Hi everyone—it’s time that I wrap up some old posts about the Higgs boson. Last December’s tantalizing results may end up being the first signals of the real deal and the physics community is eagerly awaiting the combined results to be announce at the Rencontres de Moriond conference next month. So now would be a great time to remind ourselves of why we’re making such a big deal out of the Higgs.

Review of the story so far

Since it’s been a while since I’ve posted (sorry about that!), let’s review the main points that we’ve developed so far. See the linked posts for a reminder of the ideas behind the words and pictures.

There’s not only one, but *four* particles associated with the Higgs. Three of these particles “eaten” by the *W* and *Z* bosons to become massive; they form the “longitudinal polarization” of those massive particles. The fourth particle—the one we really mean when we refer to *The* Higgs boson—is responsible for electroweak symmetry breaking. A cartoon picture would look something like this:

The solid line is a one-dimensional version of the Higgs potential. The *x*-axis represents the Higgs “vacuum expectation value,” or *vev*. For any value other than zero, this means that the Higgs field is “on” at every point in spacetime, allowing fermions to bounce off of it and hence become massive. The *y*-axis is the potential energy cost of the Higgs taking a particular vacuum value—we see that to minimize this energy, the Higgs wants to roll down to a non-zero vev.

Actually, because the Higgs vev can be any *complex* number, a more realistic picture is to plot the Higgs potential over the complex plane:

Now the minimum of the potential is a circle and the Higgs can pick any value. Higgs *particles* are quantum excitations—or ripples—of the Higgs field. Quantum excitations which push along this circle are called **Goldstone bosons**, and these represent the parts of the Higgs which are eaten by the gauge bosons. Here’s an example:

Of course, in the Standard Model we know there are *three* Goldstone bosons (one each for the *W+*, *W-*, and *Z*), so there must be three “flat directions” in the Higgs potential. Unfortunately, I cannot fit this many dimensions into a 2D picture. The remaining Higgs particle is the excitation in the not-flat direction:

Usually all of this is said rather glibly:

The Higgs boson is the particle which is responsible for giving mass.

A better reason for why we need the Higgs

The above story is nice, but you would be perfectly justified if you thought it sounded like a bit of overkill. Why do we need all of this fancy machinery with Goldstone bosons and these funny “Mexican hat” potentials? Couldn’t we have just had a theory that *started out* with massive gauge bosons without needing any of this fancy “*electroweak symmetry breaking*” footwork?

It turns out that *this *is the main reason why we need the Higgs-or-something-like it. It turns out that if we tried to build the Standard Model without it, then something very nefarious happens. To see what happens, we’ll appeal to some Feynman diagrams, which you may want to review if you’re rusty.

Suppose you wanted to study the scattering of two *W* bosons off of one another. In the Standard Model you would draw the following diagrams:

There are other diagrams, but these two will be sufficient for our purposes. You can draw the rest of the diagrams for homework, there should be three more that have at most one virtual particle. In the first diagram, the two *W* bosons annihilate into a virtual *Z* boson or a photon (γ) which subsequently decay back into two *W* bosons. In the second diagram it’s the same story, only now the *W* bosons annihilate into a virtual Higgs particle.

Recall that these diagrams are shorthand for mathematical expressions for the probability that the *W* bosons to scatter off of one another. If you always include the sum of th*e *virtual *Z*/photon diagrams with the virtual Higgs diagram, then everything is well behaved. On the other hand, if you ignored the Higgs and *only* included the *Z*/photon diagram, then the mathematical expressions do not behave.

By this I mean that the probability keeps growing and growing with energy like the monsters that fight the Power Rangers. If you smash the two *W* bosons together at higher and higher energies, the number associated with this diagram gets bigger and bigger. If these numbers get too big, then it would seem that probability isn’t conserved—we’d get probabilities larger than 100%, a mathematical inconsistency. That’s a problem that not even the Power Rangers could handle.

