Presented by

Mallika

Priyadarshini

Shivam

PHY14002

M.Sc 3rd sem

THE STANDARD MODEL

) Fundamental forces are mediated by photon, gluons, W’s and Z’s (bosons)

Basic Ingredient are quarks

and the electron-like

objects (leptons

THE STANDARD MODEL

) It provides a unified

framework for 3 of 4

(known) forces of

nature.

SU(3)× 𝑆𝑈(2) ×U(1)

THE STANDARD MODEL

)

Strong (QCD)

SU(3)× 𝑆𝑈(2) ×U(1)

THE STANDARD MODEL

)

Electroweak

(=weak +QED)

SU(3)× 𝑆𝑈(2) ×U(1)

Neutrinos... Within Standard

Model Beyond Standard Model

Massless

Left handed

Three Flavours

𝜗𝑒 , 𝜗𝜇 , 𝜗𝜏

Neutrino Oscillations

𝑝(𝜗𝑒 → 𝜗𝜇) = sin2 2𝜃 sin2[∆𝑚2𝐿

4𝐸]

∆𝑚2 = 𝑚22 −𝑚1

2

must be non-zero if neutrino oscillation exists.

BSM phenomena (seesaw) explains its tiny mass.

e

.

.

What’s meant by a gauge theory?

1.A theory described by a Lagrangian having local

symmetry properties (Invariant under local transformations)

2.Associated with each gauge symmetry is a conserved

quantity and a gauge field

[The symmetry is an internal symmetry in most gauge

theories]

Example: Electromagnetism

The Lagrangian for a free electron field 𝜳(𝒙) is

𝑳 = Ѱ 𝒊𝜸𝝁𝝏𝝁 − 𝒎 𝝍(𝒙)

Considering local symmetry

𝜳(𝒙) → 𝜳/=𝒆−𝒊𝜽 𝒙 𝜳 𝒙

• 𝜳 𝒙 𝝏𝝁𝜳 𝒙 = 𝜳 𝒙 𝝏𝝁𝜳 𝒙 − 𝒊𝜳(𝒙)[𝝏𝝁𝜶 𝒙 ]𝜳(𝒙)

Not gauge invariant covariant derivative

Maxwell’s electromagnetic field appears due to the gauge invariance principle

𝑫𝝁𝜳 = (𝝏𝝁 + 𝒊𝒆𝑨𝝁)𝜳

𝑨𝝁 = 𝑨𝝁 +𝟏

𝒆𝝏𝝁𝜶(𝒙)

ABELIAN CASE

Therefore the invariant lagrangian can be written as

𝑳/ = 𝜳𝒊𝜸𝝁 𝝏𝝁 + 𝒊𝒆𝑨𝝁 𝜳 − 𝒎𝜳𝜳

We add one kinetic energy term for the photon field

𝑳 = −𝟏

𝟒𝑭𝝁𝝑𝑭𝝁𝝑

Therefore the final lagrangian is

𝑳/ = 𝜳𝒊𝜸𝝁 𝝏𝝁 + 𝒊𝒆𝑨𝝁 𝜳 − 𝒎𝜳𝜳 −𝟏

𝟒𝑭𝝁𝝑𝑭𝝁𝝑

The following features of the equation are--

The photon is massless as the term 𝑨𝝁 𝑨𝝁is not Gauge

invariant.

The Lagrangian does not have a gauge field self

coupling.

Non Abelian gauge field

Under SU(2)

𝑯𝒆𝒓𝒆 𝒘𝒆 𝒅𝒆𝒇𝒊𝒏𝒆 𝒗𝒆𝒄𝒕𝒐𝒓 𝒈𝒂𝒖𝒈𝒆 𝒇𝒊𝒆𝒍𝒅 𝒂𝒔

The gauge field here transforms as

Ѱ′(𝒙) = 𝒆𝒙𝒑[−𝒊𝝉.𝜽

𝟐]𝜳(𝒙)

𝑫𝝁 𝜳 𝒙 = 𝝏𝝁 − 𝒊𝒈𝝉. 𝑨𝝁

𝟐𝜳 𝒙

𝝏𝝁 − 𝒊𝒈𝝉.𝑨𝝁

′

𝟐𝐔 𝜽 𝜳 𝒙 = 𝐔(𝜽) 𝝏𝝁 − 𝒊𝒈

𝝉.𝑨𝝁

𝟐𝜳(𝒙)

