# Higgs mechanism and gauge theory

I'm an undergraduate in physics, I'm on my 2nd year. I have to write this assignment about the Higgs particle and gauge theory. There are quite some things that are unclear to me however. Since I'm only on my second year I don't know a lot of deep math like group theory, just basic stuff. I know the lagrangian formalism.
My first question is about gauge invariance. I can see that the lagrangian is usually invariant under a global gauge transformation, and I can see that if introduce the covariant derivative we have a local gauge invariant lagrangian. I don't understand why we would want that however. I found the explanation that a local gauge invariance means that the theory is renormalizable, however that's just words to me. What does this mean, and is that why we want a lagrangian to be local gauge invariant? Where does this local gauge invariance origin from?
My second question is about spontaneous symmetry breaking. I'm a little lost about this concept. It appears that if the potential part of the lagrangian doesn't have a minimum at the origin you have to choose one of the minima and introduce a new field with respect to this minimum. I can see that this breaks the original symmetry in the lagrangian. I don't understand why you have to do this however? Why is it you have to translate your field so it has a minimum at the origin? I'm quite confused.
I can see that if you impose these conditions on the lagrangian
$$L = D_\mu\phi(D^\mu\phi)^*-\frac{1}{4} F^{\mu\nu} F_{\mu\nu}-V(\phi)$$
with $$D_\mu$$ as the covariant derivative and $$V(\phi)=-\mu^2\phi^*\phi+\lambda(\phi^*\phi)^2$$ yields 2 fields with masses if you pick a particular gauge. Where does the potential come from however? A good guess?

And lastly, if know some good links I'd appreciate that very much. Don't hold back even if you think it's too advanced, I can always discard it if I find it too difficult.

Thank you for your time

## Answers and Replies

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Vanadium 50
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Where does the potential come from however? A good guess?
Exactly. It's chosen so that the resulting theory has the properties we want.

arivero
Gold Member
Well, the thing is that it is a minimal guess, and then one goes on to propose methods to produce at least this guess plus some other content.

The reason why we have chosen to work with gauge theories is electromagnetic field. Maxwell equations give the first example of a gauge theory and when you go on to quantum field theory, QED is a renormalizable theory, meaning this that you are able by applying a given procedure to get from it finite predictions to be compared with experiments having an agreement with experiments that is astonishingly good (I mean never seen before for a physical theory).

With this example of a perfectly working theory for electromagnetic interaction, our hope is that Nature just repeated this success for the other interactions. This is really what happened, as uncovered by several people in '60 and '70 of last century, for weak and strong interactions. These are all gauge theories. With a significant modification that the invariance group is not simply Abelian (commutative) but rather non commutative producing a so called Yang-Mills theory.

About Higgs mechanism the choice of the proper vacuum value permits you to build up a meaningful quantum field theory. Choosing the wrong one does not produce a sensible theory and you do not get the proper mass spectrum for the theory. No sensible computations can be carried on in this case.

Jon

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arivero
Gold Member
If you are arguing along the historical approach, I think you need an extra step to reach the Higgs, and it is the need of having a massive boson to explain weak force and to explain its lack of renormalizability (when disguised as Fermi interaction). This is specially interesting if we want to repeat the unitarity argument now for the LHC.

This is really what happened, as uncovered by several people in '60 and '70 of last century,...

About Higgs mechanism the choice of the proper vacuum value permits you to build up a meaningful quantum field theory. Choosing the wrong one does not produce a sensible theory and you do not get the proper mass spectrum for the theory. No sensible computations can be carried on in this case.

blechman
Science Advisor
My first question is about gauge invariance. I can see that the lagrangian is usually invariant under a global gauge transformation, and I can see that if introduce the covariant derivative we have a local gauge invariant lagrangian.
"GLOBAL GAUGE symmetry" is a contradiction: GAUGE = LOCAL. If it's a global symmetry, then it's not a gauge symmetry! But that's just semantics.

