Subnuclear Physics in the 1970s

1. Subnuclear Physics in the 1970s IFIC Valencia. 4-8 November 2013 Lecture 10 Spin and parity 1962-2012 p0 Higgs

2. The p0 If π˚ is in the same triplet as the charged pions it most be shown to have JP=0– as them Lifetime is very short (modern value) In 1960 it was already known to have the decay modes (values are the modern ones) “internal conversion” “double internal conversion”

3. Spin cannot be =1 • Theorem: • spin of a particle decaying in two photons cannot be 1, both JP=1+ and JP=1– are forbidden • The demonstration is simple in the CM frame • The matrix element M should be written using the available vector quantities. • the CM momentum q • the transverse polarisations of the photons, e1 and e2 • With these we should build a quantity that combined with the polarisation of the p0, S, makes a scalar, hence • a vector V if JP=1–MµVS • an axial vector A if JP=1+MµAS • In addition M, and consequently V or A, must be symmetric under the exchange of the photons, which are identical Bose particles, i.e. under the interchange In addition, the photon polarisation must be perpendicular to its momentum, a condition equivalent to the electromagnetic waves being transversal No vector nor axial vector can be built to satisfy these conditions The p0spin cannot be 1.

4. The same differently Work in the CM frame. Quantisation axis = common line of flight J: total angular momentum of the two photons; mJ: its third component l: orbital momentum of the two photons; ml: its third component S: orbital momentum of the two photons; mS: its third component The orbital momentum is perpendicular to the line of flight: ml=0 Possible values of S are 0, 1 and 2 The third components of the spin of each of the gammas, on its line of flight, can be +1 or –1, but not 0. Hence, a priori possible values of mS = mJare 0, +2 and –2. However, as J=1, only mS = mJ = 0 The two gammas being identical Bose particles, their state must be completely symmetric If the state of orbital momentum is antisymmetric (l odd), such must be also the spin state (S=1) symmetric (l even), the latter must be symmetric too (S=0 or 2).

5. The same differently If the parity is negative, JP=1–, l should be odd, namely l =1, 3,… Of these only l=1 can give J=1 with S=1. The quantum numbers of the di-gamma system are then completely defined: But If the parity is positive, JP=1+, l should be even: l =0, 2,… Bose symmetry, S=0, 2. Of the different couples of values of l and S, only l=2, S=2 can give a total angular momentum J=1 (l cannot be larger than 2) Again, the quantum numbers of the di-gamma system are completely defined But

6. Scalar or pseudoscalar There is one scalar and one pseudoscalar combination of the vector and axial vectors of the problem, namely aS and aP are functions of the invariant kinematical quantities of the system, called form factors Transverse polarisation vectors of the photons tend to be parallel for scalar, perpendicular for the pseudoscalar Consider the very rare decay in two virtual photons that convert internally in electron pairs They have masses m1 and m2 ≠ 0 The polarisation vectors are correlated with the normal to the plane of the pair Kroll and Wada (1955) have shown that aS and aP are differ only for the sign, (>0 for the scalar, <0 for pseudoscalar) and that the distribution of the angle f between the two planes is given by where i=S or P

7. The bubble chamber experiment p–mesons from the Nevis Cyclotron were slowed down by polyethylene absorber and stopped in a hydrogen bubble chamber 12 in. in diameter, 6 in. in depth, placed in a 0.55 T magnetic field happens for 62% of stopped pions Data: 380 000 picture with in average 15 pions per picture Then: scanning, measuring points on the three track images and spatially reconstructed about 100 000 single internal conversion 206 double internal conversion 60 events discarded, because not completely measurable 146 good events remaining In principle there is an ambiguity in defining which e– and e+ belong to each pair. It can be solved for almost all events on statistical basis (one configuration is much less likely than the other) N. P. Smios et al. Phys. Rev. 126 (1972) 1844

8. Kinematic variables Define Kroll and Wada (1955) have shown that results can be presented in the approximate form where A anda are known functions and the two signs are for the two parities. A cut needed to have the approximation valid leaves with 112 events Compare theoretical expectation of x and y with data y distribution N. Kroll and W. Wada, Phys. Rev. 98 (1955) 1355 x/pionmass distribution

9. Results The statistical value of a particular event in discriminating between the two signs in the asymmetry is proportional to a2. Hence weight each event with the corresponding a2. weighted frequency distribution of the angle between planes of polarization

