Quantization of Rotational Motion: Spherical Harmonics and Rotational Wavefunctions - Prof, Study notes of Physical Chemistry

Download Quantization of Rotational Motion: Spherical Harmonics and Rotational Wavefunctions - Prof and more Study notes Physical Chemistry in PDF only on Docsity! Lecture 6 Rotational motion NC State University Chemistry 431 Classical Rotation In a circular trajectory Jz = pr and E = Jz2/2I. I is the moment of inertia. Mass in a circle I = mr2. Diatomic I = μr2 μ = m1m2m1 + m2 m m2 m1 Reduced mass r r The 2-D rotational hamiltonian • Solutions of the 2-D rotational hamiltonian are sine and cosine functions just like the particle in a box. • Here the boundary condition is imposed by the circle and the fact that the wavefunction must not interfere with itself. • The 2-D model is similar to condition in the Bohr model of the atom. The 2-D rotational hamiltonian • The wavelength must be a whole number fraction of the circumference for the ends to match after each circuit. • The condition 2πr/m = λ combined with the deBroglie relation leads to a quantized expression,Jz = mh. • The hamiltonian is: – h 2 2I ∂2Φ ∂φ2 = EΦ Quantization of rotational motion: solution of the φ equation The corresponding wavefunctions are: with the constraint that: Since the energy is constrained to values Jz2/2I we find that m = 0, ±1, ± 2, ... Φ m = 12π 1/2 eimφ m = ± 2IE 1/2 h The volume element in spherical polar coordinates To solve the Schrodinger equation we need to integrate of all space. This is the same thing as performing a volume integral. The volume element is: This integrates to 4π, which is the normalization constant. 4π stearadians also gives the solid angle of a sphere. dV = r 2dr sinθdθ dφ Separation of variables For sake of simplicity, we have factored the term into the energy.  At the end of the calculation we Will multiply by this term to obtain the energy. The spherical harmonics arise from the product of ΘΦ after substituting Y = ΘΦ ∂2ΦΘ ∂ϕ2 + sinθ ∂∂θ sinθ ∂ΦΘ ∂θ = sin 2θEΦΘ – h 2 2I Separation of variables When we divide by Y = ΘΦ, we obtain Now, these equations can be separated. The operators in variables θ and φ operate on function Θ and Φ , respectively, so we can write Θ∂ 2Φ ∂ϕ2 + Φsinθ ∂∂θ sinθ ∂Θ ∂θ = sin 2θEΦΘ 1 Φ ∂2Φ ∂ϕ2 + 1Θsinθ ∂ ∂θ sinθ ∂Θ ∂θ – sin 2θE = 0 The spherical harmonics as solutions to the rotational hamiltonian The spherical harmonics are the product of the solutions to the θ and φ equations.  With norm‐ ‐alization these solutions are YJ M(θ,φ) = NJMPJ |M|(cos θ)eiMφ The M quantum number corresponds to Jz the  z component of angular momentum.  The norm‐ ‐alization constant is NJM = 2J + 12 (J – |M|)! (J + |M|)! 1/2 The form of the spherical harmonics Y0 0 = 1 4π Y1 0 = 34π cosθ Y1 ±1 = 38π sinθe ±iφ Including normalization the spherical harmonics are Y2 0 = 516π 3cos 2θ – 1 Y2 ±1 = 158π sinθcosθe ±iφ Y2 2 = 1532π sin 2θe±2iφ The form commonly used to represent p and d  orbitals are linear combinations of these functions Euler relation e±iφ = cosθ ± isinθ sinθ = e iφ – e–iφ 2i cosθ = e iφ + e–iφ 2 Linear combinations are formed using the Euler relation Projection along the z‐axis is usually taken using  z = rcosθ.  Projection in the x,y plane is taken using x = rsinθcosφ and y = rsinθsinφ Spherical harmonic for P0(cosθ) • Plot in polar coordinates represents |Y00|2 where Y00=(1/4π)1/2 . • Solution corresponds to rotational quantum numbers J = 0, • M or Jz = 0. • Polynomial is valid for n≥1 quantum numbers of hydrogen wavefunctions Spherical harmonic for P1(cosθ) • Plot in polar coordinates represents |Y11|2 where Y11=(1/2)(3/2π)1/2 sinθeiφ. • Solution corresponds to rotational quantum numbers J = 1, Jz = ±1. • Polynomial is valid for n≥2 quantum numbers of hydrogen wavefunctions