SU(5) and Grand Unification

Ranjan Modak&

Debabrata Pramanik

Outline of talk..

• SU(5)• Sub-algebra of SU(5)• Branching Rule for diff rep of SU(5)• Unification and why SU(5)?• Proton decay

SU(5)

• Cartan generators for SU(5): 4 rank algebra its has 4 simple roots and 4 cartan generators(total 24 generators)

12

0 0 0 0

0 12

0 0 0

0 0 0 0 00 0 0 0 00 0 0 0 0

1

23 0 0 0 0

0 1

23 0 0 0

0 0 13 0 0

0 0 0 0 00 0 0 0 0

1

26 0 0 0 0

0 1

26 0 0 0

0 0 1

26 0 0

0 0 0 3

2

20

0 0 0 0 0

1

210 0 0 0 0

0 1

210 0 0 0

0 0 1

210 0 0

0 0 0 1

210 0

0 0 0 0 25

• The weights are w(1)= (1/2 , 1/12 , 1/24 , 1/40) w(2)= (-1/2 , 1/12 , 1/24 , 1/40) w(3)= (0 , -2/12 , 1/24 , 1/40) w(4)= (0, 0 , -3/24 , 1/40) w(5)= (0, 0 ,0 ,- 4/40)From convention w(1)>w(2)>w(3)>w(4)>w(5)Simple roots are a(i)=w(i)-w(i+1) a(1)=(1 ,0,0,0) ( 2 -1 0 0) a(2)= (-1/2 , 3/12 , 0 , 0) (-1 2 -1 0) a(3)= (0, -2/12 , 4/24 , 0) (0 -1 2 -1) a(4)=(0, 0 , -3/24 , 5/40) (0 0 -1 2)

• Fundamental wt µ(i)=(x(i),y(i),u(i),v(i))Such that µ(i).a(j)=0 for ijFrom these sets of equation we find the fundamental wt of SU(5)

in p-q notation (1 0 0 0) , (0 1 0 0) , (0 0 1 0), (0 0 0 1)The number of applications of annihilation operators to generate

all states of rep is called height of rep which is dot product of dynkin indices of highest weights(hws) and level vector(R)

Ri=2 ∑j Aij-1 where A cartan matrix

For SU(5) level vector R=(4,6,6,4)

Embedding

Small lie algebras can live inside the bigger one: N dim rotation contain N-1 dim rotation SO(N) SO(N-1)

Two kinds of Maximal embedding possible• Start with Dynkin diagram of G of rank n, remove a dot

yielding a smaller algebra H rank n-1. regular non-semisimple embedding with U(1)

G H×U(1) • Remove a dot from extended Dynkin Diagram to obtain same

rank sub-algebra :regular maximal semisimple embedding. G H

Extended Dynkin Diagram: Its just a diagram where with normal dynkin diagram you just add the negative of highest root .

Example for SU(5)Cartan matrix for SU(5)=

Inverse of Cartan matrix for SU(5)=

• We know that Highest wt state in adjoint rep of SU(5) is (1001) in terms of dynkin indices

Now seeing the inverse we can say that this highest root corresponds to 1st and 4th row of the matrix.

highest root=a+b+c+d=(1 0 0 1) Where a,b,c,d are simple roots of SU(5) from cartan matix those are (2 -1 0 0) ,

(-1 2 -1 0),(0 -1 2 -1),(0 0 -1,2) in (p-q ) notation

(hr,hr)=2 and (hr,a)=1, (hr,d)=1

Where (a(i),a(i))=2 and (a(i),a(i+1))=-1, a(i) say ith simple root of any SU(N) lie algebra

Now we can associate an additional row and column with cartan B =

Det(B)=0 , so its has one zero eigenvalue Now the question is how do I know which dot should I remove to get maximal

sub-algebra??

We have level the dots: One eigenvector with zero eigenvalue its components are called co-marks(ci) and the marks are components of zero eigenvector of transposed Matrix.

