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As far as I understand a photon is produced, or "born", whenever an electron moves from a high energy state back to its normal energy state.

The photon then travels at the speed of light across space in a straight line until it hits another atom, or rather, interacts with the electron shell(s) of that atom. The energy signature of the photon can change at this point.

The photon can then bounce off that atom, and will continue to travel across space at the speed of light until it hits another atom. And so on.

(Please correct me if any of my understanding here is off.)


However what I want to know is what happens when light stops, and how this relates to the photon. I want to know what happens when a photon "dies" - not in a literal sense, just in the sense of when it has finished its journey of bouncing from atom to atom.

If you stand in a huge and pitch-black cavern, and shine a torch, the light will only carry so far. Am I right in assuming that the photons produced by the torch eventually stop bouncing from atom to atom, or does the journey of the photon continue and its just undetectable to human eyes?

Similarly, the colour black "absorbs light" - does this mean the colour black is "eating" photons? Does the energy get transferred to the electrons of the black material? What happens to this energy?

And finally, does the same "photon death" happen when a photon hits the retina in a persons eye?

In short, what happens when a photon dies?

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    $\begingroup$ Photons are produced in more ways than just de-excitation of electrons. Just fyi $\endgroup$ – Jim Dec 18 '19 at 14:54
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    $\begingroup$ Photons die by being absorbed by electrons, as you suggest, the obverse to their birth. What is the problem? $\endgroup$ – Cosmas Zachos Dec 18 '19 at 14:56
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    $\begingroup$ Black absorbs photons more efficiently than white. What does your text say? $\endgroup$ – Cosmas Zachos Dec 18 '19 at 14:59
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    $\begingroup$ A photon is not a little ball that flies in a straight line or bounces of a single atom. A photon travels in space as a wave. Each photon reflects off the whole mirror, of all electrons in the mirror at once. Each single photon is focused by a lens while crossing through the whole aperture, not just a single point. $\endgroup$ – safesphere Dec 18 '19 at 16:07
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    $\begingroup$ Old photons never die; they just fade away. $\endgroup$ – MTA Dec 19 '19 at 4:04
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This question is about the nature of the electromagnetic field. The electromagnetic field is a physical system that is most fully described by quantum field theory, and the results match those of classical field theory in certain limiting cases. The 'photon' is a physical picture which gives us a useful way to imagine certain aspects of this field. It is primarily a way to track energy movements.

The main thing you need to know is that energy is conserved, but photons are not. When energy moves from some other form to an electromagnetic form, then photons are created. When energy moves from an electromagnetic form to other forms, then photons are destroyed.

Another way of saying the same thing is to note that when an electron moves from a higher to a lower energy level in an atom, it does so through the way its charge pushes on the surrounding electromagnetic field, causing it to vibrate at a higher amplitude (the electric and magnetic parts both start to vibrate). This vibration, when it happens at a fixed frequency, can be conveniently modelled by saying it has a fixed amount of energy, equal to $h f$ where $h$ is Planck's constant and $f$ is the frequency. If this $h f$ is equal to the energy change $\Delta E$ in the atom, then we say one photon has been created. You can also find cases where two photons are produced, one at frequency $f_1$ and the other at $f_2$, and then $h f_1 + h f_2 = \Delta E$. This kind of process is much rarer but it illustrates that energy is conserved, but a given amount of energy can be expressed physically in more than one way.

Eventually a photon may arrive at some other atom and be absorbed. What happens then is that the oscillating electromagnetic field pushes on the electrons inside the atom, until one of them gains some more energy. The field vibration then falls away as the energy is transferred. We summarise the process by saying that the photon has been absorbed. Or, if you like, the photon 'dies'. This is just another way to say that the field has stopped vibrating.

