Lectures on Supersymmetry Part I: Basics...The supersymmetry-breaking terms: freedom + determine...

Lectures on Supersymmetry

Part I: Basics

Yael Shadmi

Technion

Yael Shadmi (Technion) ESHEP2014 1 / 106

Motivation

You have probably heard about the motivation for supersymmetry

Through these lectures, this will (hopefully) become clear and moreconcrete

But it’s important to state at the outset:

There is no experimental evidence for supersymmetry

The amount of effort that has been invested in supersymmetry(theory and experiment) is thus somewhat surprising


[True: There is no experimental evidence for any underlying theory ofelectroweak symmetry breaking, which would give rise to the(fundamental scalar) Higgs mechanism as an effective description]

[There is experimental evidence for BSM:dark matter, the baryon asymmetry—CP violation]

Why so much effort on supersymmetry?


It is a very beautiful and exciting idea

(I hope you’ll see this in the lectures)

And it’s something completely different from anything we know inNature


any BSM which the LHC discovers will be exciting:eg:

composite Higgs, “technicolor”: new strong dynamics: beautiful!

(but we know QCD: asymptotic freedom, confinement, chiralsymmetry breaking!)

extra space dimensions —amazing!

(but we have 3 already..)

supersymmetry is conceptually new: it relates bosons and fermions


And so is the Higgs: the first spin-zero (seemingly) fundamentalparticle

The herald of supersymmetry?

The Higgs is the first scalar we see. Only scalars have quadraticdivergences: supersymetry removes these divergences.We will see that in some sense: Supersymmetry makes ascalar behave like a fermion

Given charged fermions: supersymmetry predicts spin-0 particles


Have we been wasting our time?

Supersymmetry is NOT a specific model (certainly not mSUGRA,cMSSM, minimal gauge mediation..)

There is a wide variety of supersymmetric extensions of the SM :

Different superpartner spectra, different signatures

In thinking about them: a whole toolbox:triggers, searches, analysisEven if it’s not supersymmetry: may help discover something else


Supersymmetry supplies many concrete examples with:

new scalars (same charges as SM fermions)

new fermions (same charges as SM gauge bosons)

[and for discovery: spin is a secondary consideration]potentially leading to

missing energy

displaced vertices

(very) long lived particles

disappearing tracks

· · ·


Plan

Lecture I: Basics:Study a few simple toy models to understand the basic concepts andstructure.These toy models=“modules” for building the supersymmetrizedstandard model.De-mystify supersymmetry:

Why is it a space-time symmetry (extending translations,rotations..)

Why does it remove UV divergences (fine tuning)

Why do we care about it even though it’s clearly broken?

· · ·Why is the gravitino relevant for LHC experiments?


Lecture II: The Minimal Supersymmetric Standard Model:Here we will put to use what we learn in I.

Motivation (now that you can appreciate it..)

The field content

The interactions: NO FREEDOM (almost)

The supersymmetry-breaking terms: freedom + determineexperimental signatures.

EWSB and the Higgs mass

Spectra (the general structure of superpartner masses)


Lecture III: Superpartners in Action: LHC signatures

General properties of spectra

Decays

Production

Is it supersymmetry? (??)

The three lectures will be somewhat independent.


Spacetime symmetry:


Spacetime symmetry:

The symmetry we are most familiar with:Poincare:

Translations xµ → xµ + aµ: generator Pµ

Lorentz transformations: xµ → xµ + wµν xν generators: Jµν

(with wµν antisymmetric)

(Throughout consider global, infinitesimal transformations)


contains:

rotations:around axis k (with angle θk): w ij = �ijkθk

eg for rotations around z :

x0 → x0 ; x1 → x1 − θx2 ; x2 → x2 + θx1 ; x3 → x3

boosts:along axis k (with velocity βk): −w 0k = w k0 = βkeg for boost along z :

x0 → x0 + βx3 ; ; x1 → x1 ; x2 → x2 ; x3 → x3 + βx0


The algebra:

[Pµ,Pν ] = 0

[Pµ, Jρσ] = 0 (1)

[Jµν , Jρσ] = i(g νρJµσ − gµρJνσ − g νσJµρ + gµσJνρ)


Let’s “discover” all the above in a simple field theory:complex scalar field

L = ∂µφ∗ ∂µφ−m2 |φ|2 (2)

Symmetry: transformation of the fields which leaves the EquationsOf Motion (EOMs) invariantthis is the case if action invariant, Lagrangian can change by a totalderivative

L → L+ α∂µJ µ (3)

α=(small) parameter of the transformation

What’s the symmetry of this theory?U(1):

