A quick introduction to gauge theories∗ Michel Bonnardeau† May 13, 2018

Abstract An introduction to gauge theories for those already familiar with quantum mechanics. It covers the broken symmetry, the Higgs mechanism and the Weinberg-Salam model.

0. Prologue To understand this introduction to gauge theories it should be sufficient to be familiar with quantum mechanics, the Klein-Gordon equation and the Dirac equation. The different sections are: • • • • • • •

Lagrangian formalism; Electromagnetism and its gauge invariance; How to make gauge invariant a complex scalar field theory; Broken symmetry without gauge invariance; Broken symmetry with gauge invariance (the Higgs mechanism); Dirac Lagrangian and chirality; Presentation of the electroweak theory (the Weinberg-Salam model).

1. Lagrangian formalism 1.1. Functionals A functional transforms a function into a number. R1 A example of a functional F acting on a function f(x) is: F [f ] = 0 f (x)dx. The functional derivative of the functional F [f (x)] is defined as: F [f (x0 ) + δ(x − x0 )] − F [f (x0 )] δF = lim δf (x) →0  ∗ May

be downloaded from http://mbond.free.fr/GAUGE/gauge.htm [email protected]

† Email:

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1.2. Principle of least action The Lagrangian L(x, x, ˙ t) holds some physics. The action is defined as: S=

R t2 t1

dtL(x, x, ˙ t)

According to the principle of least action one has: δS δx(t) = 0

1.3. Euler-Lagrange equation Calculation of δS/δx: δ R t2 δS ˙ u) δx(t) = δx(t) t1 du(¸x, x, R t2 δL δx(u) ˙ δL δ x(u) = t1 [ δx(u) δx(t) + δ x(u) ˙ δx(t) ] Rt δL d δL = t12 [ δx(u) δ(t − u)+ δ x(u) ˙ dt δ(t − u)] δL d δL = δx(t) − dt δ x(t) from part integration ˙ The principle of least action gives then the Euler-Lagrange equation or equation of motion: d δL δL δx(t) − dt δ x(t) ˙ =0

1.4. Example: classical mechanics Let be a mass m subjected to a force F proportional to its position x. The Lagrangian is written as L = T − V, with T the kinetic energy T = 12 mv 2 and V the potential energy V = F x. One has: δL δL δx = −F and δ x˙ = mv which gives the Euler-Lagrange equation: δL d δL dv δx − dt ( δ x˙ )= F − m dt = 0 i.e. the usual F = mγ

1.5. Noether’s theorem δL In the above example, let us assume that δx = 0 i.e. that the Lagrangian is invariant under a space translation. Then, according to the Euler-Lagrange 2

δL equation, δ x˙ is a constant, i.e. the momentum mv is a conserved quantity. This can be generalized: if the Lagrangian has an invariance or a symmetry, there is a corresponding conserved quantity. This is Noether’s theorem.

2. Electromagnetism 2.1. Electromagnetic tensor The quadri-vector potential Aµ = (V, A) (V the electric potential) is connected to the electric field E and the magnetic field B through: ∂A B =∇×A and E = − ∂t −∇V The electromagnetic tensor is defined as Fµν = ∂µ Aν − ∂ν Aµ . One has:   0 Ex Ey Ez  −Ex 0 −Bz By   Fµν =   −Ey Bz 0 −Bx  −Ez −By Bx 0 with the metrics g00 = g 00 = 1, gii = g ii = −1, gαβ = g αβ = 0 for α 6= β.

2.2. Lagrangian for electromagnetism When dealing with fields (instead of discrete particles), one uses the Lagrangian R density L (instead of the Lagrangian), L = Ld3 x. The Lagrangian (always density from now) for electromagnetism is (with no source, i.e. no current nor charge): L = − 41 Fµν F µν ∂L ∂L One has ∂A = 0 and ∂ (∂A ) = F µν . The Euler-Lagrange equation is then: µ ν µ   ∂L ∂L µν νµ ∂Aµ −∂ν ∂ν (∂Aµ ) = −∂ν F = 0 or ∂ν F = 0. This gives the first group of Maxwell equations: ∂E ∇E= 0 and ∇×B = ∂t ∂B The second group of Maxwell equations, ∇B = 0 and ∇×E = − ∂t is obtained from the adjoint electromagnetic tensor F 0νρ = 21 λµνρ Fλµ where λµνρ are the Levi-Civita symbols.

