# (SR) The higgs and gauge invariance

#### BenTheMan

##### Dr. of Physics, Prof. of Love
Valued Senior Member
Ok. As per the title this will be (at least at times) technical. This is for temur, especially, who asked a question about this a month ago or so, in another higgs thread.

I am at a summer school these next two weeks in Princeton (which means I don't have a lot of spare time to write long threads about gauge invariance), so I will try to make posts in stages.

Also, some ground rules. This is not a place to talk about alternatives to the higgs. If you want to learn some of the maths behind the higgs mechanism, I you can ask. Please don't interrupt this thread with alternative theories to the higgs mechanism, unless you can do it with as much mathematical detail as I am doing here---in other words, unless you can point out a flaw in my (and many Nobel Lauriate's) maths, start another thread.

I'll try to put excercizes here and there, that you can try and post solutions to if you're interested. For me, at least, I never learn anything unless I actually CALCULATE something, so it may help you to think about some of these questions (which I have struggled with in the past, too).

Finally, if I am wrong, point it out. If you read some of my responses on this Forum, you'll know that I often screw things up (even things I know well, like what holds nucleons together, or how a bubble chamber works). I will try to keep these posts mistake free, but I can only try!

Ok. The higgs.

It may help to know about representations of the algebra $$SU(N)$$, so that is what I will do before anything else. Next we'll move to Young's tableaux, and then I'll show you how to make mass terms in the standard model lagrangian. Then I will be able to state the problem, in full detail. After that, I'll show you how the higgs mechanism works, and why it generates mass. Hopefully this will convince you that the higgs mechanism is very elegant, and at least a very clever solution to a very hard problem. So clever, in fact, that no other acceptable solution has been found, in the forty years that very smart people have been thinking about it.

SU(N) and some representations.

Consider some set of $$n \times n$$ matrices which obey:
$$\det U = 1,\\ U^{\dagger} U = 1,$$ (1a)
where the dagger means take the complex conjugate of the transpose, specifically:
$$U^{\dagger} = (U^T)^*$$ (1b)
The matrices are unitary, in that they preseve the length of inner products. For example, consider two vectors, $$\left|\eta\right\rangle$$ and $$\left|\xi\right\rangle$$, which transform under $$SU(N)$$ as:
$$\left|\eta\right\rangle \rightarrow U\left|\eta\right\rangle,\\ \left\langle\eta\right| \rightarrow \left\langle\eta\right| U^{\dagger},$$(2)
with a similar expression for $$\left|\xi\right\rangle$$. The inner product of those two states then transforms as
$$\left\langle\xi\left|\eta\right\rangle \rightarrow \left\langle\xi\right|U^{\dagger}U\left|\eta\right\rangle = \left\langle\xi\left|\eta\right\rangle.$$(3)

[Aside]
Here is where I will pause and appologize to the mathematicians---if you are unfamiliar with the Dirac notaion, I will direct you to the Wikipedia page http://en.wikipedia.org/wiki/Bra-ket_notation. If you are just trying to follow along, the |>'s are called kets and the <|'s are called bras. They are vectors in some abstract space described by (in this case) $$SU(N)$$. What I did above amounts to taking a dot product, or scalar product, if you like. The main thing I wanted to show (and if you believe me then good) is that the unitary matrices we're talking about leave the inner products alone. This is crucial to understand, otherwise we wouldn't be assured of things like a good interpretation of probability.
[/Aside]

Now I will appeal to your geometrical intuition. Suppose you wanted to describe your location, somewhere on the face of the Earth, to someone else. The easiest way to do this is to use lattitude and longitude---for example, the US GPS system can tack things down to within a meter or so, I think. But either way, you only need two numbers to describe your position. Another example is locating points in a plane---you need an x coordinate and a y coordinate, or r and theta. But either way, you've expressed things in terms of basis vectors. There is a vector of unit length that points in the x direction, often called $$\hat{i}$$ and a vector of unit length that points in the y direction, called $$\hat{j}$$ which obey:
$$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = 1,\\ \hat{i} \cdot \hat{j} = 0.$$(4)
By taking linear combinations of the basis vectors, you can generate any other vector in the space, which is described by the group $$O(2)$$.

