:19) ji = X j (cid:18) ∂ X ∂ Q (cid:19) ji (cid:16) ˆ P X (cid:17) j + X j "(cid:16) ˆ P X (cid:17) j , (cid:18) ∂ X ∂ Q (cid:19) ji = (cid:16) ˆ P (cid:17) i + X j "(cid:16) ˆ P X (cid:17) j , (cid:18) ∂ X ∂ Q (cid:19) ji , (5)4here the notation [ ˆ A, ˆ B ] means the commutator of two operators. Eq. (5) shows thenon-Hermiticity of generalised momenta associated with non-Cartesian coordinates. Addi-tional terms appear due to the non-commutability of the momenta and the transformationmatrix. Using the basic commutator [ ˆ X j , ( ˆ P X ) k ] = i ~ δ jk , Eq. (5) can be simplified to (seealso Ref. [24]) ( ˆ P † ) k = ( ˆ P ) k − i ~ X j (cid:18) ∂∂X j ∂X j ∂Q k (cid:19) ◦ . (6)Here, the superscript ◦ means a differential operator inside the bracket can not operateon functions outside, in other words, the result is just normal function of coordinates.Eq. (6) clearly shows the relation between a momentum operator and its Hermitian con-jugate. Apparently, any set of functions of coordinates, { f j ( Q j ) } , can be multiplied fromthe left to the derivative operator in Eq. (2) yielding different schemes of momenta quanti-sation, i.e., ( ˆ P ) j = − i ~ · f j ( Q j ) · ∂/∂Q j constitutes also an acceptable scheme. Since thesemomentum operators are in general non-Hermitian, we cannot simply set f j ( Q j ) = 1. Fordifferent schemes of momenta quantisation, the HCMOs will vary correspondingly to keepthe KEO invariant. But, this enables us to optimise momenta quantisation schemes. Inother words with proper functions one can obtain desired forms of momentum operators,e.g., symmetric forms.With Eq. (4) and Eq. (5) we get the general scheme of constructing KEOs in terms ofarbitrary coordinates. However, the final structure seems to be complicated and actuallyit can be simplified. To this end we compare Eq. (4) and Eq. (1) and take into accountthe invariance of the classical kinetic energy T = 12 ˙ Q † M ˙ Q = 12 ˙ X † M X ˙ X = 12 ˙ Q † (cid:18) ∂ X ∂ Q (cid:19) † M X ∂ X ∂ Q ˙ Q . (7)Since ˙ Q can be any vector we must have the relation M = ( ∂ X /∂ Q ) † M X ∂ X /∂ Q .Calculating the inverse of both sides we immediately see that the complicated centralpart ( ∂ Q /∂ X ) M − ( ∂ Q /∂ X ) † in Eq. (4) is exactly M − in Eq. (1). The final KEO istherefore simplified as ˆ T = 12 ˆ P † M − ˆ P . (8)This means we can directly exploit the result of Eq. (1) and quantise the generalisedmomenta without the knowledge of the Cartesian kinetic energy. All we need are thegeneralised mass matrix M and the definition of coordinates which both appear in Eq. (1).5herefore we can identify a concise and physically transparent scheme to construct theKEO:1. Get the classical kinetic energy and make sure the mass matrix is symmetric, sothat one has the same structure as Eq. (1), T ( ˙ Q ) = ˙ Q † M ˙ Q .2. Exploit Eq. (5) or Eq. (6) to express HCMOs ˆ P † in terms of ˆ P and some functionsof Q .3. Calculate the inverse matrix of M hence the formal quantum KEO reads ˆ T = ˆ P † M − ˆ P .4. Replace ˆ P by − i ~ ∂/∂ Q for obtaining the coordinate representation.The first step is quite familiar and one can choose arbitrary coordinates to get the classicalkinetic energy. Actually we will introduce a useful partition method in Section II C whichwill significantly simplify this issue. The remaining three steps are quite straightforwardto follow. Apparently, the major effort is in the second step, i.e. determining the HCMOsaccording to Eq. (6). However, for the familiar spherical coordinates the HCMOs arewell-known which leads to a great simplification as shown in Section II D.Apart from the KEO the volume element of integration is also of importance. Since westart from Cartesian coordinates and all that we have done is a coordinate transformation,the Euclidean normalisation remains correct: dτ = dτ X = Y i dX i = | Det (cid:18) ∂ X ∂ Q (cid:19) | Y i dQ i . (9) B. Systems with Constraints
Reduced dimensional descriptions are always necessary for large systems. In the follow-ing we will give the general description for systems with constraints. Considering a systemwith some active coordinates Q and some frozen coordinates Q , the full dimensionalcoordinates and corresponding conjugate momenta read Q = Q Q P = P P . (10)6he constraining conditions are given by ˙ Q = 0. Thus we can obtain the constrainedclassical kinetic energy T = 12 ˙ Q † M ˙ Q = 12 (cid:16) ˙ Q † ˙ Q † (cid:17) M M M M ˙ Q ˙ Q = 12 ˙ Q † M ˙ Q , (11)where M ij are corresponding sub-matrices.To get the quantum KEO we have to rewrite the constraint conditions in terms ofmomenta. According to the definition of momenta it is not difficult to find the followingrelation ˙ Q ˙ Q = M M M M − P P = A BC D P P , (12)where A BC D is the inverse of matrix M . Thus one can rewrite the constraint condi-tions as C P + D P = 0 or P = − D − C P . After quantization we get the constraintrelation for the corresponding quantum operators, i.e., ˆ P = − D − C ˆ P . Based on thisobservation we can obtain the quantum KEOˆ T = 12 ˆ P † M − ˆ P = 12 (cid:16) ˆ P † ˆ P † (cid:17) M M M M − ˆ P ˆ P = 12 (cid:16) ˆ P † ( − D − C ˆ P ) † (cid:17) A BC D ˆ P − D − C ˆ P = 12 ˆ P † (cid:0) A − BD − C (cid:1) ˆ P = 12 ˆ P † M − ˆ P . (13)The final result is quite compact. Comparing Eq. (11) with Eq. (13) we observe thatthe same procedure mentioned in the full dimensional case can be followed provided weonly consider the active coordinates and completely ignore the frozen ones from the very7eginning when we generate the classical Lagrangian. This is an attractive point sincewhen we need to deal with constrained systems the present scheme only requires theclassical Lagrangian which can be obtained by traditional methods. C. A General Method for Partitioning the Classical Kinetic Energy
