Molecular vibrations and rotations
Prof. dr. Ad van der Avoird
January 28, 2010 Contents
0.1 Books ...... 2 0.2 Introduction...... 2 0.3 Vibrations of polyatomic molecules (harmonic) ...... 3 0.3.1 Harmonic oscillator in one dimension x ...... 3 0.3.2 Twocoupledoscillators...... 5 0.3.3 Harmonic vibrations of molecules in mass-weighted co- ordinates...... 8 0.3.4 Harmonic vibrations of molecules in cartesian (displace- ment)coordinates...... 12 0.3.5 Invariance conditions on the matrix F ...... 14 0.3.6 Vibrations in internal coordinates (3n-6) ...... 18 0.4 Rotations of rigid (non-linear) molecules ...... 21 0.4.1 Rotation matrices and infinitesimal rotations ...... 22 0.4.2 The rigid rotor in classical mechanics ...... 24 0.4.3 TherigidrotorHamiltonian ...... 31 0.4.4 The rigid rotor in quantum mechanics ...... 34 0.5 The molecular vibration-rotation Hamiltonian ...... 38 0.5.1 Introduction...... 38 0.5.2 The (classical) kinetic energy of a semi-rigid molecule . 39 0.5.3 Semi-rigid molecules in quantum mechanics ...... 44 0.5.4 Floppy molecules and Van der Waals molecules . . . . 46
1 0.1 Books
1. Molecular vibrations, E.B. Wilson, J.C. Decius, and P.C. Cross, McGraw- Hill, New York (1955)
2. Vibrational states, S. Califano, Wiley, London (1976)
3. Molecular Vibrational-Rotational spectra, D. Papousek and M.R. Aliev, Elsevier, Amsterdam (1982)
4. (Angular momentum in quantum physics, L.C. Biedenharn and J.D. Louck, Addison-Wesley, Reading (1981))
0.2 Introduction
Nuclear motion — Born-Oppenheimer approximation, second step
For n nuclei with coordinates rk,k = 1,...,n the Schr¨odinger equation for the nuclear motion problem is given by: ˆ H(r1,...,rn)Ψ(r1,...,rn)= EΨ(r1,...,rn) (1) with the Hamiltonian Hˆ consisting of the kinetic and potential energy oper- ators: ˆ ˆ ˆ H(r1,...,rn)= T(r1,...,rn)+ V(r1,...,rn). (2)
The potential energy V (r1,...,rn) can be obtained from the electronic Schr¨odinger equation (Born-Oppenheimer approximation, first step). It is an energy eigenvalue of the electronic Hamiltonian with coordinates of the nuclei fixed at rk,k =1,...,n. Variation of the nuclear coordinates rk yields the poten- tial energy surface V (r1,...,rn). Solution of the nuclear motion Schr¨odinger equation for molecules translations, rotations, vibrations →
Spectroscopy: microwave, submilimeter, IR / Raman
0 30 cm−1 rotations − 100 5000 cm−1 vibrations −
2 Lasers very high resolution (in principle) (7 significant→ digits)
Doppler and collision broadening occurring in the gas phase are avoided in molecular beams (“supersonic nozzle beams”: very low T)
Globally:
rotational spectrum molecular structure → vibrational spectrum intramolecular forces (“force field”) associated with chemical bonds → spectra of van der Waals molecules intermolecular (noncovalent) → forces
0.3 Vibrations of polyatomic molecules (harmonic)
0.3.1 Harmonic oscillator in one dimension x
Momentum operator, in atomic units (¯h = 1):
1 d ˆp = (3) x i dx The Hamiltonian of a harmonic oscillator is: ˆp2 1 1 d2 1 H=ˆ T+ˆ V=ˆ x + f ˆx2 = + f ˆx2, (4) 2m 2 −2m dx2 2 d2V dV where f = is the force constant. It is assumed that = 0 (potential dx2 dx minimum) and that higher than second derivatives of V can be neglected. It is convenient to introduce mass-weighted coordinates:
q = √mx (5) 1 d 1 d ˆp= = (6) i√m dx i dq
3 and write: 1 1 H=ˆ ˆp2 + λ ˆq2 (7) 2 2 f where λ = m The frequency of the harmonic oscillator is ω0 = √λ. 1 Eigenvalues are the energies E = ω (v + ) v =0, 1, 2,... (8) v 0 2 Eigenfunctions are Ψ (q)= H (√ω q) exp 1 ω q2 , (9) v v 0 − 2 0 where Hv are Hermite polynomials.
4 0.3.2 Two coupled oscillators
Example 1:
Figure 1: Linear CO2 with two internal coordinates.
Example 2:
Figure 2: Diatomic molecule in laboratory coordinates.
2 2 ˆ ˆp1 ˆp2 1 2 1 2 H= + + f11 ˆq1 + f22 ˆq2 +f12 ˆq1 ˆq2 (10) 2m1 2m2 2 2
Let us assume symmetry, so that m1 = m2 = m and f11 = f22. Define new coordinates:
Q1 =(q1 + q2)/√2 q1 =(Q1 + Q2)/√2 (11) Q =(q q )/√2 q =(Q Q )/√2 (12) 2 1 − 2 2 1 − 2 The associated momentum operators are: 1 ∂ 1 ∂q ∂ ∂q ∂ Pˆ = = 1 + 2 = (ˆp + ˆp )/√2 (13) 1 i ∂Q i ∂Q ∂q ∂Q ∂q 1 2 1 1 1 1 2 ˆ 1 ∂ P2 = = = (ˆp1 ˆp2)/√2 (14) i ∂Q2 −
5 Hence:
1 ∂ 1 ∂Q1 ∂ ∂Q2 ∂ ˆ ˆ ˆp1 = = + = P1 + P2 /√2 (15) i ∂q1 i ∂q1 ∂Q1 ∂q1 ∂Q2 1 ∂ ˆ ˆ ˆp2 = = = P1 P2 /√2 (16) i ∂q2 − Substitution of these new coordinates gives:
1 2 2 1 2 H=ˆ Pˆ + Pˆ + Pˆ Pˆ + f Qˆ + Qˆ + 4m 1 2 1 − 2 4 11 1 2 (17) 1 1 2 + f Qˆ + Qˆ Qˆ Qˆ + f Qˆ Qˆ 2 12 1 2 1 − 2 4 22 1 − 2 1 ˆ 2 ˆ 2 1 ˆ 2 = P1 + P2 + (f11 + f22 +2f12) Q1 + 2m 4 (18) 1 2 1 + (f + f 2f ) Qˆ + (f f ) Qˆ Qˆ 4 11 22 − 12 2 2 11 − 22 1 2
The last term vanishes because f11 = f22 and we get:
1 2 1 2 1 2 1 2 H=ˆ Pˆ + (f + f ) Qˆ + Pˆ + (f f ) Qˆ = Hˆ + Hˆ . (19) 2m 1 2 11 12 1 2m 2 2 11 − 12 2 1 2 As one can see, our system separates into two uncoupled oscillators, with:
coordinates Q1 =(q1 + q2)/√2 Q2 =(q1 q2)/√2 (20) 1 − 1 f + f /2 f f /2 frequencies ω = 11 12 ω = 11 − 12 (21) 1 m 2 m The energy of the system is:
1 1 E = Ev1 + Ev2 = ω1 v1 + 2 + ω2 v2 + 2 (22)
6 Example 1: CO2
Figure 3: Q1 and Q2 for CO2.
Q1 =(q1 + q2)/√2 — symmetric stretch vibration Q =(q q )/√2 — asymmetric stretch vibration 2 1 − 2
Q1 and Q2 are called “normal mode” coordinates
Example 2: Diatomic molecule
Figure 4: Q1 and Q2 for a diatomic molecule.
Q1 =(q1 + q2)/√2 — overall translation (center of mass motion) Q =(q q )/√2 — stretch vibration (relative motion) 2 1 − 2
In Example 2 the molecule (center of mass) freely moves in space, the en- ergy does not depend on Q1, and we get ω1 = 0. This corresponds to the invariance condition:
f11 + f12 (= f12 + f22) =0. (23)
7 For the stretch vibration Q =(q q )/√2 we find: 2 1 − 2 1 1 1 f f /2 2f /2 f /2 ω = 11 − 12 = 11 = 11 , (24) 2 m m µ m where µ = is the reduced mass. 2 Conclusion: Both examples show that the simple transformation Q = (q q )/√2 1,2 1 ± 2 separates the potential energy, while the kinetic energy remains separated. As a result we get two uncoupled harmonic oscillators.