2 1 0 0 11 2 1 0 00 1 2 1 00 0 1 2 11 0 0 1 2

Now for SU(5) we have found marks and comarks are same :( 1,1,1,1,1)So, you cant single out any individual root which imply that if you erase any dot you will get back the original SU(5)Cant get SU(5) subalgebra in this method even its clear from the diagram also

To get maximal embedding we have to use 1st trick. Original dynkin diagram we have to remove one dot removing d simple root we find SU(4) ×U(1) and removing c we find SU(3) × SU(2) × U(1) subalgebra

SU(5) fundamental Rep

(1 0 0 0) (1 0) (0)

-a (-1 1) (0) Su(3) triplet and Su(2) singlet(-1 1 0 0) remove 3rd col

-b(0 -1 1 0)

-c (0 -1) (0)(0 0 -1 1) (0 0) (1)

-d Su(3) singlet and Su(2) doublet(0 0 0 -1) (0 0) (-1)

5= (3,1)+(1,2)

10=(3c,1)+(3,2)+(1,1)

• (0 1 0 0) (0 1) (0) -b (1 -1) (0) (1 -1 1 0) (-1 0) (0) -a -c removing 3rd col (-1 0 1 0) (1 0 -1 1) (1 0) (1) -c -a -d (-1 1)(1) (-1 1 -1 1) (1 0 0 -1) (0 -1) (1) -b -d -a (0 -1 0 1) (-1 1 0 -1) (1 0)(-1) -d -b (-1 1)(-1) (0 -1 1 -1) (0 -1) (-1) -c (0 0 -1 0) (0 0) (0)

• Why U(1) is coming in the embedding?We know that: 3 × 3c = 8(adjt) + 1In general: Adj(SU(n))=n× nc - Trace nothing but the trace termNow say n=m (n-m)AdjSu(n) = [m (n-m)] [mc (n-m) c] –TraceIn the product we get 2 adjoint each of them involves removing a

trace ,We have subtracted only one trace in AdjSU(n) while we have subtracted 2 traces in AdjSu(m) and Adj(Su(n-m))

Left hand side must have an extra trace that is just the U(1): which imply we can assign a charge

• We want to consider the stander splitting of SU(n) to subgroup SU(m) ×SU(n-m) ×U(1) into a block diagonal form, the schematic is :

• Extra U(1) embedded as

• 5=(2,1)(3)+(1,3)(- 2) and • 10=(1,1)(2)+(1,3c )(-4/3)+(2,3)(1/3)• you can use any normalization factor as long as sum is zero

Adj rep(24 dim)

(1001) (-1101) (101-1) (0-111) (-111-1) (11-10)(00-12) (0-12-1) (-12-10) (2-100)

(0000) (0000) (0000) (0000)

(-2100) (1-210) (01-21) (001-2) (-1-110) (1-1-11) (01-1-1) (-10-11) (1-10-1)

(-100-1)

• From the previous diagram we can conclude few things• Identify the positive roots (a+b+c+d)=(1001)[hws]• a+b=(11-10)• b+c=(-111-1)• c+d=(0-111) • a+b+c=(101-1)• b+c+d=(-1101)• and 4 simple roots a,b,c,dThse are the roots given in the upper half of the previous diagram• Weyl vector: half of the sum of all positive roots

=1/2(4a+6b+6c+4d)=(1 1 11)

• These coefficients are same as level vector and weyl vetor components are also same which is a general feature of simple lie algebra

• Coexter number(h)=(1/2)L +1

• Height of Adj rep here 8 for SU(5)

• So,h=5 for SU(5)• D=r(h+1)=24 for SU(5)• [D dim of rep and r rank=4 here]

• Now say delete 3rd col you will get• (11)(0) • (-12)(0)• (2-1)(0)• (00)(0) This is the 8dim adj rep of SU(3) and SU(2) singlet• (00)(0) (8,1) (11)• (-21)(0) (-12) (2-1)• (1-2)(0) (00) (00)• (-1-1)(0) (-21) (1-2) (-1-1)

