Light traveling through a medium that slows it. Does the same photon emerge?

Submitted by TheGandPTurtle t3_111g7s9 in askscience

Suppose that we had a laser that could release a single photon, and imagine that the photon has a particular "ID number."

We use this laser two times to send a photon from point A to point B. We do this once in empty space and then once through a medium like sugar water, glass, or diamond.

It is my understanding that the light is slowed through the medium because photons are absorbed and then re-emitted repeatedly.

First question: So my question is, in the case of the diamond, glass, or water does the same photon that entered at point A exit at point B? E.g. Would our imaginary ID number be the same?

I presume it would be for the photon traveling empty space.

Second question: Are all photons at the same wavelength identical so that it just doesn't make any sense to ask this question or are there some properties that are effectively randomized each time it is re-emitted?

Probably a basic question, but I never see discussion of refraction or mediums slowing light down make this clear to me.

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Weed_O_Whirler t1_j8fo188 wrote

So, first answering your main question- elementary particles are all fungible. That means, they are truly identical, and they are impossible to label. So, if a photon is absorbed and then remitted, it doesn't really make sense to say "is it the same photon or a different one?" There aren't really "same" or "different" photons, there's just photons, unlabeled.

And it's not just photons. Any time you have a particle collision which results in some different elementary particles (like the ones from particle accelerators), if one of the products and reactants are the same elementary particle, you can't answer "is this the same or a different particle?" It's a particle. That's all you can say.

Now, to get into the can of worms you opened, and probably didn't even know it. It is this line:

> It is my understanding that the light is slowed through the medium because photons are absorbed and then re-emitted repeatedly.

I always say, if you want to get some physicists to fight, ask them why light propagates slower through a non-vacuum. You'll get a different answer from each one, and they will pretty aggressively defend their position and discount the others. I always find it fascinating, because it seems like a pretty simple question (why does light travel slower in a non-vacuum?) but the answer is quite complex, and our models for it all work, but tell slightly different stories.

The easiest to understand model is the one you mentioned- and it does work. The most common complaint is that an atom can only absorb very specific wavelengths, but light of all wavelengths is slowed down by materials. But, this is handled by understanding that collections of particles will have nearly an infinite number of modes of excitation- you can cause groups of particles to vibrate or rotate, you can cause vibrations between groups of 2 particles, or groups of 3 (or 4, or 5....). There's a ton of different excitation modes, and for a dense medium, you can absorb and re-emit any wavelength of light.

Other people will express a model where light actually takes many paths through the medium, and that superposition actually results in it appearing as if the light is traveling under 'c'. Still others will talk about how photons become a quasiparticle when in a dense medium, and that particle doesn't travel at 'c'. And I'm sure there are others out there. All of these explanations "work" and I won't say one is right over the other.

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taphead739 t1_j8fv6kd wrote

One thing I am wondering after reading your (really good) reply: Is there exchange interaction between photons?

I‘m a theoretical chemist and well familiar with exchange interactions between electrons, in magnetic materials, and in superfluid helium-4. Is there an equivalent attractive force between photons since they are bosons?

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D_jrivera t1_j8gzkvh wrote

Electrons are fermions and cannot occupy the same quantum state, hence why there is exchange energy. Photons are bosons and are able to occupy the same quantum state, so there is no exchange energy for photons. Hope this answers your question :) -another quantum chemist

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9966 t1_j8h0yvh wrote

But the total spin of helium 4 is whole, which makes the whole system a boson

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ChemicalRain5513 t1_j8g08qp wrote

I don't know if this completely what you mean, but laser works by stimulated emission of photons, since they like to be in the same quantum state.

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taphead739 t1_j8g478u wrote

This is unfortunately not what I mean, but thanks anyway. Technically speaking I am talking about the energy contribution in the Hamiltonian integral of a system of multiple identical particles that arises from the requirement that the total wave function must not change its sign upon exchange of particle labels (in the case of bosons). Does this exist for photons?

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TheoryOfSomething t1_j8i3q57 wrote

I believe that the phenomenon known as "photon bunching" could also be described as an exchange interaction (along with the associated anti-bunching effect for fermions). If you consider two points, a and b, in some source that is emitting photons and you set up 2 detectors, A and B, to detect those photons, then for photons (and all bosons) you will see an increase in the probability of simultaneous detection at A&B compared to distinguishable particles and a decrease of simultaneous detection for Fermions: https://en.wikipedia.org/wiki/Hanbury_Brown_and_Twiss_effect

Because photons are non-interacting at the tree-diagram level (that is to say that in the Hamiltonian Lagrangian for QED there is no photon-photon interaction term), this does not lead to the same energy consequences as it does for electrons or alpha particles. Both the Coulomb integral and the exchange integral are proportional to the interaction term in the Hamiltonian, except that for photons there is no interaction term! Metaphorically speaking (because there are technical problems with assigning a wavefunction to single photons), the product basis remains the diagonal basis. As a result, you see interference effects that cause the bunching behavior mentioned earlier, but not the same consequences for energy or spin correlation as with electrons.

A careful reader may object at this point and say, "Ah! But you have neglected the higher-order QED effects. Sure, at tree-level there is no photon-photon interaction, but what about the scattering mediated by virtual electron-positron pains? Surely that gives rise to some interaction which turns out to be either attractive or repulsive." And the careful read is almost correct, almost. There is very weak (starting at 4th order) photon-photon interaction and photon-photon scattering in QED, but it turns out that the effective potential that this interaction gives rise to has zero range (one might describe it as "a delta function" although there are technical problems with making this formulation precise in >1 spatial dimension; a complication that comes up repeatedly in my PhD thesis in the context of ultracold atomic physics) and therefore cannot really be described as attractive or repulsive. To expand, to first order in perturbation theory the interaction terms is quartic in the E and B fields, and if you work out what kind of potential it takes to create that you get an operator proportional to a delta function. Of course this is just the effect at first order, the effects at all higher orders will also give rise to interactions that look like they are "zero range", although if it were possible to do the full resummation and go beyond all orders, you might get a spatially-extended potential. I can't think of any reason that that potential should be always attractive or always repulsive, but who knows. Experimentally speaking, attraction/repulsion between beams of photons has not been observed or measured without coupling them to some massive interacting medium.

In the case of photons, there is a purely classical explanation for all of this. Essentially, because the classical limit of a multi-particle system of photons is not a system of interacting distinguishable particles but rather an electromagnetic wave (a coherent state superposition of all numbers of photons with a well-defined average number of photons and phase), you can do the math and arrive at correct predictions purely from considering the classical problem of detecting the signal from a spatially extended EM source with 2 nearby detectors.

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taphead739 t1_j8i66hg wrote

That was very informative and probably what I was looking for. Thanks a lot!

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hahahsn t1_j8htczx wrote

hey this is a super interesting question! I don't have the answer but in case someone does i'll leave this comment to find it at some point.

Based on my limited understanding I don't see where such an energy contribution would arise. I get it in the case of fermions but boson statistics are unfamiliar. The normalisations are weird.

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eva01beast t1_j8gga2w wrote

Photons fall under the category of bosons. They have a spin of 1.

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[deleted] t1_j8j3oup wrote

[deleted]

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taphead739 t1_j8jbc0g wrote

Most of the time: using computer programs based on quantum-mechanical equations to predict the outcome of chemical reactions. You can also predict the properties of molecules and materials.

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dack42 t1_j8ghb5l wrote

If it was absorbed and re-emitted, then surely the direction of the outgoing photon would be different. Or am I missing something?

