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Revised 5/2/2014 Lecture Notes for Chemistry 543, Part III LECTURE 30: SPECTROSCOPY OF RIGID ROTORS AND HARMONIC OSCILLATORS. Reading: Chapters 6 and 7 of Bernath; Levine pp 142-174. As in any branch of spectroscopy, we need to determine (1) the eigenvalues of the Hamiltonian in order to know the frequencies of the possible transitions, (2) the selection rules to determine what transitions actually occur, (3) the transition dipole moments to determine the strengths of the transitions, (4) degeneracies, and (5) relative populations in order to determine the intensity of transitions. The simplest case beyond an atom is a diatomic molecule. To the lowest order of approximation, we may treat such a molecule as a rigid rotor and a harmonic oscillator and consider its rotational, vibrational, and electronic degrees of freedom as completely independent quantities. At the next level of approximation, we consider the interactions of these degrees of freedom as perturbations. We start with the simplest problem of pure rotations. The Hamiltonian for a rigid rotor is Ĥrot = Jˆ2 Jˆ2 = , 2µRe2 2I (743) where Jˆ is the rotational angular momentum operator, Re is the equilibrium internuclear distance, and I is the moment of inertia. The eigenfunctions are the well known spherical harmonics, 1 |M | YJM (θ, ϕ) = √ PJ (cos θ)eiM ϕ , 2π |M | where PJ (744) is a modified Legendre Polynomial. The eigenvalues are EJM = J(J + 1)~2 , 2I (745) and the degeneracy is gJ = 2J + 1. (746) 2 The energy in wave numbers is given by EJ = BJ(J + 1), hc (747) where the rotational constant is Be = h . 8π 2 cI (748) The selection rule is obtained by evaluating the matrix element of the transition dipole operator. The integral over ϕ is elementary. For µx , which is proportional to cos ϕ, ∫ 2π ∫ 1 2π i(M1 −M2 )ϕ iϕ −iM2 ϕ iM1 ϕ e cos ϕe dϕ = e (e + e−iϕ )dϕ = πδM1 ,M2 ±1 2 0 0 (749) vanishes unless M2 − M1 = ±1. (750) The same selection rule results from the integral over ϕ for µy , which is proportional to sin ϕ: ∫ ∫ 2π 1 2π i(M1 −M2 )ϕ iϕ −iM2 ϕ iM1 ϕ e (e − e−iϕ )dϕ. (751) e sin ϕe dϕ = 2i 0 0 In the case of µz , which does not depend on ϕ, ∫ 2π ei(M1 −M2 )ϕ dϕ = δM1 M2 , (752) 0 which yields M2 = M1 . Thus the selection rule for M is therefore ∆M = 0, ±1. (753) For randomly aligned molecules, all three transitions are equally likely, whereas for molecules aligned along the axes of the electric field, the various allowed transitions correspond to different polarizations of the electric field. The selection rule for J follows from the recursion relations for Legendre Polynomials, ] 1 [ |M |+1 |M | |M |+1 sin θPJ (cos θ) = PJ+1 (cos θ) − PJ−1 (cos θ) 2J + 1 ] 1 [ |M |−1 |M |−1 = (J + |M |)(J + |M | − 1)PJ−1 (cos θ) − (J − |M | + 1)(J − |M | + 2)PJ+1 (cos θ) , 2J + 1 (754) and the orthogonality property, ∫ 1 |M | |M | PJ (x)PJ ′ (x)dx = −1 2(J + M )! δJJ ′ . (2J + 1)!(J − M )! (755) 3 For µz we found that M2 = M1 . Since the θ integral is the same as Eq. (755), we find that J2 = J1 . For a pure (i.e., non-vibrational and non-electronic) rotational transition, this is not a transition at all. For µx and µy we found that M2 = M1 ± 1. For this case, Eqs. (754) and (755) imply that J2 − J1 = ∆J = ±1. (756) From the selection rule we deduce that the pure rotational spectrum is a series of evenly spaced lines with frequencies ν̃J = Be [J + 1)(J + 2) − J(J + 1)] = 2Be (J + 1) (757) ∆ν̃ = 2Be . (758) and spacing of To get the total transition probability (the “line strength”), we need to sum over all 2J + 1 values of M and all three polarization directions. The result is called the HönlLondon factor, which is a function of J (and more generally of K and Λ, which we will learn about later). These factors are tabulated in Herzberg I, page 208 [2] and Zare, page 286 [7]. Using the same methods may be used to derive the selection rules for rotational Raman transitions. In this case we evaluate the matrix elements of the polarizability operator, which is proportional to products of the Cartesian coordinates.The result is: ∆J = ±2. (759) The Raman frequency shift for ∆J = 2 is ν̃J = Be (J + 2)(J + 3) − Be J(J + 1) = 4Be J + 6Be = 4Be (J + 3/2), (760) with an equivalent expression for ∆J = −2. The spacing between lines is therefore 4Be . Next we consider pure vibrational transitions. We will start with the case of a harmonic oscillator. The Hamiltonian in the center of mass coordinate system is Ĥvib = − or ~2 d2 1 2 + kx , 2µ dx2 2 Ĥvib ~ d2 1 k 2 =− + x, 2 ~ω 2µω dx 2 ~ω (761) (762) 4 where x = R − Re (763) is the deviation of the bond length from its equilibrium value, and the vibrational frequency in radians/s is ω= Defining the reciprocal scale length, √ k/µ. (764) √ β, by β 2 = µk/~2 , β = µω/~ = k/~ω, and the dimensionless distance, y= √ βx, (765) (766) the reduced Hamiltonian becomes Ĥvib 1 d2 1 =− + y2. 2 ~ω 2 dy 2 (767) Note that y = 1 corresponds to the turning point of a classical oscillator having an energy of ~ω/2. The harmonic oscillator eigenfunctions are ψv = |v⟩ = Nv e− 2 y Hv (y), 1 2 (768) where Hv (y) is a hermite polynomial, and [√ Nv = β/π v 2 v! ]1/2 We derive the selection rule for v by evaluating the integral ∫ ∞ ⟨n|ey|m⟩ = e ψn yψm dy. (769) (770) −∞ We evaluate this and any integral of a polynomial of y using the recursion relation 1 yHm = mHm−1 + Hm+1 , 2 (771) which is equivalent to 1 Nm Nm |m − 1⟩ + |m + 1⟩ Nm−1 2 Nm+1 √ √ = m/2 |m − 1⟩ + (m + 1)/2 |m + 1⟩. y|m⟩ = m (772) 5 In deriving the previous equation, we used the relations 2 ey /2 Hm = |m⟩, Nm (773) Nm 1 =√ . Nm−1 2m (774) ⟨m|n⟩ = δmn (775) v2 − v1 = ∆v = ±1. (776) and Applying the orthogonality condition, we deduce that It follows that the infrared for a pure vibrational spectrum of a harmonic oscillator consists of a single line. LECTURE 31: THE ANHARMONIC OSCILLATOR AND NON-RIGID ROTOR Now let’s consider the anharmonic oscillator. We replace the parabolic potential by a Taylor series in x = R − Re , 1 1 1 V (R) = V (Re ) + V ′ (Re )x + V ′′ (Re )x2 + V ′′′ (Re )x3 + V (IV ) (Re )x4 + · · · . 2 3! 4! (777) We set V (Re ) = 0, and from the definition of Re the first derivative is also zero. We are left with 1 V (R) = kx2 + cx3 + dx4 + · · · 2 (778) V (R) 1 = y 2 + c′ y 3 + d ′ y 4 + · · · ~ω 2 (779) or equivalently We next use time-independent perturbation theory to calculate the effect of these terms on the eigenvalues and eigenfunctions. To first order, (1) Ev = ⟨v|c′ y 3 + d′ y 4 |v⟩ = d′ ⟨v|y 4 |v⟩. ~ω (780) 6 The y 3 term vanishes because the integral of the product of an even and odd functions is zero. We apply the recursion relation twice to both the bra and the ket: √ √ v v+1 |v − 1⟩ + |v + 1⟩ y|v⟩ = 2 2 √ √ v v+1 2 y |v⟩ = y|v − 1⟩ + y|v + 1⟩ 2 2 [ [√ ] √ ] √ √ √ √ v v−1 v v+1 v+1 v+2 = |v − 2⟩ + |v⟩ + |v⟩ + |v + 2⟩ 2 2 2 2 2 2 1 1√ 1√ v(v − 1) |v − 2⟩ + (2v + 1) |v⟩ + (v + 1)(v + 2) |v + 2⟩. = 2 2 2 (781) We deduce that Ev d′ d′ = [v(v − 1) + (2v + 1)2 + (v + 1)(v + 2)] = [2v 2 + 6v + 3]. ~ω 4 4 (1) (782) To evaluate the c′ term, we need to go to second order perturbation theory, En(2) = ∑ |Hmn |2 m̸=n En − Em (1) (0) (0) ∑ [c3 ⟨m|y 3 |n⟩]2 = (m − n)~ω m̸=n (783) From the recursion relation we realize that this sum has four terms, [ ] (2) Ev 1 ′ 2 1 3 2 3 2 3 2 3 2 = (c ) ⟨v|y |v + 3⟩ + ⟨v|y |v + 1⟩ − ⟨v|y |v − 1⟩ − ⟨v|y |v − 3⟩ . (784) ~ω 3 3 ∫∞ Problem 69. Evaluate the integral −∞ ϕ0 x3 ϕ3 dx, where ϕn (x) is the nth harmonic oscillator eigenfunction. The matrix elements may be evaluated using the methods developed previously. We first evaluate ⟨v|y 3 and project out terms with v ± 1 and v ± 3. Putting all the pieces together we find Ev = G(v) = ωe (v + 1/2) − ωe xe (v + 1/2)2 + ωe ye (v + 1/2)3 + ωe ze (v + 1/2)4 · · · hc (785) Terms in ωe ye , ωe ze etc. require going to higher order perturbation theory. The quantity G(v) is called the term value, where ωe = ν̃ is the frequency in wave numbers measured with respect to the minimum of the potential energy. Note the minus sign before the ωe xe term. The zero point energy is given by 1 1 1 = G(0) = ωe − ωe xe + ωe ye . 2 4 8 (786) 7 We may also define the energy with respect to the ground state (v = 0), G(v) = ω0 v − ω0 x0 v 2 + ω0 y0 v 3 + · · · (787) Until now we were interested only in the shifts of the energy levels produced by an anharmonic potential. Our larger goal is to determine which transitions are allowed and to calculate the transition strengths. Do do this, we use first order perturbation theory to calculate the change in the wave function caused by the lowest order perturbation, namely c′ y 3 . |n⟩(1) = |n⟩(0) + = |n⟩(0) + c′ [1 ∑ (1) |Hmn | (0) m̸=n En − (0) Em |m⟩(0) ⟨n|y 3 |n + 3⟩|n + 3⟩(0) + ⟨n|y 3 |n + 1⟩|n + 1⟩(0) ] 1 − ⟨n|y 3 |n − 1⟩|n − 1⟩(0) − ⟨n|y 3 |n − 3⟩|n − 3⟩(0) 3 3 (788) The superscript (0) is there to remind you that the expansion is in the unperturbed basis set. Once we know what the wave function looks like, we may calculate the transition dipole moments for different final states: 0 → 1 transitions are called fundamentals; 0 → 2 terms are called overtones or second harmonics, etc. The calculation of µ12 using the perturbed wave function also involves use of the recursion relation. For example, |4⟩ contains contributions from |1⟩(0) ,|1⟩(0) ,|3⟩(0) ,|4⟩(0) , |5⟩(0) , and |7⟩(0) . The a transition |0⟩(0) → |4⟩(0) is allowed through the |1⟩(0) component of |4⟩. Problem 70. Use first order perturbation theory to calculate the strength of the first overtone of an anharmonic oscillator and compare it with the strength of the fundamental. Express your answer in terms of the single parameter c′ . (This is a long problem and will count double.) From the line spacings we can determine the anharmonicity constants. Using the ωe term 8 values and neglecting cubic and higher contributions, ( ) ( ) 1 1 2 G(v) = ωe v + − ωe xe v + v + 2 4 ( ) ( ) 3 9 2 G(v + 1) = ωe v + − ωe xe v + 3v + 2 4 ∆Gv+1/2 = G(v + 1) − G(v) = ωe − 2ωe xe v − 2ωe xe (789) ∆Gv+3/2 = G(v + 2) − G(v + 1) = ωe − 2ωe xe v − 4ωe xe ∆2 Gv+1 = ∆Gv+3/2 − ∆Gv+1/2 = −2ωe xe Higher order anharmonicity constants may be obtained by looking at higher overtones. Where does the anharmonic potential come from? The short answer is that it is obtained either experimentally or by solving the Schrodinger equation for the electronic energy as a function of R. It is also convenient to fit this potential energy curve to a model function such as the Morse function, which has the form ( )2 V (R) = De 1 − e−α(R−Re ) . (790) Here De is the dissociation energy and has nothing to do with the centrifugal distortion constant. It is straight forward to extract the force constant and therefore ωe from a Taylor series expansion of the potential, V (R) = De [(1 − 1 + α(R − Re )]2 = 1 De kx2 , 2 (791) or by simply taking the second derivative of V (R) and evaluating it at R = Re . The result is k = 2α2 De . (792) It turns out that the Schrodinger equation can be solved exactly for this potential, yielding eigenvalues ~ω Ev = v + 1/2 − (v + 1/2)2 . ~ω 4De (793) Some books and data bases define the Morse potential by the equation [6] ( )2 V = 1 − e−x , (794) x = 2β(R − Re )/Re , (795) where 9 and ωe β= √ . 