1. Alessandro Fois Detection of  particles in B meson decay

2. What are  particles?  (pronounced “eta”) particles are mesons consisting of a quark anti-quark pair: i.e. uu, dd, ss We will try to detect them here in simulated data as a product of the rare decay B +   l +

3. How do we detect the  particles?  particles have a very short mean lifetime (approximately 10 -18 seconds). They decay too soon to reach the detector!!! We reconstruct them from their products when they decay: the main decay mode is   2  (approximately 28%)

4. A typical B decay Which are the right photons? Does the sum have to have the eta’s mass? Energy? In which frame of reference? Particle : 1 id: 13 Px: 1.6650 Py: 0.2804 Pz: 2.6981 E : 3.1847 mass: 0.1057 Particle : 2 id: 211 Px: 0.5850 Py: 0.3135 Pz: -0.5261 E : 0.8584 mass: 0.1396 Particle : 3 id: -211 Px: 0.0012 Py: -0.4991 Pz: 1.1310 E : 1.2441 mass: 0.1396 Particle : 4 id: 211 Px: 0.2274 Py: 0.1597 Pz: 0.0386 E : 0.3133 mass: 0.1396 Particle : 5 id: 211 Px: 0.2504 Py: -0.0573 Pz: 0.2075 E : 0.3585 mass: 0.1396 Particle : 6 id: -211 Px: 0.1456 Py: -0.0747 Pz: 0.1264 E : 0.2495 mass: 0.1396 Particle : 7 id: 22 Px: 0.0110 Py: -0.0160 Pz: 0.0173 E : 0.0260 mass: 0.0000 Particle : 8 id: 22 Px: 0.0033 Py: -0.1451 Pz: 0.1128 E : 0.1838 mass: 0.0000 Particle : 9 id: 22 Px: 0.0178 Py: -0.1115 Pz: 0.0870 E : 0.1426 mass: 0.0000 Particle : 10 id: 22 Px: 0.0283 Py: 0.0084 Pz: 0.0206 E : 0.0360 mass: 0.0000 Particle : 11 id: 22 Px: 0.0398 Py: 0.1004 Pz: 0.0636 E : 0.1254 mass: 0.0000 Particle : 12 id: 22 Px: -0.1659 Py: 0.3464 Pz: 0.1286 E : 0.4051 mass: 0.0000 Particle : 13 id: 22 Px: -0.0383 Py: 0.0631 Pz: 0.0012 E : 0.0738 mass: 0.0000 Particle : 14 id: 22 Px: -0.0240 Py: 0.0235 Pz: 0.0021 E : 0.0336 mass: 0.0000 Particle : 15 id: 22 Px: -0.0756 Py: 0.0514 Pz: 0.0050 E : 0.0916 mass: 0.0000 Particle : 16 id: 22 Px: -0.4086 Py: -0.3953 Pz: -0.1300 E : 0.5832 mass: 0.0000 Particle : 17 id: 22 Px: -0.1085 Py: -0.1177 Pz: -0.0594 E : 0.1707 mass: 0.0000 Particle : 18 id: 22 Px: 0.0103 Py: -0.0236 Pz: -0.0076 E : 0.0269 mass: 0.0000 Particle : 19 id: 22 Px: -0.3842 Py: -0.1175 Pz: -0.1586 E : 0.4319 mass: 0.0000 Particle : 20 id: 22 Px: -0.5621 Py: -0.1833 Pz: -0.3246 E : 0.6745 mass: 0.0000 Particle : 21 id: 22 Px: -0.0249 Py: 0.0285 Pz: -0.0193 E : 0.0425 mass: 0.0000 Particle : 22 id: -14 Px: -1.0172 Py: 0.3658 Pz: 1.0798 E : 1.5279 mass: 0.0000

5. Invariant Mass We use the quantity “invariant mass” which is conserved across frames of reference: inv. mass = (E 2 – p 2 c 2 ) 1/2 The  particle has invariant mass 0.547GeV: hence we seek photon pairs with combined invariant mass of this value.

