Why doesn't the atom absorb heat energy when it is low?

[thaiqi]

(In my last thread)
Mentor Dale said:
"
An atom in the ground state can absorb energy from the environment including thermal radiation.
Once it has done so it will be excited and will no longer be in the ground state. An excited atom
can radiate and go to a lower energy state, but an atom in the ground state cannot radiate since
there is no lower energy state.
"
I then asked:
"
But how if the energy it absorbed is lower than that of the gap between the first excited state and ground state?
"
Dale said:
"
Then it is transparent to that radiation and cannot absorb it.
"
Now I have my question: Why cannot it absorb it ?

 

[Dale]

Just to confirm, I did recommend asking this question in the public forums

 

[thaiqi]

Sorry I am new to this forum. Did you mean public forum as another sub-forum different from General Physics? If so, please move it.

 

[Nugatory]

Now I have my question: Why cannot it absorb it ?

Because if it absorbed it, then its total energy would be neither the ground state energy nor the energy of the lowest excited state but somewhere in between, and that's physically impossible.
If you are trying to ask why the atom can only have those energy levels and not something in between, we'll need some quantum mechanics and even the most minimum answer is going to go beyond a B-level thread: google for "Schrodinger hydrogen atom" to get started.

But note that we're talking about radiation here. In your thread title you asked about heat energy, and it never makes sense to talk about a single atom absorbing heat energy. Heat energy is (loosely speaking) the kinetic energy of a large number of molecules moving at random.

 

[Dale]

Sorry I am new to this forum. Did you mean public forum as another sub-forum different from General Physics? If so, please move it.

No worries, you did correctly!

 

[thaiqi]

No worries, you did correctly!

Thanks.

 

[thaiqi]

Because if it absorbed it, then its total energy would be neither the ground state energy nor the energy of the lowest excited state but somewhere in between, and that's physically impossible.
If you are trying to ask why the atom can only have those energy levels and not something in between, we'll need some quantum mechanics and even the most minimum answer is going to go beyond a B-level thread: google for "Schrodinger hydrogen atom" to get started.

But note that we're talking about radiation here. In your thread title you asked about heat energy, and it never makes sense to talk about a single atom absorbing heat energy. Heat energy is (loosely speaking) the kinetic energy of a large number of molecules moving at random.

Thanks. You are more correct about the title should be "thermal radiation energy". By my question, I mean why I can't regard that the atom absorbs the energy and at the same period radiates out the same amount?

 

[Dale]

Why cannot it absorb it ?

When you solve the Schrodinger equation for an atom, one of the things that naturally falls out of the solution is that only discrete energy levels are possible. If an atom is exposed to radiation for which it has a corresponding energy level then it can absorb it and transfer to an excited state. Conversely, if no corresponding state exists then it cannot transfer and therefore cannot absorb the radiation. This is the cause of the absorption lines in gas spectra.

 

[thaiqi]

When you solve the Schrodinger equation for an atom, one of the things that naturally falls out of the solution is that only discrete energy levels are possible. If an atom is exposed to radiation for which it has a corresponding energy level then it can absorb it and transfer to an excited state. Conversely, if no corresponding state exists then it cannot transfer and therefore cannot absorb the radiation. This is the cause of the absorption lines in gas spectra.

Can I regard that the atom absorbs the energy and at the same period radiates out the same amount?

 

[anorlunda]

Can I regard that the atom absorbs the energy and at the same period radiates out the same amount?

Are you trying to ask about scattering of light?

 

[thaiqi]

Are you trying to ask about scattering of light?

Here I didn't relate it to the scattering of light. Why do you think there is connection between them?

 

[Dale]

Can I regard that the atom absorbs the energy and at the same period radiates out the same amount?

No. The atom cannot absorb the energy. There is nowhere in the atom for it to go. Such a scheme would therefore not conserve energy.

 

[thaiqi]

Do you mean you regard that if it absorbs and radiates in the same period, the energy it absorbs and radiates out cannot balance ?

 

[Dale]

Do you mean you regard that if it absorbs and radiates in the same period, the energy it absorbs and radiates out cannot balance ?

If it absorbs energy then that means that the energy enters into the atom. There is nowhere for the energy to go, so it cannot be absorbed.

It sounds like you are trying to redefine the word “absorb” such that it is possible for an atom to “absorb” energy without that energy entering the atom. This is not what “absorb” means. When an atom absorbs energy its energy level increases.