4 B e De (796) In this expression, β is dimensionless, ωe is the vibrational frequency, Be is the rotational constant, and De is the dissociation energy. The latter three quantities are all expressed in wave numbers. Problem 71. Derive Eq. (796). Next, let’s look at vibration and rotation together. The Hamiltonian becomes Ĥvib−rot = ~2 d2 1 Jˆ2 − + kx2 , 2 2 2µ(Re + x) 2µ dx 2 (797) plus the anharmonicity terms. We expand the rotational term in a Taylor series. Consider the function f (ϵ) = 1 . (1 + ϵ)2 (798) Its Taylor series expansion is 1 f (ϵ) = f (0) + f ′ (0)ϵ + f ′′ (0)ϵ2 + · · · 2 2 f ′ (ϵ) = − (1 + ϵ)3 6 f ′′ (ϵ) = (1 + ϵ)4 (799) f (ϵ) = 1 − 2ϵ + 3ϵ2 + · · · The rotational term in the Hamiltonian is therefore ) ( Jˆ2 Jˆ2 2x 3x2 Ĥrot = = 1− + 2 + ··· . 2µRe2 (1 + x/Re )2 2µRe2 Re Re (800) We see that the effect of vibrational-rotational coupling is to add the term 3Be J(J +1)x2 /Re2 to the first order perturbation calculation and the term −2Be J(J + 1)x/Re to the second order perturbation calculation. The detailed calculation is straight forward, though tedious. The linear term makes the rotational constant depend on the vibrational quantum number, Bv = Be − αe (v + 1/2). (801) This is called “vibrational-rotational” coupling. The quadratic term adds an extra term to the rotational energy, Erot /hc = Bv J(J + 1) − Dv J 2 (J + 1)2 . (802) 10 This effect is called “centrifugal distortion.” The centrifugal distortion constant is weakly dependent on the vibrational level because the average bond length increases with vibrational energy. When this effect is ignored, the constant is written as De , which is unrelated to the De in the Morse potential. Transitions in which J increases by 1 are called R-branch transitions, and those in which J decreases by 1 are called P-branch transitions. Letting ν0 indicate the change in vibrational energy, the frequencies of these two types of transitions are given by νR = ν0 + Bv′ (J + 1)(J + 2) − Bv′′ J(J + 1) = ν0 + 2Bv′ + (3Bv′ − Bv′′ )J + (Bv′ − Bv′′ )J 2 ; J = 0, 1, 2, · · · νP = ν0 + Bv′ J(J − 1) − Bv′′ J(J + 1) = ν0 − (Bv′ + Bv′′ )J + (Bv′ − Bv′′ )J 2 ; J = 1, 2, · · · (803) Primes indicate the upper state and double primes the lower state. The contribution from Dv was neglected. The lines are labeled according to greater of J ′ and J ′′ . For example, a P3 line corresponds to J ′′ = 3 → J ′ = 2, and an R3 line corresponds to J ′′ = 2 → J ′ = 3. Transitions with ∆J = 0 are called Q-branch transitions. We will learn later that a Q-branch can only exist when there is a component of the electronic orbital angular momentum parallel to the bond axis. A similar derivation yields O and S branches for Raman transitions, with ∆J = ±2. Problem 72. Look up the rotational constants Be , αe , De for 1 H 35 Cl in the NIST database http://physics.nist.gov/PhysRefData/MolSpec/Diatomic/Html/sec6.html and compare them with the values in Huber and Herzberg. Problem 73. Use the constants in Huber and Herzberg to calculate as accurately as possible the frequency of the transition v ′′ = 0, J ′′ = 0 → v ′ = 1, J ′ = 1 of 1 H 35 Cl in the ground electronic state. LECTURE 32: SYMMETRY EFFECTS IN ROTATIONAL SPECTRA Reading: Levine [1], pages 176-185 and Herzberg [2], pages 128-140. Our goal in this lecture is to discover how the symmetry of a diatomic molecule affects its rotational spectrum. We saw examples of this already in the requirement that a molecule must have a permanent dipole moment in order to have a pure rotational transition and that a homonuclear molecule does not have a pure vibrational spectrum because the dipole 11 moment does not change with stretching. We also found a rotational selection rule that ∆J = ±1. It is interesting that this rule does not require the molecule to be homonuclear. All that matters is the symmetry of the wave function with respect to the alignment (not orientation) of the molecular axis. Here we will investigate the consequences of different symmetries and show that the previous results are special cases of more fundamental principles. The two primary symmetries that we will be concerned with are parity and nuclear exchange. In cataloging different types of symmetry of a linear molecule, we must distinguish between molecular and laboratory (i.e., space-fixed) coordinates and also between properties of just electrons, just nuclei, and both. We summarize in the following table four types of symmetry that affect the wave function of a diatomic molecule. Symmetry Frame Particles Dipole rule Raman operation rule + ←→ − Parity space Π̂ θ → π − θ, and (−1)J ϕ → ϕ + π (all diatomic does electrons + ←→ + − ←→ − nuclei not molecules) affect nuclear spin Exchange space, nuclei P̂ab including (charges and nuclear spin masses must be 16 Inversion molecule equal) electrons (only charges must be 16 equal) O18 O, HD Reflection molecule Σ± , electrons only σ̂v Hund’s cases (a) and (b) for Λ > 0 s ←→ s, a ←→ a O2 , H2 only î only g ←→ u g ←→ g u ←→ u 12 Parity refers to inversion of the spatial coordinates of all particles (electrons and nuclei). We wish to discover the parity of a particular rotational level. If we write the total wave function as a product of a nuclear and an electronic wave function, the parity of the nuclear part is (−1)J , and the parity of the electronic part is ±1. The latter depends on the reflection symmetry of the electron wave function in a plane containing the bond axis. To explain this idea, we need to digress to define the term symbol for an electronic state, 2S+1 Λ± g,u . In this symbol, Λ is the component of the electronic orbital electronic momentum along the bond axis. A state with Λ = 0 is called a Σ state; a state with Λ = 1 is called a Π state; etc. States with Λ > 0 have electrons orbiting around the bond axis (like a doughnut), and are therefore symmetric tops. The 2S + 1 superscript refers to electron spin. The ± superscript refers to reflection symmetry of the electronic wave function. Although it applies to any value of Λ, it only appears in Σ states because the electronic eigenfunctions for Λ > 0 are actually superpositions of Λ+ and Λ− states. The ± superscript should not be confused with the ± parity, which unfortunately uses the same sign notation. Inversion of the nuclear coordinates is equivalent to replacing θ by π − θ and ϕ by ϕ + π. Problem 74. Prove this last statement. That is to say that Π̂YJM = (−1)J YJM . This is just the result we found when we determined the selection rule ∆J = ±1 for rotational transitions. An additional factor of −1 is introduced for Σ− states. This is because inversion of all three coordinates is equivalent to a rotation about the bond axis, followed by a reflection in a plane containing the bond; i.e. σ̂v (yz)Ĉ2 (x). The C2 rotation multiplies the nuclear wave function by (−1)J , while the reflection multiplies the electronic wave function by ±1. That is, Π̂Ψtot = Π̂Ψtot ψel = Ĉ2 (x)YJM σ̂v (yz)ψel = ±ψΣ± (−1)J YJM . (804) Let’s digress to give examples of reflection symmetry. An example of a state with + symmetry is the ground state of the hydrogen molecules, which has the term symbol 1 Σ+ g. Since the atomic orbitals are 1s, the molecular orbital must have + symmetry. An example of − symmetry is the 3 Σ− g ground state of oxygen. The electronic configuration is (1σg )2 (1σu )2 (2σg )2 (πu )4 (πgx )1 (πgy )1 , which yields a σ bond from the two electrons πz orbitals pointing along bond axis and one π bond from the 4 electrons in bonding orbitals and 2 electrons in anti-bonding orbitals perpendicular to the axis. Reflection in either the xz or yz planes changes the sign of the orbitals perpendicular to that plane. We may express this 13 FIG. 1: Molecular orbital energies of O2 . 14 analytically by looking at the Slater determinants for the unpaired electrons, sin θ1 eiϕ1 α(1) sin θ1 e−iϕ1 α(1) . Ψ= iϕ2 −iϕ2 sin θ2 e α(2) sin θ2 e α(2) (805) The minus sign in the right column is there because the projection of the L1 + L2 along the bond axis is zero for a Σ state. Expanding the determinant gives ( ) Ψ = sin θ1 sin θ2 ei(ϕ1 −ϕ2 ) − e−i(ϕ1 −ϕ2 ) α(1)α(2) = 2i sin θ1 sin θ2 sin(ϕ1 − ϕ2 )α(1)α(2). (806) Reflection in the xz plane leaves sin θ1 sin θ2 unchanged but flips the sign of sin(ϕ1 − ϕ2 ). Note that if we were to increase the dimension of the determinant include the sigma π bond, the overall reflection symmetry would not change because the additional pair of electrons would appear in each term as a multiplicative factor with positive symmetry. The bottom line is that Σ+ states with even J have + parity, and those with odd J have − parity. The rule is reversed for Σ− states. The molecule does not have to be homonuclear to possess parity. The selection rule for a dipole transition is that the parity must change, while for a Raman transition it must remain the same. More generally, if the orbital angular momentum quantum numbers of the two electrons are L1 , M1 and L2 , M2 , then the orbital wave functions for the Σ states are |Σ+ ; M ⟩ = |L1 , M, L2 , −M ⟩ + |L1 , −M, L2 , M ⟩ − |Σ ; M ⟩ = |L1 , M, L2 , −M ⟩ − |L1 , −M, L2 , M ⟩ (807) The reason these equations are true is that reflection flips the sign of M . Problem 75. The example given above is for the MS = 1 component of the triplet state (i.e., for α(1)α(2)). Clearly we would get the same result for MS = −1 (i.e., β(1)β(2)) of the triplet state.. Show that the MS = 0 component of the triplet state also has − symmetry, whereas the MS = 0 component of the singlet state has + symmetry. (Hint: For MS = 0 you need to take the sum of two Slater determinants.) A second type of symmetry that affects rotational transitions is the exchange of two identical nuclei, P̂ab . The exchange operation also occurs in the laboratory frame and includes nuclear