The Rigid Rotor in Classical and Quantum Mechanics Paul E.S

Home , Rigid rotor

The rigid rotor in classical and quantum mechanics paul e.s. wormer Institute of Theoretical Chemistry, University of Nijmegen, Toernooiveld, 6525 ED Nijmegen, The Netherlands

Contents

1 Introduction 1

2 The mathematics of rotations in R3 2

3 The algebra of real antisymmetric matrices 8

4 The kinematics of a rigid body 11

5 Kinetic energy of a rigid rotor 14

6 The Euler equations 20

7 Quantization 21

8 Rigid rotor functions 23

9 The quantized energy levels of rigid rotors 29

10 Angular momenta and Lie derivatives 34 10.1 Inﬁnitesimal rotations of functions f(r) ...... 35 10.2 Inﬁnitesimal rotations of functions of Euler angles ...... 37

1 Introduction

The following text contains notes on the classical and quantum mechanical rigid rotor. The classical part is based on the books of H. Goldstein Classical Mechanics [Addison-Wesley, Reading, MA, 1980, 2nd Ed.] and V. I. Arnold, Mathematical Methods of Classical Mechanics, [Springer-Verlag, New York, 1989, 2nd Ed.]. In the following pages Goldstein’s exposition is condensed, whereas Arnold’s terse mathematical treatment is expanded. The quantum mechanical part is in the spirit of L. C. Biedenharn and J. D. Louck Angular Momentum in Quantum Physics: Theory and Application. [Addison-Wesley, Reading, MA 1981].

1 2 The mathematics of rotations in R3

Consider a real 3 3 matrix R with columns r , r , r , i.e., R =(r , r , r ). × 1 2 3 1 2 3 The matrix R is orthogonal if

r r = δ , i, j =1, 2, 3. i · j ij

The matrix R is a proper rotation matrix, if it is orthogonal and if r1, r2 and r3 form a right-handed set, i.e.,

3 r r = ε r . (1) i × j ijk k Xk=1

Here εijk is the antisymmetric (Levi-Civita) tensor,

ε123 = ε312 = ε231 =1 (2) ε = ε = ε = 1 213 321 132 − and εijk = 0 if two or more indices are equal. The matrix R is an improper rotation matrix if its column vectors form a left-handed set, i.e., r r = ε r . (3) i × j − ijk k Xk Equations (1) and (3) can be condensed into one equation

3 r r = det(R) ε r (4) i × j ijk k Xk=1 by virtue of the following lemma.

Lemma 1. The determinant of a proper rotation matrix is 1 and of an improper rotation 1. − Proof The determinant of a 3 3 matrix (a, b, c) can be written as a (b c). × · × Now, for a proper rotation, we ﬁnd by Eq. (1) and remembering that the rk are orthonormal,

r (r r )= ε r r = ε =1, 1 · 2 × 3 23k 1 · k 231 Xk and likewise we ﬁnd 1 for an improper rotation by Eq. (3). − 2 The Levi-Civita tensor allows the following compact notation for the vector product (a b) = ε a b . × i ijk j k Xj,k For instance, (a b) = ε a b + ε a b = a b + a b . × 2 213 1 3 231 3 1 − 1 3 3 1 Theorem 1

A proper rotation matrix R =(r1, r2, r3) can be factorized thus

R = Rz(ω3) Ry(ω2) Rx(ω1) (the“xyz-parametrization”) or also

R = Rz(α) Ry(β) Rz(γ) (the “Euler parametrization”) where cos ϕ sin ϕ 0 − Rz(ϕ) := sin ϕ cos ϕ 0  0 0 1  cos ϕ 0 sin ϕ R (ϕ) := 0 1 0 (5) y   sin ϕ 0 cos ϕ − 10 0  R (ϕ) := 0 cos ϕ sin ϕ . x   0 sin ϕ −cos ϕ   Proof We first prove the xyz-parametrization by describing an algorithm for the factorization of R. Consider to that end cos ω cos ω sin ω cos ω sin ω 3 2 − 3 3 2 R (ω ) R (ω )= sin ω cos ω cos ω sin ω sin ω =: (a , a , a ). z 3 y 2  3 2 3 3 2  1 2 3 sin ω 0 cos ω − 2 2   (6) Note that the multiplication by Rx(ω1) on the right does not affect the first column, so that r1 = a1. Solve ω2 and ω3 from the first column of R,

cos ω3 cos ω2 r1 = sin ω3 cos ω2 .  sin ω  − 2   This is possible. First solve ω for π/2 ω π/2 from 2 − ≤ 2 ≤ sin ω = R (r ) . 2 − 31 ≡ − 1 3 3 Then solve ω for 0 ω 2π from 3 ≤ 3 ≤ R11 cos ω3 = cos ω2 R21 sin ω3 = . cos ω2

This determines the vectors a2 and a3.

Since a1, a2 and a3 are the columns of a proper rotation matrix [Eq. (6)] they form an orthonormal right-handed system. The plane spanned by a2 and a is orthogonal to a r and hence contains r and r . Thus, 3 1 ≡ 1 2 3 cos ω sin ω (r , r )=(a , a ) 1 1 . (7) 2 3 2 3 sin ω −cos ω 1 1

Since r2, a2 and a3 are known unit vectors we can compute

a r = cos ω (8) 2 · 2 1 a r = sin ω . 3 · 2 1 These equations give ω with 0 ω 2π. Augment the matrix in Eq. (7) 1 ≤ 1 ≤ to Rx(ω1), then

R (r , r , r )=(r , a , a )R (ω ) ≡ 1 2 3 1 2 3 x 1 = (a1, a2, a3)Rx(ω1)= Rz(ω3) Ry(ω2) Rx(ω1).

This concludes the proof of the xyz parametrization.

The Euler parametrization is obtained by solving ω2 and ω3 from r3 = a3 and then considering

cos ω sin ω (r , r )=(a , a ) 1 1 (9) 1 2 1 2 sin ω −cos ω 1 1 or, a r = cos ω , a r = sin ω . 1 · 1 1 2 · 1 1 Equation (9) can be written as

(r1, r2, r3)=(a1, a2, r3) Rz(ω1)= Rz(ω3) Ry(ω2) Rz(ω1) , which proves the Euler parametrization. Note. Some confusion exists about the Euler angles of an improper orthogonal matrix S. One can write S = S′R, where R is proper and has a unique Euler parametrization and S′ is another improper rotation matrix. Diﬀerent

4 choices of S′ are possible. Some workers choose S′ = 1 (space inversion), − while others choose a reflection, for instance in the xy plane: 10 0 S′ = 01 0 .   0 0 1 −   Since the choice of S′ is usually implicit and clouded by physical arguments, it is not always clear that the choice is only a matter of convention. In any case, one needs an extra convention, added to the Euler convention, to uniquely parametrize an improper rotation matrix. Yet another parameterization of proper rotation matrices, the (n,ϕ) parameterization, is useful. In order to introduce it, we first prove the existence of a rotation (invariant) axis. Theorem 2 (Euler’s theorem) A rotation matrix R has at least one invariant vector n, i.e., R n = n. If R has more than one invariant vector, R = 1 (the unit matrix) and any vector is an invariant vector. Proof We show that the matrix R has an eigenvalue λ = 1. Since det(R)−1 = det(R−1) = 1, we find, using the rules det(AT ) = det(A) and det(A B) = det(A) det(B),

det(R 1) = det (R 1)T = det(R T 1) = det(R−1 1) − − − − = det R−1(R 1) = det( 1) det(R−1) det(R 1) − − − − = det(R 1). − − Hence, det (R 1) = det (R 1) so that det (R 1) = 0, and we con- − − − − clude that the secular equation det (R λ1) = 0 has the root λ = 1. The − corresponding eigenvector is n. From linear algebra we know the general result that an m m matrix A × has m orthogonal eigenvectors if and only if it is normal, that is, if AA† = A†A. That is, a normal matrix is unitarily equivalent to a diagonal matrix. Its eigenvectors and eigenvalues may be complex. In the case at hand R, which obviously is normal, is equivalent to a matrix of the form eiφ 0 0 0 e−iφ 0 ,   0 0 1   because, as we just saw, it has at least one eigenvalue 1. Further the diagonal matrix is unitary since R is orthogonal. The diagonal elements of a diagonal

5 unitary matrix lie on the unit circle in the complex plane. Finally det(R)=1 (is the product of the diagonal elements), so that the two complex eigenvalues are each others complex conjugate. The two corresponding eigenvectors are in general complex. The matrix R has only two real eigenvectors, other than n, if φ = π or φ = 0. For φ = π the eigenvectors change sign and are not invariant. For φ = 0 we have the unit matrix. This proves Theorem 2. Often one writes a proper rotation matrix as R(n,ϕ), where the invariant vector n is the rotation axis and ϕ is the angle of rotation around n. It is not difficult to give an explicit expression for R(n,ϕ). Consider to that end an arbitrary vector r = an in R3, (a R), and decompose it into a component 6 ∈ parallel to the invariant unit vector n and a component x⊥ perpendicular to it: r =(r n) n + x with x = r (r n) n. (10) · ⊥ ⊥ − · The vectors r, x and n are in one plane, while y n r is perpendicular ⊥ ⊥ ≡ × to this plane. The vectors n, x⊥ and y⊥ form a right-handed frame. The vector n has unit length by definition and the vectors x⊥ and y⊥ both have the length r 2 (n r)2 (which is not necessarily unity). When we | | − · rotate r around n its component along n is unaffected and its perpendicular component transforms as

x cos ϕx + sin ϕy . (11) ⊥ → ⊥ ⊥ Hence,

R(n,ϕ)r = cos ϕ r (r n) n + sin ϕ n r +(r n) n. (12) − · × · It is easily veriﬁed that forn (n , n , n ) ≡ 1 2 3 0 n n − 3 2 n r = n 0 n r =: Nr. (13)  3 1 × n n −0 − 2 1   The dyadic product n n is a matrix with i, j element equal to n n . Evi- ⊗ i j dently, (r n) n = n nr. (14) · ⊗ By direct calculation one shows that

