Ch121a Atomic Level Simulations of Materials and Molecules

1. Ch121a Atomic Level Simulations of Materials and Molecules BI 115 Hours: 2:30-3:30 Monday and Wednesday Lecture or Lab: Friday 2-3pm (+3-4pm) Lecture 7, April 15, 2011 Molecular Dynamics – 3: vibrations William A. Goddard III, wag@wag.caltech.edu Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants Wei-Guang Liu, Fan Lu, Jose Mendoza, Andrea Kirkpatrick

2. Outline of today’s lecture • Vibration of molecules • Classical and quantum harmonic oscillators • Internal vibrations and normal modes • Rotations and selection rules • Experimentally probing the vibrations • Dipoles and polarizabilities • IR and Raman spectra • Selection rules • Thermodynamics of molecules • Definition of functions • Relationship to normal modes • Deviations from ideal classical behavior

3. Simple vibrations Starting with an atom inside a molecule at equilibrium, we can expand its potential energy as a power series. The second order term gives the local spring constant We conceptualize molecular vibrations as coupled quantum mechanical harmonic oscillators (which have constant differences between energy levels) Including Anharmonicity in the interactions, the energy levels become closer with higher energy Some (but not all) of the vibrational modes of molecules interact with or emit photons This provides a spectroscopic fingerprint to characterize the molecule

4. Vibration in one dimension – Harmonic Oscillator No friction Consider a one dimensional spring with equilibrium length xe which is fixed at one end with a mass M at the other. If we extend the spring to some new distance x and let go, it will oscillate with some frequency, w, which is related to the M and spring constant k. To determine the relation we solve Newton’s equation  M (d2x/dt2) = F = -k (x-xe) Assume x-x0=d = A cos(wt) then –Mw2 Acos(wt) = -k A cos(wt) Hence –Mw2 = -k or w = Sqrt(k/M). Stiffer force constant k  higher w and higher M  lower w E= ½ k d2

5. Reduced Mass M2 M1 Put M1 at R1 andM2 at R2 CM = Center of mass Fix Rcm = (M1R1 + M2R2)/(M1+ M2) = 0 Relative coordinate R=(R2-R1) Then Pcm = (M1+ M2)*Vcm = 0 And P2 = - P1 Thus KE = ½ P12/M1 + ½ P22/M2 = ½ P12/m Where 1/m = (1/M1 + 1/M2) or m = M1M2/(M1+ M2) Is the reduced mass. Thus we can treat the diatomic molecule as a simple mass on a spring but with a reduced mass, m

6. For molecules the energy is harmonic near equilibrium but for large distortions the bond can break. (n + ½)2 Successive vibrational levels are closer by (n + ½)3 (n + ½)2 Exact solution Real potentials are more complex; in general: (Philip Morse a professor at MIT, do not manufacture cigarettes) The simplest case is the Morse Potential:

7. Now on to multiple atoms Eigenvalue problem or N atoms => 3N degrees of freedom However, 3 degrees for translation, get l = 0 3 degrees for rotation is non-linear molecule, get l = 0 2 degrees if linear (but really a restriction only for diatomic The remaining (3N-6) are vibrational modes (just 1 for diatomic) Derive a basis set for describing the vibrational modes by solving the eigensystem of the Hessian matrix

8. Vibration for a molecule with N particles There are 3N degrees of freedom (dof) which we collect together into the 3N vector, Rk where k=1,2..3N The interactions then lead to 3N net forces, Fk = -(∂E(Rnew)/∂Rk) all of which are zero at equilibrium, R0 Now consider that every particle is moved a small amount leading to a 3N distortion vector, (dR)m = Rnew –R0 Expanding the force in a Taylor’s series leads to Fk = -(∂E(Rnew)/∂Rk) = -(∂E/∂Rk)0 - Sm (∂2E/∂Rk∂Rm) (dR)m Where we have neglected terms of order d2. Writing the Hessian as Hkm = (∂2E/∂Rk∂Rm) with (∂E/∂Rk)0 = 0, we get Fk = - Sm Hkm (dR)m = Mk (∂2Rk/∂t2) To find the normal modes we write (dR)m = Am cos wt leading to Mk(∂2Rk/∂t2) = Mkw2 (Ak cos wt) = Sm Hkm (Amcos wt) Here the coefficient of cos wt must be {Mkw2 Ak - Sm Hkm Am}=0