spin. The result of the exchange operation is to change the sign of the total wave function, with plus and minus signs corresponding to symmetric and antisymmetric wave functions. This process is related to the Pauli Exclusion Principle, which 15 refers to exchange of two electrons (in general, any two Fermions) and always causes a sign change of the electronic wave function. Nuclear exchange is important for two reasons. First, there is a selection rule that prohibits the interconversion between symmetric and antisymmetric states by any process (radiative, collisional, etc.). The rule applies to any operator, M (such as the dipole operator), ∫ that is unaffected by exchanging identical nuclei. It follows that the integral ψ1∗ M ψ2 dτ must vanish unless ψ1 and ψ2 have the same exchange symmetry. This rule is rigorous if the nuclei have zero spin. If they have non-zero spin, the rule is not rigorous (although it still is approximately correct) because the spins of the nuclei may point in different directions so that the nuclei are not truly identical. The second consequence of nuclear exchange symmetry stems from the Pauli principle. Particles with half integer spins are called Fermions, and particles with integer spin (including zero) are called Bosons. The operating principle is that the total wave function is antisymmetric with respect to the exchange of a pair of identical Fermions and is symmetric with respect to the exchange of a pair of identical Bosons. The total wave function for a molecule may be written as the product of vibrational, rotational, electronic, and nuclear factors. The vibrational part is always symmetric because it depends only on internuclear distances. We will assume for now that the electronic part is also symmetric with respect to nuclear exchange. Rotational states that change sign upon nuclear exchange are called asymmetric (denoted by the symbol a), and rotational states that do not change sign upon nuclear exchange are called symmetric (denoted by the symbol s.) In the special case that the nuclear spin is zero (e.g. for 16 O2 ,12 C2 , and C 16 O2 ), only the rotational wave func- tion determines the exchange symmetry. Since nuclei with zero spin are Bosons, it follows that the rotational wave function must be even, and hence odd rotational states for such homonuclear molecules do not exist. Lines for such states are are simply absent in the spectrum. If the nuclei do have spin, we must include the symmetry of the nuclear spin wave function in determining the symmetry of the total wave function. If the nuclei are distinguishable, the Pauli principle does not apply. If the molecule is homonuclear, there are four possibilities: (1) The nuclei are Fermions and the spin functions are symmetric; (2) the nuclei are Fermions and the nuclear spin function is antisymmetric; (3) the nuclei are Bosons and the spin functions are symmetric; (4) the nuclei are Bosons and the spin functions are antisymmetric. 16 An example of cases (1) and (2) is the H2 molecule. The spin of each nucleus is I = 1/2, and the total nuclear spin of the molecule is T = I1 + I2 = 0 or 1. T = 0 is a singlet and is antisymmetric; T = 1 is a triplet and is symmetric. The ground nuclear state “modification” is a singlet. For H2 this modification is called para hydrogen or p − H2 , and only even rotational states exist. (Odd states would violate the Pauli principle because they would make the total wave function symmetric, which is forbidden for Fermion nuclei.) The excited modification has a triplet nuclear spin wave function, which is symmetric, and accordingly the rotational quantum numbers are all odd. This case is called ortho hydrogen or o − H2 . In general, the modification with a symmetrical nuclear spin wave function is always called ortho. It is easy to figure out which states are symmetric by noticing that the highest spin state has all the spins lined up and is therefore symmetric. For H atoms, the nuclei are Fermions with I = 1/2, resulting in triplet and singlet spin states for the diatomic molecule. The triplet has s symmetry, and the singlet therefore has a symmetry. Since the total wave function for Fermionic nuclei must have a symmetry, it follows that the even rotational states are paired with the a singlet nuclear state. For H2 molecules, even J states are called para-hydrogen or p − H2 , and odd J states are called ortho hydrogen or o − H2 . For deuterium atoms, the nuclei are Bosons with I = 1, resulting in singlet, triplet, and quintet spin states for the diatomic molecule. The quintet has s symmetry. The other two states alternate in symmetry, with the triplet being a and the singlet s. Since the total wave function for Bosonic nuclei must have s symmetry, it follows that the even rotational states are paired with the s singlet and quintet nuclear states. It follows that for deuterium even J is ortho and odd J is para. In the case of the H atom, the nuclear spin wave functions are 1 ψns = √ [α(1)β(2) − α(2)β(2)] 2 (808) for the singlet state and ψns = α(1)α(2) 1 = √ [α(1)β(2) + α(2)β(2)] 2 (809) = β(1)β(2) for the triplet state. The nuclear degeneracies, g = 2T + 1, are 1 for the singlet and 3 for 17 the triplet. It follows that for a thermal equilibrium mixture of para and ortho states, the intensities of the rotational transitions alternate between factors of 1 and 3. The partition function for an equilibrated mixture of hydrogen molecules is given by ∑ Q(T ) = = ∑ (2J + 1)e−BhcJ(J+1)/kT + 3 evenJ ∞ ∑ (2J + 1)e−BhcJ(J+1)/kT oddJ (810) −BhcJ(J+1)/kT [2 − (−1) ](2J + 1)e J . J Examples of cases (3) and (4) are D2 and N2 . The D nucleus has a spin of I = 1, making it a Boson. The possible values of T are 0, 1, and 2. The T = 0 and 2 states are symmetric with degeneracies 1 and 5, while the T = 1 state is antisymmetric and has a degeneracy of 3. The even rotational levels are now ortho. The partition function is given by Q(T ) = 6 ∑ (2J + 1)e−BhcJ(J+1)/kT + 3 evenJ ∑ (2J + 1)e−BhcJ(J+1)/kT . (811) oddJ Problem 76. Calculate the rotational population distribution for H2 in thermal equilibrium at 100K. The rotational constant is B = 60.8 cm−1 . Now let’s take into account the exchange symmetry of the electronic wave function. When we exchange the nuclei in space-fixed coordinates, we also invert the axis that defines the molecular frame. This has the effect of inverting the coordinates of the electrons in the molecular frame. If the two nuclei have the same charge (although not necessarily the same mass, as for example 16 O18 O or HD), the electronic wave function will be either even (gerade, g) or odd (ungerade, u) with respect to inversion of electronic coordinates. This means that the rotational wave functions for u electronic states have the opposite exchange symmetry of the g states. The effects of changing both g → u and Σ+ → Σ− on nuclear exchange symmetry cancel each other out. Changing just Σ+ → Σ− flips both the exchange symmetry (s vs. a) and the inversion symmetry (+ vs.− parity). Putting all the pieces together, the lowest rotational level (J = 0) is s+ for a 1 Σ+ g state, 1 − 1 + a− for a 1 Σ− g state, a+ for a Σu state, and s− for a Σu state. See figure 4.14 in Levine and 6.4 in Herzberg I. The inter-relations between reflection and exchange symmetry are treated systematically in what are known as the Wigner-Witmer rules, which we will discuss in a later lecture. These rules give a general recipe for determining all possible term symbols arising from a pair of electrons with given L and M quantum numbers. 18 LECTURE 33: ROTATIONAL SPECTRA OF POLYATOMIC MOLECULES Reading: This lecture is based on Chapters 3 and 6 of Zare [7]. We begin by examining the classical treatment of rotation of a rigid body about a fixed axis. We then construct the Hamiltonian and solve for its eigenvalues and eigenfunctions. Consider first the very simple case of a single point mass at the end of a massless rod. The rod rotates about an axis perpendicular to it. The kinetic energy of the mass is 1 1 1 1 T = mv 2 = m(ωR)2 = mR2 ω 2 = Iω 2 . 2 2 2 2 (812) In these equations, R is the length of the rod, ω is the angular frequency, m is the mass of the particle, and I is its moment of inertia. Likewise, the rotational angular momentum is J = mvR = m(Rω)R = Iω. (813) We may extend these results to a collection of N particles of mass mi , 1∑ 1 ∑ 1 mi vi2 = ω 2 mi ri2 = Iω 2 , 2 i=1 2 i=1 2 N T = N and J =ω N ∑ mi ri2 = Iω, (814) (815) i=1 where I= N ∑ mi ri2 . (816) i=1 We may generalize this result further to rotation about an arbitrary axis, J= N ∑ mi (ri × vi ). (817) i=1 But vi = ω × r i , so that J= N ∑ mi [ri × (ω × ri )]. (818) (819) i=1 Expanding the cross product, using the vector identity, A × (B × C) = B(A · C) − C(A · B). (820) 19 gives J= = N ∑ i=1 N ∑ mi [ωri2 − ri (ri · ω)] (821) mi [ri2 (ωx î + ωy ĵ + ωz k̂) − (xi ωx + yi ωy + zi ωz )(xi î + yi ĵ + zi k̂)]. i=1 The Cartesian components of J are Jx = ωx N ∑ mi (ri2 − x2i ) − ωy i=1 N ∑ mi xi yi − ωz N ∑ i=1 mi x i z i , (822) i=1 with analogous expressions for Jy and Jz . We define coefficients Ipq such that Jx = Ixx ωx + Ixy ωy + Ixz ωz (823) Jy = Iyx ωx + Iyy ωy + Iyz ωz Jz = Izx ωx + Izy ωy + Izz ωz . The diagonal terms are called the moments of inertia coefficients; e.g., Ixx = N ∑ mi (ri2 − x2i ), (824) i=1 and the off-diagonal terms are called the products of inertia; e.g., Ixy = − N ∑ mi xi yi . (825) i=1 The angular momentum is now written as a tensor product, J = I · ω, or I I I J x xx xy xz Jy = Iyx Iyy Iyz Izx Izy Izz Jz (826) ω x ωy . ωz (827) We see that I is a symmetric, Hermitian matrix, which we may diagonalize. That is to say that we may find an orthogonal matrix, U such that such that I 0 0 aa −1 Id = U · I · U = 0 Ibb 0 0 0 Icc (828) 20 The eigenvectors of the original inertia tensor are the principle axes of rotation along directions (n̂a , n̂b , n̂c ), such that J = Iaa ωa n̂a + Ibb ωb n̂b + Icc ωc n̂c , (829) and the eigenvalues are the moments of inertia about those axes. In what follows we will drop the subscript d from the diagonalized intertia tensor. We may likewise write the kinetic energy as a tensor product, T = N N N 1∑ 1∑ 1 ∑ mi vi2 = mi (vi · vi ) = mi vi · (ω × ri ). 2 i=1 2 i=1 2 i=1 (830) We invoke another vector identity A · (B × C) = B · (C × A) (831) to get ∑ 1 1 1 1 T = ω· mi (ri × vi ) = ω · J = ω · I · ω = (ωa2 Iaa + ωb2 Ibb + ωc2 Icc ). 