What does this have to do with $$SU(N)$$ representations? Well, it turns out that you can also find a basis of the space described by $$SU(N)$$---the basis vectors are called generators. And for physics, we're interested in generating unitary matrices, so the basis vectors we want to use are going to be matrices.

The first trick is to see that we can always write a unitary matrix as
$$U = e^{i H},$$(5)
where $$H$$ is an $$n \times n$$ matrix.

Excercize 1: What does Equation (1) mean for $$H$$?

The matrices $$H$$ must be traceless and hermitian, which you should work out and convince yourself of. To a physicist, hermitian means
$$H^{\dagger} = H.$$ (6)

Now we need some knowledge of the dimension of the space we're dealing with. Rather than derive this, I'll just tell you. But first, keep in mind that the meaning of the word dimension'' hasn't really changed. It just means the number of basis vectors we need to cover the whole space---just as above, we needed $$\hat{i}$$ and $$\hat{j}$$ to describe the x-y plane, we'll need enough basis vectors to get all of the $$SU(N)$$ matrices we can possibly think of.

Excercize 2: What is the dimension of $$SU(N)$$? Hint: How many degrees of freedom does a traceless, hermitian matrix have?

The dimension of $$SU(N)$$ is $$N^2 - 1$$---this means that we need this many basis vectors. The canonical exaple is $$SU(2)$$, and I'm not one to break with tradition. Plus, $$SU(2)$$ is one of the most important groups to particle physics, and studying it will help us down the road.

The dimension of SU(2) is three, so let's try and find three traceless, hermitian matrices. To further simplify things, let's look for 2 x 2 matrices, although this is by no means necessary. We could do the same thing with 50 x 50 matrices, but not quite as easily.

First, traceless. This means that the main diagonal entries have to be either all zeros, or they have to be equal and opposite.

Exercize 3: Why can't there be any i's on the main diagonal?

The matrices must also be hermitian, as per Eq. (6). After some thinking, you should be able to convince yourself that a good basis (called the Pauli basis) of SU(2) is
$$\sigma_1 = \left(\begin{array}{cc}0&1\\1&0\end{array}\right), \sigma_2 = \left(\begin{array}{cc}0&-i\\i&0\end{array}\right), \sigma_3 = \left(\begin{array}{cc}1&0\\0&-1\end{array}\right).$$ (7)

Exercize 4: Verify that these matrices are traceless and hermitian.

Now we can make any 2x2 unitary matrix, with determinant 1, by simply doing this:
$$U = e^{i \vec{\alpha} \cdot \vec{\sigma}}.$$
This is shorthand notation for
$$U = e^{i \left(\alpha_1\sigma_1 + \alpha_2\sigma_2 + \alpha_3\sigma_3\right)}.$$

What we have done is to find the fundamental representation of the algebra SU(2). In this context, fundamental representation'' just means that the $$\sigma$$ matrices we found have the same size as the number in the parenthesis of the group. SU(2) ---> 2x2 matrices.

Next, I will talk about Young's Tableaux, and an easy way to figure out what happens when we deal with MANY representaitons of differing sizes.

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Awww why did you change the title?

I changed the title to remove the TeX tags that don't work in titles. That's all.

Is this (above, in its TEX form) exactly what was intended?
The matrices are unitary, in that they preseve the length of inner products. For example, consider two vectors, and , which transform under as:
(2)
with a similar expression for . The inner product of those two states then transforms as
(3)

Fixed. Thanks iceaura.

Thanks for starting this thread! I am sure it will be very interesting!

Is it correct that the vectors $$|\eta\rangle$$ and $$|\xi\rangle$$ are in $$\mathbb{C}^n$$?

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Thanks for starting this thread! I am sure it will be very interesting!

Is it correct that the vectors $$|\eta\rangle$$ and $$|\xi\rangle$$ are in $$\mathbb{C}^n$$?

Hmm. Yes I believe so. In Physics, I would say that they live in some hilbert space of states, and that they are eigenvectors of an hermitian operator (i.e. real eigenvalues). This hilbert space can be real or complex, so in general, I think that you are correct. (The language is a bit more formal than I'm used to!)

PS---Sorry for not doing this sooner...I totally missed the request in the other higgs thread.

Ok, no one has asked any questions, and no one has asked for solutions to the exercizes. I will take this as yall having figured everything out for yourselves, and move on.

I was going to talk aobut Young's Tableaux, but it is not really essential to the discussion. They're easy to learn anyway, so if you're REALLY interested you can look it up on Wikipedia and ask me some questions if you get stuck figuring it out.

More on Representations

So last time we encountered the fundamental representation of SU(2), the 2. In general, representations are tensors that transform under the specific Lie group that we're working with.

Now I'll state, without proof, some tensor multiplication rules (the only ones we'll need to understand mass terms). Commit to memory the following complicated rules.

For SU(2): $$2 \otimes 2 = 3 \oplus 1$$,
For SU(3): $$3 \otimes \bar{3} = 8 \oplus 1$$.

And that's what theoretical physicists get paid the big bucks for.

The thing that you want to keep in mind is that anytime you take the tensor product of the fundamental representaiton with it's conjugate representation, you end up with a direct sum of an adjoint representation and a singlet. Adjoint representation'' just means that it has the same dimension as the dimension of the group (remember $$N^2 - 1$$), and singlet'' means that it transforms trivially under the action of the algebra. The fundamental representation of SU(2) is self-conjugate (i.e. real).

I've never actually derived the above tensor rules, because I don't know enough math. I do know that smart people have done these things in the past, and if I ever need any of these rules, I know where to look them up.

Actually I lied. The first tensor rule is something that every student who has taken Quantum mechanics has done---it is chapter 3 in Griffiths. The algebra SU(2) happens to describe spin 1/2 particles. And you know (if you have studied QM) that there are two ways in which two spin one half particles can align. They can combine symmetrically, or anti-symmetrically. In terms of spins, the first rule can be written as follows:
$$\left| \uparrow\right\rangle \otimes \left| \uparrow\right\rangle = \bigg{\frac{1}{\sqrt{2}}\left(\left| \uparrow\downarrow\right\rangle - \left| \downarrow\uparrow\right\rangle\right)\bigg} \oplus \bigg{\left| \uparrow\uparrow\right\rangle \oplus \left| \downarrow\downarrow\right\rangle \oplus \frac{1}{\sqrt{2}}\left(\left| \uparrow\downarrow\right\rangle \oplus \left| \downarrow\uparrow\right\rangle\right)\bigg}$$
This is the common excercize of computing Clebsch-Gordan coefficients. The first {} is the anti-symmetrization (if you exchange the spins, i.e. 1<-->2, you end up with an overall minus sign), and the second {} is symmetric (exchange the first and second spins and the state comes back to itself).

This brings me to another brief point about representations---they're not necessarily matrices. They're objects that have certain properties. If I had to explain it to my mother, I'd say that representations are like pictures. Suppose I say, draw a dog. Well, ostencibly your dog and my dog would LOOK different. But they have the same general features... My dog may have long hair, and your dog may have short hair, but they still have four legs and a tail, and they both (presumably) bark. Representations are THINGS that have certain properties and obey certain rules, just like $$\left| \uparrow \right\rangle$$.

For SU(3), there is no real intuition that you are familiar with---sorry.

And now the kicker---we know, to a very high degree of certainty, that particles transform under Lie groups. Particles form (irreducible) representations, and obey the same sorts of multiplication rules, as matrices or spins or anything else that is a representation of a Lie group. In fact, there is absolutely no doubt that Nature (at least at the quantum mechanical level) obeys a gauge theory. (Let's not discuss gravity in this thread!!!)

The first clue that this was the case was Pauli's isospin''. He tried to put the proton and neutron in the 2 of SU(2)---called (and still called) the isospin(1) doublet(2). Murray Gell-Mann invented the eight-fold way(3), which explained the light meson spectrum.

The Standard Model of Particle Physics

Now we are faced with assigning (or understanding how nature assigned) representations for the fundamental matter that we see in nature. After thirty years of experiments and around 20 Nobel Prizes, we have discovered that nature obeys the Lie group
$$SU(3) \times SU(2) \times U(1)$$.
The only one of these that we haven't encountered is U(1), which essentially just gives us a charge. Sometimes, you will see this written as
$$SU(3)_c \times SU(2)_L \times U(1)_Y$$.
C is for color, L is for left, and Y is for hypercharge. SU(2)xU(1) forms what is called the Electroweak theory (for which Glashow, Weinberg and Salam won the Nobel Prize for in 1979), and the SU(3) is QCD, for which Gell-Mann is famous, and Gross Politzer and Wilczek won the Nobel Prize for in 2005.

