Investigating Two-Photon Decay in Hydrogen Atoms and Hydrogen-Like Ions

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Discussion Overview

The discussion revolves around the two-photon decay process in hydrogen atoms and hydrogen-like ions, specifically focusing on transitions from the 2s to 1s and 2p to 1s states. Participants explore theoretical frameworks, conservation laws, and energy considerations related to these transitions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant references a paper discussing the probabilities of spontaneous two-photon decay and expresses uncertainty about the implications, particularly regarding photon wavelengths and electron transitions.
  • Another participant explains that the 2s to 1s transition cannot occur via a single photon due to conservation of angular momentum and parity, suggesting that two photons with opposite polarizations are required for the transition.
  • It is noted that the wavelengths of the emitted photons vary, but their energies must sum to the energy difference between the 2s and 1s states.
  • A participant cites a conclusion from a linked paper indicating that two-photon decay occurs over a wavelength range from 1216 Angstroms to infinity, with a maximum at 1420 Angstroms, and questions whether this implies a total energy greater than the energy difference between the 1s and 2s states.
  • Another participant calculates the energy difference between the 2s and 1s states, noting that the energy difference corresponds to a wavelength of approximately 1220 Angstroms and discusses the implications of photon energies being less than this value.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the implications of the two-photon decay process, with some agreeing on the necessity of two photons for the transition while others remain uncertain about energy interpretations and calculations. No consensus is reached on the interpretation of energy levels and photon emissions.

Contextual Notes

There are unresolved aspects regarding the distribution of photon energies and the implications of energy levels in the context of ionization energies. Participants reference specific calculations and theoretical frameworks without reaching definitive conclusions.

zincshow
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Papers such as http://arxiv.org/PS_cache/arxiv/pdf/0904/0904.1503v2.pdf talk of 2 photon decay from 2s to 1s or 2p to 1s.

"The probabilities for the spontaneous two-photon decay in hydrogen atoms and hydrogen-like ions are under investigation since the theoretical formalism has been introduced by G¨oppert-Mayer [9] and the first evaluation for the two-photon E1E1 transition 2s → 2
(E1) + 1s has been presented by Breit and Teller [10]. A highly accurate calculation of the E1E1 - transition probability has been performed by Klarsfeld [11]. Recently Jentschura [12] performed a complete evaluation of the radiative corrections and presented more accurate value of the E1E1 two-photon decay probability. The double- and triple-photon decays
of metastable 3P0 atomic state were considered in [13]. The present paper is devoted also to evaluation of the probabilities for two-photon decays 2p → (E1) + (M1) + 1s and 2p → (E1) + (E2) + 1s."

I am unsure what it means. What are the wavelengths of the photons in question? Are they talking about the electron jumping to the 3rd level and falling twice? TIA.
 
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The 2s to 1s transition cannot occur via emission of a single photon for 2 related reasons. First, a photon has a polarization and therefore angular momentum, while both the 1s and 2s states have zero angular momentum. So a one photon transition would violate angular momentum conservation. A related conservation rule is that of parity. The 1s and 2s states are parity invariant, P=1, while a photon has parity P=-1, so parity would not be conserved either.

However, both conservation laws are satisfied if the final state has two photons of opposite polarizations. In this case, the transition occurs via a superposition of intermediate states.

2s \rightarrow \psi + \gamma \rightarrow 1s + \gamma + \gamma.

All higher energy states with nonzero angular momentum contribute to \psi, including the 2p state. The wavelengths of the photons are not the same for every transition, but energy conservation requires that the sum of their energies is equal to the energy difference between the 2s and 1s states:

\frac{h}{\lambda_1} + \frac{h}{\lambda_2} = E_{2s} - E_{1s}.

The paper you cite doesn't seem to calculate the distribution of photon energies, but it's done in other places, such as http://adsabs.harvard.edu/full/1984A&A...138..495N
 
Thank you for the info and link. In the conclusions of the link it says "two-photon decay extends over the wavelength range 1216 Angstroms to infinity and has its max at 1420 angstroms"

I read this to mean "the most likely event is 2 photons are ejected in opposite directions with total energy (converted to wavelength) adding to 1420 angstroms". In other words, there is considerably more then 1.89 evolts (energy difference of 1S to 2S levels) of energy released (1420A = 8.73eV).

Does that sound correct or do I remain confused?
 
zincshow said:
Thank you for the info and link. In the conclusions of the link it says "two-photon decay extends over the wavelength range 1216 Angstroms to infinity and has its max at 1420 angstroms"

I read this to mean "the most likely event is 2 photons are ejected in opposite directions with total energy (converted to wavelength) adding to 1420 angstroms". In other words, there is considerably more then 1.89 evolts (energy difference of 1S to 2S levels) of energy released (1420A = 8.73eV).

Does that sound correct or do I remain confused?

Ignoring fine structure, the 2s -> 1s difference is the first term in the Lyman series:

E_{2s} - E_{1s} = -13.6~\text{eV} \left( \frac{1}{2^2} - 1 \right) = 10.2~\text{eV}.

This corresponds to a wavelength of \sim 1220~\AA. When the photons have less energy than this, the atom's final state has some nonzero momentum.

I think the energy that you quote is the ionization energy for the 2s state, which we'd have to compare to the ionization energy for the 1s state to get the energy difference.
 
Thanks a lot. Its amazing how difficult things are when you are on the wrong energy level...
 

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