BARYON OCTETS   When we rotate the baryon hexagon so as to align Y with horizontal axis, the rotation is by angle x clockwise about x-axis. Then the  coordinates are

p (√5/2,0) , n (3/2√5 , -2/√5) ,Σ+(1/√5, 2/√5) ,Σ-(-1/√5, -2/√5),Ξ- (√5/2, 0) , Ξ0(-3/2√5 ,2/√5)

Name Quark composition S B  Y/2

= (B+S) / 2

Iτ Q=Iτ + Y/2 spin mass*e lifetime: τ

in sec

Σ+ uus -1 1 0 1 1 1/2 2328 0.8*10-10
Σ0 uds -1 1 0 0 0 1/2 2334 0.8*10-10
Σ- dds -1 1 0 -1 -1 1/2 2343 10-14
Ξ- dss -2 1 -1/2 -1/2 -1 1/2 2586 1.7*10-10
Ξ0 uss -2 1 -1/2 1/2 0 1/2 2573 3*10-10
p+ udu 0 1 1/2 1/2 1 1/2 1836.1
n0 udd 0 1 1/2 -1/2 0 1/2 1836.6 960
Λ0 uds -1 1 0 0 0 1/2 2183 2.5*10-10
no. of possible configurations : 10 [ uuu,sss,ddd], [uus,uss,uud,udd,dds,dss,uds]

First set marked blue  do not exist in octets , but in decuplets; Δ++=uuu,

Δ-=ddd; Ω-=sss ;

* 3 quarks involved. No anti-quark.

* S for strangeness, B for Baryon number, Iτ for iso-spin projection, Y for hyper charge and Q for charge. Q=Iτ + Y/2 is the Gell-Mann Nishijima formula for strong interaction.

* Up quark -> ( 1         down quark -> ( 0       strange quark -> ( 0

0                                   1                                     0

0 )                                0 )                                   1 )

in 3-D Vector Space.

* The laws of Physics are invariant under application of unitary transformation to this space i.e  ( x             (x

y    =  A   y

z )            z )

Where A is a 3x3 unitary matrix under SU(3)

* if one takes A = ( 0  1  0

-1  0  0

0  0  1 ) and applies the transformation on up quark, it becomes  a down quark and vice versa. This is known as flavour rotation.

* when a proton is transformed by every possible flavour rotation A, it turns out that it moves around in an 8-dimensional vector space. These 8 dimensions correspond to 8 particles in the so called baryon octet.

* Every Lie Group has a corresponding Lie Algebra and each group representation of the Lie group can be mapped to a corresponding representation of the Lie Algebra in the same vector space. The Lie Algebra su(3) can be written as the set of 3x3 traceless hermitian matrices. We normally discuss the representation theory of Lie Algebra su(3) in stead of Lie Group SU(3) since the former is simpler and both are equivalent.

* The abstract group SU(3) is represented by a set of eight 3×3 matrices of complex elements which have determinant of unity. These elements of the group can be generated by eight special matrices. These matrices must be Hermitian; i.e., the transpose of their complex conjugates is the same as the matrix. These matrices do not have determinants of unity; instead all have traces (sums of elements on the principal diagonal) of zero.

* If Gell-Mann matrices are represented as λi, in su(3) algebra, the generators are gi = λi  /2 . These matrices are traceless, hermitian & can generate unitary matrix group elements of SU(3) through exponentiation. They obey the rule of trace orthonormality where the orthonormality is 2 instead of 1. Any element of SU(3) can be written as exp(iθj gj) where 8  θj are real numbers.

* In math, orthonormality means a norm which is unity. Gell-Mann matrices are however normalized to a value of 2. Thus the trace of the pair of products results in orthonormalization condition ->Tr(λi λj ) =2δij

* In representing baryon octets, We can migrate from I3 -Y basis to U3-Yu basis where U is a new mathematical concept called U- spin or magnetic spin quantum number and corresponding hypercharge is Yu. The z-axis projection of U-spin is U3.

Uτ=-(1/2)Iτ +(3/4)Y or 4Uτ = -2Iτ + 3Y

and

Yu =-Q which simplifies to

Uτ = Y + Yu /2   analogous to

Q=   Iτ  + Y/2

Similar to conservation of charge, there is conservation of U spin corresponding to magnetic moment.

Q=   Iτ  + Y/2

Q=  2Y - 2Uτ

Q =4/3 * (Iτ + Uτ/2)

Migration from Q to Uτ

Q=   Iτ  + Y/2

Uτ= (-1/2)Iτ +(3/ 4)Y

Here , the coefficient of  Iτ has decreased by 150% and the coefficient of Y has increased by 150%.

+Baryon Octets : U- Spin
 particle Y Yu Yu /2 Uτ Σ- 0 1 1/2 1/2 Ξ- -1 1 1/2 -1/2 n0 1 0 0 1 Ξ0 -1 0 0 -1 Σ0, Λ0 0 0 0 0 p+ 1 -1 -1/2 1/2 Σ+ 0 -1 -1/2 -1/2 *(Σ-,Ξ-) form doublets as regards U-spin and have same magnetic moment .  (p+,Σ+)form doublets as regards U-spin and have same magnetic moment .   (Σ0, Λ0,Ξ0,n0) form triplets as regards U-spin and have same magnetic moment .Concept of Magnetic Moment of Nucleons:If nucleons are assumed as Dirac particles, the magnetic moment of a nucleon denoted by μ(N) is given by eℏ/(2mp) in SI units where ℏ is reduced planck constant and mp is the rest mass of the proton.  μ(N) is called Nuclear Magneton in analogy to Bohr Magneton for electron. But in reality, the nucleons are not dirac particles and henceμ(p) =2.793μ(N)μ(n)=-1.913μ(N)  where p,n denote proton and neutron respectively. Magnetic moment of a nucleon μ is due to the combined effects of Orbital Angular Momentum (l) and Spin Angular Momentum(s). μ =gl*l + gs*s and l,s,μ are vectors and gl, gs are linear coefficients. In the units of nuclear magneton μ(N), ✑  gl=0 for neutron  ✑ gl=1 for proton  ➲ gs=-3.8263 for neutron  ➲ gs=5.5858 for proton   More on U spin : Q =   1         1/2   *  Iτ  Uτ    -1/2      3/4     Y Where A= 1       1/2 -1/2  3/4 det A =1 if we take sin x =1/2 and cos x1=  1       cos x2=3/4 then cos2 x = cosx1*cosx2 we can substitute A by A1= cos x      sin x -sin x     cos x which is a rotational matrix, we get A1= cos 30     sin 30 -sin 30    cos 30 and Q'    =     (√3/2) Iτ    +    (1/2)Y U'τ   =   (-1/2) Iτ   +     (√3/2)Y we have to find the physical meaning of transformed Q and U .  * You shall observe that

u = Y  +   Yu/ 2

Q = ( Y' -Yu'  ) / √2

where

Y'    = R  *  Y

Yu'              Yu

and

R=  cos 45    -sin  45

sin  45      cos 45

*Below we consider two applications of the U-spin: SU(3) predictions for the magnetic moments of the octet and the transition magnetic moments of the anti-decuplet. It follows from the assumption of the U-spin conservation that the magnetic moments (and electric charges) of all members of the same U-spin multiplet are equal. From the left panel of Fig. 4, one then immediately obtains that µΣ− = µΞ− , µΞ0 = µn , µp = µΣ+

* Gellmann-Okubo G-O formula for mass of baryons : link :1

M =a1 +a2Y +a3[I(I+1)-Y2 /4] where a1,a2, a3 are experimentally determined. The formula works for baryons within 0.5% of measured value.

* We observe that charge of u quark is 2/3 whereas charge of d quark is -1/3. Question arises why there is so much asymmetry between charges of both quarks while p,e have same magnitude of charge. It may be that charge is not an indivisible attribute and consists of multiple more fundamental attributes which are symmetric. Supposing that there are 2 attributes a , b such that a+b=2/3 and  a-b=-1/3 . solving them, a=1/6, b=1/2 . So now u,d are more symmetric in their charges since a is same for u and d and same is the case for b attribute. In fact, it seems b is I3  and a is Y/2.

* It is quite likely that u is fractional because of rotation of charge q which is distributed on the surface of a sphere with radius r=1 which gives effective charge as (2/3)qr2 =2q/3 similar to the moment of inertia of a hollow sphere about an axis passing through the center.

MESON OCTETS

(Pseudo Scalar Mesons) Name Quark composition S B   Y/2=

(S+B) / 2

Iτ Q=Iτ + Y/2 spin mass*e lifetime: τ

in sec

π+ ud' 0 0 0 1 1 0 273 2.6*10-8
π0 uu' 0 0 0 0 0 0 264 0.8*10-16
π- u'd 0 0 0 -1 -1 0 273 2.6*10-8
K- u's -1 0 -1/2 -1/2 -1 0 966 1.2*10-8
K0 ds' 1 0 1/2 -1/2 0 0 974
K+ us' 1 0 1/2 1/2 1 0 966 1.2*10-8
K0' d's -1 0 -1/2 1/2 0 0 974
η0 dd' 0 0 0 0 0 0 1074 10-17
η' ss' 0 0 0 0 0 0

no. of possible configurations : 21 [ (uu,ss,dd),(ud,us,ds),(u'u',d'd',s's'),(u'd',u's',d's')], [(uu',dd',ss'),(ud',u'd,us'.u's,ds',d's)] . First set numbering 12 marked blue  do not exist because they shall result in fractional charges and they are either all quarks or all anti-quarks.

*From(u,d,s) , the combination of 2 quarks is 3c2 =3 (ud,us,ds)

*From(u',d',s') , the combination of 2 quarks is 3c2 =3 (u'd',u's',d's')

*From (u,d,s) to (u',d',s'), combination of 2 is  3 x3 =9 [(uu',dd',ss'),(ud',u'd,us'.u's,ds',d's)]

*With repetition of (u,d,s) , combination is  3 (uu,ss,dd)

*With repetition of (u',d',s') , combination is  3 (u'u',d'd',s's')

First 2 sets do not occur , because the charges are not integers. 3rd set of 9 is fully exhausted in meson octet with spin 0 and vector meson octet of spin 1. 4th and 5th sets again do not occur because charges are not integers.

If we consider the combination of non integer charges, total is 12 out of which 6 are negative of another 6. The frequency distribution of fractional charges is

4/3  ---- 1

1/3 -----2

-2/3--- 3

weighted average= [(4/3)*1+(1/3)*2 +(-2/3)*3] / 6 = 0/6 =0

Same is true for -4/3,-1/3.2/3 .

* Only 2 quarks / anti-quarks involved in mesons whereas in baryons , the number is 3. In mesons, each particle has 1 quark and 1 anti-quark.

*Every meson is the bound state of a quark and anti-quark unlike baryons where anti-quarks are not involved in octet. Hence the blue marked combinations excluded. Moreover, in the Baryon octet, there are no anti-baryons whereas in the meson octet, there are anti mesons.(π+-) ,(K+,K- ), (K0 ,K0') are anti-particles of each other. Thus in the meson octet, there are only 6 particles- 3 having their anti-particles and 3 (π0 ,η0 ,η')having no separate anti-particles

* η'  is called the eta prime meson.

*Only Gauge invariant objects are observable. Since quarks and gluons are not gauge invariant, they are not observable. Hadrons, being gauge invariant, are observable.

* Meson nonet can be put in matrix form

uu'  du'  su'

ud' dd'  sd'

us' ds'   ss'

In terms of charge Q, the matrix is as under

0 -1 -1

1 0   0

1 0  0

Since 2 rows are identical, this is singular, traceless , skew symmetric matrix.

* In a st. line along  Iτ , if we take the maximum absolute value of  Iτ and denote it by n, then total no. of particles in that line will be 2n+1. Exa- If n=1, then no. of slots in that line =2*1+1=3

Taking only 1st generation quarks into consideration, total no. of arrangement is 10 out of which 4 have integer charge(π+.π-.π00) and 6 fractional charge . The ratio of fractional to integer charge is 3:2:

(u,u',d,d') no. of combination - 4 C 2 =6

uu'    Q =0

dd'    Q=0

ud'    Q=1

u'd   Q=-1

ud    Q= 1/3 (this is equal to the charge of s' quarks with  Iτ=0 but spin 1/0)

u'd'  Q=-1/3 (this is equal to the charge of s quarks with  Iτ=0 but spin 1/0)

Then with repetition (no. of arrangements=4)--

uu    Q=4/3

u'u'  Q=-4/3

dd   Q=-2/3

d'd'  Q= 2/3

Combinations with blue and red color are either all quarks or all anti quarks and have fractional charges. we can arrange the charges as

0/3. 1/3. 2/3,3/3,4/3 with plus and minus sign and 4 combinations (2 each) of black color have integer charges.

Below combination (4 in number in red) will have only spin zero and spin 1 is ruled out because all quantum no. of fermions cannot be identical. Of course, this paved the way for a new quantum no. called color quarks. We make a table of quarks of 1st generation with fractional charges--

 combination spin Iτ Y Y/2 Q uu 1/0 1 2/3 1/3 4/3 u'u' -1/0 -1 -2/3 -1/3 -4/3 d'd' 1/0 1 -2/3 -1/3 2/3 dd -1/0 -1 2/3 1/3 -2/3 ud 1/0 0 2/3 1/3 1/3 u'd' -1/0 0 -2/3 -1/3 -1/3 Suppose above combinations add 1 quark and anti quark to make a triplet . then there will be total 24 combinations out of which 8 will be degenerate thereby eliminating 4 and hence left out will be 20. again, the ratio of fractional charges to integer charges combination shall be in the ratio 3:2. So 12 combinations will have fractional charges and 8 shall have integer charges which include 4 degenerate triplets as under:-- udu , uud -- p+,∇+ u'd'u',u'u'd' =p-,∇- ddu,udd = n0,∇0 d'd'u' , u'd'd' =n',∇0' uuu 3/2 -1/2 3/2 1 1/2 2 (∇++) uuu' 3/2- 1/2 1/2 1/3 1/6 2/3 uud 3/2- 1/2 1/2 1 1/2 1 uud' 3/2- 1/2 3/2 1/3 1/6 5/3 u'u'u -1/2 - -3/2 -1/2 -1/3 -1/6 -5/3 u'u'u' -1/2 - -3/2 -3/2 -1 -1/2 -2(∇- -) u'u'd -1/2 - -3/2 -1/2 -1/3 -1/6 -2/3 u'u'd' -1/2 - -3/2 -1/2 -1 -1/2 -1(∇+') d'd'u 3/2 -1/2 3/2 -1/3 -1/6 4/3 d'd'u' 3/2- 1/2 1/2 -1 -1/2 0 d'd'd 3/2- 1/2 1/2 -1/3 -1/6 1/3 d'd'd' 3/2- 1/2 3/2 -1 -1/2 1(∇'- -) ddu -1/2 - -3/2 -1/2 1 1/2 0 ddu' -1/2 - -3/2 -3/2 1/3 1/6 -4/3 ddd -1/2 - -3/2 -3/2 1 1/2 -1(∇- ) ddd' -1/2 - -3/2 -1/2 1/3 1/6 -1/3 udu 3/2 -1/2 1/2 1 1/2 1(p+) udu' 3/2- 1/2 -1/2 1/3 1/6 -1/3 udd 3/2- 1/2 -1/2 1 1/2 0(n0) udd' 3/2- 1/2 1/2 1/3 1/6 2/3 u'd'u -1/2 - -3/2 1/2 -1/3 -1/6 1/3 u'd'u' -1/2 - -3/2 -1/2 -1 -1/2 -1 u'd'd -1/2 - -3/2 -1/2 -1/3 -1/6 -2/3 u'd'd' -1/2 - -3/2 1/2 -1 -1/2 0(n'0)

BARYON DECUPLET Name

Quark Composition

S B Y/2=(S+B)/2 Iτ Q=Iτ + Y/2 spin mass*e lifetime: τ

in sec

Δ++ uuu 0 1 1/2 3/2 2 3/2
Δ- ddd 0 1 1/2 -3/2 -1 3/2
Δ+ uud (similar to p+ ) 0 1 1/2 1/2 1 3/2
Δ0 ddu (similar to n0 ) 0 1 1/2 -1/2 0 3/2
Σ*+ uus  (similar to Σ+ ) -1 1 0 1 1 3/2
Σ*- dds  (similar to Σ- ) -1 1 0 -1 -1 3/2
Ξ*- dss   (similar to Ξ-  ) -2 1 -1/2 -1/2 -1 3/2
Ξ*0 uss   (similar to Ξ0 ) -2 1 -1/2 1/2 0 3/2
Σ*0 uds  (similar to Σ0  ) -1 1 0 0 0 3/2
Ω- sss -3 1 -1 0 -1 3/2   8.21*10-11

Transformation from one Type to Another

 n0 (K0 ) p+ (K+ ) udd (ds') duu(us') Ξ0 (K0' ) Ξ-(K- ) uss (d's) dss(u's) Σ+( π+  ) Σ-  ( π- ) uus(u'd) dds(d'u) Σ0  ( π0) Λ0 ( η0 ) uds(uu') dus(dd') η'(ss') η'(ss') * in baryon octet, no anti-quark is involved*consists of 3 quarks *Strange quarks are absent in nucleons.   * All 8 are particles and there is no anti-particle in the octet.   * out of 10 possible combinations, 7 find their place in octet; 3 in ducuplet of resonance baryons- sss for Ω-  , uuu for Δ++  , ddd for Δ- . * Baryon no. is 1 * there is degeneracy of quarks in Σ0 &Λ0 . Both consist of u+d+s. However, their masses are different.Σ0  is heavier than Λ0 by 151me- =  77 Mev & decays into the latter. WHY MASS DIFFERENCE WITH SAME QUARKS?? Σ0  -> Λ0 + γ * particles transform by interchange of u-d   * total u -8, d-8,s-8 sum 24 * 5u-> 5d    3d->3u    4s-4s  * No. of Baryons with 0 strange quark : 2   No. of Baryons with 1 strange quark : 4  No. of Baryons with 2 strange quark :  2 * all spin 1/2 particles with exception of Ω-  which has 3/2 spin. * In meson octet, in every particle, there is one quark and one anti-quark.* consist of 2 particles , 1 quark & 1 anti quark * strange quark/anti quark absent in pions, neutral eta meson. * there are 3 particles(K+,π+,K0) and 3 antiparticles, total=6    and 3 particles which have no separate anti-particle, total=3   * out of 21 possible combinations, 9 find their place in meson octet. Other 12 ruled out because they have fractional charges and moreover they are made up of either only quarks or only anti-quarks. * Baryon no. is 0 * There is no such degeneracy in mesons.   * particles transform by interchange of  u-d and u'-d'. * ' represents anti-quarks. *total u-3,u'-3,d-3,d'-3,s-2,s'-2 sum 16 3u->3d 3u'->3d' 1s-1s 1s'-1s' η' is excluded . * if η' is included,   total u-3,u'-3,d-3,d'-3,s-3,s'-3 sum 18* No. of Mesons with 0 strange quark : 4   No. of Mesons with 1 strange quark : 4  No. of Mesons with 2 strange quark :  1* all spin 0 particles but vector mesons have spin 1 with odd parity.