So far we have given the general theory of kinetic energy quantization starting fromthe classical Lagrangian. In the following we will introduce a method for obtaining theclassical kinetic energy with a partitioning technique which will greatly simplify the prob-lem in most cases. In general it is quite convenient to divide a large system into smallsubsystems especially when a subsystems has certain symmetry. If we arbitrarily dividea system into N parts the kinetic energy is a sum of the N subsystems. According to theK¨onig theorem, T = N X i =1 T i = T C ) + N X i =1 T ( C ) i , (14)where T BA is the kinetic energy of the part A with respect to the reference frame definedby B . Here 0 is the laboratory reference frame and ( C ) is the centre of mass referenceframe.In the special case when N = 2 the following relation can be obtained N X i =1 T ( C ) i = T ( C )1 + T ( C )2 = T ( C )1 + T ( C )2 + T ( C )( C ) , (15)where ( C i ) is the centre of mass of the i th part and T ( C )( C ) is the kinetic energy of thecentre of mass of the first part with respect to the centre of mass of the second part. Ifthe two parts are both mass points Eq. (15) can be simplified as follows T ( C )1 + T ( C )2 = T ( C )( C ) = T ( C )( C ) , (16)which is a familiar result from two-body mechanics. By exploiting Eq. (15) repeatedlywe can easily express the total kinetic energy in terms of kinetic energies of subsystemswhich are much easier to obtain as will be seen in the following. D. Hermitian Conjugates of Momentum Operators
As mentioned in Section II A, the most tedious task for generating a KEO is to derivethe expressions for the HCMOs. As an example let us consider a system described by the8D spherical coordinates { R , θ , ϕ } . The coordinate transformation between 3D sphericalcoordinates and 3D Cartesian ones { x , y , z } is defined as x = R sin θ cos ϕy = R sin θ sin ϕz = R cos θ. (17)Here, we only give the corresponding HCMOs leading to the well-known KEO (for details,see Appendix V A)ˆ P † R = ˆ P R − i ~ R = − i ~ R ∂∂R R ˆ P † θ = ˆ P θ − i ~ cot θ = − i ~ θ ∂∂θ sin θ ˆ P † ϕ = ˆ P ϕ = − i ~ ∂∂ϕ ˆ T = − ~ m R ∂∂R R ∂∂R − ~ mR θ ∂∂θ sin θ ∂∂θ − ~ mR sin θ ∂ ∂ϕ . (18)So far our considerations have been concerned with the LRF. However, in order todescribe a molecule in terms of its natural motions, e.g., bond lengths and bond angles,it is usually more convenient to use one or more molecular reference frames (MRFs). Inthe following we will discuss this point and derive results for HCMOs associated withMRF spherical coordinates. Consider the LRF and a MRF defined by sets of unit vectors { e x , e y , e z } and { e x ′ , e y ′ , e z ′ } , respectively. The relation between the LRF and the MRFis just an orthogonal transformation characterised by the three Euler angles { ϑ , φ , χ } e α ′ = U z ( φ ) U y ( ϑ ) U z ( χ ) e α , (19)where α = x, y, z and U α is a rotation around e α . (The expression for the rotationaltransformation matrix U α and more details on the following derivations can be found inthe Appendix V A.)Now let us consider a vector R j characterised by three spherical coordinates { R j , θ j , ϕ j } in the MRF. According to Appendix V A, Eq. (46), we can express R j as R j = R j U z ( φ ) U y ( ϑ ) U z ( χ ) U z ( ϕ j ) U y ( θ j ) e z . (20)This is a vector equation and we can obtain the Cartesian coordinates of R j in LRF byprojecting the equation onto each LRF axis. Based on Eq. (20) we can exploit Eq. (6) toderive the expressions for HCMOs associated with MRF coordinates. The final results9re quite concise and they actually have the same form as those associated with LRFspherical coordinates. That is to say, that for the momentum operators associated with R j ( R j , θ j , ϕ j ) the following relations are validˆ P † R j = ˆ P R j − i ~ R j = − i ~ R j ∂∂R j R j ˆ P † θ j = ˆ P θ j − i ~ cot θ j = − i ~ θ j ∂∂θ j sin θ j ˆ P † ϕ j = ˆ P ϕ j = − i ~ ∂∂ϕ j . (21)In Appendix V A we also confirm that the momentum operators associated with MRFCartesian coordinates are Hermitian (cf. Eq. (63)). This result is quite important as itshows that one does not need to take the effort to derive the expressions for HCMOsprovided that one uses spherical coordinates, Cartesian coordinates or combinations ofboth, no matter whether they are defined in the LRF or MRFs. E. Angular Momentum and Rotation
For a system with N atoms we have 3 N degrees of freedom (DOFs). They are normallydivided into three translational DOFs, three rotational and 3 N − N − N − Q rot defined as Q † rot = ( ϑ φ χ ) (22)containing the three Euler angles which connect the LRF and MRF according to Eq. (19).The kinetic energy and total angular momentum J are defined as2 T = N X i =1 m i ˙ R † i ˙ R i J = N X i =1 m i R i × ˙ R i . (23)With ˙ Q rot being the angular velocity of MRF the velocities can be re-expressed as˙ R i = ˙ Q rot × R i + ˙ R ′ i , (24)10here ˙ R ′ i is the velocity of R i measured in the MRF. With the help of Eq. (24) and somevector algebra we can rewrite the kinetic energy and angular momentum as2 T = N X i =1 m i (cid:16) R i ˙ Q † rot ˙ Q rot − ( ˙ Q † rot R i ) + ˙ R ′ i † ˙ R ′ i + 2 ˙ Q † rot ( R i × ˙ R ′ i ) (cid:17) J = N X i =1 m i (cid:16) R i ˙ Q rot − ( ˙ Q † rot R i ) R i + ( R i × ˙ R ′ i ) (cid:17) . (25)Using Eqs. (25) it is straightforward to derive the following relation J = ∂T∂ ˙ Q rot , (26)which is exactly the definition of the generalised momentum vector P rot associated withthe three Euler angles.Now we can draw the following important conclusion. If the set of coordinates containsthe three Euler angles { ϑ , φ , χ } , the total angular momentum vector is just the generalisedmomentum vector associated with the three Euler angles J