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2.3. Gauge invariance If any gradient function is added or subtracted to the quadri-vector potential: Aµ −→ Aµ − ∂µ χ then the electromagnetic tensor Fµν is unchanged. And so is the Lagrangian. Electromagnetism is said to be local gauge invariant (local because ∂µ χ may depend upon x).

2.4. Massive electromagnetism Let be the Lagrangian (still with no source): L = − 41 Fµν F µν + 12 m2 Aµ Aµ The Euler-Lagrange equation is ∂ν F νµ + m2 Aµ = 0 The ∂µ derivative of the Euler-Lagrange equation gives: m2 ∂µ Aµ = 0. So ∂µ Aµ = 0 because m 6= 0. The Euler-Lagrange equation becomes: (∂ 2 + m2 )Aµ = 0 This is the Klein-Gordon equation. So the Lagrangian describes a photon with a mass m.

3. Generalization of the gauge invariance 3.1. Introduction As explained in §2.3., electromagnetism has a local gauge invariance. Is this a feature of electromagnetism only, or is this something more general?

3.2. Complex scalar field Let be the complex scalar field ψ = √12 (φ1 + iφ2 ). One has then ψ † = √12 (φ1 − iφ2 ). And let be the Lagrangian: L = (∂ µ ψ † )(∂µ ψ) − m2 ψ † ψ = 12 (∂ µ φ1 )(∂µ φ1 ) + 12 (∂ µ φ2 )(∂µ φ2 ) −

m2 2 2 φ1

The Euler-Lagrange equation is: ∂L ∂L ∂φi −∂µ ∂(∂µ φi ) = 0 with i = 1, 2 4

−

m2 2 2 φ2

that is the Klein-Gordon equation: ∂ 2 φi + m2 φi = 0 Therefore the Lagrangian describes a complex scalar field with quanta of mass m.

3.3. About the sign of the mass term With massive electromagnetism, the mass term in the Lagrangian is +m2 Aµ Aµ (see §2.4). Here it is −m2 ψ † ψ with a − sign instead of a +. Generally, with a vector boson (spin 1), the sign is +, with a scalar (spin 0) it is −. And with a fermion (spin 1/2) it is also − (see §6.1).

3.4. U(1) invariance of the Lagrangian The Lagrangian is invariant when ψ is transformed the following way: ψ −→ eiα ψ (and ψ † −→ ψ † e−iα ) with α a constant. This is a U(1) symmetry with a global (α does not depend upon x) gauge invariance. What if α depends upon x? This is a local gauge transformation but the Lagrangian is obviously not invariant. Is it possible to fiddle the Lagrangian so has to make it invariant?

3.5. Making the Lagrangian locally gauge invariant To make the Lagrangian invariant under the local gauge transformation ψ −→ eiα(x) ψ it is necessary to introduce a gauge field Aµ and to replace the derivative ∂µ with the covariant derivative Dµ = ∂µ + iqAµ where q is a charge. Futhermore, the gauge field has to transform the following way: Aµ −→ Aµ − 1q ∂µ α. The modified Lagrangian is then: L = (Dµ ψ † )(Dµ ψ) − m2 ψ † ψ = (∂ µ ψ † − iqAµ ψ † )(∂µ ψ + iqAµ ψ) − m2 ψ † ψ One can check that L is indeed invariant under a local gauge transformation: ψ −→ eiα(x) ψ L −→ [−i(∂ µ α)ψ † e−iα + (∂ µ ψ † )e−iα − iq(Aµ − 1q ∂ µ α)ψ † e−iα ] [i(∂µ α)eiα ψ + eiα ∂µ ψ + iq(Aµ − 1q ∂ µ α)eiα ψ] − m2 (ψ † e−iα )(eiα ψ) =L 5

3.6. Conclusion It is possible to have a local gauge invariance for groups other than U(1). With SU(N) one has then a Yang-Mills theory (see §7.3). A global gauge conservation implies a conserved charge (Noether’s theorem). A local gauge conservation implies a bosonic gauge field which is to mediate an interaction.