0.3.3 Harmonic vibrations of molecules in mass-weighted coordinates
Coordinates of atoms:
rk = ak + dk (k =1,...,n) (25) where ak denotes equilibrium positions and dk stands for displacements. The components of the coordinates are:
xk akx dkx y = a + d (26) k ky ky zk akz dkz Altogether:
rkα = akα + dkα k =1,...,n; α =1, 2, 3(x, y, z) (27)
ri = ai + di i =1,...... , 3n (28)
Atomic masses: mk, k =1,...,n, or mi, i =1,..., 3n
m1 = m2 = m3 m = m = m 4 5 6 etc. Mass-weighted coordinates (for displacements):
qi = √midi (29) Kinetic energy Classical expression:
1 2 1 1 T = m d˙ = q˙2 = p2 (30) 2 i i 2 i 2 i i i i X X X 8 Quantum expression:
1 1 ∂2 T=ˆ ˆp2 = (31) 2 i −2 ∂q2 i i i X X Potential energy:
∂V 1 ∂2V V = V + q + q q + . . . (32) 0 ∂q i 2 ∂q ∂q i j i i 0 i j i j 0 X X X 1 = V + f q + f q q + . . . (33) 0 i i 2 ij i j i i j X X X Forces in equilibrium position are equal to zero:
∂V f = =0 forall i =1,..., 3n (34) i ∂q i 0
V0 is a constant (we will omit it). We use the harmonic approximation: 1 V = f q q (35) 2 ij i j i,j X Matrix - vector notation:
q1 p1 . . f ... f . . 11 1,3n . .. . q = qi p = pi f = . . . (36) . . . . f3n,1 ... f3n,3n q p 3n 3n We use column-vectors, f is the force constant matrix. Now the Hamiltonian may be written: 1 1 H = pT p + qT f q (37) 2 2 We transform the coordinates as follows:
Q = L−1q q = L Q (38)
For the momenta we get:
−1 P = LT p p = LT P (39) 9 Proof (comment by PG): we know that qk = LkmQm. For coordinate Qj the conjugate momentum is (we use the Einstein’s summation convention and the chain rule) 1 ∂ 1 ∂q ∂ 1 ∂ ∂ Pˆ = = k = (L Q ) = j i ∂Q i ∂Q ∂q i ∂Q km m ∂q j j k j k (40) 1 ∂ 1 ∂ = L δ = L = L ˆp = LT ˆp km jm kj kj k jk k i ∂qk i ∂qk Substitution of the new coordinates and momenta into H yields: 1 1 H = P T L−1 L−1 T P + QT LT f L Q (41) 2 2 Let us choose L such that it is real and:
L−1 L−1 T = E (42)
Where E is the identity matrix. Taking the inverse, it follows that:
LT L = E (43)
And for non-singular L it also follows that:
LT = L−1 (44) which means that L is an orthogonal (and unitary, since it is real) matrix. We may further choose L such that:
LT f L =Λ (45)
λ1 0 .. where Λ = . is a diagonal matrix. Multiplying by L gives 0 λ3n f L = L Λ (46)
th For i column of L, denoted by Li, we have:
f Li = Liλi (47) which shows that Li is an eigenvector of f with the eigenvalue λi. Thus to find L we have to diagonalize f, i.e., to find eigenvectors and eigenvalues of f.
10 Substitution of these results into H yields: 1 1 1 1 1 H = P T P + QT Λ Q = P 2 + λ Q2 = P 2 + λ Q2 (48) 2 2 2 i 2 i i 2 i i i i i i X X X In other words the transformation q Q, p P gave us 3n uncoupled harmonic oscillators. The frequencies are→ →
ωi = λi (49) p The Qi are called “normal coordinates”:
Q = L−1 q = L q (50) i ij j ji j j j X X or: Q = L−1q = LT q (51)
We used mass-weighted coordinates qi = √midi, and we can write:
m1 0 1 m /2 2 q = M d with M = . (52) .. 0 m3n 1 1 /2 /2 M has mi on the diagonal. The normal coordinates are then given in terms of atomic displacements by:
1 Q = LT M /2 d (53)
Example: H2O normal modes.
Figure 5: Normal modes of water molecule.
Q — symmetric stretch vibration · 1 11 Q — asymmetric stretch vibration · 2 Q — bending vibration · 3 + 3 overall translations (frequencies = 0)
+ 3 overall rotations (frequencies = 0)
0.3.4 Harmonic vibrations of molecules in cartesian (displacement) coordinates d — coordinates, F — force constants
∂2V 1 ∂2V 1 1 1 F = = m /2 m /2 = m /2 f m /2 (54) ij ∂d ∂d i ∂q ∂q j i ij j i j 0 i j 0 1 1 1 1 F = M /2 f M /2 f = M − /2 F M − /2 (55) The Hamiltonian of the system in simple cartesian coordinates (d and p , d 1 1 ∂ 1 /2 ∂ pd,i = = mi ) is: i ∂di i ∂qi
1 1 (p )2 1 H = pT M −1p + dT F d = d i + F d d (56) 2 d d 2 2m 2 ij i j i i i,j X X To uncouple H we return to normal coordinates Q:
− Q = L 1d d = L Q (57)
−1 p = L T P (58) d Applying this to H:
1 − − T 1 H = P T L 1M −1 L 1 P + QT L T F L Q (59) 2 2 which becomes: 1 1 H = P T P + QT Λ Q (60) 2 2 if:
L T M L = E (61)
12 and
L T F L =Λ (62)
From the last equation we get:
−1 F L = L T Λ (63)
Substituting the previous one gives:
F L = M L Λ (64) so L must satisfy a generalized eigenvalue problem of F with M as the metric (or “overlap”) matrix. 1 1 From the equation F = M /2 f M /2 follows directly the relation between the eigenvectors L and L :
1 1 L = M − /2 L L = M /2 L (65)
1 (It can be also easily seen from q = L Q, d = L Q and q = M /2 d.) Because L−1 = LT we have:
− L 1 = L T M (66) and the normal coordinates are given in terms of atomic displacements by:
− Q = L 1d = L T M d (67) or: L T L Qi = i M d = jimjdj (68) i X L th L where i is the i column of . Remember:
the index i runs over eigenvalues λ and the associated normal coordi- • i nates Qi the index j = kα, k =1,...,n, α =1, 2, 3 (x, y, z) • Thus
Qi = Lkα,imkdkα (69) α Xk X
13 0.3.5 Invariance conditions on the matrix F
We will now derive invariance conditions which show that six eigenvalues λi of the matrix F are equal to zero. The associated normal coordinates correspond to overall translations and rotations of the whole molecule. ∂V 1 ∂2V V = V + d + d d = 0 ∂d i 2 ∂d ∂d i j i i 0 i j i j 0 X X X n ∂V 1 n (70) = V + d + F d d 0 ∂d kα 2 kα,lβ kα lβ α=1,2,3 kα 0 Xk=1 X k,lX=1 Xα,β =0 The force acting on atom k in| the{z direction} α = x, y, z is: ∂V ∂V ∂V n = = + Fkα,lβdlβ (71) ∂di ∂dkα ∂dkα 0 Xl=1 Xβ =0 Translational invariance: this| force{z should} not change if we move all atoms by the same distance: dlβ = δβ for l =1,...,n, hence n n
0= Fkα,lβδβ = δβ Fkα,lβ (72) Xl=1 Xβ Xβ Xl=1 As this should apply to any δ we must require that
n
Fkα,lβ =0 (73) Xl=1 for all k =1, . . . ,n; α =1, 2, 3; β =1, 2, 3(x, y, z) This is the first set of invariance conditions. Note that summation here is over l, with fixed k, α and β. Different k, α and β give different conditions, so indeed we have here a set of invariance conditions. Equation (64) (F L = M L Λ) written in components is:
FijLjr = miLijΛjr = miLirλr (74) j j X X th The matrix elements Ljr for j =1,..., 3n form the r column of L i.e. the th L r eigenvector r n i =(k, α) F L = m L λ (75) j =(l, β) kα,lβ lβ,r k kα,r r Xl=1 Xβ 14 Now suppose that
1 forall l =1,...,n and for β =1 Llβ,r = (76) (0 for β =2, 3
It follows immediately from the invariance condition, eq. (73) that the left hand side of equation (75) equals 0. Thus also the right hand side must be L equal to 0 and, since not all components of r are equal to zero, λr must be equal to zero. So we have found one eigenvalue λr = 0 and the associated L eigenvector r with components = 1 for β = 1 and = 0 for β = 2, 3. ′ ′′ L Similarly we find two more eigenvalues λr = λr = 0 with eigenvectors r′ L and r′′ that have components in the directions β = 2 and β = 3. Let us call these three eigenvalues:
λ1 = λ2 = λ3 =0 (77)
The associated normal coordinates are:
Qr = Lkα,rmkdkα (78) Xkα L Substituting the conditions for r into this general formula leads to:
n
Q1 = mkdk1 k=1 Xn
Q2 = mkdk2 (79) k=1 Xn
Q3 = mkdk3 Xk=1 This can be written in a more compact way:
Q1 n Q = m d (80) 2 k k Q k=1 3 X These are the normal coordinates for the overall translations of the molecule (λ1 = λ2 = λ3 = 0). In fact, they are the coordinates of the center of mass of the molecule. Rotational invariance
15 We will now see how all the atoms move when we rotate the whole molecule and we will again require that the forces on the atoms do not change during this movement.
