Subnuclear Physics in the 1970s

1A. Bettini LSC, Padova University and INFN

Subnuclear Physics in the 1970s

19 Apr 2023

IFIC Valencia. 4-8 November 2013Lecture 10

Spin and parity 1962-2012

Higgs


The

19 Apr 2023

0 → γγ BR= 98.823± 0.034( )%

τ =85.2 ±1.8 asIf π˚ is in the same triplet as the charged pions it most be shown to have JP=0– as themLifetime 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”


Spin cannot be =1

19 Apr 2023

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 frameThe 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 0, S, makes a scalar, hence

•a vector V if JP=1– MVS•an axial vector A if JP=1+ MAS

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 e1 ↔ e2 q↔−q

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 conditionsThe 0 spin cannot be 1.

e1 ⋅q= e2 ⋅q=


The same differently

19 Apr 2023

The orbital momentum is perpendicular to the line of flight: ml=0Possible values of S are 0, 1 and 2The 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 = mJ are 0, +2 and –2. However, as J=1, only mS = mJ = 0

Work in the CM frame. Quantisation axis = common line of flightJ: total angular momentum of the two photons; mJ: its third componentl: orbital momentum of the two photons; ml: its third componentS: orbital momentum of the two photons; mS: its third component

The two gammas being identical Bose particles, their state must be completely symmetricIf 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).


The same differently

19 Apr 2023

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

l,ml S,mS =2, 2, to give J ,mJ =1,

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:

l, ml S, mS =1, 1, to give J ,mJ =1,

1,0;1,0 1,0 =But

2,0;2,0 1,0 =But


Scalar or pseudoscalar

19 Apr 2023

There is one scalar and one pseudoscalar combination of the vector and axial vectors of the problem, namely

M =aSe1 ⋅e2 scalar

M =aPe1×e2 ⋅q pseudoscalar

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 pseudoscalarConsider the very rare decay in two virtual photons that convert internally in electron pairsThey have masses m1 and m2 ≠ 0

The polarisation vectors are correlated with the normal to the plane of the pairKroll 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 between the two planes is given by

dN

dφdm1dm2

∝1±ai m1,m2( )cos2φ

where i=S or P


The bubble chamber experiment

19 Apr 2023

– 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

−p → π 0n happens for 62% of stopped pions

Data: 380 000 picture with in average 15 pions per pictureThen: scanning, measuring points on the three track images and spatially reconstructed

about 100 000 single internal conversion206 double internal conversion60 events discarded, because not completely measurable146 good events remainingIn 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


Kinematic variables

19 Apr 2023

Define xi ≡ Ei+ +Ei

−( )

2− pi

+ +pi−

( )2⎡

⎣⎢⎤⎦⎥ yi ≡

Ei+ −Ei

−

pi+ +pi

−

Kroll and Wada (1955) have shown that results can be presented in the approximate form

Pt =A x1,x2 ,y1,y2( ) 1±α x1,x2 ,y1,y2( )⎡⎣ ⎤⎦cos2φ

where A and α are known functions and the two signs are for the two parities. A cut needed to have the approximation valid leaves with 112 eventsCompare theoretical expectation of x and y with data

y distribution

x/pionmass distribution

N. Kroll and W. Wada, Phys. Rev. 98 (1955) 1355


Results

19 Apr 2023

weighted frequency distribution of theangle between planes of polarization

The statistical value of a particular event in discriminating between the two signs in the asymmetry is proportional to α. Hence weight each event with the corresponding α.

αexp =2

nα i

2

1

n

∑ cos2φi = −0.41± 0.24

αtheoS = +0.47

αtheoP = −0.47

scalardn

d∝ cos2φ=1+cos2φ

Pt =A x1,x2 ,y1,y2( ) 1±α x1,x2 ,y1,y2( )⎡⎣ ⎤⎦cos2φ


KTeV

19 Apr 2023

KTeV was an experiment at Fermilab dedicated to CP violation in the K0 system0 s come from the fully reconstructed KL

0 →


KTeV pio spin parity

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KTeV on

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The analysis relies on two core systems of the detector•a drift chamber-based charged particle spectrometer•a cesium iodide (CsI) electromagnetic calorimeter.

x1 =m2 e1

+e1−

( )m

2

x2 =m2 e2

+e2−

( )m

2

E. Abuzaid et al. Phys.Rev.Lett.100:182001,2008

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.


KTeV has JP=0– indeed

19 Apr 2023

E. Abuzaid et al. Phys.Rev.Lett.100:182001,2008

Scalar contribution <3%

more than 30 000

0 → e+e−e+e−


CMS

19 Apr 2023


ATLAS

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CMS idea

19 Apr 2023


CMS Ys

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CMS 4 muons

19 Apr 2023


ATLAS 4 gammas

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Higgs in 2 gammas

19 Apr 2023


Higgs in ZZ*

19 Apr 2023

Analysis is done for the ZZ*. One at the Z mass, one at lower, depending on the event, massAs in the 0 case, but let us be more general


Spin parity J=0

19 Apr 2023

In this case we have L=SIf 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

M P =a3e1×e2 ⋅q

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 MP of a pseudoscalar, both are present if parity is violated


H spin parity

19 Apr 2023


H spin parity

19 Apr 2023

dΓdM * ∝β2 J +1 β = mH −mZ( )

2 −M *2


Results

19 Apr 2023

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 HiggsParity is conserved, hence a3=0In addition, a2=0 at the first order, with small correctionsThe 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 a3 or 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 a3 and a2 consistent with 0, in accordance with the SM


Spin 1

19 Apr 2023

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

A =b2 E1 −E2( )e1×e2

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


0+ vs. 1– and 1+

19 Apr 2023


Spin 2

19 Apr 2023

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)

Form factors are not foreseen by the SM. Need much more statics to conclude

Subnuclear Physics in the 1970s

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