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Anna Phan | USLHC | USA

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Needle in a haystack

We are back to discussing B physics today, with the observation of the rare decay: \(B^- \rightarrow \pi^- \mu^+ \mu^-\). So what is this decay? It’s a \(B^-\) meson (made of a b and an anti-u quark) decaying into a \(\pi^-\) meson (made of a d and an anti-u quark) and two muons. And why is it so rare? Well, it’s a flavour changing neutral current decay. Which means that there’s a change in quark flavour in the decay, but not charge. This type of decay is forbidden at tree level in the Standard Model and so has to proceed via a loop, which can be seen in the centre of the Feynman diagram below.

If you look closer at the loop, you can see that for the decay to occur, a b quark needs to change flavour to a t or c quark, which then needs to change to a d quark. This is another reason why this decay is so rare. Transitions in quark flavour are governed by the CKM matrix, which I illustrate on the right, where the larger squares indicate more likely transitions. So while the transition from b to t is likely, the transition from t to d is very unlikely, and the b to c and c to d transitions are both fairly unlikely. This means, that whichever path is taken, the b to d quark transition is very very unlikely.

Okay, now to the LHCb result. Below I have a plot of the fitted invariant mass for selected \(\pi^-\mu^+ \mu^-\) candidates, showing a clear peak for \(B-\) decays (green long dashed line). Also shown are the backgrounds from partially reconstructed decays (red dotted line) and misidentified \(K^-\mu^+ \mu^-\) decays (black dashed line). Candidates for which the \(\mu^+ \mu^-\) pair is consistent with coming from a \(J/\psi\) or \(\psi(2S)\) are excluded.

We see around 25 \(B^- \rightarrow \pi^- \mu^+ \mu^-\) events and measure a branching ratio of approximately 2 per 100 million decays. This result makes this decay the rarest \(B\) decay ever observed!

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7 Responses to “Needle in a haystack”

  1. Hey Anna, as a BaBarian and a B-physics enthusiast I just want to say that this in amazing result! To see a branching fraction that is of the order of 10^-8 means that LHCb is well and truly past the reach of the B factories. I can’t wait to see what else LHCb has to offer.

  2. Xezlec says:

    Neat. But why does the intermediate quark have to be a top or a charm? Why not up?

    • Anna Phan says:

      Yes, you are right, the intermediate quark can also be an up quark. However, this transition is just as unlikely as the other two.

  3. Nikola Nikolov says:

    Is the rate of that decay compatible with the SM prediction?

    • Anna Phan says:

      Hi,

      Thanks for taking an interest! Yes, the measured branching ratio [(2.4 ± 0.6 (stat) ± 0.2 (syst))×10−8] is compatible with the Standard Model prediction [(1.96 ± 0.21)×10−8]. I would have made a point of it if it wasn’t!

      Cheers,
      Anna

  4. Joe Prokop says:

    Hi Anna,

    With regard to the B+ decay, should not the B+ consist of an UP quark and an anti B quark to give a charge of + 1. I enjoy your Blog.

    Thanks

    • Anna Phan says:

      Hi,

      Thanks for noticing that the Feynman diagram didn’t match the text. I’ve changed everything to be B-.

      Cheers,
      Anna

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