Sciencemadness Discussion Board - Absorption of Light Mechanism

Absorption of Light Mechanism

fluorescence - 29-2-2016 at 05:54

Hey there,

How does the Absorption of Light actually work ?

I know there is the Drude Model which says that the Dipols are basically damped oscillations caused by the electromagnetig field and if you calcuate x(t) for the motion you get a sum of two components, one that is moving in phase which is the Dispersion and the other one which is moving off phase which is the absorption. And together Absorption and Dispersion form the Polarisability.

So Light or electromagnetic waves intereact with either permant dipols in molecules or they induce the formation of them.

An induced Dipole has the equation:

µ = a * E

with the Tensor a as polarisability (polarisation ? ) and E as vector for the electromagnetic field. One could also say that the electric part of the Photon interacts with the electric component of the dipole.
So if I insert the equation for E I get to:

µ = a * E * cos (2pi) t * v

So the dipol will oscillate with frequency of the light like an antenna and will also give off light with the same frequency. This is basically Rayleigh-scattering. The dipol oscillating and giving off the same energy as it absorbed.

Now for the scattering, if the light causes another motion like rotation or vibration the periodic functions will overlap and if you calculate a now you will come across a part which is

(da/dq) at q=0 which must not be 0. So the selection rule for the Raman Scattering will be that the polarisability will have to change.

If you insert all those information back into the first equation you will get 3 parts, the rayleigh part, and the stokes and anti-stokes part.

Now I don't want to focus on Raman that much but there was this nice little sentence I found which is:

"Light can interact with matter, it can be absorbed or it can be scattered. " So there is always that Absorption and that Dispersion (refractive index, scattering, ....) part.

So the question is how does absorption work.

I know about all the quantum mechanics for the effects like harmonic oscialltor or rigid rotor and the rest of the associated spectroscopy. And all those who use Absorption have these resonance conditions since the energy levels in quantum mechanics can only have certain values and the light needs to have exactly that energy.

If you use more energy you will get that scattering effects as well this is why the energy levels of Raman are all virtual ones that don't really have conditions.

So let's say I want to exite a vibration of the molecule. I need IR light for that, will likely cause rotation as well so we asume we are in a liquid layer and just don't have any rotations at the moment but only oscillations.

For all Absorption based methods the thing that has to change is the transition dipole moment so the dipole has to change. This can explained by Fermi's golden rule for example, but basically the Bra-Ket
<psi 2 | x | psi1> to excite from psi 1 to psi 2 needs an interaction hamiltonian which is here:

H = -µ * E and since E is part of the field its the transition dipol moment operator µ. And µ is an odd function so each energy level in the harmonic oscillator is either odd or even and µ is odd and to calculate whether something is possible or probable in quantum mechanics you need |x|². Well the integral is here over the total space and if the resulting function is an odd one the integration will cause it to become 0 so |x|² = 0 and it doesn't happen. So exictation can only be between odd and even and so its delta V = 1 as a general rule.

For vibrations we also have to use the dipole moment close to the equilibrium since the dipole changes with the oscialltion itself.

So that is basically it ? I mean this is the general rule, the dipole moment and there can be dipols or some are induced and they are manipulated by light but how exactly is light absorbed. I mean for Raman scattering this is a bit easier to understand. Dipols are forced to oscillate with the frequency of the light and they will give the same photons again (the frequency of light does not change in matter).

Can anyone tell me if the thing I wrote above would be okay, since I have an oral exam on that soon and might need to explain the basic rules there and does anybody know more about the absorption part ?

Thank you in advance.

[Edited on 29-2-2016 by fluorescence]

[Edited on 29-2-2016 by fluorescence]

fluorescence - 1-3-2016 at 01:54

I think I found something:

I havent really considered that all vibrational excitations are just higher energy levels of the same stationary Hamiltonian. So the Eigenfunction itself just changes by the vibtrational quantum number and all vibrational levels are just eigenvalues. So since the Hamiltonian is not developing in time the Eigenvalue cant become higher on its on you need another bit of energy which is a time dependend perturbation to the system.

