Now we move from U(1)xU(1) to the Standard Model’s full SU(2)xU(1) gauge symmetry. This brings us to a new and more complete view of the Higgs field, which up to now I have hidden. The Higgs field, in fact, is not a single real number at each point in space and time, as in my first cartoons. Nor is it a single complex number (equivalently, two real numbers) at each point in space and time, as I gradually snuck into my discussion of U(1) and U(1)xU(1) theories. It is in fact two complex numbers at each point in space, which for totally non-obvious reasons we will label HA(x,y,z,t), HB(x,y,z,t). (In most books they would be labeled H+ and H0; but these labels make sense only after you already know how things are going to work out, and not when you see this for the first time.)
The presence of two complex fields is why the Standard Model’s Higgs field is referred to as a “complex doublet” field. The reason we need two is that the SU(2) symmetry transforms these two fields to one another, much as an ordinary rotation of an x-y plane mixes the x and y coordinates together.
The SU(2) x U(1) Symmetry Transformations
Let’s now look in more detail at how the symmetry transformations affect the Higgs field. The pattern which we will see in a moment is sketched in the Figure below, showing how HA and HB are affected by the symmetry transformations corresponding to the W and X gauge bosons. We’ll see how the math corresponds to the picture.
The U(1) inside SU(2)
First, within the SU(2) sits a U(1) symmetry that multiplies HA and HB by a complex phase, just as we saw last time. Specifically, HA is multiplied by one phase and HB by the opposite phase:
- U(1) inside SU(2):
- HA → HA exp[ +iα/2 ]
- HB → HB exp[ -iα/2 ]
For a very small α, we can write this (because ex is approximately 1+x) as
- HA → HA (1+iα/2),
- HB → HB (1-iα/2) .
The gauge boson associated with this rotation is W3.
The rest of SU(2)
The rest of the SU(2) transforms HA into HB and vice versa, in much the way that a rotation of a plane rotates the x and y coordinates of a point on that plane. For instance, part of the SU(2), associated to W1, does this
- W1 part of SU(2):
- HA → HA cos δ/2 + HB sin δ/2,
- HB → HB cos δ/2 – HA sin δ/2
For a very small rotation angle δ we can write this (because sin x is approximately x and cos x is approximately 1 – ½ x2 , or just 1) as
- for small angle:
- HA → HA + ½ HB δ
- HB → HB – ½ HA δ
The rest of the SU(2), corresponding to W2, does something similar, but with complex numbers; we won’t need the details, so I won’t show it explicitly.
The Stand-alone U(1)
Of course the Standard Model has an SU(2) and an additional U(1). What does the latter U(1), with gauge boson X, do to the Higgs field? It multiples HA and HB by the same phase.
- the separate U(1):
- HA → HA exp[ +iβ/2 ]
- HB → HB exp[ +iβ/2 ]
For a very small β, we can write this as
- HA → HA (1+iβ/2)
- HB → HB (1+iβ/2)
Now we know how the symmetries affect the Higgs. Notice the two U(1)’s are similar to those considered last time for the U(1)xU(1) theory: they transform HA the same way and they transform HB opposite ways. Meanwhile the W1 and W2 have the feature that they rotate HA into HB and vice versa; but they are related to W3 by SU(2) symmetry.
The Higgs and Gauge Boson Potential
What do we do next? We convert these transformations into an interaction between the gauge fields and the Higgs fields, and then the potential is built from the square of these interactions. We already know the rule for how to do this for the U(1)xU(1) inside of SU(2)xU(1); we replace the angles in the exponentials with gauge bosons to get the interactions that we will have to square.