Mathematics doesn’t actually break down in this scenario—what really happens in our “no Higgs” theory is something more subtle but also disturbing: the theory becomes **non-perturbative** (or “strongly coupled”). In other words, the theory enters a regime where Feynman diagrams fail. The simple diagram above no longer accurately represents the *W* scattering process because of large corrections from additional diagrams which are more “quantum,” i.e. they have more unobserved internal virtual particles. For example:

In addition to this diagram we would also have even more involved diagrams with even more virtual particles which also give big corrections:

And so forth until you have more diagrams than you can calculate in a lifetime (even with a computer!). Usually these “very quantum” diagrams are negligible compared to the simpler diagrams, but in the non-perturbative regime each successive diagram is almost as important as the previous. Our usual tools fail us. Our “no Higgs theory” avoids mathematical inconsistency, but at the steep cost of losing predictivity.

*effective*theory of QCD. The old adage is true: when nature closes a door, it opens a window.

So if we took our “no Higgs” theory seriously, we’d be in an uncomfortable situation. The theory at high energies would become “strongly coupled” and non-perturbative just like QCD at low energies. It turns out that for *W* boson scattering, this happens at around the TeV scale, which means that we should be seeing hints of the substructure of the Standard Model electroweak gauge bosons—which we do not. (Incidentally, the signatures of such a scenario would likely involve something that behaves somewhat like the Standard Model Higgs.)

On the other hand, if we had the Higgs and we proposed the “electroweak symmetry breaking” story above, then this is never a problem. The probability for *W* boson scattering doesn’t grow uncontrollably and the theory remains well behaved and perturbative.

## Goldstone Liberation at High Energies

The way that the Higgs mechanism saves us is somewhat technical and falls under the name of the **Goldstone Boson Equivalence Theorem**. The main point is that our massive gauge bosons—the ones which misbehave if there were no Higgs—are actually a pair of particles: a massless gauge boson and a massless Higgs/Goldstone particle which was “eaten” so that the combined particle is massive. One cute way of showing this is to show the *W* boson eating Gold[stone]fish:

Indeed, at low energies the combined “massless *W* plus Goldstone” particle behaves just like a massive *W*. A good question right now is “low compared to what?” The answer is the Higgs vacuum expectation value (*vev*), i.e. the energy scale at which electroweak symmetry is broken.

However, at very high energies compared to the Higgs *vev*, we should expect these two particles to behave independently again. This is a very intuitive statement: it would be very disruptive if your cell phone rang at a “low energy” classical music concert and people would be very affected by this; they would shake their heads at you disapprovingly. However, at a “high energy” heavy metal concert, nobody would even hear your cell phone ring.

Thus at high energies, the “massless *W* plus Goldstone” system really behaves like two different particles. In a sense, the Goldstone is being liberated from the massive gauge boson:

Now it turns out that the massless *W* is perfectly well behaved so that at high energies. Further, the set of all four Higgses together (the three Goldstones that were eaten and *the* Higgs) are also perfectly well behaved. However, if you separate the four Higgses, then each individual piece behaves poorly. This is fine, since the the four Higgses come as a package deal when we write our theory.

What electroweak symmetry breaking really does is that it mixes up these Higgses with the massless gauge bosons. Since this is just a reshuffling of the same particles into different combinations, the entire combined theory is still well behaved. This good behavior, though, hinges on the fact that even though we’ve separated the four Higgses, all four of them are still in the theory.

*This* is why the Higgs (the one we’re looking for) is so important: the good behavior of the Standard Model depends on it. In fact, it turns out that any well behaved theory with massive gauge bosons must have come from some kind of Higgs-like mechanism. In jargon, we say that the Higgs **unitarizes** longitudinal gauge boson scattering.

**For advanced readers**: What’s happening here is that the theory of a complex scalar Higgs doublet is perfectly well behaved. However, when we write the theory nonlinearly (e.g. chiral perturbation theory, nonlinear sigma model) to incorporate electroweak symmetry breaking, we say something like:

*H(x) = (v+h(x)) exp (i π(x)/v)*. The

*π’s*are the Goldstone bosons. If we ignore the Higgs,

*h,*we’re doing gross violence to the well behaved complex scalar doublet. Further, we’re left with a non-renormalizable theory with dimensionful couplings that have powers of 1/

*v*all over the place. Just by dimensional analysis, you can see that scattering cross sections for these Goldstones (i.e. the longitudinal modes of the gauge bosons) must scale like a positive power of the energy. In this sense, the problem of “unitarizing

*W*boson scattering” is really the same as UV completing a non-renormalizable effective theory. [I thank Javi S. for filling in this gap in my education.]

## Caveat: Higgs versus Higgs-like

I want to make one important caveat: all that I’ve argued here is that we need something to play the role of *the* Higgs in order to “restore” the “four well behaved Higgses.” While the Standard Model gives a simple candidate for this, there are other theories beyond the Standard Model that give alternate candidates. For example, the Higgs itself might be a “meson” formed out of some strongly coupled new physics. There are even “Higgsless” theories in which this “unitarization” occurs due to the exchange of new gauge bosons. But the point is that there needs to be *something *that plays the role of *the* Higgs in the above story.

Long chain of doubtful reasonings. As it is smartly arranged!

Hi

Thanks for the post, when you call it eat a particel and gain mass, i see a bound state between 2 particals.

My questions doesn’t you need an extra pratical to bind the massless W+-Boson with the H+ (Higgs).

Same with W- and Z boson ofcouse.

/Björn

Hi Björn! There’s something very subtle here: when a gauge boson “eats” a Goldstone, i.e. when the W or Z “eats” part of the Higgs, one should *not* think about this as a bound state. The pictures are perhaps a little misleading, and I apologize for this.

The massive W particle has three polarizations (see this post), while the massless W has two polarizations. The additional polarization of the massive W (that the massless W does not have) is precisely the Goldstone boson that we say is “eaten.”

This should be contrasted with a meson, which is really a bound state of two quarks. For the meson you can really see that it simultaneously has a quark and an antiquark inside it at the same time. This quark and antiquark are bound together by gluons.

For the massive W boson, there’s no additional particle binding the Goldstone to the [massless] W. If you want, you can say that it’s the Higgs “vacuum expectation value” that is causing the W to recruit the Goldstone to be its longitudinal polarization.

Does that help? I’ll try to think if I can explain it better.

Great question!

F

One of the reasons I kind of stopped understanding things in this blog at some point was this. I never knew what “eaten” meant, and I couldn’t find any word besides “eaten” used anywhere on the internet to describe the relationship between the higgs and the W and Z bosons. So, from what I gather, both particles always exist at the same point. Why? What sticks the massless version of the Z to the H0? Why doesn’t it just sail along at the speed of light and leave the massive H0 behind? And why when we create a Z do we always get a H0 too? And why when we destroy a Z does the H0 get destroyed too? You say this is caused by the vacuum expectation value, but “caused” in what way?

Hi Xezlec–thanks for the feedback, I’ll try to write up another post about this to clarify things.

A good example to start with is a photon, where there’s no Higgs mechanism at all. Let’s say the photon is moving directly towards you. It has two polarizations: it can be wiggling (1) up and down or (2) left and right. It can even wiggle in some combination of these, but whatever direction it wiggles, it is perpendicular to the direction of motion (towards you) and some combination of up/down and left/right.

If we stopped the photon and looked at it at a specific moment, it would either be wiggling up and down or left and right. (I may be taking some liberties here about the observation of a single photon’s polarization, but this is irrelevant to the point of the example.) We would say “aha! I looked at this photon and it is an up/down wiggler.”

Alternately, we could have observed it to be a “left/right wiggler.” In fact, in some sense up/down wigglers and left/right wigglers are two different particles. If we have something like a circularly polarized beam of photons, then the up/down wigglers keep turning into left/right wigglers and vice versa. In this sense, the up/down and left/right wigglers mix.

A photon is a massless spin-1 (“vector”) particle with two polarizations. These two polarizations are the two types of wigglers. And it’s possible for these polarizations to mix with one another, meaning they can go from one polarization to another. (Why should they do this? It has to do with angular momentum—photons have quantum spin.) At any given time we can use our “up/down or left/right” observing machine to determine the polarization of the photon, and we’ll get one polarization or the other, never both simultaneously. (There’s some quantum mechanics here that I won’t go into.)

We would never observe a smaller “sub-photon” that is and up/down wiggler that is somehow bound to another sub-photon that is the left/right wiggler. That’s *not* what’s happening.