ϑ = P ϑ = ∂T∂ ˙ ϑJ φ = P φ = ∂T∂ ˙ φJ χ = P χ = ∂T∂ ˙ χ J = e ˙ ϑ P ϑ + e ˙ φ P φ + e ˙ χ P χ . (27)However, since the direction of angular velocities { e ˙ ϑ , e ˙ φ , e ˙ χ } are complicated, it is betterto transform the expressions to the Cartesian components in the LRF [12], i.e. J x = sin χP ϑ − csc ϑ cos χP φ + cot ϑ cos χP χ J y = cos χP ϑ + csc ϑ sin χP φ − cot ϑ sin χP χ J z = P χ . (28)Using Eq. (28) it is straightforward to express the KEO in terms of total angular mo-mentum and generalised momenta associated with vibrational DOFs. In other words,starting from Eq. (25) the KEO can be readily expressed as contributions of rotationalpart, vibrational part and their coupling.If the focus is on the vibrational spectrum, there are several arguments for neglectingrotational excitations. There is usually a clear time scale separation, that is, rotationalexcitation energies are much smaller than vibrational ones. If we are not aiming at high11esolution spectroscopy, rotational excitation will show up merely as a broadening of thevibrational bands and the band shifts of vibrational transitions due to the coupling canbe considered as being much smaller than the accuracy achievable by the quantum chem-istry and quantum dynamics methods for obtaining the spectra. Therefore the followingconsiderations will assume that rotational motion can be neglected, that is, we will setˆ J = 0 or ˆ P ϑ = ˆ P φ = ˆ P χ = 0. III. APPLICATIONS
In the following we will give two applications of the present approach. First, wewill consider the full-dimensional KEO for H O − employing polyspherical coordinatesas introduced in more detail in Appendix V B. Second, we consider N H +7 under theassumption of certain constraints and using bond length and angle coordinates. A. Full Dimensional KEO for H O − In this section we will give the full dimensional KEO for the monohydrated hydroxideion, H O − , which had been used in the study of different isotopomers in Ref. [25]. Thecoordinates are defined in the MRF given by Eq. (19). The four Jacobi vectors shown inFig. 1 are adopted. The three Euler angles are chosen in the same way as introduced inSection V B, i.e., the MRF spherical coordinates R and R are ( R , ,
0) and ( R , θ , R j , θ j , ϕ j ) ( j = 2 , G matrix according to Eq. (76)and the required matrix elements are shown in Eq. (78). Therefore, we can directly writedown the exact 9D KEO for total angular momentum J = 0 as 2 ˆ T vib = ˆ P † vib G vib ˆ P vib with the HCMOs given in Section II D. However, this KEO will contain a large numberof terms describing the angular momentum coupling between the shared Hydrogen andthe (OH) fragments. In order to simplify the KEO we will derive an approximate KEOwhich has the advantage that these couplings do not appear.The basic idea is to introduce a new MRF and express R with Cartesian coordi-nates ( x, y, z ). The remaining six coordinates are spherical coordinates in the old MRF( R , R , θ , θ , R , ϕ = ϕ ). The new MRF is associated with the old MRF by a rotationof an angle ηϕ around the R (the z -axis). The spherical coordinates for R in the newMRF are ( R , θ , ˜ ϕ ) with ˜ ϕ = ϕ − ηϕ , where η = µ / ( µ + µ ) is defined to minimise12he Coriolis type couplings involving the central Hydrogen. Here, µ ( µ ) is the reducedmass associated with the Jacobi vector R ( R ). After transforming the spherical coor-dinates ( R , θ , ˜ ϕ ) to the Cartesian ones ( x, y, z ) we can obtain the KEO for the centralHydrogen T = − ~ µ (cid:18) ∂ ∂x + ∂ ∂y + ∂ ∂z (cid:19) , (29)where µ is the reduced mass associated with the Jacobi vector R and Coriolis typeterms are ignored. Since the e ′ z is defined along the direction of R , the z coordinatecorresponds to the shared Hydrogen stretching vibration.For the other 6 DOFs of the (OH) fragment, the 6D KEO is written in terms ofthe G vib matrix elements Eq. (78). Here we take the angular momentum of the (OH) fragment as total angular momentum since the angular momentum of the shared Hydrogenis negligible. Combining both parts we finally obtain the 9D KEO where we introducethe new coordinates u i = cos θ i ( i = 1 , T = T + T + T (30)with T = − ~ µ ∂ ∂R − ~ µ ∂ ∂R − ~ µ ∂ ∂R (31) T = − X i =1 , (cid:18) µ i R i + 12 µ R (cid:19) ∂∂u i (1 − u i ) ∂∂u i − X i =1 , (cid:18) µ i R i − u i + 12 µ R u i − u i (cid:19) ∂ ∂ϕ + 1 µ R u p − u u p − u ∂∂ϕ cos ϕ ∂∂ϕ − µ R (cid:18)q − u ∂∂u ∂∂u q − u + ∂∂u q − u q − u ∂∂u (cid:19) − µ R u p − u (cid:18) ∂∂ϕ sin ϕ q − u ∂∂u + ∂∂u q − u sin ϕ ∂∂ϕ (cid:19) − µ R u p − u (cid:18) ∂∂ϕ sin ϕ q − u ∂∂u + ∂∂u q − u sin ϕ ∂∂ϕ (cid:19) . (32)The non-Euclidean normalisation according to the volume element is d τ = dR dR dR dxdydzdu du dϕ . The reduced masses for different isotopomers are defined asfollows: HOHOH − – µ = µ = m H m O / ( m H + m O ), µ = 2 m H ( m H + m O ) / (3 m H + 2 m O ), µ = ( m H + m O ) /
2; HODOH − – µ = 2 m D ( m H + m O ) / ( m D + 2 m H + 2 m O ); HOHOD − –13 = m D m O / ( m H + m O ) and µ and µ change correspondingly. For DODOD − the cor-responding masses of HOHOH − are modified by replacing m H by m D . In the same waywe get the masses for DOHOD − by exchanging m H and m D in HODOH − , and similarlyone can obtain DODOH − from DOHOH − . B. Reduced Dimensional KEO for N H +7 In this section we give an example for a KEO for a system with constraints stressingagain that our method is invariant for reduced dimensional descriptions. Specificallywe focus on the protonated ammonia dimer, N H +7 , whose IR spectrum in the range < − had been discussed in Ref. [26] on the basis of a 5D model including onlythe asymmetric stretching vibration of the shared proton. Subsequently, the model wasextended to account for the two degenerate bendings as well in Ref. [27]. Here