On spontaneous symmetry breaking and Higgs’s mechanism of mass production

**agkyriak** · Mar27-08, 06:54 AM

On spontaneous symmetry breaking and Higgs’s mechanism of mass production

From lectures: L. Peak and K. Varvell. The Physics of the Standard Model.

Full Lagrangian for fermion and photon

Combine the gauge-invariant Lagrangian density describing a fermion field in the presence of an electromagnetic field with that for the EM field itself
$LaTeX Code: \\begin{array}{l} L=\\bar {\\psi }\\left[ {\\gamma ^\\mu \\left( {i\\partial _\\mu -qA_\\mu } \\right)-m} \\right]\\psi -\\frac{1}{4}F_{\\mu \\nu } F^{\\mu \\nu }-j^\\mu A_\\mu = \\\\ =\\bar {\\psi }\\left[ {\\gamma ^\\mu i\\partial _\\mu -m} \\right]\\psi -\\frac{1}{4}F_{\\mu \\nu } F^{\\mu \\nu }-\\left( {j^\\mu +q\\bar {\\psi }\\gamma ^\\mu \\psi } \\right)A_\\mu \\\\ \\end{array} $
Note that the term coupling to the photon field $LaTeX Code: A_\\mu$ consists of two parts:
1) The external current density $LaTeX Code: j^\\mu$
2) A term corresponding to the fermion field itself $LaTeX Code: q\\bar {\\psi }\\gamma ^\\mu \\psi$ . This is called the electromagnetic current (think flow of the fermion charge) and when coupled to $LaTeX Code: A_\\mu$ describes the interaction vertex.

**agkyriak** · Mar27-08, 06:58 AM

Massive photons?

What would a mass term for the photon look like in the Lagrangian density?
We can use an analogy with the Klein-Gordon case
$LaTeX Code: L=\\frac{1}{2}\\left[ {\\partial _\\mu \\phi \\partial ^\\mu \\phi -m^2\\phi ^2} \\right] $
where the mass of the scalar field $LaTeX Code: \\phi$ comes in the term $LaTeX Code: m^2\\phi ^2$ .
Perhaps for the (vector) photon field $LaTeX Code: A^\\mu$ we could introduce a term $LaTeX Code: m^2A_\\mu A^\\mu$ .

Unfortunately this is not gauge invariant, since

$LaTeX Code: \\begin{array}{l} m^2Asingle-quote_\\mu Asingle-quote^\\mu =m^2\\left( {A+\\partial _\\mu \\chi } \\right)\\left( {A+\\partial ^\\mu \\chi } \\right)= \\\\ m^2\\left( {A_\\mu A^\\mu +\\left( {\\partial _\\mu \\chi } \\right)A^\\mu +\\left( {\\partial ^\\mu \\chi } \\right)A_\\mu +\\partial _\\mu \\chi \\partial ^\\mu \\chi } \\right)\\ne m^2A_\\mu A^\\mu \\\\ \\end{array} $

The term containing $LaTeX Code: \\partial _\\mu \\chi \\partial ^\\mu \\chi$ is harmless (it does not contribute to the equations of motion) but the terms linear in $LaTeX Code: A^\\mu$ do. This is not a problem for electromagnet ism since the photon is massless, but it will be a problem for the weak interaction.

**agkyriak** · Mar27-08, 07:05 AM

Spontaneous symmetry breaking

Suppose we generalise the Klein-Gordon Lagrangian density to the case where the scalar field $LaTeX Code: \\Phi$ is complex, or equivalently is a pair of real scalar fields $LaTeX Code: \\phi _1$ and $LaTeX Code: \\phi _2$ such that

$LaTeX Code: \\Phi =\\frac{1}{\\sqrt 2 }\\left( {\\phi _1 +i\\phi _2 } \\right) \\quad \\Phi ^+=\\frac{1}{\\sqrt 2 }\\left( {\\phi _1 -i\\phi _2 } \\right)$ or
$LaTeX Code: \\Phi =\\left| \\Phi \\right|e^{i\\theta } \\quad \\Phi ^+=\\left| \\Phi \\right|e^{-i\\theta } $
The Lagrangian density would then be
$LaTeX Code: L\\left( {\\Phi ,\\Phi ^+} \\right)=\\partial _\\mu \\Phi ^+\\partial ^\\mu \\Phi -m^2\\Phi ^+\\Phi . $
We can take $LaTeX Code: \\Phi$ and $LaTeX Code: \\Phi ^+$ a.s independent, just as easily as $LaTeX Code: \\phi _1$ and $LaTeX Code: \\phi _2$ .

