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

[zincshow]

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.

 

[fzero]

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

 

[zincshow]

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?

 

[fzero]

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.

 

[zincshow]

Thanks a lot. Its amazing how difficult things are when you are on the wrong energy level...