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Transitions

Transitions. Some questions. Rotational transitions Are there any? How intense are they? What are the selection rules?. Vibrational transitions Are there any? How intense are they? What are the selection rules? Can the rotational state change in a vibrational transition?.

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Transitions

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  1. Transitions

  2. Some questions... • Rotational transitions • Are there any? • How intense are they? • What are the selection rules? • Vibrational transitions • Are there any? • How intense are they? • What are the selection rules? • Can the rotational state change in a vibrational transition? • Electronic transitions • How intense are they? • Can the rotational state change in an electronic transition? • Can the vibrational state change in an electronic transition?

  3. Some preliminary considerations The intensity of an electric dipole transition is proportional to the square of the matrix element of the dipole operator between initial and final states: Recall that, for atoms, in the absence of an external field because |1> is a parity eigenstate, and d is odd However, if |1> and |2> are of opposite parity The expectation value of the electric dipole operator is zero => There are no permanent electric dipole moments For molecules ... it’s exactly the same.

  4. Some more preliminary considerations The electronic wavefunction satisfies the electronic Schrodinger equation – the one you get by clamping the nuclei. In the Born-Oppenheimer approximation the total wavefunction is a product of electronic, vibrational and rotational parts: Usually referred to as “the dipole moment” or sometimes “the permanent dipole moment”. The name is a bit misleading, unless you are clear about what it means. The expectation value of the electric dipole operator taken between these electronic wavefunctions is usually not zero (unless there is a symmetry that makes it zero, e.g. homonuclear).

  5. Yet more preliminary considerations: vectors! Cartesian components of a vector: Vx, Vy, Vz Rank 1 spherical tensor Often more convenient to work instead with the components V-1, V0, V+1 The two forms are related in a simple way: Scalar product between two vectors: We care about To make things simpler, let’s take the light to be plane polarized along z. Then, only need d0 The other polarization cases follow by analogy.

  6. The intensity of an electric dipole transition is proportional to the square of the matrix element of the dipole operator between initial and final states: For a molecule, the electric dipole operator is In the Born-Oppenheimer approximation, we write the total molecular wavefunction as a product of electronic, vibrational and rotational functions. So, we need to evaluate: Rotate the coordinate system so that the rotated z-axis lies along the internuclear axis. In this rotated system, let the electric dipole operator for the molecule be m. Its components are related to those of d in a simple way: Components in the “molecule frame” Component in the lab frame Angular functions These are simply elements of the rotation matrix that rotates from one coordinate system to another. e.g. D00 = cosQ

  7. With the help of this transformation, we have Which separates into a part that depends only on Q, F, and a part independent of these:

  8. Transitions within an electronic state Consider rotational and vibrational transitions within an electronic state, i.e. n’=n By symmetry, the only non-zero component of m is along the internuclear axis, i.e. m0 Define a rotational factor cosQ Also define Angular momentum eigenstates The electric dipole moment function for electronic state n The vibrationalwavefunctions are only large close to R0. So, Taylor expand the dipole moment function around this point: We are left with Its value at R0 is the dipole moment of the molecule in electronic state n

  9. Transitions within an electronic state - Rotational Taking just the first term in this expansion (a constant), the integral is zero unless v’=v. So the first term gives us pure rotational transitions. If the dipole moment is zero (e.g. homonuclear ), there are no rotational transitions This matrix element is straightforward to evaluate. We will do it when we study the Stark shift. For now, it is sufficient to know that it is zero unless DJ = J’-J = 0,±1 and DM = M’-M = 0 The selection rule on M is due to our choice of polarization. If we choose left/right circular light instead we get DM = -1/+1 Pure rotation: Recall that the angular factor Mrot is Selection rules for rotational transitions: DJ = 0,±1 , DM = 0,±1

  10. Transitions within an electronic state - Vibrational Have seen that the first term in our Taylor expansion cannot change the vibrational state The second term can. Vibrational transition: If the dipole moment is zero (e.g. homonuclear ), there are no vibrational transitions The vibrational integral is zero unless v’ = v ,±1 (see problem sheet) Selection rule for vibrational transitions: Dv = ±1 The rotational state can also change in a vibrational transition, with selection rules DJ = 0,±1 , DM = 0,±1 The intensity of a vibrational transition is proportional to the gradient of the dipole moment function at the equilibrium internuclear separation

  11. Transitions that change an electronic state Angular factor – same selection rules as before Define a transition dipole moment function: In the same way as before, Taylor expand about R0: Then, to lowest order, we have the result: Transition dipole moment, n->n’ Angular factor Vibrational overlap integral

  12. Franck-Condon factor James Franck (1882-1964) Edward Condon (1902-1974) The square of the overlap integral between vibrationalwavefunctions in the two potential wells. N.B. They come from two different wells – not orthogonal (unless wells are identical).

  13. Some answers... • Rotational transitions • Are there any? • How intense are they? • What are the selection rules? Yes, provided the dipole moment men is non-zero Proportional to the dipole moment, men , squared DJ = 0,±1 , DM = 0,±1 • Vibrational transitions • Are there any? • How intense are they? • What are the selection rules? • Can the rotational state change in a vibrational transition? Yes, provided the dipole moment men is non-zero Proportional to the gradient of the dipole moment, dmen/dR, squared Dv = ±1 Yes, with DJ = 0,±1 , DM = 0,±1 • Electronic transitions • How intense are they? • Can the rotational state change in an electronic transition? • Can the vibrational state change in an electronic transition? Proportional to the transition dipole moment, men’,n , squared Yes, with DJ = 0,±1 , DM = 0,±1 Yes, proportional to FC factor

  14. Laser cooling (usually) fails for molecules

  15. For some atoms, this is possible due to a “closed” energy level structure np e.g. Alkalis: Absorption Spontaneous emission ns Why laser cooling (usually) fails for molecules Laser cooling relies on repeated absorption – spontaneous-emission events How many cycles are required? Example – Rb-87 atom with initial speed of 100m/s. This situation is special. For most atoms, the energy level structure is more complex. Laser cooling may only be possible with the help of several ‘repump’ lasers, or may be practically impossible. For molecules, it’s much worse!

  16. Cold neutral atomic gases Imperial College London 1st December 2008

  17. Why laser cooling (usually) fails for molecules Following excitation, the molecule can decay to a multitude of other vibrational states. Virtually impossible to find a “closed” transition Note – it’s the vibrations that cause all the trouble. The rotations are governed by selection rules Need to scatter ~10,000 photons for laser cooling. Most molecules scatter 1, start to vibrate, and decouple from the laser N.B. Some molecules DO have Franck-Condon factors very near unity (e.g. CaF, SrF) Laser cooling should be possible for these molecules. We are doing this!

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