Vibrational levels of water -Tutorial

Truman State University

Informations

• Publié par: Odri
• Nombre de lectures: 24
• Langue: English

Extrait

Vibration frequencies and force
constants of water molecule
-classical treatment
M. Samiullah
Suchi’s Lab Group, 12/01/2009 (?)
References:
1. Wilson, Decius, and Cross, “Molecular Vibrations: The Theory of Infrared and
Raman Vibrational Spectra,” McGraw Hill, NY (1955).
rd2. Cotton, “Chemical Applications of Group Theory,” 3 edition, John Wiley &
Sons, NY (1990).
3. Landau and Lifshitz, “Quantum Mechanics : Non-relativistic Theory,” Addison-
Wesley , NY (1958).Classical Mechanics Background
• Simple Harmonic Oscillator
2 2Kinetic energy, T = ½ m (dx/dt) ; Potential energy, V = ½ k x .
2 2Equation of motion: m d x/dt + k x = 0. (1)
iωtHarmonic solution: x(t) = A e with A and ω unknown. Plug into 1.
2 iωt m(ω - k/m) A e = 0.
2 Either A = 0 or ω - k/m= 0.
kSince A ≠ 0, we must have . m
>>> So, if you know ω, you can learn about force constant k and
vice-versa.
>>> Frequency ω for a molecular system is normally found by IR
and/or Raman vibrational spectrum.Mass-weighted Cartesian Coordinates
{q}
• Simple Harmonic Oscillator Revisited
q x mLet and let over dot represent d/dt.
1 22  2T qT qKinetic energy: , often written as .
2
k 2Potential energy: 2V q
m k
 Equation of motion: q q 0
m
>>> The angular frequency of oscillation:
k
mFrequency of oscillation of
a diatomic molecule (1)
2 2Kinetic energy = ½ m (dx /dt) ½ + ½ m (dx /dt) ; 1 1 2 1
2Potential energy = ½ k (x -x ) .1 2
Equations of motion:
2 2m d x /dt + k (x -x ) = 0. (1a)1 1 1 2
2 2m d x /dt - k (x -x ) = 0. (1b)2 2 1 2
Seek harmonic solution:
iωt iωtx (t) = A e , x (t) = A e . Plug into (1).1 1 2 2
• Simultaneous equations for A and A .1 2
2(-m ω + k )A -k A =01 1 2
2-k A + (-m ω + k )A =01 2 2Diatomic molecule (2)
Write in matrix form:
2 A(-m ω + k ) -k 1 = 01
A2 2-k (-m ω + k )2
For nontrivial solution, the determinant of matrix must be
2 2zero → ω [m m ω - k (m + m )] = 0 [Secular Equation]1 2 1 2
Two solutions for ω :
ω = 0 [Translation] [A = A ] (2)1 2
and ω = √*k/μ] [Vibration] [m A = m A ] (3)2 2 1 1
where μ = m m / (m + m ), the reduced mass.1 2 1 2Small vibrations in Classical Mech
General Treatment (slide 1)
Let us use mass-weighted coordinates {q , q , 1 2
q , …, q }.3 3N
3N
22T qiKinetic energy:
i 1
3N 3N
2V f qqPotential energy: ij i j
i 1 j 1 3N
 q f q 0 i 1, ,3Ni ij jEquations of motion:
j 1
i tHarmonic solutions: q Ae i 1, ,3N; A complexi i i
3N
2f A 0 i 1, ,3NSimultaneous equations: ij ij j
j 1
• Gives normal modes of vibration, translation and rotationSmall vibrations (slide 2)
• Simultaneous equations for Amplitudes {A }i
3N
2f A 0 i 1, ,3Nij ij j
j 1– Eigenvalue/Eigenvector problem
– 3N eigenvalues are normal mode frequencies, 3 of which correspond
to translation and 3 or 2 to rotation: ω (k= 1, 2, …, 3N).k
– 3N eigenvectors Q corresponds to the normal coordinates expressed k
in terms of the {q} coordinates: Q = Sum[A q , i = 1, …, 3N+ if k ik i
properly normalized.
2 2 2• 2T = ∑(dQ /dt) and 2V = ∑ [w Q ].k k k
>>> Every problem turns into sum of simple harmonic oscillations for each
normal mode. Quantum states labeled with occupation number n of k
each mode w and energies are simply (n + ½) hbar w .k k k
*Solution of more complicated molecules immensely helped by
an application of the. Group Theory We will do H O next.2Symmetry of H O2
y• Place molecule in xz-plane.
Δy• Four symmetry operations 3
Δy1
1. Do nothing: E
0 Δy2. Rotate 180 about z-axis: C 22 Δx3Δz33. Reflect in xz-plane: σ (xz)v Δx1Δz14. Reflect in yz-plane: σ (yz) xv Δx2z Δz2
• The character table: Four 1-dimensional irreps A , A , B , B .1 2 1 2
C E C σ (xz) σ (yz) Basis2v 2 v v
A 1 1 1 1 z A and B reps1
distinguished A 1 1 -1 -1 R2 z
by action of C2
B 1 -1 1 -1 x, Ry1
B 1 -1 -1 1 y, R2 xFull Representation in the basis of the
nine Cartesian displacements
(Read: ΔX for X1, etc.)1
Displacement E C2 σ(xz) σ(yz)
X1 X1 -X2 X1 -X2
Y1 Y1 -Y2 -Y1 Y2
Z1 Z1 Z2 Z1 Z2
X2 X2 -X1 X2 -X1
Y2 Y2 -Y1 -Y2 Y1
Z2 Z2 Z1 Z2 Z1
X3 X3 -X3 X3 -X3
Y3 Y3 -Y3 -Y3 Y3
Z3 Z3 Z3 Z3 Z3
Characters 9 -1 3 1Reducing full rep into irreps
• By inspection or by using Great Orthogonality
theorem:
Г = 3 A + A + 3 B + 2 B1 2 1 2
• Look into the table to identify translation and
rotation from the list.
– Translation basis are x, y, z: B1, B2, A1
– Rotation basis are Rx, Ry, Rz: B2, B1, A2
• The remainder reps in Г are purely vibs:
Г = 2 A + B . vib 1 1