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From quantum mechanics, a photon of energy exactly equal $h\nu=E_2-E_1$ could be captured by an atom that has the energetic level levels $E_1$ and $E_2$ (with $E_1<E_2$). This corresponds to "absorption lines" for the photons in the material made of a single type of atoms.

How to explain that in nature, for example in plants, we see absorption spectrum : that is : there is no "Dirac-like" missing lines, but we have really a smooth curve of absorption spectrum.

Example :

enter image description here
(Source)

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    $\begingroup$ I think it is because plants are made up of very many atoms which can have slightly different spectra. They kind of blend into a smooth spectrum. $\endgroup$
    – Jonas
    Mar 23, 2021 at 20:22
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    $\begingroup$ @QuantumEyedea: Doppler broadening is not the cause here. Many organic molecules have discreet spectra. $\endgroup$
    – Gert
    Mar 23, 2021 at 20:38
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    $\begingroup$ @Jonas These substances are made up of molecules and it's the molecular bonding orbitals that bond the atoms that do the absorbing/emitting of photons. Not atoms or atomic orbitals. $\endgroup$
    – Gert
    Mar 23, 2021 at 20:39
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    $\begingroup$ @MathieuKrisztian [...] but we have really a smooth curve of absorption spectrum. No, not really: the spectra are very 'peaky'. The peaks are somewhat broader than in typical atomic spectra though... $\endgroup$
    – Gert
    Mar 23, 2021 at 20:43
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    $\begingroup$ @MathieuKrisztian You ask a simple question about a complex phenomenon. Crucial is to understand what technique was used to record the spectra you put on display. For organics, Raman specroscopy is often used, which does lead to peak broadening. See e.g. EtOH/MeOH spectra: open-raman.org/… I strongly suggest to familiarise yourself with Raman et al before enquiring about why these molecular peaks appear a little different from atomic spectra. $\endgroup$
    – Gert
    Mar 23, 2021 at 21:01

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I'm going to go with an intentionally provocative answer. Atoms don't have discrete absorption lines. Nothing has discrete absorption lines. Any state that can decay or be excited necessarily has a finite lifetime $\Delta t$, and thus its energy is uncertain by an amount $\Delta E$ that can be estimated from a simple Fourier argument to be $$ \Delta E = \hbar\Delta \omega \sim \frac{\hbar}{\Delta t}.$$ In the case of atoms, the typical line widths are $\Delta \omega \lesssim$ GHz and the typical frequencies can be hundreds of THz, meaning that the lines look very sharp on a coarse-grained resolution. But "Dirac delta" peaks just don't occur in Nature. Anything that can be measured couples to the outside world and therefore has a finite lifetime/linewidth.

In reality, the broad spectra of typical materials are of course explained by the points raised in Vadim's answer: you have "classical" broadening by mixing the spectra from different sources, as well as the truly "quantum" broadening that arises from coupling between systems with the same spectra. Note that the latter is essentially the same physics as I described above, since localised states of a composite system have a finite lifetime due to the coupling between subsystems.

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  • $\begingroup$ Atoms and molecules do have discrete excitations, but of a finite width. $\endgroup$
    – my2cts
    Mar 26, 2021 at 13:41
  • $\begingroup$ Thank you for your great explanation. I'll give you the credit tomorrow, since technically speaking, one needs 16 hours more. A detail, you may replace introduce the word Dirac close to delta, because some readers who may read this thread may not understand this part. $\endgroup$ Mar 26, 2021 at 19:34
  • $\begingroup$ @MathieuKrisztian Thank you! I have modified the answer as you suggest. $\endgroup$ Mar 27, 2021 at 20:18
  • $\begingroup$ This answer does not mention relevant facts such as the fact that these molecules have many spectral lines and that their spectra are measured in solution. $\endgroup$
    – my2cts
    Mar 27, 2021 at 20:28
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The two basic reasons are:

  • Materials made of atoms do not have the same spectra as the atoms themselves, due to the interactions between the constituent atoms. E.g., crystals, even made of one type of atoms, have whole spectral bands defining ranges of frequencies where they can or cannot absorb. For the amorphous materials the spectral boundaries are even less clearly defined. Molecules have discrete lines, but more complex than just a combination of those of the constituting atoms: due to interactions, as well as due to the addition of the rotational and vibrational degrees of freedom.
  • Planets are made of many types of materials and atoms, which all have very different spectra.
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  • $\begingroup$ Another way of saying things: Absorption lines can be collision- and/or doppler-broadened. In gas state, they remain narrow, but in liquid state (as in biological tissue) collisions turn them into broad bands. In solid state, phonons have much the same effect as collisions in liquid state. $\endgroup$ Mar 26, 2021 at 11:49
  • $\begingroup$ @BertBarrois in solid state, before speaking of phonons (even in adiabatic approximation), we should already note the broadening due to overlap of single-atomic orbitals and resulting hybridization of the orbitals into delocalized states. $\endgroup$
    – Ruslan
    Mar 26, 2021 at 11:51
  • $\begingroup$ @BertBarrois Yes, broadening is important, even in gaseous phase, where it comes via doppler effect. However, collisions, phonons, etc. are pretty much covered as interactions. $\endgroup$ Mar 26, 2021 at 12:49
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A typical single molecule spectrum would look as shown in this paper. These absorption spectra consist of many lines corresponding to many vibronic transitions. As spectra are usually are taken in solution the lines are inhomogeneously broadened by interaction with solvent molecules. They can also be homogeneously broadened by lifetime effects or by coupling to the phonon continuum.

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There are multiple reasons for the width observed in a spectrum.

First of all there is always a width simply due to the precision of the measurement. If the measurement was taken with poor resolution, you will get a broad spectrum even if the underlying molecule has a stick spectrum. In that case the width is not a feature of the molecule but of your measurement apparatus and the measurement has only a very limited information content. If the resolution is bad, the sticks merge and you get one broad peak instead. You can determine if that's the case if you do the measurement with higher resolution. The peaks will get more narrow and your broad peak will separate into multiple peaks with smaller width.

Besides that, there is also a natural limit for the width of the peaks due to the limited lifetime of excited states. Mark Mitchison answer addresses that point very accurately. The lifetime of excited states limits the width of peak to a minimum and you can't decrease the width of peaks below this limit no matter how good your setup is. This width is a property of your molecule and allows you to determine/define the lifetime of an excited state.

Besides this, we have broadening due to the environment of a molecule and simply the fact that we typically measure a whole bunch of molecules at once and not a single lonely one. Measurements of single molecules/atoms are possible but require advanced experimental techniques. These measurements look alot more like stick spectra that are known from atoms but typical measurements of organic molecules are done at room temperater in solution. In in that scenario environment/solvent effects have a large effect on the shape of the spectrum.

In the shown example i would assume that the measurement was done at room temperature in solution. In that case, every molecule has a different local environment which affects the energy of the states contributing to the absorption spectrum. The broad spectrum that we see can thus be seen as sum of many(lets say we have a concentration of one milli mol ~ 10^20 molecules) slightly shifted spectra. This summation will results in a very broad spectrum even if you do have good precision for single given spectrum. Just imagine the sum of 10^20 gaussians with small width spread around a central frequency, the sum will be one broad gaussian if the spread is noticeable.

Assigning peaks to states is thus not straight forward especially not in molecules were you have electronic states, vibrational states, rotational states, spin states ... Most of the states get mashed up under one broad signal and can only be separated clearly into single peaks if the experimental setup is carefully designed to do so.

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