N 2 = n n 1. (15) ⊗ − By substituting (13), (14) and (15) into (12) we obtain ﬁnally

R(n,ϕ)r = 1 + sin ϕN + (1 cos ϕ)N 2 r. (16) − 6 Since r is arbitrary, we have

R(n,ϕ)= 1 + sin ϕN + (1 cos ϕ)N 2. (17) −

Since every rotation matrix = 1 has one, and only one, invariant vector, 6 it follows that ARA T (with orthogonal A and R) has the invariant vector A n. Indeed, A n = A R n = ARA T (A n). Since furthermore Tr(R) = Tr(ARA T ) = 2 cos ϕ +1, the rotation angles of R and ARA T are equal and we ﬁnd the useful expression,

AR n,ϕ A T = R An,ϕ . (18)

This expression enables us to cast the Euler parametrizatio n into a diﬀerent form. Switching back and forth between the notation R (ϕ) R (e ,ϕ) and z ≡ z similarly R (ψ) R (e , ψ), we ﬁnd y ≡ y T Rz(α) Ry(β) Rz(γ) = Rz(α) Ry(β) R (ez,γ) Rz(α) Ry(β) T Rz(α) R (e y, β) Rz(α) (19) × R (e , α) × z ′′ ′ = R (ez ,γ) R (ey, β) R (ez, α).

The right-hand side gives the usual deﬁnition of the Euler angles. Consider a body with an orthonormal frame (ex, ey, ez) attached to it and perform the the three consecutive rotations:

1. Rotate the body around its z-axis over an angle α, this sends ey to e′ R (α)e . y ≡ z y ′ 2. Rotate the body around the new y-axis, ey, over an angle β. This sends e to e′′ R (β) e , where R (β) R (e′ , β). z z ≡ y′ z y′ ≡ y ′′ 3. Rotate the body ﬁnally around the new z-axis, ez , over an angle γ. Thus

Rz(α) Ry(β) Rz(γ)= Rz′′(γ) Ry′(β) Rz(α). (20)

The left-hand side is useful if one wants to compute the rotation matrix in the Euler parametrization, while the right-hand side corresponds to the geometric deﬁnition of the Euler angles.

7 3 The algebra of real antisymmetric matrices

The condition on the real 3 3 matrix A × AT = A − has the consequence that A is of the form

0 a a − 3 2 a 0 a .  3 − 1 a a 0 − 2 1   (The numbering and signs appearing in this matrix will become clear later.)

At this point we only note that the three real numbers a1, a2 and a3 specify the real antisymmetric matrix A uniquely. In other words, there is a one-to- one correspondence between vectors and antisymmetric 3 3 matrices, ×

a1 a = a A. (21)  2 ←→ a3   The set of antisymmetric matrices is a vector space (linear combinations are also antisymmetric and note that the zero matrix is both symmetric and antisymmetric). Its dimension is 3 and a basis is

00 0 0 01 0 1 0 − L = 0 0 1 , L = 0 00 , L = 1 0 0 (22) 1   2   3   01− 0 1 0 0 0 0 0 −       (L ) = ε , (L ) = ε , (L ) = ε . (23) 1 ij − 1ij 2 ij − 2ij 3 ij − 3ij The linear independence follows easily,

a1 L1 + a2 L2 + a3 L3 = 0 a a 0 0 0 − 3 2 a 0 a = 0 0 0 a = a = a =0.  3 − 1   1 2 3 a a 0 0 0 0 ⇐⇒ − 2 1     The fact [Eq. (23)] that matrix elements of Lk can be written by means of the Levi-Civita tensor follows by inspection. The space is also an algebra with the commutator as the product. Let AT = A, BT = B then it follows that A, B T = A, B , and we see − − − that the space is closed under this product. This algebra is a Lie algebra, commonly denoted by so(3). The reason for the designation so(3) is the following: The set of all proper rotation matrices forms a group: SO(3) (the

8 special orthogonal group in 3 dimensions). Diﬀerentiate Rz(ϕ), Ry(ϕ) and Rz(ϕ) [see Eq. (5)],

00 0 d R (ϕ) = 0 sin ϕ cos ϕ dϕ x   0− cos ϕ − sin ϕ − 00 0  = 0 0 1 Rx(ϕ) 01− 0   00 0 = R (ϕ) 0 0 1 (24) x  −  01 0   = L1 Rx(ϕ)= Rx(ϕ) L1 .

Likewise, d R (ϕ) = L R (ϕ)= R (ϕ) L (25) dϕ y 2 y y 2 d R (ϕ) = L R (ϕ)= R (ϕ) L . (26) dϕ z 3 z z 3 Diﬀerentiation at ϕ = 0 brings us from SO(3) to so(3), a linear space tangent to SO(3) and spanned by the Li.

The matrices Li generate “inﬁnitesimal rotations”. Consider, for instance, for inﬁnitesimal ∆ϕ, [∆ϕ >> (∆ϕ)2],

dR R (ϕ + ∆ϕ) = R (ϕ ) + ∆ϕ z (two term Taylor) (27) z 0 z 0 dϕ ϕ0 = Rz(ϕ0) 1 + ∆ϕ Lz and it follows that (1+∆ϕ Lz) represents an inﬁnitesimal rotation of a vector in R3 around the z-axis. We have seen that there is a 1 1 correspondence between R3 and so(3). − To a certain extent this correspondence holds also for orthogonal transfor- mations on both spaces, as is shown in the following theorem.

Theorem 3 Consider a proper or improper rotation matrix B and let

A = ai Li, ai real. (28) i=1 X 9 Then 3 T B A B = B (Ba)i Li , B := det (B) . (29) i=1 X In other words, if A a then B A BT B Ba ; except for the presence ↔ ↔ of B , a rotates as a vector.

Proof

Write B =(b1, b2, b3) and recall from Eq. (4) that

b b = B ε b , a × b abc c c X where B = 1 gives the handedness of the set (b , b , b ). Consider ﬁrst ± 1 2 3 BT L B T L i ab = Bak ( i)kl Blb k,l X = B ε B [by Eq. (23)] − ka ikl lb Xk,l = ε (b ) (b ) − ikl a k b l Xk,l = (b b ) = B ε (b ) − a × b i − abc c i c X = B (Lc)ab Bic , c X so that T B Li B = B Bic Lc . c X This is true for any orthogonal B. Substitute B BT = B−1 then, → 3 T BLiB = B LcBci. (30) c=1 X

Since this equation holds for the basis of so(3) it holds for any element of this space.

Computations are often facilitated by the results in the following lemma.

Lemma 2. If a 0 a a 1 − 3 2 a = a A = a 0 a  2  3 − 1 a ↔ a a 0 3 − 2 1     10 and b 0 b b 1 − 3 2 b = b B = b 0 b  2  3 − 1 b ↔ b b 0 3 − 2 1 then     a b = Ab = B a . (31) × − Proof By Eq. (23),

(a b) = ε a b = a (L ) b = (A) b × i ikl k l k k il l il l Xk,l Xk,l Xl = (Ab)i = b (L ) a = B a = (B a) . − l l ik k − ik k − i Xk,l Xk The proof of the well-known fact that a b transforms as a pseudovector × now follows as an easy corollary of lemma 2. Let C be orthogonal with determinant C . Then

T C ( a b) = CAb = C A C Cb (32) × = A′ b′ = C a′ b′ = C (C a) (Cb), × × since by Theorem 3 A′ C A C T corresponds with C C a C a′, i.e., ≡ ≡ a′ = C a, and similarly b′ = Cb. If C = 1 then simultaneous rotation of a and b gives rotation of a b by C. If C is improper a b is rotated but × × not inverted.

Note This is an example of a general result: an antisymmetric tensor of rank n 1 transforms contragrediently (times determinant) to a vector under the − unitary group U(n).

4 The kinematics of a rigid body

Consider a system of n point masses mk, k = 1,...,n, moving in three- dimensional Euclidean point space (aﬃne space with inner product in coordi- 3 nate space R ). At time t the masses are at the points P1(t),P2(t),...,Pn(t). Choose a ﬁxed orthonormal right-handed (laboratory) frame with origin at the point O, −→eO := (~e1,~e2,~e3).