We rotate the molecule around the axis eγ, γ =1, 2, 3(= x, y, z) over an angle ∆ϕγ . For very small (infinitesimal) rotations it turns out that a vector r0 changes as follows: r = r + ∆ϕ e r (81) 0 γ γ× 0 (This will be demonstrated in chapter 3.) Atom k has equilibrium position ak. After the rotation of the molecule its position becomes r = a + d = a + ∆ϕ e a (82) k k k k γ γ× k Thus the displacement of the kth atom is: d = ∆ϕ e a (83) k γ γ× k The force acting on atom k in the direction α was zero when the atom was in its equilibrium position, and it must still be zero after we rotated the whole molecule (because the relative positions of the atoms did not change): ∂V 0= = Fkα,lβdlβ (84) ∂dkα Xl Xβ After substituting dlβ:
n 0= F ∆ϕ e a (85) kα,lβ γ γ× l β l=1 β X X As this should apply to any ∆ϕγ for γ =1, 2 and 3 we must require that the following invariance conditions are fulfilled:
n F e a =0 kα,lβ γ× l β β=1,2,3 l=1 (86) X X for k =1,...,n,α =1, 2, 3,γ =1, 2, 3(= x, y, z) L L It follows that the vector r with components kα,r = eγ ak α is an eigen- vector of F with eigenvalue λ = 0. There are three such vectors:× γ =1, 2, 3. r Let us label these eigenvectors with r = 4, 5, 6. Associated normal coordi- nates are: n n Q = m e a d = m e a d (87) γ+3 k γ× k α kα k γ× k · k k=1 α=1,2,3 k=1 X X X 16 The vector product satisfies the relation:
(a b) c = a (b c) (88) × · · × Using this relation we get:
n Q = m e (a d ) (89) γ+3 k γ · k× k Xk=1
The vectors e1, e2 and e3 are unit vectors along the axes of the Cartesian coordinate system:
1 0 0 e = 0 e = 1 e = 0 (90) 1 2 3 0 0 1 As a result we have: Q4 n Q = m a d (91) 5 k k× k Q k=1 6 X These are the normal coordinates of overall rotations of the molecule. The associated eigenvalues are λ4 = λ5 = λ6 = 0.
So far we have found six normal coordinates of the molecule (Q1,...,Q6), that correspond with eigenvalues λ1 = . . . = λ6 = 0. The remaining normal coordinates Qi L L T Qi = mk kα,idkα = i M d (92) α Xk X must be found as the eigenvectors of the matrix F , with “overlap matrix” M. For these eigenvectors we have:
L T M L = E (93) where E is the unit matrix. Thus
L T L i M j = δij (94)
The coordinates Q1, Q2, Q3 describe the translations of (the center of mass of) the molecule, the coordinates Q4, Q5, Q6 describe the overall rotations, the remaining Q are the (3n 6) “real” vibrations. We want to separate i − these “real” vibrations from the rotations and translations. This can be achieved by using a “molecule-fixed” coordinate system that rotates and
17 translates with the molecule. Such a system is defined by requiring that at every moment the following conditions are satisfied:
Q1 n Q = m d =0 2 k k Q3 k=1 X (95) Q4 n Q = m a d =0 5 k k× k Q k=1 6 X These are the so-called Eckhart conditions. (Compare to equations (80) and (91).) Setting Q1, Q2, Q3 equal to zero implies that the origin of the molecule-fixed coordinate system is the center of mass of the molecule. Set- ting Q4, Q5, Q6 equal to zero minimizes the coupling between rotations and vibrations. The latter coupling cannot be completely removed, however, in contrast with the coupling between vibrations and translations. All of this will be explained below.
0.3.6 Vibrations in internal coordinates (3n-6)
1. Bond lengths rkl
2. Bending angles Θkln
3. Out-of-plane wagging Θklmn
4. Torsion angles τklmn
Figure 6: Example: Internal coordinates of water molecule (s1 = ∆rkl, s2 = ∆rkm, s3 = ∆Θklm).
18 The Cartesian coordinates rk are functions of the internal coordinates, and this also applies to the displacements dk. There are 3n Cartesian coordinates dk, for k = 1,...,n, but 3n 6 internal coordinates s. Besides the internal coordinates s, there are also− 3 coordinates that describe the position of the molecule in space (actually the position of the center of mass) and 3 angle coordinates that describe the orientation of the molecule.
The force constants Fij of the molecule in Cartesian coordinates have lit- tle physical meaning. Rather, we know the force constants related to bond lengths, bond angles, etc. — the internal coordinates s — these force con- stants form the matrix F :
∂2V F = (96) ij ∂s ∂s i j The harmonic approximation of the potential in the internal coordinates s is: 1 V = sT F s (97) 2 We need to write also the kinetic energy in terms of the internal coordinates s. To achieve this, we first define the transformation s = B d (98)
d is a column vector of length 3n • s is a column vector of length 3n 6 • − B is a matrix of dimension (3n 6) (3n) • − × If s is a small (infinitesimal) change of the internal coordinates, and d is a small change of the Cartesian displacements, then:
∂si Bij = (99) ∂dj
Example H2O - see Figure 6:
s1 = ∆rkl (100) 1 /2 r = (x x )2 +(y y )2 +(z z )2 (101) kl k − l k − l k − l 1 ∂s1 1 − /2 xk xl = [. . .] 2(xk xl)= − (102) ∂xk 2 · − rkl ∂s x x ∂s 1 = l − k = 1 (103) ∂xl rkl −∂xk
19 These elements, together with derivatives with respect to yk, yl, zk and zl, form the 1st row of the matrix B. The remaining elements of this row are equal to zero.
s3 = ∆Θklm (104) rkl rkm cosΘklm = · (105) rklrkm Differentiation would give us the elements of the third row of the B matrix. In a similar way we compute all the elements of the matrix. For the usual types of coordinates 1 — 4 you can find the derivatives in the books of Wilson, Decius, and Cross and of Califano. All derivatives must be calculated for the equilibrium structure of the molecule. B is not a square matrix, so B−1 is not defined. The remedy is to add to s the 6 coordinates Q1,...,Q6 of translation and rotation of the whole molecule. These also are (already presented) linear combinations of d (see section E). After adding these 6 rows B becomes a square and non-singular matrix, so it is possible to write: d = B−1s (106) For the momenta p , conjugate to s we have: s
T p = B−1 p p = BT p (107) s d d s Now we can write the Hamiltonian of the molecule as follows: 1 1 H = T + V = pT M −1p + dT F d = d d 2 2 (108) 1 1 = pT B M −1BT p + sT F s 2 s s 2 where T F = B−1 F B−1. (109) We define the matrix G : G = B M −1BT . (110) Then we can write 1 1 H = pT G p + sT F s (111) 2 s s 2 We want to uncouple the Hamiltonian: 1 1 1 H = P 2 + λ Q2 = P T P + QT Λ Q (112) 2 i i i 2 2 i X 20 To do this we must return to the normal coordinates. We use the transfor- mation: ′ s = L Q (113) Through an analogy to what we have already seen, we derive conditions:
′ − ′ L T G 1L = E (114) ′ T ′ L F L =Λ (115) (116) We can combine these two equations into one:
− ′ ′ 1 − ′ F L = L T Λ = G 1L Λ (117) This is a generalised eigenvalue problem of the matrix F with “overlap ma- − − trix” G 1. As we can see, G 1 is indeed symmetric:
− G 1 = B−1 T M B−1 (118)
After multiplying equation (117) by G from the left side:
′ ′ G F L = L Λ (119) G F L ′ L ′ i = iλi
This is the eigenvalue problem of the matrix G F . This method is called “Wilson’s GF matrix method”. The dimension of this method is 3n 6, as we can omit the 6 rows asso- −−1 −1 T ciated with Qi in B (we don’t need B in G = B M B ) and we define 2 F as F = ∂ V . The internal coordinates do not change when the ij ∂si∂sj whole molecule is translated or rotated. That is, the Eckart conditions are automatically taken into account.