2 2 2 2 i=1 N (832) Expanding this last result, we get T = Ja2 J2 J2 + b + c , 2Iaa 2Ibb 2Icc (833) where Ja = Iaa ωa , etc. The equation Ja2 J2 J2 + b + c =1 2T Iaa 2T Ibb 2T Icc (834) describes a solid known as the “inertia ellipsoid.” By convention, we order the moments of inertia as Iaa ≤ Ibb ≤ Icc . These equations describe the motion of a classical top. Different types of tops are defined according to the equality of some of the moments of inertia: Iaa = 0, Ibb = Icc , linear Iaa < Ibb = Icc , prolate symmetric top Iaa = Ibb < Icc , oblate symmetric top Iaa = Ibb = Icc , spherical top Iaa < Ibb < Icc , asymmetric top 21 A prolate symmetric top is football shaped, such as CH3 I. Think of a cylinder, with two axes perpendicular to the curved surface and one axis along the cylinder axis. The moments of inertia about the axis perpendicular to the curved surface are equal and also larger then the moment with respect to the long axis, The latter statement follows because atoms at the ends of the cylinder are far from these axes and therefore have large moment arms. An oblate symmetric top is frisbee-shaped, such as benzene. In this case the two smallest moments correspond to the axes in the plane of the molecule because some of the atoms lie on those axes. Examples of spherical tops are methane and SF6 . Symmetric tops must have a rotational axis of at least third order. We may now construct the quantum mechanical Hamiltonian by replacing the J 2 terms by their corresponding operators, Ĥrot = Jˆa2 Jˆ2 Jˆ2 + b + c . 2Iaa 2Ibb 2Icc (835) We note that the subscripts a, b, c refer to body-fixed coordinates. Normally they are written as x, y, z, while the space-fixed coordinate are labeled X, Y, Z. We need three angles to describe the molecular orientation; θ and ϕ determine the direction of the principle axis of the molecule in the lab, and χ is the rotational angle about that axis. The wave functions for symmetric and spherical tops have three quantum numbers, J for the magnitude of the angular momentum, M for its projection along the lab Z axis, and K for its projection along the molecular z axis. The Schrodinger equation is therefore Ĥrot ψJKM (θ, ϕ, χ) = EψJKM (θ, ϕ, χ). (836) It is convenient to define three rotational constants, similar to that used for the diatomic molecule: ( ~2 hc ) Ĥrot = AJˆa2 + B Jˆb2 + C Jˆc2 , (837) where A ≥ B ≥ C. For a spherical top, where A = B = C, the Schrodinger equation simplifies to Ĥrot ψJKM (θ, ϕ, χ) = Jˆ2 ψJKM (θ, ϕ, χ) = EψJKM (θ, ϕ, χ). 2I (838) The eigenvalues are the same as for the diatomic molecule, EJ = BJ(J + 1), hc (839) 22 only that the degeneracy is now (2J + 1)2 . The extra factor of 2J + 1 comes from the projection of the angular momentum along the molecular axis. We consider next symmetric tops. In the oblate case (A = B > C), ( ) Jˆa2 + Jˆb2 Jˆc2 Jˆa2 + Jˆb2 + Jˆc2 Jˆc2 Jˆc2 Jˆ2 1 1 2 ˆ Ĥrot = + = + − = + Jc − . (840) 2Iaa 2Icc 2Iaa 2Icc 2Iaa 2Iaa 2Icc 2Iaa The eigenvalues in wave numbers are E = AJ(J + 1) − (A − C)K 2 hc (841) where 0 ≤ K ≤ J. The degeneracy is 2J + 1 for K = 0 and 2(2J + 1) for K > 0. Problem 77. The C − C bond length in benzene is 0.14 nm. Calculate the three moments of inertia. You may add the mass of H to the nearest C and otherwise ignore their presence. Problem 78. Use the result of the previous problem to calculate all the rotational energy levels of benzene for J = 0, 1, 2, 3. For a prolate top (A > B = C), the same reasoning leads to E = CJ(J + 1) + (A − C)K 2 hc (842) Figure 6.8 in Zare [7] is a drawing of the ordering of the energy levels of prolate and oblate tops. Problem 79. Derive Eq. (842). The energy levels of an asymmetric top may be obtained by expanding the wave functions of such a molecule, ΨJM (ϕ, θ, χ), in the symmetric top basis set. We will find out what the latter are in the next lecture. For now we will simply designate them by the ket |JKM ⟩.We may choose to use either the prolate or oblate basis set. Since most molecules are closer to the prolate limit, that is usually the basis of choice. The expansion is ΨJM (ϕ, θ, χ) = J ∑ AK |JKM ⟩. (843) K=−J We substitute this wave function into the rotational Schrodinger equation, Ĥrot ΨJM (ϕ, θ, χ) = EΨJM (ϕ, θ, χ) to get J ∑ K=−J AK Ĥrot |JKM ⟩ = E J ∑ K=−J AK |JKM ⟩. (844) (845) 23 Left multiplying by ⟨JK ′ M | and integrating over all angles gives a set of algebraic equations, which may be solved by finding the eigenvalues and eigenfunctions of the secular determinant, |HK ′ K − EδK ′ K | = 0, (846) HK ′ K = ⟨JK ′ M |Ĥrot |JKM ⟩. (847) where the matrix elements are Zare (see section 6.3 in [7]) develops an elegant way of block-diagonalizing the secular determinant using the D2 symmetry representation of the inertia ellipsoid. Here we will just quote the result that the energy eigenvalue lies on a monotonically smooth curve connecting the prolate and oblate limits, parameterized by the quantity κ= 2B − A − C , A−C (848) where κ = −1 corresponds to the prolate limit (B = C), and κ = +1 corresponds to the oblate limit A = B). Take a look at Figure 6.9 in Zare. The various curves are labeled JK−1 ,K1 , where the subscripts refer to the limiting values of K for τ = ±1. LECTURE 34: PROPAGATORS AND CONSERVATION LAWS The rotational wave function of a symmetric top is an eigenfunction of the total angular momentum operator as well as of the projections along both the space-fixed and bodyfixed z-axis. To find these eigenfunctions, we need to learn about the transformation from space-fixed to body-fixed coordinates. We will first discuss transformations in general, following closely the material in chapter 3 of Zare. A key concept is that a coordinate transformation is a unitary operation, which transforms one basis set into another, ψj′ = Û ψj . (849) An operator Q̂ is also transformed by Û according to the rule Q̂′ = Û Q̂Û −1 = Û Q̂Û † . (850) For a microscopic transformation of some variable (e.g., a coordinate or time) by an amount ϵ, Û may be written as Û = I + ϵV̂ . (851) 24 The requirement that Û Û † = I is equivalent to (I + ϵV̂ )(I + ϵV̂ † ) = I + ϵ(V̂ + V̂ † ) = I (852) to first order in ϵ. This condition is satisfied if V̂ = iŜ † , where Ŝ = Ŝ † is Hermitian. We may subdivide this transformation into n smaller steps, so that Û = (1 + iδ Ŝ)n , (853) where δ = ϵ/n. We may now consider the case of a finite transformation in which δ → 0 and n → ∞, so that [ ( ϵ )]n Û = lim 1 + i Ŝ = eiϵŜ . n→∞ n This last step may be proved from the binomial theorem, which states that ( ) ( ) n 2 n k n (1 + x) = 1 + nx + x + ··· + x ··· 2 k and choosing x = iϵŜ/n = iδ Ŝ. Recall that ( ) n n! n(n − 1)(n − 2) · · · (n − k + 1) = = ≈ nk /k! k k!(n − k)! k! (854) (855) (856) for n ≫ k. Since we let n → ∞, this approximation is always satisfied. Recall next that transforming a system in the forward direction is equivalent to transforming the coordinate system backwards by the same amount. Let’s examine this statement graphically, using as examples rotation, translation in space, and translation in time. We showed previously that rotating a vector clockwise by some angle, α, (see drawing (a)) is equivalent to rotating the coordinate system counterclockwise by the same angle (drawing (b)). We get the same result for a coordinate displacement by either moving a point to the right (drawing (c)) or moving the origin to the left (drawing (d)). Let’s consider next the somewhat trickier example of a translation in time. Suppose we look a sinusoidal pattern on an oscilloscope and take an old-fashioned polaroid snapshot at time t = 0. The image we get is labeled ψ0 , ψ0 (t) = sin ωt, (857) shown in drawing (e). Next, suppose we trigger the camera to take a photo at some later time t′ > 0. The resulting image is shown in drawing (f), ψt′ (t) = sin ω(t − t′ ). (858) 25 FIG. 2: Operators for rotational, translational, and temporal transformation. 26 Comparing the two photos, we see that ψt′ (t) = ψ0 (t − t′ ) = ψ0 (τ ), (859) where τ = t − t′ . But we could have gotten the same result by simply relabeling the axes in the first photo so as to shift the origin of time backwards by an amount t′ instead of moving the wave forwards by the same amount in the second photo (see drawing (g)). Now let’s find the unitary operators that do these two tasks. First let’s transform the time variable, not the wave function. We need to change the time origin by an amount −t′ , as follows: ψ0 (τ ) = ψ0 (t) + (−t′ ) ∂ 1 ∂2 ′ ∂ ψ0 (t) + (−t′ )2 2 ψ0 (t) + · · · = e−t ∂t ψ0 (t) = T̂ (t′ )ψ0 (t). (860) ∂t 2! ∂t From the time-dependent Schrodinger equation, Ĥ = i~ ∂ , ∂t (861) we recognize the unitary operator T̂ (t′ ) to be ′ T̂ (t′ ) = eit Ĥ/~ . (862) ψ0 (τ ) = T̂ (t′ )ψ0 (t) = ψ0 (t − t′ ). (863) such that Instead of propagating time backwards by t′ , we may propagate the wave function forwards in time by the same amount, ψ(t) = T̂ (−t)ψ(0) = e−itĤ/~ ψ(0) = Û (t)ψ(0). (864) where the time evolution operator Û (t) is given by Û (t) = e−itĤ/~ . (865) In the last two equations we dropped the primes on t and the subscript on the wave function. If ψ is an eigenfunction of Ĥ, we get the familiar result ψ(t) = e−iE0 t/~ ψ(t = 0). (866) 27 This result shows that the invariance of a wave function in time (apart from a phase change) is equivalent to conservation of energy. In other words, if ψ is an eigenstate of Ĥ, then operating on it with e−itĤ/~ replaces multiplies the wave function by a simple phase factor. Saying that the system is in an eigenstate of the Hamiltonian is equivalent to saying that energy is conserved, and saying that the time propagator reduces to a phase factor is equivalent to saying that the system is invariant with time. The same analysis applies to other types of transformations. For example, a spatial displacement way be written as |z ′ ⟩ = |z0 ⟩ + (−z ′ ) ∂ 1 ∂2 ′ ∂ |z0 ⟩ + (−z ′ )2 |z0 ⟩ + · · · = e−z ∂z |z0 ⟩. 