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E = L² L² L² + + 21₁ 271 213 (0, 0) = = 1/22 (H) Yo.o + (H) 1/21/11 Y₁,1 L²-L2/2 L2 + 2/₁ 213 N (H) Figure 6.16 The ammonia molecule. (Eq 1) (Eq 2)

E = L² L² L² + + 21₁ 271 213 (0, 0) = = 1/22 (H) Yo.o + (H) 1/21/11 Y₁,1 L²-L2/2 L2 + 2/₁ 213 N (H) Figure 6.16 The ammonia molecule. (Eq 1) (Eq 2)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The ammonia molecule can be treated as a symmetric rigid rotator, with its 3 hydrogen atoms lying in the xy plane and the nitrogen atom lying above the plane on the z axis, as shown in the attached image. If we call the moment of inertia about the z acis I3, and the moments about the pair of axes perpendicular to the z axis I1, the rotational energy of the molecule can be written as Eq 1 (see provided image). 

Say that at time t = 0, the wave function of the ammonia molecule is given as Eq 2 (see provided image), where the Yl, m1(θ, Φ) are the sphereical harmonics.

  • What are the energy eigenvalues for this symmetric rigid rotator? Find (θ, Φ, t) , the wave function at time t.
E = + 21₁ L² L² L² + 271 213 (0, 0) = = 1 √/22 (H) Yo.o + (H) N L²-L2²/ 27₁ 1 Y₁,1 + L2 213 (H) Figure 6.16 The ammonia molecule. (Eq 1) (Eq 2)
Transcribed Image Text:E = + 21₁ L² L² L² + 271 213 (0, 0) = = 1 √/22 (H) Yo.o + (H) N L²-L2²/ 27₁ 1 Y₁,1 + L2 213 (H) Figure 6.16 The ammonia molecule. (Eq 1) (Eq 2)
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