So H = H0 + H'(t) H0 doesnt change but we want H to become higher so H'(t) has to be high enough for the resonance condition. And for the solution of these perturbation theories we can assume a sinoidal solution which is like this:

H'(r,t) = V(r) * cos(wt)

With V(r) being a potential energy I guess. And this is now exactly what light is, or at least we can insert the osciallting potential of the photon here. And with a bit of maths we can see that the transition dipole moment has to change.

So I assume we have an Energy State 1 which has a Dipol with a certain vibration and now the electric part of the light interacts with the electric part of the dipole causing a perturbation of the dipole. And I just think that perhaps both waves kinda overlap and with the right frequency you can reach a vibration of the dipole that is usually present in the vibrational energy state 2.

I think I read somewhere that Light can be kinda reduced to an harmonic oscillator as well and then its just one oscialltion + the other one overlapping to form another state.

Could this be how it works?

Marvin - 1-3-2016 at 05:47

It may be impossible for anyone not a university lecturer to untangle and answer your question. Degree chemists are usually not taught Dirac notation or Tensors here, that's left to physics students. You've nailed together a whole heap of pain and then asked for a simple answer.

You would be best explaining your education background.

Part of the problem is that an electromagnetic wave exists but a photon is really an event. Much of the issue unifying quantum theory and wave mechanics a century ago stem from massively different expectations of the properties. I've yet to talk to anyone who likes the Copenhagen interpretation.

The parts of what you are saying that I understand seem right, the states before and after the jump are oscillations which can be described by a wavefunction and the overlap of the two wavefunctions tell you about how easily they couple. At the same time though, don't assume this interpretation takes you closer to a deeper truth. It's widely accepted there isn't one.

Edit, spelling.

[Edited on 1-3-2016 by Marvin]

fluorescence - 1-3-2016 at 06:20

Well this is for the Advanced Physical CHemisry Exam in the 5 th Bachelor Semester Chemistry, the topic of the whole semester was Spectroscopy, we looked at all the quantum mechanics for transitions and how the spectra form and so on and we also talked about dispersion, so scattering and polarizing, and inducing dipols which is fine to kinda explain all the Raman effects. But we never talked about absorption and the question is if they ask me what absorption is I need a quick answer since I only have 20 mins. So I cant calculate something there nor can I talk around the problem. I'd like to understand how they intereact with each other so I can give a really short statement that sums up most of it.

I only gathered all the information case someone starts to discuss the basics and I just wanted to skip all of this and present what I already found on that topic.

I think I understood what happens but the only thing that I'm not sure about is the interaction between Light and the Dipole not in terms of what happens but rather how does it happen ? Is it like I said 2 waves forming another one ? That is the thing I cant find anywhere. I know that the conservation of energy is not fullfilled here since we dont look at what the photon does after the absorption so we just look at the molecule.

For it would make sense to call a photon just an oscillation of energy an energy that cosnsist of the same type of matter as the oscillation energy in the dipole does and therefore they add up.

Like I dunno I want to compare mass with something else, doesnt work out I need something else that has a mass if I want to raise the weight of the overall system. Here its the same.

What do Photons and Dipols share that one can transfer the oscillation of said unknown property onto the other one ?

blogfast25 - 1-3-2016 at 06:26

annaandherdad! Help!

I searched for an answer on physics.stackexchange but nothing really useful came up. I'll ask a question there.

[Edited on 1-3-2016 by blogfast25]

Marvin - 1-3-2016 at 07:22

Light is an electromagnetic wave, so it has an electric oscillation at 90 degrees to a magnetic oscillation.

This can couple to an electric or magnetic vector of a molecule/atom etc.

I'm on shaky ground, and I last did spectroscopy formally a looong time ago but I don't think it's normal to consider the 'wave function' of light in the maths, only to consider the wavefunctions of the start state and the end state, the overlap tells you something about the probability of a transition and the energy difference between the states alone tells you the wavelength of light required.

A photon is a particle, so it probably makes less sense to talk about the oscillation of a photon than of an electromagnetic wave. Light is not confined, so you can't really do particle in a box, harmonic oscillator stuff on it.

