beyond the standard model - lunds universitethome.thep.lu.se/fytn04/lecture14.pdf · supersymmetry...

Problems with the SMGrand Unification

Supersymmetryˇ

Beyond the Standard Model

Leif Lönnblad

Institutionen för Astronomioch teoretisk fysikLunds Universitet

2018-12-17

FYTN04: BSM 1 Leif Lönnblad Lund University


Supersymmetryˇ

UnconstrainedArbitrary

ˇIncomplete

The standard model and why we hate it!

I There are too many free parameters. Twelve fermionmasses, eight mixing parameters, three couplings and thehiggs field parameters (µ, λ)⇔ (mh,mZ ). In total 25parameters (26 assuming there is CP-violation in QCD)Wouldn’t it be much nicer if we had a theory where thesecould be predicted?

I Unnaturally (?) large scale ratios:me/mν ∼ 107, mW/me ∼ 105, mPl/mW ∼ 1017



Supersymmetryˇ

UnconstrainedArbitrary

ˇIncomplete

Arbitrary

I Why are there three generations.I Why are the left-handed fermions in SU(2) doublets and

the right-handed in singlets?I Why do we just have SU(3)× SU(2)×U(1)? Nature could

have picked any symmetry!I Why is there charge quantization? In principle Y in U(1)

could be anything.



Supersymmetryˇ

ˆ ArbitraryIncomplete

ˇFine-tuned?

Incomplete

I Where is the anti-matter?I Where is gravity?I Where is dark matter?I Where is dark energy?



Supersymmetryˇ

ˆ IncompleteFine-tuned?

Fine-tuned?

There is a problem with the Higgs mass.

Just as couplings are renormalized to be scale dependent, soare masses:

m→ m

(1 +

α

3π

∫ ∞m2

0

dp2

p2 + . . .

)→ m

(1 +

α

3πln

Λ2

m20

+ . . .

)

This comes from self-energy diagrams

= + + +

For the Higgs we find that∫ dp2

p2 →∫

dp2 and we have aquadratic divergence.



Supersymmetryˇ


Assuming the Higgs mass at the scale Λ is mh(Λ), looking onlyat the top-loop we have

m2h(mZ ) ∼ m2

h(Λ)− (Λ2 −m2t )

If there is no physics below mPl, that means mh(mZ ) ≈ 125 GeVcomes from the subtraction between two huge numbers.



Supersymmetryˇ


Consensus

There must be something beyond the Standard Model!

The big questions: What? and Where?



Supersymmetryˇ

SU(5)GUTMultiplets

ˇNew bosons

Grand Unification

We know about spontaneous symmetry breaking

U(1)Y × SU(2)L → U(1)EM

Imagine that at a high scale all three forces are united into oneunder a common larger symmety group GGUT.

For some reason this group is then spontaneously broken

GGUT → SU(3)QCD × U(1)Y × SU(2)L

Let’s try G = SU(5)



Supersymmetryˇ

SU(5)GUTMultiplets

ˇNew bosons

SU(5)GUT

I Simplest possible groupI Invented by Georgi and Glashow 1974I Excluded by data — but still instructive



Supersymmetryˇ

SU(5)GUTMultiplets

ˇNew bosons

The basic multiplet is given by a colourtriplet and a weak doublet in an (anti-)quintet.

Since the weak doublet is left-handed, thequarks need to be left-handed and weaksinglets, so we use the d .

U5 =

dr

db

dg

(e−

νe

)

L

Group generators are traceless N × N matrices, where thediagonal generator will give the charge and requires

∑Qi = 0.

This gives us charge quantisation and 3Qd + Qe = 0



Supersymmetryˇ

SU(5)GUTMultiplets

ˇNew bosons

Where are the other quarks and leptons?

The quintet of right-handed fields:

U5 = (dr ,dg ,db,e+, νe)R

The anti-symmetric decuplet with ten left-handed fields

U10 =1√2

0 ub −ug−ub 0 urug −ur 0

−ur −dr−ug −dg−ub −db

ur ug ubdr dg db

(0 −e+

e+ 0

)

L

and the corresponding one for the right-handed fields.



Supersymmetryˇ

ˆ MultipletsNew bosons

ˇThe GUT scale

The gauge bosons

SU(5) has 52 − 1 = 24 generators

A =

gij − 2B√30δij

Xr YrXg YgXb Yb

Xr Xg XbYr Yg Yb

(W 3√

2+ 3B√

30W +

W− −W 3√

2+ 3B√

30

)

Giving us 12 new gauge bosons!