The state of lowest energy of a system is known as the ground state, or in field theory terminology, the vacuum.

In the above Lagrangian the potential energy density $LaTeX Code: V\\left( {\\Phi ,\\Phi ^+} \\right)=m^2\\Phi ^+\\Phi$ is a minimum when $LaTeX Code: \\phi _1 =\\phi _2 =0$
But suppose we modify it to ( $LaTeX Code: \\phi _0$ a real constant)
$LaTeX Code: \\begin{array}{l} V\\left( {\\Phi ,\\Phi ^+} \\right)=\\frac{m^2}{2\\phi _0^2 }\\left[ {\\Phi ^+\\Phi -\\phi _0^2 } \\right]^2=V\\left( {\\Phi ,\\Phi ^+} \\right)= \\\\ =\\frac{m^2}{2\\phi _0^2 }\\left( {\\Phi ^+\\Phi } \\right)^2-m^2\\Phi ^+\\Phi +\\frac{1}{2}m^2\\phi _0^2 \\\\ \\end{array} $
The vacuum state $LaTeX Code: V\\left( {\\Phi ,\\Phi ^+} \\right)=0)$ now occurs when
$LaTeX Code: \\left| \\Phi \\right|=\\phi _0$ , which defines a circle in $LaTeX Code: \\left( {\\phi _{1,} \\phi _2 } \\right)$ space, i.e. there are an infinity of vacuums. The Lagrangian density has a U(1) symmetry.

Nature chooses one of these as the physical vacuum and "breaks'" this symmetry. This phenomenon is known as spontaneous symmetry breaking.

How does spontaneous symmetry breaking help? Suppose we expand the field $LaTeX Code: \\Phi$ around the chosen vacuum state, by writing
$LaTeX Code: \\Phi =\\phi _0 +\\frac{1}{\\sqrt 2 }\\left( {\\chi +i\\psi } \\right) $
Substituting into
$LaTeX Code: L\\left( {\\Phi ,\\Phi ^+} \\right)=\\partial _\\mu \\Phi ^+\\partial ^\\mu \\Phi -\\frac{m^2}{2\\phi _0^2 }\\left[ {\\Phi ^+\\Phi -\\phi _0^2 } \\right]^2 $
and doing the algebra, the Lagrangian density now becomes
$LaTeX Code: L\\left( {\\Phi ,\\Phi ^+} \\right)=\\partial _\\mu \\chi ^+\\partial ^\\mu \\chi +\\partial _\\mu \\psi ^+\\partial ^\\mu \\psi -\\frac{m^2}{2\\phi _0^2 }\\left[ {\\sqrt 2 \\phi _0 \\chi +\\frac{\\chi ^2}{2}+\\frac{\\psi ^2}{2}} \\right]^2 $
Pick out the "free particle" pieces by writing
$LaTeX Code: L=L_{free} +L_{int} $
we have
$LaTeX Code: \\begin{array}{l} L_{free} =\\frac{1}{2}\\partial _\\mu \\chi ^+\\partial ^\\mu \\chi +\\frac{1}{2}\\partial _\\mu \\psi ^+\\partial ^\\mu \\psi -m^2\\chi ^2 \\\\ L_{int} =-\\frac{\\sqrt 2 \\chi }{\\phi _0 }\\left( {\\frac{\\chi ^2}{2}+\\frac{\\psi ^2}{2}} \\right)-\\frac{m^2}{2\\phi _0^2 }\\left( {\\frac{\\chi ^2}{2}+\\frac{\\psi ^2}{2}} \\right)^2 \\\\ \\end{array} $
$LaTeX Code: L_{int} ,$ is a complicated ''self" interaction amongst the fields, which we will leave aside.

We can interpret
$LaTeX Code: L_{free} =\\frac{1}{2}\\partial _\\mu \\chi ^+\\partial ^\\mu \\chi +\\frac{1}{2}\\partial _\\mu \\psi ^+\\partial ^\\mu \\psi -m^2\\chi ^2 $
by comparing with the Klein-Gordon Lagrangian density
$LaTeX Code: L=\\frac{1}{2}\\left[ {\\partial _\\mu \\phi \\partial ^\\mu \\phi -m^2\\phi ^2} \\right] $
We can see that we have a massive, spinless scalar boson field $LaTeX Code: \\chi$ of mass $LaTeX Code: \\sqrt 2 m$ . This is called a Higgs boson.