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Weed_O_Whirler t1_j8gn3wr wrote

While it's true if a single photon is absorbed by a single atom, you cannot predict which direction the photon will be re-emitted, so how is this different? Well, the process is not so different from light "reflecting" off a mirror- while we say light reflects, a mirror is also an absorption and re-emission situation, and obviously those don't go in a random direction.

The answer comes down to conservation of momentum and interference. Since the incoming photon has momentum, there is a higher probability that the photon will be emitted in the same direction to conserve momentum. Obviously, not all of the photons are emitted in that direction, but due to the probability there will be constructive interference in the same direction and destructive in all other directions. In general, things like Snell's Law of Refraction, and angle of incidence equalling angle of reflection occur with large numbers of photons, and they do not describe what happens when a single photon is absorbed.

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MasterPatricko t1_j8ib8hl wrote

> a mirror is also an absorption and re-emission situation

I already addressed the use of "absorption and re-emission" in a previous reply to you, so I won't repeat that -- but for anyone reading, this is another case where you must not confuse "classical particle absorption and re-emission" (photon comes in, is absorbed, is gone for a moment, is emitted) with the collective scattering of probabilities/wavefunctions that we calculate in QM.

The law of reflection does not (cannot) come from a classical particle scattering model. It can be derived from either a classical wave model or a properly quantum mechanical model (wavefunctions are waves after all).

> The answer comes down to conservation of momentum and interference. Since the incoming photon has momentum, there is a higher probability that the photon will be emitted in the same direction to conserve momentum.

We've already discussed why you should not imagine the photon being completely absorbed by an atom when trying to derive the laws of optics. But let me address the momentum part of this answer (interference part is fine) as well because it's also misleading. Why should an atom prefer to return exactly to its previous state of "zero momentum"? Answer: It doesn't. Further: First of all the momentum we are talking about is tiny when a photon is absorbed by an atom, so even the thermal motion of the atoms is already going to wash out any momentum change. Second the atom is free to exchange momentum with its neighbours via phonons, so there's no reason why it has to return to zero by itself. Third, we measure single photon/single atom absorption and re-emission behaviour experimentally and it truly is isotropic (well, kinda), while your answer effectively claims it is not (contradicting your very first statement). (Incidentally, when 1 and 2 above are not true any more, like for a very cold gas, we can do fun stuff like laser Doppler cooling.)

The real answer is that we're not dealing with classical absorption/emission of photons by one atom, we're dealing with a collective behaviour and scattered probabilities/partial waves (see previous linked discussion, etc.) -- even in the case of few photons. So though all the scattered components are random in direction, we add them coherently (interference, as you said). The newly constructed wave/wavefunction has a clear direction of travel obeying the law of reflection.

/u/dack42

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WagonWheelsRX8 t1_j8ipiy5 wrote

I've been following and trying to wrap my head around these concepts as well. Did not realize something we see every day (light passing through glass) would be such a deep topic.

Based on your description, my understanding is that if you had a laser, a piece of glass (transparent material with index of refraction) and a detector placed in a dark box in a vacuum, and

  1. you emitted a single photon, there is a high probability that photon would be detected by the detector.

  2. you emitted many photons, the photon detector on the other side of the transparent material would not detect the same quantity of photons emitted, because there is a probability that some of them would be reflected back at the emitter. However, because they are all part of a wavefront, none of them would be emitted out the sides of the transparent material?

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MasterPatricko t1_j8j3fc4 wrote

I think you're on the right track, mostly. I can try to add some detail/clarification.

> 1) you emitted a single photon, there is a high probability that photon would be detected by the detector.

Yeah, more or less -- the probability depends on the coefficient of reflection of the surface of the glass. It could be high, or low, depending on the angle and refractive index of the glass, and will match the classical calculation of the relative amplitudes of transmitted and reflected waves (Fresnel equations), because fundamentally, the photon wavefunction is obeying pretty much the same wave propagation laws. Note you only get this behaviour if the photon wavefunction spatial extent (typically many photon wavelengths) means that it has the possibility to interact with many identical atoms which are spaced in a uniform way (relative to the wavelength of the photon). This is mostly true for optical wavelengths hitting normal materials (typical atom spacing 0.1nm, optical wavelength 500nm, any random variation is too small to be "seen" by a photon) but is not necessarily true at shorter wavelengths like X-rays, or if your "glass" is only a few atoms big. Those scenarios are more complicated and you don't always end up with the "normal" ray optics rules after summing those probabilities.

The only real difference to the classical wave picture is in the moment of detection -- instead of measuring a classical wave, with some part reflected and some transmitted, we are set up for a discrete, quantum mechanical interaction in the detector. A 1 or a 0. This collapses* the wavefunction according to the probabilities mentioned before, to either interact with the detector (and so we say the photon was transmitted and then absorbed in the detector) or not (and here we can say the photon was reflected).

> 2) ... However, because they are all part of a wavefront, none of them would be emitted out the sides of the transparent material?

You can hopefully see how the previous explanation scales up to emitting many photons, the numbers of photons detected will match the probability calculations. But to address your last sentence -- in this toy example, I assumed that there was no internal absorption or scattering inside the glass. In such a case, the original+scattered wavefunctions sum up such that the only possibilities are transmission straight through or reflection right at the interface, and there is no chance for a "random" photon to emerge out of the side of the glass at some angle. This isn't to do with photons being "part of a wavefront", unless I misunderstand you. More fundamentally, photons are waves; or at least, they travel in the same ways that classical waves do, and waves moving through a uniform medium don't just randomly scatter.

Now if we make things a bit more realistic -- the glass is not going to be a perfect crystal, and the uniform background of atoms assumption is not going to be exactly true. There will now be a small possibility of the wavefunction scattering differently off some imperfection in the atomic structure. Because it's associated with an imperfection, this part of the wavefunction won't be cancelled out by all the neighbouring scattered amplitudes, and you will end up with a real (but small) probability to have a photon emerging out of the side of the glass in some random direction, in addition to the main probabilities of transmission or reflection at the surfaces.

* "The measurement problem", or what exactly wave-function collapse really means physically, is a very thorny issue to which there isn't a good answer. But we know that mathematically, it works.

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WagonWheelsRX8 t1_j8je7x8 wrote

Very interesting, thanks this is helpful! Yes, I suppose in the real world there are a lot of additional factors that need to be considered as well (such as the actual uniformity of the glass, etc.)

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ben_vito t1_j8gmoov wrote

I don't follow you on the part about them being unlabeled. If there are multiple photons, why would you not be able to follow and label them, at least from a theoretical and not real life point of view?

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Weed_O_Whirler t1_j8gnkyv wrote

Sorry, I should have been more clear. You can, right up until the point where they collide (you can think of as getting close enough their wave functions overlap), at which point you can no longer know which particle is which.

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2weirdy t1_j8ha2da wrote

>their wave functions overlap

But their wave functions always overlap?

My, understanding was that the "event" that they spontaneously switch, placesis quite low, but always non-zero, and more importantly, indistinguishable from staying still.

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terribleturbine t1_j8h4r7p wrote

Thank you for clarifying. I was looking at my left and right hand thinking "Certainly the electrons in my left hand are not the electrons in my right hand."

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John_Fx t1_j8hv4av wrote

look up the single electron theory where the universe consists of only one electron bouncing back and forth in time.

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tnaz t1_j8mc77f wrote

The single electron universe model fails once you introduce other particles - a muon can decay into an electron and neutrinos. You can't represent that if there's only one electron going back and forth in time.

