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The Journal of Chemical Physics Calculation of Energy Levels for Internal Torsion and OverAll Rotation. I. CH3BF2 Type Molecules
Calculation of Energy Levels for Internal Torsion and OverAll Rotation. I. CH3BF2 Type Molecules
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Tập:
23
Năm:
1955
Ngôn ngữ:
english
Tạp chí:
The Journal of Chemical Physics
DOI:
10.1063/1.1740512
File:
PDF, 675 KB
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Calculation of Energy Levels for Internal Torsion and OverAll Rotation. I. CH3BF2 Type Molecules E. Bright Wilson Jr., Chun Chia Lin, and David R. Lide Jr. Citation: The Journal of Chemical Physics 23, 136 (1955); doi: 10.1063/1.1740512 View online: http://dx.doi.org/10.1063/1.1740512 View Table of Contents: http://scitation.aip.org/content/aip/journal/jcp/23/1?ver=pdfcov Published by the AIP Publishing Articles you may be interested in Calculation of dipolar nuclear magnetic relaxation times in molecules with multiple internal rotations. II. Theoretical results for anisotropic overall motion of the molecule, and comparison with 13C relaxation times in n alkanes and n alkyl bromides J. Chem. Phys. 60, 2890 (1974); 10.1063/1.1681458 Energy Levels for Internal and OverAll Rotation of TwoTop Molecules. I. Microwave Spectrum of Dimethyl Silane J. Chem. Phys. 34, 498 (1961); 10.1063/1.1700973 Calculation of Energy Levels for Internal Torsion and OverAll Rotation. III J. Chem. Phys. 31, 91 (1959); 10.1063/1.1730343 Calculation of Energy Levels for Internal Torsion and OverAll Rotation. II. CH3CHO Type Molecules; Acetaldehyde Spectra J. Chem. Phys. 26, 1695 (1957); 10.1063/1.1743607 Note on the Calculation of Energy Levels in Molecules with Internal Torsion J. Chem. Phys. 24, 1072 (1956); 10.1063/1.1742680 This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34 THE JOURNAL OF CHEMICAL PHYSICS VOLUME 23, NUMBER 1 JANUARY, 1955 Calculation of Energy Levels for Internal Torsion and OverAll Rotation. * I. CHaBF 2 Type Molecules E. BRIGHT WILSON, JR., CHUN CHIA LIN, t AND DAVID R. LIDE, JR.t Department of Chemistry, Harvard University, Cambridge, Massachusetts (Received June 17, 1954) Methods are described for calculating the energy levels for the overall rotation and internal torsion of molecules consisting of a rigid symmetrical top attached to a rigid; asymmetrical framework in such a way that the symmetry axis of the top coincides with a principal axis of the molecule. Probable examples are nitromethane and CH aBF2• Matrix perturbation methods are employed to obtain finite rotational secular equations valid in each of the cases: low barrier, high barrier, low asymmetry. These secular equations are modifications of the ordinary Wang equation for the rigid asymmetric rotor and can usually be solved by the continued fraction method. The symmetry groups applicable to this problem are also discussed. OLECULES such as CH3BF 2 or nitromethane1 (CH 3N0 2) may show overall rotation, internal torsion of the methyl group relative to the N0 2 group, and vibration. For the purpose of treating the overall rotation and internal torsion, it is convenient to think of the BF 2 or N02 group as a rigid framework to which is attached a symmetrical top (CH 3) which may rotate or vibrate about an axis colinear with a principal axis of inertia of the whole molecule. Furthermore, the three principal moments of inertia of the whole molecule are all different and are uninfluenced by the internal orientation of the attached top. The calculation of the rotational and internal torsional energy levels of this class of molecules is the subject of the present paper.2 Let x, y, z be moving Cartesian axes rigidly attached to the framework part of the molecule and coincident with the principal axes of inertia of the whole molecule (origin is at center of mass of whole molecule). The z axis will coincide with the symmetry axis of the top. The three Eulerian angles fl, !p, and X of x, y, z relative to