2 ∂z 2! ∂z (867) Recognizing that p̂z = −i~∂/∂z, we obtain for the unitary displacement operator, ′ D̂(z) = e−iz p̂z /~ . (868) Invariance of the wave function with spatial displacement (apart from a phase factor) is equivalent to conservation of linear momentum. The same reasoning applies to rotations. Rotation of a wave function counterclockwise about the z-axis by an angle α is given by ψ(ϕ0 + α) = e−α ∂ϕ ψ(ϕ)ϕ=ϕ0 . ∂ (869) Recalling the angular momentum operator, ∂ Jˆz = −i~ , ∂ϕ (870) we obtain the rotational evolution operator R̂z (α) = e−iαJz /~ . ˆ (871) Invariance of the wave function with rotation (apart from a phase factor) is equivalent to conservation of angular momentum. Finally, for rotation by an angle α about an arbitrary axis n̂, R̂n (α) = e−iαĴ·n̂/~ . (872) 28 LECTURE 35: ROTATIONAL WAVE FUNCTIONS OF POLYATOMIC MOLECULES We are now in a position to use rotation operators to develop an expression for the rotation of a molecule about an arbitrary axis, and from that to derive the rotational wave function. To describe such a motion, we need three angles, two angles (θ, ϕ) to locate the orientation of the molecular axis in the lab frame and a third angle for rotation about that axis. The first two angles describe the transformation from the XY Z frame to the xyz frame, and the third angle (χ) describes rotation about the body-fixed z-axis. Note that the angles θ and ϕ define the direction of the principle axis of inertia with respect to the lab frame. For linear molecules and symmetric tops, this axis coincides with the angular momentum vector. These angles are called the Euler angles. We carry our the transformation in three steps, XY Z −−−→ x′ y ′ z ′ −−−−−→ x′′ y ′′ z ′′ −−−−−→ xyz, (873) R̂z′′ =z (χ) R̂y′ =N (θ) R̂Z (ϕ) as shown in Fig. 3.2 of Zare. At each stage one axis is the same in two frames: Z = z ′ in the first rotation, y ′ = y ′′ = N in the second, and z ′′ = z ′ in the third. The intermediate y axes are called the line of nodes. Note that ϕ is the azimuthal angle about the Z axis, and χ is the azimuthal angle about the z axis. Note also that this is a passive transformation of the coordinate system, taking 0 → ϕ, 0 → θ, 0 → χ. The total transformation operator is a product of three counter-clockwise rotational operators, R̂(θ, ϕ, χ) = e−iχJz /~ e−iθJN /~ e−iϕJZ /~ . ˆ ˆ ˆ (874) This result is exact but awkward to use because it involves rotations about axes in two coordinate systems. But now the transformation Q̂′ = Û Q̂Û −1 comes to the rescue. We recognize that the rotation by angle θ about the N = y ′ axis is equivalent to a rotation of θ about the Y axis if we first carry out the transformation y ′ → Y by the operation eiϕJZ /~ ˆ and afterwards undo that transformation. In other words, e−iθJN /~ = e−iϕJZ /~ e−iθJY /~ eiϕJZ /~ . ˆ ˆ ˆ ˆ (875) Likewise, the rotation by angle χ about the z axis is equivalent to a rotation of χ about the ˆ Z axis after the transformation of eiθJN /~ . That is, e−iχJz /~ = e−iθJN /~ e−iχJZ /~ eiθJN /~ . ˆ ˆ ˆ ˆ (876) 29 Putting the pieces together, e −iχJˆz /~ −iθJˆN /~ −iϕJˆZ /~ e e [ ] ˆ ˆ −iθJˆN /~ −iχJˆZ /~ iθJˆN /~ = e e e e−iθJN /~ e−iϕJZ /~ . = e−iθJN /~ e−iχJZ /~ e−iϕJZ /~ [ ] ˆ ˆ ˆ ˆ ˆ = e−iϕJZ /~ e−iθJY /~ eiϕJZ /~ e−iχJZ /~ e−iϕJZ /~ ˆ ˆ ˆ (877) = e−iϕJZ /~ e−iθJY /~ e−iχJZ /~ . ˆ ˆ ˆ In the last step, the rotations about the Z axis are permuted. This remarkable result shows that the Euler transformation may be carried out entirely in the lab frame by just reversing the order of the rotations. Next we observe that R̂n (α) (a rotation by angle α about an arbitrary axis) commutes with Ĵ2 because 2 −αĴn /~ [Rn (α), Ĵ ] = [e ∞ ∑ 1 , Ĵ ] = (−iα)s [Ĵsn , Ĵ2 ]/~s = 0. s! s=0 2 (878) If |JM ⟩ is an eigenfunction of Ĵ2 and ĴZ , rotation by R̂n (α) can change M but not J. In other words, it transforms |JM ⟩ into a superposition of M states, J ∑ R̂(θ, ϕ, χ)|JM ⟩ = J ′ DM ′ M (θ, ϕ, χ)|JM ⟩, (879) M ′ =−J with coefficients J ′ DM ′ M (θ, ϕ, χ) = ⟨JM |R̂(θ, ϕ, χ)|JM ⟩. (880) J The array of DM ′ M coefficients is called the rotation matrix. An important property of the rotation operator is that it can transform the rotational wave function from one orientation to another. Recall that the rotation matrix was defined for a passive transformation. That is, ψJM (θ′ , ϕ′ , χ′ ) = ∑ J DM K (θ1 , ϕ1 , χ1 )ψJM (θ, ϕ, χ). (881) K In this expression, θ, ϕ, χ are the angles of the basis functions before the transformation, θ′ , ϕ′ , χ′ are the angles after the transformation, and θ1 , ϕ1 , χ1 are the angles of the passive operator. If we choose the angles of the D matrix to be identical to those of the wave function, then the transformation simply “unwinds” the angles back down to zero. That is, ψJM (0, 0, 0) = ∑ K J DM K (θ, ϕ, χ)ψJM (θ, ϕ, χ). (882) 30 If we now invert the transformation, which is equivalent to carrying out an active transformation, ψJM (θ, ϕ, χ) = ∑ J∗ DM K (θ, ϕ, χ)ψJM (0, 0, 0). (883) K This discussion is for a general asymmetric top. For a symmetric top, the wave function is invariant with respect to rotation by χ (apart from a phase factor). This property means that only one term may be present in the sum, giving rise to [ 2J + 1 |JKM ⟩ = 8π 2 ]1/2 J∗ DM K (θ, ϕ, χ) (884) Noting that e−iαJZ /~ |JM ⟩ = e−iαM |JM ⟩, ˆ (885) we may reduce the rotational matrix element to a simpler form, J −iM ϕ J DM dM K (θ)e−iKχ . K (θ, ϕ, χ) = e (886) Zare derives an analytic formula for dJM ′ M (θ). (See his Eq.(3.66) and Table 3.1). Two limiting cases are L D00 (θ, ϕ, χ) = PL (cos θ) and [ L D0M (θ, ϕ, χ) 4π = 2L + 1 (887) ]1/2 ∗ YL,−M , (888) which reduce to the wave functions for a linear rotator. Knowing the wave functions, we may determine the selection rules and line strengths. There are nice closed form expressions in Zare Table 3.1 for the product of three rotation matrix elements (e.g., two rotational wave functions and the dipole transition operator). For a pure rotational transition with the permanent dipole moment pointing along the highest order symmetry axis, Herzberg [3] derives the selection rule ∆J = 0, ±1, ∆K = 0. For a rotational-vibrational transition, we need to compare the symmetry representation of the vibration x, y, z with the direction of the symmetry axis. For the transition dipole moment parallel to the symmetry axis, ∆J = 0, ±1, ∆K = 0 for K ̸= 0 and ∆J ± 1, ∆K = 0 for K = 0. For a perpendicular transition, ∆J ± 1, ∆K = ±1. 31 LECTURE 36: INTRODUCTION TO ELECTRONIC SPECTROSCOPY: THE BORN-OPPENHEIMER APPROXIMATION Reading: Chapter 9 of Bernath [8]. Excellent reviews of this topic may be found in references [10, 11]. In most molecular electronic structure calculations, the electrons move much more rapidly than the nuclei because of their much lighter mass. A tremendous simplification results if we assume that the Schrodinger equation may be solved separately for the electrons and nuclei. In the following notes we show what approximations are involved in this separation. The Hamiltonian for the one-electron hydrogen molecule ion is given by Ĥ = − ~2 2 ~2 2 e2 e2 e2 ~2 2 ∇a − ∇b − ∇e − − + . 2Ma 2Mb 2me ra rb Rab (889) This is the only molecular problem that can be solved exactly. For an n-electron diatomic, the Hamiltonian is Ĥ = − n n n n ~2 2 ~2 2 ~2 ∑ 2 ∑ Za e2 ∑ Zb e2 ∑ e2 Za Zb e2 ∇a − ∇b − ∇i − − + + , (890) 2Ma 2Mb 2me i=1 r r r R ia ib ij ab i=1 i=1 i̸=j and for N nuclei and n electrons, N n N n n N ∑ ~2 2 ~2 ∑ 2 ∑ ∑ Zs e2 ∑ e2 ∑ Zs Zs′ e2 Ĥ = − ∇ − ∇ − + + . 2Ms s 2me i=1 i s=1 i=1 ris rij s̸=s′ Rss′ s=1 i̸=j (891) We will use a shorthand to collect all the nuclear and electronic terms together, Ĥ = T̂N + T̂e + U (r, R), (892) where T̂N is the nuclear kinetic energy operator, T̂e is the electronic kinetic energy operator, U is the total potential energy, and r and R refer to all the electronic and nuclear coordinates, respectively. For later reference, we define the electronic Hamiltonian by Ĥe = T̂e + U (r, R). (893) Next, we write the wave function as a product Ψ(r, R) = ψe (r, R)ϕN (R). (894) So far we have not made any approximations. The complete Schrodinger equation becomes Ĥ(ψe ϕN ) = T̂N (ψe ϕN ) + T̂e (ψe ϕN ) + U (r, R)ψe ϕN = Eψe ϕN . (895) 32 We next expand the kinetic energy derivatives in T̂N and T̂e , ∇2N (ψe ϕN ) = ψe ∇2N ϕN + ϕN ∇2N ψe + 2∇N ψe · ∇N ϕN (896) ∇2e (ψe ϕN ) = ϕN ∇2e ψe . (897) and We are now ready to invoke the Born-Oppenheimer (BO) approximation. We clamp the nuclei at some geometry, R. This allows us to discard the nuclear kinetic energy term, ∇2N . This is justified because the terms in Eq. (896) are all of the same order of magnitude, and when we divide them by M they become much smaller than the term 1 ∇2 ψ m e e that appears in the electronic Hamiltonian. This is the essence of the BO approximation. We may now cancel out ϕN on both sides of Eq. (895) , leaving us with the electronic Schrodinger equation, ( ) Ĥe ψe (r; R) = T̂e (r; R) + U (r; R) ψe (r; R) = V (R)ψe (r; R). (898) Here, R is treated as a parameter. This is the starting point for all quantum calculation programs (Gaussian, etc). We solve this equation numerically to get ψe (r; R) and V (R). V (R) is the energy eigenvalue of Ĥe and is called the adiabatic potential energy surface (PES), with dimension 3N-5 for linear molecules and 3N-6 for non-linear molecules. Each PES corresponds to a different electronic configuration. We note, however, that the configuration may change as R varies, and in general a multi-configurational basis is needed. If we regard ψe (r; R) as the eigenfunction of a truncated Hamiltonian ψe (r; R), we have not really made any approximations yet. At this point, however, we use the electronic Schrodinger equation to replace the second and third terms in the complete Schrodinger equation, T̂e (r; R) + U (r; R)), by V (R) to yield T̂N (ψe ϕN ) + V (R)ψe ϕN = Eψe ϕN . (899) Now here comes the mathematical content of the BO approximation. We assume that ∇2N (ψe ϕN ) ≈ ψe ∇2N ϕN . (900) Ĥ ′ (ψe ϕN ) = ϕN ∇2N ψe + 2∇N ψe · ∇N ϕN = 0. (901) This is equivalent to setting 33 In a more complete treatment we could treat Ĥ ′ as a perturbation. What remains of the complete Schrodinger equation is the nuclear Schrodinger equation, T̂N ϕN + V (R)ϕN = EϕN , (902) from which we obtain the vibrational and rotational wave functions and energies. In the familiar harmonic oscillator - rigid rotor approximation, we may further factor ϕN into separate vibrational and rotational wave functions, ϕN (R) = χvib (Q)ψ rot (θ, ϕ, χ), (903) where χv (Q) is the harmonic oscillator wave function for quantum number v, Q = R − Re is the displacement of the oscillator coordinate from its equilibrium value, and ψrot is the rigid rotor wave function. For a polyatomic molecule, χ is a product on 3N-5 or 3N-6 independent harmonic oscillator functions, one for each normal coordinate, Qs . The rotational wave function is a 3D function of the Euler angles. We may now partition the total energy into electronic, vibrational, and rotational term values, E = Te + G(v) + F (J, K). hc (904) Since different electronic potential energy functions have different equilibrium distances and different curvatures, they also have different