Transcript Molecular Vibrations

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, [email protected]
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
Ch121a-Goddard-L07
© copyright 2011 William A. Goddard III, all rights reserved
1
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
Ch121a-Goddard-L07
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2
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
Ch121a-Goddard-L07
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3
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, 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
E= ½ k d2
–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
Ch121a-Goddard-L07
No friction
© copyright 2011 William A. Goddard III, all rights reserved
4
Reduced Mass
Put M1 at R1 and M2 at R2
CM = Center of mass
Fix Rcm = (M1R1 + M2R2)/(M1+ M2) = 0
Relative coordinate R=(R2-R1)
M1
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
Ch121a-Goddard-L07
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M2
5
For molecules the energy is harmonic near equilibrium
but for large distortions the bond can break.
The simplest case is the Morse Potential:
V ( x) = hcDe (1  e
 ax 2
)
k 1/ 2
a=(
)
2hcDe
Exact solution
En = (n  1 / 2)w  ((n
v +1½)
/ 2)22 we
a 2 Successive vibrational
we =E = (n levels
2
are
closer
by
1
/
2
)

w

(
v

1
/
2
)
we
n2m
2
a
 are more complex; in general:
Realw
potentials
e =
2 2
2
m
2
E
=
(
n

1
/
2
)

w

(
v

1
/
2
w
(
n+1/½)
12/)23)wd
wd
 ....
En =
v +1½)
/ 2) )w
(n(n
n (n  1 / 2)w  ((n
e 
e ....
e 
e 
(Philip
Morse a professor
at MIT, do not manufacture cigarettes)
Ch121a-Goddard-L07
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6
3N
(to
f ij multiple
δ ij λ ) A j = 0 atoms
for i = 1, ,3 N

Now on
3N
j =1
 λAi cos(λ1/ 2t  φ)   f ij A j cos(λ1/ 2t  φ) = 0
eigenvalueproblem
Eigenvalue problem H A = λj =1A
3N
or
( f
j =1
•
•
•
•
•
•
ij
 δ ij λ ) A j = 0 for i = 1, ,3 N
H A=>
= λ3N
A degrees
eigenvalue
problem
N atoms
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
Ch121a-Goddard-L07
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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) = Mk w2 (Ak cos wt) = Sm Hkm (Amcos wt)
Here
the coefficient of© cos
wt2011
must
be
{Mk wIII,2 A
- Sreserved
Ch121a-Goddard-L07
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William
A. Goddard
allkrights
m Hkm Am}=0
8
Solving for the Vibrational modes
The normal modes satisfy
{Mk w2 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
Ch121a-Goddard-L07
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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
Ch121a-Goddard-L07
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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
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11
Normal Modes of Vibration H2O
H2O
Sym. stretch
Bend
D2O
3657 cm-1
2671 cm-1
Ratio: 0.730
1595 cm-1
1178 cm-1
Ratio: 0.735
3756 cm-1
2788 cm-1
Isotope effect: n ~ sqrt(k/M): Ratio: 0.742
Antisym. stretch
Simple nD/nH ~ 1/sqrt(2) = 0.707:
More accurately, reduced masses
Most accurately
mOH = MHMO/(MH+MO)
MH=1.007825
mOD = MDMO/(MD+MO)
MD=2.0141
Ratio = sqrt[MD(MH+MO)/MH(MD+MO)]
MO=15.99492
~ sqrt(2*17/1*18)
= 0.728
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reserved = 0.728 12
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/
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13
© copyright 2011 William A. Goddard
III, all rights reserved
http://wwwchem.csustan.edu/Tutorials/INFRARED.HTM
Normal Modes of Vibration CH4
1
3
Sym. stretch
A1
CH4
CD4
Anti. stretch
T2
2
Sym. bend
E
3
Sym. bend
T2
2917
cm-1
3019 cm-1
1534 cm-1
1306 cm-1
1178
cm-1
2259 cm-1
1092 cm-1
996 cm-1
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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
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15
1/ 4  y / 2

=
(
)
e
0


The QM Harmonic Oscillator
 1/ 4

y2 / 2
 1 = ( )  2 =ye( )1/ 4 2 ye y / 2
1

 H=e
The Schrödinger equation
 oscillator
1/ 4 1
2
 y2 / 2
for harmonic

1
 2 2= ( 2 )  =((2 y )1/ 4 1)e (2 y 2  1)e  y / 2
   1 222 
2
 =
 kx
2
2m x 2 1
1/ 4
3 1 / 4 1  y 2 /32

 3 = ( )1 13 =(2( y )  3 y )e(2 y  3 y )e  y / 2
energy
e n =e(nn= (n)
w3)w n = 0n,31=,20, 
,1,2, 
2 2
mw
wavefunctions
 =1/ 4  y / 2  = mw y =  x y =  x
 0 = ( ) e

0
2
2
2
2

 1/ 4
1 = ( )
2 ye y / 2


1
 2 = ( )1/ 4
(2 y 2  1)e  y / 2

2
 1/ 4 1
3 = ( )
( 2 y 3  3 y )e  y / 2

3
Ch121a-Goddard-L07
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reference
mhttp://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c1
w
2
2
2
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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
Ch121a-Goddard-L07
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17
IR and Raman selection rules for vibrations
The electrical dipole moment is responsible for IR
m (t ) =   (r, t )rd r
3
The intensity is proportional to
dm/dR averaged over the
vibrational state
The polarizability is responsible for Raman
m (t ) =  (t )e (t )
where e is the external electric field
at frequency n
For both, we consider transition matrix elements of the form
  ' | m (t ) | 
 =  i ,n (Qi )
i
Ch121a-Goddard-L07
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18
IR selection rules, continued
• For IR, we expand dipole moment
m = m 0  (
i
m
) 0 Qi  ....
Qi
We see that the transition elements are
m
(
)0  ni ' | Qi | ni 
Qi
The dipole changes during the vibration
Can show that n can only change 1 level at a
time
Ch121a-Goddard-L07
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19
Raman selection rules
• For Raman, we expand polarizability

