If you are talking about anti-stokes shift you are referring to Raman spectroscopy.

Dear Adam,

Emission implies photons given off in fluorescence, phosphorescence, or luminescence. Raman spectroscopy does not involve emission of photons, rather it involves scattering of photons. Emission is a mechanism that allows a molecule in a well defined quantized excited state to relax back down to the ground state. Raman scattering on the other hand invokes a virtual state, which is not a quantized state (it is too transient to have a well defined energy, delta E * delta t > h/4pi).

Due to a Boltzmann distribution, most molecules at equilibrium are in the ground vibrational level of the ground electronic state, a few are in the next higher vibrational level. Irradiating a sample with intense laser light at a wavelength away from any absorption band will excite a molecule into virtual states whose energy differs from the energy level they started out in by hv, where h is Plank's constant and v is the frequency of the laser light. So a molecule in the ground vibrational level v0 will be excited up to a virtual state of energy h(v + v0), whereas a molecule in the higher vibrational level v1 will be excited up to a virtual state of energy h(v + v1). The difference in energy of these two virtual states is equal to the difference in vibrational energy of these two populations of molecules.

Case 1

Molecules that are transiently in the virtual state of energy h(v+v0) and relax back down to the ground vibrational level with energy hv0, or molecules that are transiently in the virtual state of energy h(v+v1) and relax back down to the higher vibrational level with hv1 will result in Rayleigh scattering (what is also called purely elastic scattering). This will yield photons that scatter at the laser frequency.

Case 2

Molecules that are transiently in the virtual state of energy h(v+v0) and relax back down to the higher vibrational level with energy hv1 will result in Stokes scattering. This will yield photons that scatter at a frequency v - v0.

Case 3

Molecules that are transiently in the virtual state of energy h(v+v1) and relax back down to the ground vibrational level with energy hv1 will result in anti-Stokes scattering. This will yield photons that scatter at a frequency v + v0.

IN ALL THREE CASES ENERGY IS CONSERVED!

I hope this helps,

John

If you are talking about anti-stokes shift you are referring to Raman spectroscopy.

Dear Adam,

Emission implies photons given off in fluorescence, phosphorescence, or luminescence. Raman spectroscopy does not involve emission of photons, rather it involves scattering of photons. Emission is a mechanism that allows a molecule in a well defined quantized excited state to relax back down to the ground state. Raman scattering on the other hand invokes a virtual state, which is not a quantized state (it is too transient to have a well defined energy, delta E * delta t > h/4pi).

Due to a Boltzmann distribution, most molecules at equilibrium are in the ground vibrational level of the ground electronic state, a few are in the next higher vibrational level. Irradiating a sample with intense laser light at a wavelength away from any absorption band will excite a molecule into virtual states whose energy differs from the energy level they started out in by hv, where h is Plank's constant and v is the frequency of the laser light. So a molecule in the ground vibrational level v0 will be excited up to a virtual state of energy h(v + v0), whereas a molecule in the higher vibrational level v1 will be excited up to a virtual state of energy h(v + v1). The difference in energy of these two virtual states is equal to the difference in vibrational energy of these two populations of molecules.

Case 1

Molecules that are transiently in the virtual state of energy h(v+v0) and relax back down to the ground vibrational level with energy hv0, or molecules that are transiently in the virtual state of energy h(v+v1) and relax back down to the higher vibrational level with hv1 will result in Rayleigh scattering (what is also called purely elastic scattering). This will yield photons that scatter at the laser frequency.

Case 2

Molecules that are transiently in the virtual state of energy h(v+v0) and relax back down to the higher vibrational level with energy hv1 will result in Stokes scattering. This will yield photons that scatter at a frequency v - v0.

Case 3

Molecules that are transiently in the virtual state of energy h(v+v1) and relax back down to the ground vibrational level with energy hv1 will result in anti-Stokes scattering. This will yield photons that scatter at a frequency v + v0.

IN ALL THREE CASES ENERGY IS CONSERVED!

I hope this helps,

John

Best example is the fluorescence in the gas-phase of relatively small molecules at high temperatures. A small molecule has a nicely resolved absorption spectrum. Excitation can be performed in the vibrationless excited state. Via collisions vibrational levels are populated. Emission can occur to the ground electronic state and the vibrationless level in this So-state. This transition has a larger energy difference than the original absorption transition. Hence this is your requested anti-Stokes shift.

My question to you: for what purpose do you want to know this

N.B. In practice the transition is more complicated by the rotational structure of the transition. Furthermore emission is only allowed from a symmetric vibration and the transition should have an appreciable Franck-Condon factor.