This rule can be more easily stated if we consider small angles. For instance, because (as noted above)
- HA exp[+iα/2 + iβ/2] ≈ HA (1 + iα/2 + iβ/2)
we can reformulate the rule this way: we drop the “1” in the parentheses and replace the angles by the corresponding gauge bosons, as in iα → gW3 , etc.:
- HA → HA (1 + iα/2 + iβ/2) ⇒ (+½ gW3 + ½ g’X ) HA
- HB → HB (1 – iα/2 + iβ/2) ⇒ (-½ gW3 + ½ g’X ) HB
Meanwhile, for W1 ,
- HA → HA + HB δ/2 ⇒ (-½ igW1)HB
- HB → HB – HA δ/2 ⇒ (+½ igW1)HA
For W2 something similar applies with some additional factors of i. When the dust settles, we end up with two interactions, one that arises from SU(2)xU(1) transformations of HA
- (+½ gW3 + ½ g’X ) HA – i (½ g[W1 + i W2]) HB
and one that arises from transformations of HB
- (+½ gW3 – ½ g’X ) HB+ i (½ g[W1 – i W2]) HA
Complex-squaring these, writing HA* to mean the complex conjugate of HA, etc. (remember the gauge boson fields are real) and using the fact that
- [W1 + i W2][W1 – i W2] = W12 + W22
we get four terms which all appear in the potential:
- + ⅛ |HA|2 (g2 [W12 + W22]+ [gW3 + g’X]2)
- + ¼ HA* HB (g2[W1+i W2]W3)
- + ¼ HB* HA (g2[W1-i W2]W3)
- + ⅛|HB|2 (g2 [W12 + W22]+ [gW3 – g’X]2)
Notice the last term is precisely the Standard Model potential that appeared at the top of webpage 4! We’ll see shortly what happens to the other terms.
To complete the potential we need the terms involving just the Higgs field. In the U(1)xU(1) case these were a function of |H|2, when H was just a single complex number. Here they have the same form but are a function of |HA|2+|HB|2 .
- – ½ μH2 (|HA|2+|HB|2) + ¼ λ (|HA|2+|HB|2)2
Minimizing the potential requires |HA|2+|HB|2 to be non-zero, so some combination of HA and HB should be non-zero. And here’s where we can put the SU(2)xU(1) symmetry to use.
(Experts will notice I have been a little cavalier with factors of square root of 2, mainly to prevent unnecessary technical clutter; but I believe I was consistent in how I normalized things, even though there are some places that differ slightly from the standard treatment. If I was consistent then the physics should come out right in the end.)
The Vacuum, using SU(2) x U(1) to make it simpler
When we make a map of our local town or city, it’s up to us what direction to put at the top. That’s because any direction is as good as any other; there is a rotation symmetry that (for local geography at least) makes one direction basically the same as any other. We can put south, or northwest, or any other direction at the top of the map; to put north there, as is conventional, is a choice..
Similarly, we can choose the vacuum expectation value (vev) of the Higgs field to be in HA, or in HB, or in any combination we like. Conventionally, the non-zero combination of HA and HB is chosen to be in the real part of HB. I say again, that’s just a choice, allowed by the fact that HA and HB are related by a symmetry. So let’s make that arbitrary choice, and work out the consequences. (If we make a different choice, the physics works out to be exactly the same, just with different labels on things; that’s guaranteed by the fact that our equations are invariant under SU(2)xU(1) symmetry transformations. The various choices are related by symmetry, so it makes no difference what we choose.)
When we do this, we can set HA to zero in the vacuum. (An even stronger reason for doing this is coming shortly.) And so the potential we had above, dropping all terms containing HA , simplifies to become
- V = – ½ μH2 |HB|2 + ¼ λ |HB|4 + ⅛|HB|2 (g2[W12 + W22] + [gW3 – g’X]2 )
This is the Standard Model potential I wrote down at the beginning of webpage 4. (In the last term I have flipped the minus sign, but since it is squared, this change is purely cosmetic and has no effect.) As usual, we will call the vev of HB by the name “v“.
As in our earlier cartoons, we see that W1 and W2 have the same rest mass ½ gv, and a combination of W3 and X have a different rest mass (which would have also been ½ gv if X were absent), while the opposite combination of W3 and X has zero rest mass. These, as we will see, are the W bosons, the Z boson, and the photon. We will work out their masses next time.
Why Only One Higgs Boson?
Before we do this, however, there’s an important conceptual puzzle to address. The Higgs field (HA, HB) contains two complex (i.e. four real) fields. But the Standard Model has only one Higgs particle — one Higgs boson. Shouldn’t there be four?
This now brings us to one of the most important aspects of the Higgs mechanism. It stems from the fact that a massless gauge boson has only two basic fields within it, while a gauge boson with non-zero rest mass has three. (In scientific lingo, we would say the former has two degrees of freedom while the latter has three.)
Any photon, with zero rest mass, is always traveling at the cosmic speed limit (a.k.a. the speed of light), and you cannot slow it down (in empty space) so it has two possible polarizations. If it is traveling in the x direction, then it can be polarized in the y direction or in the z direction, perpendicular to its motion. But if a gauge boson has a rest mass, then you can stop it, and put it stationary in front of you. Then it can equally be polarized in the x direction, the y direction or the z direction. Three polarizations for a gauge boson (or any spin-one boson) with rest mass; two for a gauge boson without any rest mass.
This means that if you want, through the Anderson-Brout-Englert-Higgs mechanism, to make a zero mass gauge boson into one with non-zero mass, you are going to have to steal a field from somewhere! You need 2 to become 3.
And since we are in the process of making four massless gauge bosons (W1, W2, W3, X) into three with mass (W+, W–, Z) and a massless one (the photon), we need to steal three fields from somewhere. Guess where we get them from?
Exactly! When we choose the non-zero vev to lie in the real part of HB, then the three fields that are stolen (or “eaten”, in physics-speak) are the imaginary part of HB (which becomes part of the Z boson) and the real and imaginary parts of HA (which become part of the W bosons.) This is the real reason that we can safely set HA to zero in the potential.
Now we see why we can start with four zero-mass gauge bosons and four real Higgs fields (4*2 + 4 real fields), and end up with three nonzero-mass gauge bosons, one zero-mass gauge boson, and one real Higgs particle from one surviving real Higgs field (3*3+2+1 real fields). Twelve fields before, twelve fields after, but rearranged!
This counting tells us it was inevitable, with a complex doublet Higgs field, that the photon would have zero rest mass. If it had gotten a mass by absorbing the fourth Higgs field, then there’d be no Higgs particles at all. But that would be inconsistent. The vev of the Higgs field can always vary (look back at how we convinced ourselves there should be a Higgs particle toward the end of webpage 2). Nothing can prevent a ripple in the vev, and that ripple is a type of Higgs particle. At least one Higgs particle (possibly very hard to observe, however) is always inevitable in the Higgs mechanism. Therefore, if we wanted the photon to have a rest mass, we would need at least one more type of Higgs field.
The fact that the W and Z bosons are secretly part-Higgs-field has many experimental consequences. It explains why the W and Z interact so readily with the Higgs boson, much more strongly that one would expect from the weak nuclear force. Furthermore, the top quark decays much more quickly than one would naively have expected just from the weak nuclear force, because it interacts so strongly with the Higgs part of the W boson.
Next time we’ll identify how the physical particles and their rest masses emerge from the W and X fields, and start looking at how the triplet model can potentially change them.
A final comment: I swept these issues under the rug in my early cartoons of the Standard Model. We did see a sign of it last time, though, where a single complex field H in a U(1)xU(1) theory was able to give mass to one of the U(1) vector bosons but not the other. One of the two real-numbered fields in H was absorbed into the massive Z boson, but the photon remained massless, and (though I didn’t point this out) there was a single Higgs boson left over. The counting is similar to the SU(2)xU(1) case: two massless gauge bosons plus two real (i.e. one complex) Higgs field (2*2+2) becomes one Z, one photon and one Higgs field (3+2+1).