11 Thus, the vector pointing from O to the point mass Pk is represented by pk(t): −−→OPk(t)= −→eO pk(t) , k =1,...,n. (33)

Clearly, the geometric quantity −−→OPk(t) is independent of the orientation of the frame at O. The center of mass C(t) of the system is given by

1 n n −→OC(t) := m −−→OP (t) with M := m . (34) M k k k Xk=1 Xk=1 The vector −→OC(t) is represented by c(t), −→OC(t)= −→eO c(t). The vector −−→OPk(t) can be decomposed as,

−−→OPk(t)= −→OC(t)+ C−−−−−−→(t)Pk(t) . (35) If the inner products

−−→CP (t) −−→CP (t) ρ , k,l, =1,...,n (36) k · l ≡ kl are time independent: dρkl/dt = 0 for all k and l, the system is a rigid body. In the case of 3-dimensional rigid bodies we can attach to the system a frame −→fC with origin in the center of mass. This frame moves with the rigid body, or in other words, all particles are represented by time-independent coordinate vectors with respect to this body-fixed frame. The position of the center of mass in Euclidean point space is fixed by three real parameters and the orientation of the frame requires another three parameters. (These could for instance be the Euler angles describing the rotation of the frame −→eC —which is parallel to the space-fixed frame—to the body-fixed frame −→fC ). In the case of planar (two-dimensional) rigid bodies we can only attach two orthogonal unit vectors to the body, say −→fx and −→fy, a third unit vector can be constructed by taking the vector product between these vectors. In the case of linear (one-dimensional) systems only one vector can be attached to the system. The two polar angles of the vector with respect to the space-fixed frame gives the orientation of the body in space. Let us consider a 3-dimensional system. It is not difficult to define an orthonormal frame −→fC(t) [for instance by diagonalization of the inertia tensor, see Eq. (58) below] and we obtain

−→fC (t)= −→eC F (t). (37)

Here −→eC is the frame obtained by parallel translation of the lab frame −→eO along −→OC. Since both −→eC and −→fC(t) are orthogonal we have F (t) F (t)T = 1 . (38)

12 Assuming that −→fC(t) is right-handed, just as −→eC , it follows that F (t) is a proper rotation matrix. As we have seen in Theorem 1, F (t) is uniquely determined by three angles, for instance the Euler angles α(t), β(t) and γ(t).

The coordinates of any point Pk(t) of the rigid body with respect to −→fC (t) are time-independent. We assume that at t = 0 the space-ﬁxed and body-

ﬁxed frame coincide, i.e., F (0) = 1. Introducing the coordinates rk(0) of the point masses at t = 0, we write

−−→CPk(t)= −→fC (t) rk(0) = −→eC F (t) rk(0) = −→eC rk(t), (39) so that the relation between space-ﬁxed [rk(t)] and body-ﬁxed [rk(0)] coordinates is,

rk(t)= F (t) rk(0) , k =1,...,n. (40)

′ ′ Returning to the space-fixed frame at O and defining the point Pk so that OP−−→k is parallel to −−→CPk and using that in Euclidean space these parallel vectors are represented with respect to parallel frames by the same column vector, i.e., ′ OP−−→k := −→eO rk(t) (41) we find,

′ −−→OPk = −→OC + −−→CPk = −→OC + OP−−→k = −→eO c (t)+ rk (t) (42)

Pk C ′ Pk

O or, cf. Eq. (33),

pk(t)= c(t)+ rk(t)= c(t)+ F (t) rk(0). (43)

Since the vectors rk(0) are ﬁxed (they determine the shape of the rigid body) the motion of all particles constituting the rigid body is known, when we know the motion of the center of mass c(t) and the rotation F (t) as a function of time. We shall concentrate on F (t) and assume that C is at rest and coincides with O. We will refer to −→eC as the lab frame. The rigid body will be referred to as a rigid rotor (also known as a top).

13 5 Kinetic energy of a rigid rotor

By diﬀerentiation we ﬁnd the velocity of particle k with respect to the lab frame, d d (−−→CP )= e r (t)= e r˙ (t)= e F˙ (t) r (0) (44) dt k dt −→C k −→C k −→C k because, as stated above, we assume the mass center to be stationary. By means of the following lemma we can rewrite the vector on the right-hand side of the following equation:

r˙ (t) F˙ (t) r (0) = F˙ (t) F (t)−1 r (t). (45) k ≡ k k

Lemma 3. If F (t) is orthogonal then F˙ (t) F (t)T is antisymmetric,

0 ω (t) ω (t) − 3 2 F˙ (t) F (t)T = ω (t) 0 ω (t) Ω(t). (46)  3 1  ω (t) ω (t)− 0 ≡ − 2 1   In the Euler parametrization we have

ω1 α˙ ω = ω = M β˙ , (47)  2   ω3 γ˙     with 0 sin α cos α sin β − M := 0 cos α sin α sin β (48)   1 0 cos β   Proof Because F (t) F (t)T = 1 we ﬁnd that

0 = 1˙ = F˙ (t) F (t)T + F (t) F˙ (t)T = F˙ (t) F (t)T + F˙ (t) F (t)T , and hence F˙ (t) F (t)T = F˙ (t) F (t)T T , − so that Eq. (46) follows. In order to prove Eq. (48) we write, by using Theorem 1, which gives the Euler parametrization of F (t), Eqs. (24)–(26) for the diﬀerentiation of rotation matrices and Eq. (30) for the transformation

14 properties of the Li,

T Ω(t) = F˙ (t) F (t) = R˙ z(α) Ry(β) Rz(γ)+ Rz(α) R˙ y(β) Rz(γ) ˙ T T T +Rz(α) Ry(β)Rz(γ) Rz(γ) Ry(β) Rz(α) T ˙ = Lz α˙ + Rz(α) Ly Rz(α) β T + Rz(α) Ry(β) Lz Rz(α) Ry(β) γ˙ sin α cos α sin β − = L α˙ +(L , L , L ) cos α β˙ +(L , L , L ) sin α sin β γ˙ z x y z   x y z   0 cos β = ( β˙ sin α +γ ˙ cos α sinβ) L +(β˙ cos α +γ ˙ sin α sin β) L  − x y +(α ˙ +γ ˙ cos β) Lz.

From this we ﬁnd ω(t), recalling from Sec. 3 that Ω(t)= i ωi(t)Li, ω 0 sin α cos α sin β α˙P 1 − ω = ω = 0 cos α sin α sin β β˙ , (49)  2     ω3 1 0 cos β γ˙

      Å which proves the lemma. | {z } Note

One may wonder: do angular coordinates y(t)=(y1(t),y2(t),y3(t)) exist such that y˙(t) = ω(t)? That is, can one solve the equations y˙(t) = M q˙(t) [Eq. (49)]? [Here we introduced a ‘vector’ consisting of three Euler angles: q := (α,β,γ).] To answer this question we recall from the theory of diﬀeren- tial equations that it is necessary that the matrix M satisﬁes the conditions ∂M ∂M jk = ik , ∂qi ∂qj for the equations to be solvable. Obviously M does not satisfy these conditions. For instance, ∂M ∂M 22 = sin α = 12 =0, ∂q1 − 6 ∂q2 and hence no coordinates are associated with the angular velocities.

Returning to the velocity of particle k we consider −→fC = −→eC F (t) [cf. Eq. (37)] and since by lemma 3 F˙ (t)= ΩF (t), we ﬁnd d d −−→CP = e F (t) r (0) dt k dt−→C k = −→eC Ω F (t)rk(0) = −→eC Ω rk(t) = e ω r (t), (50) −→C × k 15 where ω = Mq˙ and M is given in Eq. (48).

We may also write the velocity in the body-ﬁxed frame −→fC (t), d −→CP = e ΩF (t)r (0) = −→f (t) F −1ΩFr (0) dt k −→C k C k = −→f (t) ωbf r (0), (51) C × k where ωbf := F −1 ω = F −1 M q˙. (52) (See Theorem 3 for the correspondence F −1 Ω F F −1 ω.) We deﬁne ↔ N := F −1M. (53)

In the Euler parametrization we get an explicit expression for N,

sin β cos γ sin γ 0 T − N = R (α) R (β) R (γ) M = sin β sin γ cos γ 0 . (54) z y z   cos β 0 1   In order to compute the kinetic energy T of the rigid body, it is convenient to deﬁne 0 z (t) y (t) − k k X (t) := z (t) 0 x (t) . (55) k  k − k  y (t) x (t) 0 − k k It is easy to show by matrix multiplication that 

X (t)T X (t)= r (t) r (t) 1 r (t) r (t), (56) k k k · k − k ⊗ k which shows that we may deﬁne the inertia tensor I by

n T I(t) := mkXk(t) Xk(t). Xk=1 T From Theorem 3 we ﬁnd Xk(t)= F (t)Xk(0)F (t) , so that

I(t)= F (t) I(0) F (t)T . (57)

We can choose F (t) such that it diagonalizes the instantaneous inertia tensor, i.e., for each t the following relation holds

T F (t) I(t) F (T ) = diag I11(0), I22(0), I33(0) . (58) Then −→fC (t) [cf. Eq. (37)] is a principal axes frame.