 

[thaiqi]

I guess it absorbs under the thermal agitation in its environment, and radiates out as classical electrodynamics said when it is accelerating. The net result is zero. You could say the net absorption is zero.

 

[Dale]

I guess it absorbs under the thermal agitation in its environment

Only if the thermal agitation has an energy corresponding to an available transition. If not, then it cannot absorb the energy.

At this point it seems like you have an agenda you wish to push and are not here to learn. Please review the forum rules which prohibit personal speculation.

 

[thaiqi]

So you regard that contrary to classical electrodynamics, even it is accelerating, it doesn't radiate electromagnetic wave ?

 

[Dale]

So you regard that contrary to classical electrodynamics, even it is accelerating, it doesn't radiate electromagnetic wave ?

The atom is neutral. According to classical electrodynamics it won’t radiate because it accelerates.

 

[sophiecentaur]

When you solve the Schrodinger equation for an atom, one of the things that naturally falls out of the solution is that only discrete energy levels are possible.

It's worth pointing out that the familiar Hydrogen Atom equation (one proton and one electron) shows the first Energy Level above ground state to be around 10eV (iirc) which is an UV transition. Thermal frequencies are associated with molecular transitions so the H atom model is not the best to hold in your mind when discussing this. The states associated with molecular vibration are more what you want but the model is not so familiar. The same principle applies, of course.

 

[Dale]

It's worth pointing out that the familiar Hydrogen Atom equation (one proton and one electron) shows the first Energy Level above ground state to be around 10eV (iirc) which is an UV transition. Thermal frequencies are associated with molecular transitions so the H atom model is not the best to hold in your mind when discussing this

Agreed, which is also why many gasses are transparent to IR or visible light.

 

[thaiqi]

The atom is neutral. According to classical electrodynamics it won’t radiate because it accelerates.

Well, I meant it radiates through the electron outside of the nucleus.

 

[Dale]

Well, I meant it radiates through the electron outside of the nucleus.

The electrons and the nucleus accelerate together

 

[thaiqi]

The electrons and the nucleus accelerate together

I mean that the electron rotates around the nucleus.

 

[Dale]

I mean that the electron rotates around the nucleus.

The observation that this radiation does not happen is one of the observations that falsifies classical electromagnetism

 

[sophiecentaur]

I mean that the electron rotates around the nucleus.

That's a very old model by Neils Bohr - pre Quantum Theory, iirc. Read about it on Wiki.

 

[thaiqi]

The observation that this radiation does not happen is one of the observations that falsifies classical electromagnetism

So it is contrary to what classical electrodynamics said?

 

[sophiecentaur]

So it is contrary to what classical electrodynamics said?

Classical theory does not include the Quantisation of Energy and it would allow for all wavelengths to interact with all charge systems.

 

[Dale]

So it is contrary to what classical electrodynamics said?

Yes. The classical electrodynamics calculations give a lifetime for the hydrogen atom on the order of 10^-11 s. Since the hydrogen ground state is stable this is contrary to classical electrodynamics

 

[Herman Trivilino]

By my question, I mean why I can't regard that the atom absorbs the energy and at the same period radiates out the same amount?

Can you think of an experiment that would distinguish between the absorption and re-emission versus no absorption at all? If you can't, then you are talking about a distinction without a difference. You are just arguing semantics, that is, the definition of what it means to absorb.

When an atom absorbs a photon, its energy level goes up by an amount equal to the energy of the photon. There are experiments that measure this increase in energy. In every case, the increase equals the difference in energy levels of two states of the atom. If the photon energy doesn't match this energy difference, then the atom's energy level doesn't change. We describe that by saying that the photon is not absorbed. If you want to insist that we are not using the same meaning of absorb as you are, then all I can tell you is that the meaning we are using matches the meaning used in the textbooks and in the literature. The meaning that you are using doesn't, and I hope you can appreciate that the rest of the world will not change its meaning to match yours.

 

[sophiecentaur]

Can you think of an experiment that would distinguish between the absorption and re-emission versus no absorption at all?

I can. Absorption and re-emission introduces random phase delays where straight scattering will not randomise the phase. There are many experiments that could show a difference in coherence of what comes out, according to the mechanism.

 

[thaiqi]

Thanks to everyone for your opinion.