φ(x)→ e iαφ(x) (4)

L is invariant this U(1) is an Internal Symmetry (NOT space-time)Yael Shadmi (Technion) ESHEP2014 16 / 106

Spacetime:Translations:

xµ → xµ + aµ (5)φ(x)→ φ(x − a) = φ(x)− aµ∂µφ(x) (6)

orδaφ(x) = a

µ∂µφ(x) (7)

Lorentz transformations

xµ → xµ + wµνxν (8)φ(xµ) → φ(xµ − wµνxν) (9)

so

δwφ(x) = wµνxµ∂νφ(x) =

1

2wµν(xµ∂ν − xν∂µ)φ(x) (10)

action is invariantYael Shadmi (Technion) ESHEP2014 17 / 106

Algebra:2 translations (with aµ, bµ)

[δa, δb]φ ≡ δa(δbφ)− δb(δaφ) = 0 (11)

2 Lorentz: (with wµν , λρσ)

[δwµν , δλρσ ]φ = iwµνλρσ · i {g νρ(xµ∂σ − xµ∂σ) + permutations}(12)


let’s move now to a supersymmetric theory


A simple supersymmetric field theory

Free theory with massive (Dirac) fermion ψ of mass m2 complex scalars φ+, φ− of mass m

L = ∂µφ∗+ ∂µφ+−m2 |φ+|2+∂µφ∗− ∂µφ−−m2 |φ−|

2+ψ̄(i/∂−m)ψ (13)

[the labels +, − are just names, we’ll see the reason for this choicesoon]

[This isn’t the most minimal supersymmetric 4d field theory. “Half ofit” is: a 2-component (Weyl) fermion plus one complex scalar. ButDirac spinors are more familiar so start with this.]


spacetime symmetry:

translations, rotations, boosts: just as in our previous example

{only difference:ψ itself is a spinor, so transforms:

ψ(x)→ ψ′(x ′) (14)


actually, the L-handed and R-handed parts transform differentlyunder Lorentz (

ψLψR

)(15)

with

ψL → ψ′L = (1− iθiσi

2− β i σ

i

2) (16)

ψR → ψ′R = (1− iθiσi

2+ β i

σi

2) (17)

so it will be useful to write everything in terms of 2-componentspinors


recall: we can write any R-handed spinor in terms of a L-handed one:

ψR = −εχ∗L (18)

where

ε ≡ −iσ2 =(

0 −11 0

)(19)

Exercise: prove eq. (18)so we can write our Dirac spinor in terms of two L-handed spinorsψ+ and ψ−: ψL = ψ+, ψR = −εψ∗−

→ψ =(

ψ+−εψ∗−

)(20)

}Yael Shadmi (Technion) ESHEP2014 23 / 106

let’s write the Lagrangian in terms of these:

L = ∂µφ∗+ ∂µφ+ + ψ†+i σ̄

µ∂µψ+

+ ∂µφ∗− ∂µφ− + ψ†−i σ̄

µ∂µψ− (21)

− m2 |φ−|2 −m2 |φ+|2 −m(ψT+εψ− + hc)

exercise: Derive this. Show also that ψT+εψ− = ψT−εψ+, where ψ±

are any 2-component spinors.


Can the spacetime symmetry be extended?

Yes: there’s more symmetry hiding in our theory:

take a constant (anti-commuting) (L) 2-component spinor ξ

δξφ+ =√

2 ξTεψ+

δξψ+ =√

2 iσµεξ∗∂µφ+ −mξφ∗− (22)

and similarly for +→ −

the symmetry transformations take a boson into a fermion and viceversa: THIS IS SUPERSYMMETRY!


exercise: Check that this is indeed a symmetry:1. Show this first for m = 0.2. Repeat for m 6= 0. Note that this only holds if the masses of thefermion and scalars are the same.


Note: For m = 0, the φ− − ψ− and φ+ − χ+ parts decouple. eachone is super-symmetric separately

so: this theory is not the most minimal supersymmetric theory (halfof it is)this is handy if we’re to implement supersymmetry in the SM,because the SM is a chiral theory


is the symmetry we found indeed an extension of Poincare?

it’s surely a spacetime symmetry since it takes a fermion into a boson(the transformation parameters carry spinor indices)

furthermore: consider the algebra:take the commutator of 2 new transformations with parameters ξ, η:

[δξ, δη]φL = aµ∂µφL with a

µ = 2i(ξ†σ̄µη − η†σ̄µξ

)(23)

a translation!Exercise: Check eq. (23). You will have to use the EOMs.


