© copyright 2011 William A. Goddard III, all rights reservedCh121a-Goddard-L07 1 Ch121a Atomic Level Simulations of Materials and Molecules William A.

1© copyright 2011 William A. Goddard III, all rights reservedCh121a-Goddard-L07

Ch121a Atomic Level Simulations of Materials and Molecules

William A. Goddard III, [email protected] and Mary Ferkel Professor of Chemistry,

Materials Science, and Applied Physics, California Institute of Technology

BI 115Hours: 2:30-3:30 Monday and Wednesday

Lecture or Lab: Friday 2-3pm (+3-4pm)

Teaching Assistants Wei-Guang Liu, Fan Lu, Jose Mendoza, Andrea Kirkpatrick

Lecture 7, April 15, 2011Molecular Dynamics – 3: vibrations


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


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


Vibration in one dimension – Harmonic Oscillator

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, 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= = A cos(t) then –Mcos(t) = -k A cos(t) Hence –M= -k or Sqrt(k/M). Stiffer force constant k higher and higher M lower

No friction

E= ½ k 2


Reduced Mass

M1M2

Put M1 at R1 and M2 at R2

CM = Center of mass Fix Rcm = (M1R1 + M2R2)/(M1+ M2) = 0Relative coordinate R=(R2-R1)Then Pcm = (M1+ M2)*Vcm = 0 And P2 = - P1

Thus KE = ½ P12/M1 + ½ P2

2/M2 = ½ P12/

Where 1/ = (1/M1 + 1/M2) or = 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,


2

)2/1()2/1(2

2

a

vnE

e

en

(n + ½)2

For molecules the energy is harmonic near equilibrium but for large distortions the bond can break.

The simplest case is the Morse Potential:

2/1

2

)2(

)1()(

e

axe

hcD

ka

ehcDxV

Exact solution

Real potentials are more complex; in general: (n + ½)2 (n + ½)3

Successive vibrational levels are closer by

(Philip Morse a professor at MIT, do not manufacture cigarettes)


Now on to multiple atoms

• N atoms => 3N degrees of freedom• However, 3 degrees for translation, get = 0• 3 degrees for rotation is non-linear molecule, get = 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

Eigenvalue problem

or


Vibration for a molecule with N particles

Fk = -(∂E(Rnew)/∂Rk) = -(∂E/∂Rk)0 - m (∂2E/∂Rk∂Rm) (R)m

Where we have neglected terms of order 2. Writing the Hessian as Hkm = (∂2E/∂Rk∂Rm) with (∂E/∂Rk)0 = 0, we get

Fk = - m Hkm (R)m = Mk (∂2Rk/∂t2) To find the normal modes we write (R)m = Am cos t leading to

Mk(∂2Rk/∂t2) = Mk 2 (Ak cos t) = m Hkm (Amcos t)

Here the coefficient of cos t must be {Mk 2 Ak - m Hkm Am}=0

There are 3N degrees of freedom (dof) which we collect together into the 3N vector, Rk where k=1,2..3NThe 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, (R)m = Rnew – R0

Expanding the force in a Taylor’s series leads to


Solving for the Vibrational modes

The normal modes satisfy

{Mk 2 Ak - m Hkm Am}=0To solve this we mass weight the coordinates as Bk = sqrt(MkAk

leading to

Sqrt(Mk) 2 Bk - m Hkm [1/sqrt(Mm)]Bm}=0 leading to

m Gkm Bm = 2 Bk where Gkm = Hkm/sqrt(MkMm) G is referred to as the reduced HessianFor M degrees of freedom this has M eigenstates

m Gkm Bmp = kp Bk (2)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


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


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


Normal Modes of Vibration H2O

1595 cm-1

3657 cm-1

3756 cm-1

H2O D2O

1178 cm-1

2671 cm-1

2788 cm-1

Sym. stretch

Antisym. stretch

Bend

Isotope effect: ~ sqrt(k/M):Simple D/H ~ 1/sqrt(2) = 0.707:

Ratio: 0.730

Ratio: 0.735

Ratio: 0.742

More accurately, reduced massesH = MHMO/(MH+MO)D = MDMO/(MD+MO)Ratio = sqrt[MD(MH+MO)/MH(MD+MO)] ~ sqrt(2*17/1*18) = 0.728

Most accuratelyMH=1.007825MD=2.0141MO=15.99492Ratio = 0.728


13

•EM energy absorbed by interatomic bonds in organic compounds

•frequencies between 4000 and 400 cm-1 (wavenumbers)

