THE RIGID ROTOR. mrmr= + m K = I. r 2 2. I = m 1. m + m K = Diatomic molecule. m 1 r 1. r 2 m 2. I moment of inertia. (center of mass) COM K.E.

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1 5.6 Fall 4 Lecture #7-9 page Diatoic olecule THE RIGID ROTOR r r r r rr= (center of ass) COM r K.E. K = r r K = I ( = r r ) I oent of inertia I = r r = µ r µ = (reduced ass) K = µ r = µv z Prob l e reduced to a one-body prob l e wit ass µ. r µ y x

2 5.6 Fall 4 Lecture #7-9 page Angular oentu L=r p L = L = r p r p = r v r v = r r = I L K = I = I Connection between angular oentu and kinetic energy for rotation Analogous to linear oentu and kinetic energy for translation Now need to describe tis quantu ecanically. xyz H ˆ = K V (x, y, z) = V (x, y, z ) µ = xyz x y z Cange to sperical polar coordinates Cartesian sperical polar (x, y, z ) (r,,) x = r sin cos y = r sin sin z = r cos = r sin r r r r r sin r sin so in sperical polar coordinates H ˆ (r,,) = E (r,,) is

3 5.6 Fall 4 Lecture #7-9 page 3 µ r r r r r sinθ θ V ( r,θ,φ) ψ r,θ,φ r sin θ φ = Eψ r,θ,φ In our rigid rotor exaple = ( r = r ) = ( r r ) V r,θ,φ V r,θ,φ i.e. r is eld constant at r. ψ ( r,θ,φ) ψ ( r,θ,φ) r ψ ( r,θ,φ ) = Rewrite Scrodinger eq. witout variable r: µr I sinθ θ ψ r sin θ φ,θ,φ = Eψ ( r,θ,φ) Set ψ ( r,θ,φ) = BY ( θ,φ) Need to solve I sinθ θ Y θ,φ sin θ φ = EY ( θ,φ)

4 5.6 Fall 4 Lecture #7-9 page 4 SOLUTIONS TO THE RIGID ROTOR Solving for te rigid rotor proble, we are now left wit solving te diff. eq. I sinθ θ Y θ,φ sin θ φ = EY ( θ,φ) Tis is ĤY ( θ,φ) = EY ( θ,φ) wit Ĥ = ˆK since te potential ˆ V = Rearranging te diff. eq. sinθ θ IE sin θ Y θ,φ = φ Y θ,φ only θ only φ We ve separated te variables, just as in te 3D particle in a box proble. Try Y ( θ,φ) = Θ( θ)φ( φ) as a solution Define β IE ( note β E) sinθ θ β sin θ Θ( θ)φ φ Dividing by Θ( θ)φ( φ) and siplifying sinθ Θ θ θ Θ θ β sin θ = = Φ φ only θ only φ φ Θ ( θ )Φ φ φ Φ φ

5 5.6 Fall 4 Lecture #7-9 page 5 Since and are independent variables, eac side of te equation ust be equal to a constant. = I sin and sin = sin II First solve for using I = Solutions are i = A e and A e i Boundary conditions quantization ( ) = i( ) = A e i i( ) = A e i A e and A e i i e = and e = Tis is only true if =, ±, ±, ± 3,... is te agnetic quantu nuber i = A e =, ±, ±, ± 3,... Noralization: d =

6 5.6 Fall 4 Lecture #7-9 page 6 i = e =, ±, ±, ± 3,... Now let s look at. Need to solve II sin sin = sin = P x dx sin Cange variables: x = cos = d Since x Also sin = cos = x After soe rearrangeent we obtain te Legendre equation d d ( ) x P( x) P x dx dx x ( x ) P x = Te continuity of constraint leads to a quantization for. = l( l ) were l =,,,... and =, ±, ±,..., ±l Tis directly leads to quantization for te energy! IE = E = I E = I l ( l ) l =,,,... (for rigid rotor, use J instead of l) E = J J ) J =,,,... J ( I

7 5.6 Fall 4 Lecture #7-9 page 7 Solutions of Legendre eq. are associated Legendre polynoials P l P l ( x)= P l (cos) ( P cos) (3cos = P ( cos) = ( P cos) = cos P ( cos) = 3cos sin ( P cos) = sin P (cos)= 3sin etc. ) So = A (cos) A = P l l l l (l )! (l )! were A l is te noralization constant A P (cos) l l sind = So now putting it all togeter: ( r,, ) = Y l (,)= l l )! (,) l (l Y l = 4 P (cos) l e l! i Tese functions are called sperical aronics. Tey are te eigenfunctions of ĤY = EY for te rigid rotor.

8 5.6 Fall 4 Lecture #7-9 page 8 etc. SPHERICAL HARMONICS 3 Y = Y = 8 (4 ) 3 3 Y = cos Y = sine 4 8 i sine i Y (,) = ( l l ) Y (,) = l (l l l =,,,... )! 4 ( l )! P i l (cos)e =, ±, ±, ± 3,... ± l Y s are te eigenfunctions to Ĥ = E for te rigid rotor proble. l Y = (4 ) Y = 5 6 (3cos ) 3 Y = cos Y ± = 5 sin cose ±i 4 8 Y = 3 8 sinei ± 5 Y = sin e ±i 3 3 Y = i sine 8 Y Y s are ortonoral: (,)Y (,)sindd = l l l ll if l = l if = noralization Krönecker delta ll = = if l l if ortogonality

9 5.6 Fall 4 Lecture #7-9 page 9 ˆ Energies: (eigenvalues of HY l = E Y ) l l Switc l J conventional for olecular rotational quantu # IE Recall = = l l ) J J ) J =,,,... ( ( E = J J J ( I ) J = J = E = Y ±, Y (3x degenerate ) I J = E = Y (nondegenerate) Degeneracy of eac state g J = (J ) fro =, ±, ±,..., ± J Spacing between states as J I ( = E E = ( J )( J ) J J J J ) I ( J ) Transitions between rotational states can be observed troug spectroscopy, i.e. troug absorption or eission of a poton Absorption - E J - E J or - E J Eission - E J-

10 5.6 Fall 4 Lecture #7-9 page Molecules need a peranent dipole for rotational transitions. Oscillating electric field grabs carges and torques te olecule. dµ Strengt of transition I JJ ( µ) dx J J dx electric field of radiation dipole oent of rotor Leads to selection rule for rotational transitions: J = ± Recall angular oentu is quantized (in units of ). Poton carries one quantu of angular oentu. Conservation of angular oentu J = ± Angular oentu of olecule canges by quantu upon absorption or eission of a poton. (J ) I J J 4 I poton poton rot J J poton J J J J E = = E = E E = (J ) = Define te rotational constant B B (Hz) or B (c - ) 8 I 8 ci (Hz) = B J ) (c - ) = B J J J ( J J E J =3 E = Bc 3 E 3 = 6Bc J = J = E E = 4Bc = Bc E = 6Bc E = Bc

11 5.6 Fall 4 Lecture #7-9 page Tis gives rise to a rigid rotor absorption spectru wit evenly spaced lines. B Spacing between transitions is J J J J B (Hz) or B (c - ) = B (J ) B( J ) = B Use tis to get icroscopic structure of diatoic olecules directly fro te absorption spectru! Get B directly fro te separation between lines in te spectru. Use its value to deterine te bond lengt r! B = I = µr 4 ci µ = r = (B in c - ) or r = (B in Hz) 8 cbµ 8 Bµ