In general what will holding an anti-hydrogen atom for more than a 1/10th of second allow scientists to discover? Specifically, given that they can hold one for <1/10th of a second, what would they discover that have not previously been able to determine. Or if not known, what have they been able to discover to date?
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$\begingroup$ Put like that, 1/10th of a second seems a short time. In terms of particle physics, however, it is a large time as processes often happen on nanosecond (and shorter) time scale. $\endgroup$– HeatherNov 22, 2010 at 15:59
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$\begingroup$ Yes good point, in the question I was just trying to distinguish between the current level of tech/achievement, and what is currently being attempted/to be achieved. $\endgroup$– Anonymous TypeNov 22, 2010 at 21:39
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1$\begingroup$ I suspect that the reason this question was asked was due to the recent press releases surrounding this article in Nature: nature.com/nature/journal/vaop/ncurrent/full/nature09610.html . The official "Nature News" article is also informative nature.com/news/2010/101117/full/468355a.html $\endgroup$– j.c.Nov 23, 2010 at 3:27
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$\begingroup$ @j.c. - You are correct. $\endgroup$– Anonymous TypeNov 24, 2010 at 2:24
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The ultimate goal is to be able to do precision spectroscopy of antihydrogen, to make sure that the energy states are the same as in ordinary matter. If there are differences between the energy levels of ordinary hydrogen and antihydrogen, that would violate "CP" symmetry, which says that if you change the sign of all the charges in some system, and invert the parity, every interaction should be the same.
We know that CP violation occurs in nature-- it's been observed in kaon decay, among other things-- and it's related to the observed asymmetry between matter and antimatter in the universe. The known sources of CP violation are not enough to explain the matter-antimatter imbalance in the universe, though, so there have to be other forms of it out there that have yet to be discovered.
From what we known about the interactions of matter and antimatter, any differences in the antihydrogen states would have to be very small, but laser spectroscopy can be used to do measurements of astonishing precision-- there are single-ion atomic clocks that are good to a few parts in $10^{18}$. Having the target atoms trapped for only a tenth of a second complicates matters, but a group at Argonne National Lab did spectroscopic measurements of the charge radius of unstable helium isotopes that don't last very much longer than that, so it's a good step toward the goal of doing spectroscopy.
Another thing that people talk about testing with anti-atoms is the behavior of gravity. Again, you need to have trapped neutral atoms for this, because electrostatic forces are thirty-some orders of magnitude stronger than gravity. That will also require extreme precision, and many more atoms than have been trapped to date, but the recent experiments are a good start, and the remaining issues are mostly technical, not fundamental.
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1$\begingroup$ CPT symmetry, do you mean? Charge and parity are only conserved together with time in general. $\endgroup$– NoldorinNov 21, 2010 at 22:27
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6$\begingroup$ @Noldorin: ALL Quantum field theories exhibit CPT symmetry. A violation of CPT would be a violation of Lorentz invariance. CP symmetry is approximately observed by most quantum field theories, and until the '70s or '80s (I forget when the experiment was done), was believed to be an exact symmetry of the standard model. You need CP violation to get matter and not antimatter cosmologically. $\endgroup$ Nov 21, 2010 at 23:04
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1$\begingroup$ Note that we already do scattering spectroscopy with anti-protons, and those experiments agree with the proton ones. The big advantage of bound spectroscopy on neutral anti-hydrogen is that the measurements can be made to mind-numbing precision. $\endgroup$ Nov 21, 2010 at 23:15
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1$\begingroup$ I think what is being tested is CPT, violation of which would be quite a bit more surprising than CP violation (every Lorentz invariant theory you can write down is automatically CPT invariant). $\endgroup$– user566Nov 22, 2010 at 2:37
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1$\begingroup$ @Jerry, Moshe: Ah yes, that makes more sense. Thank you. $\endgroup$– NoldorinNov 22, 2010 at 10:52