•Useful for resolving molecular vibrations

http://webbook.nist.gov/chemistry/

http://wwwchem.csustan.edu/Tutorials/INFRARED.HTM

The Infrared (IR) SpectrumCharacteristic vibrational modes


Normal Modes of Vibration CH4

2917 cm-1 3019 cm-1 1534 cm-1CH4

CD4 1178 cm-1 2259 cm-1 1092 cm-1

Sym. stretch

1

Anti. stretch

3

Sym. bend

2

Sym. bend

3

A1 T2 E T2

1306 cm-1

996 cm-1


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)

H2CO

MC: Morse bond stretch and cosine angle bendMCX: include 1 center cross terms


The Schrödinger equation H =

for harmonic oscillator22

22

2

1

2kx

xm

The QM Harmonic Oscillator

energy

wavefunctions

reference http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c1


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


IR and Raman selection rules for vibrations

The electrical dipole moment is responsible for IR

rrr 3),()( dtt

The polarizability is responsible for Raman

)()()( ttt For both, we consider transition matrix elements of

the form

iini Q

t

)(

|)(|'

,

The intensity is proportional to d/dR averaged over the vibrational state

where is the external electric field at frequency


IR selection rules, continued

• For IR, we expand dipole moment

....)( 00

ii i

QQ

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

iiii

nQnQ

||')( 0


Raman selection rules

• For Raman, we expand polarizability

....)( 00

ii i

QQ

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

=


21

•center of mass translation x= x y=0 z=0

x=0 y=y z=0

x=0 y=0 z=z•center of mass rotation (nonlinear molecules) x=0 y=-cx z=bx

x= cy y=0 z=-ay

x= -bz y=ax z=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 and V are constant =0

Translation and Rotation Modes

Both K and V are constant =0


Classical Rotations

• The moment of inertia about an axis q is defined as

)(2 qxmIk

kkqq 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

qqqq

q qq

q

qqqqrot

IJ

I

JqIE

2)(

2

12

2

kk

kkk

m

mrR0


Quantum Rotations

The rotational Hamiltonian has no associated potential energy

zz

z

yy

y

xx

x

I

J

I

J

I

JH

222

222

JJJM

M

JJJK

J

KIII

JJMKJE

J

J

JJrot

,...,1,

,...,1,

,...2,1,0

)2

1

2

1(

2

)1(),,( 22

2

zJIII

JH )

2

1

2

1(

2

2

For symmetric rotors, two of the moments of inertia are equivalent, combine:

Eigenfunctions are spherical harmonic functions YJ,K or Zlm with eigenvalues


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


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


Raman selection rules

• For Raman, we expand polarizability

....)( 00

ii i

QQ

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

=


Another simple way of looking at Raman

)cos()cos(2

1)cos(2)(

)cos()cos(2

12)(

)cos()(2)()()(

intint00

0int0

0

tttttt

ttt

ttttt

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 substractThe peak of the incident light is called the Rayleigh line


28

•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

)ωcos(ε)( 0 tEtE

)(μ)( tEt

)]ωω(δ)ωω(δ[|με|2

π)ω(

220 fififi if

EP

fffi ωωω

)ω(ωρ)ω( fifii f

irad PE

20π8

Ecn

S c: speed of lightn: index of refraction of the medium

The Sorption lineshape - 1


29

•Absorption cross section ()

•Define absorption linshape I() as

•It is more convenient to express I() in the time domain

S

Erad )ω()ω(α

)ωω(δ|με|ρ3)1(ω4π

)ω(α3)ω(

2

ωβ2

fi

i fi if

e

cnI

dtet

dteiffiI

dte

ti

tEE

i

i fi

ti

if

ω

ω]-[

ω

)(μ)0(μ2π

1

|με||με|ρ2π

3)ω(

2π

1ω)(δ

I() is just the Fourier transform of the autocorrelation function of the dipole moment

ensemble average

Beer-Lambert law Log(P/P0)=bc

The Sorption lineshape - II


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

),()1(),( ,,

JJ MJJ

MJ

NVEN

YY


31

Fig

ure

taken fro

m S

treitw

iser &

Heath

cock,

Intro

ductio

n to

Org

anic C

hem

istry, C

hapte

r 1

4, 1

97

6Electromagnetic Spectrum

How does a Molecule response to an oscillating external electric field (of frequency )? Absorption of radiation via exciting to a higher energy state ħ ~ (Ef - Ei)

© copyright 2011 William A. Goddard III, all rights reservedCh121a-Goddard-L07 1 Ch121a Atomic Level Simulations of Materials and Molecules William A.

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