I kind of pulled a fast one on you. Nature doesn't actually LOOK like $$SU(3) \times SU(2) \times U(1)$$ untill you do experiments at around 100 GeV, which is a pretty huge energy. If you're doing experiments and lower energies, nature looks more like $$SU(3)_c \times U(1)_{em}$$. Note that THIS U(1) is not the same U(1) as before, this U(1) is electromagnetism, or QED.

What happened, you ask?(4) Somehow we have to find a way to do this
$$SU(3)_c \times SU(2)_L \times U(1)_Y \rightarrow SU(3)_c \times U(1)_{em}$$.
We have to break the symmetry in such a way as to not screw up any of the things we've workes so hard to figure out. Breaking'' just means we take one group ( $$SU(2)_L \times U(1)_Y$$ ) and make it smaller by taking away some of the generators.

Exercize 1: Calculate the dimension of the Standard Model (SM) before and after symmetry breaking. Hint: The dimension of U(1) is 1.

Understanding how the symmetry is broken is not very difficult, and I may continue with it in the future. But before I get TOO far ahead of myslef, let me tell you about the matter that lives in the standard model.

Quarks
There are three generations (or families) of quarks (called up, down, charm, strange, top, and bottom(5) ). Under SU(2), they live in doublets:
$$\left(\begin{array}{c}u\\d\end{array}\right), \,\,\left(\begin{array}{c}c\\s\end{array}\right), \,\,\left(\begin{array}{c}t\\b\end{array}\right).$$
What does this mean? This means that, if we perform an SU(2) transformation of a u quark, we end up with a d quark, etc. etc. Also, u, c, and t are usually called up-type'' quarks, and d, s, and b are called down-type''.

The quarks transform as triplets under SU(3)
$$\left(\begin{array}{c}u^r\\u^g\\u^b\end{array} \right), \cdots$$
The r is for red, g for green, and b for blue. These are charges of the group SU(3)---this is why SU(3) is hard to understand. Instead of positive and negative, SU(3) has red, green(6), and blue.(7)

As for hypercharge, the quarks have a charge of 1/3.

So the quarks all transform as a triplet (3) under SU(3) and a doublet (2) under SU(2), and a hypercharge eigenvalue of 1/3. It is customary to write this as
$$Q = (3,2)_{1/3}.$$

What about the anti-quarks? Well, under SU(3) they transform as an anti-triplet (because they carry anti-color(7) ), and they are singlets under SU(2). What is the significance of them being singlets under SU(2)? This just means that they don't interract under this group. All of the up-type quarks and down-type quarks have the same hypercharges:-4/3 and 2/3 respectively. Usually we write
$$\bar{u} = (\bar{3}, 1)_{-4/3}, \\ \bar{d} = (\bar{3}, 1)_{2/3}.$$
The bars on top of u and d just remind us that they are anti-quarks.

We're ALMOST done with the background. Next I need to tell you about leptons and how they transform, and then we can tackle mass terms.

============
(1) http://en.wikipedia.org/wiki/Isospin
(2) Note: We will often call the 2 of SU(2) a doublet and the 3 of SU(3) the triplet.
(3) http://en.wikipedia.org/wiki/Eightfold_way_(physics), not to be confused with http://en.wikipedia.org/wiki/Noble_Eightfold_Path
(4) Why, the higgs, of course!
(5) For some reason, some physicists (Europeans mostly, I think) call the top quark truth'' and the bottom quark beauty''. Like so many other things, they are wrong about this.
(6) After 9-11, the US legislature passed a bill to have green'' renamed as white''.
(7) Because the dimension of U(1) (which, recall, describes electromagnetism) is 1, it is better to say positive'' and anti-positive''. Likewise, SU(3) also has anti-red, anti-blue, and anti-green.

Ok. Because I'm procrastinating, I'll finish the leptons section before I actually go and do some honest work.