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tpolakov1 t1_j8jirfo wrote

> "Certainly the electrons in my left hand are not the electrons in my right hand."

You'd be surprised. The states are certainly different, but it makes no sense to talk about electrons, other than the states being filled or not. There is no such thing as "that electron", only "that electron state" and "a electron".

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terribleturbine t1_j8jnj6o wrote

But isn't it true to say "that electron state" in my left hand and "that electron state" in my right hand are states of two separate electrons?

It was my understanding that electrons had a ton of possibilities states/superpositions that they only "chose" one when they became entangled, is it wrong to think of all the possible positions as the "electron" and it's current configuration in my hand as the electron state in this branch of the wave function?

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tpolakov1 t1_j8ju8fx wrote

> But isn't it true to say "that electron state" in my left hand and "that electron state" in my right hand are states of two separate electrons?

They are distinct state, but you can't say that electron 1 is in the left hand and electron 2 is in the right hand. You can say that an electron has filled (or not) the left state and an electron has filled the right state.

> It was my understanding that electrons had a ton of possibilities states/superpositions that they only "chose" one when they became entangled

They stay in superposition until measured, there's no need to bring entanglement into that.

> is it wrong to think of all the possible positions as the "electron" and it's current configuration in my hand as the electron state in this branch of the wave function?

Nope. You can't describe many-particle states by the individual identities of constituent particles as you do in classical physics. When I have three electrons in a bucket, I describe the state of the bucket with electrons in it, i.e., two electrons in the n=1 level and one electron in the n=2 level. But there's no way of knowing which electron is which in those levels.

In the same way, when we're talking about two electrons in your two hands, we describe the state as one electron in the left state and one in the right state, but there's no such thing as left and right electron. This effect is what ultimately leads to things like the Pauli's exclusion principle or the Gibb's paradox, so we know that the electrons fundamentally don't have identities and it's not just a limit of our measurement.

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Purplestripes8 t1_j8h5094 wrote

Location is indeed a way to distinguish identical particles, however if two identical particles share the same location then there is no way to distinguish them. You can not say "which is which", only that two particles came in and two went out.

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drc500free t1_j8j35ct wrote

Think of the individual photons as different locations where a single universal wave function has high density, not as distinct particles. The idea that they are distinct only holds while those locations don't interact and interfere significantly.

Once they interact, the resulting waveform will have other regions of high density. Mapping one of them back to one of the original ones and saying this new photon is the "same" as that original photon is something that might make your brain understand it better by pretending they are Newtonian objects. But it's just a model for understanding, and the further the interaction is from Newtonian collisions the more wrong it will be.

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MasterPatricko t1_j8i7f5l wrote

There's some good info here but also a lot of misleading or easily misread info.

You're correct on the fungibility of photons. Nothing to add there.

But you are dangerously vague for the rest.

> The easiest to understand model is the one you mentioned- and it does work.

Careful. The confusion you describe (even among physicists) comes at the root from not being specific on what we mean by absorption and emission, and unless you clarify you have not helped the situation. Let me try to be extremely clear about the possible models.

  1. Classical scattering of waves. Here, there is a main EM wave which can partially scatter off atoms, since they are charged -- and importantly it's a collective scattering from all of the atoms in the path. The superposition (combination) of the original wave and the small scattered waves leads to a new, slower wave, with the same frequency as the original. Note that at no point is the incoming wave completely extinguished (absorbed), it's only ever a partial effect. This model works mathematically.

  2. Classical scattering of particles. Here, you imagine a billiard ball photon travelling along that happens to hit an atom and be absorbed. For a moment there is no more photon anywhere. Then the atom relaxes, and a photon is re-emitted to continue on its journey. This model does not work mathematically in any way to explain the speed of light in a medium. You cannot assign the slowing down of light to a random time delay between classical particle absorption and emission nor to a particle taking a 'longer zigzag route' through the medium.
    When people ask this question on the internet, it's usually with a foundation of only classical mechanics, not quantum mechanics, and so this is usually what they are imagining when they say "absorption and re-emission" and it is wrong.

  3. Quantum mechanical scattering. This one is tricky to understand without at least an intro to quantum mechanics but fundamentally we are scattering probabilities, not whole particles. There are several ways to do the maths -- you can consider the propagation, partial scattering, and interference of the photon wavefunction as it interacts with virtual energy levels in the atoms (looks very similar to the classical wave math -- superpositions of the original and partially scattered probability waves). Virtual energy levels are guaranteed to be unstable and exciting the atom to one is fundamentally different to exciting to a stable energy level. Or you can consider a path integral approach like that of Feynman (again, summing probable paths, not definite or discrete ones), or you can start calculating the collective excitation of the photon and all neighbouring atoms as a dressed quasiparticle with new properties (specifically, mass -- so travels less than c). The important thing here is that all the behaviour is collective, never discretely with a single atom -- and at no point is the photon (with its original frequency and direction) completely gone. When physicists say "absorption and re-emission" they are often thinking of this model because we use the same terms for classical scattering and scattering of a QM wavefunction -- but that does not mean it is the same model as 2) above. It is not, it is fundamentally different.

> The most common complaint is that an atom can only absorb very specific wavelengths, but light of all wavelengths is slowed down by materials. But, this is handled by understanding that collections of particles will have nearly an infinite number of modes of excitation

Your answer doesn't work. It is true that "excitations are always discrete" is an oversimplification -- the existence of black-body radiation proves that. But you don't get to claim that all materials absorb at all frequencies and therefore slow light through absorption and reemission, without also explaining why the transparency of a material doesn't have anything to do with the speed of light through it (cf. glass at optical wavelengths). Again this confusion you introduced comes from not clearly differentiating between the QM scattering of probabilities -- which may involve virtual energy levels and whole lot of behind-the-scenes stuff -- and complete, classical absorption of a photon to a new stable energy level. The real answer is simply that you don't need stable energy levels (which cause the material to become opaque) to exist to do the QM scattering math. Though transitions to virtual energy levels are low probability and necessarily temporary, their collective sum, including the original photon as well, gets us to where we want to go.

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pavilionaire2022 t1_j8gz3bp wrote

It's not possible to label a photon, but not all photons are indistinguishable. So can we rephrase the question as, are the photon properties, such as spin, conserved? If the photon that enters is entangled with another, is the photon that emerges also entangled?

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tpolakov1 t1_j8jji5t wrote

Yes and no. The entanglement will be most certainly lost if it flies through a room temperature chunk of material consisting of billions of atoms.

In a more idealized case of it interacting with just a single cold atom, it can be preserved and will entangle with the atom also.

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agentchuck t1_j8gihur wrote

Can fungible particles be discerned through quantum entanglement?

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karantza t1_j8gpnha wrote

Nope, and this is actually what leads to some of the weirdnesses of entanglement. Since in terms of information, if you have two particles A and B, that are identical, AB and BA are the same states.

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binarycow t1_j8gxbpr wrote

>Nope, and this is actually what leads to some of the weirdnesses of entanglement. Since in terms of information, if you have two particles A and B, that are identical, AB and BA are the same states.

💡That makes a ton of sense. I should have thought of that sooner.

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xenneract t1_j8j579k wrote

> AB and BA are the same states

For bosons, fermions pick up a -1 on exchange

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masterofshadows t1_j8gp7z2 wrote

>elementary particles are all fungible. That means, they are truly identical, and they are impossible to label. So, if a photon is absorbed and then remitted, it doesn't really make sense to say "is it the same photon or a different one?" There aren't really "same" or "different" photons, there's just photons, unlabeled.

