5.61 Lecture #17 Rigid Rotor I

Transcription

1 561 Fall 2017 Lecture #17 Page Lecture #17 Rigid Rotor I Read McQuarrie: Chapters E and 6 Rigid Rotors molecular rotation and the universal angular part of all central force problems another exactly solved problem The rotor is free, thus Ĥ = T (V = 0) Once again, we are more interested in: * E and Ĵ t * effects of Ĵ2, Ĵz, Ĵ± = Ĵx ± iĵy ( raising and lowering or ladder operators) * qualitative stuff about shape (nodal surfaces) of ψ M (θ, φ) than we are in the actual form of ψ M (θ, φ) and in the method for solving the rigid rotor Schrödinger Equation Next Lecture no differential equation, no ψ, just: * [ i, j ] = iħ k ε ijk k definition of an angular momentum operator * 2 ψ M = ħ 2 ( + 1)ψ M * z ψ M = ħmψ M * ± = x ± i y * ± ψ M = [( + 1) M(M ± 1)] 1/2 ψ M±1 The standard problem is for the motion of a mass point constrained to the surface of a sphere However, we are really interested in the rotational motion of a rigid diatomic molecule Mass point is the same as one end of a rotor (length r 0, mass µ) with bond axis extending from the coordinate origin at the Center of Mass

2 561 Fall 2017 Lecture #17 Page 2 Z M is projection of on Z M θ m 1 r 1 r cm r 2 m 2 at origin, r 0 = r 1 + r 2 Y X precesses about Z Why do we say this when there is no motion in solutions to TISE? For a molecule, there are two coordinate systems: the body fixed system denoted by x, y, z and the laboratory-fixed system denoted by X, Y, Z What are we going to want to remember? 1 E(, M) 2 What do, M mean about { location of NOT the same! location of internuclear axis 3 qualitative stuff about shape of ψ,m (θ, φ) # of nodes locations of nodes] surfaces where is the internuclear axis? reduced pictures of ψ,m (θ, φ) } distributions 4 Selection rules and values of integrals of 2, Z, ±

3 561 Fall 2017 Lecture #17 Page 3 Rotor is free bond length V (r = r 0, θ, φ) = 0, V (r r 0, θ, φ) = Need to keep track of only T (θ, φ) because V = 0 when r = r 0 Ĥ = T for 1 D T = ˆp2 2µ linear momentum For rotor we want angular rather than linear momentum From classical mechanics: L = r p ω p 1 r 1 p 1, r 2 p 2 m 2 r 2 r 1 m 1 r 0 = r 1 + r 2 r 0 for rotation on surface of sphere r p v 1 = r 1 ω, p = m v = mr ω r p = L = m 1 r 2 1ω + m 2 r 2 2ω Iω I = i m i r 2 i moment of inertia (for a diatomic or any linear molecule) Use m 1 r 1 = m 2 r 2 center of mass constraint to get I = µr 2 0 Now write T as analogous to p2 2µ p L µ I T = L 2 2I Now we have a mass point µ (at r 0 from origin), where bond axis points at the mass point, µ

4 561 Fall 2017 Lecture #17 Page 4 Z looks bad! X φ ASA: get XY Z rθφ Ĥψ = T ψ = Eψ solutions are ψ(r 0, θ, φ) Y (θ, φ) Solve for θ, φ dependence of Y (θ, φ) θ µ project onto the X, Y plane r 0 Y [ ( ) ( ( )) T = ħ2 1 sin θ + 1 ( )] 2 2I sin θ θ θ sin 2 θ φ 2 We really do not want to do this in detail because we have little reason to actually look at the Y (θ, φ) eigenfunctions THIS SHOULD BOTHER YOU! Outline of Steps separation of variables Y (θ, φ) = Θ(θ)Φ(φ) seen this before standard process arrange θ stuff on LHS, φ stuff on RHS Then divide both sides by Θ(θ)Φ(φ) A lot of algebra