 

[DanMP]

Absorption and re-emission introduces random phase delays

Why? Please elaborate.Huygens principle states that "each point on wavefront act as a fresh source of distribution of light". Are those "fresh sources" introducing random phase delays?

 

[sophiecentaur]

Why? Please elaborate.Huygens principle states that "each point on wavefront act as a fresh source of distribution of light". Are those "fresh sources" introducing random phase delays?

When an atom goes into an excited state, there is a random element in the time it takes to re-emit its photon so the coherence is destroyed. The only time this effect doesn't happen is in a laser, when there is stimulated emission. But that only happens (to any significant degree) when you have a Population Inversion. It's a long time since I learned this but it only happens under the 'right conditions, only in some molecules and when vast numbers of the atoms are in the excited state.
So this doesn't constitute Huygens sources, which is when the whole wave interacts with the bulk of the substance, rather than with individual atoms. These Huygens sources are not 'real particles' but a mathematical construct, invented before Calculus was available for diffraction calculations and are much the same thing (afaiaa).

 

[sophiecentaur]

If it worked by absorption and re-emission, there would be no Rectilinear Propagation, except in a vacuum.

 

[DanMP]

When an atom goes into an excited state, there is a random element in the time it takes to re-emit its photon so the coherence is destroyed.

True, but what if the atom/molecule goes into a virtual state (more probable than into an excited state, when the material is transparent ...)? In this case it is not a real absorption but a failed one, always and promptly followed by the re-emission ... It would be almost like with the Huygens sources ...

If it worked by absorption and re-emission, there would be no Rectilinear Propagation, except in a vacuum.

You are forgetting the Fresnel part in Huygens–Fresnel principle ...

 

[sophiecentaur]

True, but what if the atom/molecule goes into a virtual state (more probable than into an excited state, when the material is transparent ...)? In this case it is not a real absorption but a failed one, always and promptly followed by the re-emission ... It would be almost like with the Huygens sources ...
.

You are assuming that the propagation is based on single atom interactions. That only happens in low density gases, in which photons are actually absorbed by specific atomic energy transitions. When that happens, the re-emitted light goes in all directions and the result is a dark 'absorption line'. Why does it go in all directions? Because there is no longer phase coherence in the original propagation direction.
PS is all this stuff strange to you? It's pretty basic. Your above argument contains a "What if" and an "almost", as if there were some doubt about the validity of what we get in basic textbooks.

 

[DanMP]

You are assuming that the propagation is based on single atom interactions. That only happens in low density gases, in which ...

And what happens (with the photons) in air and water? How they propagate? They don't interact with the atoms/molecules? [Please exclude the absorption due to atomic/molecular energy transitions, because I'm talking about transparent materials with no (or very few) absorption lines.]

 

[sophiecentaur]

And what happens (with the photons) in air and water? How they propagate? They don't interact with the atoms/molecules? [Please exclude the absorption due to atomic/molecular energy transitions, because I'm talking about transparent materials with no (or very few) absorption lines.]

They interact with the whole lattice of atoms by Elastic Scattering and not by Absorption.
Personally, I find the wave model much easier to cope with here. If you want to talk in terms of individual photons then, if a photon isn't absorbed by a particular atom (= measured / observed) then it has no location and then you have to consider all the possible paths it could take through the medium (being elastically scattered off all the atoms in the material). This boils down to the same calculation as if you started off considering waves in the first place.

 

[Herman Trivilino]

And what happens (with the photons) in air and water? How they propagate? They don't interact with the atoms/molecules? [Please exclude the absorption due to atomic/molecular energy transitions, because I'm talking about transparent materials with no (or very few) absorption lines.]

They very much do interact with the molecules.

As an aside, can you explain why the sky is blue?

 

[DanMP]

They interact with the whole lattice of atoms by ...

What "lattice of atoms" in air? Can you provide a link about this?

They very much do interact with the molecules.

I know. My job is to record [IR &] Raman spectra ... That's why I suggested virtual states ...

As an aside, can you explain why the sky is blue?

Yes, blue light is scattered more than red light. It is also the reason for the red sun at the horizon.

 

[sophiecentaur]

What "lattice of atoms" in air? Can you provide a link about this?

I was being sloppy and referring to transparent solids and liquids. Gases are transparent and very low density, nonetheless, the scattering is still elastic and involves many atoms and not just one atom per photon (that's a sort of definition of the word elastic, I think you could say.)