I don't understand why we would want that however. I found the explanation that a local gauge invariance means that the theory is renormalizable, however that's just words to me. What does this mean, and is that why we want a lagrangian to be local gauge invariant? Where does this local gauge invariance origin from?
there's a lot going on here. first of all, to say that something is "renormalizable" is related to the fact that it is unitary (that is: probabilities stay below 1.0). If a theory is "non-renormalizable" it means that there is an energy scale at which this breaks down. Examples of this are pion interactions (which break down around 1 GeV when you can resolve the quarks and gluons), and Fermi (4-fermion) weak interactions (which break down when you can resolve the W-boson). Gauge theories are renormalizable, meaning that they don't break down like this. This is a good thing, but as our understanding of QFT improves, it is not a required thing.

The next point is that there is a theorem somewhere that says that the ONLY way to write down a RENORMALIZABLE theory of massless spin-1 bosons is with a gauge theory, and the only way to write down massive gauge bosons is to spontaneously break said gauge symmetry ("Higgs mechanism"). This is why these theories are so popular. They are (in a certain sense) unique.

My second question is about spontaneous symmetry breaking. I'm a little lost about this concept. It appears that if the potential part of the lagrangian doesn't have a minimum at the origin you have to choose one of the minima and introduce a new field with respect to this minimum. I can see that this breaks the original symmetry in the lagrangian. I don't understand why you have to do this however? Why is it you have to translate your field so it has a minimum at the origin? I'm quite confused.
if you are doing perturbation theory, you have to start at the minimum of the potential. otherwise your results don't make sense. To put it another way, you need to start in equilibrium. If your zeroth-order solution is out of equilibrium, then the solution is not stable and you cannot apply perturbation theory. Think of it this way: if you want to talk about small fluctuations in a gravitational field acting on a massive body, you could consider these effects on the body while it's rolling down a hill, or you consider these effects on it while it's stationary at the bottom of the hill, or rolling around near the bottom? The first problem cannot be solved with perturbation theory since the solution is not stable (the ball is rolling away even without the small gravity fluctuations), while the second problem can be solved with perturbation theory, since the ball will not run away without help from the fluctuations.

More technically, what this means is that you can write out your potential as:

$$V(\phi)=V_0 + \frac{1}{2}\omega^2(\phi-v)^2+\cdots$$

where $\omega^2=\left.\frac{d^2V}{d\phi^2}\right|_{\phi=v}$ and v is the vev of the field (this is just a Taylor series) - with this you can use the simple harmonic oscillator approximation (which is stable), with higher terms being perturbations if they're small. If you are not at the minimum, then the linear term does not vanish and you are not in any sort of equilibrium, so your "zeroth-order solution" is not stable and you cannot say anything (at least with perturbation theory)!

I can see that if you impose these conditions on the lagrangian
$$L = D_\mu\phi(D^\mu\phi)^*-\frac{1}{4} F^{\mu\nu} F_{\mu\nu}-V(\phi)$$
with $$D_\mu$$ as the covariant derivative and $$V(\phi)=-\mu^2\phi^*\phi+\lambda(\phi^*\phi)^2$$ yields 2 fields with masses if you pick a particular gauge. Where does the potential come from however? A good guess?
yes and no: it's generic in the sense that it's a Taylor expansion of a general potential with given symmetries. The only "guess" part is that the higher-order terms ($\phi^6$ for example) are irrelevant or not there (you can use renormalization arguments for why this is the case). Also, it is a guess in the sense that you are only using one field - that's the "minimum" model, but there could in general be more than one.

nrqed
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there's a lot going on here. first of all, to say that something is "renormalizable" is related to the fact that it is unitary (that is: probabilities stay below 1.0). If a theory is "non-renormalizable" it means that there is an energy scale at which this breaks down. ...
Just a question...
It is my understanding that renormalizability is a different issue than unitarity. One does not imply the other. Isn't that correct?