5 561 Fall 2017 Lecture #17 Page 5 ( ) ( ( )) sin θ sin θ Θ(θ) + β sin 2 θ Θ(θ) θ θ }{{} only θ β 2IE ħ 2 collecting constants = 1 ( 2 Φ(φ) φ }{{ 2 } only φ both sides must equal same constant: call it m 2 ) Φ(φ) 2 separate equations: ( ) separation constant φ: Φ(φ) = m 2 Easy to solve Φ(φ) ( φ) 2 ( ( )) sin θ 2 θ: sin θ Θ(θ) + β sin 2 θ = m 2 Leads to Legendre Equation Θ(θ) θ θ Most of the useful insight comes from # 1! ( ) 2 Φ = m 2 Φ(φ) φ 2 Φ(φ) = A m e imφ m = 0, ±1, ±2, A m is normalization constant Φ(φ) = Φ(φ + 2mπ) A m = (2π) 1/2 Notice that e imφ is complex periodic boundary condition HOW DO WE GET THIS? Can convert to real functions by 2 1/2 [e imφ ± e imφ ] Are these eigenstates? Turns out we exchange the all-real functions of the geometric picture for a picture of the eigenstates of L z # nodes in XY plane φ = 0 m = 1 cos φ = π = mφ φ = π 2 3π ( π ) same nodal plane if m = 1 = mφ φ = 3 2 2m m is # of nodal planes thru XY plane

6 561 Fall 2017 Lecture #17 Page 6 2nd differential equation for Θ(θ) is more complicated Legendre Equation: can look at notes Note that the Θ DE depends on the value of m 2nd boundary condition the requirement of continuity in θ gives energy quantization Notice that the form of Θ(θ) eigenstates depends on both L, M Final result for energy levels of rotor is E L = ħ2 L(L + 1) L = 0, 1, 2, 2I Notation: L general orbital angular momentum l angular momentum of one electron angular momentum of molecular rotation several places where rigid rotor appears Same equations, L, l, often used interchangeably Also other angular momenta S, I, and N The beauty is that once we understand one angular momentum we understand all angular momenta Pictures ψ LM (θ, φ) = YL M (θ, φ) spherical harmonic [( ) ] 1/2 2L + 1 (L M )! = P M L (cos θ) 4π (L + M )! }{{} eimφ }{{} Legendre Polynomial normalization constant (solution of Legendre Equation) But we want memorable pictures and intuition guiding cartoons! See McQuarrie, page 286 for table of Y m L (θ, φ) [at end of these notes] Vector Model - relates body frame (x, y, z) properties to lab frame (X, Y, Z) observations

7 561 Fall 2017 Lecture #17 Page 7 1 = r p is a vector to plane of rotation 1 CM 2 2 θ, φ are the lab frame coordinates of a fictitious particle of mass µ moving on the surface of a sphere of radius r 0 They are also the coordinates of one end of the rotor axis where the rotation is about the center of mass along this axis 3 Key questions: a Where does point in lab? b How long is? c Where does internuclear axis, r, point in lab? ( r) Z α precesses about laboratory Z at constant α Z Y M = }{{} ħm Y M length of projection of on Z * Length of 2 Y M = [ ] 1/2 = 2 1/2 = ħ 2 ( + 1)Y M = ħ[( + 1)] 1/2 ħ( + 1/2) * What is cos α?

8 561 Fall 2017 Lecture #17 Page 8 Z α cos α = ħm ħ[( + 1)] 1/2 M CRUCIAL INSIGHT Where is the internuclear axis in lab for a specified tilt of from laboratory Z axis? If = M, is almost exactly along lab Z (Why almost?) Z molecular internuclear axis is rotating in XY plane If M = 0, is precessing exactly in the XY plane: when is along X, molecular axis is in Y Z plane for M = 0 when is along Y, molecular axis is in XZ plane so molecular axis is maximally (but not exclusively) along Z Molecular Beam in Z direction M = 0, in XY helicopter M = ± molecular axis is in XY plane frisbee molec axis is more along Z Z different kinds of collisions

9 561 Fall 2017 Lecture #17 Page 9 Polar Plots Shown without explanation in most introductory texts (for atomic orbitals) distance of plotted point from origin r(θ, φ) = 1 2 [ Y M (θ, φ) + Y M (θ, φ) ] real function of θ, φ These are not the solutions of the (r, θ, φ) full 3D Schrödinger Equation They are a way of representing a 3D object in 2D xy plane etc > 1 s spherical p 3 dumbbells with axis along X, Y, and Z Notice that # of nodal surfaces is L (or ) L # surfaces 0(s) 0 1(p) 1 2(d) 2 Number of nodal lines in XY plane is M Now you should be able to recognize, M (or L, M L ) from the number and placement of nodes know how is located in a laboratory frame know how internuclear axis is located in a laboratory frame

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