Spherical harmonic for P1(cosθ) • Plot in polar coordinates represents |Y10|2 where Y10=(1/2)(3/π)1/2 cosθ with normalization. • Solution corresponds to rotational quantum numbers J = 1, Jz = 0. • Polynomial is valid for n≥2 quantum numbers of hydrogen wavefunctions. Spherical harmonic for P2(cosθ) • Plot in polar coordinates of |Y20|2 where Y20=1/4(5/π)1/2(3cos2θ-1) • Solution corresponds to rotational quantum numbers J = 1, Jz = 0. • Polynomial is valid for n≥3 quantum numbers of hydrogen wavefunctions Rotational Wavefunctions J = 0 J = 1 J = 2 These are the spherical harmonics YJM, which are solutions of the angular Schrodinger equation. The degeneracy of the solutions • The solutions form a set of 2J + 1 functions at each energy (the energies are E = h2 J(J +1)/2I. • A set of levels that are equal in energy is called a degenerate set. J = 0 J = 1 J = 2 J = 3 Orthogonality of wavefunctions • For the theta integrals we can use the substitution • x = cosθ and dx = sinθdθ • For example, for s and p-type rotational wave functions we have < s | p > ∝ cosθ sinθ dθ 0 π = x dx 1 – 1 = x 2 2 1 – 1 = 12 – 1 2 = 0 Question Which of the following statements is true: A. The number of z-projection of the quantum numbers is 2J+1. B. The spacing between rotational energy levels increases as 2(J+1). C. Rotational energy levels have a degeneracy of 2J+1. D. All of the above. Question Which of the following statements is true: A. The number of z-projection of the quantum numbers is 2J+1. B. The spacing between rotational energy levels increases as 2(J+1). C. Rotational energy levels have a degeneracy of 2J+1. D. All of the above. ΔE ~ (J +2)(J + 1) – J(J + 1) = 2(J + 1) The moment of inertia The kinetic energy of a rotating body is 1/2Iω2. The moment of inertia is given by: The rigid rotor approximation assumes that molecules do not distort under rotation. The types or rotor are (with moments Ia , Ib , Ic) - Spherical: Three equal moments (CH4, SF6) (Note: No dipole moment) - Symmetric: Two equal moments (NH3, CH3CN) - Linear: One moment (CO2, HCl, HCN) (Note: Dipole moment depends on asymmetry) - Asymmetric: Three unequal moments (H2O) I = miri2Σi = 1 ∞ Pure rotational spectra • A pure rotational spectrum is obtained by microwave absorption. • The range in wavenumbers is from 0-200 cm-1. • Rotational selection rules dictate that the change in quantum number must be ΔJ = ± 1 and ΔMJ = 0. • A molecule must possess a ground state dipole moment in order to have a pure rotational spectrum. The Dipole Moment Expansion The permanent dipole moment of a molecule oscillates about an equilibrium value as the molecule vibrates. Thus, the dipole moment depends on the nuclear coordinate Q. where μ is the dipole operator. μ Q = μ0 + ∂μ ∂Q Q + ... Interaction with radiation An oscillating electromagnetic field enters as E0cos(ωt) such that the angular frequency hω is equal to a vibrational energy level difference and the transition moment is Mrot = μ0 YJ+1,M * cos θ YJMsin θ dθdφ 0 π 0 2π Interaction with radiation The choice of cos(θ) means that we consider z-polarized microwave light. In general we could consider x- or y-polarized as well. x sin(θ)cos(φ) y sin(θ)sin(φ) z cos(θ) μ0 = μ Xi + μY j + μZk μ0 = μ0 sinθcosφi + sinθsinφ j +cosθk Energy level spacing Energy levels EJ = h2 2I J(J + 1) Energy Differences of ΔJ = ± 1 EJ+1 – EJ = 2h2 2I (J + 1) Question Which molecule found in the atmosphere has a pure rotational spectrum? A. Diatomic oxygen B. Diatomic nitrogen C. Water D. Carbon Dioxide Question Which molecule found in the atmosphere has a pure rotational spectrum? A. Diatomic oxygen B. Diatomic nitrogen C. Water D. Carbon Dioxide A typical rovibrational spectrum Note that the rotational spectrum is centered a vibrational frequency