It is my understanding that renormalizability is a different issue than unitarity.
A perharps related paper : Gauge Theories on an Interval: Unitarity without a Higgs

One does not imply the other. Isn't that correct?
It is clear that unitarity is defined in elementary quantum mechanics, while renormalizability issues appears in QFT, so one cannot imply the other. However, they can still be related for the standard model. From the above paper
One of the main arguments for the existence of the Higgs is that without it the scattering amplitude for the longitudinal components of the massive W and Z bosons would grow with energy as ∼ E2, and thus violate unitarity at energies of order 4πMW/g ∼ 1.5 TeV.
So basically, without the Higgs you seem to end up with a theory that is not only non-renormalizable (which we understand today, is not that bad) but also non-unitary at a reasonable scale (which is a showstopper).

blechman
Science Advisor
Just a question...
It is my understanding that renormalizability is a different issue than unitarity. One does not imply the other. Isn't that correct?
no. the two are definitely related. nonrenormalizable theories break down at a given energy scale. how is it that they break down? well, they break down because cross sections blow up!

another handwavy way to see it is by the optical theorem: cross sections are related to imaginary parts of loop diagrams. if the loop diagrams contain divergences that are not under control, then the cross sections blow up. reverse the argument works, so renormalizability and unitarity are joined at the hip.

the standard model sans higgs is nonrenormalizable AND nonunitary - you can write the theory in terms of a nonlinear sigma model containing the longitudinal modes of the gauge bosons. of course, there are other things you can do to fix it (a single scalar field is just the simplest possibility).

as to humanino's comments: again, nonrenormalizable means the theory breaks down at some scale (specifically, it means the cross sections get too large) - again, this is not a "showstopper" - it simply is. it means that you cannot apply the theory in a place where it makes no sense. you cannot talk about dynamical pions in several-GeV collisions, for example.

as to humanino's comments: again, nonrenormalizable means the theory breaks down at some scale (specifically, it means the cross sections get too large) - again, this is not a "showstopper" - it simply is. it means that you cannot apply the theory in a place where it makes no sense. you cannot talk about dynamical pions in several-GeV collisions, for example.
I said non-unitary is a showstopper, but non-renormalizable is fine.

I am not convinced franckly. Unitarity is much more general and important that renormalizability. Perturbative amplitudes blow up to infinity in QFT and non-renormalizable means that you can not extract a finite part in a sensible manner. But you can do your calculations on a lattice and you do not need to worry about renormalizability. Besides, renormalizability is not really a "quantum" concept. It makes sens even outside the quantum domain, where unitarity is not defined.

nrqed
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no. the two are definitely related. nonrenormalizable theories break down at a given energy scale. how is it that they break down? well, they break down because cross sections blow up!
They break down because at that scale an infinite number of terms is required to absorb all the divergences. I thought this was unrelated to unitarity

another handwavy way to see it is by the optical theorem: cross sections are related to imaginary parts of loop diagrams. if the loop diagrams contain divergences that are not under control, then the cross sections blow up. reverse the argument works, so renormalizability and unitarity are joined at the hip.

the standard model sans higgs is nonrenormalizable AND nonunitary - you can write the theory in terms of a nonlinear sigma model containing the longitudinal modes of the gauge bosons. of course, there are other things you can do to fix it (a single scalar field is just the simplest possibility).
My point was that I thought that for a genreal field theory, unitarity and renormalizability did not necessarily imply one another (i.e. you could have one without having the other).
I recently read something that seemed to imply this. I think it was an article by a former student of Bryce DeWitt who ha dworked on quantized gravity stuff. I will try to find the reference.

Regards

Haelfix
Science Advisor
You definitely can violate unitarity if you do a calculation in a strongly coupled sector of a nonrenormalizable theory. For instance, if you remove the Higgs from the straight forward standard model and naively extrapolate WW collision to some high energy scale, you will see unitarity is explicitly broken.