Supersymmetryˇ

ˆ New bosonsThe GUT scaleInteractions

The GUT scale

At some large scale SU(5) is an exact symmetry with a singlecoupling g5.

We expect all SM couplings to come together at some highscale MGUT.

Remember:1

αi(M2)=

1αi(µ2)

+bi

4πln

M2

µ2

with b3 = 11− 4nF/3, b2 = 22/3− 4nF/3, b′1 = −4nF/3



Supersymmetryˇ


We should have eg.

1α3(µ2)

+b3

4πln

M2GUTµ2 =

1α2(µ2)

+b2

4πln

M2GUTµ2

using 1/α2(m2Z ) ≈ 30 and 1/α3(m2

Z ) ≈ 10 we get

MGUT ∼ 1018 GeV



Supersymmetryˇ


Even if we haven’t specified the way the GUT is broken weshould be able to estimate e.g. the ratio between g1 and g2 atlower energies.

Let’s look at the covariant derivative of SU(5)

Dµ = ∂µ − ig5TaUµa

And Pick out the parts relevant to the electro-weak sector using

Bµ = Aµ cos θW + Zµ sin θW

Wµ3 = −Aµ sin θW + Zµ cos θW



Supersymmetryˇ


Dµ = ∂µ − ig5(T3Wµ3 + T1Bµ + . . .)

= ∂µ − ig5 sin θW (T3 + cot θW T1)Aµ + . . .

= ∂µ − ieQAµ + . . .

Identify e = g5 sin θW and Q = T3 − cot θW T1 ≡ T3 + cT1.

Now, for any representation, R, of a group we haveorthogonality and equal normalization of the generators Ta, sothat

TrRTaTb = NRδab



Supersymmetryˇ


TrQ2 = Tr(T3 + cT1)2 = (1 + c2)TrT 23

since TrT 23 = TrT 2

1 and

sin2 θW =1

1 + c2 =TrT 2

3TrQ2 =

0 + 0 + 0 +(1

2

)2+(1

2

)2(13

)2+(1

3

)2+(1

3

)2+ 1 + 0

=38

Including the running of the couplings we can get close tosin2 θW ≈ 0.23 at around mZ .



Supersymmetryˇ


Interactions

How does the new gauge bosons interact?

looking at the terms when we sanwich the gauge boson matrix,A, between the fundamental representations

U5AU5, U10AU5 and U10AU10

we get eg.

X → uu, X → e+d , Y → ud , Y → d νe, Y → e+u

Giving the charges 4/3 and 1/3 for X and Y .



Supersymmetryˇ


Decaying protons!

p = u ud → u Y → u ue+ → π0e+

Remembering the muon width we estimate

Γp ∝ α25

m5p

m4Y

and with mY ∼ mGUT ∼ 1015 GeV we get τp ∼ 1031±2 years.

(The current limit p → π0e+ is τp > 1033 years.)


Grand UnificationˆSupersymmetry

(Super) String Theory

New particlesR-parity

ˇFine-tuning solved?

Supersymmetry

Postulate there being a symmetry between fermions andbosons, with an operator Q changing one into the other

|bi〉 = Q|fi〉 and |fi〉 = Q|bi〉

but leaving any other quantum number unchanged.

The transformation is actually defined in terms of an algebrawhere

{Q, Q} = QQ + QQ = 2σµPµ

where Pµ = i∂µ is a translation.






normal partner spinqL qL 0 squarksqR qR 0 (can mix)lL lL 0 sleptonslR lR 0 (also mix)νL νL 0 sneutrinos






g g 1/2 gluino

γ (γ) 1/2 (photino zino higgsino)Z 0 (Z ) 1/2 all mix together intoh0 1/2 neutralinos χ0

iH0 (H) 1/2 2 + 3 + 1 + 1 + 1 = 8 spin states for the bosonsA0 1/2 gives four neutralinos with two spin states each.

W± W± 1/2 (wino higgsino) mix togetherH± H± 1/2 charginos χ±i

We need an extra higgs doublet (4 new higgs particles):






In the simplest version of SUSY (MSSM) we can derive massrelations for the Higgs particles, and get mh < mZ < mH .

But there are many ways of constructing SUSY.

If SUSY was an exact theory we would have mq = mq and itwould be easy to find the new particles.

Since we have not found any sparticles, SUSY is broken.

There are many ways of breaking SUSY.






R-parity

We can define a “parity” relating to SYSU, called R-parity

R = (−1)L+3B+2S

where L is lepton number, B is baryon number and S is spin.

All ordinary particles have R = +1 and their super-partnershave R = −1.

If R-parity is conserved, sparticles can only be produced inpairs.