we willderive the 5D model and comment on its extension. A reasonable reduced description forthe spectrum which is influenced by the shared proton motion should take into accountthe following two conditions: (i) experimental data suggest that the relevant energy rangeis presumably below 1000 cm − , but might extend into the < − range due tocombination bands and (ii) symmetry selection rules dominate the anharmonic couplingsespecially in this low-energy range. Therefore, we first assume that the C symmetry ofthe N H fragment, i.e., excluding the central proton, will not be broken. Second, thelength of the N-H covalent bonds shall be fixed. Notice that this neglects the bendingmotion of the shared proton perpendicular to the line connecting the two nitrogen atoms.This is justified by the different symmetry and the lower anharmonicity of these twomodes as compared to the proton motion along the N-N line. These constraints leave fiveinternal coordinates to describe the system as shown in Fig. 2, i.e., the shared protonstretching with respect to the center of mass of the rest N H fragment, z , the relativemotion of the centres of mass of the two ammonia, R , the umbrella type motion of thetwo ammonia, θ and θ , and the rotation (torsion) of the NH fragments with respect toeach other, ϕ .For these five active coordinates the classical kinetic energy with all other internal14oordinates frozen can be obtained by exploiting Eq. (15) repeatedly T = T (N H )(H) + T (N H )N H = T (N H )(H) + T (NH )NH + T (NH ′ )NH ′ + T (NH ′ )(NH ) = T (N H )(H) + T (NH ′ )(NH ) + (cid:16) T (H )H + T (H )(N) (cid:17) + (cid:16) T (H ′ )H ′ + T (H ′ )(N ′ ) (cid:17) , (33)where (AB) is the center of mass of AB. It is straightforward to obtain each term inabove equation T (N H )(H) = 12 µ p ˙ z T (NH )(NH ) = 12 µ R ˙ R T (H )H = 32 m H (cid:18) d ( R sin θ i ) dt (cid:19) + 32 m H ( R sin θ i ) ˙ ϕ i T (H )(N) = 12 µ (N − (cid:18) d ( R cos θ i ) dt (cid:19) , (34)where µ p = 2 m H (3 m H + m N ) / (7 m H + 2 m N ), µ R = (3 m H + m N ) and µ (N − =3 m H m N / (3 m H + m N ). The orientation angle of each individual ammonia is denoted by ϕ , and only the difference between them is the torsion shown in Fig. 2 and R is the freeN-H covalent bond length. Thus the final kinetic energy is T = 12 µ p ˙ z + µ R ˙ R + 3 m H m N m H + m N ) R (cid:16) ˙ θ sin θ + ˙ θ sin θ (cid:17) + 32 m H R X i =1 , ( ˙ θ i cos θ i + ˙ ϕ i sin θ i )= 12 µ p ˙ z + 12 µ R ˙ R + 12 I vib ( θ ) ˙ θ + 12 I vib ( θ ) ˙ θ + 12 I rot ( θ ) ˙ ϕ + 12 I rot ( θ ) ˙ ϕ , (35)where I vib ( θ ) = I (cos θ + m N sin θ ) / (3 m H + m N ), I rot ( θ ) = I sin θ and I = 3 m H R .The last two terms in above equation can be rewritten and after separation of the globalrotation we finally simplify the kinetic energy as T = 12 µ p ˙ z + 12 µ R ˙ R + X i =1 , I vib ( θ i ) ˙ θ i + 12 I tor ( θ , θ ) ˙ ϕ , (36)where the reduced moment of inertia for the torsion is I tor ( θ , θ ) = I rot ( θ ) I rot ( θ ) / ( I rot ( θ ) + I rot ( θ )).So far we did not consider the Coriolis type couplings with total angular velocity.However, due to the C symmetry, the total angular momentum equals to zero is equivalent15o total angular velocity equals to zero for this specific case since none of the adoptedcoordinates contributes to total angular momentum. Therefore the Coriolis type couplingsare zero.According to the general procedure detailed in Section II A we can obtain the followingquantum KEO ( ϕ = ϕ − ϕ )ˆ T = 12 µ p ˆ P † z ˆ P z + 12 µ R ˆ P † R ˆ P R + ˆ P † ϕ ˆ P ϕ I tor ( θ , θ )+ 12 ˆ P † θ I − ( θ ) ˆ P θ + 12 ˆ P † θ I − ( θ ) ˆ P θ . (37)Inspecting the adopted coordinates we notice that one of them is Cartesian and the otherfour are spherical coordinates defined in MRF. Therefore, according to Section II D theHCMOs are given by ˆ P † z = ˆ P z ˆ P † R = ˆ P R − i ~ R ˆ P † θ j = ˆ P θ j − i ~ cot θ j , j = 1 , P † ϕ = ˆ P ϕ . (38)Hence the final KEO and the Euclidean normalisation condition are given as followsˆ T = − ~ µ p ∂ ∂z − ~ µ R R ∂∂R R ∂∂R − ~ I tor ( θ , θ ) ∂ ∂ϕ − ~ X i =1 , θ i ∂∂θ i sin θ i I vib ( θ i ) ∂∂θ i dτ = R sin θ sin θ dzdRdθ dθ dϕ. (39)In Ref. [27] we have extended this model by including the shared proton bendingcoordinates x and y as shown in Fig. 2. In terms of the KEO this is straightforward sincein Eq. (34) we simply have to set T (N H )(H) = 12 µ p ( ˙ z + ˙ x + ˙ y ) (40)which is carried through the derivation to give a corresponding contribution to Eq. (39).Compared to the experiment this gives a rather good description of the bending funda-mental transitions as well as an improvement concerning the shared proton fundamentalstretching transition. 16 V. SUMMARY
In summary we have presented a new method for obtaining kinetic energy operatorswhich involves a straightforward quantisation of the classical Lagrange/Hamilton function.It is based on the notion of hermitian conjugate momentum operators whose derivationpresents the major effort for practical applications. An important point of our approach isits validity for systems with constraints. This makes it particularly attractive for studyinglarger systems in reduced dimensionality.We have applied our approach to the case of a general system described by polysphericalcoordinates and showed the equivalence of the kinetic energy operator with that obtainedby the method of Gatti and coworkers. Afterwards we discussed two specific applicationsto charged cluster systems having strong hydrogen bonds. The strong coupling betweendifferent coordinates as well their structural floppiness makes it necessary to perform highdimensional quantum dynamics simulations as done in Refs. [25, 26, 27] on the basis ofthe kinetic energy operators derived in Section III.