• (10)(1)• (-11)(1)• (0-1)(1) Su(3) triplet and SU(2) doublet:(3,2)• (10)(-1)• (-11)(-1)• (0-1)(-1)• (01)(1)• (1-1)(1)• (-10)(1) Su(3) triplet and SU(2) doublet:(3c,2)• (01)(-1)• (1-1)(-1)• (-10)(-1)

• (00)(2)• (00)(0) Su(3) singlet and SU(2) triplet(adj):(1,3)• (00)(-2)• (00)(0) (1,1): SU(3) and SU(2) singlet this is nothing but actually U(1)

So,24=(8,1)+(3,2)+(3c

,2)+(1,3)+(1,1)

Adj of SU(3) adj of SU(2)This is also a property of embedding Adj rep of G must contain adj of sub

algebras

UNIFICATION

Theory of Everything

• Four fundamental forces of Nature- Strong, Electro-magnetic, Gravitational, Weak.

• Why not there are only one force and these forces are the different manifestations of that single force?

• Gravitational force is separate type. We are interested in unifying the Strong, Weak and Electro-magnetic forces in a single thread.

What is Higgs?

• Higgs is the cause of mass.• Everywhere there exist a Higgs field and the

elementary particles acquire masses due to interaction between the Higgs field.

• Without mass, the universe would be a chaotic sea of particles moving at the speed of light.

Higgs• Mediator of electromagnetic force i.e. photon is

massless.• W & Z bosons carry masses.• Thus, electroweak symmetry is broken.• Higgs is the particle that hides the symmetry of the

Standard Model.• Higgs shifts the equations in such a way that once-

massless particles can have mass without other changes in the equations.

Why SU(5)?

• Symmetry group or standard model SU(3) SU(2) U(1)• The unifying group must be atleast of rank 4 lie group which

can involve one coupling strength• Possibilities are [SU(2)]4,[O(5)]2,[SU(3)]2,

[G2]2,O(8),O(9),Sp(8),F4,SU(5)• 1st two groups unacceptable because don’t contain SU(3)• Under the subgroup SU(3) SU(2) leptons are SU(3) singlet

and SU(2) doublet• Anti-leptons are singlet under both groups • Quarks are SU(3) triplet and SU(2) doublet and Anti-quarks

are SU(3) 3c and SU(2) singlet

So, SU(3)×SU(2) content of the 30 fields is 2(1,2)+2(1,1)+2(3,2)+4(3c,1)

• The representation must be complex, not equivalent to complex conjugate

• This criteria is satisfied only for SU(5) and [SU(3)]2

• But [SU(3)]2

Does not explain Hadron: generator corresponding to electric charge does not admit fractional charges

We are left with only SU(5)

Projection Matrix

• We can derive a projection matrix which will project highest weights(hws) of an irreps of SU(5) onto hws of SU(2)×SU(3)

• (1 0 0 0), hws of 5, branches to (2,1)+(1,3) and color 3 state of hw can be defined to be hws of representation

• So, P(1 0 0 0)=(0)(1 0)• And same way, P(0 0 0 1)= (0) (0 1) hws of 10 is (0 1 0 0) branches to (1,1)+(2,3)+(1, 3c) hws of the Sub group is (1)(1 0)

So, P(0100)=(1)(1 0)

And for P(0010)=(1)(0 1)

P(SU(5) SU(2) ×SU(3))=

• A family of left handed fermions is assigned to 5c+10 of SU(5)