It doesnt mean that quantum mechanics has broken down (like other unitarity violating effects, eg theories with gauge anomalies, many of which persist to the full nonperturbative sector), it just means you've made a mistake in where you made the calculation.

blechman
Science Advisor
my statement comes from straightforward dimensional analysis: a nonrenormalizable theory has operators with dimension larger than 4 (and therefore coupling constants that go like $M^{-a}$ for some positive integer a. Then the 2-2 scattering amplitude must go like $s^{a/2}/M^{a}$ which is greater than 1 for s large. Thus unitarity is violated in nonrenormalizable theories!

I assumed perturbativity, since i don't know what "nonrenormalizable" means in a nonperturbative context. This is the fundamental idea behind "Effective Field Theory" where you have a description that fails at some scale. The fact that it is nonrenormalizable is just a mathematical byproduct of that failing. the statement about infinite number of counterterms is also related to this breakdown, but never mind that: this is a tree-level analysis, in principle! Of course, the two statements are related due to the optical theorem as i said above.

My first question is about gauge invariance. I can see that the lagrangian is usually invariant under a global gauge transformation, and I can see that if introduce the covariant derivative we have a local gauge invariant lagrangian. I don't understand why we would want that however.
There is no reason to force local gauge symmetries onto our theories, much like there is no reason we are forced to impose Lorentz symmetry. We are driven to do that by the beauty of symmetries and by experimental support. blechman has also mentioned some theorems that require gauge symmetry in our theories.

Another bonus of gauge theories is that it allows a straightforward way to couple massless vector fields to our fields of interest. In the case of QED, we couple the Maxwell field to the fermions by virtue of imposing a local U(1), and electroweak and strong forces arise from similar considerations (but with different groups).

I found the explanation that a local gauge invariance means that the theory is renormalizable, however that's just words to me. What does this mean, and is that why we want a lagrangian to be local gauge invariant? Where does this local gauge invariance origin from?
There are various reasons to look for gauge symmetries in our theories. Along with the reasons presented above and elsewhere in this thread, it provides additional structure to our theory, e.g. the standard model has gauge symmetries of SU(3) x SU(2) x U(1). Indeed I could list the symmetries of the standard model more succinctly than writing out the full Lagrangian (or, more correctly, the symmetries provide a better way of organising the terms in the Lagrangian).

My second question is about spontaneous symmetry breaking. I'm a little lost about this concept. It appears that if the potential part of the lagrangian doesn't have a minimum at the origin you have to choose one of the minima and introduce a new field with respect to this minimum. I can see that this breaks the original symmetry in the lagrangian. I don't understand why you have to do this however? Why is it you have to translate your field so it has a minimum at the origin? I'm quite confused.
Along with the requirement of being at the minimum for perturbation as presented elsewhere in the thread, also consider that at low energies, you will be at the vacuum and exploring the local area only.

Where does the potential come from however? A good guess?
The Lagrangian density you have presented with normal derivatives would be the most general locally U(1) [i.e. multiplying by a complex phase factor] invariant renormalizable Lagrangian involving one complex scalar field. The renormalizable condition arises from requiring no coupling constants with dimensions of negative mass in the theory. Since the action, which is the Lagrangian integrated over a measure involving 4 length dimensions, we can see we are only allowed up to the 4th power of the field in our Lagrangian.

arivero
Gold Member
My second question is about spontaneous symmetry breaking. I'm a little lost about this concept. It appears that if the potential part of the lagrangian doesn't have a minimum at the origin you have to choose one of the minima and introduce a new field with respect to this minimum. I can see that this breaks the original symmetry in the lagrangian. I don't understand why you have to do this however? Why is it you have to translate your field so it has a minimum at the origin? I'm quite confused.
Well, the field is going to be expanded as perturbation of the state with minimal energy, so...