My job is to record [IR &] Raman spectra

And Raman scattering is Inelastic, so it does involve interaction with individual molecules. It involves absorption and re-emission so that would introduce random phase shifting which would reduce the coherence of the idealised plane wave. The fact that you are looking at Spectral detail tells you that things are different from Rayleigh scattering, which is just dependent on particle size, with a small 'tilt' over the optical band.

 

[DanMP]

And Raman scattering is Inelastic, so it does involve interaction with individual molecules. It involves absorption and re-emission so that would introduce random phase shifting which would reduce the coherence of the idealised plane wave. The fact that you are looking at Spectral detail tells you that things are different from Rayleigh scattering, which is just dependent on particle size, with a small 'tilt' over the optical band.

In wikipedia there is a diagram:

showing/suggesting that Rayleigh scattering (the most dominant of the 3 scatterings above) involves also a short "transition" to a virtual energy state, meaning that there is a very short-lived "absorption" immediately and always followed by re-emission. That's why I wrote this.

 

[sophiecentaur]

In wikipedia there is a diagram:

showing/suggesting that Rayleigh scattering (the most dominant of the 3 scatterings above) involves also a short "transition" to a virtual energy state, meaning that there is a very short-lived "absorption" immediately and always followed by re-emission. That's why I wrote this.

I guess the proof of this would be in the shape of the emerging beam pattern which would have to be affected by even a small phase uncertainty. It sounds an interesting phenomenon.

 

[DrClaude]

showing/suggesting that Rayleigh scattering (the most dominant of the 3 scatterings above) involves also a short "transition" to a virtual energy state, meaning that there is a very short-lived "absorption" immediately and always followed by re-emission.

Be careful here. Virtual transitions are virtual, meaning they are not real. It is a coherent process where "absorption" and "emission" take place at the same time. It comes from a perturbative approach, just as virtual particles do (a subject that has been discussed endlessly on PF).

 

[Herman Trivilino]

Virtual transitions are virtual, meaning they are not real. It is a coherent process where "absorption" and "emission" take place at the same time. It comes from a perturbative approach, just as virtual particles do (a subject that has been discussed endlessly on PF).

But it provides the quantum mechanical explanation of why the sky is blue?

 

[sophiecentaur]

But it provides the quantum mechanical explanation of why the sky is blue?

There is a perfectly good explanation that doesn't involve QM so why introduce something like virtual particles if they are not necessary? There is the temptation to ask 'but what is really happening?' but that's not Physics. Sometimes you can get a good prediction from calculations using virtual particles so it clearly works but are you any nearer to 'the truth'?
"The TRUTH? You can't handle the TRUTH" - brilliant film.

 

[DanMP]

I guess the proof of this would be in the shape of the emerging beam pattern which would have to be affected by even a small phase uncertainty. It sounds an interesting phenomenon.

As far as I know, there is no problem with the phase/coherence through air :)

If you find the idea/phenomenon interesting and want/need a proof for it, I suggest to try/test it for the Fizeau experiment and the Sagnac effect. [I did it and I was amazed ... and very happy with the results]

why introduce something like virtual particles if they are not necessary?

What virtual particles? It was about virtual states ... And they are both already introduced and useful.

There is the temptation to ask 'but what is really happening?' but that's not Physics.

Well, for me it is. I'm interested to find "what is really happening".

Be careful here. Virtual transitions are virtual, meaning they are not real. It is a coherent process where "absorption" and "emission" take place at the same time. It comes from a perturbative approach, just as virtual particles do

Can you elaborate on this and/or provide a link? In wikipedia I found:

a virtual state is a very short-lived, unobservable quantum state

they still have lifetimes derived from uncertainty relations

 

[DrClaude]

I looked through many sources, but I didn't find any I could cite that would give a clear justification of what I said. But it is clear that these virtual states (or virtual transitions, see below) appear only through perturbation theory (hence to comparison with virtual particles). That said, I will try to cite different sources that I hope will clarify things.

(I will emphasise in red. All other emphasis is in the original.)

[Section 11.6 Raman spectroscopy] In the energy level scheme of Fig. 11.69 the intermediate state $E_v = E_k + \hbar \omega_0$ during the scattering process is often formally described as a virtual state, which, however, is not a real stationary eigenstate of the molecule. If this virtual state coincides with one of the molecular eigenstates one speaks of the "resonance Raman effect".