0.4 Rotations of rigid (non-linear) molecules
See also: Paul Wormer, Courses and teaching material http://www.theochem.ru.nl/ pwormer/teachmat.html The rigid rotor in classical and quantum mechanics
21 0.4.1 Rotation matrices and infinitesimal rotations
Choose a coordinate frame that “rotates with the molecule” — such a frame is called a body-fixed (BF) frame. So the orientation of the molecule is defined by the orientation of the BF frame relative to the laboratory = space fixed (SF) frame. Hence, the orientation of the molecule, i.e., the orientation of the BF axes, with respect to the SF frame is described by a 3 3 matrix of direction cosines: the rotation matrix C. × Any orientation (rotation) can be described by three Euler angles ϕ, ϑ, χ:
rotate around the z-axis over ϕ x′, y′, z′ axes (z′ axis = z axis) → rotate around the y′-axis over ϑ x′′, y′′, z′′ axes (y′′ axis = y′ axis) → rotate around the z′′-axis over χ
The corresponding rotation matrices (containing direction cosines) are: cos ϕ sin ϕ 0 cos ϑ 0 sin ϑ − C (ϕ)= sin ϕ cos ϕ 0 C (ϑ)= 0 1 0 z y 0 0 1 sin ϑ 0 cos ϑ − (120) For rotations around the x-axis we have: 10 0 C (α)= 0 cos α sin α (121) x 0 sin α −cos α The Euler (three-angle) rotation matrix is:
C(ϕ, ϑ, χ)= C ′′ (χ)C ′ (ϑ)C (ϕ), (122) z y z which can also be written as (prove this!): C(ϕ, ϑ, χ)= C (ϕ)C (ϑ)C (χ) (123) z y z We define the so-called Lie derivatives or infinitesimal rotations J , J , x y J : z sin ϕ cos ϕ 0 0 1 0 dC (ϕ) − − − J = z = cos ϕ sin ϕ 0 = 1 0 0 z dϕ − ϕ=0 0 00 0 0 0 ϕ=0 (124) 00 0 0 01 J = 0 0 1 J = 0 00 x − y 01 0 1 0 0 − 22 Any vector can be decomposed as: x r = y = xe + ye + ze (125) 1 2 3 z in term of unit vectors: 1 0 0 e = 0 e = 1 e = 0 (126) 1 2 3 0 0 1 The infinitesimal rotations of the vector r are: 0 J r = z = e r x 1 −y × z J r = 0 = e r (127) y 2 x × − y − J r = x = e r z 3 0 × ωx Any infinitesimal rotation ω = ω can be described as y ωz J = ω J + ω J + ω J = ω x x y y z z 0 ω ω − z y (128) = ω 0 ω = Ω z − x ω ω 0 − y x — this is the most arbitrary antisymmetric 3 3 matrix (any 3 3 antisym- metric matrix can be written in this way). × × It is easy to verify that J r = Ω r = ω r (129) ω × Furthermore, it follows that: sin ϕ cos ϕ 0 dC (ϕ) − − z = cos ϕ sin ϕ 0 = J C (ϕ)= C (ϕ)J (130) dϕ z z z z 0− 00 23 0.4.2 The rigid rotor in classical mechanics
A set of particles (atoms) has the coordinates rk(t) in the laboratory (SF) frame. Their masses are mk. The position vector of the center of mass is:
n −1 R(t)= M mkrk(t) (131) Xk=1 with M denoting the total mass, M = k mk. In a coordinate frame which moves withP R(t), but has axes parallel to the axes of the laboratory (SF) frame, the atoms have coordinates:
ρ (t)= r (t) R(t) (132) k k − It follows that: m ρ (t)=0 at each time t (133) k k Xk (prove this). In other words: the position vector of the center of mass is 0 in the center of mass frame. We assume that the particles (atoms) belong to a moving rigid body (molecule). We define a rotating (BF) coordinate frame so that any time t:
ρ (t)= C(ϕ(t), ϑ(t), χ(t))a (134) k k
The vectors ak are the constant position vectors of the atoms in the BF coordinate frame that rotates with the molecule. The angles ϕ(t), ϑ(t), χ(t) are the Euler angles describing the orientation of the BF frame with respect to the SF frame at time t. Finally, we can write:
rk(t)= R(t)+ C(ϕ(t), ϑ(t), χ(t))ak (135) with
r (t) — coordinates of atom k in the laboratory (SF) frame · k R(t) — coordinates of the center of mass in the laboratory (SF) frame · C(ϕ(t), ϑ(t), χ(t)) — rotation matrix containing direction cosines of the · rotating BF axes in the laboratory (SF) frame
24 a — constant position vector of atom k in BF coordinates · k The velocities of the atoms are: ∂r (t) v (t)= k =r ˙ (t) (136) k ∂t k ˙ ˙ r˙k(t)= R(t)+ C(ϕ(t), ϑ(t), χ(t))ak (137) = R˙ (t)+ C˙ (ϕ, ϑ, χ)C−1(ϕ, ϑ, χ)ρ (t) k Rotation matrices are orthogonal:
C−1 = CT (138)
Theorem: The matrix C˙ C−1 = C˙ CT is an antisymmetric matrix. Proof: C CT = E identity matrix T (139) C˙ CT + C C˙ =0 Hence T T C˙ CT = C C˙ = C˙ CT (140) − − Let us assume that:
0 ωz ωy T − C˙ C = Ω = ω 0 ω (141) z − x ω ω 0 − y x Theorem: The relation between the angular velocity ω and the time derivatives ϕ˙ ϕ ∂ q˙ = ϑ˙ = ϑ of the Euler angles is given by: ∂t χ˙ χ ϕ˙ ω = U ϑ˙ = U q˙ (142) χ˙ with 0 sin ϕ cos ϕ sin ϑ − U = 0 cos ϕ sin ϕ sin ϑ (143) 1 0 cos ϑ 25 Proof: C˙ (ϕ, ϑ, χ)= C˙ (ϕ)C (ϑ)C (χ) z y z (144) + C (ϕ)C˙ (ϑ)C (χ)+ C (ϕ)C (ϑ)C˙ (χ) z y z z y z
∂ ∂C (ϕ) ∂ϕ C˙ (ϕ)= C (ϕ)= z = J C (ϕ)ϕ ˙ (145) z ∂t z ∂ϕ ∂t z z
Ω = C˙ CT = J C (ϕ) C (ϑ) C (χ)CT (χ) CT (ϑ) CT (ϕ) ϕ˙ z z y z z y z
(146) + C (ϕ)J C (ϑ) C (χ)CT (χ) CT (ϑ) CT (ϕ) ϑ˙ z y y z z y z
+ C (ϕ)C (ϑ)J C (χ)CT (χ) CT (ϑ)CT (ϕ)χ ˙ z y z z z y z Each “box” yields the identity matrix E (because rotation matrices are or- thogonal), so we have:
Ω = J ϕ˙ + C (ϕ)J CT (ϕ) ϑ˙ + C (ϕ)C (ϑ)J CT (ϑ)CT (ϕ)χ ˙ (147) z z y z z y z y z
Substitution of eq. (120) for the matrices C and C and eq. (124) for J z y z and J , and using the relation in eq. (128) between Ω and ω proves eqs. (142) y and (143). As a result we can express the velocities of the atoms as:
˙ ˙ T r˙k(t)= R(t)+ C C ρ (t) k (148) = R˙ (t) + Ω ρ (t) k One may remember that multiplication of the vector ρ (t) by the antisym- k metric matrix Ω can be replaced by taking the vector product ω ρ (t): × k
r˙ (t)= R˙ (t)+ ω ρ (t) (149) k × k Because ρ (t) is the velocity of atom k in a coordinate frame that moves with k the center of mass, but does not rotate, it follows that ω is the angular velocity in the laboratory (SF) frame
ωSF = ω (150)
26 We may further write
r˙ (t)= R˙ (t)+ ω ρ (t) k × k = R˙ (t)+ C C−1 ω ρ (t) × k (151) − − = R˙ (t)+ C C 1ω C 1ρ (t) × k h i The last line follows from the properties of the vector product under rotation:
C (a b)=(C a) (C b). × × Here we can substitute a = C−1ρ (t) and we get: k k
r˙ (t)= R˙ (t)+ C ωBF a (152) k × k The angular velocity with respect to the rotating (BF) frame is:
ϕ˙ ϕ˙ − − − ωBF = C 1ω = C 1ωSF = C 1U ϑ˙ = V ϑ˙ (153) χ˙ χ˙ with sin ϑ cos χ sin χ 0 − − V = C 1U = sin ϑ sin χ cos χ 0 (154) cos ϑ 0 1 This can be easily shown, if one remembers eq. (143):
0 sin ϕ cos ϕ sin ϑ − − V = C 1U = CT (χ)CT (ϑ)CT (ϕ) 0 cos ϕ sin ϕ sin ϑ = z y z 1 0 cos ϑ = CT (χ)CT (ϑ)CT (ϕ) e , C (ϕ)e , C (ϕ)C (ϑ)e = (155) z y z 3 z 2 z y 3 sin ϑ cos χ sin χ 0 T T T − = C (χ)C (ϑ)e , C (χ)e , e = sin ϑ sin χ cos χ 0 z y 3 z 2 3 cos ϑ 0 1
We have now expressed the velocities of the atomsr ˙k in terms of the velocity ϕ˙ of the center of mass R˙ and the angular velocity ωSF , or ωBF ,orq ˙ = ϑ˙ . χ˙ 27 We want to derive the expression for the kinetic energy, but first we define:
x (t) 0 z (t) y (t) k − k k ρ (t)= y (t) X (t)= z (t) 0 x (t) (156) k k k k k z (t) → y (t) x (t)− 0 k − k k — an antisymmetric matrix with the cartesian (SF) coordinates of atom k. We define a similar matrix for the BF frame: a 0 a a kx − kz ky a = a A = a 0 a (157) k ky k kz kx a → a a −0 kz − ky kx When we consider the elements of the product matrix XT X : k k
XT X = y2 + z2 =(x2 + y2 + z2) x2 k k k k k k k k 11 − (158) T X X = xkyk k k 12 − we see that the matrix
ISF (t)= m XT (t)X (t) (159) k k k Xk is the inertia tensor of the molecule in the laboratory frame. Similarly, we see that: IBF = m AT A (160) k k k Xk is the inertia tensor in the BF frame. It is obvious that IBF is constant in time. The kinetic energy
T = 1 m v2 = 1 m r˙ r˙ = 1 m r˙T r˙ (161) 2 k k 2 k k · k 2 k k k Xk Xk Xk can be written in different ways. In the laboratory (SF) frame we get:
T T T T = 1 m R˙ R˙ + m R˙ ω ρ + 1 m ω ρ ω ρ 2 k k × k 2 k × k × k k k k X X X (162)
28 The second therm vanishes, because
T T T mkR˙ ω ρ = R˙ ω mkρ = R˙ (ω 0)=0 (163) × k × k! × Xk Xk see eq. (133). We may further write that:
Ω ρ = ω ρ = ρ ω = X ω (164) k × k − k × − k and the result becomes:
1 ˙ T ˙ 1 T T T = 2 MR R + 2 ω mkX X ω (165) k k! Xk 1 ˙ T ˙ 1 T SF T = 2 MR R + 2 ω I ω (166) The first term corresponds to the translation of the center of mass, the second to the rotation. As we see, there is an exact separation between translation and rotation. In the BF frame we obtain (also here the cross term vanishes):
T T T = 1 MR˙ R˙ + 1 m ωBF a CT C ωBF a 2 2 k × k × k k X (167) T T = 1 MR˙ R˙ + 1 ωBF m AT A ωBF 2 2 k k k k ! X 1 ˙ T ˙ 1 BF T BF BF T = 2 MR R + 2 ω I ω (168) ϕ Let us now express the kinetic energy in terms of the Euler angles q = ϑ . χ To this end we must substitute: ωSF = ω = U q˙ (169) ωBF = C−1U q˙ = V q˙
Ater this substitution we get:
T T = 1 MR˙ R˙ + 1 q˙T U T ISF U q˙ 2 2 (170) 1 ˙ T ˙ 1 T T BF T = 2 MR R + 2 q˙ V I V q˙
29 which we can also write as
1 ˙ T ˙ 1 T T = 2 MR R + 2 q˙ g q˙ (171) where g = U T ISF U = V T IBF V (172) is the metric matrix (metric tensor — “overlap matrix”) belonging to the Euler angles q (N.B. we have seen this previously, in Wilson’s GF method, where instead of q we used internal coordinates s. The matrix G , defined there, is the inverse − of the metric tensor, G 1 = g . Of course, g and g are different matrices, s s but they play a similar role.) Finally, we present an expression for the kinetic energy in terms of angular momenta J = J = m r r˙ (173) k k k × k Xk Xk In SF coordinates: rk = R + ρ k (174) r˙ = R˙ + ω ρ k × k J = m R R˙ + m ρ R˙ k × k k × Xk Xk (175) + m R ω ρ + m ρ ω ρ k × × k k k × × k Xk Xk Because m ρ = 0, the second and the third term vanish. The fourth k k k term can be rewritten as follows P m ρ ω ρ = m ρ ρ ω = k k × × k − k k × k × Xk Xk (176) = m X X ω = m XT X ω = ISF ω − k k k k k k Xk Xk so that we get J = MR R˙ + ISF ω (177) × ˙ In this equation J CM = MR R is the angular momentum due to the center of mass motion, and ISF ω is× the rotational angular momentum.
30 In BF coordinates: rk = R + C ak r˙ = R˙ + C ωBF a (178) k × k BF BF J = J CM + C I ω Summarizing:
SF SF SF SF J = J CM + J rot J rot = I ω BF BF BF BF (179) J = J CM + C J rot J rot = I ω
Apart from the term related to the motion of the center of mass, we then get for the kinetic energy: − 1 SF T SF 1 SF T = 2 J I J − (180) 1 BF T BF 1 BF T = 2 J I J 0.4.3 The rigid rotor Hamiltonian
To go from classical mechanics to quantum mechanics, we must first write the classical kinetic energy in terms of the coordinates q and conjugated momenta p.