6. Photon pairs We consider the invariant masses of all possible pairs of photons in an event: lots of background! Two salient peaks: - 0.1 GeV  0  - 0.55 GeV 

7. Reconstructing the B As, mentioned before, B +   l + Now consider invariant masses of all pairs of photons combined with the lepton and neutrino for each event:

8. Reconstructing the B There is no peak! Maximum is at about 4 GeV: B has invariant mass 5.29 GeV. Clearly too much background: can we isolate the real decays?

9. Cuts To isolate the real decays, we can take “cuts”: set criteria for the decay to be accepted as “real”. Obviously, the photon pairs must have invariant mass of about 0.5-0.6 GeV to be  candidates. Energy cuts: there are many photons with tiny energies. We only accept photon pairs with each photon above a certain energy, E. The result?

10. Energy Cuts: before and after E > 0.2 GeV Peak at 5.3 GeV! No energy cut

11. Further Cuts: Momentum Since the  only has invariant mass 0.55GeV, while the B has invariant mass 5.27 GeV, the  s must be fast for invariant mass (and energy) to be conserved in the decay. So, we cut on momentum in the centre of mass frame. But what is the best cut for momentum?

12. Determining the best momentum cut Consider the 2D histogram of momentum of  vs. invariant mass of  (no cuts):

13. There is no obvious clustering around the “  window” of invariant mass near 0.55GeV. But, if we use our previous energy cuts…

14. Determining best  momentum E > 0.2 GeVE > 0.1 GeV Clustering between 0.5 and 0.6 GeV!

15. Select a window Accept only photon pairs in this window: i.e. combined momentum > 1.0 GeV/c, combined invariant mass between 0.5 and 0.6 GeV.

16. The Result Spike at 5.3 GeV!!!

17. How good are the cuts? Do our cuts filter well when applied to random data? Run our cuts on Phil’s data (no  s): Looks similar, but: Much smaller and broader distribution (maximum only 250 (c.f. 500 when run on  s), and there are 3 times as many events in Phil’s file. Distribution is centred about 4.9 GeV (c.f. 5.3 GeV when run on  s)

18. Another Trick: Beam Constrain We know that our reconstructed B will have 0 momentum and 5.27 GeV of energy in its own frame of reference (it is stationary in its own reference frame!!) In fact, we set E = 5.29 (a correction, since the B’s frame is not the CoM frame), and calculate the invariant mass of our reconstructed B in the centre of mass frame. Upon plotting a frequency histogram of this quantity…

19. Another Trick: Beam Constrain Data with no  particles, with cuts Without cutsWith cuts In all cases, we get a distribution around about 5.27GeV, but the less background, the sharper the spike.

20. One Last Trick:  E Since energy is conserved, we can also make use of the quantity  E =  E i – 5.29, where the E i are the energies in the centre of mass frame of the pair of photons, lepton and neutrino that reconstruct the B. Since the E i should add to the centre of mass frame energy of the B, 5.29 GeV, for the actual decay products, we should get a distribution with a spike at  E = 0.

21. One Last Trick:  E Data with no  particles, with cuts: More background, peak at  E = -0.3GeV With cuts: Background cut, peak at  E = 0 Without cuts: Background distribution, no spike

22. The real candidates If we plot a 2D histogram of  E vs. the beam constrained invariant mass, we should have the “real” Bs separating from the background and clustering about  E = 0 and inv. mass = 5.29 GeV Data with no  particles, with cuts With cutsWithout cuts

23. Conclusions and Beyond Our cuts yield a peak at the required value for the reconstructed B, but only 7304 candidates remain from an original sample of 163384 h s, thus giving a 2% efficiency. Our cuts also isolate the real decays using the 2D plots. Our cuts preferentially select  s, filtering out 4 times as much background as  s. To go further, more work would need to be done making the cuts more precise and efficient. For example, we could select only the fastest photon in each event as one candidate for the reconstructed .

24. Acknowledgements: Paul Harrison, “Natures flawed mirror”, Physics World, July 2003 K. Hagiwara et al. (Particle Data Group), Phys. Rev. D 66, 010001 (2002) Young and Freedman, University Physics, 10 th ed., 2002, p. 1032-42 Kevin Varvell, for putting (and keeping) me on the right track.