V. APPENDIXA. Hermitian Conjugate Momentum Operators for Spherical Coordinates
The derivation of the expressions for the HCMOs in LRF is the most tedious part ofthe present method and it becomes even more involved if coordinates defined in MRFsare used. In this Appendix we will first summarise the relevant relations between theLRF and MRFs and subsequently turn to the derivation of HCMOs.Consider the LRF defined by three orthogonal unit vectors { e x , e y , e z } and a MRFdefined by three orthogonal unit vectors { e x ′ , e y ′ , e z ′ } . The orientation angles of e z ′ inthe LRF are ( ϑ, φ ). To obtain the connection between the LRF and the MRF we firstapply two excessive rotations U y ( ϑ ) and U z ( φ ) to the LRF, where U y ( θ ) means rotating ϑ around e y and U z ( φ ) means rotating φ around e z . The matrix representation of them17re U y ( ϑ ) = cos ϑ ϑ − sin ϑ ϑ U z ( φ ) = cos φ − sin φ φ cos φ
00 0 1 . (41)The new reference frame generated by applying U y ( ϑ ) and U z ( φ ) to the LRF has thesame z axis as the MRF, i.e., the only difference between the two reference frames isjust a rotation by χ around e z ′ . Therefore, the MRF can be obtained by applying threesuccessive rotations U y ( ϑ ), U z ( φ ) and U z ′ ( χ ) to the LRF, namely e α ′ = U z ′ ( χ ) U z ( φ ) U y ( ϑ ) e α , α = x, y, z, (42)where U z ′ ( χ ) means rotating χ around e z ′ . The matrix representation for U z ′ ( χ ) in theMRF is the same with U z ( χ ) in the LRF. Since the third rotation U z ′ ( χ ) in Eq. (42)has no effects on e z ′ we actually have e z ′ = U z ( φ ) U y ( ϑ ) e z . Exploiting the rules betweenvector and operator transformations we can obtain the expression for U z ′ ( χ ) in the LRF U z ′ ( χ ) = [ U z ( φ ) U y ( ϑ )] U z ( χ ) [ U z ( φ ) U y ( ϑ )] − . (43)Consequently Eq. (42) can be rewritten as e α ′ = U z ( φ ) U y ( ϑ ) U z ( χ ) e α α = x, y, z. (44)Eq. (44) tells us that an equivalent way to obtain the MRF is to apply three successiverotations U z ( χ ), U y ( ϑ ) and U z ( φ ) to the LRF.Now let us consider vectors R j characterised by spherical coordinates ( R j , θ j , ϕ j ) inthe MRF ( j = 1 − N ). With Eq. (44) and the rule between basis vectors and componentstransformations we can obtain the Cartesian components of these vectors in LRF R jx R jy R jz = U z ( φ ) U y ( ϑ ) U z ( χ ) R jx ′ R jy ′ R jz ′ , (45)where R jα and R jα ′ are Cartesian components of R j in the LRF and the MRF, respec-tively. We can rewrite Eq. (45) in more formally as R j = R j U z ( φ ) U y ( ϑ ) U z ( χ ) U z ( ϕ j ) U y ( θ j ) e z , (46)18q. (46) is a vector equation therefore it is also valid for an arbitrary reference frame.Now we turn to the main task, that is, deriving the Hermitian conjugate of momentumoperators. This can be done step by step according to Eq. (6) with the help of thecoordinate transformation Eq. (45). From Eq. (45) we can see that ( θ j , ϕ j ) only appear inthe Cartesian components of R j . This greatly simplifies the expression for the Hermitianconjugates of momentum operators as followsˆ P † R j = ˆ P R j − i ~ X α = x,y,z (cid:18) ∂∂R jα ∂R jα ∂R j (cid:19) ◦ ˆ P † θ j = ˆ P θ j − i ~ X α = x,y,z (cid:18) ∂∂R jα ∂R jα ∂θ j (cid:19) ◦ ˆ P † ϕ j = ˆ P ϕ j − i ~ X α = x,y,z (cid:18) ∂∂R jα ∂R jα ∂ϕ j (cid:19) ◦ . (47)Here the three Euler angles in Eq. (45) are purely parameters. To calculate the partialderivatives we first give some useful relations following from the orthogonality of thetransformation Eq. (45) R jα = X β ′ ∂R jα ∂R jβ ′ R jβ ′ , R jβ ′ = X α ∂R jβ ′ ∂R jα R jα X β ′ (cid:18) ∂R jα ∂R jβ ′ (cid:19) = X α (cid:18) ∂R jα ∂R jα ′ (cid:19) = 1 ∂R jα ∂R jβ ′ = ∂R jβ ′ ∂R jα (48)where α = x, y, z and β ′ = x ′ , y ′ , z ′ . Note here all the derivatives { ∂R jβ ′ /∂R jα } areparameters only depending on the three Euler angles.Let us first derive the expression for ˆ P † R j . It is straightforward to get the followingderivatives of MRF Cartesian coordinates ∂R jx ′ ∂R j = sin θ j cos ϕ j = R jx ′ R j ∂R jy ′ ∂R j = sin θ j sin ϕ j = R jy ′ R j ∂R jz ′ ∂R j = cos θ j = R jz ′ R j (49)where R j = R jx + R jy + R jz = R jx ′ + R jy ′ + R jz ′ . Based on Eq. (49) we can obtain the19erivatives of LRF Cartesian coordinates with respect to R j and derive the final result X α (cid:18) ∂∂R jα ∂R jα ∂R j (cid:19) ◦ = X α ∂∂R jα X α ′ R jβ ′ R j ∂R jα ∂R jβ ′ ! ◦ = X α,β ′ R j ∂R jβ ′ ∂R jα + R jβ ′ ∂ (cid:0) R jx + R jy + R jz (cid:1) − ∂R jα ∂R jα ∂R jβ ′ = X α,β ′ (cid:18) R