𝑨𝝁𝒊′ = 𝑨𝝁

𝒊 + 𝜺𝒊𝒋𝒌𝜽𝒋𝑨𝝁𝒌 −

𝟏

𝒈(𝝏𝝁𝜽𝒊)

𝑭𝝁𝝑𝒊 = 𝝏𝝁𝑨𝝑

𝒊 − 𝝏𝝑𝑨𝝁𝒊 + 𝒈𝜺𝒊𝒋𝒌𝑨𝝁

𝒋𝑨𝝑

𝒌

The complete gauge invariant lagrangian is

But we again got massless bosons because there is no mass term.

THEN HOW DO PARTICLES GET MASS???

𝑳 = 𝜳 𝒊𝜸𝝁𝑫𝝁𝜳 − 𝒎𝜳 𝜳 −𝟏

𝟒𝑭𝝁𝝑

𝒊 𝑭𝝁𝝑𝒊

Higgs Field and Symmetry Breaking

The presence of particle masses in the Standard model Lagrangian is prohibited by the SU(2)L × U(1)Y gauge symmetry of the electroweak interaction.

The Higgs mechanism has been suggested which leads to spontaneous breakdown of the electroweak symmetry by condensation of a scalar Higgs field.

Particles acquire momentum (mass) by interacting with this field.

Particles that interact strongly with the Higgs field are heavy, while those that interact weakly are light.

We consider the simple case of abelian U(1) Gauge theory

𝑳 = 𝑫𝝁𝝋∗𝑫𝝁𝝋 − 𝝁𝟐𝝋∗𝝋 − 𝝀(𝝋∗𝝋)𝟐 −𝟏

𝟒𝑭𝝁𝝑𝑭𝝁𝝑

There will be two cases 𝝁𝟐 > 𝟎 𝒂𝒏𝒅 𝝁𝟐 < 𝟎.

But since we want to generate the mass we are interested

in 𝝁𝟐 < 𝟎

Shifting the origin to 𝝋𝟏(𝒙) = 𝒗, 𝝋𝟐(𝒙) = 𝟎,

And expanding the lagrangian in terms of 𝜼 and ξ

𝝋 =𝟏

𝟐(𝒗 + 𝜼 𝒙 + 𝒊𝝃 𝒙 )

Then the Lagrangian will be 𝑳 =𝟏

𝟐(𝝏𝝁𝝃)𝟐 +

𝟏

𝟐(𝝏𝝁𝜼)𝟐 −

𝒗𝟐𝝀𝜼𝟐 +𝟏

𝟐𝒆𝟐𝒗𝟐𝑨𝝁𝑨𝝁 − 𝒆𝒗𝑨𝝁𝝏𝝁𝝃 −

𝟏

𝟒𝑭𝝁𝝑𝑭𝝁𝝑 + ⋯ 𝐨𝐭𝐡𝐞𝐫 𝐭𝐞𝐫𝐦𝐬

To remove this Goldstone boson we need to make the

following Gauge corrections.

𝝋 =𝟏

𝟐[𝝑 + 𝜼]𝒆𝒊𝝃/𝝑

And, 𝑨𝝁 = 𝑨𝝁 +𝟏

𝒆𝝑𝝏𝝁𝝃

So the final Lagrangian after these transformations

becomes

𝑳 =𝟏

𝟐(𝝏𝝁𝜼)𝟐 − 𝒗𝟐𝝀𝜼𝟐 +

𝟏

𝟐𝒆𝟐𝒗𝟐𝑨𝝁𝑨𝝁 − 𝝀𝝑𝜼𝟑 −

𝟏

𝟒𝝀𝜼𝟒 +

𝟏

𝟐𝒆𝟐𝑨𝝁

𝟐 + 𝝑𝒆𝟐𝑨𝝁𝟐𝜼 −

𝟏

𝟒𝑭𝝁𝝑𝑭𝝁𝝑

Thus we see

Massless vector boson + Goldstone boson = Massive

Vector Boson

This is called the Higgs mechanism

• The symmetry we use here is the

SU(2)×U(1) Gauge symmetry.