16 By Eqs. (50) and (31)

d −−→CP = e Ω r (t)= e r (t) ω = e X (t) ω. (59) dt k −→C k −−→C k × −−→C k To show that we do not loose any kinetic energy terms by choosing the origin O of the lab frame in the center of mass C, we consider the general expression, cf. Eq. (42),

n ˙ ˙ ˙ ˙ ˙ ˙ T = 1 m −−→OP −−→OP = 1 m −→OC + −−→CP −→OC + −−→CP 2 k k · k 2 k k · k k=1 k X X = 1 Mc˙(t)2 + c˙(t) m p˙ + 1 ωT I(t) ω 2 · k k 2 k 1 2 1 T X = 2 Mc˙(t) + 2 ω I(t) ω, (60) where we used that ˙ 0= mk −−→CPk = mk −−→CPk = −→eC mk p˙k. (61) Xk Xk Xk 1 2 The kinetic energy consists of a translational part, 2 M c˙(t) , and a rotational 1 T part 2 ω I(t)ω. We see that the rotation and translation energy are strictly additive, there is no coupling between the two. Recall q1 α q q2 β , ω = M q˙ and ωbf = N q˙. (62)     ≡ q3 ≡ γ     Then 1 T T Trot = 2 q˙ M I(t) M q˙ . (63) Using Eqs. (53) and (57), we also ﬁnd

1 T T Trot = 2 q˙ N I(0) N q˙. (64)

In general one deﬁnes the metric tensor by (summation convention is used)

i j T 2T = gij q˙ q˙ , or 2T = q˙ g q˙, (65) so that we have here

g = M T I(t) M = N T I(0) N. (66)

17 Equations (63) and (64) give the kinetic energy in the Lagrange formalism. In order to write it in the Hamilton formalism we deﬁne the generalized (covariant) momentum,

∂T p := = g q˙j or p = g q˙. (67) k ∂q˙k kj

i ij −1 Thenq ˙ = g pj or q˙ = g p, so that

T −1 T T −1 T T −1 2Trot = p g p = p M I(t) M p = p N I(0) N p. (68)

We can rewrite this expression by introducing the angular momentum L~ with respect to C, ˙ −→L = m L~ m −−→CP −−→CP k k ≡ k k × k Xk Xk = e m r (t) Ω r (t) −→C k k × k k X = e m r (t) X (t) ω −−→C k k × k k X = e m X2(t)ω −−→C k k k X sf = −→eC I(t)ω = −→eC I(t) M q˙ =: −→eC L . (69)

Hence Lsf = I(t) M q˙ = M T Lsf = g q˙ = p. (70) ⇒ ¨ If the external torque m −−→CP −−→CP = 0 then dL/dt~ = 0, and Lsf is a k k k × k constant of the motion. Substitution of p = M T Lsf into (68) gives P sf T −1 sf 2Trot = L I(t) L . (71)

We repeat the same calculation in the body-ﬁxed frame using Eq. (51) and Theorem 3,

−→L = −→f (t) m r (0) F −1 Ω F r (0) C k k × k k X = −→f (t) m r (0) X (0) F (t)−1 ω − C k k × k k X = −→f (t) m X (0)2 F (t)−1 ω − C k k Xk −1 = −→fC (t) I(0) F (t) ω = −→fC (t) I(0) N q˙, (72)

18 where in the last step we have used Eqs. (52) and (53). Hence, we may deﬁne

Lbf := I(0) N q˙, (73) and by (66) N T Lbf = g q˙ = p, (74) so that from the rightmost side of Eq. (68)

bf T −1 bf 2Trot =(L ) I(0) L . (75)

Since in Eq. (72) −→fC(t) depends on time we cannot conclude from dL/dt~ =0 that dLbf /dt = 0. Thus, Lbf is in general not a constant of the motion. Note that Lbf and Lsf have the usual relationship existing between body- and space-ﬁxed coordinates, since from Eqs. (73), (53), (57) and (70) we have

F (t) Lbf = F (t) I(0) F −1(t) M q˙ = I(t) M q˙ = Lsf . (76)

In summary, we have written the classical kinetic energy as,

T T T T 2Trot = q˙ M I(t) M q˙ = q˙ N I(0) N q˙ (Lagrange) −1 −1 = pT M T I(t) M p = pT N T I(0) N p (Hamilton) sf T −1 sf bf T −1 bf = L I(t) L = L I(0) L (angular momentum) “space-ﬁxed” “body-ﬁxed” | {z } | {z } (77) The angular momentum L~ is,

−→L = −→eC I(t) M q˙ = −→fC (t) I(0) N q˙ T −1 T −1 = −→eC (M ) p = −→fC (t)(N ) p. (78)

The angular velocity is:

ω = M q˙ and ωbf = N q˙. (79)

k The linear momentum pk = ∂T/∂q˙ is

p1 α˙ α˙ p = M T I(t) M β˙ = N T I(0) N β˙ . (80)  2     p3 γ˙ γ˙       It is of interest for statistical mechanics to have the explicit form of the classical body-ﬁxed Hamilton function. Using N [Eq. (54)] and taking a

19 body-ﬁxed frame for which I(0) is diagonal with the inertia moments I1, I2 and I3 on the diagonal, we get

T T −1 2Trot = p N I(0)N p 1 2 = 2 ((pα pγ cos β) cos γ pβ sin β sin γ) (81) I1 sin β − − 1 2 1 2 + 2 ((pα pγ cos β) sin γ + pβ sin β cos γ) + pγ , I2 sin β − I3 where p p , p p , p p . α ≡ 1 β ≡ 2 γ ≡ 3 6 The Euler equations

The classical equations of motion for the rigid rotor in the absence of an external torque (Euler’s equations) follow directly from the following results (all proved above):

dLsf = 0 (no external torque) dt F˙ (t) = Ω F (t) (lemma 3) Ω Lsf = ω Lsf (lemma 2) × F −1(ω Lsf ) = ωbf Lbf (corollary to lemma 2). × × Now,

dLsf d 0 = = F (t) Lbf = Ω F (t) Lbf + F (t) L˙ bf dt dt = ω Lsf + F (t) L˙ bf . × And so, L˙ bf = Lbf ωbf. (82) × Noting that [Eqs. (78) and (79)]

Lbf = I(0) ωbf, (83) we can rewrite these equations as follows

I (0)ω ˙ bf = I (0) I (0) ωbf ωbf 1 1 2 − 3 2 3 I (0)ω ˙ bf = I (0) I (0) ωbf ωbf (84) 2 2 3 − 1 3 1 I (0)ω ˙ bf = I (0) I (0) ωbf ωbf. 3 3 1 − 2 1 2

20 In the case of a spherical top, where by deﬁnition I1(0) = I2(0) = I3(0), the right-hand sides of Eq. (84) vanish, so that ωb and Lbf are constants of the motion.

bf bf In the case of a symmetric top, I1(0) = I2(0), only ω3 and L3 are constants of the motion. Note ﬁnally that

Lbf Lbf = Lsf Lsf , (85) · · so that (Lbf )2 is always a constant of the motion, irrespective of the symmetry of the top. The Euler equations (84) cannot be integrated as they stand, since ωbf and ω are not time derivatives of certain coordinate vectors. Transformation to Euler angles and their time derivatives is necessary before integration can be performed.

7 Quantization

Using the quantization rule ∂ p i~ (86) i → − ∂qi we ﬁnd from Eq. (78) that the angular momentum operator becomes, ∂/∂α ∂/∂α L~ i~ e (M T )−1 ∂/∂β = i~ −→f (t)(N T )−1 ∂/∂β −→C   C   → − ∂/∂γ − ∂/∂γ Jˆ  ˆ    1 P1 =: eC Jˆ =: −→fC ˆ . (87) −→  2 P2 Jˆ3 ˆ3   P  Matrix inversion gives cos γ sin γ 0 − 1 − N 1 = sin β sin γ sin β cos γ 0 (88) sin β   cos β cos γ cos β sin γ sin β −   and so the body-ﬁxed angular momentum operators become, cos γ ∂ ∂ ∂ ˆ = i~ sin γ cot β cos γ P1 sin β ∂α − ∂β − ∂γ sin γ ∂ ∂ ∂ ˆ = i~ cos γ + cot β sin γ (89) P2 −sin β ∂α − ∂β ∂γ ∂ ˆ = i~ . P3 − ∂γ

21 From cos α cos β sin α cos β sin β − 1 − − M 1 = sin α sin β cos α sin β 0 (90) sin β −  cos α sin α 0 we ﬁnd the space-ﬁxed angular momentum operators, 

∂ ∂ cos α ∂ Jˆ = i~ cos α cot β + sin α 1 ∂α ∂β − sin β ∂γ ∂ ∂ sin α ∂ Jˆ = i~ sin α cot β cos α 2 ∂α − ∂β − sin β ∂γ ∂ Jˆ = i~ . (91) 3 − ∂α In order to obtain the kinetic energy operator we cannot simply depart from any of the classical expressions (77) for Trot. We must use the Laplace- Beltrami operator, ∂ ∂ 2Tˆ = ~2 2 = ~2 g−1/2 g1/2 gij , (92) rot − ∇ − ∂qi ∂qj

ij −1 where g is the determinant of the metric tensor g. Further g (g )ij. ≡ T −1 Note that Eq. (92) is equivalent to the classical expression 2Trot = p g p, because classically g1/2 commutes with p. By Eq. (66) we ﬁnd for g,

g det(g) = det(N T I(0) N)= I I I det(N)2 = I I I sin2 β, (93) ≡ 1 2 3 1 2 3 where I1, I2 and I3 are the principal moments of inertia [the eigenvalues of I(0)]. Hence

