In the steps toward a cartoon of the triplet model, we had to first explore a cartoon of the Standard Model itself (see pages 1, 2 and 3 in this series.) When considering just the Higgs field H and the W field, the potential looked something like this

- V = – ½ μ
_{H}^{2}H^{2}+ ¼ λ H^{4}+ ⅛ g^{2}H^{2}W^{2}

But where did this potential come from? Why were these three terms present, and why weren’t there others?

In fact a more complete cartoon of the Standard Model has the following form

**V = – ½ μ**_{H}^{2}|H|^{2}+ ¼ λ |H|^{4}+ ⅛ |H|^{2}( g^{2 }[W_{1}^{2 }+^{ }W_{2}^{2}] + [g W_{3}– g’X]^{2})

What is all this mess? That will be partly answered on this page, and continued in the next one.

**SYMMETRIES (REAL AND OTHERWISE)**

Symmetries are a major player in the physicist’s toolbox. However, I want to warn you right up front that there are several classes of symmetries.

- There are symmetries of the real world — these will be reflected in our equations because, well, they’re real. An example would be the 90-degree rotation symmetries of a square.
- And there are symmetries that are
**useful**to add to our equations. They are**not**symmetries of nature, and they appear in our equations simply because we don’t know how to deal with the equations unless we put them in. These not-really-symmetries are “gauge” symmetries. They help us solve the equations, but in the end, we take them out.

**The Standard Model has almost no real symmetries at all.** **But it does have gauge symmetries**, and so its equations are going to reflect them. The gauge symmetry of the Standard Model is often said to be SU(3) x SU(2) x U(1), terminology which I’ll partially explain. The SU(3) part has to do with the strong nuclear force, so we won’t need it. Our focus will just be on SU(2) x U(1), the electroweak part that gives rise to the weak nuclear force and electromagnetism.

It turns out that gauge symmetries of this type always come with gauge bosons — particles of spin 1. Or really, **it’s the other way round:** particles of spin 1, once discovered experimentally, can often be hard to describe mathematically unless you put gauge symmetries into the equations which you use to predict their behavior. In the Standard Model these gauge bosons include the W bosons, the Z boson and the photon, and these are most easily described using an SU(2) x U(1) gauge symmetry. (The gluons of the strong nuclear force are associated with the SU(3) gauge symmetry.)

But the W boson, Z boson and photon, and their corresponding fields, are not so simple in terms of the SU(2)xU(1) language. In the red-colored potential I wrote at the end of the introduction, there are four “gauge fields” which appear: a triplet of fields W_{1}, W_{2}, and W_{3}, associated with SU(2), and the X field, associated with U(1). We’ll soon see how the particles I mentioned arise from these fields, though not yet today.

**THE POTENTIAL AND SYMMETRIES**

Under a symmetry (whether real or gauge) the potential V has to be unchanged, or “invariant”. (After all, when we say that a square is symmetric under 90 degree rotations, we mean that ** it doesn’t change when we do that rotation**.)

For instance, as already mentioned here, if there is a symmetry where H can be replaced everywhere with -H, that ensures that the potential cannot contain a term proportional to H, because if one were present, then the potential would change under this symmetry. Suppose that

- V = a H + b H
^{2};

then the transformation H → -H would change V by flipping the sign of the first term

- V = + a H + b H
^{2}→ V =**–**a H + b H^{2}

and so **a theory with this potential would not be symmetric under exchanging H with -H.**

H^{3} terms are similarly disallowed. But the potential can contain an H^{2} term, an H^{4} term, an H^{6} term, etc., because they will not change when H is replaced with -H. The exchange H –> -H is one among the infinite number of SU(2)xU(1) symmetries, so indeed SU(2)xU(1) gauge symmetry forbids the potential from containing any terms H^{n} for any odd n.

*What about H ^{6}? why didn’t I write it too, along with H^{8} and so forth? Actually, I could have written these terms down too, and in some contexts they might have been important. But usually they can be ignored. Typically, all terms with more than four fields in them will be highly suppressed in their effects, and so unless we are doing a calculation to extreme precision, they won’t matter at all. This would be strictly true if the Standard Model were a fundamental theory, but it is even true as long as it is valid to energy scales at least five to ten times higher than the ones we currently probe experimentally. (The underlying reason has to do with a more subtle symmetry — scaling symmetries — that I’ll try to get back to someday.)*

So we only have to worry about H^{2} and H^{4} terms. Why are their coefficients negative and positive, respectively? We don’t have a deep theoretical reason to expect their coefficients to be one sign or the other. It is experiment that tells us that

- the Higgs field has a vacuum expectation value;
- the universe is stable on multi-billion-year time scales.

These two requirements imply that the potential likely looks like a “W” shape — with a positive H^{4} and a negative H^{2} term (i.e., λ > 0 and **μ _{H}^{2} > 0 [not a typo]**) — putting us in the “broken phase”, with <H> not equal to zero.

That takes care of the first two terms in the potential. What about the terms in which H appears along with gauge bosons? We won’t answer that entirely yet today, but let’s understand their origin in SU(2) x U(1).

