Question about Photons causing Electron Transitions in Atoms

  • #1
Joe Prendergast
18
4
An electron requires an "exact" wavelength photon to transition from one level of an atom to another. Yet the wavelength of a photon has a a continuous probability distribution, implying that the point probability of achieving an exact wavelength is zero. One can only talk meaningfully about the probability that the photon's wavelength lies between two values. How is it then that electron transitions occur?
 
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  • #2
It doesn’t have to be exact. There is a time-energy uncertainty relation that causes the absorption spectra to not be perfectly sharp peaks, even in ideal conditions. Real conditions also include Doppler broadening and several other similar effects that broaden the absorption peaks.
 
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  • #3
Ok, thank you. I've always wondered about this.
 
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  • #4
Joe Prendergast said:
Ok, thank you. I've always wondered about this.
It's a good question and essentially cannot be answered by basic QM. In order to explain the process fully, you need to consider both the atom and the quantized elecromagnetic field.

Within the framework of QFT (Quantum Field Theory), you can calculate the probability that an atom absorbs a photon of a given energy and find that it is actually a sharply-peaked distribution and not a single energy value.
 
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  • #5
PeroK said:
the probability that an atom absorbs a photon of a given energy and find that it is actually a sharply-peaked distribution and not a single energy value.
Isn't a difference between the energies of two states of the atom a single value?
 
  • #6
Hill said:
Isn't a difference between the energies of two states of the atom a single value?
It is in the basic QM model. But, the absorption and emission spectral lines are actually (Lorenztian) distributions.
 
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  • #7
PeroK said:
It is in the basic QM model. But, the absorption and emission spectral lines are actually (Lorenztian) distributions.
Yes, but for the energy to change by a single value, the atom has to absorb that single value of energy.

I think that the issue with the OP is a misinterpretation of this:
Joe Prendergast said:
the point probability of achieving an exact wavelength is zero
The point probability is indeed zero, but it does not mean that a point result cannot be achieved. It is a general case of probability distribution of a random real variable: if we randomly pick a real number from an interval, probability to get any specific exact real number is zero, but when it is picked, it is some specific exact real number.
 
  • #8
Hill said:
Yes, but for the energy to change by a single value, the atom has to absorb that single value of energy.
That depends to some extent how you interpret the mathematics of QFT.
Hill said:
The point probability is indeed zero, but it does not mean that a point result cannot be achieved. It is a general case of probability distribution of a random real variable: if we randomly pick a real number from an interval, probability to get any specific exact real number is zero, but when it is picked, it is some specific exact real number.
Although this argument is common, the probability distribution of a random real variable is a mathematical construction. In practical terms, it's impossible to pick a (uniformly) random real number from an interval. Not least because it's impossible to describe most real numbers.

It's also not possible to test the theory that an atom has an infinitely precise energy level (corresponding to some well-defined real value), because infinitely precise measurements are not possible. (I should say, there was a long debate about this recently and not everyone agrees that infinitely prescise measurements are not possible.)
 
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  • #9
Hill said:
Isn't a difference between the energies of two states of the atom a single value?
Hill said:
Yes, but for the energy to change by a single value, the atom has to absorb that single value of energy.
The energy doesn’t change by a single value. It has an uncertainty too. Energy is conserved, but not certain.
 
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  • #10
PeroK said:
It is in the basic QM model. But, the absorption and emission spectral lines are actually (Lorenztian) distributions.
...and that's because there are fluctuations of the electromagnetic field as soon as you use the full QED, where the electromagnetic field is quantized. If you have just an atom in an excited eigenstate of the atomic Hamiltonian and no photon then there is a certain probability due to the coupling of the electrons in the atom with the fluctuating electromagnetic radiation field (treated as a perturbation) to emit a photon and leaving the atom in another lower energy eigenstate. The transition energy is not sharp due to the fluctuating em. field, which provides a continuum of possible eigenstates of the full Hamiltonian, i.e., atomic Hamiltonian + coupling to the quantized radiation field (usually described in the dipole approximation). Another result of the corresponding perturbation theory is that the atomic energy levels (discrete when neglecting the coupling to the radiation field) get "broadened", i.e., the corresponding spectrum consists not of a sum of perfect sharp lines but somewhat broadened "Lorentzian distributions". The corresponding "width" of the Lorentzian in energy is the inverse lifetime of the excited state.
 
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  • #11
PeroK said:
It's also not possible to test the theory that an atom has an infinitely precise energy level (corresponding to some well-defined real value), because infinitely precise measurements are not possible. (I should say, there was a long debate about this recently and not everyone agrees that infinitely prescise measurements are not possible.)
What are you referring to? That sounds interesting. Of course you always have a finite energy resolution and the minimum "natural linewidth" due to quantum fluctuations (see my previous posting) but afaik atomic energy levels can be determined with an accuracy almost at this natural limit. Only think about frequency combs which make time measurements so accurate that you can measure the gravitational redshift due to local variations in the gravitational field of the Earth!
 
  • #12
vanhees71 said:
Of course you always have a finite energy resolution and the minimum "natural linewidth" due to quantum fluctuations (see my previous posting) but afaik atomic energy levels can be determined with an accuracy almost at this natural limit.
You have confused me. How does one measure the energy levels of an "isolated atom"? One can perhaps draw inferences, but a direct measurement would require decoupling the atom from the EM field (at least in my conception of the world). Because this seems unlikely, I am confused by the notion, as perhaps is the OP.
 
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  • #14
But I quibble with the notion of an isolated atom as an actual realizable object. I think it an abstraction and perhaps that is a more direct answer to the OP. When looking for the central frequency of the transition (e.g. an atomic clock), one averages over both an (~isolated) ensemble and over time to obtain the central value. I think this needs emphasis here. I also recommend the article cited by @vanhees71
 
  • #15
It's not since the achievements by Aspect, Harouche, Cohen-Tanoudji (Nobel prize 1997).
 
  • #16
vanhees71 said:
Aspect, Harouche, Cohen-Tanoudji (Nobel prize 1997).
You left out Steven Chu, Mr. Obama's Secretary of Energy (back when competence mattered....)
 
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  • #17
Yes, he got the Nobel prize together with Cohen-Tanoudji.
 
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  • #18
PeroK said:
It's a good question and essentially cannot be answered by basic QM. In order to explain the process fully, you need to consider both the atom and the quantized elecromagnetic field.

Within the framework of QFT (Quantum Field Theory), you can calculate the probability that an atom absorbs a photon of a given energy and find that it is actually a sharply-peaked distribution and not a single energy value.
 
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  • #19
Thanks for the explanation. I was wondering if you could recommend an introductory textbook on QFT.
 
  • #21
(An off-topic thread hijack has been excised from this thread, along with the replies to it.)
 
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