22_580ln_fa08

22_580ln_fa08 - MIT OpenCourseWare http://ocw.mit.edu 5.80...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.80 Lecture #22 Fall, 2008 Page 1 of 8 pages Lecture #22: Rotation of Polyatomic Molecules I A diatomic molecule is very limited in how it can rotate and vibrate. �� * R is to internuclear axis * only one kind of vibration �� � A polyatomic molecule can have R oriented along any body fixed direction — symmetric and † asymmetric tops A polyatomic molecule can stretch any bond or bend any bond pair — Normal modes of vibration A lot of very complicated classical mechanics. � TODAY: Derive H ROT �� 2 �� 2 �� 2 Rx Ry Rz and evaluate matrix elements in |KJM basis set, where Ix, Iy, = + + 2Ix 2Iy 2Iz and Iz are called principal components (i.e. eigenvalues) of the 3 tensor, I, and are analogous to μR 2 in an AB diatomic. AB 3 moment of inertia 1. Center of mass. 2. rigid body rotation � ROT in terms of � (angular velocity), m , (x , y , z ) T i i i i (positions of atom i in center of mass body frame) 1/2 † � � Ia 0 0 � † T IT principal axes � 0 Ib 0 � 0 0 Ic � ROT � 3. H and matrix elements in |KJM basis. 4. Symmetric tops - prolate and oblate energy level formulas. � Consider a rigid N-body system. Each atom has mass mi and body-fixed coordinate q (defined relative to an arbitrary body-fixed origin). Our first task is to locate the center of mass, because we expect to separate the 3N degrees of freedom into 3 center of mass translations, 3 rotations about the center of mass, and 3N – 6 vibrations. Center of Mass: 3 Cartesian component equations. �� m i ( q i � q CM ) � 3N 0= i =1 atoms † In fact, the definition of body fixed axis system is not even obvious for vibrating molecule. 5.80 Lecture #22 Fall, 2008 Page 2 of 8 pages Example: 120° NH3 (projected onto xy plane) Take advantage of symmetry whenever possible! pick C3 (3-fold rotation axis) axis as z axis locate origin at N atom (a convenient way to start) locate H1 at � = 0 (i.e. in xz plane) bond length C3 axis H1 Carte ���������� �����������sian � � q H1 = ( r , �, � = ( R, �, 0 ) � ( x1, y1 ,z1 ) = ( R sin �, 0, R cos � ) ) ������������ spherical polar � z �1 2� 3 � � R sin �, R cos �� q H 2 = � R , �, � � ( x 2 ,y 2 , z 2 ) = � � R sin �, � �2 2 3 �1 4� 3 � � q H 3 = � R , �, � � ( x 3 ,y 3 ,z 3 ) = � � R sin �, � R sin �, R cos �� �2 � 2 3 � q N = (0, 0, 0 ) R N x Now solve for center of mass. xCM = yCM = 0 are trivial z equation 0 = � m i ( z i � zCM ) = 3m H ( R cos � � zCM ) + m N ( 0 � zCM ) i zCM = 3m H R cos � 3m H + m N � �� �� =M So we have coordinates of all atoms relative to new origin now at center of mass, expressed in terms of 2 unknown bond coordinates, R and �. 5.80 Lecture #22 Fall, 2008 Page 3 of 8 pages Next we need to write out HROT and put it into a convenient form. � ROT = T ROT + VROT H �� � ��� � � � � free rotor, thus��VROT = 0 � ROT 1 T = � m i v2 i 2i � � � Want to re-express all vi’s in terms of q i and where specifies the direction and magnitude of the � angular velocity of the rigid body rotations. (All atoms experience the same .) � � mi qi � center of mass � � � vi = � q i � � � qi|| � � q i ·� q i � cos � q i|| = = � � need velocities for TROT � parallel to unit vector qi� (right hand rule requires minus sign) |vi| = |qi| | | sin �i qi, known. Must solve for sin �i. qi|| 2 � � q i ·� � qi � 2 � � �� qi � sin � i = = qi qi so 2 v2 = q 2�2 sin 2 �i = �q 2�2 � ( q i ·�) � �i � i i 1/ 2 (sin2�i = 1 – cos2�i) 2 � ROT 1 H = � m i �q 2�2 � ( q i ·�) � �i � 2i Go to Cartesian coordinates (always safe for setting up quantum mechanical Hamiltonian operator). 