Our simple theory is supersymmetric.We have an extension of spacetime symmetry that involvesanti-commuting generators.The supersymmetry transformations relate bosons and fermions.If the bosons and fermions had different masses: no supersymmetry.And let’s count the physical dof’s: on-shell we havefermions: 2 + 2 = 4bosons: 2 + 2 = 4(off shell: bosons same, but fermions: 2× 4)


The vacuum energy

Global symmetries → Noether currentsjµ with ∂µj

µ = 0 so that there is a conserved charge:

Q =

∫d3x j0(x) with

d

dtQ = 0 (24)

translations: conserved charge = Hamiltonian H

so what we found above means that the anti-commutator of twosupersymmetry transformations gives the Hamiltonian:

{SUSY, SUSY} ∝ H (25)


consider the vacuum expectation value of this:

〈0| {SUSY, SUSY} |0〉 ∝ 〈0|H |0〉 (26)

if SUSY unbroken:SUSY|0〉 = 0 (27)

therefore〈0|H |0〉 = 0 (28)

The vacuum energy vanishes!!

In a supersymmetric theory: the ground state energy is zero!


but remember: the vacuum energy usually divergesand it’s the worst divergence: quartic

(just like in the case of the harmonic oscillator: the infinite constantthat we choose to be the zero energy in “Normal Ordering”)

here: supersymmetry completely removes this divergence!

so this gives us hope that supersymmetry can help with other UVdivergences


the next worst divergence: quadraticwhere? scalar mass squared:

δm2 ∝ Λ2 (29)


This is why we are worried about fine tuning in the Higgs mass

But no one ever worries about the electron mass!Why?because fermion masses have no quadratic divergences!only log divergences!This is a very important result so we will see it in 3 ways(words+math, just words, symmetrydimensional analysis!)


why is there no quadratic divergence in the fermion mass? (1)if start from some m0 in the Lagrangian

L = ψ̄(i/∂ −m0)ψ (30)= ψ̄(i/∂)ψ −m0(ψ†LψR + ψ

†RψL)

so if m0 = 0 ψL, ψR don’t talk to each other:a mass (=L-R coupling) is never generatedthen

δm ∝ m0 (31)

with m0 = 0 we have 2 different species: ψL—call it a “blue”fermion, and ψR , a red fermion, and they don’t interact at all


Another way to see this (2):consider m0 6= 0:take a L fermion (spin along p̂) the blue fermionwe can run very fast alongside: p̂ → −p̂, spin stays same:L becomes Rthe blue fermion turns into a red fermion(helicity is not a good quantum number)but if m0 = 0: can never run fast enough..→ L and R are distinctthe blue fermion and the red fermion are decoupled


we learn:δm ∝ m0 (32)

How can the cutoff Λ enter?dimensional analyis:

δm ∝ m0 logm0Λ

(33)

soδm = 0 · Λ + # m0 log

m0Λ

(34)


why is there no divergence in the fermion mass? (3)CHIRAL SYMMETRY

L = ψ̄(i/∂)ψ −m0(ψ†LψR + ψ†RψL) (35)

when m0 = 0: U(1)L × U(1)Rthis forbids the mass term (would break to the diagonal U(1)V )


This is an example of what you all know already:The quantum world is tough:if something can diverge, it will divergeand in the worst possible way..unless there’s a symmetry protecting it..


supersymmetry: boson mass = fermion mass+chiral symmetry: no quadratic divergence in fermion massso:in a supersymmetric theory: no quadratic divergence in bosonmass

but there’s more:


You could say: we know there is no supersymmetry in Nature:there is no scalar with mass = electron massso why should we care?

supersymmetry is so powerful that even when it’s broken by massterms the quadratic divergence doesn’t reappear!all we need to see this is dimensional analysis:


what if we take a supersymmetric theory and change the scalar mass:

m02scalar = m0

2fermion + m̃

2 (36)

will there be a quadratic divergence in the scalar mass?

