# I Can a Photon be Absorbed Without the Exact Energy Level?

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1. May 2, 2016

### NaiveTay

In my education of QM, I've heard countless times how energy is quantized by Planck's Constant, and how radiation is only emitted and absorbed in these discrete steps. Recently I've heard that that's not the full picture of energy, and I was hoping you could draw some clarity for me.

In reading Feynman's lectures, he mentions the quantization formula E = hf, where Planck's Constant, h, is multiplied by the frequency of radiation. He then says that though the constant is a certain step size, since the frequency of light can be anything whatsoever, then the energy of a photon can have any value whatsoever.
This is all good and well, but my question is along the lines of: if radiation from atoms is only ever created and only ever absorbed at very specific energy levels (because of the finite shells for an electron to jump), then what happens to light that doesn't exactly "fit the bill"? If radiation can be created by other means outside of electron jumps, then what are the chances this radiation will be the exact amount required for an electron jump of any of the finite number of elements?
Or, perhaps more encompassing, if a photon's energy is affected by collisions, gravity, and other interactions, then how does even a single photon maintain the precise energy needed to be reabsorbed? Clearly my picture is a little skewed.

My guess would be that when people say that energy can only be absorbed in particular amounts, it's not that there's only certain kinds of photons that can be absorbed atomically, but instead that atoms can only absorb a set amount but they reflect the excess energy as a lower frequency photon(s).
Is this correct? What are the limitations of absorption for photon's that excessively "fit the bill"? Obviously the only criteria for atomic absorption isn't just having the required amount of energy, otherwise all light with sufficient energy would be partially absorbed and the remainder reflected. Certain materials have certain colors, which wouldn't be the case (as far as I can tell) if there were really no limitations for absorption other than a photon having adequate energy. But, equally, it doesn't make sense to me to say that there are exact absorption energies which must be matched exactly by the energy of a photon (so that the photon is completely absorbed, with no excess energy remaining) if the energy of a photon is dynamic and can scale an infinite range.

Summarized: if a photon can have any energy level, and if that energy is continually nudged by interaction, how can a photon ever have the exact energy any electron jump demands? And if the photon's energy doesn't exactly have to match to be absorbed, then what are the limitations that allow for colors to be isolated in their reflections?

2. May 2, 2016

### ZapperZ

Staff Emeritus
I don't quite understand the issue here, so I'm going to shoot out a few points:

1. If a photon does not have the right energy to be absorbed by a photon, it doesn't get absorbed. It will simply passes through the atom without any interaction. This is what we see in absorption spectrum measurements... a continuous spectrum, with only several "dark lines" signifying the frequencies that happen to match the allowed absorption.

2. Just because a broadband photon energies/frequencies can be produced, doesn't mean that a particular system must have all of the same broadband absorption. I can produce a wide band of frequencies of photons from an undulator at a synchrotron light source. However, this is a different system than an isolated atom with discrete energy levels. So why would one absorb the same way as the other is transmitting?

There has been at least one other similar thread on this:

Zz.

3. May 2, 2016

### Staff: Mentor

This is a good question. I will move it to the quantum mechanics section where the best experts will see it easier.

I am not an expert, but my understanding is that even though energy is conserved it is uncertain. The faster a transition happens the more uncertain the energy. So emission and absorption spectra are not pristine spikes, but rather broad peaks where a range of energies will be absorbed or emitted characteristic of the time scale of the interaction.

4. May 2, 2016

### Buzz Bloom

Hi Naive:

Another factor that broadens the width of a line is the velocity difference between the emitting particle and the measuring device due to Doppler effects.

Regards,
Buzz

5. May 4, 2016

### NaiveTay

Thank you for moving this to the proper section, I'm still very new here so I appreciate the hand.

Ask a quantum mechanical question, get a quantum mechanical answer. It seems my fault in logic was in trying to place a classical picture on a quantum system (as is still so easy for me to do without realizing). If I'm hearing you correctly, the required absorption energy for a certain spectral line doesn't have to be exactly matched by the incoming photon energy in order to cause the electron to jump to a higher orbit. There's an intrinsic... variability, or "uncertainty" in all quantum systems which allows for a range of possibilities.
That must mean, as you mentioned, that the spectral lines aren't so much sharp lines as bell-shaped probability curves, where photons closer to the precise absorption energy have a higher probability of being absorbed, but they don't have to match exactly. I'd be curious to understand the falloff of these spectral lines if this is the case.
Would you say that's on the right track? I find your explanation quite interesting and satisfying, so thank you for taking the time to enlighten me.

