N we obtain mn d r e dr a m a n h m h n e 2 d using

Info icon This preview shows pages 24–26. Sign up to view the full content.

n ( ? ) , we obtain ? mn = ( d ? ( r e ) dr ) A m A n -∞ H m ? H n e - ? 2 d ? Using recursive relation, ? H n = 1 2 H n + 1 + nH n - 1 , we obtain ? mn = ( d ? ( r e ) dr ) A m A n [ 1 2 -∞ H m H n + 1 e - ? 2 d ? + n -∞ H m H n - 1 e - ? 2 d ? ] To simplify expression we rearrange A m A n -∞ H m ( ? ) H n ( ? ) e - ? 2 d ? = 𝛿 m , n to -∞ H m ( ? ) H n ( ? ) e - ? 2 d ? = 𝛿 m , n A m A n Substitute into expression for ? nm gives ? mn = ( d ? ( r e ) dr ) [ 1 2 A n A n + 1 𝛿 m , n + 1 + n A n A n - 1 𝛿 m , n - 1 ] P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 24 / 26
Image of page 24

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Harmonic Oscillator Recalling A n 1 2 n n ! 𝜋 1 2 we finally obtain transition dipole moment for harmonic oscillator ? mn = ( d ? ( r e ) dr ) [ n + 1 2 𝛿 m , n + 1 + n 2 𝛿 m , n - 1 ] For absorption, m = n + 1 , transition is n n + 1 and ? mn 2 gives R n n + 1 = 𝜌 ( ? mn ) ? mn 2 6 𝜖 0 2 = 𝜌 ( ? mn ) 6 𝜖 0 2 ( d ? ( r e ) dr ) 2 n + 1 2 For emission, m = n - 1 , transition is n n - 1 and ? mn 2 gives R n n - 1 = 𝜌 ( ? mn ) ? mn 2 6 𝜖 0 2 = 𝜌 ( ? mn ) 6 𝜖 0 2 ( d ? ( r e ) dr ) 2 n 2 P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 25 / 26
Image of page 25
Harmonic Oscillator Selection rule for harmonic oscillator is Δ n = ± 1 . Also, for allowed transitions ( d ? ( r e )∕ dr ) must be non-zero. For allowed transition it is not important whether a molecule has permanent dipole moment but rather that dipole moment of molecule varies as molecule vibrates. In later lectures we will examine transition selection rules for other types of quantized motion, such as quantized rigid rotor and orbital motion of electrons in atoms and molecules. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 26 / 26
Image of page 26
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern