Higgs mechanism and Goldstone's bosons - - Instytut Fizyki

Spontaneous symmetry breaking Higgs for Dummies Goldstone’s theorem Higgs Mechanism

Higgs mechanism and Goldstone’s bosons

Remigiusz Durka

Instytut Fizyki TeoretycznejWroclaw

March 15, 2008

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Spontaneous symmetry breaking

In physics spontaneous symmetry breaking takes place when asystem, that is symmetric with respect to some symmetrygroup, goes into a vacuum state that is not symmetric. At thispoint the system no longer appears to behave in a symmetricmanner.

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Real scalar field

L = 12(∂µφ)2 − (1

2µ2φ2 + 1

4λφ4)

V (φ) = 12µ2φ2 + 1

4λφ4 symmetry: φ(x) → −φ(x)

Minimum:

−ν = −√−µ2

λ

+ν = +√−µ2

λ

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Interesting case arising if we take µ2 < 0. The potential termwith a negative value of µ2 is odd. It appears to represent aparticle with imaginary mass.

It is easy to see why it is so: it gives a ”negative resistance”to any attempts of moving it from the origin, where the field iszero.

The potential decreases in both directions, so it isenergetically favorable for our field to roll down to one of thetwo saddle points.

These lie at the value:

φ = ±ν = ±√−µ2/λ

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We like to call the minimum of the potential our ”vacuum”:→ you cannot have less energy than that.

In the case of our potential with negative µ2, the vacuumdoes not correspond to zero value of the field! Rather, thefield takes the value ν.

There is something wrong here: imaginary mass, vacuumcontaining non-zero fields We can try to fix it by redefinitionof our scalar field.

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We shift it to the minimum at ±ν by introducing:

φ(x) = ν + η(x)

The physics cannotdepend on the shift ofthe scalar field by aconstant value ν, andnow the lagrangian takesa different form.

L = 12(∂µη)2 − λ(ν2η2 − νη3 − 1

4η4)

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L = (12(∂µη)2 − λν2η2)− λνη3 − λ

4η4)

In terms of the shifted field η(x), there is nothing wrong withthe lagrangian any more: the vacuum has zero value for thefield (it is a real vacuum!), and the field has a mass term ofthe right sign: −1

2m2η2 = −λv 2η2.

So the mass of the scalar is now m =√

2λv 2 =√−2µ2, a

positive value (forget the rest of terms, they describeself-interactions, not the mass).

However L is not symmetric for the operation η → −η anymore!

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Complex scalar field

L = (∂µΦ)∗(∂µΦ)− µ2Φ∗Φ− λ(Φ∗Φ)2

Now we wanted to write a lagrangian which is symmetricunder a continuous transformation law of the field, not justthe simple mirroring as before. (That will allow us to stateGoldstones theorem).

Field Φ = 1√2(φ1 + iφ2) is a complex scalar field.

Lagrangian has global U(1) symmetry introduced bymodification of the field by a phase transformationΦ → Φ′ = exp(iα)Φ (with α the constant phase shift).

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Again we have two parameters in the potential energy terms,λ > 0 and µ2 < 0 and we’re getting a field with an imaginarymass term.

Worse! We now have not just two, but a full circle of minimafor the potential, lying at the values of the field satisfying:

φ21 + φ2

2 = −µ2/λ.

An infinity ofchoices for thevacuum!

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Having previously worked out the simpler example of onesingle real scalar field, we are not impressed by thecompication, since we know how to get things straight.

We choose one of the vacua for a translated field by writing:

Φ(x) = 1√2

(ν + ξ(x) + iη(x)

)In terms of the shifted fields L becomes:

L = 12

((∂µξ)

2 + (∂µη)2)

+ µ2η2

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If we examine L we recognize kinetic terms (the ones with twoderivatives of the field) for the scalar fields ξ and η.

But while we also have a mass term (the one quadratic in thefield) for η.

Field ξ gets no mass term! That means that the field ismassless.

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The two fields correspond to orthogonal oscillations about thevacuum we have chosen.

Massless field corresponds to oscillations along the directionwhere the potential remains at a minimum - along the circle ofminima. Because of that, it encounters no resistance - noinertia, no mass.

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The spontaneous breaking of the symmetry of the originallagrangian for φ has generated a mass for one of the twoscalars, and a further massless scalar has appeared in thetheory.

This is the Goldstone theorem in a nutshell:

The spontaneous breaking of a continuous symmetry of thelagrangian generates massless scalars. They correspond tofluctuations around the chosen vacuum in the directiondescribed by the neighboring vacua.

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Massless scalar particles do not belong to any reasonabletheory of nature. Our world would be a quite different place ifthere were massless scalars around.

We do not observe such particles. Indeed, there is a trick,called the Higgs mechanism, which gets rid of the masslessGoldstone bosons.

The degrees of freedom of the theory associated to theGoldstone bosons reappear as the mass terms for the weakvector bosons as will be discussed later.

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We saw what is Goldstones theorem and why should webother knowing it.

The theorem is a crucial preliminary to understand the needfor a Higgs boson in the Standard Model.

Once agian Goldstone’s theorem can be stated as follows:

If a continous symmetry of the Lagrangian is spontaneouslybroken, and if there are no long-range forces, then exists azero-frequency excitation at zero momentum.

