1000 Solved Problems in Modern Physics

(Romina) #1

10.1 Basic Concepts and Formulae 539


Parity(porπ): The concept of parity was mentioned in Chap. 3. The absolute
intrinsic parity cannot be determined. Parity of a particle can be stated only rela-
tive to another particle. By convention baryons are assigned positive parity. All the
antifermions have parity opposite to the fermions. On the other hand, bosons have
the same parity for particle and antiparticle. Pions and Kaons have odd parity.
Parity is a multiplicative number, so that the parity of a composite system is equal
to the parities of the parts. Thus, for a system comprising of particles A and B,


P(AB)=p(A).p(B).p(orbital motion) (10.3)

wherep(orbital motion)=(−1)lis the parity associated with the relative motion of
the particles,lbeing the orbital angular momentum quantum number (0, 1 , 2 ,...).
Overall parity is conserved in strong and em interactions but is violated in weak
interactions.
Charge – conjugation: (C-parity) is the process of replacing a particle by an antipar-
ticle or a system of particles by the anti particle (s).
In general, a system whose charge is not zero cannot be an eigen function ofC.
However ifQ=B=S=0, the effect ofCis to produce eigen value±1.
Cis conserved in strong and em interactions but not in weak interactions. For
π^0 ,C=+1. For photonC=−1 and for n-photons


C=(−1)n (10.4)

G-parity: The operation G consists of rotation of 180^0 about they-axis orz-axis in
isospin space followed by charge conjugation.
G-parity for the pion is−1 and for baryon it is zero. It is a multiplicative quantum
number. For a system of n pions


G=(−1)n (10.5)

G - parity is conserved in strong interaction and is a good quantum number for
non-strange mesons. For aN−Nsystem,


G=(−1)l+S+I (10.6)

Time reversalmeans changing the sign of time. Strong interactions are invariant
under time reversal as evidenced by the absence of electric dipole moment of neu-
tron and verification of the predicted ratio of forward and backward reactions at the
same energy in the CMS.
IfA+B→C+D,


then

σAB→CD
σCD→AB

=

(2sA+1)(2sB+1)PC∗^2
(2sC+1)(2sD+1)PA∗^2

(10.7)

wheresis the spin of the particles, andp∗is the momentum, the particle beams
being unpolarised.

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