j ∂R jβ ′ ∂R jα − R jβ ′ R jα R j (cid:19) ∂R jα ∂R jβ ′ = X α (cid:18) R j − R jα R j (cid:19) = 3 R j − R j = 2 R j , (50)where Eq. (48) has been used in the fourth step. Finally we can get the HCMO associatedwith R j as follows ˆ P † R j = ˆ P R j − i ~ R j = − i ~ R j ∂∂R j R j . (51)Next we will derive the expression for ˆ P † θ j following the same procedure. The derivativesof MRF Cartesian coordinates read ∂R jx ′ ∂θ j = R j cos θ j cos ϕ j = R jx ′ cot θ j ∂R jy ′ ∂θ j = R j cos θ j sin ϕ j = R jy ′ cot θ j ∂R jz ′ ∂θ j = − R j sin θ j = − R jz ′ cot θ j . (52)Based on Eq. (52) we can obtain the derivatives of Cartesian coordinates in LRF withrespect to θ j ∂R jα ∂θ j = R jx ′ cot θ j ∂R jα ∂R jx ′ + R jy ′ cot θ j ∂R jα ∂R jy ′ − R jz ′ cot θ j ∂R jα ∂R jz ′ (53)Furthermore, we obtain (cid:18) ∂∂R jα ∂R jα ∂θ j (cid:19) ◦ = cot θ j ∂R jα ∂R jx ′ ∂R jx ′ ∂R jα + cot θ j ∂R jα ∂R jy ′ ∂R jy ′ ∂R jα − θ j ∂R jα ∂R jz ′ ∂R jz ′ ∂R jα + (cid:18) R jx ′ ∂R jα ∂R jx ′ + R jy ′ ∂R jα ∂R jy ′ + R jz ′ cot θ j ∂R jα ∂R jz ′ (cid:19) ∂ cot θ j ∂R jα = cot θ j − θ j cos θ j ∂R jα ∂R jz ′ ∂R jz ′ ∂R jα + (cid:18) R jα + (tan θ j − R jz ′ ∂R jα ∂R jz ′ (cid:19) ∂ cot θ j ∂R jα . (54)20ince cot θ j = R jz ′ / q R jx ′ + R jy ′ we can get the following derivatives ∂ cot θ j ∂R jx ′ = − R jx ′ R jz ′ q R jx ′ + R jy ′ ∂ cot θ j ∂R jy ′ = − R jy ′ R jz ′ q R jx ′ + R jy ′ ∂ cot θ j ∂R jz ′ = 1 q R jx ′ + R jy ′ . (55)Based on Eq. (55) we have ∂ cot θ j ∂R jα = 1 q R jx ′ + R jy ′ (cid:18) − R jx ′ R jz ′ ∂R jx ′ ∂R jα − R jy ′ R jz ′ ∂R jy ′ ∂R jα + ( R jx ′ + R jy ′ ) ∂R jz ′ ∂R jα (cid:19) = 1 q R jx ′ + R jy ′ (cid:18) − R jz ′ R jα + R j ∂R jz ′ ∂R jα (cid:19) . (56)Combining Eq. (54) and Eq. (56) we can now derive X α (cid:18) ∂∂R jα ∂R jα ∂θ j (cid:19) ◦ = X α (cid:18) cot θ j − θ j cos θ j ∂R jα ∂R jz ′ ∂R jz ′ ∂R jα (cid:19) + X α q R jx ′ + R jy ′ (cid:18) R jα + (tan θ j − R jz ′ ∂R jα ∂R jz ′ (cid:19) (cid:18) − R jz ′ R jα + R j ∂R jz ′ ∂R jα (cid:19) = 3 cot θ j − θ j cos θ j + 1 q R jx ′ + R jy ′ (cid:0) − R jz ′ R j + R j R jz ′ + (tan θ j − − R jz ′ R jz ′ + R j R jz ′ ) (cid:1) = 3 cot θ j − θ j cos θ j + (tan θ j −
1) cot θ j = cot θ j (57)The Hermitian conjugate of each ˆ P θ j associated with the corresponding MRF polar angle θ j can be finally expressed asˆ P † θ j = ˆ P θ j − i ~ cot θ j = − i ~ sin θ j ∂∂θ j sin θ j . (58)Finally we come to ˆ P † ϕ j . The procedure is the same, however, it is much simpler as21ompared to ˆ P † θ j . Again we start with the derivatives of MRF Cartesian coordinates ∂R jx ′ ∂ϕ j = − R j sin θ j sin ϕ j = − R jy ′ ∂R jy ′ ∂ϕ j = R j sin θ j cos ϕ j = R jx ′ ∂R jz ′ ∂ϕ j = 0 (59)to obtain the final X α ∂∂R jα ∂R jα ∂ϕ j ! ◦ = X α ∂∂R jα (cid:18) R jx ′ ∂R jα ∂R jy ′ − R jy ′ ∂R jα ∂R jx ′ (cid:19)! ◦ = X α (cid:18) ∂R jx ′ ∂R jα ∂R jα ∂R jy ′ − ∂R jy ′ ∂R jα ∂R jα ∂R jx ′ (cid:19) = 0 (60)The momentum operator associated with each orientation angle ϕ j is Hermitian accordingto Eq. (60), namely ˆ P † ϕ j = ˆ P ϕ j = − i ~ ∂∂ϕ j . (61)Let us recall the above detailed procedure. The only condition we need is that thetransformation matrix between Cartesian coordinates in the LRF and those in a MRFis orthogonal. First, if we set all the three Euler angles equal to zero we can obtain theHCMOs associated with spherical coordinates in the LRF. The final results are the samewith Eq. (51), Eq. (58), and Eq. (61) since a unit matrix is also an orthogonal matrix. Sec-ond, we can use more rotations to define more MRFs and the HCOMs associated with thespherical coordinates in each different MRF obey Eq. (51), Eq. (58), and Eq. (61). As animportant consequence the Hermitian conjugates of all the momentum operators associ-ated with real bond lengths, bond angles and dihedral angles obey Eq. (51), Eq. (58), andEq. (61), respectively, irrespective of how complicated the molecule might be. Similarly,the result holds for any reference frame provided there exists an orthogonal transforma-tion to transform it to the LRF. Based on the above conclusions we can see the HCMOsassociated with the three Euler angles obey the same relationsˆ P † ϑ = ˆ P ϑ − i ~ cot ϑ = − i ~ sin ϑ ∂∂ϑ sin ϑ ˆ P † φ = ˆ P φ = − i ~ ∂∂φ ˆ P † χ = ˆ P χ = − i ~ ∂∂χ . (62)Finally, we can obtain the HCMOs ˆ P † jα ′ associated with MRF Cartesian coordinatesfollowing the same procedure. Since ∂R jα /∂R jα ′ is just a parameter which does not22epends on R jx , R jy or R jz we can easily getˆ P † jα ′ = ˆ P jα ′ − i ~ X α (cid:18) ∂∂R jα ∂R jα ∂R jα ′ (cid:19) ◦ = ˆ P jα ′ , (63)where α ′ = x ′ , y ′ , z ′ . Apparently, the momenta associated with MRF Cartesian coordi-nates are Hermitian. B. KEO in Terms of Polyspherical Coordinates
