# Raman Scattering - what emits the photon?

• n0_3sc

#### n0_3sc

A raman scattered photon occurs when you drive the bonds of a molecule causing oscillations between the bonded molecules right?

What exactly emits the photon? If its the electrons, then shouldn't multiple photons be emitted with the same wavelength?

Raman scattering is an example of inelastic scattering. The atom or molecule scatters the photon. The wavelength of the scattered photon is usually lesser than the wavelength of the incident photon. The scattering results in a change in the vibrational, rotational or electronic level of the atom or molecule.

Raman scattering is an example of inelastic scattering. The atom or molecule scatters the photon. The wavelength of the scattered photon is usually lesser than the wavelength of the incident photon. The scattering results in a change in the vibrational, rotational or electronic level of the atom or molecule.

When you say the atom "scatters" the photon, does the atom absorb the photon and re-emit it or does the photon literally bounce off the atom?

When you say the atom "scatters" the photon, does the atom absorb the photon and re-emit it or does the photon literally bounce off the atom?

You have to keep in mind the different ways a photon is scattered.

Let an atom have 2 energy states: initial state 'a' and final state 'b'. Let $\omega$ be the angular frequency in state 'a' and $\omega '$ be the angular frequency of state 'b'. The energies of states a & b are $E_a = \hbar \omega$ & $E_b = \hbar \omega '$ respectively.
The frequency of state a & b are related by the equation:
$$\omega ' = \omega + {(E_a - E_b)\over \hbar}$$
$$= \omega - \omega_{ba}$$

If $\omega =\omega '$ i.e. emitted frequency is same as incident frequency, we have elastic scattering or Rayleigh scattering.

Raman Scattering is an example of inelastic scattering, in which case $\omega ' < \omega$ i.e. the scattered radiation has a lesser frequency.

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Let an atom have 2 energy states: initial state 'a' and final state 'b'.

I now understand those scattering techniques, but what's confused me even more is the energy states... You say the atom has 2 energy states where you described Rayleigh and Raman scattering - those energy states are the "Virtual" states and not the discrete quantum states of the electrons in the atom correct? So then what are these virtual states? Are they just energy levels that aren't high enough to be a discrete quantum state?

No, they are discrete quantum states. I just gave an example of 2 energy states without specifying their type. It could be a ground state or an excited state of the atom or between 2 excited states of an atom. If you are familiar with the selection rules, you know for an atom to emit or absorb a photon the selection rule is $\Delta l = \pm 1$ where 'l' is the angular momentum quantum number.

No, they are discrete quantum states.
...If you are familiar with the selection rules, you know for an atom to emit or absorb a photon the selection rule is $\Delta l = \pm 1$ where 'l' is the angular momentum quantum number.

If they're discrete quantum states then how are they defined?
Yes I am familiar with the selection rules but only for excitation to an electronic state. Sorry I didn't learn much at all from my Q.M. courses...

Some sites and books say virtual levels are not discrete levels and some say they are discrete levels...