δm2scalar = #Λ2 + #m0

2scalar log

m02scalar

Λ2?? (37)

NO: again because of dimensional analysis:for m̃2 = 0: supersymmetry restored: there shouldn’t be a quadraticdivergenceso Λ2 term must be proprtional to m̃2

but there’s nothing we can write in perturbation theory that wouldhave the correct dimension


so: if supersymmetry is broken by

m02scalar 6= m02fermion (38)

the scalar mass-squared has only log divergences

(the supersymmetry breaking does not spoil the cancellation of thequadratic divergence)

This type of breaking is called soft-supersymmetry breaking

[this is what we have in the MSSM]


[as opposed to hard breaking:

change some pure number:

take a supersymmetric theory and change the coupling of bosonscompared to the coupling of a fermion

reintroduces divergences ]


we derived all these results based on dimensional analysisnow let’s see them concretelyfor that we have to add interactions:so go back to our simple theory

L = ∂µφ∗+ ∂µφ+ + ψ†+i σ̄

µ∂µψ+ (39)

+ ∂µφ∗− ∂µφ− + ψ†−i σ̄

µ∂µψ− (40)

− m2 |φ−|2 −m2 |φ+|2 −m(ψT+εψ− + hc) (41)


our 2 fermions look like the two pieces of an electron or a quark:ψ− (like the SM SU(2)-doublet quark)ψ+ (like the SM SU(2)-singlet quark)so let’s add a complex scalar h:with the “Yukawa” interaction:

δL = −y hψT+εψ− + hc (42)

here y is the couplingof course to make a supersymmetric theory we need alsoa (L) fermion h̃ (their susy transformations are just like φ+ and ψ+)

for simplicity let’s set m = 0


(if h and h̃ remind you of the Higgs and Higgsino that’s great, buthere they have nothing to do with mass generation, we are justinterested in the interactions)it’s easy to see that if we just add this Yukawa interaction, theLagrangian is not invariant under susyso we must add more interactions:

L = ∂µφ∗+ ∂µφ+ + ∂µφ∗− ∂µφ− + ∂µh∗ ∂µh+ ψ†+i σ̄

µ∂µψ+ + ψ†−i σ̄

µ∂µψ− + h̃†i σ̄µ∂µh̃

+ Lint (43)

with

Lint = − y (hψT+εψ− + φ+h̃Tεψ− + φ−h̃Tεψ+ + hc)− |y |2[|φ+|2 |φ−|2 + |h|2 |φ−|2 + |h|2 |φ+|2] (44)


now that we have an interacting supersymmetric theorywe are ready to consider the UV divergence in the scalar mass-squared

take δm2h


there’s a φ+ loop, a φ− loop and a fermion loop


for the latter: let’s convert to Dirac fermions:

y hψT+εψ− + hc = yhψ̄PLψ + hc (45)

so fermion loop: (“top contribution to Higgs mass”)

−|y |2∫

d4p

(2π)4TrPL

i

/pPR

i

/p= 2|y |2

∫d4p

(2π)41

p2(46)

boson loop: (“stop contribution to Higgs mass”)

2× i |y |2∫

d4p

(2π)4i

p2= −2|y |2

∫d4p

(2π)41

p2(47)


before we argued that the cancellation is not spoiled by softsupersymmetry breakinglet’s see this in this example:suppose we changed the φ± masses-squared to m̃

2±:

still no quadratic divergence

δm2h ∝ |y |2∫

d4p

(2π)4

[2

p2− 1

p2 − m̃2+− 1

p2 − m̃2−

](48)

= |y |2 m̃21∫

d4p

(2π)41

p2(p2 − m̃2+)+ (m̃2+ → m̃2−)


we see:when supersymmetry is softly broken:scalar mass squared is log divergentthe divergence is proportional to the susy breaking m̃2

as opposed to a “hard breaking”: if we changed one of the 4-scalarcoupling from |y |2: quadratic divergence


we now know a lot of susy basics (more than half of what we need ..)

let’s recap and add some language:

supersymmetry is an extension of Poincare: it’s a spacetime symmetry

the basic supersymmetry “module” we know is

(a complex scalar + 2-component spinor) OF SAME MASSeg

(φ+, ψ+) (49)

these transform into each other under supersymmetry


[together they form a representation, or multiplet of supersymmetryfor obvious reasons, we call this the “chiral supermultiplet”]

on-shell: number of fermionic dof’s = number of bosonic dof’s (truegenerally)


we also saw:supersymmetry dictates not just the field content but also theinteractions:the couplings of fermions, bosons of the same supermultiplets arerelated (true generally)


starting from scalar1-fermion2-fermion3 vertex

supersymmetry requires alsofermion1-scalar2-fermion3 + fermion1-fermion2-scalar3

all with same couplingYael Shadmi (Technion) ESHEP2014 56 / 106

+ 4 scalar (same coupling squared)

you see that the structure of supersymmetric theories is veryconstrainedand that as a result it’s less divergent


