## Monday, 17 May 2010

### New Physics Claim from D0!

Tevatron not dead, or so it seems. Although these days all eyes are turned to the LHC, the old Tevatron is still capable to send the HEP community into an excited state. Last Friday the D0 collaboration presented results of a measurement suggesting the standard model is not a complete description of physics in colliders. The paper is out on arXiv now.

The measurement in question concerns CP violation in B-meson systems, that is quark-antiquark bound states containing one b quark. Neutral B-mesons can oscillate into its own antiparticles and the oscillation probability can violate CP (much as it happens with kaons, although the numbers and the observables are different). There are two classes of neutral B-mesons: $B_d$ and its antiparticle $\bar B_d$ where one bottom quark (antiquark) marries one down antiquark (quark), and $B_s,\bar B_s$ with the down quark replaced by the strange quark. Both these classes are routinely produced Tevatron's proton-antiproton collisions roughly in fifty-fifty proprtions, unlike in B-factories where mostly $B_d,\bar B_d$ have been produced. Thus, the Tevatron provides us with complementary information about CP violation in nature.

There are many final states where one can study B-mesons (far too many, that's why B-physics gives stomach contractions). The D0 collaboration focused on the final states with 2 muons of the same sign. This final state can arise in the following situation. A collision produces a $b \bar b$ quark pair which hadronizes to B and $\bar B$ mesons. Bottom quarks can decay via charged currents (with virtual W boson), and one possible decay channel is $b \to c \mu^- \bar \nu_\mu$. Thanks to this channel, the B meson sometimes (with roughly 10 percent probability) decays to a negatively charged muon, $B \to \mu^- X$, and analogously, the $\bar B$ meson can decay to a positively charged antimuon. However, due to $B \bar B$ oscillations B-mesons can also decay to a "wrong sign" muon: $B \to \mu^+ X$, $\bar B \to \mu^- X$. Thus oscillation allow the $B, \bar B$ pair to decay into two same sign muons a fraction of the times.

Now, in the presence of CP violation the $B \to \bar B$ and $\bar B \to B$ oscillation processes occur with different probabilities. Thus, even though at the Tevatron we start with the CP symmetric initial state, at the end of the day there can be slightly more -- than ++ dimuon final states. To study this effect, the D0 collaboration measured the asymmetry
$A_{sl}^b = \frac{N_b^{++} - N_b^{--}}{N_b^{++} + N_b^{--}}$.
The standard model predicts a very tiny value for this asymmetry, of order $10^{-4}$, which is below the sensitivity of the experiment. This is cool, because simply an observation of the asymmetry provides an evidence for contributions of new physics beyond the standard model.

The measurement is not as easy as it seems because there are pesky backgrounds that have to be carefully taken into account. The dominant background comes from ubiquitous kaons or pions that can sometimes be mistaken for muons. These particles may contribute to the asymmetry because the D0 detector itself violates CP (due to budget cuts the D0bar detector made of antimatter was never constructed). In particular, the kaon K+ happens to travel further than K- in the detector material and may fake a positive value of asymmetry. We have to cross our fingers that D0 got all these effects right and carefully subtracted them away. At the end of the day D0 quotes the measured asymmetry to be
$A_{sl}^b = -0.00957 \pm 0.00251(stat) \pm 0.00146 (syst)$,
that is the number of produced muons is larger than the number of produced antimuons with the statistical significance estimated to be 3.2 sigma. The asymmetry is some 100 times larger than the value predicted by the standard model!