I can see that if you impose these conditions on the lagrangian
$$L = D_\mu\phi(D^\mu\phi)^*-\frac{1}{4} F^{\mu\nu} F_{\mu\nu}-V(\phi)$$
with $$D_\mu$$ as the covariant derivative and $$V(\phi)=-\mu^2\phi^*\phi+\lambda(\phi^*\phi)^2$$ yields 2 fields with masses if you pick a particular gauge. Where does the potential come from however? A good guess?
It could be a good idea to write down both lagrangians, the original and the, hmm, "phenomenological", for some particular example, and explicitly check that one of them is gauge invariant, but no the other (and see what gets modified in the second case). Is there any volunteer able to do it by heart, from memory ?

nrqed
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my statement comes from straightforward dimensional analysis: a nonrenormalizable theory has operators with dimension larger than 4 (and therefore coupling constants that go like $M^{-a}$ for some positive integer a. Then the 2-2 scattering amplitude must go like $s^{a/2}/M^{a}$ which is greater than 1 for s large. Thus unitarity is violated in nonrenormalizable theories!

I assumed perturbativity, since i don't know what "nonrenormalizable" means in a nonperturbative context. This is the fundamental idea behind "Effective Field Theory" where you have a description that fails at some scale. The fact that it is nonrenormalizable is just a mathematical byproduct of that failing. the statement about infinite number of counterterms is also related to this breakdown, but never mind that: this is a tree-level analysis, in principle! Of course, the two statements are related due to the optical theorem as i said above.
Ok, I see what you mean by the statement that nonrenormalizability implies lack of unitarity.

But I was under the impression that the converse is not true. That renormalizability does not imply unitarity. What's your take on that?

Thanks for the informative posts by the way.

blechman
Science Advisor
hi Pat.
I see what you're trying to say. as humanino said before, unitarity is a larger issue than renormalizability (although there is such a thing as "renormalization" in QM!). however, there is a tempting relationship due to the optical theorem: if you look at nonunitary scattering cross sections, they are related to forward scattering amplitudes! this makes me think seriously about a strong connection going the other way, although i admit that it's not a "proof" (!)

but all i was ever trying to do was to answer the question "Why do we care about renormalizability?" that was what NewGuy was asking, and the answer is that IF a theory is nonrenormalizable, then NECESSARILY it violateds unitarity. that's all i ever claimed. of course, nowadays we aren't so upset about such things (you are an EFT expert, after all ). but renormalizable theories have more than aesthetic appeal.

Thanks for the informative posts by the way.
Yes, I see that I have not expressed my own gratefulness. Thanks everybody, it is indeed very informative.

My current prejudice is that I am trying hard to understand Connes' geometrical approach, like in noncommutative geometry and physics, and I am forced to admit that it baffles me quite a bit. So for now I should stick to the points presented in this very thread I guess.

nrqed
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hi Pat.
I see what you're trying to say. as humanino said before, unitarity is a larger issue than renormalizability (although there is such a thing as "renormalization" in QM!).
Hi Ben....

Yes, I almost wrote up a post on this point...that even in QM one has to renormalize!
however, there is a tempting relationship due to the optical theorem: if you look at nonunitary scattering cross sections, they are related to forward scattering amplitudes! this makes me think seriously about a strong connection going the other way, although i admit that it's not a "proof" (!)

but all i was ever trying to do was to answer the question "Why do we care about renormalizability?" that was what NewGuy was asking, and the answer is that IF a theory is nonrenormalizable, then NECESSARILY it violateds unitarity. that's all i ever claimed. of course, nowadays we aren't so upset about such things (you are an EFT expert, after all ). but renormalizable theories have more than aesthetic appeal. Ok, then we are in phase completely. (I know a little about efts but am far from an expert :shy:)

Thanks for your input.

Patrick

blechman
Science Advisor
Hi Ben....
Andrew.

I should emphasize, btw, that when I say "renormalizability implies nonunitarity", I should technically be saying "PERTURBATIVE unitarity". personally, i think this is semantics (like i said earlier, i don't know what renormalizable means in a non-perturbative context), but when discussing this post with some of my colleagues, i got yelled at until i admitted that it is the (perturbative) THEORY that breaks down and not physics itself! I guess some people are really picky... in any event, i think this concession makes peace with humanino! 