6. May 4, 2016

### Staff: Mentor

Right.
Perfectly motionless atoms show a Breit-Wigner distribution (which has various names). In most experimental setups, thermal motion of the particles is more relevant, which leads to a Gaussian distribution. Collisions with other atoms can be relevant as well.

Usually you need lasers and/or very sensitive diffraction gratings to note those effects. Shorter-living excited states lead to larger energy uncertainties and broader lines. This is not limited to spectroscopy, it applies to all short-living states, including particles. If you measure the mass of short-living particles, you won't always get exactly the same result. For particles, this is called "decay width".

7. May 5, 2016

### stevendaryl

Staff Emeritus
Putting aside the broadening of the energy levels, it's not completely clear to me whether the rule that an atom can only absorb discrete amounts of energy is a rigorous result, or whether it's just the dominant effect.

If you model the interaction between an atom and the electromagnetic field by a time-dependent potential, considered as a perturbation to the hamiltonian (using non-relativistic quantum mechanics), then the first-order amplitude for transitions from state $|n\rangle$ to state $|m\rangle$ is given by:

$T_{nm} = \frac{-i}{\hbar} e^{-i \omega_m t} \int^t dt' V_{mn}(t') e^{i \omega_{mn} t'}$ (see equation 8.42 in http://web.mst.edu/~parris/QuantumTwo/Class_Notes/TDPT.pdf) where $\omega_m$ is the frequency corresponding to $|m\rangle$ and $\omega_{mn}$ is the difference in frequencies, and $V_{mn}$ is the matrix element of the interaction potential: $\langle m|V(t)|n\rangle$

This transition is time-dependent. If you let the interaction perturbation last forever, then the integration becomes a Fourier transform, and you only get a nonzero result for the Fourier component of $V_{mn}$ with frequency $\omega_{mn}$. However, there are two buts to this:
1. If the perturbation doesn't last forever, then there will be a nonzero amplitude even if the potential doesn't have frequency $\omega_{mn}$
2. This is only the first-order effect, so getting zero here doesn't imply that the full transition amplitude is zero.
So this makes me think that "atoms can only absorb energy at discrete frequencies" is more of a rule of thumb than a rigorous result.

8. May 5, 2016

### ZapperZ

Staff Emeritus
But you can't just say that and then turn a blind eye on the numerous absorption spectra results! Bad theorists do that! How does what you have described above fit in with all of the experiments and analysis that have been done?

Zz.

9. May 5, 2016

### stevendaryl

Staff Emeritus
I was just asking whether it was a rule of thumb, or a rigorous result. I don't understand why you want to be belligerent about it.

There are two different issues: (1) What do the experiments show? (2) How (and if) the experimental results are explained by the theory. Yelling doesn't seem an appropriate. I consider your post to be a violation of the standards of civility for Physics Forums.

10. May 5, 2016

### stevendaryl

Staff Emeritus
I'm going to have to make myself calm down a little, rather than impulsively putting ZapperZ into my "ignore" list. But I am very irritated by the incivility.

11. May 5, 2016

### ZapperZ

Staff Emeritus
You are right. I was being a bit "belligerent" in my response. But I find it rather incredulous that you would present such an analysis, but never once mention anything about all these absorption spectra. I am sure you will agree that, no matter what you come up with, it has to be compared against any and all experiments that have been performed. And in this area, there have been a huge body of experimental evidence. UV-VIS experiments, for example, are so common, I have undergraduates doing this as part of their advanced labs.

Now, considering that absorption spectra in atomic gas show discrete absorption spectrum, and that these are then used the same way as emission spectrum to identify elements, I do not see it as being simply a mere "rule of thumb", unless you also think that emission spectra are also a rule-of-thumb.

Zz.

12. May 5, 2016

### Staff: Mentor

If the perturbation has a finite time, it does not have a well-defined single frequency.