moments of inertia and vibrational frequencies. The vibrational selection rules that we derived for pure vibrational transitions no longer hold. This is true because the transition dipole moment may be written as ∫ ∫ ∫ ∫ ∗ ∗ ∗ 3 rot∗ rot 2 vib µ12 = e ψe2 ϕN 2 rψe1 ϕN 1 dτ = e ψe2 rψe1 d τe ψJ ′ K ′ ψJ ′′ K ′′ d Ω χvib v ′ χv ′′ dR, (905) vib where ϕN = ϕrot∗ JK ϕv . The square of the last integral is called a Franck-Condon factor, qv′ v′′ = |⟨v ′ |v ′′ ⟩|2 . In general, closure requires that ∞ ∑ qv′ v′′ = 1. (906) (907) v ′ =0 If the equilibrium constants of the lower and upper states are nearly equal (i.e. if B ′ ≈ B ′′ ), qv′ v′′ will be largest for v ′ = v ′′ = 0. If the equilibrium constants are very different, the transition probability is greatest for a vertical transition that maximizes the overlap of the 34 vibrational wave functions. For v ′′ = 0, the transition connects the equilibrium position of the lower potential energy curve to the turning point of the upper curve. In the limiting case that oscillator is harmonic and that the frequencies of both electronic states are equal, one may show that qv0 1 = v! ( (∆R̄e )2 2 where ∆R̄e = )v e−(∆R̄e ) 2 /2 , √ β(R̄e′ − R̄e′′ ). (908) (909) Problem 80. Derive Eq. (908). Hint: use the recursion property of the harmonic oscillator eigenfunctions. Problem 81. Show that Eq. (908) satisfies the closure condition. An interesting thing also happens to the rotational spectrum. Consider for example the P-branch. The transition energy in wave numbers is given ν̃P = ν̃0 + B ′ J(J − 1) − B ′′ J(J + 1) = ν̃0 − (B ′ + B ′′ )J + (B ′ − B ′′ )J 2 , (910) where J is the rotational quantum number of the lower electronic state (J ′′ ). We see that if B ′ − B ′′ is positive, the frequency passes through a minimum. In this case the rotational band turns back on itself and is said to have a “band head.” A similar thing may happen for an R-branch. In this case, ν̃R = ν̃0 + B ′ (J + 1)(J + 2) − B ′′ J(J + 1) = ν̃0 + 2B ′ + (3B ′ − B ′′ )J + (B ′ − B ′′ )J 2 . (911) Here, if B ′ − B ′′ is negative, the frequency passes through a maximum. We may derive a uniform expression by defining m = J ′′ + 1 for the R-branch and m = −J ′′ for the P-branch. The transition wave number is then ν̃P,R = ν̃0 + (B ′ + B ′′ )m + (B ′ − B ′′ )m2 . (912) When plotted as ν̃P,R vs. m, the function is known as a Fortrat parabola, with m ≤ 0 corresponding to the P-branch and m > 0 corresponding to the R-branch. The function has a turning point at dν̃ = 0 = (B ′ + B ′′ ) + 2m(B ′ − B ′′ ), dm producing a band head at mH = − B ′ + B ′′ . 2(B ′ − B ′′ ) (913) (914) 35 Substituting mH into Eq. (912) gives for the distance of the turning point ν̃H − ν̃0 = − (B ′ + B ′′ )2 . 4(B ′ − B ′′ ) (915) Problem 82. Verify Eqs. (912) and (915). If B ′ > B ′′ the band head will be in the P-branch, and the band is said to blue-degraded because the spectrum has a sharp edge at longer wavelengths (νH < ν0 ) and tails off at short wavelengths. If B ′ < B ′′ the band head will be in the R-branch, and the band is said to red-degraded because the spectrum has a sharp edge at shorter wavelengths (νH > ν0 ) and tails off at long wavelengths. Problem 83. Calculate the first few transition frequencies (in wave numbers) for the Pand R-branches of the B(v ′ = 40) ← X(v ′′ = 0) transition of I2 . Calculate enough lines to construct the Fortrat parabola and find the location of the band head. Use Herzberg’s spectroscopic parameters for I2 . Problem 84. Use the Table from Huber and Herzberg [5] to calculate the wavelengths of the transitions X 1 Σ+ v ′′ = 0, J ′′ = 1 → B 1 Σ+ v ′ = 1, J ′ = 2 and X 1 Σ+ v ′′ = 0, J ′′ = 1 → B 1 Σ+ v ′ = 1, J ′ = 0 of the HF molecule. Include the effects of anharmonicity, vibrationalrotation interaction, and centrifugal distortion. LECTURE 37: BREAKDOWN OF THE BORN-OPPENHEIMER APPROXIMATION We now turn to the interesting question of what happens when the adiabatic approximation fails. When we wrote down the adiabatic Schrodinger equation, we ignored the nuclear kinetic energy operator, T̂N . This is a reasonable approximation so long as the energies of the electronic states are far apart. If two states are degenerate (i.e. if they cross), this approximation fails completely. Let us see how this can happen. There is a famous principle in quantum mechanics called the “non-crossing” rule, which states that the potential energy curves of two electronic states of the same symmetry cannot cross. It is an example of a general result known as the adiabatic theorem. (Caution: As we shall see shortly, this is a rigorous result only for diatomic molecules, although it has more wide spread applications.) In simple terms, it says that if we have two approximate wave functions and we take a linear combination of them to get a better estimate of the energy, 36 the resulting energies will repel each other; i.e., the lower of the two energies gets still lower and the higher one gets even higher. The proof, derived by von Neumann and Wigner, is as follows. Suppose we know all but two of the energies and their eigenfunctions. We choose two additional wave functions, ψ1 and ψ2 , that are orthogonal to each other and to all the other wave functions. We construct the missing eigenfunctions as linear combinations of these two, ψ = c1 ψ1 + c2 ψ2 , (916) and solve for the roots of the resulting secular equation (which comes from the variational principle). In other words, we diagonalize the Hamiltonian, H −E H12 c 11 1 = 0 H21 H22 − E c2 (917) Each of the matrix elements is a function of R. The roots (eigenvalues) of the equation are ( ) √ 1 2 2 E± = H11 + H22 ± (H11 − H22 ) + 4H12 . (918) 2 Clearly, E+ > H22 (the energy of the unperturbed upper state), and E− < H11 . The only way the two energies could be equal is if simultaneously H11 (R) = H22 (R) (919) H12 (R) = 0. (920) and In the case of a diatomic molecule, there is only one independent degree of freedom (R), and it is impossible to satisfy both conditions at the same location. Hence the potential energy curves repel each other. (This does not hold if the two states are of different symmetry, because then H12 = 0.) Let us take as an example the ionic and covalent states of NaCl, both of which have 1 Σ+ symmetry. If the neutral atoms approach each other very slowly (“adiabatically”), at some point their interaction switches from covalent to ionic. What happens physically is that at some distance an electron jumps from the Na atom, flies through space, and attaches itself to the Cl atom. The potential energy curves calculated for the ionic and covalent states, based on electrostatics and short range repulsion, intersect at some point, Rx . In the adiabatic approximation, if the neutral atoms approach each other sufficiently slowly, the electrons adjust 37 their configuration continuously until R = Rx , at which point an electron flies through space and lands on the more electronegative atom. As R gets still smaller, the two ions approach each other along the same adiabatic potential energy curve, which is now ionic. This is called a “harpoon reaction” [12] The crossing point occurs where the attractive Coulomb potential energy matches the the difference between the ionization potential of the Na atom the electron affinity of the Cl atom, e2 = IP − EA Rx (921) or Rx (Å) = 14.4 . (IP − EA) (eV ) (922) The reaction cross section is πRx2 . For the reaction N a + Cl → N a+ + Cl− , (923) IP (N a) = 5.13 eV , EA(Cl) = 3.6 eV , Rx = 9.6Å, and σ = 290 Å2 . Other examples of harpoon reactions are K + CH3 I → KI + CH3 (924) Xe + Cl2 → XeCl + Cl. (925) and Noting that the non-crossing rule predicts that the two lowest adiabatic potential energy curves repel each other, we expect the adiabatic path to involve a switch from covalent to ionic character at some point. On the other hand, we also expect that if the atoms collide with very high velocity, they will not have time to transfer an electron, and the molecule remains in its initial (covalent) configuration. In other words, it “jumps the track.” Another way of saying this is that it is possible to calculate the ionic and covalent potential energy curves separately without invoking configuration interaction. The resulting states, which are not eigenstates of the electronic Hamiltonian, cross at R = Rx . These are called “diabatic states.” What is the probability of the system jumping across the gap (i.e. to remain on the same diabatic state)? The answer is given by the famous Landau-Zener formula, which is based on the assumptions that the diabatic states cross at t = 0 and that the energy difference between the states varies linearly in time. This is equivalent to saying that the potential 38 energies vary linearly in the region of their intersection and that the atoms move at constant velocity in the interaction region (say in a collision). If the system is initially in adiabatic state 1, the probability of it jumping to adiabatic state 2 (i.e., remaining on the same diabatic state) after passing the interaction region may be calculated using time-dependent perturbation theory. We write the wave function as a superposition of the diabatic wave functions, ψ d , ψ(t) = A(t)ψ1d e−iϕ1 (t) + B(t)ψ2d e−iϕ2 (t) , where i ϕs (t) = − ~ ∫ t Es (t′ )dt′ . (926) (927) 0 The phases may be derived from the time-dependent Schrodinger Eq. for the individual states, i~ ∂ψs = H̄ψs = Es (t)ψs . ∂t (928) The diabatic energies of the two states are E1 and E2 , and the energy difference is E2 − E1 = E12 = v(F1 − F2 )t = αt, (929) where v is the velocity, and forces F1 and F2 are the slopes of the diabatic potential energy curves. This last expression comes from the definition of the force, dE dE dx = = −F v dt dx dt (930) and the assumption that the potential varies linearly with distance in the vicinity of the crossing. Time-dependent perturbation theory gives for the evolution of the coefficients [13] H12 B(t) ϕ(t) e ~ (931) H21 A(t) −ϕ(t) e ~ (932) Ȧ = − and Ḃ = − where ∫ ϕ(t) = ϕ1 (t) − ϕ2 (t) = t E12 dt/~ = 0 αt2 2~ (933) The quantities H12 = H21 are matrix elements of the Hamiltonian in the diabatic representation. We can eliminate one of the unknowns, say A(t), by differentiating Eq. (932) and 39 using Eq. (931) to replace Ȧ by B, giving [13] B̈ = − 2 2 |H12 |B(t) iE12 Ḃ(t) |H12 |B(t) iαtḂ(t) − = − − . ~2 ~ ~2 ~ (934) This equation can be solved with some difficulty [14], and the result as t → ∞ is the Landau-Zener transition probability of a diabatic (i.e, non-adiabatic) transition, P = |B(t → ∞)|2 = e−π ξ , 2 (935) where ξ is called the adiabaticity parameter [12] ξ = t∆E/h, (936) and ∆E(Rx ) is the change in nuclear kinetic energy at the distance where the electron jumps. A large value of ξ means that the system remains on one adiabatic curve. In this case the nuclei do not jump from diabatic curve to another. This is equivalent to saying that the electron jumps in order to follow the adiabatically changing potential. Suppose the electron jumps in a localized region of width b. Then the time the atoms spend in this region is t = b/v. ∆E(Rx ) has the value dV1 dV2 = b|∆Fx |. ∆E(Rx ) = b − dR dR Rx (937) It follows that the adiabaticity parameter has the value ξ= ∆E 2 . hv|∆Fx | (938) A large value of χ favoring the electron jump (i.e., remaining an a single adiabatic potential curve) is the result of a large ∆E, a small v, and a small |∆F |. The last condition is equivalent to having the two “tracks” nearly parallel. A net reaction occurs if a non-adiabatic transition occurs as the atoms approach and not as they recede or the reverse. The total reaction cross section is therefore given by σ(E) = 2P (1 − P )σx , (939) σx = πRx2 . (940) where The non-adiabatic Harpoon reaction is a kinetic example of non-BO behavior. Much more famous (and ubiquitous) are spectroscopic consequences of non-BO behavior [4, 8]. 