spacefixed axes will describe the overall orientation of the molecule, while the angle 0: will give the relative orientation of top and framework. MODEL AND COORDINATES in which I a is the momen.t of inertia of the top about its symmetry axis, lx, I y , I. are the principal moments of the entire molecule, and W x , W y , W z are components of angular velocity of the framework along x, y, z. It has been usua15 to eliminate the crossproduct terms by means of a transformation, but this will not be done in this treatment. To obtain the Hamiltonian form, use the definitions of the momenta: M The model then consists of two connected rigid3 bodies. One (the top) has two equal principal moments of inertia about axes perpendicular to a principal axis of the whole molecule. There are four degrees of freedom, three for overall rotation and one for rotation of the top about its unique axis (see Fig. 1). THE HAMILTONIAN The kinetic energy has previously been written as4 2T=f:.w,,?+IyWl+l zwz2+1 a«+2I aaw. p=aT/aa, Px=aT/awx,etc. FIG. 1. A model of the CHaBF 2 type molecule. (1) (2) This leads in the usual way to the form (3) in which all angular momenta are in units h=h/27r and * The research reported in this paper was made possible by support extended Harvard University by the Office of Naval Research under U. S. Office of Naval Research Contract N50ri 76, Task Order V. t Socony Vacuum Predoctoral Fellow, 19531954. t Now at National Bureau of Standards, Washington, D. C. 1 Microwave spectra due to these motions have been recently reported for nitromethane by Tannenbaum, Johnson, Myers, and Gwinn, J. Chern. Phys. 22, 949 (1954). 2 The class treated here is in some ways simpler than the case of methyl alcohol which has been thoroughly studied by Dennison and coworkers [see Ivash and Dennison, J. Chern. Phys. 21, 1804 (1954)J but the approach differs somewhat. 3 For a discussion of vibrations in molecules of this type, see B. L. Crawford, Jr., and E. B. Wilson, Jr., J. Chern. Phys. 9, 323 (1941). C=th2/ (1. I a), F= th 2I./[I ,,(1.I a)]. (4) V (0:) is the potential energy restricting the internal rotation. In the model used here, the coefficients are all constants. Note that C involves the moment of inertia of the framework part alone while F contains the reduced moment of the two parts of the molecule. The quantities P x , P y , and p. were defined in Eq. (2) but by using the basic definition of angular momentum • B. L. Crawford, Jr., J. Chern. Phys. 8, 273 (1940). /; See K. S. Pitzer and W. D. Gwinn, J. Chern. Phys. 10, 428 (1942). 136 This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34 CALCULATION OF ENERGY LEVELS, it is easy to show that they are equal to the components of the total angular momentum of the molecule (in units h/27r), including the contributions arising from any internal rotation of the top. Similarly it can be shown that p is the total contribution of the motion of the top atoms to the z component of angular momentum, including both the internal and overall motions. This classical Hamiltonian becomes a quantummechanical operator or alternately a matrix by regarding P x, P y, P z , and p as operators or matrices. The commutation rules 6 are PjPkPkPj=iPl, (j,k,l=x,y,zincyclicorder) (5) and (6) The first rule is the standard one for the components of angular momenta along moleculefixed axes. The second follows from the fact that, as an operator, p= i(a/aa)e, '1', (7) x whereas the P;'s do not involve a. By inserting in H the expressions for the Pi in terms of pe, P'l" and Px and in turn treating these latter momenta as differential operators, the Schrodinger wave equation in the variables fJ, cp, x, a could be written down, but this is not necessary for the solution of the problem. To simplify the later equations, let X'=[HD(P x2+P,l+P z2)]/(CD) (8) D=HA+B). (9) X'=b(PxLPi)+PldpP z+ jp2+ V' (a) (to) in which Then with b=HAB)/(CD) (11) d=2C/(CD) j=F/(CD) V'= V ICCD). Then if A' is an eigenvalue of X', the energy is given by W=1(J+1)D+ (CD)A/. x', as well as H, is diagonal in 1 and M (12) so only one 1M block need be considered at a time. TABLE E C. Cu C, p,>p, Pz>Pz p.,.