So then do we know for sure photons actually move and don't just vibrate or something causing the next one in the chain to vibrate or something like that? Kind of how AC current works except with photons?

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HoldingTheFire t1_j8gzz8b wrote

Because photons aren’t traveling through a medium of photons. It’s a propagating electromagnetic field.

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mckulty t1_j8gsewc wrote

> So then do we know for sure photons actually move and don't just vibrate or something

That's like asking "is it a particle or is it a wave?"

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silent_cat t1_j8hwizj wrote

Think of the ripples on a lake made by throwing a stone. The water is only going up and down, but the waves move forward. Is that the same wave, or is it a new one?

How would you label a wave to distinguish it from a different wave?

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BlazeOrangeDeer t1_j8hnofd wrote

They are vibrations of quantum fields (in this case the electromagnetic field), so you can say that the fields are passing energy, momentum, angular momentum, etc from one place to another, and this "bucket brigade" of physical quantities is what we call a particle.

You could technically describe this as photons continuously being destroyed as they create new photons in adjacent locations. But it's not that physically meaningful, it's like adding +1 and -1 to the same side of an equation. It doesn't really do anything but use more ink.

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binarycow t1_j8gxe0t wrote

>So then do we know for sure photons actually move and don't just vibrate or something causing the next one in the chain to vibrate or something like that? Kind of how AC current works except with photons?

Does it matter if individual photons move?

All that matters is the end result.

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IonizedRadiation32 t1_j8h4epi wrote

Your first point is new to me, and weirdly offputting. The idea that "identity" stops being a thing when talking about subatomic particals is oddly disconcerting. Why is that the case? Is it part of the uncertainty principle? What words should I google if I want to learn more?

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Derice t1_j8hdqj8 wrote

The Wikipedia article on this is quite good. If you want some intuition for this, you have encountered things that work kind of like this in your daily life: cups of water.

  1. Take two different cups of water, cup A and B. As long as they are kept separate you can kind of label them by the cup they are in.
  2. Pour them into the same container. Now you know that you have two cups of water, but it does not make sense to ask which is which.
  3. Pour the water out into the two cups again. You can not say whether the current cup A is the same as the first cup A.

That particles act like this has huge consequences for any physics that depend on how many different states are available to the system. Consider two distinguishable particles: 1 and 2, that each can be in one of two states: up or down. There are four possible states:

  1. 1 up, 2 up
  2. 1 up, 2 down
  3. 1 down, 2 up
  4. 1 down, 2 down

If they are indistinguishable there are only three states:

  1. Two particles are up
  2. One is up and one is down
  3. Two are down

This is a pretty big thing, and has macroscopic consequences! Distinguishable particles follow Maxwell-Boltzman statistics while indistinguishable particles follow either Fermi-Dirac or Bose-Einstein statistics. Identical particles of the first group are called fermions and the second bosons. Fermions have the property that two fermions can not be in exactly the same quantum state, and since protons, neutrons and electrons are fermions this is partly what gives matter "solidity". Bosons can all be in the same state, allowing for things like laser light and Bose-Einstein condensates.

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IonizedRadiation32 t1_j8hkskc wrote

1, thank you for the detailed reply! I can't wait to understand this better.

2, at least for me, the water cup analogy doesn't quiiite work, because the reason they become indistinguishable when you mix them is because they are made of a bunch of the same stuff and it gets mixed, but subatomic particles are made from distinct units, so in theory even if you "mix" them you should be able to follow where each part goes, at least in theory, no?

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Derice t1_j8ho1g1 wrote

> subatomic particles are made from distinct units, so in theory even if you "mix" them you should be able to follow where each part goes

Actually no. Subatomic particles are all excitations of the same underlying quantum field, and if we are using quantum field theory, they are not really things in themselves.
If you use quantum field theory to model e.g. sound waves you find that you can describe them with particles called phonons. However, if you have a sound wave in a material and pause time, no matter how much you zoom in on the sound wave you will never find it to be made of little balls flowing through the material.
In quantum field theory particles are less the water in my cup analogy, and more the abstract volume measurement of "a cup". You can add or remove 1 particle's worth of excitation, but when you do you do not add a "real thing", you add an amount of excitation to a real thing: the field.

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jazekers t1_j8hm11l wrote

>subatomic particles are made from distinct units

Then we enter into the particle vs wave interpretation. If you think of them as rigid particles then you would indeed think that you could follow them (keeping out the fact that observing means interactions, which means altering the state). My particle physics professor said it like this "subatomic particles are spatiotemporal fluctuations of quantum fields", which is a very abstract but interesting way to put it.

A proton for example is made up of three quarks, kind of. In fact, it also contains virtual quark pairs that exist for a ridiculously short amount of time, being fluctuations in the strong nuclear field.

But some things are still conserved. Meaning that if I have two particles with one being spin up, and one being spin down. Then when I measure them I will still find one spin up, and one spin down. But that doesn't mean that the particle remained "intact" and rigid along the way. What is conserved is the total spin of the system. Not that of the individual particles.

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Treczoks t1_j8h82nx wrote

This makes me wonder. So if photons travel through a non-vaccum medium by being absorbed and re-emitted, how the heck does the information travel through that medium? Who tells the emitting atom to generate photons of exactly this frequency and polarisation in exactly that direction? How does it actually generate that frequency, e.g. the 432.1THz of a ruby laser when passing through a pane of glas? If one adds unspecific energy to the same piece of glass, i.e. melts it, it glows in yellow or white. Is there any way to make that glass emitting photons of a certain frequency except shining the right frequency into it?

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MasterPatricko t1_j8j943n wrote

As far as I understand you, you're asking two different questions.

> So if photons travel through a non-vaccum medium by being absorbed and re-emitted, how the heck does the information travel through that medium? Who tells the emitting atom to generate photons of exactly this frequency and polarisation in exactly that direction? How does it actually generate that frequency, e.g. the 432.1THz of a ruby laser when passing through a pane of glas?

This is the wrong type of "absorbed and re-emitted". Photons are not completely absorbed and then re-emitted by a single atom, like you get when you cause fluorescence or something. See my longer explanation. So while you are correct to worry about random direction or energy in the case of classical particle absorption and re-emission, that's not what's happening.

> If one adds unspecific energy to the same piece of glass, i.e. melts it, it glows in yellow or white. Is there any way to make that glass emitting photons of a certain frequency except shining the right frequency into it?

If you had just a tiny amount, like a few atoms, of glass, you would pretty much only see photon emissions at their characteristic energy levels (associated with electron shells, vibrational modes, etc). But as you add more and more atoms, the modes get washed out and photons get absorbed and re-emitted within the glass itself many times before finally emerging (here I am talking about complete absorption and re-emission) such that the final spectrum always looks like black-body radiation. That's why objects of a certain temperature end up looking like certain colors.

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Treczoks t1_j8jvsaf wrote

> This is the wrong type of "absorbed and re-emitted". Photons are not completely absorbed and then re-emitted by a single atom, like you get when you cause fluorescence or something. See my longer explanation. So while you are correct to worry about random direction or energy in the case of classical particle absorption and re-emission, that's not what's happening.

That was exactly what I was wondering about. Thank you for the long explanation. So, basically, if a laser goes into the glass pane here it comes out there because this "there" is the most quantum-probably place, and the same with other parameters. Interesting approach, and it actually makes sense.

It is amazing to see the path that physics traverses through mathematics on the different layers. Basic algebra for laws of leverage, calculus when it comes to the relativistic stuff, and probability and information theory down below when things go quantum.