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    $\begingroup$ This made me wonder--could two atoms theoretically "Ping-pong" a photon back and forth forever without loss? I'm so used to the macro world where perpetual motion is considered--well at best unlikely :) $\endgroup$ – Bill K Dec 18 '19 at 23:12
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    $\begingroup$ The optical arrangement required to achieve this ping-pong would be very difficult in practice, requiring mirrors, and they would not be perfect. But quantum physics offers other kinds of perpetual motion: the current in a superconductor, and the orbits of electrons in atoms. There are many such examples. $\endgroup$ – Andrew Steane Dec 18 '19 at 23:44
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    $\begingroup$ @BillK That's sort of kind of what happens when two particles with the same charge repel each other. $\endgroup$ – Acccumulation Dec 19 '19 at 1:36
  • $\begingroup$ To explore a little further in the direction of the OP: If I understand correctly, some of the absorbed light does not excite electrons (which would later emit the same energy as photons, so long-term no photons would be destroyed) but instead is scattered, creating excitations of the molecules, e.g. phonons or smaller-scale molecular oscillations which eventually emerge as heat on the macroscopic scale. Now I would assume that usually only a part of the photon's energy is absorbed during inelastic scattering -- does that mean that we have a lot of increasingly low-frequency photons around? $\endgroup$ – Peter - Reinstate Monica Dec 19 '19 at 7:40
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    $\begingroup$ @tsj This is a v. good question. It concerns the measurement problem or interpretation problem. If one asks 'what is the probability a photon which left A arrives at B' then there is a single unambiguous answer. The wavefunction $\psi$ is a tool allowing one to obtain the answer to this question for a whole range of locations B. The ripple that spreads outwards for 1 photon is in that photon's $\psi$; I would not regard $\psi$ as a physical thing, but rather as a mathematical tool to allow probabilities of physical events to be expressed. It is the physical events which conserve energy etc. $\endgroup$ – Andrew Steane Dec 20 '19 at 17:08
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A photon is just a wave of change in the background electromagnetic field. Or, to be more precise, it is a packet of electromagnetic energy that is irreducible without altering its frequency of oscillation.

Knowing this, a photon is "born" when one form of energy is converted into electromagnetic energy (e.g. from an electron's potential energy when transitioning from the excited state to a lower state or from thermal energy in blackbody emission or from mass energy in particle annihilation). A photon, conversely, "dies" when its electromagnetic energy is transformed into another form of energy. Some examples of this could be through the excitation of an electron in an atom, it could be absorbed and turned into thermal energy, it could be used in particle pair production, or it could be turned into mass when falling into a black hole.

That's it. Knowing a photon is just a packet of energy means that the rules of energy conservation apply. Energy cannot be created or destroyed*, merely transformed into another form. So a photon can be transformed into other energy and, thus, "die" by any means we might use to convert energy from one form to another.


$^{\text{* this is locally true}}$

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just wanting to learn more about it.

Start with the electromagnetic spectrum which has freqencies from very small to very high. The visible spectrum, color you call it, is a small part of the spectrum. Electromagnetic waves are described by the classical Maxwell equations.

Then we found that everything at the microscopic level follows quantum mechanics, and the particle photon builds up in superposition with zillion others of the same energy to make up the classical electromagnetic wave , art of which is the color spectrum our eyes see.

Then biology comes in, and color perception. What our eyes call red, is not the red in the spectrum ( the rainbow for example has the pure frequencies). It is called color perception. Our eyes call black the absence of the perceivable colors, but the object called black still radiates photons according to its temperature.

Photons are created in two ways, and always quantum mechancis holds.

a) when a charged particle accelerates or decelerates, interacting with a field, magnetic or electric, a photon comes out , with a probability given by quantum mechanical calcualtions

b) what you describe, individual atoms in excited energy levels given by the quantum mechanical solutions, can deexcite and give photons. When photons of that energy meet an atom, they can scatter and excite it to a higher level, thus the photon is absorbed and "dies".

c)more generaly in matter which is composed out of quantum mechanical entities, atoms, molecules, lattices of molecules, the solution of the equations defines fixed energy levels for the electrons/nuclei/atoms/ molecules/ lattice positions. Because all matter has a specicif temperature, and temperature is connected with kinetic energy, the motion of these quantum charged entities generates a spectrum of photons from excitations and deexcitations, called black body radiation. The absorption is the "death" of that photon.

The color we see as black, means that it is absorbing visible photons, picking up energy. Thus a black surface in the sun is hotter than a white one, which reflects visible light.

So it is not just bouncing but also interacting in various ways with the matter on its way that a photon "dies" .

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As stated in the comments there's more than one way to create a photon. But the description you give is pretty accurate for the process you are describing. The photon generated by the first electron transition meets with another atom and is scattered. In that process an electron in the second atom could, in theory, absorb all the energy of the photon, move into an excited state, then decay down to a lower state emitting photon(s) in the process. Or in the more general case of scattering the incident photon imparts some momentum and energy to the electron and bounces off with less momentum and energy, all in such a way to conserve both quantities. You are using the word "die" I think to describe the state where the photon is no longer part of the picture, but then paint a picture where some phone returns. One could say that in either of the process I described the initial photon is gone, ceases to exist, and a new photon is generated. Nothing has to happen to the original photon. The important thing is that energy and momentum are conserved in the process. At a quantum field level photons undergo a process where they spontaneously generate particle anti-particle pairs that recombine to make "the photon" again. And likewise electrons are emitting and reabsorbing photons. These processes are combined to produce corrected values of charge, mass, and possibly other quantities. This is called renormaliztion in QFT. A collection of charged particles and photons can be thought of as a system and the individual components elements on that system. What matters is that all the relevant mechanical properties stay conserved. So we sometimes refer to the configuration of all these components as the state of the system rather than focusing on each particle as if it had an identity. In cases where light is completely absorbed and not emitted again in free space the energy is not lost but trapped in the material as thermal energy, or some other type of mechanical energy like acoustic vibrations. Some emission still occurs in the form of heat but one cannot attribute a specific component of the system to the original photon as so much is going on inside. This type of interaction is not a fundamental process. We typically handle this statistically.