4. Broken symmetry (without gauge invariance) 4.1. Introduction What is the physical implication of the local gauge invariance for the complex scalar field Lagrangian? To see this, an interaction term between the fields is added, λ(ψ † ψ)2 with λ > 0. In this section, the local gauge invariance is set aside (it will be back in §5). The Lagrangian is then: L = (∂ µ ψ † )(∂µ ψ) − V (ψ † ψ) with V (ψ † ψ) = m2 ψ † ψ + λ(ψ † ψ)2 and still ψ =

√1 (φ1 2

+ iφ2 )

The potential V has a 0 minimum for ψ † ψ = 0 (φ1 = φ2 = 0). This is the ground state and the system tends to settle there. The quanta of the φ1 and φ2 fields have the mass m.

4.2. Broken symmetry Let be a potential U of the form: U (ψ † ψ) = −m2h ψ † ψ + λ(ψ † ψ)2 where mh is a new parameter with the dimension of a mass, and still λ > 0. This potential has a minimum, U = 0, for ψ † ψ = 0 (or φ1 = φ2 = 0) and m2 m2 m2 another one, U = − 4λh , for ψ † ψ = 2λh (or φ21 + φ22 = λh ), see Figures 1 and 2. The system described by the Lagrangian L = (∂ µ ψ † )(∂µ ψ) − U (ψ † ψ) has an instable minimum at φ1 = φ2 = 0 and tends to settle to the lower m2 minimum at the circle φ21 + φ22 = λh (the ground state). This a broken symmetry.

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Figure 1: A 2D cut across of the potential U with ρ =

p ψ † ψ.

Figure 2: A 3D view of the potential U, showing the minimum shaped like a circular gutter.

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4.3. Mass of the quanta for the broken symmetry m2

φ1 and φ2 may be chosen any way along the circle φ21 + φ22 = λh . Let us choose q 2 mh φ10 = λ and φ20 = 0 and let us make a Taylor development of U around that point: dU 2 3 2 dφ1 = −mh + λφ1 + λφ1 φ2 dU and the symmetric equation for dφ . These derivatives are 0 at φ10 , φ20 , of 2 course; d2 U 2 2 2 dφ21 = −mh + 3λφ1 + λφ2 d2 U 2 dφ21 (φ10 , φ20 ) = 2mh d2 U 2 2 2 dφ22 = −mh + 3λφ2 + λφ1 d2 U dφ22 (φ10 , φ20 ) = 0 So, around φ10 , φ20 , the potential U may be written: m4

U (φ1 , φ2 )|φ10 ,φ20 = − 4λh +

1 2 2! 2mh (φ1

− φ10 )2 + ...

Let be φ01 = φ1 − φ10 and φ02 = φ2 − φ20 . Ignoring the constant term in the development of U (no effect on the Euler-Lagrange equation), this gives the Lagrangian: L = 12 ∂ µ φ01 ∂µ φ01 + 12 ∂ µ φ02 ∂µ φ02 − 12 2m2h (φ01 )2 + ... √ The mass of the quantum associated with φ01 is 2mh . There is no −(φ02 )2 term, so the quanta associated with φ02 is massless.

4.4. Conclusion There are two possible excitations: a massive quantum that oscillates up and down the gutter (φ01 ), and a massless one that roll around in the gutter (φ02 ). The massless quantum is called a Goldstone boson. The Goldstone boson has never been observed (this would be easy as it is massless), so this is an unwanted feature of the theory. The same calculation can be done for the instable potential minimum at φ1 = φ2 = 0. The two quanta are then found to be massless.