[this is the real reason theorists like supersymmetryit’s easier..(the more supersymmetry, the easier it getsless divergences, more constraints, can calculate many things, even atstrong couplingby the time you get to maximal supersymmetry in 4d: a finite theory,scale invariance)]


we also know a great deal about supersymmetry breaking:

unbroken supersymmetry: vacuum energy = 0

so: the vacuum energy is an order parameter for supersymmetrybreaking

and therefore supersymmetry breaking ↔ scale: Evac


supersymmetry (breaking) and UV divergences:

ubroken supersymmetry: only log divergences

soft breaking (=by dimensionful quantities): only log divergences

hard breaking (by pure numbers): reintroduces quadratic divergences[ so not that interesting]


let’s pause and talk about language:this will be useful when we supersymmetrize the SMour simple example2 chiral supermultiplets:each: complex scalar + L-handed fermion

φ+ ψ+ φ− ψ− (50)

in the SM each fermion, eg the top quark, comes from a fusion of 2Weyl fermions:an SU(2) doublet + an SU(2) singletthese are the analogs of ψ+, ψ−


When we supersymmetrize: must add 2 scalars (stops) these are theanalogs of φ+, φ−but one often refers to the doublet and singlet fermions as“L-handed” and “R-handed”. This is bad language (we can alwayswrite a left handed spinor using a right-handed spinor)if we used this bad language, we could call our fermions ψL, ψR andthe accompanying scalars: φL, φRjust like stop-left stop-rightof course the stops are scalars, so have no chiralitythe names just refer to their fermionic partnersstop-left stop-right etc are bad names but we’ll use them anywaybecause everyone does..


Spontanous supersymmetry breaking: the vacuum

energy, UV divergences

if supersymmetry is realized in Nature it’s realized as a brokensymmetrywe already saw that even explicit (soft) supersymmetry breaking canbe powerfulbut that’s not very satisfying: we don’t want to put in the parameterm̃2 by handwe want it to be generated by the theory itselfwe want the theory to break supersymmetry sponatenously


we saw that with unbroken supersymmetry the vacuum energyvanishesthe potential V ≥ 0supersymmetry is unbroken if there are solution(s) of the EOMs withV = 0

[This followed from

〈0| {SUSY, SUSY} |0〉 ∝ 〈0|H |0〉 (51)

and with unbroken SUSY:

SUSY|0〉 = 0 (52)

]


but if susy is spontaneously broken:

SUSY|0〉 6= 0 (53)

then the ground state has nonzero (positive) energy!let’s think about this classically:in the SM, the only scalar is the Higgs, so the only potential is theHiggs potential, and we’re not that used to thinking about scalarpotentials


but in susy theories, fermions are always accompanied by scalars andany fermion interaction results in a scalar potential: eg

V (φ+, φ−, h) = |y |2[|φ+|2 |φ−|2 + |h|2 |φ−|2 + |h|2 |φ+|2

](54)

Note: indeed: V ≥ 0 —unbroken supersymmetryTo break supersymmetry spontaneously all we need is to find asupersymmetric theory with a potential which is always above zerominimum should have nonzero energyneed a scaleclassically: put in scale by hand


The O’Raifeartaigh model

The simplest supersymmetric theory with chiral supermultiplets thatbreaks supersymmetry spontaneouslyit has 3 chiral supermultiplets:

(φ, ψ) , (φ1, ψ1) , (φ2, ψ2) (55)

and two mass parametersI will only write the scalar potential (the kinetic terms andfermion-fermion-scalar interactions are there too, but there’s nothinginstructive in them at this point. You could check that the fullLagrangian is supersymmetric, but that would be unnecessarilypainful—and painfully unnecessary!—there are much easier ways tocheck that)


V = |yφ21 − f |2 + m2|φ1|2 + |2φ1φ + mφ2|2 (56)

here m is a mass, f has dimension mass2, y is a dimensionlesscouplingbecause of the special form of this potential, it is VERY easy to seethat there is no supersymmetric minimum:the first two terms cannot vanish simultaneouslysupersymmetry is broken!


note that need f 6= 0 for that (must push some field away fromorigin) (and also m 6= 0)finding the ground state requires a bit more effortlet’s assume f < m2/(2y)the ground state is at φ1 = φ2 = 0 with φ arbitrary (a flat directionof the potential)

V0 = |f |2 (57)

the spectrum:fermions: one massless Weyl fermionone Dirac fermion of mass m(real) bosons: 2 massless, 2 with m2, one with m2 + 2yf , and onewith m2 − 2yf ,[for f = 0 susy restored: degeneracy restored]


why massless bosons?because φ is arbitrary: flat direction (2 real dof’s)

why a massless Weyl fermion?normally a sponatneously broken global symmetry → masslessGoldstone bosonhere: spontaneously broken supersymmetry → massless Goldstonefermiongenerated by broken SUSY transformation

SUSY|0〉 6= 0 (58)


recap:susy broken spontaneouslyvacuum energy nonzerofermion, boson masses split


note: needed a scale (we put it in by hand)suppose we started with no scale in Lagrangiansusy unbrokenperturbation theory: no scale generatedso: susy unbroken to all orders in perturbation theory!!