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BishoxX t1_j8ha8kg wrote

I saw an explanation which said since light is an EM wave it moves electrons when it passes. The wave of the electrons moving is a slower wave which when combined with light makes it slower.

Or something similar to that

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xanthraxoid t1_j8hw5p2 wrote

I think I'd take your statement on the fungibility of photons further to say that the "particle" is more illusory than the "wave".

I tend to think of seeing a particle as somewhat analogous to seeing an eddy in a flow of water*. The water (/electromagnetic field) is the "thing" and the eddy (/photon) is just an observable behaviour of that ground truth.

The wave behaviour of particles can matter for electrons, and atoms, and even surprisingly large molecules! As you look at larger and larger things, though, the downsides of treating them as particles become more and more irrelevant. One major reason for this is that the "wavelength" of a wavicle gets shorter in proportion to its momentum going up.

While a photon of visible light has a momentum in the region of 10^-27 Kg.m.s^-1 and a wavelength in the region of 10^-7 metres, Schrödinger's cat moseying along at walking pace has a momentum in the region of 1 Kg.m.s^-1 and a "wavelength" in the region of 10^-34 metres. That's about 1/1,000,000,000,000,000,000,000,000 the size of an atom - a little smaller than the size of the cat, and squirting a stream of cats through a 10^-34 metre slit would probably give you a somewhat messy version of the classic diffraction pattern (sorry, Tiddles!)

(Note, the numbers in that last paragraph are all essentially to zero significant figures and within the limits of my patience of counting zeros, but with those kinds of numbers, even a couple of orders of magnitude really doesn't make any difference :-P)

* A closer analogy would be waves in water, but eddies are easier to visualise as being distinct entities. To be fair, eddies can exhibit some properties (destroying / combining with each other / splitting) that can be entertainingly analogous to colliding subatomic particles, but probably not in ways that help my use of the analogy :-P

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Durable_me t1_j8i86ex wrote

>So, first answering your main question- elementary particles are all fungible. That means, they are truly identical, and they are impossible to label. So, if a photon is absorbed and then remitted, it doesn't really make sense to say "is it the same photon or a different one?" There aren't really "same" or "different" photons, there's just photons, unlabeled.
>
>And it's not just photons. Any time you have a particle collision which results in some different elementary particles (like the ones from particle accelerators), if one of the products and reactants are the same elementary particle, you can't answer "is this the same or a different particle?" It's a particle. That's all you can say.

So how does that coincide with the entanglement of two particles.?
These two particles are identified for sure....

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Atlantic0ne t1_j8h2ah0 wrote

A quality answer and even some better added mystery. Nice! Well done. So, why do physicists fight about that? Is it actually unsettled science?

Are there any other mysterious things in physics you can share?

My favorite thing I’ve been learning about lately (as a layman) is the double slit experiment, quantum mechanics, how particles behave differently when observed. I’ve been reading people say that one possible answer is the many worlds theory, and I haven’t heard other respected theories yet. Any thoughts?

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MasterPatricko t1_j8jak1s wrote

> A quality answer and even some better added mystery. Nice! Well done. So, why do physicists fight about that? Is it actually unsettled science?

It's not as mysterious or unsettled as portrayed -- rather it is a case of using non-specific language when trying to simplify for students or laypeople, leading to confusion. The math is exact and well tested. Have a look at my direct reply.

As for interpretations of quantum mechanics -- that one is unsettled. We know the math of QM works very well, but we have little idea what physical meaning (if any) to assign to a lot of the intermediate operations we do in a calculation. Ars Technica recently published a decent article on the topic.

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Atlantic0ne t1_j8jczlx wrote

Another quality reply, thank you.

So… if you’re in the mood, what do you rank as the #1 plausible reason for QM/particle behavior when observed? Have you heard of the many worlds theory?

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rexregisanimi t1_j8h6s8s wrote

>I always say, if you want to get some physicists to fight, ask them why light propagates slower through a non-vacuum.

I still remember an argument in my intro QM class about why photons slow in a medium lol

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aggasalk t1_j8h60y6 wrote

> So, first answering your main question- elementary particles are all fungible. That means, they are truly identical, and they are impossible to label. So, if a photon is absorbed and then remitted, it doesn't really make sense to say "is it the same photon or a different one?" There aren't really "same" or "different" photons, there's just photons, unlabeled.

Isn't there any sense in which, say, a photon flying through space at time t and then a moment later at time t+1 is "the same photon", and in which two photons flying in opposite directions at the same moment and the same point in space (with different energies, even) are "different photons"?

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MasterPatricko t1_j8javj0 wrote

Quantum mechanics is best understood with a solid grasp of classical wave mechanics.

If there were two water waves in different locations, it's easy to keep track of them, even if they momentarily cross. But if there were two travelling together in the same direction -- is that still two waves? If they then separate, which one is which? This is what indistinguishability means.

Photons are fundamentally just bumps in the global electromagnetic field. When the bumps are well-separated, we can say this is bump 'A' and this is bump 'B'. When they are close, or moving together, or interfering ... those labels are not possible.

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Keudn t1_j8hd0dv wrote

I've always taken the explanation that "light is a propagation of a wave in the electromagnetic field, and it propagates slower through a medium than a vacuum." So from my viewpoint, its not that the photon is slowing or being absorbed and re-emitted, its simply that the speed of light in a medium is less than a vacuum. Is there a reason this interpretation would be wrong, or just yet another way to think about it?

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lucidludic t1_j8hi3wt wrote

From my perspective you have only stated the observation, this isn’t an explanation and naturally leads to questions like, “why does light propogate slower through a medium?” or “why does the speed vary depending on the medium?”

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Weed_O_Whirler t1_j8ipjyz wrote

The problem is that according to General Relativity, a massless particle must always travel at c. So when dealing with the photon nature of light, a photon cannot slow down. But of course, we know light propagates slower through a medium, so somehow you have to reconcile these things.

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MasterPatricko t1_j8j6gkn wrote

> according to General Relativity, a massless particle must always travel at c.

That's special relativity, not general relativity (yes, GR includes SR, but you get what I mean).

The fundamental wrong assumption people make is that a wavepacket of the EM field -- a photon -- in a vacuum is somehow "the same" as the wavepacket of the EM field in a complex background of charged particles, i.e. a real material made of atoms.

To make an analogy (of limited range, please don't abuse it :)): no-one is surprised that a classical gravity wave in water and a wave in honey behave differently. Why is it surprising for an EM wave? /u/Keudn 's description is classical and doesn't explain how to calculate the speed in a medium, but aside from that missing depth, strictly correct.

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Weed_O_Whirler t1_j8j8j8m wrote

I specifically talked about how there are models where the photon becomes a quasiparticle in my original answer to the question, I was simply explaining to the person who asked why there are complications when viewing EM waves moving slower.

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MasterPatricko t1_j8j9h5k wrote

Right, and I didn't contradict you on that, I'm adding to your answer there. The only mistake you made here was you have confused GR and SR.

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NudeSeaman t1_j8hd6ui wrote

So are there really different photons or are all photons the same photon that we just perceiving as different photons ?

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Greenwashingmachine t1_j8hdtb5 wrote

A related question to stump a physicist is what does a polarised photon look like

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Shammah51 t1_j8hlvwp wrote

Might this be an effect we need quantum gravity to explain? My intuition is any gravitational effect at that scale would be too small to account for the observation, but tiny relativistic effects do have a way of stacking up in many particle systems.