>p., , p>p (a>a) p.>p. Pz,.>Pz,. p>p (a>a) P.>p. Px,u>Pz,. P>P (a>a) (IJ>7rIJ) ( Cp>7r+ "') (x>27rx) (IJ>7rIJ) ( cp>1r+ cp ) (X>7rx) P>P (a>a) (IJ>8) (cp>cp ) (x>x) 6 I. Effect of fourgroup symmetry operations. O. Klein, Z. Physik 58, 730 (1929). (IJ>8) ( cp>cp) (x>7r+x) ROTATION AND TORSION TABLE II. Character table for case s=3. Class E 2C. A 1 1 1 1 1 1 1 1 1 1 0 2 1 1 1 1 1 1 1 1 1 0 2 B. E, 2 Bx B. E, 1 1 2 3C,C. C, 2C,C. 1 1 1 137 3CuC. 1 1 0 1 1 0 SYMMETRY CONSIDERATIONS It is well known that the rigid asymmetric rotor wave equation is invariant under the group of 180 rotations about each of the principal axes. These operations variously change the signs of the Pi, which leaves the energy expression unaltered. In the present problem the crossterm in pp, is not invariant unless p changes sign when P z does. If the potential energy V (a) is an even function of a, then the four operations E, Cx, Cy, C. whose effects are in Table I will still leave H invariant and will form a group (the "fourgroup") as before. Consequently every solution of the wave equation will belong to one of the four species A, B x , By, Bz of this group7 as in the case of the asymmetric rotor, but here the part of the function involving a must be included. Further, any secular equation or energy matrix can be factored into at least four factors by using the foregoing symmetry. Often V (a) will possess additional symmetry properties. For example, in nitromethane, it presumably is invariant under a7a+27rk/6, where k=O, 1, 2,3,4, S. In addition, a7(27rk/6)a should also leave V unchanged. These operations are isomorphous. with the point group C 6v but this group is not here used in the same way as when group theory is applied to the vibration problem. The symmetry of V(a) cannot be indiscriminately applied to H because p appears in the cross term. However, the "rotations" a7a+ 27rk/ s (here s= 6) do leave H invariant. So do the "reflections" a7(27rk/s)a if simultaneously P ,7 P ,. The fourgroup operations listed earlier and the s internal rotations a7a+ 27rk/ s generate a group of 4s operations. Table II shows the result when s= 3, in which case the group is isomorphous with the group C6v • The onedimensional species are here labeled so that they correspond to the species for the subgroup V. The degenerate species El would become A B" in the subgroup while E2 would yield Bx+By. The case s= 6 is easily constructed (see Fig, 2) from Table II because the new group is obtained by introducing the additional operation C63 which causes a7a+7r. This commutes with all foregoing operations so the new group has twice as many classes and twice as many species as the old (see Table III). 0 + 7 See King, Hainer, and Cross, J. Chern. Phys. 11, 27 (1943). This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34 138 WILSON, LIN, AND LI DE TABLE E 2C. 3C.C. 1 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 0 0 0 0 C, III. Character table for case s = 6. c.' 2C,C. 3C.C. 1 1 2 1 1 1 1 1 0 1 1 2 1 1 2 1 1 1 1 1 1 1 0 2 1 1 1 2 1 1 1 1 1 0 1 1 2 1 1 1 1 2 1 1 1 1 1 1 2 1 0 These higher symmetries permit further factoring of the secular equation or energy matrix if the expansion functions are chosen to have symmetries in accord with the various species. The dipole moment fJ. for this model would ordinarily be along the z axis. If this is the case, fJ. will have the symmetry B. in Table II and in the case s= 6 it will belong to B.e. Consequently, for this case, the selection rules for dipole absorption will be (s= 3): A~Bz, B",~By, E1~E1, E2~E2. The rules will apply for the case s=6 with the additional condition that e species go only into e species, and 0 species into 0 species. Nuclear permutation effects will be governed by the symmetry also. The subgroup which governs the exchange of the oxygen atoms in nitromethane or the fluorines in CHaBF2 consists of E and C.Csa