40 For example, the Ã1 B2u − X̃ 1 A1g transition of benzene is forbidden because x, y, z have E1u and A2u symmetry, so that µ12 = 0. Nevertheless, the transition is observed with vibrational transitions ∆v6 = 1. (The subscript 6 refers to a particular normal mode of the molecule.) How does this happen? When two (or more) electronic states are close in energy, the electronic Hamiltonian picks up extra terms that depend on nuclear normal coordinates. We can write this perturbation as a Taylor series, Ĥe = Ĥe0 + ∑ ( Qs s ∂ Ĥe ∂Qs ) + ... (941) Qs =0 Truncating after the linear term gives a perturbation operator, ( ) ∑ ∂ Ĥ e Ĥ ′ = Qs . ∂Qs s (942) Qs =0 First order (time-independent) perturbation theory shows that the wave function of the perturbed electronic state (in the case of benzene, ψà ) picks up a component from the perturbing state. If we write the wave function of the unperturbed state as ψf0 , the perturbed state becomes ψf = ψf0 + ∑ ckf ψk0 , (943) k where the sum is over all (usually one) perturbing electronic states. The amount of mixing (also called “intensity borrowing”) is given by the square of the coefficient, ∫ 0 ′ 0 3 ψk Ĥ ψf d r ckf = . Ef0 − Ek0 (944) In order for the integral to be non-zero, its integrand must contain a totally symmetric component. In the benzene example, the perturbing state is C̃(1 E1u ). Since the X̃ → C̃ transition is dipole allowed (because x, y has the same symmetry as the C̃), mixing of the C̃ and à state lends oscillator strength to the latter. The ν6 vibration, which induces the perturbation, has e2g symmetry. The symmetry of the matrix element in Eq. (944) is therefore (B2u ⊗ E2g ⊗ E1u ), which contains a component of A1g symmetry. (Verify this with a character table.) This effect is called a “vibronic transition,” and the non-totally vibration that allows it to happen is called the “coupling mode.” Vibronic coupling leads to many important phenomena. Possibly the most interesting is the conical intersection between potential energy surfaces. Once thought to be impossible, 41 then later to be very rare, it is now known to be very wide spread and to play a central role in processes such as vision and combustion, and indeed it is essential for life. A conical intersection is a point where two electronic states are degenerate. This degeneracy may be visualized by looking at a 2D subspace of the PES spanned by two normal coordinates. (It is conventional to use the normal coordinates of the ground state.) One of these coordinates is chosen to be totally symmetric. Variation along this coordinate has the effect of sliding the energies of the two states up and down. For this reason it is called a “tuning coordinate,” Qt . The other coordinate, Qc , corresponds to a vibronic coupling mode. We will use pyrazine as an example [15, 16] and investigate the intersection of the first two excited singlet states, S1 (B3u ) and S2 (B2u ). S1 is reached from the ground state by an n → π ∗ transition and S2 by a π → π ∗ transition. The pyrazine molecule as 24 normal modes, 5ag , 4b1u , 4b2u , 4b3g , 2au , b1g , 2b2g , 2b3u (Students should verify this). For this molecule there are five possible tuning modes of symmetry ag . For the S1 and S2 states there is only one coupling mode, which must have b1g symmetry for the coupling matrix element ⟨S1 |Qc |S2 ⟩ not to vanish (Students should verify this). This mode is an out-of-plane deformation of the ring. Calculation of the two PESs shows that they intersect for Qc = 0. We note that for Qc ̸= 0 the symmetry of the molecule is lowered from D2h to C2h . When that happens, the S1 and S2 states have the same Bu symmetry and do not cross, as predicted by the non-crossing rule. Although this rule applies only to diatomics, it gives the correct result for this 2D subspace. For Qc ̸= 0, however, the states have different symmetry, and the coupling matrix element is not zero. For the proper values of the tuning coordinate the surfaces are able to cross. A laser tuned to excite a vibronic level in the S2 manifold populates what is called a “bright state,” namely one which has oscillator strength for the S0 → S2 transition. If there was no coupling between S2 and S1 , the bright states would be eigenstates of He . Such a state would live for a very long time, until the molecule either fluoresces or interacts with its environment. In fact, there is very strong vibronic coupling, resulting is extremely rapid (10 fs) internal conversion. This happens because the initially excited bright state is not an eigenstate of the total Hamiltonian, but rather is a superposition of many eigenstates that are embedded in the quasicontinuum of S1 . This process of internal conversion allows the molecule to survive intact (without fragmenting) in the presence of UV radiation. Such UV photostability is critical for survival of biological systems and plays a key role in evolution. 42 LECTURE 38: ANGULAR MOMENTUM COUPLING: THE WIGNER-WITMER RULES Reference: Chapter 2 of Zare In atoms and molecules there are usually many sources of angular momentum, including electron spin, nuclear spin, orbital angular momentum, and rotation. Just as it is impossible to know all three components of a single angular momentum vector, similarly there are constraints on what we may know (i.e., measure simultaneously) about sums of angular momentum vectors. To give a concrete example, suppose L is the orbital angular momentum, S is the spin, and J = L + S is the vector sum. It is possible to measure either the set of observables L̂2 , Ŝ 2 , Jˆ2 , Jˆz or the set L̂2 , L̂z , Ŝ 2 , Ŝz . We must choose between them. In other words, we may construct a complete basis set out of the eigenfunctions of one set of operators or the other. The defining characteristic of an angular momentum vector operator is its commutation property, [ĵx , ĵy ] = i~ĵz . (945) It is easy to show that if ĵ1 and ĵ2 are each angular momentum vectors, then their sum ĵ = ĵ1 + ĵ2 is also an angular momentum vector. The individual operators have known eigenfunctions and eigenvalues: ĵ12 |j1 m1 ⟩ = j1 (j1 + 1)~2 |j1 m1 ⟩ ĵ22 |j2 m2 ⟩ = j2 (j2 + 1)~2 |j2 m2 ⟩ ĵ 2 |jm⟩ = j1 (j + 1)~2 |jm⟩ ĵ1z |j1 m1 ⟩ = m1 ~|j1 m1 ⟩ (946) ĵ2z |j2 m2 ⟩ = m2 ~|j2 m2 ⟩ ĵz |jm⟩ = m~|jm⟩ Out of these eigenfunctions we may construct two representations: 1) Uncoupled representation: |j1 m1 ⟩|j2 m2 ⟩ which is also written as |j1 m1 j2 m2 ⟩ 2) Coupled representation: |j1 j2 jm⟩ which is also written as |jm⟩ In the uncoupled representation, ĵ1 and ĵ2 are each allowed to precess about the z-axis. By this is meant that ĵ1x , ĵ1y , ĵ2x , ĵ2y are unknown. This in turn means that the length of the ĵ vector is unknown. If it were known precisely, it would put constraints on the components 43 ĵ1x , ĵ1y , ĵ2x , ĵ2y that violate the Uncertainty Principle. In fact we can show that j is restricted by the triangle rule, j1 − j2 ≤ j ≤ j2 + j1 . (947) In the coupled representation, ĵ1 and ĵ2 precess about ĵ. This means that m1 and m2 are not known individually; only their sum is known: m = m1 + m2 . (948) For given values of j1 and j2 we therefore have a choice of two representations: In the uncoupled representation basis |j1 m1 ⟩|j2 m2 ⟩ there are (2j1 +1)(2j2 +1) eigenstates. ∑ 2 +j1 In the coupled representation basis |j1 j2 jm⟩ there are jk=j (2k+1) = (2j1 +1)(2j2 +1) 2 −j1 eigenstates. In this sum, k is the number of projections of each value of j sum. We may transform one basis set into the other using ladder operators, ĵ± |jm⟩ = √ j(j + 1) − m(m ± 1)~|j, m ± 1⟩. (949) That these operators are defined by ĵ± = ĵx ± iĵy , (950) ĵ± = ĵ1± + ĵ2± . (951) from which we deduce that It is easiest to see how this is done by working out a specific example. Let’s use the case of j1 = 3, j2 = 1, j = 3. We will use the shorthand |3m⟩ for the coupled representation and |2m1 ⟩|1m2 ⟩ for the uncoupled representation. Clearly, |33⟩ = |22⟩|11⟩. (952) Using the lowering operator for the coupled representation, ĵ− |33⟩ = √ 6~|32⟩. (953) Using the lowering operator for the uncoupled representation, [ĵ1− + ĵ2− ]|22⟩|11⟩ = Comparing these two results gives |32⟩ = √ √ √ 4~|21⟩|11⟩ + 2~|22⟩|10⟩. 2 |21⟩|11⟩ + 3 √ 1 |22⟩|10⟩. 3 (954) (955) 44 Note that the sum of squares of the coefficients is 1, as expected for a unitary matrix. We repeat this process as far as possible: |31⟩ = c1 |20⟩|11⟩ + c2 |21⟩|10⟩ + c3 |22⟩|1, −1⟩ |30⟩ = c4 |2, −1⟩|11⟩ + c5 |20⟩|10⟩ + c6 |21⟩|1, −1⟩ |3, −1⟩ = c7 |2 − 2⟩|11⟩ + c8 |2, −1⟩|10⟩ + c9 |20⟩|1, −1⟩ (956) |3, −2⟩ = c10 |2, −2⟩|10⟩ + c11 |2, −1⟩|1, −1⟩ |3, −3⟩ = |2, −2⟩|1, −1⟩ The coefficients in this transformation are called Clebsch-Gordan coefficients, which are calculated by using the lowering operator repeatedly. They form a unitary matrix: |jm⟩ = ∑ C(j1 j2 j; m1 m2 m)|j1 m1 , j2 m2 ⟩ (957) m1 ,m2 and the inverse |j1 m1 , j2 m2 ⟩ = ∑ C(j1 j2 j; m1 m2 m)|jm⟩. (958) j,m These are actually single sums because m = m1 + m2 . The coefficients are defined by C(j1 j2 j; m1 m2 m) = ⟨j1 m1 , j2 m2 |jm⟩ = ⟨jm|j1 m1 , j2 m2 ⟩. They are related to the Wigner 3 − j symbols by ⟨j1 m1 , j2 m2 |j3 m3 ⟩ = (−1)j1 −j2 +m3 √ 2j3 + 1 (959) j1 j2 j3 m1 m2 −m3 (960) Our first application of angular momentum coupling is to figure out what are all the possible terms formed when two atoms combine to form a diatomic molecule. Suppose the atoms have spins S1 , S2 and orbital angular momenta L1 , L2 . We will assume that the spins are coupled to give a total of S, while the orbital angular momenta are described by the uncoupled representation, |L1 M1 ⟩|L2 M2 ⟩. Our goal is to discover all possible term symbols, 2S+1 Λ, including ± reflection symmetry and g, u inversion symmetry, when they exist. Recall that Λ = |M1 + M2 |. Note that in the uncoupled representation, the length of the resultant vector L is not well defined (i.e., is not known precisely). 