• Spontaneous symmetry breaking

makes SU(2)×U(1)→ 𝑼(𝟏)𝒆𝒎

• From SU(2), we get 3 gauge bosons

and from U(1) we get one Gauge Boson,

• Higgs mechanism gives mass to 3 of the

4 Gauge bosons.

HIGGS

MECHANISM

IN THE

STANDARD

MODEL

Under SU(2)×U(1) local Gauge transformation

𝝋 → 𝒆𝒊𝜽𝒂𝑻𝒂+𝒊

𝟐𝜶𝒀

𝝋

Now the Lagrangian of Higgs field can be written as

𝑳𝑯𝑰𝑮𝑮𝑺 =𝟏

𝟐𝑫𝝁𝝋

ϯ(𝑫𝝁𝝋) − 𝝁𝟐(𝝋+𝝋)

Where, we define

𝑫𝝁 = (𝝏𝝁 − 𝒊𝒈𝑾𝝁𝒂𝑻𝒂 −

𝒊

𝟐𝒈/𝑩𝝁𝒀)

A simple and useful form of the Higgs field is Φ=𝟎𝒂

To generate masses we need to give a fluctuation to a

Φ= 𝟎

𝒂 + 𝜼

We do in steps, first we don't take the fluctuation and

generate the gauge boson masses as follows

𝑫𝝁 𝟎𝒂

= (-ig𝑾𝝁𝒂𝑻𝒂-

𝒊

𝟐𝒈′𝑩𝝁Y)

𝟎𝒂

= 𝑫𝝁𝟎𝒂

= -i𝒂

𝟐

𝒈𝑾𝝁+

−𝒈𝑾𝝁𝟑 + 𝒈,𝑩𝝁

𝟏

𝟐𝑫𝝁𝝋

𝟐=

𝒂𝟐

𝟖(𝒈𝟐𝑾𝝁

+𝑾𝝁− + −𝒈𝑾𝝁

𝟑 + 𝒈/𝑩𝝁𝟐) = 𝒎𝒘

𝟐 𝑾𝝁+ +

𝟏

𝟐𝒎𝒛𝒛

𝟐

Where we define,

𝑧 =−𝒈𝒘𝝁

𝟑+𝒈/𝑩𝝁

𝒈𝟐+𝒈/𝟐 and 𝑾𝝁

+𝑾𝝁𝟏 = 𝑾𝝁

𝟏𝑾𝝁𝟏 + 𝑾𝝁

𝟐𝑾𝝁𝟐

We generated the masses of 3 bosons which are 𝑊+

. , Z.

𝒎𝑾± =𝒈𝟐𝒂𝟐

𝟖 𝒎𝒛 =

(𝒈𝟐+𝒈′𝟐)𝒂𝟐

𝟒

𝐴𝜇 field is orthogonal to Z

𝑨𝝁=𝒈/𝑾𝝁

𝟑+𝒈𝑩𝝁

𝒈𝟐+𝒈/𝟐

where , sin 𝜃𝑤 =𝑔/

𝑔2+𝑔/2 andcos 𝜃𝑤 =

𝑔

𝑔2+𝑔/2

Since there is no Mass term for the 𝐴𝜇 field So photon

remains massless in this theory also.

• Fermion masses

• For Fermion masses we consider the interaction Lagrangian

𝑳𝒊𝒏𝒕 = -𝑮𝒆(𝑳 𝜱𝑹 + 𝑹 𝜱+𝑳)

• 𝜳𝑳Φ= 𝝑𝒆 𝒆 𝑳

𝟎

𝜱𝟎 +𝒉(𝒙)

𝟐

𝜳 𝑳Φ𝜳𝑹 =𝒆 𝑳 𝜱𝟎 +𝒉(𝒙)

𝟐𝒆𝑹

Similarly 𝜳 𝑹𝜱+𝜳𝑳= 𝒆 𝑹 𝜱𝟎 +𝒉(𝒙)

𝟐𝒆𝑳

• 𝑳𝒊𝒏𝒕= -𝑮𝒆𝜱𝟎(𝒆 𝑳𝒆𝑹 + 𝒆 𝑹𝒆𝑳)- 𝑮𝒆𝒉(𝒙)

𝟐(𝒆 𝑳𝒆𝑹 + 𝒆 𝑹𝒆𝑳)

• Thus electron acquires a mass m = 𝑮𝒆𝜱𝟎

• Thus STANDARD MODEL is a powerful synthesis that successfully explains all the masses of gauge bosons and

fermions, but failed in the problem of neutrino mass !!!!