Hypercharge (Y) mapping [Bijective]

Baryon - Mesons

 particles Y mapping particles Y p+, n0 1 K+, K0 1 Σ+, Σ0 , Σ-, Λ0 0 π+,π0 ,π- ,η0 0 Ξ0 ,Ξ- -1 K0', K- -1

Strangeness (S) mapping [Bijective]

Baryon - Mesons

 particles S mapping particles S p+, n0 0 K+, K0 1 Σ+, Σ0 , Σ-, Λ0 -1 π+,π0 ,π- ,η0 0 Ξ0 ,Ξ- -2 K0', K- -1

Quark Triplet particle spin I3 type B S Y=B+S Y/2 Iτ Q=Iτ+Y/2 U spin=-Iτ /2  + 3Y/4 u 1/2 fermion 1/3 0 1/3 1/6 1/2 2/3 0 u' -1/2 antifermion -1/3 0 -1/3 -1/6 -1/2 -2/3 0 d 1/2 fermion 1/3 0 1/3 1/6 -1/2 -1/3 1 d' -1/2 anti fermion -1/3 0 -1/3 -1/6 1/2 1/3 -1 s 1/2 fermion 1/3 -1 -2/3 -1/3 0 -1/3 -1 s' -1/2 anti fermion -1/3 1 2/3 1/3 0 1/3 1 1+5+2+3+2+5+1=19

AB=AC=DF=EF=√13/6

BC=DE=1

BD=CE=2/3

BE=CD=√13/3

AF=4/3

sin DAF =1/√5

DAF=26.5650 °

AOD=121.7175 °

Contribution (%) of Hypercharge (Y) & Iso-Spin Projection (Iτ)

to

the Charge Q of particles

 particle type Y  (%) contribution Iτ (%) contribution (%) total contribution u quark fermion 25 75 100 d quark fermion 25 75 100 s quark fermion 100 0 100 e- lepton (fermion) 50 50 100 μ- lepton (fermion) 50 50 100 τ- lepton (fermion) 50 50 100 (K+) p+ (us')udu nucleon/baryon/hadron(meson/hadron) 50 (+) 50 (+) 100 ( π+) Σ+ (ud')uus hyperon/baryon/hadron(meson/hadron) 0 100 (+) 100 ( π-)Σ- (u'd)dds hyperon/baryon/hadron(meson/hadron) -100 (-) 0 -100 (K-) Ξ- (u's)dss hyperon/baryon/hadron(meson/hadron) -50 (-) -50(-) -100 (K0) n0 (u's)dud nucleon/baryon/hadron(meson/hadron) 50 (+) -50 (-) 0 (K0') Ξ0 (d's)uss hyperon/baryon/hadron -50 (-) 50 (+) 0 ( π0) Σ0 (uu')uds hyperon/baryon/hadron(meson/hadron) 0 0 0 (η0)Λ0 (dd')uds hyperon/baryon/hadron(meson/hadron) 0 0 0 Δ++uuu baryon/hadron 25 75 100 Δ-ddd baryon/hadron 50 -150 -100 Ω-sss hyperon/baryon/hadron 100 0 100

Vector Mesons (nonet) Vector mesons are mesons with total spin 1 and of odd parity. Pseudo vector mesons

have total spin 1 and are of even parity.