45 The first thing we need to do is count the number of possible terms, with degenerate states counted only once. The possible values of Λ are (L> − L< ) ≤ Λ ≤ (L> + L< ), where L> is the larger of L1 , L2 and L< is the smaller of L1 , L2 . The states with Λ > 0 are all doubly degenerate because the energy is unchanged upon changing the sign of M1 + M2 . Note that a change in the sign of M is equivalent to a reflection in a plane containing the bond axis because that reverses the direction of the precession. A complication is associated with the Σ states, which are not degenerate and are characterized by a reflection parity: |Σ+ ; M ⟩ = |L1 , M, L2 , −M ⟩ + |L1 , −M, L2 , M ⟩ |Σ− ; M ⟩ = |L1 , M, L2 , −M ⟩ − |L1 , −M, L2 , M ⟩ (961) Our first task is to count the number of Σ states (or terms) and the number of Λ > 0 terms. It is easy to show that the former is just (2L< + 1)(2S< + 1). For each value of S, a Σ state state contains components of the type |L1 , −M2 , L2 , M2 ⟩, where L2 < L1 . But the number of spin states is 2S< + 1, giving the claimed result. It takes a bit more work to show that the total number of Σ and non-Σ states is (2S< + 1)(2L< + 1)(L> + 1). The remaining job is to find the number of Σ± states and the number of g, u states; the latter occurs if the molecule is homonuclear. The answer is given by the so-called WignerWitmer rules, which we shall only cite here. For reflection symmetry, recall that σ(xz) = c2 (y)i. (962) The parity of an atom having electrons with orbital angular quantum numbers li is given by p = (−1)q where q= ∑ li . (963) (964) i The total inversion parity for both atoms is p1 p2 . The parity associated with C2 (y) is (−1)L1 for atom 1 and (−1)L2 for atom 2. Hence the total reflection parity is p1 p2 (−1)L1 +L2 . If the number of Σ states is even, half are Σ+ and half are Σ− . If this number is odd, the reflection symmetry of the extra state is p1 p2 (−1)L1 +L2 . This applies for each spin multiplicity. Homonuclear molecules also have exchange symmetry. The rules for g, u assignment depend on the value of S, as follows: 46 For Σ states, if S is even, all Σ+ are g and Σ− are u. If S is odd, all Σ+ are u and Σ− are g. For Λ > 0, if Λ is odd, are an even number of states with the same Λ value, half of which have g symmetry and half u symmetry. If Λ is even, there are an odd number of states, with even values of S for a given Λ lambda value having g symmetry and odd S values having u symmetry. Example: Find all the terms for an oxygen molecule built up from two O(3 P ) atoms. The possible values of Λ are 0, 1, 2, denoted as Σ, Π, ∆. The possible values of S are 0,1,2. The total number of terms is (2 × 1 + 1)(2 × 1 + 1)(1 + 1) = 18. Of these, the number of Σ states is 9. These fall into three sets of single, triplet, and quintets. In each set, there is one extra Σ+ state because p1 p2 (−1)L1 +L2 = 1. The Π states are half g and half u because Λ = 1. The ∆ states are either g or u, depending on the value of S. The extra Σ± states are also g or u, depending on the value of S. The final result is 1 1 + 1 − 1 1 1 Σ+ g , Σg , Σu , Πg , Πu , ∆g 3 3 + 3 − 3 3 3 Σ+ u , Σu , Σg , Πg , Πu , ∆u 5 5 + 5 − 5 5 5 Σ+ g , Σg , Σu , Πg , Πu , ∆g (965) LECTURE 39: ANGULAR MOMENTUM COUPLING: HUND’S CASES In the previous lecture we discussed the effects of electronic angular momentum on the electronic states of a diatomic molecule. We are now in a position to ask how rotational angular momentum couples to the electronic part and explore its effect on the rotational spectrum. The various types of angular momentum (excluding nuclear spin) are: L = electronic orbital angular momentum S = electronic spin angular momentum R = nuclear rotational angular momentum J = R + L + S = total angular momentum It is also useful to identify the total angular momentum excluding spin, N = J − S. 47 We denote the projections of these operators on the molecular axis by ⟨L̂z ⟩ = Λ~ ⟨Ŝz ⟩ = Σ~ ⟨R̂z ⟩ = 0 (966) ⟨Jˆz ⟩ = Ω~ = (Λ + Σ)~ In addition, the projection of the total angular momentum operator on the space-fixed axis is ⟨JˆZ ⟩ = M ~. (967) Note that Herzberg uses N instead of R, and K instead of N. We cannot know the expectation values all of the angular momentum operators. Fortunately, real molecules have several limiting types of behavior known as the Hund cases. You may think of these different cases as distinct representations or basis sets. When we say that a state corresponds to a particular case, we mean that its total wave function may be well described by one of the eigenfunctions is a particular basis. The most common case is Hund’s case (a). In this case, the interaction of nuclear rotation with electronic angular momentum is very weak, while the electronic motion is strongly coupled to the nuclear axis. We may think of the molecule as a dumbbell with a flywheel attached, with the inertial axis of the electrons along the bond axis and perpendicular to R, forming a prolate symmetric top. The good quantum numbers are Λ, Σ, ΩJ, S. We may denote the eigenfunctions as |ΛΣΩJSn⟩, where Ω = Λ + Σ, and n refers to all other quantum numbers such as vibration. The rotational term energy is the same as for a prolate symmetric top (A > B = C), with quantum number K replaced by Ω, Fv (J) = Bv J(J + 1) + (A − Bv )Ω2 . (968) Because of the very light mass of the electron, A ≫ Bv . By convention, this term is absorbed into the electronic term energy, Te , so that the rotation term becomes Fv (J) = Bv [J(J + 1) − Ω2 ]. (969) with J = Ω, Ω + 1, Ω + 2, · · · Examples illustrating the rotational levels of 2 Π1/2 ,2 Π3/2 and 3 ∆1 ,3 ∆2 ,3 ∆2 are given in Figure 98 of Herzberg, Vol. I [2]. 48 Since the energy depends only on the absolute values of Λ and Ω, the wave function is a linear combination of eigenfunctions with ±Λ and ∓Σ, of the form |nΛ⟩|SΣ⟩|JΩM ⟩, giving rise to states of definite parity, p± . For example, for a Π state with S = 1/2, 1 |n2 Π1/2 JM p± ⟩ = √ [|n, 1⟩|1/2, −1/2⟩|J, 1/2, M ⟩ ± |n, −1⟩|1/2, 1/2⟩|J, −1/2, M ⟩] 2 (970) 1 2 ± |n Π3/2 JM p ⟩ = √ [|n, 1⟩|1/2, 1/2⟩|J, 3/2, M ⟩ ± |n, −1⟩|1/2, −1/2⟩|J, −3/2, M ⟩] 2 Each state is doubly degenerate, with the ± sign corresponding to the two parities. The rotational Hamiltonian for case (a) is given by ĤRot = 1 (Jˆ − L̂ − Ŝ)2 . 2 2µRe (971) We may simplify this expression by noting that R is perpendicular to the molecular axis, so that in the molecular frame ĤRot = 1 [(Jˆx − L̂x − Ŝx )2 ] + [(Jˆy − L̂y − Ŝy )2 ] 2µRe2 1 ˆ2 [J + L̂2x + Ŝx2 + Jˆy2 + L̂2y + Ŝy2 − 2Jˆx L̂x − 2Jˆy L̂y − 2Jˆx Ŝx − 2Jˆy Ŝy + 2L̂x Ŝx + 2L̂y Ŝy ] 2µRe2 x 1 = [(Jˆ2 − Jˆz2 ) + (L̂2 − L̂2z ) + (Ŝ 2 − Ŝz2 ) + (L̂+ Ŝ− + L̂− Ŝ+ ) − (Jˆ+ L̂− + Jˆ− L̂+ ) − (Jˆ+ Ŝ− + Jˆ− Ŝ+ )]. 2µRe2 = (972) Since the wave function is not an eigenfunction of L̂2 , the expectation value of L̂2 − L̂2z is lumped into the electronic term energy. The ladder operator terms have only off-diagonal matrix elements, which couple different electronic states. These are perturbations, which are omitted in the pure case (a). The origin of the ladder operators in the Hamiltonian may be understood by examining a typical pair of terms: L̂+ Ŝ− + L̂− Ŝ+ = (L̂x + iL̂y )(Ŝx − iŜy ) + (L̂x − iL̂y )(Ŝx + iŜy ) = 2L̂x Ŝx + 2L̂y Ŝy . (973) These terms are part of the spin-orbit operator HSO = Av L̂ · Ŝ, (974) where Av is the spin-orbit coupling constant for vibrational level v. The effect of these terms is to mix in states with ∆Λ = ±1 and ∆Σ = ∓1. The resulting perturbation is called “homogeneous” because Ω is unchanged. In a similar fashion, the terms Jˆ+ L̂− + Jˆ− L̂+ 49 are called the “L-uncoupling operator,” which induces an inhomogeneous perturbation that mixes in states with ∆Λ = ±1 and ∆Ω = ∓1. Likewise, the terms Jˆ+ Ŝ− + Jˆ− Ŝ+ is the “S-uncoupling operator,” which induces an inhomogeneous perturbation that mixes in states with ∆Σ = ±1 and ∆Ω = ∓1. Hund’s case (a) is built on the assumption that L and S are uncoupled and precess about the molecular axis. This assumption breaks down either if L = 0 or if the molecule rotates very rapidly. The criterion of rapid rotation is that Bv J ≫ Av Λ. In the limit of high J or for the case of Σ states (i.e., for S = 0), the molecule is described by Hund’s case (b). In this case, L still precesses about the molecular axis. Its projection, Λẑ couples to R to produce an angular momentum vector, N. This vector than adds to S to produce the total angular momentum, J. The quantum numbers that describe case (b) are contained in |nJSN Λ⟩. The rotational lines are then split by the coupling of S with N. For example, for S = 1/2, every J state is split into states labeled F1 and F2 . We can see the transition from case (a) to case (b) by diagonalizing Ĥrot in a suitable basis set. Using the case (a) basis, the eigenvectors of the Hamiltonian for a 2 Π state are |F2 ⟩ = aJ |2 Π1/2 vJ⟩ − bJ |2 Π3/2 vJ⟩ |F1 ⟩ = bJ |2 Π1/2 vJ⟩ + aJ |2 Π3/2 vJ⟩ (975) and the eigenvalues are ERot (2 Π, v, J) = Bv [(J − 1/2)(J + 3/2) ± X/2], (976) where X = [4(J − 1/2)(J + 3/2) + (Y − 2)2 ]1/2 Y = Av /Bv √ (977) X + (Y − 2) aJ = 2X √ X − (Y − 2) bJ = 2X 2 In the case (a) limit, (Y − 2) ≫ (J − 1/2)(J + 3/2), bJ = 0, and the energy is split by 2ESO . In the case (b) limit, (Y − 2)2 ≪ (J − 1/2)(J + 3/2), aJ = bJ =, and the energy is split by (2J − 1)(2J + 3). [1] I. N. Levine, Molecular Spectroscopy, (Wiley-Interscience, New York) 50 [2] G. Herzberg, Molecular Spectra and Molecular Structure. I. Spectra of Diatomic Molecules, (Van Nostrand, New York) [3] G. Herzberg, Molecular Spectra and Molecular Structure. II. Infrared and Raman Spectra of Polyatomic Molecules, (Van Nostrand, New York) [4] G. Herzberg, Molecular Spectra and Molecular Structure. III. 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