As for the torch example it is not clear that your assumption is true. Stars are torches in the pitch black cavern of empty space (to some approximation). The light, in theory, will go on forever but intensity will decrease since photons from the source are traveling in different directions. In an atmosphere like air there will be some attenuation that will cause the light to diminish due to absorption and diffuse scattering. From a particle physics point of view the photon does not have a lifetime, it does not decay on its own. It contributes to processes by interacting with other particles and vie those processes may cease to exist as a component of the whole system.

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As far as I understand a photon is produced, or "born", whenever an electron moves from a high energy state back to its normal energy state.

It would be reasonable to think that the exact opposite happens, and that is indeed the case (with some optional extra details that aren't important).

Photosynthesis is one of nature's applications where this effect is directly used (in a very complicated process that scrapes off tiny amounts of energy in a long chain of electron transmissions, and that finally does an oxyhydrogen gas reaction without blowing nearby things to smithereens, which is darn cool) to build up high-energetic chemical stuff from low-energetic components. Though, regardless, electrons get excited anyway, whether someone makes "good use" of that or not. And then, some time later, something happens (we cannot tell what). One thing that may happen is a different photon being emitted, another thing is some unknown, random chemical reaction that needs energy taking place. Often, that whatever unknown reaction is a source of radicals. This is one of the reasons why we get skin cancer from UV light, by the way.

what happens when light stops [...]
If you stand in a huge and pitch-black cavern, and shine a torch, the light will only carry so far.

That is not what really happens. Three things occur here. First of all, photons get scattered in space, and space tends to "consume" stuff very eagerly. The mathematical formulation of that is "distance attenuation". While one might think that being away twice as far halves the amount of photons, in reality it cuts them down to one quarter ("inverse squares"). Evidently, something that works this way very quickly smothers anything that is "very finite" such as e.g. light coming from a torch. It doesn't matter quite so much for "practically infinite" things like the sun, but in principle, the same is of course true. So, the amount of light shed by a torch in a large cave isn't terribly huge.
The second thing is that "somewhat close to zero" and "zero" are the exact same thing. Your eyes are unable to see single photons (well your eyes are technically able to receive a single photon, but neither does the biochemical pathway, nor the processing work that way). There is plenty of light remaining in that pitch black cave (well, plenty is maybe somewhat of an exaggeration), only you are not able to see it.
Lastly, there is air in your pitch black cavern, and there is dust and vapor in the air. All of these will absorb and/or reflect photons to some extent. The "reflect" part is why you can often "see" the light orb when in fact actually that's not possible at all (what exactly is it that one would expect to see!). On the other hand, light that is reflected away isn't going to hit your eye (other than incidentially, after having been reflected at least one more time). What's absorbed is gone, one way or the other, so it doesn't illuminate the rest of the pitch black cavern.

the colour black "absorbs light" - does this mean the colour black is "eating" photons?

The opposite is the case. All materials absorb light to some extent. Some only absorb very little of it, and only in a very narrow frequency range. Some absorb huge amounts, and in a large frequency range. Those materials appear black to you because black is your conception of no light meeting your eye. It's not the black absorbing photons, but you see black because they've been absorbed. Note by the way that something can very well appear black and emit lots of photons at the same time (you are only able to see a relatively small range).
Things can be quite deceptive. Glass appears to absorb no light at all (look out your window!) but that isn't true at all. It only absorbs a relatively small (~8-10%) amount of the light that you can see. If you consider e.g. UV light or infrared, things look completely different!

does the same "photon death" happen when a photon hits the retina in a persons eye

Yes. The photon excites an electron in a rhodospin molecule (there are a few variants of these) and it's "gone" after that. The transferred energy causes a structural change in the protein which activates a G-Protein. That one kicks off a certain amount of the second messenger cGMP. When there's enough of that around (not the case for a single photon), the cell decides to fire, and then a neural network on the back of the retina which clusters some areas together in some obscure way gets to decide whether or not to forward an impulse to your brain. Only then, after another few thousand iterations, you have a chance of actually seeing something.

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