In order to connect our approach to existing one we consider the KEO in terms ofso called polyspherical coordinates defined in the LRF and a MRF. This will lead us toexpression which have been reported in Ref. [17]. Consider a molecular system composedof N + 1 atoms. After separating the total centre of mass motion we can describe it with N vectors R , R , · · · , R N . The three Euler angles ( ϑ, φ, χ ) are chosen in such a way thatthe spherical coordinates of R N in the LRF are ( R N , ϑ, φ ) and the spherical coordinatesof R in the MRF are ( R , θ , ϕ = 0). That is to say the e z ′ axis of the MRF is definedto be along the direction of R N . The rest N − { R j , θ j , ϕ j } in the MRF ( j = 2 , · · · , N − j = 1 , · · · , N and θ N = ϕ N = ϕ = 0, we can derive the LRFcomponents of the N velocity vectors˙ R j = R j ˙ ϑ U z ( φ ) U ′ y ( ϑ ) U z ( χ ) U z ( ϕ j ) U y ( θ j ) e z + R j ˙ φ U ′ z ( φ ) U y ( ϑ ) U z ( χ ) U z ( ϕ j ) U y ( θ j ) e z + R j ˙ χ U z ( φ ) U y ( ϑ ) U ′ z ( χ ) U z ( ϕ j ) U y ( θ j ) e z + ˙ R j U z ( φ ) U y ( ϑ ) U z ( χ ) U z ( ϕ j ) U y ( θ j ) e z + R j ˙ θ j U z ( φ ) U y ( ϑ ) U z ( χ ) U z ( ϕ j ) U ′ y ( θ j ) e z + R j ˙ ϕ j U z ( φ ) U y ( ϑ ) U z ( χ ) U ′ z ( ϕ j ) U y ( θ j ) e z , (64)where U ′ is the first order derivative of the corresponding rotational transformation ma-trix. The above equations are still vector equations, however, they are only valid in theLRF since we have used the relation ˙ e z = 0. To make them valid in an arbitrary referenceframe we only need to add the corresponding terms containing ˙ e z to the right hand side.Before proceeding we will discuss the orthogonality of the terms appearing in Eq. (64). The last three terms in Eq. (64) are just the spherical velocity components measuredin the MRF while the first three terms are components of the velocity caused by the23on-inertial MRF. Each term in Eq. (64) can be expressed in a formally simple way interms of angular velocity vectors˙ R j = ˙ ϑ e ˙ ϑ × R j + ˙ φ e ˙ φ × R j + ˙ χ e ˙ χ × R j + ˙ R j R j /R j + ˙ θ j e ˙ θ j × R j + ˙ ϕ j e ˙ ϕ j × R j , (65)where e ˙ ϑ is the direction of the angular velocity ˙ ϑ , and similar for e ˙ φ , e ˙ χ , e ˙ θ j , and e ˙ ϕ j .With the help of Eq. (65) one can immediately see that the fourth term is parallel to R j while the other terms are perpendicular to R j . Recalling the velocity in sphericalcoordinates we know that the last three terms are orthogonal to each other. Therefore, inEq. (64) (or Eq. (65) ), the fourth term (namely the ˙ R j term) is orthogonal to the otherterms and the last three terms are orthogonal to each other.Having at hand all the velocities it is quite straightforward to write the classical kineticenergy [28] according to T = 12 N X i,j =1 µ ij ˙ R † i ˙ R j , (66)where the matrix { µ jk } combines the reduced masses and the transformation matrixbetween adopted and Jacobi vectors [28]. The final results in LRF can be written in asymmetric form as in Eq. (1) T = 12 ˙ Q † M ˙ Q , Q = Q rot Q vib , (67)where the coordinates are separated as rotational, Eq. (22), and vibrational DOFs Q † vib = (cid:16) R · · · R N θ · · · θ N − ϕ · · · ϕ N − (cid:17) . (68)The next step is to calculate the inverse matrix of M to generate the quantum KEO.Aiming at a separation of the rotational and vibrational DOFs, we divide the matrix M into the following four blocks M = M rot M Cor M † Cor M vib , (69)where the subscripts “ rot ”, “ Cor ”, and “ vib ” correspond to rotational, Coriolis, and vibra-tional terms, respectively. Therefore, M rot , M Cor , and M vib are 3 ×
3, 3 × (3 N −
3) and(3 N − × (3 N −