Leptons

Leptons are light particles which carry no color. This means that they are singlets under the group SU(3). Further, we have three charged leptons (electron, muon, tau), and three neutral leptons (three neutrinos).(1) Under SU(2), the electrons are doublets, which we usually just call L:
$$L=\left(\begin{array}{c}\nu_e\\e\end{array}\right), \,\,\left(\begin{array}{c}\nu_{\mu}\\ \mu\end{array}\right), \,\,\left(\begin{array}{c}\nu_{\tau}\\\tau\end{array}\right).$$

There are also the anti-leptons---$$\bar{e}, \bar{\mu}, \bar{\tau}$$.

The same generational pattern follows us into the lepton sector, and the quantum numbers of the leptons are as follows:
$$L=(1,2)_{-1},\\ \bar{e} = (1,1)_{2}.$$

So now we have the full SM, in all of it's glory.

Next time: I will show you how to make a mass term in the lagrangian. For the ambitious, a homework excercize:

How do you make a mass term in the lagrangian? Hint: we wish to find combinations of fields that are gauge invariant. That is, we wish to build tensors that transform trivially under all of the SM gauge groups.

Have fun!

=========
(1) It has been discovered that neutrinos have a small mass. This is not a feature of the original SM, but can be incorporated rather easily. We won't concern ourselves with that here, because it is not essential to the discussion. The neutrino masses are generally generated using something called the see-saw mechanism, as opposed to the higgs mechanism.

I understand it is impossible to build the theory from ground up in a post, but I am lacking some basic understanding of these things.

* What does it mean to say something is triplet (singlet, doublet) under, say, SU(3)?

* Particles are transformed under some Lie group. But what does it represent physically?

temur---

These two questions are closely related. The first one is technical and the second one is physical.

As to the first question, a triplet is the fundamental of SU(3) and a doublet is the fundamental of SU(2). A singlet just means that the tensor (see below) transforms trivially (i.e. not at all). I think that this is what you were asking.

The second question is a bit trickier. I am by no means a mathematician, so forgive me if I mangle anything. At a purely mathematical level, all of these things are tensors. So, for example, the fundamental representation of SU(N) can be represented as an N dimensional vector, or a tensor with one leg. The conjugate representation, N* or $$\bar{N}$$ is always constructed such that
$$N \otimes \bar{N} = 1 \oplus Adj(G)$$,

What we're really interested in is how one can combine these tensors to form invariant objects under the algebra. This is where the spin example is useful: what you're doing when you combine spins is looking at the ways that you can possibly combine SU(2) tensors. We'll take this one step further (I don't want to spoil the fun!) in the next few posts---but essentially what I'll show is that it's IMPOSSIBLE to form invariant objects (under the SU(3)xSU(2)xU(1) SM gauge group) from the tensors that I've given you, if you place a requirement called renormalizability'' on the theory.

So, in short, one should take the fields (or particles'', if you must) of the SM to be tensors which transform under the (semi-simple) Lie algebra su(3)xsu(2)xu(1). These tensors are irreducible representations of the Lie group SU(3)xSU(2)xU(1).

A similar statemtent can be made about the Lorentz group, SO(4). The fields that we see in nature MUST be irreducible representations of this group (i.e. SO(4) tensors), if we are to maintain Lorentz invariance. This is why we don't see fractional spins---fractional spins violate Lorentz invariance, because the SO(4) groups only have spinor representations (fermions) and vector representations (bosons). This is what so many people, like Farsight and Uclock, clearly don't understand, and won't bother to try and learn---if you don't treat time like a direction, then you lose chirality in nature (there is only one chirality in three dimensions, so this would mean we would have no chiral fermions).

So, I hope I haven't meandered too far affield, and that I have actually (somewhat) answered your questions.

Ok. So, we have looked at the transformation properties of the Standard Model fields under the Standard Model gauge group. I hope, temur, that I sufficiently answered your question earlier. The generators of the symmetry are called gauge bosons, and they transform in the adjoint representation. In SU(2), using the spin example, this is easy to see. If anyone is interested then I will explain this in a bit more detail. But I can motivate the need for a higgs just by looking at the fermion sector of the theory.

Mass Terms

We have outlined a theory of quarks and leptons in the past few posts. Now we need a way to describe the way that those quarks and leptons interact. The important thing to remember is that the lagrangian must transform as a scalar. This is the punchline. But to get here, I first want to take a foray into classical physics, which is a beautiful subject to study for it's own sake if you've ever the time.