9. Solving for the Vibrational modes The normal modes satisfy {Mkw2 Ak - Sm Hkm Am}=0 To solve this we mass weight the coordinates as Bk = sqrt(Mk)Ak leading to Sqrt(Mk) w2 Bk - Sm Hkm [1/sqrt(Mm)]Bm}=0 leading to Sm Gkm Bm = w2 Bk where Gkm = Hkm/sqrt(MkMm) G is referred to as the reduced Hessian For M degrees of freedom this has M eigenstates Sm Gkm Bmp = dkp Bk (w2)p where the eigenvalues are the squares of the vibrational energies. If the Hessian includes the 6 translation and rotation modes then there will be 6 zero frequency modes

10. Saddle points If the point of interest were a saddle point rather than a minimum, G would have one negative eigenvalue. This leads to an imaginary frequency

11. For practical simulations We can obtain reasonably accurate vibrational modes from just the classical harmonic oscillators N atoms => 3N degrees of freedom However, there are 3 degrees for translation, n = 0 3 degrees for rotation for non-linear molecules, n = 0 2 degrees if linear The rest are vibrational modes

12. Normal Modes of Vibration H2O H2O D2O Sym. stretch 3657 cm-1 2671 cm-1 Ratio: 0.730 1595 cm-1 1178 cm-1 Bend Ratio: 0.735 Antisym. stretch 3756 cm-1 2788 cm-1 Ratio: 0.742 Isotope effect: n ~ sqrt(k/M): Simple nD/nH ~ 1/sqrt(2) = 0.707: More accurately, reduced masses mOH = MHMO/(MH+MO) mOD = MDMO/(MD+MO) Ratio = sqrt[MD(MH+MO)/MH(MD+MO)] ~ sqrt(2*17/1*18) = 0.728 Most accurately MH=1.007825 MD=2.0141 MO=15.99492 Ratio = 0.728

13. The Infrared (IR) Spectrum Characteristic vibrational modes • EM energy absorbed by interatomic bonds in organic compounds • frequencies between 4000 and 400 cm-1 (wavenumbers) • Useful for resolving molecular vibrations 13 http://webbook.nist.gov/chemistry/ http://wwwchem.csustan.edu/Tutorials/INFRARED.HTM

14. Normal Modes of Vibration CH4 1 3 2 3 Sym. stretch Anti. stretch Sym. bend Sym. bend A1 T2 E T2 3019 cm-1 CH4 2917 cm-1 1534 cm-1 1306 cm-1 CD4 2259 cm-1 1092 cm-1 996 cm-1 1178 cm-1

15. Fitting force fields to Vibrational frequencies and force constants • Hessian-Biased Force Fields from Combining Theory and Experiment; S. Dasgupta and W. A. Goddard III; J. Chem. Phys. 90, 7207 (1989) MC: Morse bond stretch and cosine angle bend MCX: include 1 center cross terms H2CO

16. The QM Harmonic Oscillator The Schrödinger equation H = e for harmonic oscillator energy wavefunctions reference http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c1

17. Raman and IR spectroscopy • IR • Vibrations at same frequency as radiation • To be observable, there must be a finite dipole derivative • Thus homonuclear diatomic molecule (O2 , N2 ,etc.) does not lead to IR absorption or emission. • Raman spectroscopy is complimentary to IR spectroscopy. • radiation at some frequency, n, is scattered by the molecule to frequency, n’, shifted observed frequency shifts are related to vibrational modes in the molecule • IR and Raman have symmetry based selection rules that specify active or inactive modes

18. IR and Raman selection rules for vibrations The intensity is proportional to dm/dR averaged over the vibrational state The polarizability is responsible for Raman where e is the external electric field at frequency n • For both, we consider transition matrix elements of the form The electrical dipole moment is responsible for IR