Anyway, if you’re happy with that, then the case of the massive bosons is almost analogous. Now instead of the two polarizations of the massless photon, the massive Z has three polarizations. We know that the third polarization is essentially one of the four Higgses (I’ll follow your convention and call it H0). Let me now address your questions:

1. Do the Z and the H0 exist at the same point? No, in the same way that the left/right wiggler and up/down wiggler don’t exist at the same point: if we observe the polarization of the Z, then we either see it wiggling up/down, left/right, or forward/backward. Quantum mechanically when we’re not looking at it, it can be any of them and is indeed all of them. But when we actually look at it, we only see one polarization. At this level we don’t have to distinguish between the “original massless Z” and the H0, you just see polarizations of the massive Z and if you want you can interpret these as pieces of the original Z and the H0.

2. What sticks the massless Z and the H0 together? In some sense it is also conservation of angular momentum, the same thing which “sticks” the up/down photon polarization to the left/right photon polarization. Probably a more satisfying answer is the Higgs vacuum expectation value. The “massless Z” (it’s not really massless, see below) bounces off of the Higgs vev and produces an H0. This is the same way in which an anti-positron can bounce off of the Higgs vev to produce an electron. You can think of this as a Feynman rule of the theory.

3. I should also clarify that as soon as the Higgs obtains a vacuum expectation value, both the Z and the H0 obtain the same mass. So it’s a bit silly for me to say “massless Z.” A better description is the ‘transverse Z’ (meaning the up/down and left/right polarizations only). This is also why the Z doesn’t run away from the H0, they both have the same mass and travel at the same velocity.

4. Do we create a Z and an H0 simultaneously? If you want you can say that you don’t. You can produce either the “transverse Z” or the H0 (the “longitudinal Z”). In fact, these are produced in slightly different ways. However, once you produce any component of the Z, it can wiggle into any other component as long as that’s allowed by conservation of angular momentum.

5. In the same way, if you destroy the “transverse Z” you also destroy the H0 because they’re *one* particle. If you destroy the transverse Z there’s nothing to wiggle into an H0 down the road.

I think the confusion is that there is *one* particle that’s moving around. It’s not like a meson where you essentially have two particles stuck together. This is one particle that can change its identity.

Maybe an example is Michael Jordan. At some point in Michael Jordan’s career he retired [temporarily] from basketball to play baseball. During that time he wasn’t Michael-the-basketballer, he was Michael-the-baseballer. In some sense are two different Michaels: one showed up in NBA games and the other showed up in MLB games. However, in another sense they’re the SAME Michael, just doing different things. And that same Michael could go from being a basketballer to being a baseballer and then back to being a basketballer. It’s not that the baseballer and basketballer were somehow glued together, they’re the SAME person manifesting himself in different ways.

Does that help? In the above example, basketball and baseball are two ‘polarizations.’ You could have had a Michael Jordan that only had one polarization (basketball.) However, once you go to a Michael that has two polarizations (basketball and baseball), Michael had to “eat a Goldstone boson” — in this case this is something like taking a roster spot on a baseball team.

(This analogy breaks down at some point… I’ll try to think this through more carefully for a future post.)

Thanks for the great questions,

Flip

Doesn’t Have the Higgs boson (h) a wiggling particle? Maybe fotons have not time to wiggle into them.

Or maybe I have a mess in my head.

Hola Cristóbal… the “wiggle” is meant to be somewhat poetic. The Higgs boson has one polarization, which becomes the “back and forth wiggle” of the massive gauge boson.

Yes, h has “back and forth wiggle” and photons can be “up/down wiggler” or “left/right wiggler”. Which I was trying to ask is if they could be the same particle in some sense but the conversion between them never happening.

Hi Cristóbal: I see, if I understand correctly, you are asking about whether “THE Higgs” (not one of the three Goldstones) can be understood as the would-be longitudinal polarization of the photon (except for some reason the photon doesn’t talk to the Higgs)?

This is indeed a tempting thing to say, but unfortunately it doesn’t work out. In fact, the properties of The Higgs don’t allow it to be the longitudinal mode of a gauge boson (this is due to how it behaves under parity, a spacetime symmetry). Another way of saying this is that “electroweak symmetry breaking” preserves electromagnetism, the Higgs really does not talk to the photon.

Good question, though!

F