Electronic Transitions

As for atoms molecules have different energy states according to different electronic configurations. These states are also described by Spin (Singlet, Doublet, Triplet,...) and kinetic moment; or to put it more exactly, by kinetic moment component LZ that is parallel to molecular axis since this is the only chosen direction in a molecule. The corresponding eigenvalues Λ are assigned by greek letters:

S, Π, Δ, F, G, ...

Λ = 0, ±1, ±2, ±3, ±4, ...

Comparing it with an atom:

l = 0, 1, 2, 3, 4, ...

S, P, D, F, G, ...

We can write down the following approximate expression for a diatomic molecule:

E = Eel + Evib + Erot

E = Te + we(v + ½) + BJ(J+1),

where Eel is the potential curve minimum. When having electronic transition all three energies can be changed; moreover one should pay attention on the fact that we and B should be different for both electronic states. The lower energy state is marked by two strokes (superscript index) on the quantum numbers (v'', J''); and the higher energy state is marked by one stroke (subscript index) on the quantum numbers (v', J'). Therefore the transition lays on the following energy difference:

ΔE = ΔEel + ΔEvib + ΔErot

ΔEel = E'el - E''el = T'e - T''e
ΔEvib = w'e(v'+ ½) - w''e(v''+½)
ΔErot = B'J'(J'+1) - B''J''(J''+1)

What transition are possible now; one can answer on this question by using the selection rules which are true for atoms:

DL = 0, ± 1 und ΔS = 0

Example:
allowed | forbidden
1Π ←1Σ | ΔS = 1, 2, ...
1Π ←1Π | DL = 2, 3, ...
3Δ ←3Δ | ΔS = 1, DL = 2

The following is true for rotation (kinetic moment conservation):

ΔJ = 0, ± 1 | but J'' = 0 ↔ J' = 0 is forbidden

ΔJ = 0 is allowed since the configuration of molecule changes during electronic transition, if Λ'' or L' ¹ 0, i.e. only for S« Σ there is no any ΔJ = 0.

ΔJ = J' - J''
= +1 R-Branch
= 0 Q-Branch (not S-S and not J' = 0 ↔ J'' = 0)
= - 1 P-Branch

There are no selection rules for vibrations since i.A. of the equilibrium distance re' and re'' and also the vibrational frequencies we' and we'' are different. The wavefunctions for both vibrational levels v' and v'' (yv' and yv'') are solutions of different motion problems since potentials are not the same for both electronic states. However, we can calculate according to Franck-Condon-Principle the intensity of transitions into any vibrational state of another electronic state. The (v', v'')-dependence is the given overlapping:

½ò yv'* · yv'' · dt½2

The above-mentioned factors are also referred to as Franck-London-Faktors.

n0 = Eel' - Eel'' + we'(v' + ½) - we''(v'' + ½)

If we have transition v' ← v'' then we will obtain:

ΔJ = +1 | (R-Branch): | nR = n0 + 2Bv' + (3Bv' - Bv'')J'' - (Bv'' - Bv')J''2 | J'' = 0, 1, 2, ...
ΔJ = -1 | (P-Branch): | nP = n0 - (Bv' + Bv'')J'' - (Bv'' - Bv')J''2 | J'' = 1, 2, ...
ΔJ = 0 | (Q-Branch): | nQ = n0 - (Bv'' - Bv')J'' - (Bv'' - Bv')J''2 | J'' = 1, 2, ...

Now we would like to understand the line position with the increasing rotational quantum number J''. Moreover, we will modify the equation for R-Branch in order that it would begin when J'' = 1 ( i.e. J'' ® J'' - 1) :

nR = n0 + (Bv' + Bv'')J'' - (Bv'' - Bv')J''2 ; J'' = 1, 2, 3, ...

We formally let P-Branch to have negative J'' values (i.e. J'' ® -J''):

nP = n0 + (Bv' + Bv'')J'' - (Bv'' - Bv')J''2 ; J'' = -1, -2, -3, ...

Abbreviating ΔB = Bv'' - Bv' and B = (Bv'' + Bv')/2 and variables z as ordinal numbers we will obtain:

n = n0 + 2Bz - ΔBz2 R-Branch z > 0; P-Branch z < 0
n = n0 - ΔBz - ΔBz2 Q-Branch

Since normally ½DB½< B the Q-Branch lines lay close to n0 . If now we mark horizontal axis with the transition wavenumber n and vertical axis with z then we will obtain the Fortrat parabolas:

a)
ΔB > 0 re' > re'' zband head > 0 νband head>n0 | normal behavior |
Spectrum: | ← band head |
b)
ΔB < 0 re' < re'' zband head < 0 νband head<n0 |  |

In case a) the highest transition frequency lays in the region of high rotations of R-Branch. In case b) the lowest frequency lays in the region of high rotations of P-Branch. This rotation forms the so-called band head. We can obtain it by calculating the extreme value:

dν/dx = 0 = 2B - 2ΔBzband head → zband head = B/ΔB und νband head = n0 + B²/ΔB