[Appendix E: Raman and two-photon transitions] A Raman transition involves two laser beams with frequencies $\omega_{\mathrm{L}1}$ and $\omega_{\mathrm{L}2}$ [...] A Raman transition between two atomic levels, labelled 1 and 2, involves a third atomic level, as shown in Fig. 9.20. This third level is labelled $i$ for intermediate, but it is very important to appreciate that atoms are not really excited to level $i$.
[...]
It is vital to realize that the Raman transition has a quite distinct nature from a transition in two steps, i.e. a single-photon transition from level 1 to $i$ and then a second step from $i$ to 2. The two-step process would be described by rate equations and have spontaneous emission from the real intermediate state. This process is more important than the coherent Raman process when the frequency detuning $\Delta$ is small so that $\omega_{\mathrm{L}1}$ matches the frequency of the transition between $|1 \rangle$ and $| i \rangle$.⁶ The distinction between a coherent Raman process (involving simultaneous absorption and stimulated emission) and two single-photon transitions can be seen in the following example.

[Section 5.7 Time-Dependent Perturbation Theory] We visualise that the transition due to the second-order term takes place in two steps. First $|i\rangle$ makes an energy non-conserving transition to $|m\rangle$; subsequently, $|m\rangle$ makes an energy-nonconserving transition to $|n\rangle$, where between $|n\rangle$ and $|i\rangle$ there is overall energy conservation. Such energy non-conserving transitions are often called virtual transitions. Energy need not be conserved for those virtual transitions into (or from) virtual intermediate states. In contrast, the first-order term $V_{ni}$ is often said to represent a direct energy conserving "real" transition.

[Section 6.2 Many-level systems, (c) The first-order correction to the wavefunction]
The last equation echoes the expression derived for the two-level system in the limit of a weak perturbation and widely separated energy levels (eon 6.18). As in that case, perturbation theory guides us towards the form of the perturbed state of the system. In this case, the procedure simulates the distortion of the state by mixing into it the other states of the system. This mixing is expressed by saying that the perturbation induces virtual transitions to these other states of the model system. However, that is only a pictorial way of speaking: in fact, the distorted state is being simulated as a linear superposition if the unperturbed states of the system.

I have also found a source that eschews virtual states and transitions:

[Section 4.3 Extending the perturbation approximation] It is very important to emphasise that the language of "intermediate states" or "virtual states" has no place in this discussion. Any of these non-linear processes occur while the driving radiation is not resonant with any of the states $|j\rangle$.

Finally, I found one source that tackles both Raman and Ramsey scattering at the same time. Note that explanation is semi-classical.

[Complement A_V 1.c.β The Raman effect]
Imagine that an optical wave of frequency $\Omega/2\pi$ strikes this molecule. This frequency, much higher than those considered previously, is able to excite the electrons of the molecule; under the effect of the optical wave, the electrons will undergo forced oscillation and re-emit radiation of the same frequency in all directions. This is the well-known phenomenon of the molecular scattering of light (Rayleigh scattering). What new phenomena are produced by the vibration of the molecule?

What happens can be explained qualitatively in the following way. The electronic susceptibility of the molecule is generally a function of the distance $r$ between the two nuclei. When $r$ varies (recall that this variation is slow compared to the motion of the electrons), the amplitude of the induced electric dipole moment, which vibrates at a frequency of $\Omega/2\pi$, varies. The time dependence of the dipole moment is therefore that of a sinusoid of frequency $\Omega/2\pi$ whose amplitude is modulated at the frequency of the molecular vibration $\omega/2\pi$, which is much smaller (fig. 3). The frequency distribution of the light emitted by the molecule is given by Fourier transform of the motion of the electric dipole shown in figure 3. It is easy to see (fig. 4) that there exists a central line of frequency $(\Omega - \omega) /2\pi$ (Raman-Stokes scattering) and frequency $(\Omega + \omega) /2\pi$ (Raman-anti- Stokes scattering).

