# $2s\rightarrow 1s$ transition in hydrogen

I would like to understand how the parity-violated interaction between electron and proton can provide the photon with circular polarization in the $$2s\rightarrow 1s$$ transition with single photon emission. As I understand, this interaction plays the crucial role. I only have found this problem and have read the paper by Breit & Teller, where the authors discuss metastability of $$2s$$ state.

I am looking for any references, books, etc which can be helpful for my problem.

• I do not understand what you meant by "parity-violated interaction between electron and proton". Where is it mentioned in the problem set you have linked? It just shows that single photon emission can happen by magnetic dipole transition but it's much slower than two-photon electric dipole transition. – wcc Mar 28 at 14:26
• Interesting case. I found this paper but it is behind a paywall : link.springer.com/article/10.1134%2FS0030400X0804005X – my2cts Mar 28 at 19:42
• @IamAStudent, I have corrected the question, thank you – Artem Alexandrov Mar 30 at 7:46
• "....photon with non-zero helicity" all photons have non-zero helicity as there can be only two values of $\pm1$. I am suspecting that your question really boils down to how a magnetic dipole transition works. – wcc Apr 1 at 19:33
• On the contrary. M1 transition can have circular polarizations (e.g. |2,-2> to |1,-1> hyperfine transition in rubidium atoms). – wcc Apr 2 at 14:51

It isn't particularly clear to me what you mean by

I would like to understand how the parity-violated interaction between electron and proton can provide the photon with circular polarization in the $$2s\rightarrow 1s$$ transition with single photon emission.

and particularly what you really mean by "understand how ...".

• The fact that the symmetries of this transition do allow for a nontrivial difference in the emission rates to different helicities is reasonably well explained in Problem 3 of one document you've linked to ('Final Solutions', to the 221B course at Berkeley, 2001, by Hitoshi Murayama).
• The specific meaning of 'circular polarization' is explained fairly well in footnote 18, p. 63, problem 1.14, of the book you've linked to, Atomic Physics: An Exploration Through Problems and Solutions (D. Budker, D. F. Kimball and David P. DeMille, Oxford University Press, 2004).

In short, you define a quantization axis (by convention, the $$z$$ axis), you prepare $$2S_{1/2}$$ atoms in definite-$$\hat{S}_z$$ states, and you put detectors at the positive and negative $$z$$ axis, which are able to detect the helicity of the produced photons, and which are insensitive to the two-photon radiation at half the $$2S_{1/2}$$-$$1S_{1/2}$$ energy spacing. Then preparing the $$m_s=+1/2$$ state will produce a right-circular photon ($$S_z=1$$) and preparing the $$m_s=-1/2$$ state will produce a right-circular photon ($$S_z=-1$$). Within standard QED, both of these channels do come out with circular polarizations (despite your claims to the contrary in comments), but they do so at equal rates. If you include parity-violating weak interactions, then there will be a nonzero difference in the emission rates for the two, probably at the "large" level of one part in $$10^4$$. (If I understand the literature below correctly, that is.)

If you want to go beyond that, then there is just an enormous sliding scale of difficulty in terms of how technical you want to get, with no clear cutoff point for what constitutes "understanding" $$-$$ which is in any case a personal, subjective descriptor.

Still, since you've tagged this as resource-recommendations, here is one way into that rabbit hole:

That last paper is probably where all the real action is - but maybe there are other rabbit-hole branches coming from other reference follow-ups, so do look around.

Still, given how unclear the question is about what exactly it is you want to know, I don't really see how digging into any of these specific resources with a deeper analysis would make sense. (And also: if you don't find anything in this list accessible, then I would suggest that it's too ambitious of a target for you at this stage.)

In any case, though, I have to ask: why are you so interested in this? As the problem set you've linked to lays out very clearly, the two-photon decay mechanism dominates the M1 single-photon channel by a factor of $$\alpha^6\approx 1.5 \times 10^{-13}$$, or, in other words, "this channel just doesn't happen in real life": you need to prepare $$10^{13}$$ metastable atoms for one of them to decay via the M1 route, and you need to repeat this $$10^4$$ times to get any signal at all.

This is not to say that the channel isn't interesting, but that challenge does mean that if you want to use the $$2s$$-$$1s$$ transition to test this type of mechanism, your best bet is to look at higher hydrogenic ions or similar systems $$-$$ as the literature above demonstrates.

And, moreover, this is also why the current work on using precision spectroscopy to test CPT-related effects (which is quite substantial $$-$$ for a decent recent review on work with hydrogenic systems see e.g. 'Lorentz and CPT tests with hydrogen, antihydrogen, and related systems', V. Alan Kostelecký and Arnaldo J. Vargas, Phys. Rev. D 92, 056002 (2015)) does not use this system.

• Weak neutral currents provide the circular polarization which can be estimated as $P=FA(E_1)/A(M_1)$, where $A(E_1)$ is the amplitude of $2p_{1/2}\rightarrow 1s_{1/2}$ and $A(M_1)$ is the amplitude for magnetic dipole transition, $F$ is mixing between $2p_{1/2}$ and $2s_{1/2}$ states due to parity violating interaction. Your answers was really helpful and comprehensive. Thank You! – Artem Alexandrov Apr 8 at 10:30

There are two parity-even operators that contribute to the $$2s\to 1s$$ transition: $$(\vec{p}\cdot \vec{A})^2$$ and $$\sigma\cdot \vec{B}$$. The first obviously contains two photons but the second is just a spin flip. In other words, the electron can flip its spin and remain in an S state.

• Isn't there also a quadropole interaction? – garyp Mar 31 at 14:17
• @garyp I think (p.A)^2 contains a quadrupole term – Eric David Kramer Mar 31 at 14:38
• I"m not certain, but $p$ is the dipole operator, and if I recall correctly, the quadrupole interaction involves a field gradient. Someone reading this will know the answer for sure. – garyp Mar 31 at 14:49
• By p I mean the momentum operator. – Eric David Kramer Mar 31 at 15:06
• Of course. I'll keep quiet now. – garyp Mar 31 at 15:07