When you have a little time I'd be grateful if you explained to why the selection rule for Raman is a change in polarisability instead of a the change in dipole moment it is for IR

I've been struggling to understand that for a while.

fluorescence - 1-3-2016 at 07:54

It is actually explained quite well here:

http://www.colorado.edu/chemistry/volkamer/teaching/RefMater...

As far as I understood is that basically do Rayleigh scaterring. So your Dipoles (induced) are like little antenna and the osciallte with the frequency of the irradiated liight, giving off the same frequency.

I think I heard somewhere that in the ocean the water molecules do not move, they stay at their position but just move up and down and that is what we see as waves, not sure it ifts true but something similar happens if you work with glass and light, so a prism. The dispersion is an effect that for example creates the index of refraction. And as far as I know according to Drude-Lorentz and that is how we explained absoprtion the soltuion is

Polarisability = Disperion + Absorption.

So its one contribution to the effect how light and matter interact if there is no resonance condition. The frequency of light in matter does not change but it takes a while. So the photon that hits the glass was is absorbed, and the oscillating dipole sends out another one that looks the same and so it takes a while for the photon to travel through glass for example.

Rayleigh scattering is pretty much light causing the dipole to osciallte with its frequency and the emitting it back again, fully elastic without loss. And if you now have more energy that just inducing the dipole, but enough to cause a vibration or rotation the two frequencies will interact with each other.

So you have to express your pola. a as a function of time a(t).

This is how we solved it:

a(t) = a0 + delta a cos 2 w R(t)
µ = a0 + delta a cos 2 w R(t) * E cos (wt)

So the first sum is a and the part at the end with E is the field and this is how both trigonometric functions interaact.

And if you solve this you get the stokes and anti stokes part as well.

And there is also the rule:

µ(ind) = µ(perm) + a * E

µ(perm) = 0 so to induce a dipole a mustn be 0.

As the paper shows that a1 is what changes depending on vibration or oscialltion. And that is what has to change in order to change anything in the system I guess. So if that is zero nothing will happen similar to that perturbation I showed above.

blogfast25 - 1-3-2016 at 08:17

Quote: Originally posted by fluorescence

I think I heard somewhere that in the ocean the water molecules do not move, they stay at their position but just move up and down and that is what we see as waves, not sure it ifts true [...]

Yes, that is true. A wave is a travelling oscillation, usually travelling through an elastic medium like water, air or some solid. For water waves, water molecules indeed move up and down only.

This:

http://www.acs.psu.edu/drussell/Demos/waves/wavemotion.html

... shows it wonderfully.

EMR of course doesn't need a medium and propagates through vacuum.

[Edited on 1-3-2016 by blogfast25]

annaandherdad - 1-3-2016 at 09:46

Quote: Originally posted by fluorescence

Hey there,
How does the Absorption of Light actually work ?

"Light can interact with matter, it can be absorbed or it can be scattered. " So there is always that Absorption and that Dispersion (refractive index, scattering, ....) part.

So the question is how does absorption work.

This seems to be the clearest question you have posed.

First, I would modify the above by saying, when light interacts with matter, it may be emitted, absorbed or scattered.

You're mainly interested in molecules, which have their degrees of freedom separated into electronic, vibrational and rotational sets, but the basic principles apply to any system (atom, nucleus, solid, etc).

The matter has energy states, one of which is the ground state, and all the other are excited. If the matter is in an excited state, it can emit a photon and drop into a lower energy state. The energy of the photon is the difference in energies between the two states. This called spontaneous emission.

If the matter is in any state in the presence of a light wave, the system can either be lifted into a higher energy state (absorption) or drop into a lower energy state (stimulated emission), in which the energy of the photon absorbed or emitted is the difference between the initial and final states (with the signs right to make conservation of energy). Of course if it's in the ground state then there is no lower state for it to drop into. To be clear, the photon involved in absorption and stimulated emission has the same frequency as that of the light wave; in other words, the light wave consists of photons of this energy (as well as direction and polarization), and the photon emitted or absorbed can be regarded as one of the photons of the light wave. Thus, the strength of the light wave either increases or decreases during stimulated emission or absorption. The difference between stimulated emission and ordinary (spontaneous) emission, as mentioned in the previous paragraph, is that spontaneous emission does not require a pre-existing light wave, and the emitted photon can go out into any direction or polarization state, as allowed by the selection rules.

