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They are not the same photons. They are emitted by the electron changing energy states after excitation of the incoming photon. But your intuition that there is

They are not the same photons. They are emitted by the electron changing energy states after excitation of the incoming photon. But your intuition that there is some quantum weirdness is correct. Quantum Electrodynamics is the theory you're looking for. It unifies quantum mechanics and special relatively in the context of light/matter interaction. Given you seem to be familiar with Feynman. You're in for a treat! His book on QED is meant to be an accessible explanation.[1] [1]https://en.m.wikipedia.org/wiki/QED:_The_Strange_Theory_of_Light_and_Matter https://en.m.wikipedia.org/wiki/QED:_The_Strange_Theory_of_L...

whatshisface 7y ago
Hydrogen atoms can't store photons that do not have the same wavelength as any of its spectral lines. Nonetheless, hydrogen gas has an index of refraction that is a continuous function of frequency, and as a result it can reflect light of any wavelength. Therefore, hydrogen does not store any energy for any length of time in the course of typical reflection. Even if the Feynman diagram for reflection looks like a combination of absorption and re-emission, if the absorption and emission process can't occur at separate times, then it is not right to say that one is followed by the other.
- boomboomsubban 7y ago
  Isn't hydrogen gas almost exclusively H2, giving it plenty of possible electrons to interact with photons?
  - whatshisface 7y ago
    That's a good catch, but H2's spectral lines are still discrete. It's not that gasses can't absorb photons, it's that they can only do so at very specific frequencies. To simplify the argument, you could use Helium instead. Here is a (discrete!) list of its spectral lines: https://physics.nist.gov/PhysRefData/Handbook/Tables/heliumtable2.htm https://physics.nist.gov/PhysRefData/Handbook/Tables/heliumt... And here is a chart of its (continuous!) index of refraction: https://refractiveindex.info/?shelf=main&book=He&page=Mansfield https://refractiveindex.info/?shelf=main&book=He&page=Mansfi...
empath75 7y ago
> They are not the same photons. How do you define "the same" for two photons?
- tsbinz 7y ago
  This is the question that should be higher up in the thread. This whole discussion makes no sense. Photons carry no hidden identity.
  - zamadatix 7y ago
    A photon doesn't need to carry a UUID for people to conceptualize a difference between a photon going through free space and being absorbed and re emitted. The parent comment asks the right question, what is being meant by "the same" is required to know before being able to say the discussion makes no sense. Taking your first literal interpretation as the only way the discussion could have value is no more helpful than this exchange: "I'm going to the store, do you want the same hot dogs we have in the fridge?" "That doesn't make sense, the store can't have the same hot dogs we have in the fridge".
    - izietto 7y ago
      > "I'm going to the store, do you want the same hot dogs we have in the fridge?" "That doesn't make sense, the store can't have the same hot dogs we have in the fridge" Yeah but hot dogs belong to classical mechanics, while photons to quantum mechanics, and in quantum mechanics you can actually have the same hot dogs in different places, until you observe them
      - deleted 7y ago
        [deleted]
      - zamadatix 7y ago
        That a photon can be in superposition is even a better example than my hot dog example. Nobody is impressed if you can find a corner case interpretation that makes the question sound dumb they are impressed if you can try to figure out what the person really wanted to know and answer it to the best of current understanding. In the hotdog example the point is the person is actually trying to figure out if you want them to buy you more hotdogs not whether or not the exact same hotdogs can be found at the store and fridge at the same time. In this actual discussion about the photon it's not about whether or not the photon actually carries an ID card with it it's about what actual interactions are happening to photons when the light is reflected or refracted in different ways. Sometimes what a person is trying to figure out may not line up with how things actually work, in fact take any science and add 100 years of knowledge and pretty much all discussions are this way. That does not mean it was nonsensical to talk about it just means the details are more exact and complicated than the asker knew how to phrase (and again, it's always like more exact and complicated than any of us know as we don't know everything there is to know about the universe).
BeetleB 7y ago
You didn't address the concern. A given atom cannot absorb any photon. They can only absorb photons of certain energy levels/wavelengths. Electrons can occupy only certain energy levels. So they can only absorb photons that match the difference if two energy levels.
viklove 7y ago
But that's just not true. The electron will only get excited if the incoming photon has a very specific wavelength, so most of the light that hit your house's walls, for example, is reflecting. The photoelectric effect is real, but it only applies in specific circumstances...
- chemeng 7y ago
  There will be a transfer of energy to the system if there is an interaction, but will only absorb the incident photon at specific energies. In the case of reflection in this classical view, it simply re-emits the energy back out as there are no valid energy transitions. The reality is far more complex and this thread is mixing multiple theories together, wave and particle views and simultaneously comparing single atom mechanics with those of lattices/solids. This is all causing a lot of confusion that's hard to address in comments.