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5. Broken symmetry with gauge invariance 5.1. Introduction As in §4, a complex scalar field with an interaction λ(ψ † ψ)2 (λ > 0), is considered, but this times with a local gauge invariance: L = (∂ µ ψ † − iqAµ ψ † )(∂µ ψ + iqAµ ψ) − 14 Fµν F µν − V (ψ † ψ) with Aµ the gauge field and q the charge and Fµν = ∂µ Aν − ∂ν Aµ V (ψ † ψ) = m2 ψ † ψ + λ(ψ † ψ)2 (no broken symmetry). The Lagrangian is invariant under ψ −→ eiα(x) ψ with Aµ −→ Aµ − 1q ∂µ α. It describes 2 scalar fields, φ1 and φ2 , with ψ = √1 (φ1 + φ2 ) and with quanta (2)

of mass m. It also describes a vector field Aµ with massless quanta, therefore with two components only (like electromagnetism). Polar coordinates Let us switch to polar coordinates: ψ = ρeiθ . One has: Dµ ψ = (∂µ ρ)eiθ + iqCµ ρeiθ with Cµ = Aµ + 1q ∂µ θ (Dµ ψ † )(Dµ ψ) = ∂ µ ρ∂µ ρ + q 2 ρ2 C µ Cµ This gives for the Lagrangian: L = (∂µ ρ)2 + q 2 ρ2 C 2 − 41 Fµν F µν − V (ρ) with Fµν = ∂µ Aν − ∂ν Aµ = ∂µ Cν − ∂ν Cµ

5.2. Broken symmetry As in §4.2, instead of the potential V, let us use the potential: U (ψ † ψ) = −m2h ψ † ψ + λ(ψ † ψ)2 = U (ρ) = −m2h ρ2 + λρ4 q 2 mh The ground state is the circle ρ = ρ0 = 2λ . This is a broken symmetry. The Taylor development of U (ρ) around ρ0 , up to the second order, gives: m4

U (ρ)|ρ0 = − 4λh +

1 2 2! 4mh (ρ

− ρ0 )2 + ...

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Ignoring the constant term in the development of U (ρ), this gives the Lagrangian: L = (∂µ ρ)2 − 2m2h (ρ − ρ0 )2 + q 2 ρ2 C µ Cµ − 14 Fµν F µν + ... χ Let us set ρ = ρ0 + √ . This gives: 2 √ 2 L = 12 (∂µ χ)2 − 21 2m2h χ2 − 14 Fµν F µν + 12 2ρ20 q 2 C µ Cµ + 2ρ0 q 2 χC µ Cµ + q2 χ2 C µ Cµ +... This describes a scalar field χ with a quantum q of2 mass √ mh field Cµ with 3 components of mass 2ρ0 q = q λ .

√

2mh and a vector

5.3. Conclusion: the Higgs mechanism Without the broken symmetry one has 2 quanta of mass m and 2 quanta of mass 0. With the broken symmetry one ends up with 4 massive quanta. This is the Higgs mechanism, a consequence of the local gauge invariance. ψ is a Higgs field. We no longer have the unwanted Goldstone boson of §4.4.

6. The Dirac Lagrangian 6.1. The Dirac equation Let be the Lagrangian: ¯ µ ∂µ − m)ψ L = ψ(iγ with γ µ the Dirac matrices, ψ the 4-component Dirac spinor and ψ¯ = ψ † γ 0 the adjoint spinor. One has: ∂L ∂L = iγ µ ∂µ ψ − mψ and ¯ ¯¯ = 0 ¯ ∂ψ ∂(∂µ ψ) ¯ ∂L¯ − so the Euler-Lagrange equation for ψ, ¯ ∂µ ψ

∂L ¯¯ = 0, gives: ∂(∂µ ψ)

(iγ µ ∂µ − m)ψ = 0 This is the Dirac equation. And the Euler-Lagrange equation for ψ gives an equation whose transpose is again the Dirac equation. The Dirac equation describes fermions, i.e. spin 1/2 particles.