So far I have described first order processes (in which only one photon has played a role, the one either emitted or absorbed). Scattering of light is a second order process, in which one photon is absorbed and another emitted.
If the energies of the two photons are equal, the scattering is elastic; otherwise, it is inelastic. In elastic scattering, the initial and final energy states of the matter are the same (and the energy of the matter does not change). In inelastic scattering, the matter changes from an initial energy state to a different final one, and the difference in energies between these two states is the same as the difference in energies between the absorbed and emitted photon. Again, with signs to make conservation of energy come out right.

In molecules, elastic scattering is called Rayleigh scattering, and inelastic, Raman (with Stokes and anti-Stokes indicating whether the scattered photon has less or more energy than the incident one, respectively).

These basic facts don't rely on much except the concept of energy levels and conservation of energy.

Depending on the system (atom, molecule, nucleus etc) there will be selection rules and other details about the strength, polarization etc of the emission/absorption/scattering process. For example, in a molecule, to change the rotational state of the molecule the photon has to impart or carry off some angular momentum, and the existence or lack thereof of "permanent electric dipoles" has an effect on vibrational transition rates.

Sometimes spontaneous emission can take place by a second order process, for example, in making the 2s->1s transition in hydrogen the atom emits two photons, whose energies add up to the energy difference in the two atomic states.

Hope this helps.

[Edited on 1-3-2016 by annaandherdad]

fluorescence - 1-3-2016 at 09:56

Well thank you, it kinda sums up what I wrote above. I know with the Einstein Coefficients you can go really deep into that stimulted Emission topic till you build something like a 3 level laser. As I said the mathematical part seems to be clear and also the things that happen, although you sum it up really well, I think I might take that as an explanation if I'm asked that is really short but I'm interested in the interaction itself.

blogfast25 - 1-3-2016 at 10:01

And at physics.stackexchange my question was answered here. I hope it helps...

fluorescence - 1-3-2016 at 13:55

Thank you for your effort. I will check it later.
Here is something I found:

https://www.google.de/url?sa=t&rct=j&q=&esrc=s&a...

That sums it up pretty much. Actually I'l probably print that out and keep it for later. That summary is so good and has so many topics I never head about but explained so well. And it finally explains the interaction ! Really cool, check it out, I will try to include the things mentioned in the ppt into my calculations above to make like a conclusion on the interaction.

Thank you all for your help and ideas !

Edit:

It seems the tag "RABI OSCILLATION" is an important thing as well but I dont understand the texts about it.

So I understood according to the ppt file the interaction as an dipole dipole interaction but then they only gave NMR and EPR as examples so I'm not sure if this is also for vibrations as well.

But one thing just became clear it is called resonance condition because then the two eigenstates are in resonance, I thought the field and the energy level need to be in resonance. But that would explain the whole thing and also that overlap between the states as probabillity. So the two energy states can be brought into resonance with the help of light.

The last thing we need to find out would be if its really a dipole dipole interaction than we would have our solution.

[Edited on 1-3-2016 by fluorescence]

fluorescence - 12-3-2016 at 10:06

So I thought I'll bring this up again. I was looking for a lecture on non rigid rotors last night and came across this pdf and on my way scrolling through the pages I came across to quite interesting pages:

https://www.princeton.edu/cefrc/Files/2013%20Lecture%20Notes...

It's all on page 2.

First thing is this is the first time I actually saw a potential where vibrations and rotations are both present which kinda gives a quite good summary on the topic. I think you cant really position them there since for each rotational level the mean bond lengh should stay the same for one vibrational level and here they are between 2 vibrational levels but its quite good to get the concept. And second thing is that the Morse Potential itself is the electronic excitation. That is something I never though of. For me UV/VIS excitations were always between two Morse potentials but I never associated the whole thing with the electronic part.Good to know.

