Translational Quantum Mechanical Models Vibration and Rotation of Molecules Chapter 18 Vibrational Molecular Energy Molecular Motions Rotational Electronic … Vibrations of Molecules: Frequency of vibration – classical approach Model approximates molecules to atoms joined by springs. A vibrations (one type of – normal mode of vibration) of a CH2 moiety would look like; Water: http://www.youtube.com/watch?v=1uE2lvVkKW0 CO2 http://www.youtube.com/watch?v=W5gimZlFY6I O2 http://www.youtube.com/watch?v=5QC4OVadKHs http://en.wikipedia.org/wiki/Molecular_vibration The motions are considered as harmonic oscillators. For a molecule of N atoms there are 3N-6 normal modes (nonlinear) or 3N-5 (linear). Vibrational motion - harmonic oscillator, KE and PE – classical approach Center of mass, does not move. Center of mass coordinates Diatomics: Force constant of spring = k Whether pulled apart or pushed together from the equilibrium position, the spring resists the motion by an opposing force. Amplitudes are real numbers, b1 and b2 are real or = 0. Applying the BC, set at t = 0; x(0) = 0, v(0) = v0. =1 Solutions to the DE will be of the form; Therefore: where b1 c1 c2 and b2 i (c1 c2 ) and to find T: angular velocity = Frequency = 1/T k 2 where frequency = phase angle Energy terms (KE, PE) are 1 2 1 v PE kx 2 2 2 where v and x are velocity and position. KE Harmonic oscillator potential function Morse curve 0 http://www.youtube.com/watch?v=5QC4OVadKHs No restriction on E classically. Study Example Problem 18.1 Similar for low energy situations = ground state 1. Expression for total vibrational energy, Etot ,vib KE PE x ,x 1 2 1 2 v + kx 2 2 p2 1 2 kx 2 2 2. Construct the total energy Operator, the Hamitonian 2 Equivalent to a particle of mass , vibrating around its equilibrium distance. How can a particle like that be described by a set of wavefunctions. 1 p 2 1 kx 2 1 i d 1 kx 2 H 2 2 2 dx 2 3. Set the Schrodinger equation, eigenequation E H H n ( 1/ 2 x ) - Hermite polynomials n = vibrational quantum number Solutions (normalized wavefunction, eigenfunctions) of which are; With the normalization constant The first few wavefunctions, n (n = 0,..4) The first few wavefunctions (n = 0,..3) The respective eigen energies are; Note the resemblance to 1D box. zpe Equal energy gaps ~ contrast to 1D box. The first few wavefunctions, n2 (n = 0,..4) The constraint imposed on the particle by a spring results in zero point energy. The energy values of vibrational states are precisely known, but the position of the particles (as described by the amplitude is imprecise, only a probability ] 0. density 2(x) of x can be stated, [x, H Note: the overflow (forbidden) of wavefunction beyond the potential barrier. Note the resemblance to 1D box. At high quantum numbers n (high energy limit) the system gets closer to a classical system (red probability of x in q.m. oscillator, blue probability of x classical oscillator. n = 12 n >>0 high low high high low high Blue to black, a particle in a barrel Rotational energies of a classical rigid rotor (diatomic). In vibrational motion, velocity, acceleration and momentum are parallel to the direction of motion. In the center of mass coordinate system, rigid rotor motion equivalent to a mass of moving in a circle of radius r (= bond length) with an angular velocity and a tangential velocity v. Rotation in 2D (on a plane) y r0 No opposing force to rotation – no PE (stored of energy). All energy = KE. x p = linear momentum Acceleration in the radial direction Angular velocity vector Angular velocity and acceleration, r0 Mass with constant makes = 0. RHR Tangential linear velocity, Angular velocity (~ momentum) vector normal to the plane of motion. (coming out of the plane) = I, moment of inertia The rotation on the x,y plane occurs freely. Equivalently the particle moves on the circle (radius r = r0) freely; Epot = V(x,y) = 0 QM Angular momentum (2D): p 2E I angular momentum Etot ,vib KE PE KE r p2 2 2 2 2 2 p ( x, y ) ( x, y ) H 2 2 2 x y 2 Magnitude of l = l Energy: vr SE: No restriction on l (or E) classically. In polar coordinates (variable is ): r2sin d d dr BC: ( ) ( 2 ) Rotational eigenfunction. General solution for DE (as seen before): ml = integer clockwise counter-clockwise ml = rotational quantum number, quantization Normalization constant A; 2 * ml ( ) ( ) ml ( )d 1 1 2 e iml 0 2 A2 e iml iml e d 1 0 2 A2 d 1 A2 2 1 0 A ( ) 1 2 m being an integer, rotational frequency will take finite values. 