A massless, spinless scalar boson field $LaTeX Code: \\psi$ . This is called a Goldstone boson.

The Higgs boson is like a fluctuation around the vacuum point in the direction in which the potential density increases. The Goldstone boson is like a fluctuation in the direction in winch the potential density is flat.

At this point, we seem to have introduced new fields into our toy theory and not gained a lot. However, the full theory mast be locally gauge invariant, which is not yet the case.

For local gauge invariance we require invariance under
$LaTeX Code: \\Phi \\left( x \\right)\\to \\Phi ^single-quote\\left( x \\right)=e^{-iq\\theta \\left( x \\right)}\\Phi \\left( x \\right) $
and the introduction of a gauge field $LaTeX Code: A_\\mu$ , transforming as
$LaTeX Code: A_\\mu \\left( x \\right)\\to A_\\mu ^single-quote \\left( x \\right)=A_\\mu \\left( x \\right)+\\partial _\\mu \\theta \\left( x \\right) $
with the Lagrangian looking like

$LaTeX Code: \\begin{array}{l} L\\left( {\\Phi ,\\Phi ^+} \\right)=\\left[ {\\left( {\\partial _\\mu -iqA_\\mu } \\right)\\Phi ^+} \\right]\\left[ {\\left( {\\partial ^\\mu +iqA^\\mu } \\right)\\Phi } \\right]-\\frac{1}{4}F_{\\mu \\nu } F^{\\mu \\nu }- \\\\ -\\frac{m^2}{2\\phi _0^2 }\\left[ {\\Phi ^+\\Phi -\\phi _0^2 } \\right]^2 \\\\ \\end{array} $

where as before
$LaTeX Code: F^{\\mu \\nu }=\\partial ^\\mu A^\\nu -\\partial ^\\nu A^\\mu $

Again the vacuum state is when $LaTeX Code: \\left| {\\Phi \\left( x \\right)} \\right|=\\phi _0$ , and since $LaTeX Code: \\theta \\left( x \\right)$ is arbitrary, we can choose it so that $LaTeX Code: \\Phi \\left( x \\right)$ is real, breaking the symmetry.

**agkyriak** · Mar27-08, 07:09 AM

Higgs’s mechanism of mass production

Proceeding as before, we expand about the chosen vacuum, writing
$LaTeX Code: \\Phi \\left( x \\right)=\\phi _0 +\\frac{h\\left( x \\right)}{\\sqrt 2 } $
with $LaTeX Code: h\\left( x \\right)$ real.

Substitution into the Lagrangian density now gives
$LaTeX Code: \\begin{array}{l} L\\left( {\\Phi ,\\Phi ^+} \\right)=\\left[ {\\left( {\\partial _\\mu -iqA_\\mu } \\right)\\left( {\\phi _0 +\\frac{h\\left( x \\right)}{\\sqrt 2 }} \\right)} \\right] \\left[ {\\left( {\\partial ^\\mu +iqA^\\mu } \\right)\\left( {\\phi _0 +\\frac{h\\left( x \\right)}{\\sqrt 2 }} \\right)} \\right]- \\\\ -\\frac{1}{4}F_{\\mu \\nu } F^{\\mu \\nu }-\\frac{m^2}{2\\phi _0^2 }\\left[ {\\sqrt 2 \\phi _0 h+\\frac{1}{2}h^2} \\right]^2 \\\\ \\end{array} $
and we can again write $LaTeX Code: L=L_{free} +L_{int}$

Rearranging terms
$LaTeX Code: \\begin{array}{l} L_{free} =\\frac{1}{2}\\partial _\\mu h\\partial ^\\mu h-m^2h^2-\\frac{1}{4}F_{\\mu \\nu } F^{\\mu \\nu }+q^2\\phi _0^2 A_\\mu A^\\mu \\\\ L_{int} =q^2A_\\mu A^\\mu \\left( {\\sqrt 2 \\phi _0 h+\\frac{1}{2}h^2} \\right)-\\frac{m^2h^2}{2\\phi _0^2 }\\left( {\\sqrt 2 \\phi _0 h+\\frac{1}{4}h^2} \\right) \\\\ \\end{array} $

How to interpret $LaTeX Code: L_{free}$ now?

