Stimulated Emission: Is this how we see images?

[David Waller]

I got the explanation about stimulated emission and lasers as explained here: Understanding Stimulated Emission: Got the what, how about the why?

What I'm still unsure about is if this also explains the simpler question of how light passes through a transparent medium, like glass, yet the photons still retain their original momentum vector from the source, so an image can be reconstructed spatially.

In this case we are not talking about coherent photons, like in a laser. However, that shouldn't matter in terms of a high flux of photons passing through a material. I imagine an atoms-eye view, with a stream of photons all the same frequency (color) coming from a pinpoint on the source at me. There are so many that there's never any opportunity for spontaneous emission, thereby losing the direction information from an absorbed photon.

Rather, I imagine photons being absorbed and then immediately readmitted via stimulated emission because there is a flood of other photons hitting the cross section of the atom.

IF this concept is correct, there is an interesting side-effect: When the light gets dim enough, so that the flux is on the order of photons separated by more time than most spontaneous emissions, the material effectively becomes opaque, the photons are scattered rather than retaining their momentum vector.

Keep in mind we exist in a transparent medium -- air -- that photons are passing through.

So -- experts, how does this look?

 

[Khashishi]

If light is passing through a transparent material, it isn't being absorbed and reemitted. Are you talking about a transparent material or an opaque material?

 

[David Waller]

If light is passing through a transparent material, it isn't being absorbed and reemitted. Are you talking about a transparent material or an opaque material?

Actually, it is. This is why light travels "slower" in denser media, and the entire basis of optics.

Without this absorption and re-emission -- which introduces a small delay in the "travel" of the photon -- light would not bend when transitioning different density media at an angle to the interface.

 

[Khashishi]

That's incorrect. I think there's a frequently asked question on that in these forums, but no. If light were absorbed, it would be reemitted in some random direction, and you would have scattering. For simple materials, light can only be absorbed at very specific frequencies in the absorption spectrum.

 

[jtbell]

If light is passing through a transparent material, it isn't being absorbed and reemitted. Are you talking about a transparent material or an opaque material?

Actually, it is. This is why light travels "slower" in denser media, and the entire basis of optics.

This is a common misconception. Khashishi remembers correctly that we have an FAQ on this which was recently moved to our Insights blog:

https://www.physicsforums.com/insights/do-photons-move-slower-in-a-solid-medium/

 

[Drakkith]

This is a common misconception. Khashishi remembers correctly that we have an FAQ on this which was recently moved to our Insights blog:

https://www.physicsforums.com/insights/do-photons-move-slower-in-a-solid-medium/

I have an issue with the wording in this FAQ. Specifically the underlined sentence below:

On the other hand, if a photon has an energy beyond the phonon spectrum, then while it can still cause a disturbance of the lattice ions, the solid cannot sustain this vibration, because the phonon mode isn’t available. This is similar to trying to oscillate something at a different frequency than the resonance frequency. So the lattice does not absorb this photon and it is re-emitted but with a very slight delay. This, naively, is the origin of the apparent slowdown of the light speed in the material. The emitted photon may encounter other lattice ions as it makes its way through the material and this accumulate the delay.

If the photon isn't absorbed, how is it re-emitted?

 

[jtbell]

I suspect this is simply awkward wording, and the "not" is meant to apply to the entire rest of the sentence, i.e. using parentheses for grouping:

So the lattice does not (absorb this photon and it is re-emitted but with a very slight delay).

Perhaps @ZapperZ can clarify for us.

 

[ZapperZ]

It is awkward wording. The whole paragraph should be reworked in some way. If anyone wants to join in the fun, we can certainly edit that entry.

Zz.

 

[Domullus]

To answer David Waller original question: "What I'm still unsure about is if this also explains the simpler question of how light passes through a transparent medium"

No, your understanding about stimulated emission in every day life is not correct. Just take for example the Rayleigh scattering. If your theory is right, then such thing could not exist.

You need to understand difference between absorption and scattering. Note, that scattering can occur without absorption. Compton scattering, Thomson scattering, Rayleigh scattering - all are processes where there are no photon absorption.

However it gets complicated when you look into the nature of photon. Photons can exist only at the speed of light. So you cannot just describe scattering as bouncing off the photon from the atom (in this case photon must slow down to zero velocity and then accelerate again, which is impossible). Therefore, during the interaction photon disappears and after few femtoseconds reappears again. Atom (or other particle) does not get excited during this interaction, but it can gain momentum and kinetic energy from photon (see Compton scattering for example).

If atom is not excited it cannot radiate (no spontaneous and no stimulated emission).

So why do photons fly in straight paths if all atoms scatters them? Actually, they don't. Take a look at Rayleigh scattering in atmosphere again. More energetic blue photons have higher probability to "collide" with atoms then red ones, and during "collision" they change their direction - therefore sky has blue color.
But remember, that atmospheric scattering takes place only in the upper atmosphere, where concentration of atoms are low. At sea level there is no Rayleigh scattering (it does not exist in clear water, high quality glass and etc.).