 = 0 (
) 0 Qi  ....
Qi
i
substitute the dipole expression for the induced dipole
 
( ( ) 0  e)=0(t)e n(ti )' |Q
nii| ni 
nii ' || Q
Qi Qi
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
Ch121a-Goddard-L07
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20
Translation and Rotation Modes
•center of mass translation Both K and V are constant  l=0
Dx= Dx Dy=0
Dz=0
Dx=0
Dy=Dy
Dz=0
Dx=0
Dy=0
Dz=Dz
•center of mass rotation (nonlinear molecules)
Dx=0
Dy=-cDqx Dz=bDqx
Dx= cDqy
Dy=0
Dz=-aDqy Both K and V are
Dx= -bDqz Dy=aDqx
Dz=0
constant  l=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
21
Ch121a-Goddard-L07
© copyright 2011 William A. Goddard III, all rights reserved
21
Classical Rotations
• The moment of inertia about an axis q is defined as
I qq =  mk xk2 (q)
xk(q) is the perpendicular distance to the axis q
k
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!
r m
=
m
k
R0
k
k
k
the rotational energy has the form
Erot
k
2
J
1
q
=  I qqwq2 (q ) = 
2 q
q 2 I qq
J q = I qqwq
Ch121a-Goddard-L07
© copyright 2011 William A. Goddard III, all rights reserved
22
Quantum Rotations
The rotational Hamiltonian has no
associated potential energy
J y2
J x2
J z2
H=


2 I xx 2 I yy 2 I zz
For symmetric rotors, two of the
moments of inertia are equivalent,
combine:
J2
1
1
H=
(

)J z
2I  2I  2I 
Eigenfunctions are spherical
harmonic functions YJ,K or Zlm with
eigenvalues
J ( J  1) 2
1
1
Erot ( J , K J , M J ) =
(

) K 2 2
2I 
2I  2I 
J = 0,1,2,...
K J = J , J  1,..., J
M
M J =Ch121a-Goddard-L07
J , J  1,..., J
© copyright 2011 William A. Goddard III, all rights reserved
23
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
Ch121a-Goddard-L07
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24
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
Ch121a-Goddard-L07
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25
Raman selection rules
• For Raman, we expand polarizability

 = 0 (
) 0 Qi  ....
Qi
i
substitute the dipole expression for the induced dipole
 
( ( ) 0  e)=0(t)e n(ti )' |Q
nii| ni 
nii ' || Q
Qi Qi
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
Ch121a-Goddard-L07
© copyright 2011 William A. Goddard III, all rights reserved
26
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)
m (t ) =  (t )e (t ) = 2 (t )e 0 cos(wt )


1
2


m (t ) = 2 0  D cos(wintt )e 0 cos(wt )
1
m (t ) = 2e 0 cos(wt )  De 0 cos(wt  wintt )  cos(wt  wintt )
2
We get the Stokes lines when we add the frequency and the antiStokes when we substract
The peak of the incident light is called the Rayleigh line
Ch121a-Goddard-L07
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27
The Sorption lineshape - 1


•The external EM field is monochromatic E(t ) = E0ε cos(ωt )
•Dipole moment of the
  n n 
T otal
Dipole
Dipole
μ =μ
=
μi μi
system T otal
  N Ni i i
Molecular
Molecular
Dipole
Dipole
μi μ
=i 
=
rj qrjj


•Interaction between the field and the molecules  (t ) = μ  E (t )
i =1 i =1
j =1 j =1
•Probability for a transition from the state i to the state f (the
Golden Rule)
πE
Pi f (ω) = 0
2
2
  2
f | ε  μ | i [δ(ω fi  ω)  δ(ω fi  ω)]
ω fi = ω f  ω f
•Rate of energy loss from the radiation to the system
 E rad (ω) =  ρi ω fi Pi f (ω)
i
f
•The flux of the incident radiation
c: speed of light
cn 2
S=
E0
n: index of refraction of the medium
28
8π
Ch121a-Goddard-L07
© copyright 2011 William A. Goddard III, all rights reserved
28
The Sorption lineshape - II
 E rad (ω)
α(ω) =
S
Beer-Lambert law Log(P/P0)=bc
•Define absorption linshape I(w) as
•Absorption cross section (w)
I (ω) =
3cnα(ω)
  2
=
3
ρ
f
|
ε
 μ | i δ(ω fi  ω)

i
2
βω
4π ω(1  e )
i
f
•It is more convenient to express I(w) in the time domain
1  iωt
I(w) is just the Fourier transform of the
δ(ω) =
e
dt
2π 
autocorrelation function of the dipole moment
3
 
I (ω) =
ρ
i
|
ε
μ | f

i
2π i f
=
 
f | ε μ | i



e
i[
E f  Ei

-ω]t
dt
ensemble average
1  

iωt
=
μ
(
0
)

μ
(
t
)
e
dt



© copyright 2011 William A. Goddard III, all rights reserved
2πCh121a-Goddard-L07
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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
N
E V
N

=




– Total wavefunction symmetric
for Boson and antisymmetric YJ , M (q   ,  ) = (1) J YJ , M (q ,  )
for Fermion
Ch121a-Goddard-L07
J
© copyright 2011 William A. Goddard III, all rights reserved
J
30
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)
Ch121a-Goddard-L07
© copyright 2011 William A. Goddard III, all rights reserved
Figure taken from Streitwiser & Heathcock, Introduction to
Organic Chemistry, Chapter 14, 1976
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