∂/∂α ~2 ∂ ∂ ∂ 2Tˆ = , , sin βN −1 I(0)−1 (N T )−1 ∂/∂β . (94) rot −sin β ∂α ∂β ∂γ   ∂/∂γ   This expression can be rewritten in terms of the angular momentum operators

Jˆi and Pˆi. This is possible because of the following relations i~ ∂ ∂ ∂ , , sin β N −1 = ˆ , ˆ , ˆ (95) −sin β ∂α ∂β ∂γ P1 P2 P3 i~ ∂ ∂ ∂ , , sin β M −1 = Jˆ , Jˆ , Jˆ . (96) −sin β ∂α ∂β ∂γ 1 2 3

22 Using Eq. (88) for N −1 we find differentiating the different columns,

i~ ∂ ∂ ∂ cos γ + sin β sin γ + cos β sin γ + cos β cos γ cos β sin γ −sin β − ∂α ∂β ∂γ − h i = ˆ , P1 i~ ∂ ∂ ∂ sin γ + sin β cos γ + cos β cos γ cos β sin γ cos β cos γ −sin β − ∂α ∂β − ∂γ − h i = ˆ , P2 i~ ∂ ∂ sin β = i~ −sin β − ∂γ − ∂γ h i = ˆ , (97) P3 so that we have indeed Eq. (95). The proof of Eq. (96) runs along the very same lines. Here too, some scalar terms appear that cancel mutually.

Hence from Eq. (94), and assuming that −→fC is a principal axes frame, [cf. Eq. (58)], we ﬁnd

2Tˆ = ˆ I(0)−1 ˆ (98) rot Pi ii Pi i X ˆ I −1 ˆ = Ji (t) ij Jj. (99) ij X Note that although these expressions have the same form as their classical counterparts [Eqs. (77)], one cannot simply state the quantization rules Lsf i → Jˆ and Lbf ˆ . As the derivation shows, the similarity of the classical i i → Pi and quantum forms is due to Eqs. (95) and (96), and hence to some extent coincidental.

8 Rigid rotor functions

Consider an angular momentum operator Lˆ acting on an abstract Hilbert space , [do not confuse this Lˆ with the classical vector of Eq. (69) or the H antisymmetric 3 3 matrices of Eq. (22)] ×

Lˆ1 Lˆ Lˆ with Lˆ , Lˆ = i ε Lˆ . (100) ≡  2 i j ijk k Lˆ k 3 X   Deﬁne also Lˆ := Lˆ i Lˆ . (101) ± 1 ± 2

ˆ Ä e−iÒ· (103) gives an active rotation of lm around n over an angle n . Choosing n | i | | successively along the z, y and z axis and applying (103) three times, we de- ﬁne an operator that represents a R3 rotation (in the Euler parametrization, see Theorem 1) on the Hilbert space H Uˆ(α,β,γ) := e−iαL3 e−iβL2 e−iγL3

We now introduce the so-called Wigner D-matrix by

(l) ′ D ′ (αβγ) := lm Uˆ(α,β,γ) lm . (104) mm h | | i Below we shall show that

Lˆ Uˆ(α, 0, 0) Lˆ Uˆ(α, 0, 0)† = ( sin α , cos α) 1 (105) 2 − Lˆ 2 Lˆ1 Uˆ(α,β, 0) Lˆ Uˆ(α,β, 0)† = (cos α sin β , sin α sin β , cos β) Lˆ (106). 3  2 Lˆ3   Knowing this, we consider ∂ Uˆ(αβγ) = iLˆ Uˆ(αβγ) ∂α − 3 ∂ Uˆ(αβγ) = ie−iαLˆ3 Lˆ eiαLˆ3 Uˆ(αβγ) ∂β − 2 Lˆ = i( sin α , cos α) 1 Uˆ(αβγ) − − Lˆ 2 ∂ Uˆ(αβγ) = ie−iαLˆ3 e−iβLˆ2 Lˆ eiβLˆ2 eiαLˆ3 Uˆ(αβγ) ∂γ − 3

Lˆ1 = i(cos α sin β , sin α sin β , cos β) Lˆ Uˆ(αβγ) −  2 Lˆ3   24 or

∂/∂α 0 0 1 Lˆ1 ∂/∂β Uˆ(αβγ)= i sin α cos α 0 Lˆ Uˆ(αβγ)   −  −   2 ∂/∂γ cos α sin β sin α sin β cos β Lˆ3       (107) Equation (48) shows that the matrix in Eq. (107) is M T . From Eq. (87) follows (with ~ = 1)

Jˆ1 ∂/∂α −1 Jˆ = i M T ∂/∂β (108)  2 −   Jˆ ∂/∂γ 3     so that Jˆ Uˆ(αβγ)= Lˆ Uˆ(αβγ). (109) i − i This expression relates the angular momentum operators Lˆi acting on the abstract Hilbert space to the rigid-body angular momentum operators. H Consider the action of Jˆ3 on the D-matrix,

ˆ (l) ˆ ˆ ′ ˆ ˆ ′ J3 Dmm′ (αβγ) = lm J3U lm = lm L3U lm h | (l) | i h | − | i = m D ′ (αβγ), (110) − mm where we used Lˆ† = Lˆ . Complex conjugate and use Jˆ∗ = Jˆ then 3 3 i − i ˆ (l) ∗ (l) ∗ J3 Dmm′ (αβγ) = m Dmm′ (αβγ) . (111)

Likewise we ﬁnd

(l) ∗ 1/2 (l) ∗ Jˆ D ′ (αβγ) = l(l + 1) m(m 1) D ′ (αβγ) . (112) ± mm − ± m±1,m From Jˆ2 = Jˆ Jˆ + Jˆ (Jˆ 1) we derive easily + − 3 3 − ˆ2 (l) ∗ (l) ∗ J Dmm′ (αβγ) = l(l + 1) Dmm′ (αβγ) . (113)

Similar relations can be derived for the body-ﬁxed operators ˆ . Instead of Pi modifying the above procedure, we proceed as follows. From Eqs. (87) and (37) follows R (αβγ) Jˆ = ˆ . (114) ij i Pj i X where we changed the notation from F (t) to R(αβγ). From Eq. (109) we ﬁnd R Jˆ Uˆ = R Lˆ U.ˆ (115) ij i − ij i i i X X 25 Use the relation (proved below)

† Rij(αβγ) Li = Uˆ(αβγ) Lj Uˆ(αβγ) . (116) i X Then Eq. (115) becomes

ˆ Uˆ(αβγ)= Uˆ(αβγ) Lˆ , (117) Pj − j which is the counterpart of Eq. (109). Computing matrix elements, complex conjugation and using ˆ∗ = ˆ we obtain Pj −Pj (l) ∗ ′ (l) ∗ ˆ D ′ (αβγ) = m D ′ (αβγ) (118a) P3 mm mm (l) ∗ ′ ′ 1/2 (l) ∗ ˆ D ′ (αβγ) = l(l + 1) m (m 1) D ′ (αβγ) (118b) P∓ mm − ± mm ±1 ˆ2 (l) ∗ (l) ∗ Dmm′ (αβγ) = l(l + 1) Dmm′ (αβγ) . (118c) P Note that the role of the step-up and step-down operators is interchanged. The complex conjugate of a Wigner D-matrix is an eigenfunction of a symmetric top Hamiltonian. Indeed, a symmetric top is characterized by −1 −1 I(0)1 = I(0)2 and hence we ﬁnd from Eq. (98)

2Tˆ = I(0)−1 ˆ2 + I(0)−1 I(0)−1 ˆ2. (119) rot 1 P 3 − 1 P3 From Eq. (118a) and (118c) we ﬁnd the eigenvalue equation

(l) ∗ −1 ′ 2 −1 −1 (l) ∗ 2Tˆ D ′ (αβγ) = l(l +1) I(0) +(m ) I(0) I(0) D ′ (αβγ) . rot mm 1 3 − 1 mm h i (120) (l) ∗ Since Dmm′ (αβγ) is an eigenfunction of the rigid rotor kinetic energy operator it is sometimes referred to as a rigid rotor function.

It rests us to prove Eq. (116) from which Eq. (105) and (106) follow as special cases. We will prove ﬁrst a similar result for the (n,ϕ) parametrization, cf. Eq. (17).

Theorem 4 Let nˆ be an arbitrary unit vector and ϕ be an angle, then solely from ˆ ˆ ˆ [Li, Lj]= i k εijkLk we ﬁnd P Lˆ1 Lˆ1

ˆ ˆ

Ä Ò Ä eiϕÒˆ· Lˆ e−iϕˆ· = R(nˆ,ϕ) Lˆ (121)  2  2 Lˆ3 Lˆ3     where R(nˆ,ϕ) is given by Eq. (17).