since to exchange these atoms requires that a~a+ll', X~X+ll', which is CzCi. A e, B ze , E le , B",o, B yo, E 20 are even, the other species odd to this operation. Therefore only the do dilo d~o d~o d~o b=o bJlo b=o V=o vo blO Vo dlo billo dlo b=o V»o V»O Vlo V· 0 baD czea3 2C,C.' 3C.C.C.' 1 1 1 1 1 1 0 2 1 1 1 1 0 1 1 2 I 1 +1 1 0 1 1 1 1 1 0 2C,C.C.' 2C.C.C.' 1 1 1 1 1 1 1 1 1 0 1 1 2 1 1 +1 1 1 1 1 1 1 2 1 0 1 0 1 1 0 first set can occur in nitromethane (with ISO). In CHaBF2 these species will have onethird the fluorine spin weight of the others. The operations C 3, C3\ exchange two pairs of H atoms in the foregoing molecules. All the nondegenerate species are of species A in this subgroup, all the degenerate species will be degenerate in the subgroup. The two kinds of levels will then have equal proton spin weight. THE ENERGY MATRIX If the asymmetry (b) were zero and there were no barrier (V'), the reduced Hamiltonian X' would be PldpP z + jp2 (13) which yields a diagonal matrix with the basis functions S.T K .If (0, 'P )e iK xeim.a in which S is the 0, wave function and 'P (14) factor of the symmetric rotor K=O,±I,±2,···±J; m=O,±I,±2,···<Xl. (15) These functions may be used to set up a matrix for the general form of X', i.e., the true wave functions may be expanded in terms of the foregoing functions. The asymmetry term in b will give offdiagonal elements the same as those which occur with the rigid asymmetric rotor (with C calculated for framework only), The barrier potential V' is usually assumed to be of the form V' =! V o'(1 cossa)=! Vo't Vo'(ei''''+e i ,,,,) (16) in which s is the number of equivalent minima and (CD)V o'= Vo is the barrier height. The barrier will also introduce offdiagonal elements. The matrix for X' will then have the nonvanishing elements8 XKm,Km=X'Km,Km!V o'=K2dmK+jm 2 XK, m; K±2, m = bK, K±2 =!b{[J2 (K±1)2J[(J+l)L (K±I)2J}! FIG. 2. Correlation of energy levels corresponding to various limiting cases as specified at the head of each column, for even values of K for the case s=6. (17) XK,m;K,m±s=tVo'=~ 8 See the following for the matrix elements of P z , P u, and P •. This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34 CALCULATION OF ENERGY LEVELS, Here the constant! V o' has been incorporated so that the eigenvalues of Je are related to those of Je' by Unless the barrier height is zero, this form of Je corresponds to an infinite secular equation. Note, however, that there are no elements connecting even with odd K values so that the secular equation factors into one for even and one for odd K values. (This is part of the factoring into different symmetry species.) CASE OF FREE INTERNAL ROTATIONS. o o o 6VJb 4+2md;>.. (210)!b o o (KIK±2)=bK ,K±2 =!b{[J2 (K±1)2J[(J+l)2 (K±1)2J}!. The energy W is then related to the roots equation (see Eq. 12). ;>.. (19) by the W=J(J+l)D+Fm 2+(CD)X. (20) This secular equation is similar to the Wang equation for the ordinary rigid asymmetric rotor except for the added term  dmK on the diagonal. This term spoils the additional factoring possible in the rigid case. As an example, consider the case J = 5, K even. This factor becomes o o o X (21O)!b 42mdX 6VJb (210)!b o (22) as for the rigid asymmetric rotor. The symmetry requirements discussed earlier must also be met. If p, is along the z axis, K cannot change parity. The rule for m is t.J.m=O. (KIK)=K2dmK;>..; (210)!b For each nonvanishing value of Im I, there will be two identical such equations, so all levels are doubly degenerate unless m=O. In the latter case, the ordinary rigid rotor levels are obtained, except that the moment of inertia about the z axis is the moment of the framework group (I z  I a) as already seen. The energy levels in the free rotation case must of course conform to the symmetry restrictions given for the general case. Thus, with s=3 or 6, if m is not a multiple of 3, the symmetry is one of the degenerate E species. If m is a multiple of 3, the levels involve A or B species which would be split into nondegenerate components if there were a sufficient barrier. If K is even, the species may be A, B z , or E 1 ; if K is odd, B x , By, or E 2 • If s= 6, the species further divide into even or odd (denoted by subscripts e and 0) according as m is even or odd. The selection rules for dipole radiation involve t.J.J=O, ±1 139 TORSION 9 When the barrier height is zero, ~ =  t V o' = 0 and the energy matrix becomes diagonal in m. The secular equation therefore factors into one block for each value of m, as well as into even and odd K factors. The secular equation for a given value of J, M, m has 16+4md;>.. 