Franck-Condon factors have been used to predict / assign the vibrational structure of photo-electron spectra.

$FCF=|\langle\psi''_{vib}|\psi'_{vib}\rangle|^2 = |\langle\psi''_{vib,1}\psi''_{vib,2},\ldots,\psi''_{vib,n}|\psi'_{vib,1}\psi'_{vib,2},\ldots,\psi'_{vib,n-1}, \psi'_{vib, emitted}\rangle|^2=|\langle\psi''_{vib,1}|\psi'_{vib,1}\rangle|^2\ldots|\langle\psi''_{vib,n}|\psi'_{vib,emitted}\rangle|^2$,

Question: what is $\psi'_{vib,emitted}$? Is it just a free particle, with its wavefunction spanning all space, hence it doesn't have any effect on the integral?


$\psi^\prime_{vib, emitted}$ is an energy normalized continuum wavefunction. The Franck–Condon factors involving bound ($\psi_v$) and continuum states ($\psi_E$) are actually Franck–Condon densities that are functions of the energy.

The Franck–Condon factors and densities are related by a closure relation $$ \sum_{v^{\prime\prime}}|\langle \psi_{v^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 + \int |\langle\psi_{E^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 dE^{\prime\prime} = 1$$


[1] J. Tellinghuisen, The Franck-Condon Principle in Bound-Free Transitions, in Photodissociation and Photoionization, edited by K.P. Lawley (John Wiley & Sons, Ltd, 1985).

[2] R.J. Le Roy, Comput. Phys. Commun. 52, 383 (1989).


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