13. May 5, 2016

### stevendaryl

Staff Emeritus
I was not giving a course on the subject. I was asking a question, and giving some background to my question. It's not usually required that a post on a topic contain a survey of everything known about that topic.

14. May 5, 2016

### NaiveTay

Experimental results are always limited by the rudimentary measuring apparatus we have available, so I don't see anything wrong with conjecturing beyond our current limitations to experiment, as long as it's clear it's just a conjecture. Now I understand that if an idea does not fit with direct observational results, it is wrong, but our examination of spectral lines is limited to a certain degree of accuracy, and to say that we have them completely understood is saying more than the experiments have shown. But that might not be what you're saying, so my apologies if experiments have proven that it cannot be just "a rule of thumb".
That's not to say I understand spectral lines or have anywhere near your expertise in this area. I'd be very interested to know how accurately we know the energy range of these spectral lines. However, brushing off stevendaryl as a "bad theorist" seems like a truncation of curious reasoning. We all gain an understanding from discussing things openly and kindly, and being harsh just shuts off the flow of ideas with a wall of egotism, and I find that intolerable.

Back to the topic at hand. You all have a deeper knowledge of spectral lines than I do, and it seems I'm missing something because from my viewpoint, you all disagree. Yes, there's a width to a spectral line, so clearly there's a range of permitted energies, but then I read "...atomic gas show discrete absorption spectrum", and I'm uncertain all over again. The meaning of "discrete" seems to be different to us. A discrete quantity to me would be something like pi, an exact numeric value with unlimited precision. Now if the absorption energy must "discretely" be matched by the radiation energy, there must be something else going on because the emission of a "discrete" wavelength of radiation is instantly varied due to many factors, e.g., gravitational influences. So that "discrete" quantity is changed, and because there's an infinite range of these "discrete" quantities on any numeric scale, it would be impossible for any photon to ever have any "discretely" predefined amount of energy. Clearly that's bullocks, so I return to my original question about the range of permitted absorption energies. What is the falloff of spectral lines? In other words, if the bands have a width to their dark portion where radiation within that range is completely absorbed, then is there also a partially darkened area on either side of these bands where the radiation has a probability of being absorbed? That's the question I'd really like answered.

15. May 5, 2016

### Staff: Mentor

Within the width of the line, which is often so small that you don't care about it. Just a sloppy use of words. The opposite would be a continuous absorption spectrum where relevant absorption happens over a large frequency/wavelength range.

There is no frequency with 100% absorption. The strongest absorption is in the center of the line, towards the sides absorption goes down.

16. May 6, 2016

### vanhees71

Well, the first-order time-dependent perturbation-theory analysis in the dipole approximation explains a lot but for sure not all phenomena of photon scattering on atoms. You can have as well elastic and inelastic scattering of single photons or two-photon absorption from a classical em. wave (coherent state in the quantum treatment of the em. field) etc. etc. It's a rich field covered in books on atomic physics, e.g., H. Friedrich, Theoretical Atomic Physics, Springer (2006)

For the most simple standard analysis of the photoeffect (absorption of a single "photon" out of a classical em. wave), see my Insights article

https://www.physicsforums.com/insights/sins-physics-didactics/

17. May 6, 2016

### stevendaryl

Staff Emeritus
The original poster's question can be interpreted (I think) as a question about inelastic scattering of photons. Is it possible for an electron in a bound state with energy $E_n$ to interact with a photon of energy $\hbar \omega$, with the result being that the electron transitions to state $E_m$ and a photon with energy $\hbar \omega'$ is produced? (Where $\hbar \omega' + E_m = \hbar \omega + E_n$) That would be sort of like Compton scattering, but involving bound electron states.

18. May 6, 2016

### Staff: Mentor

But they don't show discrete absorption spectra if you look close enough. That is the point of the OPs question and stevendaryl's response. It is important to keep the theory relevant to measurements, but careful measurements also support the question and the reaponse. Stevendaryl is not turning a blind eye, but rather giving a closer scrutiny than you see in undergraduate laboratory exercises.

19. May 6, 2016

### Staff: Mentor

This is called Raman scattering.

20. May 6, 2016

### stevendaryl

Staff Emeritus
Thanks for the words of support.