MODE NORMAL ANALYSIS OF THE VIBRATIONS OF MOLECULES & CLUSTERS
I.
Thoery of "small amplitude" vibrations Classical Quantum
II.
Examples HCO Si(100)-(2x1)
III.
What about anharmonicity and coupling? Thoery Some examples
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I.
Theory of Small Amplitude Vibrations
z 3 2
r3 r2 r4
1 y 4
r1
x
Let ri be the position vector of atom i relative to a space-fixed coordinate system. In general T (kinetic energy) =
i
Px2i + Py2i + Pz2 i 2Mi
(1)
where Mi is the mass of atom i.
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V (potential energy) = V(r1 , r2 , r3,K)
(2)
but typically V depends only on the distances between the atoms. Thus, for a fixed geometry of the atoms, V does not depend on a rigid rotation or translation of one entire collection of atoms. (This means there are no external fields.) Expand V about minimum (only one?) in a Taylor series:
V =V
u ,n ;
{ }
u 0i
0 1 + 2 =x,y ,x ij u
2 ,i
V (u u p, j
,i
- u 0,i )
= x , y , o r e.g., ux,1 = x1 , z
uz,29 = z29
and the coordinates of the minimum are
0 0 0 0 (x1 , y10 , z10 Lx N y N , z N ) for N atoms
Define mass-scaled cartesian coordinates ~ xi Mi x i , etc., where Mi is the mass of atom i. Thus, 1 N T = (mi x i2 + mi yi2 + miz i2 ) 2 i=1 1 N 2 2 2 ~ ~ = ( xi + yi + zi ) ~ 2 i=1 and Pxi = ~ T ~ = xi , etc. ~ xi
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Define mass-scaled cartesian displacements qi that are indexed from 1 to 3N as follows ~ ~0 ~ ~0 ~ ~N q1 x1 - x1 ,q 2 = y1 - y1 ,K,q3N = z N - z0 Thus, (to second order)
0
1 3N V = V ( 0 , 0 , 0K0) + , 2 i=1
qi
i= j
3N
V qi q j
2
qj
0,0,K
(Why are the first derivative terms "missing"?) and 1 3N 2 T = qi 2 i=1 The sum T + V is the Hamiltonian. 3N 2 3N 3N H = 1 qi + 1 qiFi,j q j , 2 i=1 2 i=1 i= j 2V where Fi , j is the mass scaled Force Constant Matrix. In qi q j 0,0,K matrix notation q1 q2 t q= ; q = [ q1q2 Lq3N ] M q3N and thus
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H = 1 q tq + 1 qt Fq 2 2
1 3 2
T
1 24 4 3
V
Example 3 atoms q1 q2 ~ ~0 q = ; q1 = x1 - x1 , q9 = z 3 - z3 ~ ~0 M q9 q1 q2 T = 1 [q1q2 L q9 ] = 2 M q9 F 11 F21 V = 1 [ q1q2 Lq9 ] M 2 M F91 F12 F22 L L (q12 + q22 +L + q92 ) F12 q1 F29 q2 M = M F99 q9
1 2
1 2
qi Fij q j
i j
F92
Note Fij = Fji , i.e., F is a symmetric matrix "Proof" Also, note
1 T is separable, i.e., T = t i , where ti = 2 qi2 , i 2 V V = qi q j q j qi 2
but V is not separable, i.e., V = Vi .
i
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Thus, even to lowest (2nd) order there is coupling. However, because V is bilinear in qiq j there exists a transformation to a new set of variables, Qi such that T and V are both separable in the Qi. Let q = CQ , where C is a 3N x 3N matrix. Consider a special class of matrices called orthogonal matrices. These have the property that C -1 = C t , where Ct is the transpose of C. That is,
(C t ) i,j = C j,i
Thus, q t = (CQ) = Q t Ct = Q t C -1
t
Aside--if D = EF then Dt = Ft Et Proof-- Dij = Eik Fkj ,
k t t t Dij = Fik Ekj k ?
= Fki Ekj = E jk Fki
k k
= D ji Thus,
1 T = 1 qt q = 2 Q t C t C Q = 1 Qt Q 2 { 2
=
1 2
Qi2 = t i
i i
V = 1 q t Fq = 1 Q t C t FC Q 1 Qt Q 2 2 2
(
)
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The new force const matrix
= C t FC and we determine C by requiring
to be diagonal. This is a example of an eigenvalue/eigenvector problem. Before discussing that, what advantage is there in requiring to be diagonal? Ans:
H = (ti + Vi ) = 1 Qi2 + 1 i Qi2 2 2
i i
Now the system looks like 3N uncoupled harmonic oscillators with normal mode harmonic frequencies
i =
i .