3) dimensional matrices, respectively. Suppose the inverse matrix of M is divided in the same spirit as M − = G rot G Cor G † Cor G vib , (70)24e can express the quantum KEO as a sum of rotational, vibrational, and Coriolis termsˆ T = ˆ T rot + ˆ T vib + ˆ T Cor = 12 ˆ P † rot G rot ˆ P rot + 12 ˆ P † vib G vib ˆ P vib + 12 (cid:16) ˆ P † rot G Cor ˆ P vib + h . c . (cid:17) . (71)According to Section II E, the generalised momentum vector ˆ P rot associated with therotational DOFs is just the total angular momentum vector J . If we are interested in therotational DOFs only, we obtain upon setting the vibrational momenta equal to zero:ˆ T rot = 12 ˆ P † rot G rot ˆ P rot = 12 ˆ J † G rot ˆ J , (72)where the components of J should be along the directions of e ˙ ϑ , e ˙ φ and e ˙ χ according toEq. (27). In passing we note that one can also transform the components to Cartesianones according to Eq. (28). On the other hand, if we are interested in the vibrationalDOFs, the KEO for the total angular momentum J = 0 readsˆ T vib = 12 ˆ P † vib G vib ˆ P vib . (73)Finally, we give the details on how to calculate the different blocks of the inversematrix of M . One can see that the matrix M can be congruently block-diagonalised inthe following way − M Cor M − M rot M Cor M † Cor M vib − M − M † Cor = M rot − M Cor M − M † Cor
00 M vib . (74)Calculating the inverse of both sides of Eq. (74) leads to M − = G rot G Cor G † Cor G vib = M rot M Cor M † Cor M vib − = − M − M † Cor M rot − M Cor M − M † Cor
00 M vib − − M Cor M − . (75)Calculating the matrix product in above equation leads to the final results G rot = (cid:16) M rot − M Cor M − M † Cor (cid:17) − G Cor = − G rot M Cor M − G vib = M − + M − M † Cor G Cor . (76)25he matrix G is essentially the same as the one reported in Ref. [17]. There, however,the essential point has been to express the MRF Cartesian components of the angularmomentum associated with each vector { R i } in terms of J and { P θ i , P ϕ i } .For the special case of Jacobi vectors, M vib is diagonal which greatly simplifies thecalculation of Eq. (76). It is straightforward to obtain the diagonal elements according toEq. (64) ( M vib ) R j R j = µ j ( M vib ) θ j θ j = µ j R j ( M vib ) ϕ j ϕ j = µ j R j sin θ j . (77)The inverse matrix of M vib can be obtained quite easily. We only need to calculate theinverse of a matrix of dimension 3 × G vib (Hermitian) which will be used inSection III: G R i R j = δ ij µ j , G R i θ j = 0 , G R i ϕ j = 0 ,G θ i θ j = δ ij µ j R j + cos( ϕ i − ϕ j ) µ N R N ,G θ i ϕ j = cot θ j sin( ϕ i − ϕ j ) − cot θ sin ϕ i µ N R N ,G ϕ i ϕ j = δ ij µ j R j sin θ j + 1 µ R sin θ + cot θ i cot θ j cos( ϕ i − ϕ j ) + cot θ − cot θ (cos ϕ i cot θ i + cos ϕ j cot θ j ) µ N R N . (78) Acknowledgments
We gratefully acknowledge financial support by the Deutsche Forschungsgemeinschaftthrough the GK 788. 26
1] X. Huang, B. J. Braams, S. Carter, and J. M. Bowman, J. Am. Chem. Soc. , 5042(2004).[2] X. Huang, B. J. Braams, and J. M. Bowman, J. Chem. Phys. , 044308 (2005).[3] W. H. Miller, N. C. Handy, and J. E. Adams, J. Chem. Phys. , 99 (1980).[4] T. Carrington and W. H. Miller, J. Chem. Phys. , 3942 (1984).[5] N. Shida, P. F. Barbara, and J. E. Alml¨of, J. Chem. Phys. , 4061 (1989).[6] D. P. Tew, N. C. Handy, and S. Carter, Phys. Chem. Chem. Phys. , 1958 (2001).[7] K. Giese and O. K¨uhn, J. Chem. Phys. , 054315 (2005).[8] B. Podolsky, Phys. Rev. , 812 (1928).[9] P. R. Bunker, Molecular symmetry and spectroscopy (Academic, New York, 1979).[10] R. Mecke, Z. Phys. , 405 (1936).[11] C. Eckart, Phys. Rev. , 552 (1935).[12] E. B. Wilson, J. C. Decius, and P. C. Cross, Molecular Vibrations (Dover, New York, 1955).[13] J. K. G. Watson, Mol. Phys. , 479 (1968).[14] J. K. G. Watson, Mol. Phys. , 465 (1970).[15] J. M. Bowman, S. Carter, and X. Huang, Int. Rev. Phys. Chem. , 533 (2003).[16] M. J. Bramley, W. H. Green, Jr., and N. C. Handy, Mol. Phys. , 1183, (1991).[17] F. Gatti, C. Munoz, and C. Iung, J. Chem. Phys. , 8275 (2001).[18] F. Gatti and C. Iung, J. Theor. Comp. Chem. , 507 (2003).[19] C. Iung and F. Gatti, Int. J. Quant. Chem. , 130 (2006).[20] O. Vendrell, F. Gatti, D. Lauvergnat, and H.-D. Meyer, J. Chem. Phys. , 184302 (2007).[21] F. Gatti, Y. Justum, M. Menou, A. Nauts, and X. Chapuisat, J. Mol. Spectr. , 403(1997).[22] A. Nauts and X. Chapuisat, Chem. Phys. Lett. , 164 (1987).[23] D. Lauvergnat and A. Nauts, J. Chem. Phys. , 8560 (2002).[24] W. S. l’Yi, Phys. Rev. A , 1251 (1996).[25] Y. Yang and O. K¨uhn, Z. Phys. Chem. , 1375 (2008).[26] K. R. Asmis, Y. Yang, G. Santambrogio, M. Br¨ummer, J. R. Roscioli, L. R. McCunn, M. A.Johnson, and O. K¨uhn, Angew. Chem. Int. Ed. , 8691 (2007).[27] Y. Yang, O. K¨uhn, G. Santambrogio, D. J. Goebbert, and K. R. Asmis, submitted.[28] C. Iung, F. Gatti, A. Viel, and X. Chapuisat, Phys. Chem. Chem. Phys. , 3377 (1999). igure Captions FIG. 1: The four Jacobi vectors of the H O − anion which are used to define the 9D KEO.FIG. 2: Definition of active coordinates of the reduced N H +7 model. R R R Figure 1, Yang and K¨uhn 29 x θ θ ϕ zyzy