It is very simple to interpret thesis lines in terms of photons. Consider an optical photon of energy $\hbar \Omega$ which strikes the molecule when it is in the state $|\varphi_v\rangle$ (fig. 5-a). If the molecule does not change vibrational state during the scattering process, the scattering is elastic. Because of conservation of energy, the scattered photon has the same energy as the incident photon (fig. 5-b; Rayleigh line). However, the molecule, during the scattering process, can make a transition from the state $|\varphi_v\rangle$ to the state $|\varphi_{v+1}\rangle$. The molecule acquires an energy $\hbar \omega$ at the expense of the scattered photon, whose energy therefore is $\hbar(\Omega - \omega)$ (fig. 5-c): the scattering is inelastic (Raman-Stokes line). Finally, the molecule may move from the state $|\varphi_v\rangle$ to the state $|\varphi_{v-1}\rangle$, in which case the scattered photon will have an energy of $\hbar(\Omega + \omega)$ (fig. 5-d; Raman-ani-Stokes line).

(note the description in terms of the scattering of a photon, not absorption-emission)

 

[DanMP]

I looked through many sources, but I didn't find any I could cite that would give a clear justification of what I said. But it is clear that these virtual states (or virtual transitions, see below) appear only through perturbation theory (hence to comparison with virtual particles). That said, I will try to cite different sources that I hope will clarify things.

Thank you very much for your interest and effort.

After reading all the material you offered, I'm not yet convinced that

"absorption" and "emission" take place at the same time

(note the description in terms of the scattering of a photon, not absorption-emission)

True, but this is probably because we usually say absorption/emission when we deal with transitions to/from a real energy state, not a virtual one. That's why I/we wrote "absorption"/"emission" (and "transition").In your last quote I found:

the molecule, during the scattering process, can make a transition from the state ...

This suggests that there is a process that takes time, probably very close to zero, but still different from zero, as my quote from wikipedia (in my previous post) also suggested: "they still have lifetimes derived from uncertainty relations".

By the way, when a photon interacts with a molecule or with an atom, there is the possibility, if the energy is right, to be absorbed, but most of the time the energy is not right and the photon is not absorbed. The question is: there is a trial and error process (that may take time), or the "decision" requires no time at all? Keep in mind that atoms and molecules are systems, made of smaller parts ...I couldn't ignore (again, in you last quote):

Imagine that an optical wave of frequency Ω/2π strikes this molecule. This frequency, much higher than those considered previously, is able to excite the electrons of the molecule; under the effect of the optical wave, the electrons will undergo forced oscillation and re-emit radiation of the same frequency in all directions.

If the 'optical wave' is "caused" by one photon and 'the electrons will undergo forced oscillation and re-emit radiation [photons?] of the same frequency', this process, without the "absorption" of the incident photon, would produce many similar photons, and this is true only for lasers (with a different mechanism).

 

[DrClaude]

This suggests that there is a process that takes time, probably very close to zero, but still different from zero, as my quote from wikipedia (in my previous post) also suggested: "they still have lifetimes derived from uncertainty relations".

I think Wikipedia is completely wrong here. If there would be any instant, however short, where the system could be found in the virtual state, then the state would be real (since we could measure the system in that state), not virtual.

It is important to note that the time-energy uncertainty principle is not Heisenberg's uncertainty principle, since time is not an observable in QM. And that idea of employing the uncertainty principle to justify what is happening is again reminiscent of the problem with virtual particles, where some will say that the energy needed to create the particle can be "borrowed" from the vacuum, so long as the lifetime of the particles is short, which is a very questionable statement.

By the way, when a photon interacts with a molecule or with an atom, there is the possibility, if the energy is right, to be absorbed, but most of the time the energy is not right and the photon is not absorbed. The question is: there is a trial and error process (that may take time), or the "decision" requires no time at all? Keep in mind that atoms and molecules are systems, made of smaller parts ...

QM doesn't work that way. The interaction of the photon with the atom (or molecule) will result in the atom and the electromagnetic field being in a superposition of non-excited atom + one photon and excited atom + no photon. Whether the photon was absorbed or not can only be discovered by means of a measurement. That measurement can be done at any time, and in that sense the transition is instantaneous (there is no intermediate state, so the atom can not be at any time "in the process" of absorbing a photon).

If the 'optical wave' is "caused" by one photon and 'the electrons will undergo forced oscillation and re-emit radiation [photons?] of the same frequency', this process, without the "absorption" of the incident photon, would produce many similar photons, and this is true only for lasers (with a different mechanism).

I did point out that this is a semi-classical explanation. The optical wave is here a classical electromagnetic wave, which is being scattered by molecules. There are no "new" photons coming out.