6VJb AND the elements (1S) A'=A+!Vo'. ROTATION (23) This follows from the fact that the dipole moment does not depend on a whereas the wave function involves a only through the factor o o o =0. (21) 6VJb 164mdX For small asymmetry the energy levels may be expanded in powers of b. For large values of the quantum number n, the diagonal elements of the secular equation will be large and the first terms in the expansion in powers of b may be adequate. Secondorder perturbation theory then gives X=KL mdK+(b 2/S){[J2(KIFJ. X [(J + l)L (K 1)2J/ (2K md 2)}  (b 2/S) {[J2 (K+l)2J X[(J+l)2 (K+IFJ/(2Kmd+2)} + .... (24) For very large values of m, X?K2mdK. (25) For transitions with t.J.J=l, t.J.K=O, t.J.m=O, the transitions with large values of m would then tend to converge (from both sides) to a band head. Of course the intensities of these higher members would tend to decrease because of the unfavorable Boltzmann factors. CASE OF LOW BARRIER If the barrier height Vo is small but not negligible (compared with the quantity F), a useful solution can be obtained with the Van Vleck perturbation method.lO The reduced Hamiltonian matrix Je of Eq. (17) is split into an unperturbed, completely diagonal matrix JeO with diagonal elements (26) S. C. Wang, Phys. Rev. 34, 243 (1929). See E. C. Kemble, The Fundamental Principles of Quantum Mechanics (McGrawHill Book Company, Inc., New York, 1937), p.394. 9 10 8. This case was treated by H. B. G. Casimir, Z. Physik 59, 623 (1930). This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34 140 WILSON, LIN, and a perturbation X~ with elements AND LIDE Am=O is no longer exact and must be replaced by the rigorous symmetry selection rules given earlier. X~K,m;K±2,m=bK,K±2 HIGH BARRIERS X~K,m;K m±8=t= tVo'= tVo/(CD). The Van Vleck transformation reduces the elements offdiagonal in m to order t2. If these are then neglected, the secular equation becomes (KJK)=K2dmK + (l/s)t 2 { (1/[Kd (2m+s)j])  (1/[Kd (2ms)j])}A (28) If the barrier is sufficiently high so that the torsional levels of a given symmetry are widely spaced compared with the rotational levels, a different perturbation treatment is possible, in which the asymmetry and the coupling term are treated as perturbations by the Van Vleck procedure. This leads as before to a rotational secular equation. The unperturbed operator for the energy is H o=DF2+ (CD)P z2+Fp2+ V (a) (30) while the perturbation operator is Hl=!(AB) (P:r?Pl)2CpP z • (K K±2) = bK, K±2. J (31) H 0 can be diagonalized with the basis functions This secular equation is similar in its properties to that for free rotation. It is not valid, however, for the case m = ±!s, s even, because then a near degeneracy occurs. For this case, the unperturbed states with m= ±!s must both be taken into consideration. The equation then has the nonvanishing elements (;,Ki;,K) =K2_!dsK + (1/s) [t2/(Kd 2sf)]A (;,KI;,K )=K2+!dSK  {l/s)[t 2/(Kd+2sf)]A (29) (;,KI~,'K )=t (;,KI;,K±2) =h,K±2 ( ;,KI;,K±2)=b K,K±2' Because of the symmetry (s/2,K J ±s/2,K') = ( s/2,  K J =Fs/2,  K') this larger secular equation can be factored into two factors, similar to the way in which the Wang equation is factored. This leads to a splitting of the degeneracy. When m is not a multiple of !s, the levels are inherently doubly degenerate for all barriers. For m a multiple of !s the degeneracy is ultimately split at high enough barriers. With the approximation given here this splitting will not appear except for m=!s. The higher the multiple of ts the higher the order of perturbation required to demonstrate the splitting. The quantum number m is no longer a good quantum number when there is a barrier but if Vo is small it can still be used to label the levels. The selection rule (32) in which symmetric rotor functions are used as before and the U's are eigenfunctions of [Fp2+ V(a)]Uv.