**THE U(1) GAUGE BOSON AND THE HIGGS**

Electromagnetism is an example of a force whose particle, the photon, is associated with a gauge field (usually called A, by convention) and a U(1) gauge symmetry. You might think that’s the U(1) of the Standard Model, but it’s not; things are a little more subtle than that, as we’ll see.

Nevertheless, electromagnetism does give us a nice example of how a U(1) gauge symmetry works. A fact (whose explanation requires a little more math, and so I’ll save it until the end of this series) is that if you have a spin-zero field F whose particle has electric charge q under electromagnetism, then the potential must contain the term

**½ e**^{2}q^{2}|F|^{2}A^{2}

where

- e is the basic strength of the electromagnetic force
*(although sometimes one calls e*^{2}/(4π) the strength; this is a matter of semantics) - q is the electric charge of the field F (where the electron’s charge is -1 and the proton’s charge is +1)
- A is the gauge field for electromagnetism, whose particle is the photon
- F is a complex field (i.e. F, which as for any field is shorthand for F(x,y,z,t), is a complex number at each point in space and time),
- and |F|
^{2}means F* F, where F* is the complex conjugate of F;*equivalently, |F|*^{2}is the square of the absolute value of the complex field F

Although this structure doesn’t appear in the Standard Model, a very similar one does. The U(1) symmetry in the Standard Model is usually called “hypercharge” (for historical reasons that are not very interesting here). The Higgs boson has hypercharge +½, and the interaction strength of hypercharge is usually called g’. By analogy with the previous equation, the potential must contain the term

- ½ (g’)
^{2}(½)^{2}|H|^{2}X^{2}= ⅛ g’^{2}|H|^{2}X^{2}

The g’ has to be present because all physics involving the hypercharge U(1) must have it. The coefficient is ⅛ because H has q=½, and so **½ q ^{2} = ⅛**. The |H|

^{2}and the X

^{2}all have to be there by gauge symmetry, and gauge symmetry also forces the overall sign to be positive.

If you look back at the red-colored potential in the introduction, you’ll see this term is present; just set all the W fields to zero, and you’ll find it there. *(Note that the minus sign in front of X cancels when you square it.)* So if the Standard Model had no SU(2), and thus no W fields, we’d now understand its potential completely!

Well, almost. Why aren’t there any terms involving X only, perhaps an X term, or an X^{2}, or an X^{4}? These are all forbidden by the U(1) gauge symmetry; again, to see why requires a little more math than I want to do today.

The first two terms in the potential now assure that H has a vacuum expectation value, which we’ll call “v”. When we replace H with v, the quadratic terms for X in the potential tell us the X boson’s mass. Since the X^{2} term in the potential has coefficient

- ⅛ g’
^{2}v^{2}= ½ [**(g’)**],^{2}(½)^{2}v^{2}

we see that **the X boson has mass m _{X} = ½ g’ v** .

**THE SU(2) GAUGE BOSONS AND THE HIGGS**

The reason U(1) has a single gauge boson is that it represents a class of symmetries that are similar, mathematically, to the set of symmetries that rotate a circle. You can only rotate a circle by **one** angle in **one** plane, and that math determines the number of gauge bosons.

SU(2), by contrast, is a set of symmetries that are mathematically like the rotations of a two-dimensional sphere. You can rotate a sphere around any of **three** planes (the x-y plane, the y-z plane, and the z-x plane), using three angles (“Euler angles”). That gives you **three** gauge bosons, from three W fields, often simply called W_{1}, W_{2} and W_{3}.

The Higgs field, under SU(2), has (again) charge ½. *[This is not precise, but to give the full story at this stage would take us far afield, so again, let me put this off for later.]* That ensures that if the Standard Model only had an SU(2) and no U(1), so that we could erase the X field, the terms in the red-colored potential that combine the H and W fields would be

- ½ g
^{2}(½)^{2}|H|^{2}(W_{1}^{2}+W_{2}^{2}+W_{3}^{2})

You can see these terms are indeed present if we set X to zero. Compared to the potential for the U(1) case, g replaces g’, the (½)^{2} is the same because H has charge ½ again, and the W fields all must have the same potential because they are related to each other by the SU(2) symmetry (just as the x-y, y-z, and z-x planes around which one can rotate a sphere are themselves related to each other by symmetry.)

Again, in the broken phase where H has a vacuum expectation value v, we can read off the W bosons’ masses: **they all have mass** **m _{W} = ½ g v .**

*By the way, although W and W ^{2} terms in the potential are still forbidden by gauge symmetry, certain specific W^{3} and W^{4} terms are allowed and in fact required. But they don’t affect any calculations of the properties of the vacuum, because the equations for the vacuum set W = 0 whether these terms are accounted for or not.*

Now that we’ve done U(1) and SU(2) separately, we’ll next start trying to put them together next time. You might guess that all we have to do is add their potentials together, but if you look at the red-colored potential, you’ll see it’s more complex than that.

## One Response

Matt,

Your series on W-mass using only high school math is excellent. I have a suggestion. You might inquire to some teaching journal like American Journal of physics if it is suitable for pedagogic purpose. In that case it will be available to wider audience and permanently stored in libraries. Also high school and undergraduate teachers who normally are busy teaching and doing research and usually do not read blogs may benefit from that. I know that hardly any of my former colleagues read such blogs!