{this is a very strong resultit’s a consequence of the constrained structure of supersymmetry(holomorphy of superpotential)it’s true generally:if susy is unbroken at tree level, it can only be broken bynon-perturbative effects, with a scale that’s generated dynamically:just as in QCD:

Λ = MUV exp

(−8π2

bg 2

)(59)

exponentially suppressed compared to cutoff scalethis is called: dynamical supersymmetry breakingwe will come back to this when we discuss the standard model butleads to a beautiful scenario:the supersymmetry breaking scale can naturally be 16 or so orders ofmagnitude below the Planck scale

}Yael Shadmi (Technion) ESHEP2014 73 / 106

The chiral and vector multiplets

We have the chiral supermultiplet:

(φ, ψ) (60)

with φ a complex scalar, ψ a 2-component fermionin the real world we have spin-1 gauge bosons Aaµso there must also be “vector supermultiplets:”

(Aaµ, λa) (61)

gauge field + gaugino


on-shell: Aaµ has 2 dof, so λa is a 2-component spinor

Aaµ is real, so if we want to write λa as a 4-component spinor it must

be a Majorana spinor

(λ−ελ∗

)[compare to the Dirac fermion : ψ =

(ψ−−εψ∗+

)(62)

under a supersymmetry transformation

Aaµ λa transform into each other

and we can construct supersymmetric Lagrangians from them as wedid for the chiral supermultiplet


Supersymmetric Lagrangians

We now have the gauge module (gauge field + gaugino)and the chiral module (scalar + fermion)What are the Lagrangians we can write down?With a theory of such a constrained structure, you expect to havemany limitationsindeed: all the theories we can write down are encoded in 2 functions:The Kähler potential (K): gives the kinetic and gauge interactions(as we will see, there is no freedom there at the level of 4d terms sowe won’t even write it down)The superpotential (W): gives the non-gauge (Yukawa like)interactions of chiral fieldsLet’s start with the gauge part: after all, gauge interactions arealmost all we measure


A pure supersymmetric gauge theory

we want a gauge-invariant supersymmetric Lagrangian for

(Aaµ, λa) (63)

gauge symmetry (and supersymmetry) determine it completely!

Lgauge = −1

4F aµνF aµν + λ

a†i σ̄ · Dλa (64)

exercise: Check that this Lagrangian is supersymmetric.

(there can be higher dimension terms of course)


A supersymmetric theory with matter fields and

only gauge interactions

We also want to couple “matter fields” to the gauge fieldso add our chiral modules (φi , ψi)here too: no freedom because of gauge symmetry + supersymmetry

L = Lgauge +Dµφ∗i Dµφi + ψ†i i σ̄

µDµψi

−√

2g (φ∗i λaTT aεψi − ψ†i ελ

a∗T aφi)−1

2DaDa (65)

whereDa = −gφ†i T

aφi (66)

as in the chiral theory, supersymmetry dictates a “new” coupling:


in non-susy theories havegauge field - fermion -fermionnow also have:gaugino - fermion - scalar(and of course there’s gauge field -scalar-scalar)


and we also get a scalar potential with a 4-scalar interaction:

V =1

2DaDa (67)

withDa = −gφ†i T

aφi (68)

(note: summed over i : all charged scalars contribute)a quartic scalar potentialbut coupling isn’t arbitrary: it’s the gauge couplingthis will be very important when we discuss the Higgs!


note what happened:starting with a non-supersymmteric gauge theory with a gauge fieldAaµ and a fermion ψinteraction g Aµ − ψ − ψwhen we supersymmetrize:field content: gauge field + gaugino, fermion + scalar

interactions: Aµ − φ− φ (nothing new, φ is charged)but also: λ− φ− ψ : gaugino-scalar-fermionall with same coupling gand in addition: 4 scalar interaction, with coupling g 2

and NO FREEDOM: ALL DICTATED BY SUSY + GAUGESYMMETRY


Yukawa like interactions:

we also want Yukawa-like interactions of just the chiral scalars andfermions

(φi , ψi) (69)

a recipe for writing down the most general supersymmetricinteraction Lagrangian:choose an analytic function W (φ1, . . . , φn) the “superpotential”(no daggers!)the interactions are given by