=Ev.U v ., (33) The torsional states are described by the quantum numbers VK, where v is the principal quantum number of the vibrational level, and K is a degeneracy index. For the potential function of Eq. (16), the nondegenerate eigenvalues of the torsional Eq. (33), which is now related to the Mathieu equation, can be obtained from publishedll tabulations through the connection (34) in which s= (4/ s2)(Vo/F) (35) and the eigenvalues be r and bO r are given in tables as functions of the parameter S. The values of the quantum number v=O, 1, 2, etc. are identified respectively with the eigenvalues beo, bo l ; bel, bo 2; be a, etc. The degenerate levels have not been tabulated but can be calculated to any desired degree of accuracy by expansion of the eigenfunction in a Fourier series, However, another method is easier and is reasonably accurate for high barriers. Make the substitution (37) in Eq. (33), where K=!s+1, !s+2, ... , !s. The equation for u is then [Fp2+2FKP+FK 2+ VJu=Ev.u. (38) If K= ±1, ±2, ... , ±ts1, the degenerate levels are obtained. But the terms involving K in Eq. (38) can be considered as perturbations on the K=O case, if the barrier is high enough. The term in K2 obviously just subtracts from Ev •. To the second order the effect of 11 Tables Relating to Mathie14 Functions (Columbia University Press, New York, 1951). This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34 CALCULATION OF ENERGY LEVELS, the terms in " and ,c. is Ev.=Evo+F,,2+4F2,,2L'ip.v,i2/(EvoE.,o) (39) .' in which the sum is over all the nondegenerate levels and 2r p••' =  i f° Uvo*(iJU.'ojaa)da Here we have expanded the nondegenerate eigenfunctions as m=tXl These Fourier coefficients may be found from the tabulations of reference 11 by the identification k,=O for even v, Av. 0= 2Deo(S) A V.  8k!8= Av. 8kHs= (I)k+IDo 2k+1 (5) (even v), A V .  8k!s=  Av.skt.= (I)k+IDe2k+1 (5) (K i K) = K2 d'"K  A" This result can be shown to follow from Eq. (41) and (42), since 4Fptl (45) for high barriers in which harmonic oscillator functions can be used for U vo in Eq. (40). The energy levels of rotation are then calculable by the standard methods used for the rigid asymmetric rotor. The torsional levels will be degenerate, threefold if n=3, sixfold if n=6. However, some of this degeneracy may be eliminated when permutation symmetry occurs, as in nitromethane, whose levels will be only threefold degenerate. In cases for which the approximations so far described are all inadequate, another approach can be used l2 if the asymmetry of the molecule is small. Let the reduced energy matlix x' of Eq. (10) be set up in terms of the basis functions (46) in which 5 is a symmetric rotor function and /'vK(a) is a periodic function of a which is an eigenfunction of the equation (47) with eigenvalue WvK. K enters as a parameter. The nonvanishing matrix elements for X' then become X'vK.vK=K2+W vK where (49) which enters because /,'s for different K's are not orthogonal. The existence of the terms offdiagonal in v spoils the factoring into v blocks but these terms are normally small and can be reduced to lower order by a Van Vleck transformation. This yields a rotational secular equation X[(J+l)L (K±I)2]}l W =J(J + I)D+ Evo+ F',,2+A"(C' D) C'=C(I+4Cp) d' = 2C(1 +4Fp)/ (C'  D) (KiK)=K2+W vK A' + ibK.K+2i 2L'·i ~ •. K;v'.K+2i2/ (42) v' v' (WvK  W.'K+2) 2 + i bK.K_2i L' i ~v.K;v'.K_2i2/ b'=tcAB)/(C'D). .' VERY HIGH BARRIERS If the barrier V is sufficiently high, the lower energy levels should approximate those of a rigid rotor with moments of inertia I x, I y, I z (not I z I a) and a