What happens to photons after they hit objects?

Question

If I am not wrong when light hits for example white wall most of the photons are absorbed and transformed into heat and few of the photons at certain wavelength are reflected from the object. So white wall reflects photons at wavelength corresponding to white color so we can see it.

1.Can someone clarify if it is correct?

2.What happens if that photons reflected from white wall hit another white wall (are they reflected once? If that so can they be reflected forever?)

3.What happens when I "shoot" photons of certain wavelength into object which color corresponds to that photons wavelength e.g. Blue light into blue wall

Answer 1

It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%.

There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds like a mirror, but the difference is that a mirror reflects light according to the "the angle of incidence equals the angle of reflection" rule, whereas a white wall or other white surface exhibits diffuse reflection, which means that it reflects light in all different directions.

A photon reflected from a white wall can certainly be reflected from another white wall. However, the reflection can’t occur forever, because no real wall can reflect 100% of the light shone on it.

If a wall is blue, that means that it reflects blue photons well, and doesn't reflect as many photons of colors that aren't blue.

Answer 2

Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone electron (i.e. the simple Feynman diagram for Compton scattering from a free electron), it gets a little bit moot and not very meaningful to decide whether we have a "reflexion" or an "absorption-re-emission"). If you look at the possibilities below, you should be able to find yours amongst them: "blue photons" against a "blue wall", for instance, is possibility number 4 below, so you simply have a higher amplitude for scattering than you would for incident white light.

1. If the absorber truly returns to exactly the same state as it was before the photon absorption, the emitted photon must have exactly the same linear momentum, energy and angular momentum as the incident photon. That means it must be travelling in the same direction as before, it must have precisely the same wavelength as before. We have a perfectly transparent material. If the photon's angular momentum is conserved, it must remain circularly polarised if initially so, but there can be a rotation of linear polarisation, through a quantum version of the excited absorber shape deformation that a falling cat (or astronaut) undergoes to rotate themselves in space whilst conserving angular momentum see this question here or my article "Of Cats and Their Most Wonderful Righting Reflex" here. That is, the photon represented by the quantum superposition $\alpha_L \psi_L + \alpha_R\,\psi_R$, where $\psi_L$ and $\psi_R$ are the amplitudes for the pure left and right handed circular polarisation states can be changed to any superposition of the form $e^{i\,\phi_L}\,\alpha_L \psi_L + e^{i\,\phi_R}\,\alpha_R\,\psi_R$, i.e. the superposition weights can shift in phase, but not in magnitude. This is NOT the same as birefringence, however;

2. If the absorber is as in 1), but angular momentum is transferred between the absorber and the surrounding lattice, the material is perfectly transparent and the quantum state $\alpha_L \psi_L + \alpha_R\,\psi_R$ can undergo any unitary transformation. We are describing a perfectly transparent, birefringent material. Light passing through a quarter wave plate of such a material will exert a measurable torque on it, as in this famous experiment:

Richard A. Beth, "Mechanical Detection and Measurement of the Angular Momentum of Light", Phys. Rev. 50, 15th July 1936

3. If the absorber is as in 1) but can transfer impulse to its surrounding lattice, the re-emitted photon will like one that is elastically scattered from the lattice: the wavelength is the same but the photon's direction is different. We are describing a perfect mirror;

4. If the excited absorber has a nonzero amplitude to transfer energy to the surrounding lattice, then the lattice heats up. There may still be reflected photons, but the light intensity may be less. Moreover the absorption - re-emission process's amplitude will be colour dependent, so some photons are likelier to be absorbed without re-emission than others. In that case, white light comprising photons of different colours, or (more esoterically) identical photons in a pure "white" quantum state (as described in my answer here) will be re-emitted with a different spectral makeup: the scatterer is thus coloured;

5. 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.

Answer 3

1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body).

2) It will be reflected back and forth, but only a finite amount of time. This is because reflectivity (or, equivalently, the probaility of a photon being not absorbed) is not 100%. Mathematically and experimentally one observes an exponential decay. (-> integrating speheres).

Example: Integrating spheres are hollow spheres and made of very high reflective material and have one small opening. When you look into the opening it will appear white. When you shine light into the sphere and turn it off suddenly, you will observe that it takes some time until the light has completely decayed. This is because the photons which are stored inside it, are reflected back and forth in it until they have escaped or are absorbed by the non perfeclty reflecting material. However, even for very high reflectivity spheres this time is typically below a microsecond, so you cannot see this effect with the naked eye. It is well measurable though, and basically the ring down time is so short because the speed of light is so high (and you have many many roundtrips in a time interval).

3) They will be reflected (and with higher probability than photons of other wavelength, otherwise it would not appear blue).