which is the moment of inertia mm -- the center of mass is given by: m11 r m2r 2

Transcription

1 Chapter 6: The Rigid Rotator * Energy Levels of the Rigid Rotator - this is the odel for icrowave/rotational spectroscopy - a rotating diatoic is odeled as a rigid rotator -- we have two atos with asses of and, which are located r and r fro the center of ass, respectively -- it is rigid because we are keeping the bond length fixed --- we leave the changes in the bond length to the haronic oscillator odel for vibrations --- the agnitude of vibration is sall copared to the eq bond length which eans are rigid rotator is a good first approx n - Classical Equation of otion for a rigid rotator K v v r r r r I where v r, v r, I r r which is the oent of inertia -- the center of ass is given by: r r & r r r I r -- Angular Moentu, L, of our rotator I L --- L I K I I I - Tie for Quantu -- no external forces acting on our rotator so the Hailtonian will coprise only the K ter ˆ ˆ H K where is the Laplacian -- it is ore convenient to work with spherical coordinates since this syste is spherically syetric --- therefore, our Laplacian is r sin r r r, r sin r, r sin r, --- since we are keeping r fixed our Laplacian becoes

2 sin r sin r, r sin r, -- our Hailtonian is then described by ˆ H sin r sin r r sin ˆ H sin r sin sin r, r,, r, ˆ H sin I sin r, sin r, with the angular oentu operator: ˆ L sin sin r, sin r, -- since r is a constant our wavefunction depends only on & or Y(, ) HY ˆ, EY, sin Y,, I sin r, sin r, if we ultiply both sides by sin we get EY sin sin Y, Y, sin EY, I r, r, IE if we let then sin sin Y, Y, sin Y, 0 r, r, -- we will address the for of the wavefunction in a bit, here is the solution: JJ J 0,,, EJ JJ J 0,,, I where the degeneracy of the rotational levels is g J J * Application to a rotating diatoic olecule - like the haronic oscillator, the selection rule for a rigid rotator is J = - also to be detectable the diatoic ust have a peranent dipole (change in Electronegativity) -- e.g. Br is not active but HBr is - for absorption, J = +

3 E E E JJ J J EJ J J J JJ J 3J J J I I I h EJJ J J J I I 4 I Since h E hv v J J 0,,, 4 I - why do we say this is a good odel for icrowave spectroscopy? -- the order of the reduced ass and the eq bond length is: kg r 00 p 0 Therefore, the oent of inertia is: I r 50 kg 0 50 kg plugging this into our frequency relationship gives a range of x 0 0 to x 0 Hz which is in the icrowave range - usually, the frequency is written as: h v BJ B J 0,,, where B is called the rotational 8 I constant of the olecule in question In wavenubers the frequency becoes h v BJ B J 0,,, 8 ci Therefore the transitions between states is: 0 v B v 4B 3 v 6B So, using all these ters we can deterine I and therefore r for a diatoic

4 olecule. * Sections Skip for now * Wave Functions of the Rigid Rotator & Spherical Haronics - once again we need to use separation of variables: Y (, ) ( ) ( ) - if we substitute and divide this relationship into our SE: sin sin ( ) ( ) ( ) ( ) sin ( ) ( ) 0 ( ) ( ) sin sin ( ) ( ) sin 0 ( ) ( ) - As always we can separate these two equations into their independent coponents: sin sin ( ) ( ) sin 0 ( ) ( ) 0 ( ) ( ) sin sin ( ) sin ( ) - Let s look at phi first: ( ) ( ) ( ) ( ) 0 -- we have seen the solution to this equation any ties however, we ust take into consideration that these are spherical coordinates and what that eans for our wavefunction i i Ae A e --- to be single valued we need to set since phi is the coponent that circle the z-axis and so it will be at the sae exact place for & i i i i Ae Ae Ae Ae i for these relationships to be true e with 0,,, i --- Noralization of phi leads to N e - Theta s Turn

5 sin sin ( ) sin ( ) we rearrange the equation: d d d d Now, we change the variables: xcos, ( ) P( x) sin sin ( ) sin ( ) ( ) 0 sin sin d d or sin d where 0 x and sin cos x d d d d sin sin Px ( ) xpx ( ) 0 d d x x Px ( ) xpx ( ) 0 sin sin sin Px ( ) x Px ( ) 0 dp( x) d P( x) x x x x P( x) 0 d P( x) dp( x) x x x x P( x) 0 d P( x) dp( x) x x P( x) 0 x -- due to the continuity of theta we have a quantization constraint for β l l where l 0,,, and 0,,, -- therefore we also have quantization of energy: IE E l l l 0,,, l I I this is very siilar to rigid rotator: E JJ J 0,,, I -- our P(x) is actually a set of polynoials called Legendre (lə zhän drə) polynoials, Pl ( x ), which arise fro central force probles in classical physics d 0 Pl ( x) ( x ) Pl ( x) Pl (cos ) ---- the first few P l0 ( x ) P0 ( x) P ( x) xcos P ( x) 3x 3cos ---- Application of our forula for the first few P l ( x )

6 d 0 d P ( x) ( x ) P ( x) ( x ) x( x ) (cos ) sin d 0 d P( x) ( x ) P ( x) ( x ) 3x ( x ) 3x 3sin cos d 0 d P ( x) ( x ) P ( x) ( x ) 3x 3sin --- Noralizing these obnoxious polynoials l!! l l Pl ( x) Pn ( x) ln or Nl l l! l! -- Putting it all together Spherical Haronics: ll! l! l Yl, ( ) Pl (cos ) e Pl (cos ) e l! 4 l! where l 0,,, and 0,,, --- let s get a few spherical haronics: ! 0 i0 Y0 P0 (cos ) e 4 0 0! 4 i i 0 0! 0 i0 3 Y P (cos ) e cos 4 0! the rest on in Table 6.5 in your text on page 86 * RR Selection Rule is J = - J will later be replaced by QN l (our angular oentu quantu nuber) * The Three Coponents of Angular Moentu do not play nice together - they cannot be easured siultaneously - recall the angular oentu operator by coponents: ˆ L ˆ ˆ x ypz zpy iy z isin cotcos z y Lˆ ˆ ˆ y zpx xpz iz x icos cotsin x x Lˆ ˆ ˆ z xpy ypx ix y i y x -- ˆ i i Le i i z i e i i e e --- since the Legendre coponent of the spherical haronic has no dependency: ˆ LY z l, Yl, --- therefore the spherical haronics are also eigenfunctions of the z-

7 coponent of the angular oentu operator -- it turns out that the spherical haronics are not eigenfunctions of the x- and y- coponents of the angular oentu operator - tie to prove l -- ˆ ˆ LYl, l l Yl, & LY z l, Yl, l0,, -- Also, by definition: Lˆ Lˆ ˆ ˆ ˆ ˆ ˆ ˆ x Ly Lz L Lz Lx Ly -- ˆ ˆ ˆ ˆ L Lz Yl, l l Yl, Lx LyYl, -- since Lˆ ˆ x Ly is a su of two squares, ll 0ll -- finally, since l and are both integers l 0,,, l --- which is what we can the agnetic quantu nuber *Appendix - Skip