Lint = −1

2

∂2W

∂φi∂φjψTi εψj + hc−

∑i

|Fi |2 (70)

where

F ∗i = −∂W

∂φi(71)

This is supersymmetric! (there is an elegant way to see this)Yael Shadmi (Technion) ESHEP2014 82 / 106

let’s see our previous interacting examples in this language:(by interactions I also mean masses)start with the theory containing h, φ+, φ−:take:

W = y h φ+ φ− (72)

so

F ∗h = −∂W

∂h= yφ+φ−

∂2W

∂h ∂φ+= yφ− (73)

and similarly for the rest


and we indeed recover

Lint = − y (hψT+εψ− + φ+ h̃Tεψ− + φ− h̃Tεψ+ + hc)− |y |2(|φ+|2 |φ−|2 + |h|2 |φ−|2 + |h|2 |φ+|2) (74)


exercise: Check that the massive theory with φ± is obtained from

W = mφ+φ− (75)

exercise: Check that the O‘Raifeartaigh model is obtained from

W = φ (yφ21 − f ) + mφ1φ2 (76)


The R-symmetry

we know how to write the most general supersymmetric theory in 4d(minimal supersymmetry)there are 2 things I want you to notice about it:1) there is a global U(1) symmetry:charges:gaugino=+1; chiral fermion=0 and scalar=-1orgaugino=+1; chiral fermion=-1 and scalar=0

this is called a U(1)R symmetry: does not commute withsupersymmetry(members of the same supermultiplet have different charges)this symmetry (or it’s remenant) is crucial in LHC SUSY searches!


F -terms and D-terms

2)in writing the theory, we defined F -terms and D-terms, (whichseemed very innocent, just a way to write the potentials)but let’s look at this some more:for each vector multiplet (λa,Aaµ) we have a

Da = −gφ†i Taφi (dim− 2) (77)

(all the charged scalars contribute)for each chiral multiplet (φi , ψi) we have a

F ∗i = −∂W

∂φi(dim− 2) (78)

and the scalar potential is

V = F ∗i Fi +1

2DaDa (79)


with this language, we can revisit supersymmetry breaking:first: indeed V ≥ 0 !second: supersymmetry is broken if some Fi 6= or Da 6= 0!so the Fi ’s and D

a’s are order parameters for supersymmetrybreaking!


Local supersymmetry: the gravitino mass and

couplings

so far we thought about supersymmetry as a global symmetry (andsame for spacetime)

but translations and Lorentz transformations are local symmetries(general coordinate transformations)the “gauge theory” of that is gravityso we have no choice: supersymmetry is a local supersymmetry too(supergravity)

the graviton (spin 2) has a supersymmetric partner: the gravitino(spin 3/2)

supersymmetry is broken → gravitino should get massYael Shadmi (Technion) ESHEP2014 89 / 106

if you’re only interested in collider experiments: should you careabout this?normally, the effects of gravity are suppressed by the Planck scale andwe can forget about themhere: not necessarilythe gravitino mass is related to a broken local symmetry(supersymmetry)→ it gets mass by “eating” the Goldstone fermion→ a piece of the gravitino (the longitudinal piece) is some “ordinary”field (which participates in supersymmetry breaking)→ gravitino couplings to matter are not entirely negligibe(and they are dictated by the supersymmetry breaking)


Supersymmetry breaking and the gravitino mass

suppose supersymmetry is broken by some F term: F 6= 0we know the gravitino should acquire masswhat is this mass?we must have

m3/2 ∝ F (80)

and it should decouple when MP →∞so

m3/2 = #F

MP(81)


we now understand the elements of supersymmetryin the next lecture, we will supersymmetrize the SM, and feed insupersymmetry breaking


Additional material


The SUSY Higgs mechanism: A U(1) toy model

for those of you with QFT background (non-susy Higgs mechanism)here’s a useful exercise in preparation for tomorrow:go thru the next two slides and work out the exercise at the end.consider a theory with U(1) gauge symmetrywant to break the symmetry using the Higgs mechanismwe will need 2 Higgs fields of opposite charges:φ+ and φ−with

〈φ+〉 = 〈φ−〉 (82)

so we have double the number of Higgses compared to the non-susycasedouble the number of would-be Nambu-Goldtone-Bosons ??


resolution:one combination of φ+, φ− is the usual Goldstone: it is eaten byphoton (gives longitudinal polarization)so now the massive vector multiplet: 3 dof’ssusy unbroken: must have a fermion of the same mass: 4 dof’s (thiscomes from a combination of the gaugino and the Higgsinos)to balance the numbers of fermion and boson dof’s: need the secondcombination of φ+, φ− to have the same massso the massive photon supermultiplet:gauge boson (3)Dirac fermion (4)real scalar (1)


Exercise: Work out the details of the susy Higgs mechanism in thisexample. That is, expand around the vacuum (82)

φ+(x) = v + iπ(x) + iA(x) + h(x) + H(x) (83)

φ−(x) = v − iπ(x) + iA(x) + h(x)− H(x) (84)

and find the spectrum.