And here comes the second part of your question: why scattering is absent in dense material? It is due to the coherence. When there are a lot of atoms situated close to each other, all photons that interact with them become coherent. This means that constructive-destructive coherence will start, and it can be shown that only forward propagation is not forbidden in this case. This is the outcome of modern QED theory, and also can be explained by Huygens–Fresnel principle from old times.

Interesting outcome is this: you are right about the material becoming "opaque" for dim light. However, the light must be really dim- only few photons. If you let single photon in the glass, photon will move like Brownian like particle - each atom will scatter it.

 

[Drakkith]

Interesting outcome is this: you are right about the material becoming "opaque" for dim light. However, the light must be really dim- only few photons. If you let single photon in the glass, photon will move like Brownian like particle - each atom will scatter it.

I don't think this is true. I take photos of the night sky through a refractor telescope and many of the targets are extremely dim. Some are only bright enough to send around one photon per second (or less) through the glass lenses and yet I have no problem imaging faint targets.

 

[Domullus]

Drakkith,

My point was, that a material can become "opaque" (note quotation-marks) for a single photon :) I do not imply that material will absorb this photon. There exist a higher probability for the photon to be scattered in single-photon case if compared to a collective behaviour of coherent photons. All antireflective coatings of your telescope optics will stop working for single photons. Therefore the actual path of a photon through the glass can be very complex.

Also, you cannot get correct image just from one photon impacting your super-sensitive CCD camera (photomultiplier). Therefore a statistical acquisition is needed. You do this by exposing object for a sufficient long time, gathering many photons, and the resulted image is the statistical outcome of glass/photon interaction :)

 

[sophiecentaur]

If the photon isn't absorbed, how is it re-emitted?

The whole of the structure is involved in the interaction with the photon. The arrangement of atoms is large, compared with the wavelength so it could be regarded as a temporary absorption that's shared with many atoms (forget the discrete energy levels associated with a gas atom) and then re-emitted. But the structure is massive and the phases of all the contributions from all the atoms are such that (as with a large, end-fire transmitting array) the energy gets transmitted only in the forward direction. Interference from all the sources will cancel energy flow in other directions. For small particles (many atoms but still not large compared with the wavelength) you will get scattering (as with water droplets and tiny ice crystals) but a large volume of water or ice will not scatter light - just refract it / slow it down.

 

[Drakkith]

My point was, that a material can become "opaque" (note quotation-marks) for a single photon :) I do not imply that material will absorb this photon. There exist a higher probability for the photon to be scattered in single-photon case if compared to a collective behaviour of coherent photons. All antireflective coatings of your telescope optics will stop working for single photons. Therefore the actual path of a photon through the glass can be very complex.

I was pretty certain that a photon would interfere with itself on an anti-reflective coating, similar to how it behaves in the double slit experiment. Is that incorrect?

Also, you cannot get correct image just from one photon impacting your super-sensitive CCD camera (photomultiplier). Therefore a statistical acquisition is needed. You do this by exposing object for a sufficient long time, gathering many photons, and the resulted image is the statistical outcome of glass/photon interaction :)

Maybe it's just me, but here you appear to be using 'single photon' to mean 'only one photon. ever. for all time.' Would it not be better to stick to something more along the lines of 'one photon passing through at a time'?

 

[sophiecentaur]

one photon passing through at a time'?

At what rate, though? How would you actually identify the 'separateness' of each photon? What would the interval between them need to be? (To answer that, you'd need to say how long each one lasted for - so you could not get an overlap or coincidence.)

 

[Drakkith]

At what rate, though? How would you actually identify the 'separateness' of each photon? What would the interval between them need to be? (To answer that, you'd need to say how long each one lasted for - so you could not get an overlap or coincidence.)

No idea. I was under the assumption that a single photon was the same as multiple photons for the purposes of interference, diffraction, etc.

 

[sophiecentaur]

Stimulated Emission is a very special phenomenon and it only happens in a very small fraction of cases of emission. That's why lasers are so special and we waited for decades for one to be made. What goes on when EM is transmitted through a medium is due to many atoms working together and not a single atom handling each photon.

 

[sophiecentaur]

No idea. I was under the assumption that a single photon was the same as multiple photons for the purposes of interference, diffraction, etc.

Well yes. The statistics of many photons (either a lot at once or a long period of a slow rate) is the same as the conventional wave behaviour. (This is deja view with the other thread, just down the page on wave-particle duality.)

 

[Domullus]

I was pretty certain that a photon would interfere with itself on an anti-reflective coating, similar to how it behaves in the double slit experiment. Is that incorrect?

Yes, you are right here. My comment regarding coatings is not correct, it would work for single-photon. You have to add complex amplitudes off all possible photon paths in material and then to take it square for real probability (standard quantum mechanics rule). And in the result you get same reflectivity as for a wave. Therefore and scattering probability should also be the same for single photon as for all photons, but you just has to gather statistics. Thanks Drakkhit for critical thinking ;)