1 1 ( ) 2 Stationary state 1 e iml 2 1 2 1 2 Non-stationary state unacceptable 2 real part e iml = 1 2 (cos m l + i sin m l ) 1 cos 2 1 cos 3 2 1 cos 5 2 cos 2 cos 4 cos 6 Energy of rotational states: Note; No non-zero ZPE. ZPE appears if the potential energy confinement exists, not here (V()=0) Degenerate states - two fold (equal energy) 2 Eml I = 2E I 2 2 2 ml2 ml Eml I I 2I I m being an integer, rotational frequency will take finite values. – discrete set of rotational frequencies!! The wavefunctions, …. of a rigid rotor are similar to the wavefunctions of a free particle restricted to a wall of a circle (particle in a ring)! ,l z ] 0 Note: for 2D rigid rotor [ H Angular Momentum – 2D rigid rotor. Operator: p l z i l z i Both operators has the same eigenfunctions. angular analogue Hence angular momentum (z component) for this case can be precisely measured. But position, here , cannot be measured precisely; ,l z ] 0 probability for any interval of d is the same [ r Angular momentum (in z direction) is quantized!! ,l z ] 0 Note: for 2D rigid rotor [ H QM Angular momentum (3D): Particle on a circle. r r0 r0 2D 3D Particle on the surface of a sphere. Classical 3D rotor KE operator in polar coordinates = Just the steps in solving the SE a. collect the constants: b. multiply by sin2, rearrange to get PE confinement – none; V(,) = 0 PE operator set to zero. E = KE + PE = KE SE: Differentiation only w.r.t. Therefore Wavefunction; ;separation of variables possible. spherical harmonic functions c. substitute for Y(,) divide by ()(), we get solutions Because each side of the equation depends only one of the variables and the equality exists for all values of the variables, both sides must be equal to a constant. BC leads to q.n.; spherical harmonic functions written in more detail would take the form: solutions BC leads to more q.n.; Energy of degenerate states of q. n l : For a given l there are (2l +1), ml values. That is the state described by a l quantum number is comprised of (2l+1) sub-states. Degeneracy of state with l, is 2l+1. 2 r02 E 2 I 2 E l (l 1) 2 El 2 l (l 1) 2I quantization of rotational energy Eigen-energy: El 2 l (l 1) 2I Quantization of Angular Momentum As will be seen later, the shapes, directional properties and degeneracy of atomic orbitals are dependent on the l and ml vales associated with these orbitals. Verifiable in SE: Spherical harmonics are eigen functions of the total energy operator. Eml :in a 3D rotor is undefined; the q. n. ml here determines the z-component of the angular momentum vector I. For 3D rotor; Etot 2 l2 l H 2I 2I constant Therefore both operators have a common set of eigenfunctions. The operators should commute! ,l 2 ] 0 [H Components of the angular momentum in x, y and z. Now; The respective operators are (Cartesian): and therefore Note that the above operators differ by a multiplication constant (1/2I) thus the eigen values differ by (1/2I). Also note the calculated (and measurable) angular momentum quantity (of precise (eigen) value) is that of | l2 | (therefore I). I l (l 1) The respective operators are in spherical coordinates would be: The following commutator relationships exist. f ( , ) The component operators do not commute with one another f ( ) The direction of I cannot be specified (known) ‘precisely’ for rotations in 3D in QM systems as opposed to classical rigid rotors. Need all vectors to specify direction of I. However, angular momentum wavefunction is an eigenfunction of the Hamiltonian and therefore, Conclusion – spherical harmonics are eigenfunctions of l2 and lz hence magnitudes of both |I| and Iz can be known simultaneously, but not the x and the z component values of the angular momentum. Examining; !!! lz ml , l2 and l commute and l , l and l do not commute. H z x y z Spatial quantization with one another. Example l 2 case. ml = +2 ml = +2 Vector model of angular momentum Vector model of angular momentum Example l 2 case. Spherical harmonic wavefunction shapes: l=0 l=1 l=2 Complex functions - visualization – not possible. Under such situations alternate wave functions with same eigenvalues can be constructed from the ‘complete set’ of wave functions available, i. e. proper linear combinations from the complete set of eigenfunctions. px = < 90 180 <90 0 90 0 surface on x-z plane, for simplicity all p all d

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