But to continue with page 2, it took me a while to see that the upper picture shows rotations and the lower one vibrations. The question is just what does the picture tell us ( having in mind all the stuff I talked about in the previous posts) ???

So for the rotation it seems to me like the the dipole itself is rotating so µ becomes + , - , + , -
And then there is this µ(x) which is I believe the Dipol Moment in one direction. The same thing is there for vibrations, too. The vibrational picture seem easier here the Dipole µ or I think this should be |µ| is "oscillating" als the vibration happens and if you mark longest part as +1 and the lowest one as -1 you get this trigonometric function that looks like a wave. That seems okay to me.

The question is what this other figure for the rotation marked with E ? Should be the electromagnetic wave but rotations are caused by microwaves... and a microwave has let's say the wavelenght of 1 mm. This is the smallest value I found for the part that starts to cause rotations. Of cause they are excited all the time for higher frequencies but lets stick to a pure microwave spectrum. Ok now we take the size of a molecule. Actually I have no clue how the lenght of a molecule is defined if the bond lengh is between both surfaces or centers so I googled for molecule sizes and its between 1 and 10 A. So Lets say the molecule is 1 nm long and the wave is 1 mm long. So a factor 1000000. Still both waves are equal here which kinda bothers me.

I know there is the concept of Rabi Oscillations and so on but if the dipole has a wave and the radiation has a wave that causes a rotational transition, are they equally long ? Or what does this picture acutally tell me ?

I feel like at this moment the concept of a photon as particle with a dipol and the molecule as particle with a dipol coupling via a dipol-dipol force seems much easier to understand but I can't find more evidence on that besides that pdf I had further above ....

Any suggestions ?

fluorescence - 16-3-2016 at 11:50

So for those who are still reading these posts

I passed the exam with a 1.0 or A or however you call that in your Country and my Prof didn't ask me for
that but quite much on Dipole moments and Fourier Transormations and Raman as well so good that I focued on that here as well.

After we were done I asked him about the problem. And indeed the wave just passes by it doesnt mean that the oscillation of the molecule mus be the same as the one of the wave otherwise you would stretch it for like a meter or so. It just passes by and the bond gets a bit longer but not as long as the wave is. It absorbs some of the energy the rest just passes along. That would finally solve the problem.

I you do this with energy there are resonance conditions if you look at it mechanically or visually you dont need to think of sizes and compare them but rather the the wave as a big field and depending on the bond of the molecule it will be streched for a certain value that is all.

aga - 16-3-2016 at 12:16

Congratulations !

Any Pass is good.

So, now you're a qualified expert, kindly explain why most OC products (ignoring organo-metallics) are clear liquids or white solids.

In your own time if you please.

Edit:

and the brown tarry gunk.

[Edited on 16-3-2016 by aga]

fluorescence - 17-3-2016 at 05:04

There is a nice calculation if you get the concept of the particle in the box to
calculate the color of Carotene I think. Well usually you calculate bonds with that
and not the other way around but if you look for color its more like an electronic transition.

If I remember correctly from bonding molecular orbitals into others.
Like a pi -> pi* transition. This takes a certain amount of energy and if the reflected part
is not in the visible spectrum you won't see anything. So in order for them to be colored
you need a gap that has the right size.

This is usually done with for example long conjugared pi-Systems. You can apply particle in the box here and calculate the Energy difference between two states and that can be in the visible spectrum I can look that up for you cause this was in the first semester when I had to do this.

The other possibility is Azo-Compounds. What most people don't see is yes azo compounds have that Azo-Bridge but that isnt what causes the color. The Azo connects two very different molecules usually, one that takes electrons and one that donates them. Therefore you need a smaller "box" for the same result causing even small azo compounds to become colored.

Or why Iodine changes color when starch is added. Those relative positions of the energy levels change when complexes form and so does the energy required to excite it so the color shifts.

Here is a good picture that shows it:
https://260h.pbworks.com/f/1321207827/Transitions.png