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Fluglichkeiten t1_j8i6skc wrote

Great explanation! I never knew there were so many competing explanations for light slowing down in a medium. I’d always just thought of it as the mass of the material warping space a little, meaning that the photon is effectively travelling further. Of course I never studied physics properly.

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avcloudy t1_j8l55un wrote

I know this is literally exactly what you talked about, but the best thing you can say is that the real answer is complex. Light as little balls being absorbed and re-emitted is a surprisingly good solution mathematically, but we know it has to be an incomplete picture because light is also a wave.

There are also good practical examples of why this is an incomplete description: it completely fails to explain refraction and reflection. The interesting part to me is that a) the real mean path model is a really good mathematical model of the speed of light in a medium b) despite that not being the only thing that’s happening c) even though it is happening. It’s a very inertial mass = mass situation.

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Krail t1_j8v6zuf wrote

I've never managed to get a straight answer I can wrap my head around for why light slows down in refractive media. Turns out "we don't know for sure and there's lots of heated disagreement on it" was the actual answer I needed. Thank you.

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Leemour t1_j9110bn wrote

>The most common complaint is that an atom can only absorb very specific wavelengths, but light of all wavelengths is slowed down by materials

Yeah, but not slowed at the same rate (dispersion), which is where I think quantum treatment of the system may be more insightful. Truth is that there is a fundamental problem of trying to account for everything on the very small scale (scattering, absorption, phonon interactions etc.) to match observations on the large scale ( "simple" intensity and spectral measurements, maybe with a clock).

It's like having a large container of 100+ tennis balls, and then trying to predict where they all land if the container is flipped. I think it's unfair to dismiss the insights of the dynamics of a single tennis ball, but clearly it's not enough to predict how the group collectively behaves.

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Lorien6 t1_j8gz8bn wrote

What if photons have tiny little “fingerprints” where you could, with sensitive enough equipment, identify them?

Or like a serial number…

Would that work for identifying them each and seeing if it was the same?

−1

lucidludic t1_j8hiihq wrote

If you assume photons are uniquely identifiable, then of course you could identify them. But why should that assumption be correct?

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boxdude t1_j8gmxjg wrote

The physics of light slowing through a medium is well explained by Feynman in this lecture:

https://www.feynmanlectures.caltech.edu/I_31.html

It is a result of moving charges in the material and its effect on the field.

The photon absorption and re-emission model doesn't hold up as well as Feynman's explanation.

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gr7ace t1_j8hil80 wrote

Such a good physicist, critical thinker and author. Feynman books are an amazing read.

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leftoutoctopus t1_j8hn1ed wrote

This block of text felt too complex for a leyman like me to fully understand it as I read through, is there a "for idiots" version of it?

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common_sensei t1_j8hpj47 wrote

I like this Fermilab video for explaining light slowing in a medium: https://youtu.be/CUjt36SD3h8

You can think of it as a wave moving through water with a bunch of ping pong balls. As the wave lifts and drops the ping pong balls they resist the acceleration, and that makes a little inverse ripple within the bigger wave. The big wave and the little ripples stack together into a slower wave, but the energy doesn't change, so once the wave moves past the ping pong balls it goes back to the same speed and height.

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leftoutoctopus t1_j8hvp8a wrote

Incredible video, but I still don't understand how does a wave not loses energy as it passes through mediums, or the energy it loses is negligible since it has so little? What happens? I've understood something and now I understand even less.

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KillerCodeMonky t1_j8hzclh wrote

A wave doesn't contain energy, it is energy. And it would only lose that energy by doing work, same as anything else. So what work do you think it is doing when it passes through a medium?

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Stillcant t1_j8i24m2 wrote

Well the person above said the wave is resisted by the medium, and the wave moves the medium (the ping pong balls) as it passes. That sounds like work in the analogy used at least.

And could a photon not shift to a lower energy wavelength? Different photons have different energy I thought?

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KillerCodeMonky t1_j8i455z wrote

Here on earth, a ball resists being lifted due to gravity. It also returns all that energy when it's dropped again.

A wave that lifts a bunch of ping pong balls on it's leading edge, then drops them on it's trailing edge where the energy is returned to the wave, has done net zero work.

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grahampositive t1_j8ici68 wrote

This is where the analogy breaks down though. A water wave lifting a ping pong ball and returning it to it's initial position has lost energy. The ping pong ball pushes against air as it moves upwards, and the resulting ball-air collisions generate heat which is lost to entropy.

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KrombopulosThe2nd t1_j8i7auz wrote

Well here on earth, is it possible to lift anything with 100% efficiency? Shouldn't there be some loss of energy

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KillerCodeMonky t1_j8ia5jn wrote

Yes of course these are not perfect processes. Otherwise an undersea earthquake would create a tsunami on all of its coastlines. Energy is lost or made non-coherent in a variety of imperfections, including heat and scattering. In the case of ocean waves, they are typically created and recharged by the wind as they move along.

I mostly wanted to make the point that a wave is not displacing its medium, but simply moving through it. A wave is nothing but water and energy. There's nothing there displacing the water to do work. The water moves around due to the energy, but in a way which is generally neutral in terms of work actually done. A wave hitting the shoreline really just transfers from moving through the water to moving through the land.

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DamionFury t1_j8ijw6m wrote

Work can turn out to be one of the less intuitive aspects of physics. For example, magnetic fields cannot do any work because they act orthogonally to the direction of motion, yet it certainly looks like work when you use an electromagnet to lift an object and make it float. I wish I could remember the explanation my Electromagnetism professor gave me for what is actually doing the work in that scenario.

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Parrek t1_j8j1x62 wrote

Usually the battery or another power source maintaining the magnetic field. Otherwise, the back current induced by the object being picked up would cancel everything out

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DamionFury t1_j8j6cgr wrote

Ah yes. That's right. I also remember we talked about the case of permanent magnets and objects suspended above them. I don't remember what his answer was and I wouldn't be surprised if he told me to try to work it out for myself because my questions were holding the class up.

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kkngs t1_j8l4jhj wrote

If the object is just being suspended by the permanent magnet, being held still against the force of gravity, then no work is being done. It’s not conceptually different than being held up by a shelf.

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agoodpapa t1_j8jpira wrote

Not sure. The ping pong balls have been moved and therefore have absorbed energy.

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KillerCodeMonky t1_j8jr7mn wrote

Movement perpendicular to and away from the earth will convert to potential gravitational energy, which is then released when the ball moves back towards the earth.

In addition, you're assuming that the balls are passive participants. In actuality, from the waves perspective, they are actively resisting being lifted, and actively attempting to fall, limited by their buoyancy in the water. When the wave causes the water to fall away from the balls, they are falling into the water on that trailing edge.

Finally, conservation of energy dictates that that energy does not just disappear. If the energy goes into the ball as movement, something has to stop that movement. That something is going to be either gravity if moving up, or the water itself in any other direction. So the energy is moving from the water, into the ball, back into the water as it resists the ball displacing it to move.

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Monadnok t1_j8ij84r wrote

Someone below pointed out the gravity analogy. Another is a spring, where an object can do work on the spring, and then the spring can return the energy by doing work on the object.

As Feynmann's lecture points out, the electrons that are worked by the applied electric field of the photon can be considered to be "fastened" to the atoms of the materials by a "spring". So the incoming photon moves the electron against a "spring", and the spring returns the energy by working on the electron, which produces a new oscillating electric field to add to the field of the incoming photon.

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Dirty_Virmling t1_j8inomo wrote

Worth pointing out that electromagnetic waves do lose energy when traveling through naturally occurring materials.

