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How does an atomic transition between ground and excited states depend upon the direction of polarisation of incident light?

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2 Answers 2

An electronic transition is characterized by its Transition dipole moment. To put it simple, it is a vector that shows the direction and magnitude of the electron cloud displacement. The probability of interaction with a photon is proportional to a scalar product of the transition dipole moment and the photon polarization. Electronic transitions with zero dipole moment are thus called forbidden.

Since atom is spherically symmetric, there is no given direction and the transition dipole moment points everywhere. Hence, there will be no preference to the polarization of light. However, you can impose a direction, for example, by applying electric field. Then the transition dipole moment will be oriented and the interaction with light will depend on its polarization.

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You can't really talk about the relationship between excited atom states and incoming light polarization: the atomic states are what they are, independent of the incoming light. However, different transitions between atomic states can have probability amplitudes that depend on the incoming light's polarization: it's easy to see that the electrons in a molecular bond will interact more strongly with an electric field aligned with the bond.

If you're talking about incoming light provoking photon emission, well that's stimulated emission, and basic symmetry considerations show that the emitted light's polarisation must be aligned with that of the incoming light. See the Wikipedia page on stimulated emission and also the hyperphysics pages on the same topics and also on the Einstein A and B coefficients.

To think about what happens when light puts atoms/molecules into excited states and then the atoms/molecules spontaneously emit light sometime afterwards, one applies the principles of conservation of energy, momentum, and angular momentum to the light/atom/environment system as a whole. Polarisation relationships are about conservation of angular momentum. To understand this last statement, see, for example, the chapter called "Angular Momentum" in the third volume of "the Feynman lectures on physics". What does this mean practically? We think of three different cases:

  1. In an optically transparent medium, photons are absorbed by electrons and the latter emits new photons in their place almost instantaneously (within femtoseconds or less). There is little no time for the electrons to interact with the atoms and molecules around them between emission and absorption. Therefore, the emitted photons must have precisely the same wavelength (corresponding to energy conservation), direction (corresponding to momentum conservation ) and polarisation (corresponding to angular momentum conservation)as the incoming photons, and the medium's effect on the light is simply one of delay, so the medium is modelled by a refractive index, i.e. a lightspeed scaling factor.

  2. There is a slight exception to process (1) in birefringent mediums. The process is almost the same, but there is some transfer of angular momentum by the absorbing electrons to the surrounding atom/molecules/environment. The light's polarisation state changes, and in turn the light and medium exert a minute torque - the angular impulse - on one another.

  3. Lastly we have fluorescence. The absorbing, excited atom/molecule "waits" a long time before spontaneous emission: typically nanoseconds, may be as long as milliseconds In that time it can interact the medium it is steeped in - its environment. So the energy, momentum, angular momentum relationships are complicated:

    • Energy: There is almost always a Stokes shift with fluorescence: state that the atom/molecule fluoresces from may be lower than the state the atom/module was first raised to by the incoming light. Moreover the atom/molecule may not fluoresce fully to the ground state. (See my drawing below, which typifies fluorescein i.e. "green fluoro pen ink"). The energy losses mean that the absorption/fluorescence process transfers vibrational energy and heat to the atom/environment system.

    • Linear momentum: during a fluorescence lifetime the atom/molecule can bear against its surrounding medium and vice versa, so impulse is transferred to the environment. Therefore there is almost always little relationship between the incoming light's direction and that of the fluorescence;

    • Angular momentum: interactions between atoms, molecules and their surrounding environment tend not to involve torque as much as they do impulse transfer. This is intuitively reasonable: forces tends to be directed along the line between centres of mass. However, there is some torque and angular impulse, especially for long lifetime fluorophores. Therefore fluorescence polarisation tends to be quite strongly correlated with that of the incoming light, but there is definite depolarisation as well. A further factor that tends to keep fluorescence and incoming light polarisation quite well correlated is that angular momentum is quantized, whereas linear momentum is not. Simple mindedly you can think along the lines that angular impulses must reach a definite $\pm\hbar$ threshold before they become real transfers.

Fluorescein fluorescence

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