harmonic torsional vibrator with a reduced moment of inertia (43) (Ki K±2)=bK. K±2~v.K;v. K±2 (K i K±4) = hK. K±2hK±2.K±4 XL' and potential energy v' Hd 2 V /da 2)oa2 • (48) X'v. K; v'. K±2 = bK. K±2~v. K;v'. K±2 (41) (KiK±2)=W{[J2 (K±I)2] F'=F(I+4Fp) p=L'iPvv,i 2/(EvoEv'0) 141 TORSION (odd v). These coefficients require normalization. The perturbation HI of Eq. (31) now needs to be introduced by the Van Vleck procedure. However, for the degenerate torsional states, it is convenient to combine this perturbation with that given above, yielding the secular equation with AND CASE OF SMALL ASYMMETRY '" Avmeima . Uvo(a)= L A •.  sk = Av. sk = (I)kDe2k(S), ROTATION (44) 12 ~v. K;v'. K±2~.', K±2;v. K±4 W vK  W v 'K±2 . This is very close to that introduced by Dennison, reference 1. This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34 142 WILSON, LIN, AND LIDE Koehler and Dennison13 have shown that the quantities W~K can be obtained from a continued fraction of the form 111 ~=Mo M_l~M_2~M_3~" . (51) 1 1 in which 4s2F 1 1=0, ±1, ±2, ... M z=(1g)2, Vo g=[(CK/F)K)]/S, For each root sought, there will be an appropriate choice of the leading term M I which will cause the most rapid convergence. The various values of K yield the different component levels which in the limit of high barriers come together to form a (degenerate) harmonic oscillator level. However, Koehler and Dennison13 showed that the roots are periodic functions of K [as may also be seen from Eq. (51)J with period Fs/C. Furthermore, if this periodic function is known for the case K=O, it can be used to calculate the roots for the other values of K by a "phase shift," i.e., by reading off the value of ~ for K'=KFK/C K=O, 1,2, .. ·s1 (52) WnK=!VO'_[C2K2/F(CD)]+tVo'~. To calculate the .1's continued fractions may also be used. The function 'YnK is expanded in a Fo "rier series: from the same periodic curve used for K=O. The dependence of ~ on K and K vanishes for sufficiently high barriers. This suggests expanding ~ as a Fourier series (only cosine terms will appear) ~= ~O+~l COS211'g+~2 Q() 'YnK= L AnK.mei""". This is substituted in the differential Eq. (47) and leads to a set of simultaneous equations for the coefficients A : where These simultaneous equations yield the continued fraction 1 1 AnK.m 1 cos41l'g+' . '. For high enough barriers only a few terms are required. The coefficients ~o, ~1, ~2 can be found by solving the continued fraction for the necessary number of values of g, conveniently chosen to make it easy to solve for the required number of es. The value of ~o can also be obtained from available tablesll by means of the relation: s2F ~=berCS)2 Vo or s2F bo r CS)2 Vo The normalization condition with S=4V o' / s2f=4V o/ s2F. Then the .1's may be written m=Q() enables the determination of the A's to be completed. There will of course be an infinite number of roots for each value of K and K. These roots are most conveniently calculated if none of the denominators Mz~'" are small. If one of these should turn out to be small, the continued fraction should be transformed so that the offending Mz appears in the leading position; i.e., 1 1 ~=Mz Mll~Ml2~" 1 (53) 1 Ml+l~Ml+2~" 13 . '. J. S. Koehler and D. M. Dennison, Phys. Rev. 57, 1006 (1940). .1 n 'K.'nK=211' L'" An'K'.mAnK.m. m=oo In some cases, especially when b is sufficiently low, the CK IK±4) elements of the secular Eq. (50) can be neglected altogether. However, this procedure is not always valid. Under such circumstances the secular equation can be made easier to solve by various transformations which reduce the order of the troublesome K,K±4 elements so that they can be ignored and the remaining equation treated by the continued fraction method. These transformations are similar to Van Vleck transformations but have to be adapted to the several cases which arise because of neardegeneracies and will therefore not be detailed here. This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 199.212.66.33 On: Thu, 27 Nov 2014 12:31:34