whatshisface 7y ago

Hydrogen atoms can't store photons that do not have the same wavelength as any of its spectral lines. Nonetheless, hydrogen gas has an index of refraction that is a continuous function of frequency, and as a result it can reflect light of any wavelength. Therefore, hydrogen does not store any energy for any length of time in the course of typical reflection. Even if the Feynman diagram for reflection looks like a combination of absorption and re-emission, if the absorption and emission process can't occur at separate times, then it is not right to say that one is followed by the other.

boomboomsubban 7y ago
Isn't hydrogen gas almost exclusively H2, giving it plenty of possible electrons to interact with photons?
- whatshisface 7y ago
  That's a good catch, but H2's spectral lines are still discrete. It's not that gasses can't absorb photons, it's that they can only do so at very specific frequencies. To simplify the argument, you could use Helium instead. Here is a (discrete!) list of its spectral lines: https://physics.nist.gov/PhysRefData/Handbook/Tables/heliumtable2.htm https://physics.nist.gov/PhysRefData/Handbook/Tables/heliumt... And here is a chart of its (continuous!) index of refraction: https://refractiveindex.info/?shelf=main&book=He&page=Mansfield https://refractiveindex.info/?shelf=main&book=He&page=Mansfi...

empath75 7y ago

> They are not the same photons. How do you define "the same" for two photons?

tsbinz 7y ago
This is the question that should be higher up in the thread. This whole discussion makes no sense. Photons carry no hidden identity.
- zamadatix 7y ago
  A photon doesn't need to carry a UUID for people to conceptualize a difference between a photon going through free space and being absorbed and re emitted. The parent comment asks the right question, what is being meant by "the same" is required to know before being able to say the discussion makes no sense. Taking your first literal interpretation as the only way the discussion could have value is no more helpful than this exchange: "I'm going to the store, do you want the same hot dogs we have in the fridge?" "That doesn't make sense, the store can't have the same hot dogs we have in the fridge".
  - izietto 7y ago
    > "I'm going to the store, do you want the same hot dogs we have in the fridge?" "That doesn't make sense, the store can't have the same hot dogs we have in the fridge" Yeah but hot dogs belong to classical mechanics, while photons to quantum mechanics, and in quantum mechanics you can actually have the same hot dogs in different places, until you observe them
    - deleted 7y ago
      [deleted]
    - zamadatix 7y ago
      That a photon can be in superposition is even a better example than my hot dog example. Nobody is impressed if you can find a corner case interpretation that makes the question sound dumb they are impressed if you can try to figure out what the person really wanted to know and answer it to the best of current understanding. In the hotdog example the point is the person is actually trying to figure out if you want them to buy you more hotdogs not whether or not the exact same hotdogs can be found at the store and fridge at the same time. In this actual discussion about the photon it's not about whether or not the photon actually carries an ID card with it it's about what actual interactions are happening to photons when the light is reflected or refracted in different ways. Sometimes what a person is trying to figure out may not line up with how things actually work, in fact take any science and add 100 years of knowledge and pretty much all discussions are this way. That does not mean it was nonsensical to talk about it just means the details are more exact and complicated than the asker knew how to phrase (and again, it's always like more exact and complicated than any of us know as we don't know everything there is to know about the universe).

BeetleB 7y ago

You didn't address the concern. A given atom cannot absorb any photon. They can only absorb photons of certain energy levels/wavelengths. Electrons can occupy only certain energy levels. So they can only absorb photons that match the difference if two energy levels.

viklove 7y ago

But that's just not true. The electron will only get excited if the incoming photon has a very specific wavelength, so most of the light that hit your house's walls, for example, is reflecting. The photoelectric effect is real, but it only applies in specific circumstances...

chemeng 7y ago
There will be a transfer of energy to the system if there is an interaction, but will only absorb the incident photon at specific energies. In the case of reflection in this classical view, it simply re-emits the energy back out as there are no valid energy transitions. The reality is far more complex and this thread is mixing multiple theories together, wave and particle views and simultaneously comparing single atom mechanics with those of lattices/solids. This is all causing a lot of confusion that's hard to address in comments.