26 Proof Recursively we introduce a notation for multiple commutators

A, B (0) := A (122)

A, B (k+1) := A, B (k), B . (123) Thus, e.g. A, B (2) = A, B (1) , B = A, B , B . In this notation we can write ∞

ˆ ˆ 1

Ä Ò Ä e−iϕÒˆ· Lˆ eiϕˆ· = ϕk Lˆ, inˆ Lˆ . (124) k! · (k) k=0 X This famous relation (sometimes erroneously referred to as the Baker-Campbell- Hausdorﬀ theorem) follows easily from the Taylor expansion of F (ϕ)

ˆ ˆ ≡

Ä Ò Ä e−iϕÒˆ· Lˆ eiϕˆ· and F (0) = Lˆ. Indeed,

∞ 1 dkF F k (ϕ)= ϕ k (125) k! dϕ ϕ=0 Xk=0 and dF = (inˆ Lˆ) F (ϕ)+ F (ϕ)(inˆ Lˆ)= F (ϕ), inˆ Lˆ dϕ − · · · so that dF = Lˆ, inˆ Lˆ . dϕ · ϕ=0 Likewise from d2F dF (ϕ) = , inˆ Lˆ = F (ϕ), inˆ Lˆ , inˆ Lˆ dϕ2 dϕ · · · it follows that d2F = Lˆ, inˆ Lˆ , inˆ Lˆ , dϕ2 · · ϕ=0 and similarly for the higher terms. The multiple commutators satisfy

Lˆ, inˆ Lˆ = ( 1)k Lˆ, inˆ Lˆ (126) · (2k+1) − · Lˆ, inˆ Lˆ = ( 1)k Lˆ (nˆ Lˆ)nˆ . (127) · (2k) − − · Note that Eq. (127) only holds for k > 0, whereas Eq. (126) holds for k 0. ≥ Equation (126) is trivially true for k = 0. We prove Eq. (127) for k = 1 by

27 using ε ε ′ ′ = δ ′ δ ′ δ ′ δ ′ : a abc ab c bb cc − bc cb P Lˆ , inˆ Lˆ , inˆ Lˆ = (i)2 n n [Lˆ , Lˆ ], Lˆ a · · b c a b c b,c X = i n n ε Lˆ , Lˆ − b c abd d c b,c,d X = nbnc εabd εdce Lˆe b,c,d,eX = n n δ δ δ δ Lˆ b c ac be − ae bc e b,c,e X = n n Lˆ n n Lˆ b a b − b b a Xb Xb = n (nˆ Lˆ) Lˆ . a · − a Hence Lˆ, inˆ Lˆ = Lˆ, inˆ Lˆ , inˆ Lˆ = nˆ (nˆ Lˆ) Lˆ, (128) · (2) · · · − so that Eq. (127) holds for k= 1. We prove Eq. (126) for k 1 as follows ≥ Lˆ, inˆ Lˆ = Lˆ, inˆ Lˆ , inˆ Lˆ · (2k+1) · (2k) · = ( 1)k (Lˆ (nˆ Lˆ) nˆ), inˆ Lˆ (by induction) − − · · = ( 1)k Lˆ, inˆ Lˆ (since nˆ Lˆ, nˆ Lˆ = 0). − · · · In order to prove Eq. (127) for any k, we write

Lˆ, inˆ Lˆ = Lˆ, inˆ Lˆ , inˆ Lˆ · (2k) · (2k−1) · = ( 1)k−1 Lˆ , inˆ Lˆ , inˆ Lˆ (by induction) − · · = ( 1)k Lˆ nˆ(nˆ Lˆ) by Eq. (128) . − − · Next we split the sum on the right-hand side of Eq. (124) as follows,

∞ ∞ ϕ2k ϕ2k+1 Lˆ + Lˆ, inˆ Lˆ + Lˆ, inˆ Lˆ (2k)! · (2k) (2k + 1)! · (2k+1) k=1 k=0 X X∞ ( 1)k = Lˆ + Lˆ (nˆ Lˆ) nˆ − ϕ2k − · (2k)! k=1 ∞ X ˆ ( 1)k + Lˆ , inˆ Lˆ − ϕ2k+1 · (2k + 1)! k=0 X = Lˆ + Lˆ (nˆ Lˆ) nˆ (cos ϕ 1) + Lˆ, inˆ Lˆ sin ϕ − · − · = Lˆ cos ϕ +(nˆ Lˆ) nˆ (1 cos ϕ)+ Lˆ, inˆ Lˆ sin ϕ. · − · 28 Finally, Lˆ , inˆ Lˆ = n ε L = (nˆ Lˆ) (129) a · − b abc c − × a b,c X or Lˆ , inˆ Lˆ = Lˆ nˆ . (130) · × Hence

F (ϕ)= Lˆ cos ϕ + nˆ(nˆ Lˆ)(1 cos ϕ)+(Lˆ nˆ) sin ϕ. · − × Recalling from the proof of Eq. (17) that

R(n,ϕ)Lˆ 1 + sin ϕN + (1 cos ϕ)N 2]Lˆ ≡ − = Lˆ cos ϕ + sin ϕ(n Lˆ)+(1 cos ϕ)(n Lˆ)n × − · and noting that sin ϕ(n Lˆ) = sin( ϕ)(Lˆ n), while the cosine is even, we × − × see that Theorem 4 follows.

Equation (116) follows as an easy corollary. First take nˆ = (0, 0, 1). Write

Uˆ := exp[iϕ Lˆ3]. Then from Theorem 4:

Lˆ1 cos ϕ sin ϕ 0 Lˆ1 † − Uˆ Lˆ Uˆ = sin ϕ cos ϕ 0 Lˆ (131)  2    2 Lˆ3 0 0 1 Lˆ3       By taking nˆ = (0, 1, 0) we get the analogous relation for rotation around the y-axis.

Note

Provided ϕ is not any of the Euler angles contained in Jˆi, Theorem 4 holds for (Jˆ , Jˆ , Jˆ ) as well. Deﬁning ¯ = the theorem holds also for ( ¯ , ¯ , ¯ ). 1 2 3 Pi −Pi P1 P2 P3 9 The quantized energy levels of rigid rotors

For more details about the material in this section see the book by W. Gordy and R. L. Cook Microwave Molecular Spectra, 3rd Ed. Wiley, New York 1984. Let us order the inertia moments (the eigenvalues of the inertia tensor) of the rigid rotor: I I I or A B C, a ≤ b ≤ c ≥ ≥ where the rotational constants A, B, and C are half the inverse of the corresponding inertia moments, thus A =1/(2Ia), etc. The kinetic energy of the

29 rotor takes the form [Eq. (98)]

2 2 2 T = Pa + Pb + Pc (132) 2Ia 2Ib 2Ic = A 2 + B 2 + C 2 Pa Pb Pc One can distinguish four kinds of rotors:

1. A = B = C, the so-called spherical top. Clearly T = B 2. Recall spher P that by Eq. (118c) the Wigner D-matrix is an eigenfunction of 2, P hence (j) ∗ (j) ∗ Tspher Dmk(αβγ) = Bj(j + 1)Dmk(αβγ) . (133) An example of a spherical top is a tetrahedrally or octahedrally shaped body.

2. A > B = C, the so-called prolate top. Clearly T = B 2+(A B) 2. prol P − Pa We deﬁne Euler angles αp, βp and γp by expressing the principal axes of inertia (the eigenvectors of the inertia tensor) with respect to the frame −→eC with origin at the center of mass of the top and parallel to a space-ﬁxed frame,

(−→fb, −→fc, −→fa)=(−→ex, −→ey, −→ez ) R(αpβpγp). (134)

The vector −→fa plays the role of −→fz and hence from Eqs. (118a) and (118c)

T D(j) (α β γ )∗ = B j(j +1)+(A B) k2 D(j) (α β γ )∗. (135) prol mk p p p − mk p p p A prolate top is cigar shaped with the ‘cigar axis’ being at least a 3-fold rotation axis.

3. A = B>C, the so-called oblate top. Clearly T = B 2 +(C B) 2. obl P − Pc We deﬁne Euler angles αo, βo and γo by expressing the principal axes of inertia (the eigenvectors of the inertia tensor) with respect to the frame −→eC (−→fa, −→fb , −→fc)=(−→ex, −→ey, −→ez ) R(αoβoγo). (136)

The vector −→fc plays the role of −→fz and hence from Eqs. (118a) and (118c)

T D(j) (α β γ )∗ = B j(j +1)+(C B) k2 D(j) (α β γ )∗. (137) obl mk p p p − mk p p p An oblate top is frisbee shaped with its rotational axis being at least a 3-fold rotation axis.

30 4. A>B>C, the so-called asymmetric top. The kinetic energy can be rewritten in different ways. The choice depends on the relative size of B C and A B, because it is common to stay as close as possible − − to the symmetric top. Assuming that B C < A B (the top is − − prolate-like), we write 1 T = 1 (B +C) 2 + A 1 (B +C) 2 + (B C)( 2 + 2 ), (138) asym 2 P − 2 Pa 4 − P+ P− where = i . If the last term were not present in Eq. (138) the P± Pb ∓ Pc Wigner D-matrix would be an eigenfunction. In the oblate-like limit 1 T = 1 (A+B) 2 + C 1 (A+B) 2 + (A B)( 2 + 2 ), (139) asym 2 P − 2 Pc 4 − P+ P− where = i . P± Pa ∓ Pb The rotational energy levels of the first three tops are simply their eigenvalues. Note that the levels of a prolate top increase with increasing k , while | | those of an oblate top decrease with increasing k . This is because A B | | − and C B are positive and negative, respectively. The levels with k are − ± degenerate. In the case of an asymmetric top one must numerically diagonalize the kinetic energy operator in order to obtain the energy levels. A suitable basis is a row of the D-matrix characterized by a certain fixed m. When the top rotates in isotropic space m is conserved and the elements in row m span a 2j + 1-dimensional space that is strictly invariant under , and , so Pa Pb Pc that no further approximation is introduced by restricting the basis to this row. It is common to label the asymmetric top energy levels by their correlation with neighboring prolate and oblate symmetric top energy levels. This correlation is on basis of symmetry. In order to explain this we use Theorem

4. From this theorem we ﬁnd that rotation over ϕ = π around −→fj (j =1, 2, 3) leaves invariant and maps the other two angular momentum components Pj as . The rigid rotor kinetic energy being quadratic in the ’s Pk 7→ −Pk Pj is invariant under these three rotations. The three rotations (plus identity) form a group V4 (the four group) which is isomorphic to D2.