Of course, it's too early to start dancing and celebrating the downfall of the standard model, as in the past the bastard have recovered from similar blows. Yet there are reasons to get excited. The most important one is that the latest D0 result goes well in hand with the anomaly in the $B_s$ system reported by the Tevatron 2 years ago. The asymmetry measured by D0 receives contributions from both $B_s$ and $B_d$ mesons. The $B_d$ mesons are much better studied because they were produced by tons in BaBar and Belle, and to everyone's disappointment they were shown to behave according to the standard model predictions. However BaBar and Belle didn't produce too many $B_s$ mesons (their beams were tuned to the Upsilon(4s) resonance which is a tad too light to decay into $B_s$ mesons), and so the $B_s$ sector can still hold surprises. Two years ago CDF and D0 measured CP violation in $B_s$ decays into $J/\psi \phi$, and they both saw a small, 2-sigma level discrepancy from the standard model. When these 2 results are combined with all other flavor physics data it was argued that the discrepancy becomes more than 3 sigma. The latest D0 results is another strong hint that something fishy is going on in the $B_s$ sector.

Both the old and the new anomaly prompts introducing to the fundamental lagrangian a new effective four-fermion operator that contributes to the amplitude of $B_s \bar B_s$ oscillations:
$L_{new physics} \sim \frac{c}{\Lambda^2}(\bar b s) ^2$ + h.c.,
with a complex coefficient $c$ and the scale in the denominator on the order of 100 TeV. At this point there are no hints from experiment what could be the source of this new operator, and the answer may even lie beyond the reach of the LHC. In any case, in the coming weeks theorists will derive this operator using extra dimensions, little Higgs, fat Higgs, unhiggs, supersymmetry, bricks, golf balls, and old tires. Yet the most important question is whether the asymmetry is real, and we're dying to hear from CDF and Belle. There will be more soon, I hope...

Lumo said...

Thanks, Jester, interesting!

Jester said...

Of course I had a typo in the operator, and now you have one too ;-)

Lumo said...

Huh, so this is really meant to change the strangeness by two? ;-)

I don't have a good model for that, do I?

Jester said...

There are models. This sort of flavor changing operators is even generated at 1 loop in the SM when you exchange W bosons in a box diagram, but obviously this contribution alone does not explain the D0 result. So you need another contribution from KK particles, susy particles, 4th generation, etc...this could be just anybody.

Lumo said...

Right, thanks. I realized that it can be just obtained from two simple cubic vertices.

It could be anybody but do you really have doubts that if it's true, it's supersymmetry?

Anonymous said...

Looks like Dennis Overbye likes to read your blog. lulz. Quick turnaround too. http://www.nytimes.com/2010/05/18/science/space/18cosmos.html

Lumo said...

Dear Anon, I don't want to make too vague bets but I think that he was more likely to get it from my blog. We're Facebook friends, too. ;-)

Anonymous said...

Question: Would an hitherto underscovered Bs resonance have the potential of introducing an asymmetry?
(eg, amplifying the standard model asymmetry?)

Jester said...

Naively, it's unlikely. We have found no surprises in Bd mesons which have been studied in more detail, and the spectroscopy of Bs mesons should be similar (a look at the PDG shows that analogous states have been identified in both sectors). But I'm not an expert in this, so maybe there is a caveat...

Anonymous said...

This is probably a really stupid question, but wouldn't the effective Lagrangian you've written down contribute to

$\bar B_s -> B_s$

oscillations, rather than the reverse (which is what you want)?

Jester said...

In the formula i wrote h.c. stands for the hermitian conjugate of the term i wrote explicitly, and that conjugated term corresponds to Bs -> Bsbar. In fact, this formula should not be taken literally as in reality things are more complicated (there are 8 relevant operators with different lorentz structures and color contractions).

chris said...

I heard some rumors, that today CDF will present a 5-sigma result for new physics from B_s->J/Psi Phi.

interesting stuff...

Jester said...

it's exactly opposite :-)

Anonymous said...

Indeed, the beta_s CDF result is perfectly consistent with the SM. Also, their dimuon asymmetry is of opposite sign compared to that of D0,
although with larger error bars. If one takes the D0 dimuon result and tries to explain it with new physics in B_s mixing, one gets nonsense like sin x = -2.5 (since Delta M_s is very well constrained), see 1005.4238. The bottom line: the D0 result is just another fluke.

coraifeartaigh said...

Super post, many thanks
Cormac