In classical mechanics, if qi are the (possibly curvilinear) cordinates, then the conjugated momenta are defined as: ∂T pi = (181) ∂q˙i
x In cartesian coordinates, r = y , the expression for the kinetic energy is: z 1 2 1 2 2 2 T = 2 mv = 2 m(x ˙ +y ˙ +z ˙ ) ∂T This is consistent with eq. (181), because p = = mx˙, etc., and p = x ∂x˙ mr˙ = mv. ϕ Let us now apply this to the Euler angles q = ϑ , and first derive the χ conjugated momenta p. We return to eq. (171):
1 T 1 T = 2 q˙ g q˙ = 2 gijq˙iq˙j (182) i,j X 31 with the metric tensor given by g = U T ISF U = V T IBF V , see eq. (172). The momenta are: ∂T p = = g q˙ p = g q˙ q˙ = g−1 p (183) i ∂q ij j i j X and the classical Hamiltonian for the kinetic energy is:
T T 1 T −1 −1 1 T −1 T = 2 p g g g p = 2 p g p (184) 1 T −1 T = 2 p g p (185)
According to the postulates of quantum mechanics, the momentum operators conjugated to a set of (curvilinear) coordinates qi are:
h¯ ∂ pi = (186) i ∂qi
The quantum mechanical momentum operators conjugated to the Euler an- gles are: ∂/∂ϕ h¯ p = ∂/∂ϑ (187) i ∂/∂χ From here on, we will use atomic units withh ¯ = 1. We want to quantize equation (185). But we must be careful, because the matrices U and V are functions of ϕ, ϑ, χ, and so is the matrix g. So, p does not commute with g−1. B. Podolski [Quantum-mechanically correct form of Hamiltonian function for conservative system, Phys. Rev., 32, 812 (1928)]: We cannot simply replace the classical momenta in eq. (185) by operators, but we have to use the general formula for the Laplace operator
∂2 ∂2 ∂2 ∆= 2 = + + (188) ∇ ∂x2 ∂y2 ∂x2 in curvilinear coordinates q. This yields:
1 1 1 − /2 T /2 −1 T = 2 g p g g p (189)
32 where g = det(g). We see that, if g did not depend on q, then g would commute with p and we 1 T −1 would have the simpler expression: T = 2 p g p. The expression for g is known in terms of the Euler angles, g = V T IBF V , (Euler angles appear in V , IBF is constant for a rigid rotor). So, we can derive an explicit quantum mechanical formula for T . We start by calculating
T BF 2 BF 2 g = det g = det V I V = det V det I = sin ϑ I1I2I3 (190) I1, I2, I3 are the so-called principal moments of inertia — the eigenvalues of IBF . We can then write:
2 ∂/∂ϕ h¯ ∂ ∂ ∂ −1 −1 T = , , sin ϑ V −1 IBF V T ∂/∂ϑ (191) −2 sin ϑ ∂ϕ ∂ϑ ∂χ ∂/∂χ Further we should plug in the explicit formula for V , eq. (154). It appears that the result can simply be written in terms of the angular momenta J SF or J BF , but we must first quantize these. In classical mechanics: J SF = ISF ω = ISF U q˙ −1 −1 (192) q˙ = g−1p = U −1 ISF U T p
Combining these two equations gives:
−1 J SF = U T p (193)
∂/∂ϕ 1 After inverting U T and substituting p = ∂/∂ϑ we derive the expressions i ∂/∂χ for J SF in terms of the Euler angles:
∂ ∂ cos ϕ ∂ J SF = i cos ϕ cot ϑ + sin ϕ x ∂ϕ ∂ϑ − sin ϑ ∂χ ∂ ∂ sin ϕ ∂ J SF = i sin ϕ cot ϑ cos ϕ (194) y ∂ϕ − ∂ϑ − sin ϑ ∂χ ∂ J SF = i z −∂ϕ 33 Analogously, we find −1 −1 J BF = V T p = C−1 U T p = CT J SF (195) After a similar elaboration we get: cos χ ∂ ∂ ∂ J BF = i sin χ cot ϑ cos χ x sin ϑ ∂ϕ − ∂ϑ − ∂χ sin χ ∂ ∂ ∂ J BF = i cos χ + cot ϑ sin χ (196) y −sin ϑ ∂ϕ − ∂ϑ ∂χ ∂ J BF = i z −∂χ Now we know that ∂/∂ϕ −1 −1 J BF = V T p = V T ∂/∂ϑ (197) ∂/∂χ and we can also show (via substitution and differentiation) that
ih¯ ∂ ∂ ∂ − T − , , sin ϑ V 1 = J BF ,J BF ,J BF = J BF (198) sin ϑ ∂ϕ ∂ϑ ∂χ x y z When we apply this to equation (191), we get simple quantum mechanical expressions for the kinetic energy:
− 1 BF T BF 1 BF T = 2 J I J (199) and similarly − 1 SF T SF 1 SF T = 2 J I J (200) These are the exact quantum mechanical expressions in which the operators J BF and J SF are known. It appears that in terms of the angular momenta, the classical and quantum mechanical expressions are the same! To some extent this is a coincidence.
0.4.4 The rigid rotor in quantum mechanics
It is simpler to work in the BF frame, because (in contrast with ISF , which depends on the Euler angles) IBF is a constant matrix
IBF = m AT A k k k Xk 34 depending on the constant position vectors ak. It is a symmetric 3 3 matrix, so we can easily diagonalise it and its eigenvalues are real. It is× customary to choose a rotating coordinate frame that makes IBF diagonal I 0 0 I−1 0 0 1 − 1 BF BF 1 −1 I = 0 I2 0 I = 0 I2 0 (201) − 0 0 I 0 0 I 1 3 3 and define the rotational constants: 1 1 1 A = B = C = (202) 2I1 2I2 2I3 Then, we obtain the well known kinetic energy expression:
BF 2 BF 2 BF 2 T = A Jx + B Jy + C Jz (203) The components of J SF satisfy the following commutation relations:
SF SF SF Jx ,Jy = ihJ¯ z (204) (with cyclic permutations of the indices x, y, z). The components of the BF angular momentum operators, J BF , obey anomalous commutation re- lations: J BF ,J BF = ihJ¯ BF (205) x y − z (cyclic permutations of the indices x, y, z are allowed; pay attention to the minus sign!) Since J SF = C J BF and CT C = E it follows that
2 T T 2 J SF = J SF J SF = J BF CT C J BF = J BF = J 2 (206)
2 All components of J SF commute with J SF and, similarly, all components BF BF 2 of J commute with J . ∂ ∂ Furthermore, J SF = i commutes with J BF = i . So, we have a set z − ∂ϕ z − ∂χ of commuting operators: 2 SF BF J ,Jz ,Jz (207) The question is: are these operators constants of the motion, i.e., do they commute with the Hamiltonian? BF 2 BF 2 BF 2 T = A Jx + B Jy + C Jz
The answer is:
35 J 2 commutes with T , because it commutes with J BF , J BF , and J BF . · x y z ∂ J SF also commutes with T , because J SF = i and the operators · z z − ∂ϕ ∂ J BF , J BF do not contain ϕ (they contain only ) x y ∂ϕ
BF Jz does not commute with T , because it does not commute with · BF 2 BF 2 Jx , and Jy Let us now consider some special cases.
Symmetric top
A = B = C 6 prolate A = B
Examples of a symmetric top: benzene, NH3 (we will use the notation J BF ≡ J ) H = T = A J 2 + J 2 + CJ 2 = AJ 2 +(C A) J 2 (208) x y z − z BF This Hamiltonian does commute with Jz = Jz . Hence, there are three constants of the motion in this case, and three quantum numbers: j, k, and m.
H j, k, m = E j, k, m | i j,k | i J 2 j, k, m = j(j + 1) j, k, m | i | i (209) J BF j, k, m = k j, k, m j k j z | i | i − ≤ ≤ J SF j, k, m = m j, k, m j m j z | i | i − ≤ ≤ The energy eigenvalues are
E = Aj(j +1)+(C A)k2 (210) j,k − The energy does not depend on the quantum number m, so it is (2j +1)-fold degenerate. BF Since T commutes with Jz , the eigenfunctions of T have to be also eigen- ∂ functions of J BF = i , with eigenvalue k. So, they contain a factor z − ∂χ
36 ∂ exp(ikχ). Similarly, they have to be eigenfunctions of J SF = i , with z − ∂ϕ eigenvalue m, which yields the factor exp(imϕ). Thus, they can be written as: (j) (j) ∗ j, k, m =Ψj,k,m(ϕ, ϑ, χ) = exp(imϕ)dmk(ϑ) exp(ikχ)= Dmk(ϕ, ϑ, χ) | i (211) (j) The functions Dmk(ϕ, ϑ, χ) are called Wigner rotation matrices. The func- (j) tions dmk(ϑ), which are called Wigner (small) d-matrices, are the solutions of a differential equation in cos ϑ; they can be obtained by means of recursion relations.
Spherical top
A = B = C, examples: CH4, SF6 H = T = AJ 2 (212)
Ej = Aj(j +1) (213) The energy does not depend on the quantum numbers k and m, so it is (2j + 1)2-fold degenerate. The eigenfunctions of the spherical top Hamiltonian are the same functions (j) ∗ Dkm(ϕ, ϑ, χ)
Asymmetric top
BF The three rotational constants are different, A>B>C. Jz is not a constant of the motion, thus k is not a good quantum number. But j and m remain good quantum numbers. The eigenfunctions of H can be written as:
j (j) ∗ Ψj,m,i(ϕ, ϑ, χ)= Dmk(ϕ, ϑ, χ) cki (214) − kX= j The coefficients cki follow from diagonalization of the matrix of the Hamil- (j) ∗ tonian H in the basis of Wigner D-functions, Dkm(ϕ, ϑ, χ) , for fixed j and m, and k = j,...,j The dimension of this matrix is 2j + 1. − The matrix elements can be easily derived by writing T in terms of raising (or step-up) and lowering (or step-down) operators: BF BF BF J± = J iJ (215) x ∓ y 37 These operators act on the basis functions as:
1 BF /2 J± j, k, m = [j(j + 1) k(k 1)] j, k 1, m (216) | i ± ± | ± i Note the interchange of the + and signs in the definition of these operators, − eq. (215), which is due to the anomalous commutation relations between the components of J BF . The alternative is not to interchange the + and signs in the definition of the operators, but then the interchange occurs in− eq. (216). For more details we refer to Paul Wormer’s lecture notes, mentioned at the beginning of this section, and to the textbooks on angular momentum theory listed in these notes.