We ﬁnd upon rotation around −→fb over π

(−→f , −→f , −→f ) (−→f , −→f , −→f ) R (π) b c a 7→ b c a x = (−→ex, −→ey, −→ez ) R(αp, βp,γp) Rx(π) = (e , e , e ) R(α + π, π β , 2π γ ), (140) −→x −→y −→z p − p − p 31 where Rx(ϕ) is deﬁned in Eq. (5). Equation (140) can easily be proved from the relations

R (π)= R (π)R (π)= R (π)R (π) and R (π)R (φ)= R ( φ)R (π) x y z z y x z z − x and the corresponding relations with other subscripts. Hence

α α + π p 7→ p Rb(π) : β π β  p 7→ − p γp 2π γp 7→ − In this manner one derives the following table:

αp βp γp αo βo γo R (π) α β π + γ α + π π β 2π γ a p p p o − o − o R (π) α + π π β 2π γ α + π π β π γ b p − p − p o − o − o R (π) α + π π β π γ α β π + γ c p − p − p o o o

In order to adapt row m of the D-matrix to V4 we introduce Wang functions, (k 0) ≥ 1 αβγ jkǫ := D(j) (αβγ)∗ +( 1)j+ǫD(j) (αβγ)∗ . (141) h | i √2 m,k − m,−k For k = 0 there are two Wang functions, distinguished by the value of ǫ =1, 2, 6 which are degenerate in the case of prolate or oblate symmetric tops. We recall the following properties of the D-matrix

D(j) (α + π, π β, 2π γ)∗ = ( 1)jD(j) (αβγ)∗ (142a) m,k − − − m,−k D(j) (α + π, π β, π γ)∗ = ( 1)j+kD(j) (αβγ)∗ (142b) m,k − − − m,−k D(j) (α,β,π + γ)∗ = ( 1)kD(j) (αβγ)∗ (142c) m,k − m,k Writing η (α,β,γ), we have the Wang functions η jk ǫ and ≡ h p | a p i η jk ǫ , which are eigenfunctions of T and T , respectively. These h o | c o i prol obl functions have the symmetry properties given in the following table

32 ε ε p o k 1 2 k 1 2 a c

B (3 ) 0 B − c 30 c 3 B B c b 1 B B B (3 ) b a b 31

B (3 ) a 21 2 B A A (3 ) c 22 2 B A a

B (3 ) c 12

B (3 ) b 13 1 B B c b B (3 ) a 03 3 B B b a 0 B − a

Figure 1: Example of an energy level correlation diagram for j =3. On the left the prolate symmetric top limit for which the values with ǫp = 1, 2 are degenerate. On the right the oblate limit. The symmetry labels corresponding to ǫo =1, 2 are given. In the middle the asymmetric top levels with their symmetry and the jkakc designation.

The group V4 has the following character table:

V4 E Ra(π) Rb(π) Rc(π) A 11 1 1 B 1 1 1 1 a − − B 1 1 1 1 b − − B 1 1 1 1 c − − The Wang functions are of deﬁnite symmetry, their symmetry assignment is

33 k ǫ η jk ǫ η jk ǫ h p | a p i h o | c o i e e A A

e o Ba Bc

o e Bb Ba

o o Bc Bb Here e stands for even value and o for odd value of k and ǫ. By inspection of this table we ﬁnd that η j, k ,k + k (i.e., ǫ = k + k ) has the same h p | a a c i p a c symmetry as η j, k ,k . h o | c a i The correlation of asymmetric top levels with symmetric top levels goes as follows. On the left we draw prolate levels (increasing with increasing k ) of | | certain j and on the right the oblate levels (decreasing with increasing k ) of | | the same j. By means of the above table we make the symmetry assignment. In the middle we draw the energy level diagram of the asymmetric top (for instance obtained from numerical diagonalization). We assume that the V4 symmetries of the asymmetric top levels are known. Then we draw a line from the middle level to the nearest level on the left of the same symmetry, say it has quantum number ka, and to the right also with the same symmetry, which has quantum number kc. The asymmetric top level in the middle is designated by jkakc . See Fig. 1 for an example with j = 3.

10 Angular momenta and Lie derivatives

To introduce the manner in which Lie derivatives, inﬁnitesimal rotations, and their connection with angular momenta, can be derived, we consider a function f(ϕ), 0 ϕ< 2π. Active rotation of the function over α gives ≤ Rˆ(α) f(ϕ)= f(ϕ α). (143) − (The minus sign appears here to satisfy a homomorphism condition, see E.P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.) Assuming f to be analytic we can make a Taylor expansion d 1 d2 f(ϕ α) = f(ϕ) α f + α2 f + − − dϕ 2! dϕ2 ··· = e−α d/dϕ f(ϕ). (144) For inﬁnitesimal rotations over ∆α, ∆α (∆α)2, the rotation operator ≫ Rˆ(∆α) exp[ ∆α d/dϕ] can be approximated by ≡ − d Rˆ (∆α) 1 ∆α . (145) ∼ − dϕ

34 This equality holds strictly in the limit ∆α 0. The operator → ∆α d/dϕ i∆α L − ≡ z generates the inﬁnitesimal rotation of f(ϕ) overb ∆α. In group theory iLz is known as a Lie derivative. In this example the Lie derivative only diﬀers in sign from the ordinary derivative d/dϕ. In physics Lz is known as thebz component of a one-particle angular momentum operator. A Lie derivative can be computed if we take α to dependb on a parameter t, such that the curve α(t) passes through 0, i.e. α(0) = 0. The “velocity” in 0:α ˙ (0) := [dα/dt] is not equal to zero. Then writing ϕ′(t)= ϕ α(t) we t=0 − get by the chain rule

df ϕ′(t) df(ϕ′) dϕ′(t) = . dt dϕ′ dt t=0 α=0 t=0 Since df(ϕ′) df(ϕ) dϕ′(t) = and = α˙ (0), dϕ′ dϕ dt − α=0 t=0 we obtain df ϕ′(t) d = α˙ (0) f. (146) dt − dϕ t=0 If we write this as df ϕ′(t) ∆f ∆α ∆f = lim = lim , (147) dt ∆t→0 ∆t ∆t→0 − ∆t ∆ϕ t=0 or df(ϕ) ∆f = ∆α , − dϕ we see that the inﬁnitesimal generator ∆α d/dϕ can be obtained by consid- ′ − ering [df ϕ (t) /dt]t=0. This diﬀerentiation picks out the second (linear) term in the Taylor expansion of f(ϕ α), i.e., it makes a linear approximation to − the rotated function.

10.1 Infinitesimal rotations of functions f(r) After this preamble we consider an arbitrary function f(x, y, z)= f(r) and a rotation R(t) with R(0) = 1 and R˙ (0) = Ω. The matrix Ω is antisymmetric. We will show that the well-known orbital angular momentum operator is a generator of an infinitesimal rotation of the function f(r). We consider f r′(t) f R(t)−1 r , where the inverse of R(t) appears to satisfy the ≡ homomorphism condition, cf. Wigner loc. cit.. We differentiate f r′(t) with 35 respect to t at t = 0, which is equivalent to considering the second (linear) term in a Taylor expansion. Invoke the chain rule, use summation convention, and note that at t = 0: r′ = r,

′ ′ ′ df r (t) ∂f(r ) dri = ′

dt ∂r ′ dt Ö t=0 i Ö = t=0 ∂f(r) dR (t)−1 = ij r ∂r dt j i t=0 = (∇f) Ω r. (148) − · From lemma 2 we know that Ω r = ω r and by a well-known rule from × vector analysis, we find df(r′(t)) = (ω r) ∇f = ω (r ∇) f(r). dt − × · − · × t=0 The orbital angular momentum operator is defined as L := i(r ∇), so − × that ′ df r (t) b = i(ω L) f(r). (149) dt − · t=0 This result holds for an arbitrary rotation Rb(t) with tangent Ω( ω) at ↔ t = 0. To be more specific we write the matrix R(t) in the xyz-parametrization (Theorem 1)