5.80 Lecture #22 Fall, 2008 Page 4 of 8 pages 2 � ROT 1 H = � m i �( x 2 + y 2 + z 2 ) (�2 + �2 + �2 ) � ( x i�x + y i�y + z i�z ) � i i x y z �i � 2i a bit of algebra = 1 � m i �(x 2i + y 2i ) �2z + (x 2i + z2i ) �2y + (y 2i + z2i ) �2x � 2x iy i�x�y � 2x izi�x�z � 2y izi�y�z � � � 2i Reformulate as matrix diagonalization problem! Ixx � � m i ( y 2 + z 2 ) i i etc. Ixy = � � m i ( x i y i ) = Iyx etc. perpendicular distance squared from x axis i Define i � ROT 1 H= 2 † I This is a compact form for messy equation above! � � � � � � � � � y � z� Ixx Ixy I z x I � Iyx Iyy Iyz Izx x Izy I z z † = x y z I is a real, symmetric matrix. It can be diagonalized (by a coordinate transformation, a rotation about center of mass) to give Ia T †IT = 0 0 Ib 0 0 0 0 Ic Ia � Ib � Ic by definition and are called the “principal moments of inertia”. T†T = 1 � ROT = 1 H 2 † I= 1 ( 2 † T )( T †IT )( T † ) 5.80 Lecture #22 Fall, 2008 Page 5 of 8 pages � � � � x a �� � †� T � � = � b � y We find a special body fixed coordinate system � � � � � � � � z c with origin at the center of mass which causes I to be diagonal. � � � a � x �� �� T † � y � = � b � � � � � � z � � c � Usually possible to find principal axes by inspection. 1. One axis is axis of highest order rotational symmetry, called z by convention. 2. Another axis is to Cn and to a �v plane. E.g. if �(xz) exists, then � m x y = � m y z iii i iii = 0 because there is always an identical nucleus at (x, +y, z) and at i (x, –y, z). (What happens when there is no �v plane? e.g. S1 acetylene.) 3. 3rd is to first 2 axes. So when I is diagonal � ROT 1 H = ( Ia 2 + Ib a 2 the nuclear rotational angular momentum is defined as J=I 2 b + Ic 2 c ) should actually use notation of R or N � H ROT like 2 a 2 b 2 c 1 † �1 J J J = J I J= + + 2 2 Ia 2 Ib 2 Ic J 2 p2 � 2m 2I (The reciprocal or inverse of a diagonal matrix is trivial.) 5.80 Lecture #22 Fall, 2008 Page 6 of 8 pages We can now define three rotational constants h1 c 8 � 2 Ia h1 B= c 8 � 2 Ib h1 C= c 8 � 2 Ic A= A�B�C cm–1 (E/hc) Note that we will sample “rotational constants” with I–1 averaged over specific vibrational state, not at the equilibrium geometry. Want equilibrium geometry, get strange average. Note that we are eventually going to want to compute derivatives of I–1 � μ with respect to each of the 3N – 6 normal coordinate displacements. cm–1 cm–1 (again, by definition) One obtains A, B, C by picking bond lengths and angles, specifying atomic masses, and diagonalizing I. For each change in masses (isotopic substitution) or iterative change in geometry, I must be rediagonalized. Example: Principal Moments for NH3 (refer to table on page 2) C3 axis must be one principal axis y x � �� �� � �1 3� � 1 3� so � Iz = R 2 sin 2 � �m H + � + � m H + � + � m H �4 4� �4 4� � = 3m H R 2 sin 2 � 2 2 1 (the 2 3 � � � � distance2 of each atom from axis specified) existence of reflection plane �x � � 3m N m H � � �� � 3m H � � + R 2 sin 2 � � � � v (xz) �i�� Iy = R 2 cos 2 � � � mplies �2� �M� 2 principal component to the xz plane. You show that Ix = Iy (for any symmetric top). [General rule, every molecule with n ��3 Cn rotation axis has two equal moments of inertia!] Special case of D2d S4 axis: cycloctatetrene and allene allene • C3H4 cyclo-octa-tetra-ene C8H8 5.80 Lecture #22 Fall, 2008 Page 7 of 8 pages � H ROT for symmetric top. By convention, Ix = Iy, Iz is unique (for all sym. tops). � 2 �2 �2 � ROT = J x + J y + J z H 2 I 2Iy 2 Iz x � 2 � 2 � 2 � 2 J � J z = J x + J y Ix = Iy � H ROT manipulate this into a form convenient for JKM basis set. key step! � 2 �2 1 � � 2 � 2 � J z 1 ( � 2 � 2 ) Jz = = J � Jz + � J x + J y � + 2 Ix 2 Iz 2 I 2I x z 1 � 2 � 1 1 � � 2 � ROT H= J + � � � J z 2 I � 2Iz 2 I � x x Use |JKM symmetric top basis functions which are just like |J�M functions for a diatomic molecule. So � �2 �2 � 2 � = J( J + 1) + � � �K 2 I � 2Iz 2 Ix � x ������� � 2 E ROT like a diatomic molecule � projection of J onto unique (i.e. symmetry) axis of body (like �) 2 types of symmetric top: 1. Iz � Ia is unique, Ib = Ic > Ia, prolate top, like a cigar. Coefficient of K2 is > 0 because A > B by definition. E ROTte prola = BJ( J + 1) + ( A � B)K 2 hc 2. Iz � Ic is unique. Ia = Ib < Ic, oblate top, like a frisbee. Coefficient of K2 is < 0. E ROaTte obl = BJ( J + 1) � (B � C)K 2 hc 5.80 Lecture #22 Fall, 2008 J = 0, 1, 2, … possible levels K = 0, ±1, … ±J Page 8 of 8 pages denote as (J,K) or J K ...
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This note was uploaded on 11/28/2011 for the course CHEM 5.74 taught by Professor Robertfield during the Spring '04 term at MIT.

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