19. IR selection rules, continued • We see that the transition elements are • The dipole changes during the vibration • Can show that n can only change 1 level at a time For IR, we expand dipole moment

20. Raman selection rules substitute the dipole expression for the induced dipole = Same rules except now it’s the polarizability that has to change For both Raman and IR, our expansion of the dipole and alpha shows higher order effects possible For Raman, we expand polarizability

21. Translation and Rotation Modes Both K and V are constant l=0 • center of mass translation • Dxa= DxDya=0 Dza=0 • Dxa=0 Dya=Dy Dza=0 • Dxa=0 Dya=0 Dza=Dz • center of mass rotation (nonlinear molecules) • Dxa=0 Dya=-caDqxDza=baDqx • Dxa= caDqyDya=0 Dza=-aaDqy • Dxa= -baDqzDya=aaDqxDza=0 • linear molecules have only 2 rotational degrees of freedom • The translational and rotational degrees of freedom can be removed beforehand by using internal coordinates or by transforming to a new coordinate system in which these 6 modes are separated out Both K and V are constant l=0 21

22. Classical Rotations xk(q) is the perpendicular distance to the axis q Can also define a moment of inertia tensor where (just replace the mass density with point masses and the integral with a summation. Diagonalization of this matrix gives the principle moments of inertia! the rotational energy has the form The moment of inertia about an axis q is defined as

23. Quantum Rotations For symmetric rotors, two of the moments of inertia are equivalent, combine: Eigenfunctions are spherical harmonic functions YJ,K or Zlm with eigenvalues The rotational Hamiltonian has no associated potential energy

24. Transition rules for rotations • For rotations • Wavefunctions are spherical harmonics • Project the dipole and polarizability due to rotation • It can be shown that for IR • Delta J changes by +/- 1 • Delta MJ changes by 0 or +/-1 • Delta K does not change • For Raman • Delta J could be 1 or 2 • Delta K = 0 • But for K=0, delta J cannot be +/- 1

25. Raman scattering Phonons are the normal modes of lattice vibrations (thermal + zero point energy) When a photon absorbs/emits a single phonon, momentum and energy conservation  the photon gains/loses the energy and the crystal momentum of the phonon. q ~ q` => K = 0 The process is called anti-Stokes for absorption and Stokes for emission. Alternatively, one could look at the process as a Doppler shift in the incident photon caused by a first order Bragg reflection off the phonon with group velocity v = (ω/ k)*k

26. Raman selection rules substitute the dipole expression for the induced dipole = Same rules except now it’s the polarizability that has to change For both Raman and IR, our expansion of the dipole and alpha shows higher order effects possible For Raman, we expand polarizability

27. Another simple way of looking at Raman Take our earlier expression for the time dependent dipole and expose it to an ideal monochromatic light (electric field) We get the Stokes lines when we add the frequency and the anti-Stokes when we substract The peak of the incident light is called the Rayleigh line

28. The Sorption lineshape - 1 • The external EM field is monochromatic • Dipole moment of the system • Interaction between the field and the molecules • Probability for a transition from the state i to the state f (the Golden Rule) • Rate of energy loss from the radiation to the system • The flux of the incident radiation c: speed of light n: index of refraction of the medium 28

29. The Sorption lineshape - II • Absorption cross section a(w) • Define absorption linshape I(w) as • It is more convenient to express I(w) in the time domain Beer-Lambert law Log(P/P0)=abc I(w) is just the Fourier transform of the autocorrelation function of the dipole moment ensemble average 29

30. Non idealities and surprising behavior • Anharmonicity – bonds do eventually dissociate • Coriolis forces • Interaction between vibration and rotation • Inversion doubling • Identical atoms on rotation – need to obey the Pauli Principle • Total wavefunction symmetric for Boson and antisymmetric for Fermion

31. Electromagnetic Spectrum How does a Molecule response to an oscillating external electric field (of frequency w)? Absorption of radiation via exciting to a higher energy state ħw ~ (Ef- Ei) Figure taken from Streitwiser & Heathcock, Introduction to Organic Chemistry, Chapter 14, 1976 31