A technical aside: the auxiliary fields F and D, the

Kahler potential, the superpotential, and

holomorphy

F and D are auxiliary fields (=non-dynamical fields) which we add tothe theory to simplify calculations (we will see how this works now).I avoided them in order to emphasize that these are really technicaltoolsin fact we could finish these lectures and understand everything weneed about supersymmetry without them


so why bring them up?1. so that you know what people talk about when they talk about Fand D terms (especially in the context of supersymmetry breaking)2. because it’s fun: they do make the formalism much simpler

so let’s go back to our toy exampleremember that our results (the invariance of the Lagrangian, thecommutator of 2 susy transformations gives translations) requiredthat the fields satisfy the EOMsit would be nice to have a Lagrangian for which they hold off-shell too


so let’s add an auxiliary field to each chiral multiplet, say

(φ+, ψ+,F+) (85)

δξφ+ =√

2ξTεψ+ (86)

δξψ+ =√

2iσµεξ∗∂µφ+ +√

2ξF (87)

δξF+ = −√

2iξ†σ̄µ∂µψ (88)

F is a scalar of dimension 2


The Lagrangian:for now no interactions (and no mass terms)F is an auxiliary field (cannot add physical dof’s or particles) soshould have no kinetic term:so the Lagrangian:

L = ∂µφ∗+ ∂µφ+ + ψ†+i σ̄

µ∂µψ+ + F∗+F+ (89)


In this case, the modification is trivial:F+ is not dynamical so we can always solve for it using its EOM:

F+ = 0 (90)

(which restores the susy transformations we had before)

BUT:δξL = 0 (91)

and

[δξ, δη] = aµ∂µ with a

µ = 2i(ξ†σ̄µη − η†σ̄µξ

)(92)

even OFF SHELLexercise: Check these 2 statements.


Now include interactions: this is where the auxiliary fields really helpus organize and classify the possible interactions.Consider a theory of chiral supermultiplets

(φi , ψi ,Fi) (93)

choose an analytic function of φi :

W = W (φi) (94)

W is called the superpotential (because it determines the potentialassociated with the chiral fields in supersymmetric theories)note W is analytic (or holomorphic): it does not depend on any φ∗i


construct the Lagrangian

L = Lkin + Lint (95)

whereLkin = ∂µφ∗i ∂µφi + ψ

†i i σ̄

µ∂µψi + F∗i Fi (96)

and

Lint =∂W

∂φiFi −

1

2

∂2W

∂φi∂φjψTi εψj + hc (97)

This Lagrangian is supersymmetric!

exercise: Check. You only need to check this for the interactingpart. We already know that the kinetic part is invariant.


indeed: supersymmetric interactions are very constrainedwe can now derive many of the results we saw/quoted quite generallyand elegantly:first, let’s solve for Fi :

F ∗i = −∂W

∂φi(98)

substituting this back into the Lagrangian:

V =∑

i

∣∣∣∣∂W∂φi∣∣∣∣2 (99)

indeed V ≥ 0!


What about the vector supermultiplet:Here too we add an auxiliary field Da;

λa Aaµ Da (100)

Da is a Lorentz scalar, real, of dim=2The susy transformations:

δξAµa = ξ†σ̄µλa + λa†σ̄µξ (101)

δξλa = iσµνF aµνξ + D

aξ (102)

δξD = −i [ξ†σ̄µDµλa −Dµλa†σ̄µξ] (103)

here Dµ is the covariant derivative:

Dµ = ∂µ − igAaµT a (104)

and T a is the generator (in the appropriate representation)


What are the possible Lagrangians?

L = Lgauge +Dµφ∗i Dµφi + ψ†i i σ̄

µDµψi + F ∗i Fi−√

2g (φ∗i λaTT aεψi − ψ†i ελ

a∗T aφi) + gDaφ∗i T

aφi (105)

− 12

DaDa (106)

whereDa = −gφ†i T

aφi (107)


Lectures on Supersymmetry Part I: Basics...The supersymmetry-breaking terms: freedom + determine...

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Transcript of Lectures on Supersymmetry Part I: Basics...The supersymmetry-breaking terms: freedom + determine...