0.5 The molecular vibration-rotation Hamil- tonian
0.5.1 Introduction
In Chapter 2 we discussed pure vibrations (in the harmonic approximation), in Chapter 3 pure rotations. Now we will consider the vibrations and rota- tions of molecules simultaneously. The question is: how do we define the rotating (BF) axes in a vibrating molecule? Or: how can we separate vibrations and rotations? The answer to the latter question is that we cannot do this exactly. By a suitable choice of the rotating axes we can minimize the coupling between vibrations and rotations, however. Our task is then to:
1. Derive the Hamiltonian for which the rot-vib coupling is minimal.
2. Find the approximations needed to arrive at a harmonic oscillator model for the vibrations an a rigid rotor model for the rotations.
A lot of attention has been given to these issues, by Eckhart, Sayvetz, Van Vleck, Wigner & Hirschfelder, Wilson & Howard, Darling & Dennison, Wat- son, and others.
38 0.5.2 The (classical) kinetic energy of a semi-rigid molecule
The word semi-rigid implies that the vibrations involve only small displace- ments of the atoms from their equilibrium positions in the molecule. To derive the classical expression for the kinetic energy T we write the 3n cartesian coordinates rk(t) of the atoms with respect to a laboratory (SF) frame in terms of:
3 coordinates R(t) — the origin of the moving BF frame
3 Euler angles ϕ, ϑ, χ — the orientation of the rotating BF frame axes
3n 6 internal coordinates — bond lengths, angles between bonds, etc. − by means of the equation:
rk(t)= R(t)+ C (ϕ(t), ϑ(t), χ(t)) ak(t) (217)
Because of the vibrations, also the ak(t) depend on t now. However, they are not internal coordinates, but (3n, not 3n 6) cartesian coordinates of the − atoms with respect to the rotating frame. The atomic velocities are:
r˙ = R˙ (t)+ C˙ (ϕ, ϑ, χ)a + C a˙ k k k (218) = R˙ (t)+ C(ωBF a )+ C a˙ × k k When comparing this to the previously derived equations, we see an extra term C a˙ k, because ak(t) depends on time. First, let us choose R(t) as the position of the center of mass:
−1 R(t)= M mkrk(t) (219) Xk Then, it follows that at each time t
mkak(t)=0 (220) Xk so, it is also true that
mka˙ k(t)=0 (221) Xk
39 The kinetic energy can be written as
1 T T = 2 mkr˙k r˙k = k X T T = 1 MR˙ R˙ + m R˙ C ωBF a + 2 k × k k T X T (222) + m R˙ C a˙ + 1 m ωBF a ωBF a + k k 2 k × k × k k k X TX + m ωBF a a˙ + 1 m a˙ T a˙ k × k k 2 k k k k k X X We have used here that CT C = E (C is an orthogonal matrix). The second and third term vanish, because of the translational conditions for ak anda ˙ k, eqs. (220) and (221):
T T m R˙ C ωBF a = R˙ C ωBF m a = k × k × k k k k ! (223) X T X = R˙ C ωBF 0 =0 × ˙ T ˙ T ˙ T mkR C a˙ k = R C mka˙ k = R C 0 =0 (224) ! Xk Xk So, there is an exact separation between the translations in the kinetic energy and the rotations and vibrations. Because of this separation, we can now 1 ˙ T ˙ “forget” about the term 2 MR R that corresponds to the kinetic energy of the (center-of-mass) translation, as it does not interfere with the rotations and vibrations. The fourth term was already elaborated in Chapter 3, where we wrote it in BF terms of the inertia tensor I . It contains the coordinates ak(t) and is now time-dependent (because of the vibrations). The fifth term is the rotation-vibration coupling. We will also rewrite this term, with the aid of the relation (a b) c = a (b c). The result is: × · · × T T T = 1 ωBF IBF ωBF + m ωBF (a a˙ ) + 1 m a˙ T a˙ (225) 2 k k × k 2 k k k k k rotation X X rot-vib coupling vibration | {z } | {z } | {z } We cannot eliminate the rot-vib coupling term but we can minimize it, by an appropriate choice of the rotating axes.
40 Eckart’s first attempt (1934)
At each time t, choose the principal axes of the inertia tensor IBF (t) as the rotating axes, i.e., choose a BF coordinate frame, that makes IBF diagonal. BF The principal moments of inertia I1, I2 and I3 are the eigenvalues of I . This led to the so-called Eckart paradox: the inclusion of the rotation- vibration coupling (with the aid of first order perturbation theory) yields 2 2 2 a kinetic energy term that has the same form, AJx + BJy + CJz , as the rigid rotor Hamiltonian. The constants, however, are given by expressions such I1 1 2 as A = 2(I2−I3) , instead of 2I1 , as for the rigid rotor. This is a paradox, because one expects that in the limit of zero amplitude vibrations T should be precisely the rigid rotor kinetic energy. It was shown later by Van Vleck that the inclusion of second order terms restores the correct expressions. A derivation of this result was given by F. Jørgensen in his thesis.
Eckart’s second attempt (1935)
Let us assume an equilibrium structure of the molecule with atomic positions ak(0) and small displacements dk(t) a (t)= a (0) + d (t) k k k (226) ˙ a˙ k(t)= dk(t) since ak(0) is constant.
It is obvious that, if dk(t)=0, we may choose the axes of the rotating frame as the principal axes of IBF . But how do we choose the frame when d (t) =0? k 6 Eckart (1935): choose the rotating frame such that the vibrational displace- ments at each time t satisfy the Eckart conditions (see Chapter 2):
a) mkdk(t)=0 Xk (227) b) m a (0) d (t)=0 k k × k Xk The three conditions a) follow directly from putting the origin of the coordi- nate system at the center of mass, which implies that:
0 = mkak(t)= mkak(0) + mkdk(t) (228) Xk Xk Xk =0 =0
| 41{z } | {z } The first, time-independent, term in the above equation equals 0, because we assumed that dk(0) = 0. Then, it follows that also the second term must be equal to zero at all times t. We knew this already: the translations separate exactly from both the rotations and vibrations. The three conditions b) give the best possible separation of rotations and vibrations, because: T T = ωBF m a (t) a˙ (t) Coriolis coupling rot-vib k k × k k T X T (229) = ωBF m a (0) d˙ (t)+ ωBF m d (t) d˙ (t) k k × k k k × k k k X X The first term, which is linear in the vibrational displacements dk, equals 0 because of (the time derivative of) conditions b). The second term that is quadratic in dk remains. This term is (the time derivative of) the min- T imum value of k mkak ak. One can see this by looking at the change in T k mkak ak under an infinitesimally small rotation ∆ω, which is equal to T P 2 mk∆ω (a (0) a (t)). −P k k × k TheP conditions b) implicitly fix the directions of the axes of the rotating (BF) frame. The orientation of this frame relative to the laboratory (SF) frame is determined by the rotation matrix C(ϕ, ϑ, χ). An explicit formula for C(ϕ, ϑ, χ) can be derived directly from the conditions b). Write b) as: 0 = m a (0) d (t)= m a (0) a (t) (230) k k × k k k × k Xk Xk which holds because a (0) a (0) = 0. Recall that k × k rk(t)= R(t)+ C ak(t) (231) or a (t)= CT [r (t) R(t)] (232) k k − Combined with b) this gives: 0 = m a (0) CT [r (t) R(t)] (233) k k × k − Xk This equation, written in components, is 0= m a (0)CT [r (t) R (t)] k kα βγ kγ − γ k γ X X (234) m a (0)CT [r (t) R (t)] − k kβ αγ kγ − γ γ Xk X 42 Now we define a matrix Y with components:
Y (t)= m [r (t) R (t)] a (0) (235) αβ k kα − α kβ Xk or, in matrix notation
Y (t)= m [r (t) R(t)] a (0) (236) k k − ⊗ k Xk where defines the outer (or tensor) product of two vectors. We find that: ⊗ (C Y C Y )=0 (237) γβ γα − γα γβ γ X or, in matrix form Y T C = CT Y (238) C is orthogonal, CT = C−1, so:
Y T = CT Y CT (239)
Y T Y = CT Y CT Y =(CT Y )2 (240)
1 Y T Y is symmetric and positive-definite, so (Y T Y ) /2 is well defined, and we can write: 1 (Y T Y ) /2 = CT Y (241) So we get: 1 C = Y (Y T Y )− /2 (242) This equation gives the rotation matrix C, which defines the orientation of the body-fixed frame, in terms of the matrix Y (t). The matrix Y (t) contains the equilibrium positions ak(0) of the atoms in the molecule, but also depends on their instantaneous positions r (t) R(t)= ρ (t) relative to the space fixed k k frame (with the origin at the center− of mass). So, at each time t, we can determine the BF frame that satifies the Eckart conditions b) by computing the matrix C. The result of applying the Eckart conditions is that:
T T T T = 1 ωBF IBF ωBF + ωBF m d d˙ + 1 m d˙ d˙ (243) 2 k k × k 2 k k k k k X X Let us assume now that the 3n displacements dk that satisfy the six Eckart conditions depend on 3n 6 internal coordinates, namely by: − 1 d = L Q = M − /2 L Q (244)
43 (see Chapter 2). The first six Qi are zero because of the Eckart conditions (see Section 3.5). The other Qi are the normal coordinates of vibration, which in turn depend on 3n 6 internal coordinates s (bond lengths, angles) ′ −1 − as Q = L s. Substitution of these results into the kinetic energy, while remembering that LT = L−1, yields:
1 BF T BF BF BF T α ˙ 1 ˙ T ˙ T = 2 ω I ω + ωα Q ζ Q + 2 Q Q (245) α X The matrices ζα are antisymmetric Coriolis coupling matrices, with elements defined as: 3 ζα = ζα = m ǫ L L (246) ij − ji k αβγ kβ,i kγ,j Xk β,γX=1 where ǫαβγ is the Levi-Civita symbol which is +1 if (α,β,γ) is an even per- mutation of (1,2,3), 1 if it is an odd permutation, and 0 if two or three of − its subscripts are equal. The result in eq. (245) is the classical kinetic energy, in terms of the rotations ωBF and the normal coordinates of vibration Q.