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SemiDirectInsult t1_j8mqdsc wrote

How else would radiation raise the temperature of objects? Unless I’m fundamentally misunderstanding something, that’s energy transmission.

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leftoutoctopus t1_j8i04ax wrote

Well, passing through it. I mean, is anything really "free"? Changing the energy from one form (wave) to another doesn't "spend" part of itself?

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KillerCodeMonky t1_j8i2a4l wrote

Well, let's look at a wave in water. It moves some water around, but in a cyclical pattern that should result in very little actual work being done. The same way that lifting something gives you all the energy back when you drop it.

Maybe you're thinking of this from the perspective of a solid object moving through a medium, where it must spend energy on displacing the medium in order to move? This is where my note regarding the wave being energy is important. It does not have to displace anything to move through the medium, the same way that the heat from your stove doesn't displace your pan in order to heat your food. The energy simply goes through.

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DuskyDay t1_j8iio35 wrote

This can't be understood without math.

The wave doesn't lose energy as long as the crystal lattice is perfect, because the physics of our universe allows it to spread in a special way. (The photons don't interact with the particles the medium is made of.)

0

skyler_on_the_moon t1_j8iwbez wrote

I watched that whole video to see the ping pong ball experiment and am now disappointed.

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common_sensei t1_j8j1l8a wrote

Try it yourself!... although it probably won't work as described.

The wave would mostly regain the speed after the ping pong ball area, but it would be much reduced in height. A lot of energy would be lost to ripples going everywhere. My analogy just serves to illustrate the general concept of particles reacting to the approach and passing of the wave and generating interfering waves of their own.

However, a similar effect can be seen in wave tanks when you change the depth of the water: https://youtu.be/4_VejGC0DMM?t=261

In this case the interfering waves are bouncing from the new lower bottom of the tank, slowing down the wave.

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TruthOrFacts t1_j8i3b2i wrote

I don't see how this explanation is internally consistent. It is a behavior we see in other types of waves, but in those other cases the wave only appears to slow, and the actual wave proceeds at full speed.

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bandti45 t1_j8ia88q wrote

One thing to remember with photons is that while speed is consistent, intensity isn't.

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RailRuler t1_j8ixshe wrote

Intensity is a measurement of the number of photons arriving times the energy of each photo. The question is asking about a single photon, which is actually meaningful to speak of. Are you saying that the single photon's energy (wavelength/frequency) is changing?

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bandti45 t1_j8jc812 wrote

Well, the person I was replying to was talking about light in general, not a single photon. This is something I only have basic knowledge about.

Interactions with stuff does change it in one way or another. The way the sky is blue is light interacting with air, and tinted glass changes the light going through it. I don't know the mechanics, but that's a change in the photons without changing total speed.

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RailRuler t1_j9kjypo wrote

Those are both the medium filtering out certain colors of light, reducing its intensity. It doesn't change the photons.

The only thing that changes photons is https://en.wikipedia.org/wiki/Fluorescence

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bandti45 t1_j9l737f wrote

I feel like I should have gotten this concept a while ago... thank you for informing me.

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AdiSoldier245 t1_j8hue8s wrote

Honestly it's pretty abstract even if you know what's going on. Just think of it like light causing other waves that look mixed with light making it all muddled up and looking slower.

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DBNodurf t1_j8ifbxe wrote

Leyman: since ley is the Spanish word for law, is a leyman a lawman?

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shouldbebabysitting t1_j8jkzcj wrote

Feynman's book QED, The Strange Theory of Light and Matter is written without math. I highly recommend it if you want to understand light.

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ImpatientProf t1_j8hzdda wrote

Accelerating the charges in the material requires absorbing some energy from the light. Emission from the charges (the radiation field) requires depositing some energy back into the light. This is consistent with the photon absorption and re-emission model. Sure, it's not individual photons being absorbed by individual charges, but it's still an exchange of energy and the light is still quantized.

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MasterPatricko t1_j8icx22 wrote

> Sure, it's not individual photons being absorbed by individual charges,

You are correct overall but dismissing this specific point as if most people understand it easily is pedagogically dangerous -- based on my teaching experience that's exactly what a lot of people who hear "photon absorption and emission" end up thinking of.

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back_seat_dog t1_j8itvdf wrote

Exactly. I think people forget that the way the other interpret what you say is very important. Even if you are 100% correct, your wording can lead others to think something that is completely wrong. People aren't robots absorbing all information as it is, they interpret and change that information to align with preconceived notions and to make it easy to digest, and this can lead to misunderstanding.

The explanation is also problematic because this "absorption and emission" happens instantaneously, it doesn't absorb some energy, wait for a while, and release the energy. The release isn't isotropic either.

At the end of the day, it might be technically correct to say it is an absorption and re-emission, but it does more harm than good and doesn't really help clarify what is happening.

3

boxdude t1_j8i7en7 wrote

I work in optical engineering and am not an expert scientist in the nature of light. Our professional society SPIE held conferences (titled nature of light: what is a ohoton) over the last several years ( i believe there were 7 sessions over 7 years) where the experts argued out the nature of light and there is still disagreement amongst them on what a photon is and whether it exists.

Meanwhile, solving electromagnetic field components in the presence of optical components similar to Feynmans approach, whether it be a simple lens or photonic integrated circuits, well models actual behavior in all practical cases encountered with modern optical devices that I am aware of. I haven't seen any benefit to switching to a photon emission based model, especially when the experts cant even agree on the fundamental nature of the photon.

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_AlreadyTaken_ t1_j8ijqm7 wrote

If it induces the charged particles to move due to its field then those moving charged particles will also produce em waves.

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shouldbebabysitting t1_j8jkls1 wrote

>The photon absorption and re-emission model doesn't hold up as well as Feynman's explanation.

In his book QED, Feynman explained that he favored the photon absorption model but used field mathematics because it was the only method to deal with the vast quantities of photons.

"I want to emphasize that light comes in this form-particles. It is very important to know that light behaves like particles, especially for those of you who have gone to school, where you were probably told something about light behaving like waves. I'm telling you the way it does behave- like particles."

-Feynman, "QED The Strange Theory of light and Matter"

He then went on to explain everything about transmission, refraction and reflection as individual photon absorption and emissions.

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Leemour t1_j90ykbm wrote

Your explanation isn't relevant to the OP. You're using classical explanations, while OP asked how the photon behaves (i.e asked for the quantum model).

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jawshoeaw t1_j8gw48k wrote

First of all, Light doesn’t “slow down” because of absorption and re-emission. This is fairly easy to prove by simply choosing a wavelength of light that isn’t absorbed by that particular medium. It will still slow down or appear to do so. Interestingly you can in fact shoot single photons through a medium and they will do the same thing that millions of them do: change direction and appear to slow down. But what’s actually happening is that the photon is interacting with the electric fields of the atoms in the medium. A photon is still a wave and the wave is altered as it adds and subtracts with the electric fields around it. You can’t label a wave since it’s not a thing. Is the photon that comes out the other end the same photon? Sort of. But not really because light is really more imo a wave with some peculiar particle like properties. And it’s wave like properties are subject to addition and subtraction

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shouldbebabysitting t1_j8jljhy wrote

"I want to emphasize that light comes in this form-particles. It is very important to know that light behaves like particles, especially for those of you who have gone to school, where you were probably told something about light behaving like waves. I'm telling you the way it does behave- like particles."