10. KTeV KTeV was an experiment at Fermilab dedicated to CP violation in the K0 system p0 s come from the fully reconstructed

11. KTeV pio spin parity

12. KTeV on p0 • The analysis relies on two core systems of the detector • a drift chamber-based charged particle spectrometer • a cesium iodide (CsI) electromagnetic calorimeter. Electrons are identified as charged particles whose entire energy is deposited in the CsI, while photons are reconstructed from showers in the CsI with no associated charged tracks. E. Abuzaid et al. Phys.Rev.Lett.100:182001,2008

13. KTeV p0 has JP=0– indeed more than 30 000 Scalar contribution <3% E. Abuzaid et al. Phys.Rev.Lett.100:182001,2008

14. CMS

15. ATLAS

16. ATLAS

17. CMS idea

18. CMS Ys

19. CMS 4 muons

20. ATLAS 4 gammas

21. Higgs in 2 gammas

22. Higgs in ZZ* Analysis is done for the ZZ*. One at the Z mass, one at lower, depending on the event, mass As in the p0 case, but let us be more general

23. Spin parity J=0 In this case we have L=S If the parity of the final state is positive, L must be even and we have two possibilities, L=S=0 and L=S=2. Bose statistics is satisfied because, in both cases, the spin and space wave functions are symmetrical under the exchange of the two Zs, and hence the total wave functions are such. We add together the two terms to have the most general expression, namely In the first addendum the two polarisation vectors combine to make a scalar, for S=0, and q does not appear, corresponding to L=0. In the second term q appears twice, corresponding to L=2 and the polarisation vectors combine, corresponding to S=2. For the pseudoscalar final state there is only one possibility, L=S=1. We write The polarisations combine to make an axial vector, corresponding to S=1, and q appears once, corresponding to L=1. Both the spin and the space wave functions are antisymmetric, making the total one symmetric, respecting Bose statistics. If the decay conserves parity, only MS is present in the decay of a scalar and only MPof a pseudoscalar, both are present if parity is violated

24. H spin parity

25. H spin parity

26. H spin parity

27. Results The three form factors ai are functions in particular of m1 and m2, and, consequently, provide information to distinguish between different cases. In the event sets selected by the experiments one of the dilepton masses, m1, is close to the Z mass, while the smaller one, m2, varies in a wide range from one event to the other. Consequently, the momentum q varies in the even set, being larger when m2 is smaller. In the three terms q appears at different powers: 0, 1 and 2. Hence, the m2 spectrum is enhanced by a factor increasing with this power The SM predicts completely the form factors for the Higgs Parity is conserved, hence a3=0 In addition, a2=0 at the first order, with small corrections The likelihood is calculated for the event sample. The result is compared to the expected distributions of this quantity in the Higgs hypothesis versus a scalar boson decay with a larger a3or a pseudoscalar boson. Notice however, that the SM does not give information on the form factors for bosons different from the Higgs and that additional assumptions are needed. The conclusion is that the data favour a3and a2consistent with 0, in accordance with the SM

28. Spin 1 If the particle observed to decay in two gamma is the same particle spin cannot be 1, but a priori do not assume that If if JP=1–, L must be odd, hence L=1 and only one q factor appears. There is only a possibility for S to combine to make J=1, namely S=1. We write the symmetric expression For the JP=1+ case, arguments similar to those we just made would lead to L=S=2. However, the fact that E1 and E2 are different allows also a simpler choice, namely L=0, S=1 Here the factor E1–E2 compensates the antisymmetry of the spin wave function under the exchange of the Zs and Bose statistics is satisfied Notice that these expressions confirm that for massless vector bosons, like the photons and the gluons, V=0, because and , and A=0 because E1=E2. Notice also that A=0 also for two massive vector bosons when E1=E2. The form factors b1 and b2 are functions of the kinematical quantities both in production and in decay. These functions, once more, are not given by the SM. Additional assumptions are needed

29. 0+ vs. 1– and 1+

30. Spin 2 L must be even: = 0, 2, 4. S must be symmetric = 0, 2 (L,S) – (0,2); (2,0); (2,2); (2,4) 2+ L must be odd: = 1, 3. S must be antisymmetric = 1 (L,S) – (1,1); (3,1); (2,2) 2– Form factors are not foreseen by the SM. Need much more statics to conclude