We still have a Higgs boson (now denoted by

of mass $LaTeX Code: \\sqrt 2 m$ .

There is now a gauge boson $LaTeX Code: A_\\mu$ with a mass! (the term $LaTeX Code: q^2\\phi _0^2 A_\\mu A^\\mu$ represents a field with mass $LaTeX Code: \\sqrt 2 q\\phi _0 )$ .

There is no longer Goldstone boson (it has been "eaten" by the gauge field to give it a mass).

Spontaneous symmetry breaking has introduced a way of giving mass to the gauge boson of the theory, at the expense of introducing a new scalar particle (the Higgs boson).

The theory turns out to remain renormalizable (calculations give sensible results) following spontaneous symmetry breaking.

**humanino** · Mar27-08, 07:52 AM

Free lecture ?

Originally Posted by agkyriak

The theory turns out to remain renormalizable following spontaneous symmetry breaking.

Can you prove that ?

**agkyriak** · Mar27-08, 09:56 AM

Originally Posted by humanino

Free lecture ?Can you prove that ?

Proving normalization for the full Weinberg-Salam theory, t'Hooft and Veltman won the 1999 Nobel Prize.

If interested, please, see for example
http://www.slac.stanford.edu/library...nobel1999.html

**humanino** · Mar27-08, 10:15 AM

Originally Posted by agkyriak

Proving normalization for the full Weinberg-Salam theory, t'Hooft and Veltman won the 1999 Nobel Prize.

Well, since you were lecturing us, I thought maybe you could outline the proof here. That would be interesting.

**BenTheMan** · Mar27-08, 10:25 AM

Originally Posted by agkyriak

with the Lagrangian looking like

$LaTeX Code: L\\left( {\\Phi ,\\Phi ^+} \\right)=\\left[ {\\partial _\\mu -iqA_\\mu } \\right] \\left[ {\\partial ^\\mu +iqA^\\mu } \\right]-\\frac{1}{4}F_{\\mu \\nu } F^{\\mu \\nu }-\\frac{m^2}{2\\phi _0^2 }\\left[ {\\Phi ^+\\Phi -\\phi _0^2 } \\right]^2$

Hmmm I suspect a typo... (I thnk you're missing a $LaTeX Code: \\Phi$ in the kinetic term).

**Haelfix** · Mar27-08, 03:34 PM

The proof in full generality is lengthy and wonderfully unlovely. Don't be mean!

**humanino** · Mar27-08, 05:11 PM

Originally Posted by Haelfix

The proof in full generality is lengthy and wonderfully unlovely. Don't be mean!

I did not want to be mean, I apologize if I seemed mean

. I really would greatly appreciate such an outline, because everytime I attacked myself to this subject I "lost sight of the forest for the trees". There are several general proofs for instance, some of them applying to non-gauge QFTs. If you take Collins' CUP "Renormalization", it is full of examples and extremely useful when you want to learn the technics as a student, but if you want the general proof, it is barely outlined (and not obvious to me that it is straightforward

).

**agkyriak** · Mar28-08, 04:43 AM

Originally Posted by humanino

Well, since you were lecturing us, I thought maybe you could outline the proof here. That would be interesting.

Unfortunately, that would be very complex LaTex text.
I think it would be more interesting, if I told about some interpretation of the string theory, in which the generation of mass does not require renormalization.

**agkyriak** · Mar28-08, 04:44 AM

Originally Posted by BenTheMan

Hmmm I suspect a typo... (I thnk you're missing a $LaTeX Code: \\Phi$ in the kinetic term).

You are right. Thanks. I have already corrected.

**blechman** · Mar28-08, 10:37 AM

Originally Posted by humanino

I did not want to be mean, I apologize if I seemed mean . I really would greatly appreciate such an outline, because everytime I attacked myself to this subject I "lost sight of the forest for the trees". There are several general proofs for instance, some of them applying to non-gauge QFTs. If you take Collins' CUP "Renormalization", it is full of examples and extremely useful when you want to learn the technics as a student, but if you want the general proof, it is barely outlined (and not obvious to me that it is straightforward ).

Have you checked out Peskin&Schroder's chapter on R $LaTeX Code: _\\xi$ gauges? I thought it was actually pretty well-done.