Transitions. 2 Some questions... Rotational transitions  Are there any?  How intense are they?  What are the selection rules? Vibrational transitions.

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Presentation transcript:

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?

Recall that, for atoms, in the absence of an external field because |1> is a parity eigenstate, and d is odd The expectation value of the electric dipole operator is zero => There are no permanent electric dipole moments However,if |1> and |2> are of opposite parity For molecules... it’s exactly the same. 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:

In the Born-Oppenheimer approximation the total wavefunction is a product of electronic, vibrational and rotational parts: The electronic wavefunction satisfies the electronic Schrodinger equation – the one you get by clamping the nuclei. 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). Some more preliminary considerations

Yet more preliminary considerations: vectors! Cartesian components of a vector: V x, V y, V z Often more convenient to work instead with the components V -1, V 0, V +1 The two forms are related in a simple way:Rank 1 spherical tensor 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 d 0 The other polarization cases follow by analogy.

For a molecule, the electric dipole operator is 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: 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 . Its components are related to those of d in a simple way: Component in the lab frame Angular functions Components in the “molecule frame” These are simply elements of the rotation matrix that rotates from one coordinate system to another. e.g. D 00 = cos  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:

With the help of this transformation, we have Which separates into a part that depends only on , and a part independent of these:

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  is along the internuclear axis, i.e.  0 Define a rotational factorAlso define The electric dipole moment function for electronic state n cos  Angular momentum eigenstates Its value at R 0 is the dipole moment of the molecule in electronic state nWe are left with The vibrational wavefunctions are only large close to R 0. So, Taylor expand the dipole moment function around this point:

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. Pure rotation: Recall that the angular factor M rot is 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  J = J’-J = 0,±1 and  M = 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  M = -1/+1 If the dipole moment is zero (e.g. homonuclear ), there are no rotational transitions Selection rules for rotational transitions:  J = 0,±1,  M = 0,±1

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:  v = ±1 The rotational state can also change in a vibrational transition, with selection rules  J = 0,±1,  M = 0,±1 The intensity of a vibrational transition is proportional to the gradient of the dipole moment function at the equilibrium internuclear separation Transitions within an electronic state - Vibrational

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 R 0 : Then, to lowest order, we have the result: Angular factor Transition dipole moment, n->n’ Vibrational overlap integral

12 Franck-Condon factor James Franck ( ) Edward Condon ( ) The square of the overlap integral between vibrational wavefunctions 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? 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? Yes, provided the dipole moment  e n is non-zero Proportional to the dipole moment,  e n, squared  J = 0,±1,  M = 0,±1 Yes, provided the dipole moment  e n is non-zero Proportional to the gradient of the dipole moment, d  e n /dR, squared  v = ±1 Yes, with  J = 0,±1,  M = 0,±1 Proportional to the transition dipole moment,  e n’,n, squared Yes, with  J = 0,±1,  M = 0,±1 Yes, proportional to FC factor