Feynman, "QED The Strange Theory of Light and Matter"

2

cdstephens t1_j8gi364 wrote

I think it’s typically better to understand what’s happening at the classical level before the quantum level for questions like this.

Classically, when an electromagnetic wave enters a material, the material itself responds to the electromagnetic wave because it’s composed of charged particles. The collective oscillation of this macroscopic number of charged particles itself creates an electromagnetic field. The field that you can physically observe and measure is the total electromagnetic field. Through the superposition principle, the new total field will be moving slower, and you can analyze the properties of this total field.

The reason I point this out is that intuition about quantum physics breaks down. For instance, it doesn’t necessarily makes sense to label a photon a specific ID number; photon number is not conserved. Moreover, you cannot distinguish photons of the same energy from each other: to ask if it’s the “same” photon is thus not a meaningful question to ask. Not to mention that photons don’t have classical trajectories in the usual sense, and so on.

In particular, what we conceive as a “photon” is a freely propagating quantum of light in a vacuum, without undergoing interactions. But in a medium, light is clearly very strongly with the material. Indeed, the light in the medium is physically the result of the original light wave interacting with the material. So whatever quantum particle (which is really an excitation of a quantum field) you want to use to describe what’s happening won’t behave like “ordinary” photons.

Some people will even say that you shouldn’t of think of photons like physical objects you can touch and manipulate, but rather the footprint of a quantum mechanical interaction between the electromagnetic field and whatever it is you’re talking about.

Which is all to say: photons don’t act like classical billiard balls of light, and unlike electrons are purely relativistic, so ordinary non-relativistic quantum mechanics won’t work either.

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AxolotlsAreDangerous t1_j8fpyxj wrote

>It is my understanding that the light is slowed through the medium because photons are absorbed and then re-emitted repeatedly.

That doesn't line up with what we observe. If an object absorbs a photon, it doesn't "remember" the direction the photon came from and then emit a new photon continuing along the same path - it emits a photon in a completely random direction. Yet we see light travel in a straight line, only bending when moving between mediums.

I don't think you can get a satisfactory explanation for why light slows down by thinking about photons. Light is an electromagnetic wave that exerts a force on the electrons in a material, the electrons oscillate and produce their own electromagnetic waves; do the maths and you find that the sum of all of these waves is a single slower moving one.

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rayschoon t1_j8i0brp wrote

With the photon absorption and re-emission model, you’d expect that when a beam of light hits an object, that it would be scattered then

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sysKin t1_j8hzsvf wrote

When thinking of light moving, don't even think of it as individual photons. Instead, it's a probabilistic wave function, with energy corresponding to one photon and which evolves according to all possible interactions it might have.

As long as the wavefunction interacts with something that is not connected to you, it continues to evolve in its strange quantum way, possibly extending its probabilistic state to other objects. We then call it entanglement.

Finally, that wavefunction will interact with something connected to you (your eye, your detector...) and at that point, we say that wavefunction collapsed and you can count it as a photon. But before that - the only significance of "photon" is that total energy of the function was one photon.

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ImMrSneezyAchoo t1_j8hb6ub wrote

About the fungible particles thing - an interesting tidbit is that physicists had hints about this way before QM was formally developed. If you look up the derivation of the sackur-tetrode equation, they had to reduce by a factor which removed the states which could be accounted for by swapping particles (e.g. because of their uniqueness via a label). This correction turned out to be absolutely correct in deriving the entropy of an ideal gas (and more importantly, validating the extensiveness of entropy). Gibbs new about this a long time before it was formally resolved - see Gibbs' Paradox. Fascinating if you ask me

Edit: and this was roughly late 1800s, for context

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coldlasercookies t1_j8i0rpp wrote

My understanding is that the photons are not absorbed and re-emitted. The reason being that absorption and subsequent re-emission (ie the exciting of an atom and subsequent relaxing of the electron via spontaneous emission of a photon) is a random process, in that the photon is emitted in a direction completely uncorrelated to the direction it came in at. This is clearly not what the phenomena of refraction does, this would amount to light being dispersed in the material which is not what we observe.

This problem is actually a lot easier to conceptualise with classical electromagnetic waves. With classical em waves, refraction occurs due to the change in the dielectric constant in the material, which is essentially due to the ability of the molecules in the material to polarise when exposed to an electric field, which kind of "reduces" the effect of the field. Another way to think about it is the molecules in the material are producing their own fields via polarisation in response to the applied field that get summed with the incident wave (simple wave superposition) making it appear to propagate slower. However you want to conceptualise it, mathematically the result is the same, the light appears to propagate more slowly in the medium.

Okay back to photons. Photons aren't classical, but they are waves. Now I'm actually a little shaky on the best way to conceptualise this, and I may be outright wrong since it's been a while since I studied this stuff but I believe we can treat photons exactly as we would treat a classical electromagnetic wave when looking how it behaves as a wave, propagating through space and interacting with other fields. And so all of the above paragraph about em waves applied to photons as well. The only thing we have to keep in mind with photons is that they must come in discrete packets of energy, and they can interact with certain atoms to deposit this energy by exciting an electron in the atom (or more realistically interact with the bulk atom lattice to excite electrons in an essentially continuous band). Except in materials where refraction occurs (ie transparent materials), the bandgap is actually too large to allow this to occur, so absorption is not possible and the photon will not be destroyed, and instead pass through as an electromagnetic wave would.

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AxelBoldt t1_j8jn17r wrote

It's not useful to think of individual photons as having "id numbers", because they don't. Photons are accounting devices to keep track of the strength and direction of electromagnetic waves. It's like the dollars in your bank account: they don't have individual identity either.

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ScootysDad t1_j8lnw75 wrote

I'll leave it to someone else to show the math but the short answer is that light is a propagating electromagnetic wave. As it passes through an atom, an electron within feel the force of the light wave and oscillate which sets up its own electric field (albeit at a different frequency). Through superposition, we can than add up the sum of those waves to give a net wave which is slower than the originating light.

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Leemour t1_j910acq wrote

Short answer is we don't know if it is the same photon or not, because our useful models treat them as fields and waves for the most part. We treat light as a particle when it's interacting with matter, but not when it's propagating. Whether we'd find the same unique ID or not is something yet to be seen maybe one day.

>Are all photons at the same wavelength identical so that it just doesn't make any sense to ask this question or are there some properties that are effectively randomized each time it is re-emitted?

This depends on the laser, and I don't think it's wise to write a whole wall of text to explain the ways in which a laser may emit photons with the same energy, but other factors may be different and make the laser better or worse for the intended application. Given the nature of lasers (and any radiating black body) we can't produce a laser that emits photons that all have the same energy; we can roughly select for a desired wavelength and it's far more efficient than with conventional light sources, but still we have for every laser emission a bandwidth. Depending on whether you want a light pulse with extreme short time duration or extremely high temporal coherence, you'll pick broad bandwidth or narrow bandwidth respectively. Unfortunately, as of now this is not a simple switch we flip, it requires the addition or removal of many optical components to the point that it's a total overhaul of the system.

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jimnrd t1_j8hjsue wrote

In a medium that slows down the speed of light, such as diamond, glass, or water, the same photon that enters at point A does not exit at point B. The absorption and re-emission process in the medium causes the photon's direction to change, and the photon's original energy and phase can be affected.

In terms of wavelength, photons of the same wavelength are identical in terms of their properties such as energy, momentum, and frequency. However, when a photon is absorbed and re-emitted in a medium, the phase and direction can be randomized, resulting in a loss of coherence between the original photon and the re-emitted photon.

This effect is known as scattering.

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