Beyond Standard Model

But Why??

RIGHT HANDED NEUTRINOS

ARE INSERTED BY HAND..

We get three neutrino mass

terms—

1. 𝑳𝒎𝒂𝒔𝒔𝑫 =

𝟏

𝟐 (𝒎𝑫𝝑 𝑹𝝑𝑳 +

𝒎𝑫𝝑 𝑳𝒄𝝑𝑹

𝒄 ) +h.c

2. 𝑳𝒎𝒂𝒔𝒔𝑳 =

𝟏

𝟐𝒎𝑳𝝑 𝑳

𝒄𝝑𝑳 + 𝒉. 𝒄

3. 𝑳𝒎𝒂𝒔𝒔𝑹 =

𝟏

𝟐𝒎𝑹𝝑 𝑹

𝒄 𝝑𝑹 + 𝒉. 𝒄

. 𝑳𝒎𝒂𝒔𝒔 = 𝑳𝒎𝒂𝒔𝒔𝑫 + 𝑳𝒎𝒂𝒔𝒔

𝑳 + 𝑳𝒎𝒂𝒔𝒔𝑹

= 𝝑 𝑳𝒄 𝝑 𝑹

𝒎𝑳 𝒎𝑫

𝒎𝑫𝑻 𝒎𝑹

𝝑𝑳

𝝑𝑹𝒄

The above mass matrix is 𝟎 𝒎𝑫

𝒎𝑫𝑻 𝒎𝑹

𝒂𝒔 𝒎𝑳=0 .

After diagonalizing the matrix the following mass eigen states are obtained---

𝒎𝟐 ≈ 𝒎𝑹 ≈ 𝟏𝟎𝟏𝟒 𝑮𝒆𝑽

𝒎𝟏 ≈𝒎𝑫

𝟐

𝒎𝑹

𝒎𝑫𝒎𝑹−𝟏𝒎𝑫

𝑻 =𝟏𝟎𝟐×𝟏𝟎𝟐

𝟏𝟎𝟏𝟒 ≈ 𝟎. 𝟏 𝒆𝑽

.INVERSE SEESAW MODEL

• Here small neutrino masses arise as a result of new Physics at TeV scale .

• May be probed at LHC , unlike TYPE I.

• 3 right handed neutrinos 𝑁𝑅 + the three extra SM gauge singlet neutral fermions S + the three active neutrinos 𝜗𝐿

• =1

2𝜗𝐿 𝑁𝑅

𝑐 𝑆𝑐

0 𝑚𝐷 0

𝑚𝐷𝑇 0 𝑀𝑅𝑆

0 𝑀𝑅𝑆𝑇 𝜇

𝜗𝐿𝑐

𝑁𝑅

𝑆

. A diagonalisation of the 9× 𝟗 matrix leads to the

effective light neutrino mass matrix ie.

𝒎𝝑= 𝒎𝑫𝑻 𝑴𝑹𝑺

𝑻 −𝟏𝝁 𝑴𝑹𝑺

−𝟏𝒎𝑫𝑻

Or, 𝒎𝝑

𝟎.𝟏 𝒆𝑽 =

𝒎𝑫

𝟏𝟎𝟎 𝑮𝒆𝒗

𝟐 𝝁

𝟏 𝑲𝒆𝑽

𝑴𝑹𝒔

𝟏𝟎 𝑻𝒆𝑽

−𝟐

Thus we see that Standard neutrinos with mass at sub ev scale are obtained for 𝒎𝑫 at electroweak scale and 𝑴𝒔 at Tev scale .

ISS is also called DOUBLE SEESAW .

24

Dark matter-connection

[1]R. N. Mohapatra and G. Senjanovic, Phys. Rev. Lett., 44, 912,

1980.

[2] Halzen, Francis, and Alan D Martin, “Quarks and

Leptons”,John Wiley & Sons(1984

[3] Moriyasu,K., “An Elementary Primer for Gauge Theories,”

World Scientific, (1983)

[4] S. F. King, arXiv:hep-ph/0208266.

[5] Carlo Giunti, arXiv:hep-ph/020572

[6] G. Altarelli and F. Feruglio, arXiv:hep-ph/0206077

[7]Y Fukuda et al. 1998 Evidence for oscillation of atmospheric

neutrinos Phys. Rev. Lett. 81 1562–1567

References