0.5.3 Semi-rigid molecules in quantum mechanics
Before we can quantize the rotational-vibrational Hamiltonian T , we first have to switch to conjugate momenta also for the vibrations. The vibrational momenta Pi conjugated with the normal coordinates are: ∂T P = = Q˙ ωBF ζαQ (247) i ˙ i − α ij j ∂Qi j α X X The vibrational angular momentum Π is defined as:
T α Πα = Q ζ P (248) and the components of the total angular momentum are defined as: ∂T J BF = = IBF ωBF + QT ζαQ˙ (249) α α ∂ωα Substituting these expressions into the classical expression for the kinetic energy of a semi-rigid molecule (eq. 245), we obtain:
T T = 1 J BF Π µ J BF Π + 1 P T P (250) 2 − − 2 44 where − µ = I′ 1 (251) ′ and the components of I — the effective inertia tensor — are
′ BF T α β Iαβ = Iαβ + Q ζ ζ Q (252)
The effective inertia tensor depends both indirectly, through IBF , and di- rectly on the vibrational coordinates Qi. Note that equation (250) is not quantum mechanical, it is just a classical expression in terms of conjugate momenta. The quantum expression is then, according to Podolski
1 T 1 1 1 T = 1 µ /2 J BF Π µ− /2 µ J BF Π + 1 µ /2 P T µ− /2 P (253) 2 − − 2 with µ = det(µ) and the vibrational linear and angular momentum operators P and Π given by: h¯ ∂ Pi = (254) i ∂Qi T α Πα = Q ζ P (255)
The total angular momentum operator J BF was already defined in eq. (196). The total Hamiltonian is H = T + V (256) The potential V depends only on the internal coordinates, it describes the so-called molecular force field. This “Eckart” Hamiltonian was obtained by Wilson and Howard in 1936, [E.B. Wilson and J.B. Howard, The Vibration-Rotation Energy Levels of Polyatomic Molecules I. Mathematical Theory of Semirigid Asymmetrical Top Molecules, J. Chem. Phys. 4, 260-268 (1936)] who followed the pro- cedure described above, and it was further refined by Darling and Dennison in 1940 [B.T. Darling and D.M. Dennison, The water vapor molecule, Phys. Rev. 57, 128-139 (1940)]. It remained the standard until 1968, when Watson [J.K.G. Watson, Simplification of the molecular vibration-rotation Hamilto- nian, Mol. Phys. 15, 479-490 (1968)] was able to simplify it drastically by commuting the determinant of the metric tensor through the derivatives. The ro-vibrational Hamiltonian obtained by Watson, often referred to as the “Watson Hamiltonian”, is
T H = 1 J BF Π µ J BF Π + 1 P T P + U + V (257) 2 − − 2 45 The potential-like term U is the “Watson term”
1 3 U = µ (258) −8 αα α=1 X It is proportional to the trace of the effective reciprocal inertia tensor and depends only on the internal coordinates, not on the momenta. The next question of interest is how to arrive at the harmonic oscillator + rigid rotor model? Three approximations must be made: 1. Assume that the effective inertia tensor I′ is constant, i.e., does not depend on the internal coordinates Q. Then µ commutes with Π and the terms quadratic in Π vanish. Also, replace I′ with IBF (0), i.e., by the inertia tensor of the equilibrium structure. This implies a neglect of distortions of the structure by centrifugal forces. 2. Neglect Coriolis interactions, i.e., the cross terms between the total an- gular momentum J BF and the vibration angular momentum Π. 3. Neglect anharmonic terms in the potential V , i.e., terms with higher than quadratic dependence on the vibrational coordinates Q. After these three approximations we get the rigid rotor + harmonic oscillator Hamiltonian: H = T + V = − 1 BF T BF 1 BF 1 T 1 T = 2 J I (0) J + 2 P P + 2 Q Λ Q = 1 BF −1 BF 1 2 2 (259) = 2 Jα I (0)αβ Jβ + 2 Pi + λiQi α,β i X X rotations harmonic vibrations The Coriolis and| anharmonic{z terms that} were| neglected{z are so}metimes taken into account by means of perturbation theory.
0.5.4 Floppy molecules and Van der Waals molecules
The Eckart and Watson Hamiltonians were derived with the assumption that the vibrations of the molecule involve only small displacements of the atoms, with respect to a well-defined equilibrium structure. There are also many examples where this assumption is not valid, they are called floppy molecules (or molecular complexes).
46 Floppy molecules
These are molecules where some vibrations have large amplitudes. Examples:
internal rotation in ethane, CH —CH · 3 3 inversion-tunneling in ammonia, NH · 3 A generalization of the Eckart approach was given by Sayvetz: in essence, he treated the “floppy” internal coordinates in the same way as Eckart treated the external coordinates. This imposes additional (Eckart-Sayvetz) condi- tions on the remaining internal coordinates.
Van der Waals molecules
These are weakly bound complexes of “normal” molecules. The molecules are held together by noncovalent forces: hydrogen bonding or Van der Waals (dispersion) forces. All intermolecular coordinates are “floppy” in this case. The “Van der Waals vibrations” against the weak intermolecular bonds have large amplitudes and much lower frequencies than the intramolecular vibrations. Except for the mode that involves the stretch of the intermolecular bond length, they are more like (hindered) internal rotations, rather than small-amplitude vibra- tions. The potential V is strongly anharmonic and often shows multiple equivalent global (and local) minima, instead of a single equilibrium struc- ture. The barriers between these minima are relatively low and the complex can quantum mechanically tunnel through these barriers, which gives rise to additional splittings of the rotation-vibration levels. These splittings can be observed in spectra, especially if these are measured with high resolution. Also here one can define a rotating (BF) frame and transform the coordi- nates from the SF to the BF frame. Coriolis coupling between the internal and overall rotations is usually quite important, however. The potential V is simpler in BF coordinates, because it only depends on the internal coor- dinates.
47 Symmetry
In semi-rigid molecules one can use the point group symmetry of the equi- librium structure to simplify the calculation of the (normal mode) vibrations and to derive selection rules. In “floppy” molecules or complexes one may apply the permutation-inversion symmetry group, which is based on the fact that the atomic nuclei are bosons or fermions. Not all permutations of iden- tical nuclei are “feasible”, however, because some permutations may inter- convert structures that are separated by high energy barriers. Only those permutation-inversion operations are called feasible that interconvert struc- tures separated by small barriers, and give rise to observable level splittings due to tunneling through these barriers. The only permutations of identical nuclei that are feasible for semi-rigid molecules correspond to the overall rotations of the whole molecule. The permutation-inversion group is isomorphic to the point group of the equilib- rium structure in this case. For “floppy” molecules it contains additional fea- sible permutation-inversion operations corresponding to tunneling between different, but equivalent, equilibrium structures (minima in the potential surface). For more information, see for example: P. R. Bunker and P. Jensen, Molecular Symmetry and Spectroscopy, NRC Research Press, Ottawa, Second edition (1998).
Acknowledgement
I am very grateful to Piotr Gniewek, for his excellent contributions to the preparation of these lecture notes.
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