This also means that the lightest super-symmetric particle(LSP) is stable.






Even if the masses are not the same, the couplings do not careif we have particles or sparticles. Hence we have that verticessuch as eg.

W + → e+νe, W + → e+νe, W + → e+νe, W + → e+νe

all have the same coupling (g2). So as soon as we come abovethe mass-threshold for producing sparticles, they will beproduced at the same rate as their ordinary particle equivalents.




ˆ R-parityFine-tuning solved?

The Higgs mass revisited

m2h(mZ ) ∼ m2

h(Λ)− (Λ2 −m2t )

With SUSY the Higgs would also have self-energy loops fromstop squarks (t), but since they are bosons, the sign of the loopis reversed

m2h(mZ ) ∼ m2

h(Λ)− (Λ2−m2t ) + (Λ2−m2

t ) ∼ m2h(Λ)− (m2

t −m2t )

So, as long as mt is not too large the fine-tuning goes away.





A dark matter candidate

If there is a stable LSP, which only interacts weakly (eg. χ0) itwould be produced copiously at the big bang and wouldbasically still be around.

Even if R-parity is not conserved and we could have decays likeγ → νγ, it could still contribute to dark matter if the decay isslow enough.

If R-parity is not conserved we could get lepton and/or baryonnumber violation (and proton decay).





Desperately seeking SUSY

With R-parity conserved we would get characteristic decaychains of sparticles according to their mass hierarchy., eg.

u → d + [χ+1 → ντ + [τ+ → τ+χ0

1]]

Should be easy to see at the LHC

(not seen yet)





The trouble with SUSY

I (more than) double number of particlesI (more than) double number of massesI (more than) double number of mixing angles

In total more than 100 free parameters to measure




StringsExtra dimensions

ˇLarge extra dimensions

A quantum theory of gravity

How do we include gravity in the standard model?

The naive way is to take General Relativity and reinterpret it asa Lagrange density

This leads to a spin-2 graviton (possibly with a supersymmetricspin 3/2 graviton) and a theory that is not renormalisable.

This is related to the fact that Field theory assumes particles tobe point-like.






(Super) String theory

Elementary particles are not point-like but vibration modes ofone-dimensional objects – Strings.

I a point like particles describes a world line, xµ(τ)

I a string will describe a world sheet, xµ(τ, σ)

A string can be open or closed (xµ(τ, σ + 2π) = xµ(τ, σ)).






Since nothing is point-like there is a natural cutoff to ensurerenormalisability.






To combine string theory with QFT is tricky

I To avoid tachyons (m2 < 0) we need 26 space-timedimensions.

I Including SUSY we can get down to 10 if we have SO(32)or E8 × E8 symmetry.

I Good news (1):E8 ⊃ E6 ⊃ SO(10) ⊃ SU(5) ⊃ SU(3)× SU(2)× U(1)but a new zoo of particles only interacting with gravity

I Good news (2): Ony five such theories: type-I, type-IIA,type-IIB, heterotic SO(32) and heterotic E8 × E8

I Good news (3)? Only special cases of 11-dimensionalM-theory, with ∼ 10500 different string theories consistentwith (SuSY) SM.






Where are all the extra dimensions?

The universe extends only L ∼ m−1Pl in all but four dimensions –

compactification.

Have we checked that there are only four dimensions?

Current limit is L . 1µm.




ˆ Extra dimensionsLarge extra dimensions

How do we check the number of dimensions?

Look at the gravitational potential in 4 + n dimensions

V (r) ∼ mmn+2

Pl

1rn+1 ,

Now, if the n extra dimensions are of size R, for distances muchlarger than that the potential would look like

V (r) ∼ mmn+2

Pl

1Rn

1r





we have found mPl4 ≈ 1019 GeV. But if the extra dimensions arelarge this means that the true Planck scale is much smaller

mPl ∼ R−n

n+2 m2

n+2Pl4

could even be close to the scales reachable at the LHC.





BSM phenomenology

1. Make stuff up2. Check that it is consistent with the SM3. Make prediction (for the LHC)4. Convince the experiments to look for it5.





BSM phenomenology

1. Make stuff up2. Check that it is consistent with the SM3. Make prediction (for the LHC)4. Convince the experiments to look for it5. Go to Stockholm and collect prize





BSM phenomenology

1. Make stuff up2. Check that it is consistent with the SM3. Make prediction (for the LHC)4. Convince the experiments to look for it5. When they find nothing, goto 1.


beyond the standard model - lunds universitethome.thep.lu.se/fytn04/lecture14.pdf · supersymmetry...

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