Conservation of charge in Strong / weak interactions

 particles type chirality + or  right handed  chirality -      or  left handed iso-spin(I) 3rd component of iso-spin (I3) weak iso-spin(T) 3rd component of weak iso-spin (T3) Hypercharge(Y) weakhypercharge (Yw)Left handed particles massMev B-L X charge u fermion - 1/2 +1/2 1/2 +1/2(LH) 1/3 +1/3(LH) 1.4 +1(LH) d fermion - 1/2 -1/2 1/2 -1/2(LH) 1/3 +1/3(LH) 4.5 +1(LH) u' anti-fermion + -1/2 -1/2 -1/2 -1/2(RH) -1/3 -1/3(RH) -1(RH) d' anti-fermion + -1/2 +1/2 -1/2 +1/2(RH) -1/3 -1/3(RH) -1(RH) s fermion - 0 0 1/2 -1/2(LH) -2/3 +1/3(LH) 90-95 +1(LH) s' anti-fermion + 0 0 -1/2 +1/2(RH) +2/3 -1/3(RH) -1(RH) c fermion + 0 0 1/2 +1/2(LH)) 4/3 1/3(LH) c' anti-fermion - Leptons B-L X charge e- fermion - 0 0 1/2 -1/2(LH) 0 -1(LH) 0.511 -1 -3(LH) νe- fermion - 0 0 1/2 1/2(LH)0 (RH) 0 -1(LH)0 (RH) -1 -3(LH) μ- fermion - 0 0 1/2 -1/2(LH) 0 -1(LH) 52.5 -1 -3(LH) νμ- fermion - 0 0 1/2 1/2(LH)0 (RH) 0 -1(LH)0 (RH) -1 -3(LH) τ- fermion - 0 0 1/2 -1/2(LH) 0 -1(LH) -1 -3(LH) ντ- fermion - 0 0 1/2 1/2(LH)0 (RH) 0 -1(LH)0 (RH) -1 -3(LH) * X-charge is a conserved quantum number associated with the SO(10) Grand Unification Theory. It is thought to be conserved in electromagnetic, strong, weak, gravitational, Higgs interaction. Because it is associated with weak hypercharge, it varies with the helicity of a particle. A Left handed quark has an X-charge +1 where as a right handed quark has X-charge -1(up,charm,top quarks) or -3(down,strange,bottom quarks) X=5(B-L) -2Yw * X-Charge in proton decay: Proton decay is a hypothetical form of radioactive decay, predicted by many grand unification theories. During proton decay, the common baryonic proton decays into lighter subatomic particles. However, proton decay has never been experimentally observed and is predicted to be mediated by hypothetical X and Y bosons. Many protonic decay modes have been predicted, one of which is shown below. p+ → e+ + π0 This form of decay violates the conservation of both baryon number and lepton number, however the X-charge is conserved. Similarly, all experimentally confirmed forms of decay also conserve the X-charge value.   * iso spin symmetry for strong interaction involves u,d quarks and it is an approximate symmetry. The exact symmetry is broken by mass (u quark  mass is different from d quark mass) and electromagnetism(u,d quarks have different charges). Iso spin symmetry is defined as the invariance of the Hamiltonian of the strong interaction under the action of Lie Group SU(2). * For Strong Interaction, Q=I3 + Y/2  where Y=(B+S + C+t+b')/2 * For Weak Interaction, Q=T3  +  Yw /2 where  Yw =(5/2)(B-L) -X/2 Where B is baryon no., L lepton no. and X is a conserved quantity under GUT.This is applicable to left handed particles. X=1 or X +2 Yw = 5(B-L) . Pl. note that B-L is a conserved quantity in weak interaction. Weak interaction is one where neutrinos of any of the 3 (or combination/mixing of 3 flavors)  different flavors are produced. Iso-spin (for strong interaction) is similar to but should not be confused with weak iso-spin. * Weak iso-spin is the gauge symmetry of the weak interaction which connects quark & lepton doublets of left handed particles, (u,d), (t,b),(e- ,νe- ) *Iso-spin is a symmetry of strong interaction under the action of Lie group  SU(2), the 2 states being up flavor and down flavour. .Strong interaction connects only up and down quarks , acts on both chiralities (left, right) and is a global (not gauge) symmetry. * Iso spin corresponding to weak interactions is known as weak iso spin T or T3. A quark of one flavor can transform to a quark of another flavor and the matrix of interaction is called CKM matrix. * Quarks are always in a bound or confined stage but in QGP(quark gluon plasma), they occur in de-confined stage. * For massless particles, chirality and helicity are same thing whereas for massive particles, both must be distinguised. Universe prefers left handed chirality. (why???). Chirality is defined by whether the particle transforms in a right handed or left handed representation of the Poincare Group * iso-spin is conserved  only in strong interaction whereas weak iso-spin & weak hypercharge are  conserved in electroweak interactions.  It is conserved (only terms that are overall weak-hypercharge neutral are allowed in the Lagrangian). However, one of the interactions is with the Higgs field. Since the Higgs field vacuum expectation value is nonzero, particles interact with this field all the time even in vacuum. This changes their weak hypercharge (and weak isospin T3). Only a specific combination of them, Q = T3 +  YW /2(electric charge), is conserved.* Weak hypercharge corresponds  to Gauge Symmetry U(1). Hypercharge is the eigen value of the charge operator.* Gauge transformation is one where transformation of the potential leaves the field invariant.* In nature, so far we have not come across right handed neutrinos. When the left handed neutrino isa)  receding  from the observer or b) approaching the observer , its spin axis is always parallel to the direction of motion. However iso-spin angular momentum is* anti-parallel to direction of motion when it is receding and the iso-spin rotation is anti-clockwise.* parallel to direction of motion when it is approaching  and iso-spin rotation is clockwise.* A quark cannot decay in weak interaction, into another quark having the same 3rd component of weak iso spin i.e. T3. In case of leptons, something similar happens for left handed leptons (+ chirality) such as electron, muon, tao particle and corresponding right handed neutrinos.* Fermions with + chirality (LH fernions) and anti- fermions with - chirality(RH anti fermions)) form singlets with T=T3=0 and do not undergo charged weak interaction.*Strange particles can only decay through weak interaction and strangeness is not conserved in weak interaction.* Iso spin and 3rd component of isospin is not conserved in electromagnetic and weak interaction. Hence a separate term weak iso spin, weak hypercharge is coined which is conserved in weak interaction.* We consider left handed particles as spinning anti-clockwise while the particle is receding from the observer. Chirality of such particles is termed positive.* There are 8 generators of SU(3) out of which 6 correspond to 6 flavors of quarks, 1 corresponds to isospin and 1 corresponds to hypercharge. Cabibbo Angle and Mixing :There are 3 families {(u,d) , (c,s) , (t,b) }of quarks and 3 families {(e,νe),(μ,νμ),(τ,ντ)} of leptons. The coupling constant g between members of the same family is same . Coupling constant universality is a characteristic feature of quarks and leptons. The transition amplitude L is proportional to g/ 2√π. The transition probability αw  is proportional to  square of  L. The transition probability from u to d should be same as that of e to νe    . But in reality, it is slightly less. The difference is the transition probability from u to s. This is explained by Cabibbo angle. We define flavor generators T for first 2 generations of quarks as under--T(u) =                T(c) =2/3  0   0            2/3  0   00    0    0            0    0    00   0    -1           0    0    0T(d) =              T(s) =-1/3  0   0          -1/3  0   00     0    0           0     0    00    0    -1          0     0    0-sinθc =Tr [T(d)T(c) ] =-2/9  and angle is about 12.7 degree.Mixing is between flavor mass eigen states(u,c,t,d,s,b) and flavor interaction states(u',c',t',d's',b').u'  =  uc'      ct'      t d'  =Vckm  *   ds'                    s b'                   b