Return to C t FC =
(diagonal). thus
Thus, multiply by to C get CC t FC = C { FC = C Consider a 2x2 example:
F11 F21 F12 C11 F22 C21 F11 F21 F11 F21
C12 C11 C12 1 = C22 C21 C 22 0 F12 C11 C11 = 1 F22 C21 C21 F12 C12 C12 = 2 F22 C22 C22
0 2
Can show (homework) this is equivalent to: and
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Thus,
FC(1) = 1C (1) ; FC(2) = 2 C (2) ,
where C (1) , C (2) , LC (3N ) are the columns of the C-matrix. These columns are called the eigenvectors. The procedure to find the C-matrix is termed finding the eigenvalues and eigenvectors of F. FC = C (I will spare youthe details on how this is "diagonalization of F" is done.)
Remarks There are 6 zero frequency modes, i.e., 1 - 6 = 0 . These normal modes correspond to 3 translations and 3 rigid rotations of the equilibrium system, for which there is the remaining 3N-6 s (7,-3N-6 ) are not zero and they are the normal mode vibration frequencies. For each
i
there is a corresponding eigenvector, C (i) which
relates normal mode Qi to the q j, j=1,K,3 N , the mass-scaled cartesian displacements. Recall q = CQ , so Q = C t q , Q1 C11 C21 L CN1 q1 Q2 C12 C22 q2 = M M M Q3N q3N
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Q1 = c11q1 + c21q2 + L + c N1q3N To visualize the mode, place a vector on each atom with components
(c11 , c21, c31) , etc.
Equations of motion Hclassical = hic ;
i
hic =
it
Pi2 2 it
2
+1 2
2 2 i Qi
Qi (t ) = Acos Hqm = hiq ;
i
+ Bsin
hiq = - h2 d 2 + 1 2Qi2 2 i dQi
2
(hiq - (i) )(ni )(Qi ) = 0 n
( i ) = hi ni + 1 ; n = 0,1, 2,... n 2
(
)
Etotal = h
i
i
(ni + 1 ) 2
Total zero-point energy = h i / 2
i
For IR-active mode light is absorbed at fundamental frequencies
active . i
Also, obtain vibrational partition functions, x-ray structure factors, etc.
II.
Examples and practical matters III.
HCO (formyl radical) 3 vibrational. modes. Units: is typically reported in spectroscopy as cm-1!! But, should be sec-1. Let ~ = c, c
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is the speed of light. Then ~ has dimensions of cm-1. Notation should be ~ but people get lazy and just use . What about h energy. How to convert from ~ cm -1 to energy. 1 hartree 219474.6 cm-1 What about V? This was obtained by saving the electronic energy as a function of nuclear geometry. A search for the minimum was successful and the force constant matrix was obtained numerically.
(n + 1 )? This is an 2
(
)
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HCO Normal Modes
H
1=2981 cm-1
"CH-stretch"
O
C
H
2=1382 cm-1
C O
"bend"
H
3=1091 cm-1
O C
"CO-stretch"
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Si(100)--2x1 cluster (28 atoms), 2 layers. The vibrational modes of an ordered solid are termed phonons. The potential in this example was a model one based on harmonic nearest neighbor and next nearest neighbor interactions.
IV. What about anharmonicity and coupling
Anharmonicity Vi (Qi ) = 1 2
2 2 i Qi
+ ci Qi3 + diQi4 + L
anharmonic terms
All potentials contain such terms; however, for very small amplitude motion they may be very small. "Easy" to treat anharmonic terms.
Coupling V (Q1, Q2 ,L) =
3 Vi (Qi ) + aQ1Q22 + bQiQ2 Q4 + L i
Coupling terms in V (Q1, Q2 ,L,Q3N -3 ) are terms that involve two or more modes. This is hard to treat, but very important to consider. For example, H2OOH+H must involve coupling.
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The importance of anharmonicity and coupling In HCO (energies in cm-1) NMHO ZPE 1 2 3 2899 1145 1905 2748 NMAO 2857 1147 1869 2494 NMAO/C 2836 1104 1885 2448 1087 1868 2435 EXP
Conclusion: NMHO is qualitatively correct, but not quantitative. (Huge CH(3) stretch anharmonicty.)
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