• In the 5c state

• P( 0 0 0 1)=(0)(0 1) is a charge 1/3 anti-quark singlet

• P(0 0 1 -1)=(1)(0 0) is a charge 0 member of lepton doublet

• P(0 1 -1 0)=(0) ( 1-1) is a charge 1/3 anti-quark singlet

• P(1 -1 0 0)=(-1)(0 0) is a charge -1 member of a lepton doublet

• P(-1 0 0 0)=(0) (-1 0) is a charge 1/3 anti-quark singlet

0 1 1 01 1 0 00 0 1 1

10 rep of SU(5)

• P(0 1 0 0)=(1)(1 0) is a u quark with Q=2/3 and I3=1/2

• P(1 -1 1 0)=(0)(0 1) is uc with Q=-2/3 and I3=0

• P(-1 0 1 0)=(1)(-1 1) is a u with Q=2/3 and I3=1/2

• P(1 0 -1 1)=(-1)(1 0) is a d with Q=-1/3 I3=-1/2

• P(-1 1 -1 1)=(0)(0 0) is e+ with Q=1 and I3=0

• P(1 0 0 -1)=(0)(1 -1) is uc with Q=-2/3 and I3=0

• P(-1 1 0 -1)=(1)(0 -1) is a u with Q=2/3 and I3=1/2

• P( 0 -1 0 1)=(-1)(-1 1) is a d with Q=-1/3 I3=-1/2

• P(0 -1 1 -1)=(0)(-1 0) is uc with Q=-2/3 and I3=0

• P( 0 0 -1 0)=(-1)(0 -1) is a d with Q=-1/3 I3=-1/2

• Same way we can assign all right handed particles in 5+10c rep• Few additional things for completeness:Two conditions must satisfied for Higgs field giving mass to the

fermions in SU(5) theory Higgs rep must appear in the tensor product of SU(5) in which

fermions and anti-fermions appear Higgs rep must have a component that transform under SU(3)

×SU(2) ×U(1) like the Higgs field of SU(3) ×SU(2) × U(1) model----(1,2)(1/2)

• Now right handed positron and d quark in the 5 and their anti particle 10c

In SU(5): 5×10c=5c+45c

right handed u and uc both transform as 10c

10 c×10c=5+45+50 5 and 45 both contain (1,2)(1/2):Higgs field component

These are useful for giving mass to u but not 50 as it does not contain (1,2)(1/2)

• 45=(1,2)+(3,1)+(3,3)+(3c, 1)+(3c,2)+(6c,1)+(8,2)

5=(1,2)+(3,1)

50=(1,1)+(1,3)+(3,3c)+(3,6c)+(1,6)+(2,8)

Proton Decay

• In standard model Lagrangian, we can find U(1) global symmetry which corresponds to Baryon number conservation.

• p+--> e+ +2• p+--> µ+ +π0

• This does not conserve Baryon number, according to standard model Proton decay is not possible.

• In GUT[SU(5)] quarks, anti-quarks and the electrons all appear in the same irreps

• So, some of the interactions may not conserve Baryon number. Proton decay is possible according to this GUT

**Note: B-L number is conserved

• According to GUT one should observe proton decay but upto now experiments do not give us any evidence of proton decay.

• Recent experiment: experiments at the Super-Kamiokande water

Cherenkov radiation detector in Japan gave lower limits for proton half-life, at 90% confidence level, of 6.6×1033years via antimuon decay and 8.2×1033 years via positron decay

• Advantages of this Model:• Three independent gauge couplings of standard model reduce

to single coupling• It qualitatively predicts mass of b quark• It explains charge current weak interaction correctly• Ambiguity of this model:• Neutrino is massless• Proton decay • Magnetic monopole

Reference:• Group Theory For Unified Model Building by R.Slansky• “Unity of All Elementary Particle Forces” Georgi and Glashw,PRL,vol 32 ,438(1974)• Howard Georgi, Lie Algebras in Particle Physics.• Pierre Ramond, Group Theory. • Semisimple lie algebra and representation: Robert Cahn• Introduction to High energy physics by Donald .H.Perkins• Quantum field theory by Peskin and Schroedert

Acknowledgement

Dr. Aninda Sinha