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6.2. Gauge invariance The Dirac Lagrangian is globally gauge invariant under ψ −→ eiα ψ with α a constant. With the electron, this implies the conservation of the electric charge (Noether’s theorem). The Lagrangian may be made locally gauge invariant by replacing ∂µ with a covariante derivative Dµ and introducing a gauge field, as with the Lagrangian for the bosons (see §3).

6.3. Chirality One has the operators: pˆ0 = i∂0 for the enery ˆ = −i∇ for the momentum. p Let use  γ0 =  γi =

the chiral representation of the Dirac matrices:  0 I I 0  0 σi with σ i the Pauli matrices. −σ i 0   ψL Let be ψ = with ψL and ψR the 2-component Weyl spinors. This ψR gives for the Dirac equation: (ˆ p0 − σ i pˆi )ψR = mψL (ˆ p0 + σ i pˆi )ψL = mψR When one has m = 0 the components ψL and ψR are no longer mixed, so they are independent. ψL is said to be left-handed (with the spin antiparallel to the momentum) and ψR to be right-handed (with the spin parallel to the momentum). The weak interaction of particle physics interacts only with the left-handed components (so it violates parity). The neutrino νe is left-handed only (it has no right-handed component).

7. The Weinberg-Salam model 7.1. Introduction As an application of this introduction to gauge theories, the Weinberg-Sam model (or electroweak theory) is presented. This model unifies electromagnetism 11

and the weak interaction.

7.2. The symmetries at the Big Bang (before their breaking) According to the model, at the creation of the Universe, the electrons and the neutrinos are massless. Therefore the electrons have independent left-handed and right-handed components eL and eR (see §6.3). The neutrinos νe are already left-handed only. There are the following charges: the weak hypercharge Y and the weak isospin I. These charges are conserved. (There is also the electric charge Q which is related to Y and I via the Gell-Mann-Nishijima relation Q = I3 + Y2 with I3 the component of I along an arbitrary axis.) The weak charges of the electron and of the neutrino are: Y I I3

νe −1 1/2 1/2

eL −1 1/2 −1/2

eR −2 0 0

The conservation of the weak isospin I comes from a symmetry (Noether’s theorem) which is SU(2). And the conservation of the weak hypercharge Y comes from the U(1) symmetry. The theory is then said to be SU(2)×U(1).

7.3. Lagrangian for the electrons and the neutrino Still at the creation of the Universe, before any symmetry breaking, the fields are written the following way:   νe L= eL R = eR One has the Dirac Lagrangian (with no local gauge invariance yet): ¯ µ ∂µ R + Liγ ¯ µ ∂µ L L0 = Riγ We are going to make this Lagrangian gauge invariant under the two symmetries, SU(2) for the weak isospin I and U(1) for the weak hypercharge Y: U(1) symmetry This works as in §3.5. One wants the local invariance under the rotation β(x): i

L, R −→ e 2 Y β(x) L, R

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with Y = −1 for L, Y = −2 for R. For the invariance to be local, a gauge field Bµ is needed, which transforms like: Bµ −→ Bµ +

1 g 0 ∂µ β

where g 0 is the coupling to the weak hypercharge, and one needs the covariant derivative: Dµ = ∂µ − 2i g 0 Y Bµ SU(2) symmetry With SU(2), one has a Yang-Mills theory, which is slightly more complicated than with U(1). One wants the local invariance under the 3-component rotation α(x): L −→ eiI τ α(x) L R −→ R with τ the Pauli matrices. One needs theses matrices because of SU(2); it is usual to use the notation τ for the isospin, instead of σ which is for the spin. One has I=1/2 for L, I=0 for R. For the invariance to be local, a gauge field is needed. With SU(2), this field is a 3-component vector Wµ , which transforms under the gauge as: Wµ −→ Wµ + g1 ∂µ α − α × Wµ where g is the coupling to the weak isospin. There is a non-commutative term α×Wµ because SU(2) is non-Abelian. One also needs the covariant derivative: Dµ = ∂µ − igIτ Wµ SU(2)×U(1) symmetry One has then the covariant derivatives, with Y = −1 and I = 1/2 for L and with Y = −2, I = 0 for R: Dµ L = ∂µ L − 2i gτ Wµ L + 2i g 0 Bµ L Dµ R = ∂µ R + ig 0 Bµ R This gives the locally gauge invariant Lagrangian for the electrons and the neutrinos:

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¯ µ (∂µ + ig 0 Bµ )R + Liγ ¯ µ (∂µ − i gτ Wµ + i g 0 Bµ )L L0 = Riγ 2 2 (B)

(W )

− 14 Fµν F (B)µν − 41 Gµν G(W )µν with (B)

Fµν = ∂µ Bν − ∂ν Bµ (W )

Gµν = ∂µ Wν − ∂ν Wµ + gWµ × Wν It is useful to look at the dimensions of the various quantities. The Lagrangian L0 has the dimension of an energy, so we write [L0 ] = E. With ¯h = c = 1 dimensionless, one has: [R] = [L] =dimensionless, [B] = [W] = E −1/2 , [g] = [g 0 ] = E 3/2 .

7.4. The Higgs field A Higgs field φ is introduced, with Y=+1 and I=1/2. It has one component with I3 = + 12 and another one with I3 = − 21 :   +   φ3 + iφ4 φ 1 √ φ= = 2 φ1 + iφ2 φ0 φ+ has I3 = +1/2, φ0 has I3 = −1/2 and φ† φ = (φ+ )∗ φ+ + (φ0 )∗ φ0 = 12 (φ21 + φ22 + φ23 + φ24 ). One wants the Higgs field to be invariant under the local gauge transformation SU(2)×U(1):  +   +  i φ φ τ α (x) 2 SU(2): −→ e 0 φ0  + φ  +  i φ φ U(1): −→ e 2 β(x) φ0 φ0 Therefore one needs the covariant derivative: Dµ φ = ∂µ φ − 2i gτ Wµ φ − 2i g 0 Bµ φ The Lagrangian for the Higgs field is then: Lφ = (Dµ )† (Dµ φ) − U (φ† φ) with U (φ† φ) = −

m2h † λ † 2 2 φ φ + 4 (φ φ) ,

the potential that gives a broken symmetry.

Interaction of the electrons and neutrinos with the Higgs field The Lagrangian for the interaction of the electron and neutrino fields with the Higgs field is:

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¯ ¯ † L) LI = −Ge (LφR + Rφ with Ge a coupling constant. One has the dimensions: [φ] = E −1/2 , [mh ] = E, [λ] = E 3 , [Ge ] = E 3/2 .

7.5. Symmetry breaking The total Lagrangian (or Weinberg-Salam Lagrangian) is: Ltotal = L0 + Lφ + LI ¯ µ Dµ L + Riγ ¯ µ Dµ R + (Dµ φ)† (Dµ φ) + = Liγ (B)

m2h † 2 φ φ

− λ4 (φ† φ)2

(W )

¯ ¯ † L) − 1 Fµν F (B)µν − 1 Gµν G(W )µν −Ge (LφR + Rφ 4 4 10−12 s after the Big Bang, when the Universe has cooled under 1016 ◦ K, the m2 Higgs field settles the minimum of the potential U (φ† φ) at (φ† φ)0 = λh = v 2 . The new ground state for the Higgs field may be chosen as: (φ1 )0  = φ0 =

√

2vwith  (φ2 )0= (φ3 )0 = (φ4 )0 = 0 0 φ+ = v 0 φ0 0

We are interested in the variations of φ around φ0 . It can be shown that there is always a unitary gauge for which all the components of φ are 0, but one. φ may then be written as: ! 0 φ= with h(x)