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Ferromagnets:

The absence of long-range forces, which may tend to couplespins at large distances, is necessary for the existance of amode with ω → 0 as k → 0.

Superconductors:

In Bardeen-Cooper-Schrieffer (BCS) theory there is aspontaneous breaking of the phase invariance associated withthe conservation of the electron number.

However there is an energy gap (equal to the mass of theCooper pairs), so there is no Goldstone’s boson.

The reason is that there are long-range electromagneticforces.

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Superfluids (for example low-temperature Bose system):

The condenstate field at T = 0 is < Φ >= ξ, which is relatedto the particle number density by n = |ξ|2.

The phonon spectrum is:

ω2 =k2

2m(k2

2m+ 2nV (k))

A short-range potential has the property that V (k) (theFourier transform of the two-body potential) is finite andpositive. In that case, ω →

√nV (k = 0)/m as k → 0. This

not so for a long-ange potential.

For the Coulomb force, V (k) = e2

k2 and, as k → 0,

ω → e√

n/m = ωp, the plasma frequency.

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Proof of Goldstone’s theoremin context of the U(1) scalar field theory

The U(1) symmetry is Φ → Φe−iα, or δΦ = −iαΦ if |α|2 � 1.

The conserved current density in terms of the shifted field:

jµ = χ2∂µχ1 − χ1∂µχ2 −√

2ξ∂µχ2

The total charge, Q =∫

d3x j0(x) is conserved.

The change in Φ due to an infinitesimal change in phase canalso be expressed in operator form as:

δΦ = iα[Q, Φ]

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Taking the thermal, or ensemble, average of δΦ, we find< δΦ >= −iα < Φ >= −iαξ.

Taking the thermal average of previous equation (with totalcharge as the generator of the phase transition) we find anexpression for the condenstate field:

ξ = −∫

d3x < [ j0(x, t), Φ(0, 0)] >

Now we define the function:

F µ(k0, k) =

∫d4x e ik·x < T [ jµ(x), Φ(0)] >

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F µ(k0, k) =

∫d4x e ik·x < T [ jµ(x), Φ(0)] >

Since T [jµ(x), Φ(0)] = jµ(x)Φ(0)θ(x0) + Φ(0)jµ(x)θ(−x0)and ∂µ jµ = 0 we can write that:

kµF µ = −i∫

d4x ∂µ

(e ik·x < T [ jµ(x), Φ(0)] >

)+

+i∫

d3x e−ik·x < [ jµ(x), Φ(0)] >

Surface term vanishes and second term can be obtain fromcomparison with ξ = −

∫d3x < [ j0(x, t), Φ(0, 0)] >

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We end up with expression:

limk→0

kµF µ = −iξ

If ξ 6= 0, which means that the U(1) symmetry isspontaneously broken, then F has a pole at k = 0.

This pole corresponds to a zero-frequency excitation at zeromomentum.

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It’s not diffcult to determine F µ. Substituting jµ in terms ofχ1 and χ2 into F µ leads to:

F µ = −ξkµ

∫d4x e ikx < T [χ2(x)χ2(0)] >= −iξkµD2(k)

where D2 is real time Green’s function.

Conbining F µ = −iξkµD2(k) with limk→0 kµF µ = −iξ tells usthat the imaginay part of th shifted field has a dispersionrelation with the property that ω(k = 0) = 0. This is theGoldstone boson.

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Higgs Mechanism

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Glashow-Weinberg-Salam model

L = (DµΦ)†(DµΦ)− µ2Φ†Φ− λ(Φ†Φ)2 − 14bµνbµν − 1

4f µνa f a

µν

Lagrangian has SU(2)xU(1) symmetry. There is SU(2) gaugefield Aa

µ and a U(1) gauge field Bµ.

Strengths are:

f aµν = ∂µAa

ν − ∂νAaµ − gεabcAb

µAaν

gµν = ∂µBν − ∂νBµ

Covariant derivative: Dµ = ∂µ + 12igAa

µτa + 12ig ′Bµ

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which acts on a complex SU(2) field:

Φ =1√2(

φ1 + iφφ3 + iφ4

)

Let’s introduce the vacuum expectation value:

< Φ >=1√2

( 0ν

)

Φ =1√2U−1(ζ)

( 0ν + η

)

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U(ζ) = exp(−iζτ

2ν)

Φ → Φ′ = U(ζ)Φ =1√2

( 0ν + η

)After some cacululations:

L = (∂µη)∂µη)− 12µ2(ν + η)2 − 1

4λ(ν + η)4 − 1

4bµνbµν −

14f µνa f a

µν + +14Φ†(g ′Bµ + gτAµ)(g ′Bµ + gτAµ)Φ′

This can be written as the sum of classical part, a partquadratic in the fields and a part giving rise to interactionsthat is cubic and quadratic in the fields

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We define:

W±µ =

1√2

(A1

µ ± iA2µ

)

Zµ =1√

g 2 + g ′2

(g ′Bµ − gA3

µ

)

Aµ =1√

g 2 + g ′2

(gBµ + g ′A3

µ

)

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The masses:

mA = 0

mW =1

2gν

mz =1

2

√g 2 + g ′2

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