Comparison of iso-spin vrs weak iso-spin

&

hypercharge vrs Weak hypercharge

 Q= I3+ Y /2 = T3+ YW/2 (LH)2YW = 5(B-L) -   X (RH)2YW +5(B-L)for(u,c,t) -5(B-L)for(d,s,b) +X I3 T3(RH) (LH)T3 Y (RH)YW (LH)YW X B L B-L u 1/2 0 1/2 1/3 4/3 1/3 1 1/3 0 1/3 d -1/2 0 -1/2 1/3 -2/3 1/3 1 1/3 0 1/3 s 0 0 -1/2 -2/3 -2/3 1/3 1 1/3 0 1/3 c 0 0 1/2 4/3 4/3 1/3 1 1/3 0 1/3 t 0 0 1/2 4/3 4/3 1/3 1 1/3 0 1/3 b 0 0 -1/2 -2/3 -2/3 1/3 1 1/3 0 1/3 e- 0 -1/2 0 -1 -3 0 1 -1 νe- 0 -1/2 0 -1 -3 0 1 -1 μ- 0 -1/2 0 -1 -3 0 1 -1 νμ- 0 1/2 0 -1 -3 0 1 -1 τ- 0 1/2 0 -1 -3 0 1 -1 ντ- 0 1/2 0 -1 -3 0 1 -1