R(t)= Rx ϕ1(t) Ry ϕ2(t) Rz ϕ3(t) , and we get the matrix expression dR =ϕ ˙ (0) L +ϕ ˙ (0) L +ϕ ˙ (0) L Ω, dt 1 x 2 y 3 z ≡ t=0 so that ω ϕ˙ (0). The quantities L appearing in this equation are 3 i ≡ i i × 3 matrices and must not be confused with the vector operator L. Then, applying Eq. (149), we may write b df r′(t) = i ω L + ω L + ω L f(r). (150) dt − 1 x 2 y 3 z t=0 The factorization of R(t) into a productb of threeb matricesb leads to the result that any infinitesimal rotation can be expanded as a sum of three infinitesimal rotations applied to f(r): around the z-axis over ∆ϕ3, around the y-axis over ∆ϕ2 and around the x-axis over ∆ϕ1. The differential operators Li span a 3-dimensional Lie algebra isomorphic to so(3). In group theory the operator ω (r ∇) (acting on f) is knownb as − · × the Lie derivative (of f) along the tangent ω. The Lie derivatives form a Lie

36 algebra which is (isomorphic to) the Lie algebra of a transformation Lie group [in this example SO(3)]. The Lie algebra product can be easily established,

(r ∇) , (r ∇) = ε (r ∇) × i × j − ijk × k k X so that

Li , Lj = i εijk Lk. k X b b b 10.2 Infinitesimal rotations of functions of Euler angles Let us next consider a function ψ(η) of the Euler angles η (η , η , η ) that ≡ 1 2 3 specify the position of a body-fixed frame −→fC with respect to a space-fixed frame −→eC , i.e., as in Eq. (37) we have

−→fC = −→eC R(η). (151) In this case we can also define infinitesimal rotations. More specifically we will show that we can introduce two definitions: acting on the left and on the right of R(η). In (Lie) group theory these operations are known as left and right translations, respectively, on the group manifold. We will relate this concept to infinitesimal rotations of space- and body-fixed frames. Left infinitesimal rotations are generated by space-fixed rotor angular momentum operators Jî [cf. Eq. (91)] and those acting on the right by body-fixed operators ˆ [cf. Eq. (89)]. Remember that Jˆ acts on the left subscript of a rigid Pi i rotor function and ˆ on the right subscript, which mnemonically is a useful Pi fact. In order to show that Jˆ and ˆ generate infinitesimal rotations, we define i Pi an (infinitesimal) rotation by acting on −→fC or on −→eC . Writing a rotation matrix in the xyz-parametrization as S ξ(t) S (ξ ) S (ξ ) S (ξ ), with ≡ z 3 y 2 x 1 ξ =(ξ ,ξ ,ξ ), we have either 1 2 3 ′ e−→C = −→eC S ξ(t) , (152) or ′ f−→C = −→fC S ξ(t) . (153) We shall show that the infinitesimal rotation defined by Eq. (152) is generated by i ξ˙ Jˆ and that the generator arising from Eq. (153) is i ξ˙ ˆ . − i i i i iPi How doP the Euler angles transform under Eqs. (152) and (153)?PIn the case L of (152) we keep −→fC fixed and define η by

′ L −→fC = e−→C R(η ). (154)

37 From Eqs. (152) and (151)

L −→fC = −→eC S(ξ) R(η )= −→eC R(η) or R(ηL)= S(ξ)−1 R(η) (left translation). (155)

R In the case of (153) we keep −→eC ﬁxed and deﬁne η by

′ R f−→C = −→eC R(η ). (156) From Eqs. (153) and (151)

′ f−→C = −→eC R(η) S(ξ) or R(ηR)= R(η) S(ξ) (right translation). (157) The matrix giving the right translation is not inverted, while the left translation is performed by an inverted matrix. Both translations give a homo- morphic map of the group manifold. Note that the columns of R(η) represent the coordinate vectors of the body-fixed frame with respect to the space-fixed frame −→eC . Equation (155) describes a rotation of each coordinate vector individually and hence corresponds to a rotation of the rigid body, [which is the inverse of the rotation Eq. (152) of the space-fixed frame]. If no external torque acts on the body (isotropic space), Eq. (155) describes a symmetry operation. On the other hand, Eq. (157) describes a linear combination of the columns of R(η). If these are for instance principal axes, then (157) “undiagonalizes” the inertia tensor, except when 2 or 3 inertia momenta are equal. That is, in the case of a spherical top all matrices S(ξ) are symmetry operators, whereas in the case of a symmetric top rotations around one fixed axis are symmetry operators. To define the infinitesimal rotations we consider ′ ′ dψ(η ) ∂ψ(η ) ′ = ′ η˙i(0). (158)

dt ∂η ′ t=0 i i = X Let us ﬁrst compute η˙ ′(0) from the deﬁnition Eq. (155), i.e., η′ ηL. We ≡ use S ξ(0) = 1 and dS−1 dS = S−1 S−1 dt − dt so that −1 dS ξ(t) dS ξ(t) = . dt − dt !t=0 !t=0

38 Differentiate Eq. (155) and use Eqs. (24)–(26), dR(η′) dS ξ(t) = R(η)= ξ˙ (0) L R(η). (159) dt − dt − i i t=0 t=0 i X On the other hand, from lemma 3 we obtain dR(η′) = Ω R(η′) = ω L R(η) (160) dt t=0 i i t=0 i X with ω M η˙ ′(0) . Comparing Eqs. (159) and (160) and recalling that the ≡ Li are linearly independent, we find ξ˙ (0) = ω = M η˙′ (0) i − i − ij j j X or η˙ ′(0) = M −1 ξ˙(0). (161) − Substitution of Eq. (161) into Eq. (158) and use of Eq. (87) gives ′ dψ(η ) −1 ∂ = ξ˙ (0) M T ψ(η) (162) dt − j ji ∂η t=0 i,j i X = i ξ˙ (0) Jˆ ψ(η). (163) − j j j X Since η′ ηL, we find that the infinitesimal left translation is indeed gener- ≡ ated by the space-fixed angular momentum operators Jî. It can be shown in general that the left Lie derivatives

−1 ∂ := M T (164) Lj − ji ∂η i i X span the Lie algebra of the Lie group in question, i.e, SO(3). (Left and right translations constitute in fact regular representations of the group). In the case at hand this yields the so(3) Lie algebra product, , = ε (165) Li Lj ijk Lk k X so that with Jˆ = i i Li Jî , Jˆj = i εijk Jˆk. (166) k X Let us next consider the infinitesimal rotation derived from Eq. (157), i.e., η′ ηR. Differentiation of this equation yields, ≡ dR(η′) dS ξ(t) = R(η) = ξ˙ (0) R(η) L . (167) dt dt i i t=0 t=0 i X 39 From Eq. (167) and Eq. (160) we obtain

−1 ξ˙i(0) R(η) Li R(η) = ωi Li. (168) i i X X By Eq. (30) applied to the left-hand side of Eq. (168),

ξ˙i(0) R(η)ji Lj = ωj Lj. i,j j X X From which R(η) ξ˙(0) ω = M η˙ ′ (0). (169) ≡ Using N = R(η)−1 M [Eq. (53)], we ﬁnd

η˙ ′(0) = N −1 ξ˙(0). (170)

Substitute into Eq. (158),

dψ(η′) ∂ = ξ˙ (0)(N T )−1 ψ(η) (171) dt j ji ∂η t=0 i,j i X =: ξ˙ (0) ψ(η). (172) j ℜj j X The right Lie derivatives span also the Lie algebra so(3), i.e., they obey ℜj the so(3) product , = ε . (173) ℜi ℜj ijk ℜk k X From Eqs. (87) and (171) follows that the body-ﬁxed angular momentum operator is given by ˆ = i , (174) Pi − ℜi so that ˆ , ˆ = i ε ˆ , (175) Pi Pj − ijk Pk k X where we meet again the famous i in the commutation relations of the − body-ﬁxed angular momentum.

Note Theorem 4, which holds for (i =1, 2, 3), states that −Pi

1 1 − P3 P P3 P e iϕ eiϕ = R (ϕ) , P2 z P2 P3 P3    

40 where R (ϕ) R(−→f ,ϕ) is deﬁned in Eq. (5). Here ϕ is an arbitrary ﬁxed z ≡ 3 angle. Similar equations hold for rotations around −→f 1 and −→f 2. Rotation over ϕ = π around f gives , and . The 3 P1 7→ −P1 P2 7→ −P2 P3 7→ P3 kinetic energy Eq. (98) is invariant under this map. Finally we want to point out that

Jˆ, ˆ =0, i, j =1, 2, 3. (176) i Pj This follows directly from the fact that the rotations (155) and (157) commute, so that also the corresponding inﬁnitesimal rotations commute. It is of some interest to prove this more directly. Equation (114) shows that Jˆ and ˆ are related in the same way as the frames [Eq. (151)], i Pj ˆ = R (η) Jˆ . (177) Pi ji j i X It can be shown along the lines that led to Eq. (109) that

Jˆi, Rjk(η) = εijl Rlk(η). (178) l X From (177) and (178), suppressing η in the notation we get,

Jˆ , ˆ = Jˆ , R Jˆ = R Jˆ , Jˆ + Jˆ , R Jˆ k Pi k ji j ji k j k ji j j j X X = Rji εkjl Jˆl + εkjl Rli Jˆj =0. (179) j,l X Since the kinetic energy of the rigid rotor in the principal axes frame is [Eq. (98)] −1 Tˆ = 1 ˆ I 0 ˆ , (180) 2 Pi ii Pi i X where the inertia moments I(0)ii are constants, it follows directly from (176) that Jˆi commutes with this Hamiltonian.