A Bit of Classical Mechanics

Suppose we have a classical mechanical system---imagine a ball on a hill. You can add constraints (the ball rolls without slipping down the hill) if you like. Now, there is a function called a lagrangian (after Lagrange, and his multipliers) which describes the evolution of the system. It is actually what physicsts call a functional'', which means a function of functions. The lagrangian explicitly depends on the positions and velocities of the particles involved, and implicitly on time.

To understand the evolution of the system, one minimizes the lagrangian in phase space. Phase space is just the space of all accessible positions and momenta for the system---in general, for D dimensional space, there is a 2D dimensional phase space. The problem is exactly analagous to finding geodesics on a curved manifold---there you're minimizing the distance traveled in space. Understanding how a system evolves necessitates minimizing the total distance traveled in phase space, which (I'm pretty sure) just means minimizing the total energy.

How do wou build a lagrangian? This is the easy part---we write an expression for the potential energy, called V, we write an expression for the kinetic energy called T, and then the lagrangian (usually a fancy L) is defined as

$$\mathcal{L} \equiv T - V.$$

The thing to note is that T and V are scalar functions. This statement can be understood in several ways---the easiest way I know to understand this is to note that potential energy doesn't have a direction. This isn't an especially DEEP statement, just a very important one.

Now how do we deal with (classical) fields. IF you're interested, you can look up Goldstein's book and turn to the very last chapter. He does a good job of taking a continuum limit of a system of classical springs. What you eventually find is that the lagrangian is no longer a function of the coordinates of space-time, but of the fields. (This statement requires a bit of a leap, and I'm sorry. If you don't trust me, consult Goldstein!) So the Lagrangian, at a (classical) field theoretic level, is a functional of the fields that it describes.

The other features of the classical lagrangian have been preserved, though. Specifically the fact that it transforms as a scalar. I want to examine this statement a bit more closely---suppose we have a classical system which is spherical in nature, like a planet orbiting a sun, or some such. One would expect that the equations describing the evolution of the system to exhibit some of the same spherical symmetries that are inherint in the problem. So, for example, one wouldn't want to use cartesian coordinates to describe such a problem. You know, at least, that the potential is spherically symmetric (thanks to Newton).

Either way, symmetries of the problem are symmetries of the lagrangian, and the lagrangian must be unchanged under the symmetries of the problem.

So what do we know---the lagrangian is a functional in field space, which has the same symmetries of the problem. Without clouding the issue with more of my words, I will simply state that the whole formalism survives quantization. If not, there is trouble---we get something called anomalies. To understand the evolution of the lagrangian, we minimize it in field space, giving us the equaitons of motion for the fields, which are (in general) second order differential equations.

What does that mean for us? Well, I told you that the lagrangian has to transform as a scalar---this is the same as saying that it has to be invariant under the symmetries of the problem. But what are the symmetries of the problem'' here? This is where a light comes on in your head and you say Of course---the lagrangian must respect SU(3)xSU(2)xU(1)!!! It all makes sense.''

And I say Well, you're almost right.'' What I have neglected to mention, and a subject on which I could write another four pages worth of posts, AND a subject which some of the pseudo-science regulars simply don't understand, is the Lorentz symmetry. Essentially what one realizes is that there are only three ways for things to transform under the lorentz group---either a scalar, a spinor, or a vector. Fermions transform as spinors. This imposes a certain constraint on the form of the mass terms. I will say more about this in just a second.

Back to Mass Terms
---or---
The Good Stuff

Ok. So we want to build a lagrangian which is invariant under SU(3)xSU(2)xU(1), as well as something that is invariant under the Lorentz Group, SO(4).

Let's look at Lorentz Invariance first. Lorentz Invariance means that we have to form mass terms which look like

$$\mathcal{L}_{mass} \sim m\bar{f}f$$

where f is a fermion and f-bar is an anti fermion, both which have mass m. So one can think of the quarks, and see that, evidently, we must have something like

$$\mathcal{L}_{mass} \sim m_u \bar{u}u + m_d \bar{d} d + \cdots.$$

Ok, now we ask ourselves, is this part of the lagrangian symmetric under the symmetries of the theory? Well, by construction it is lorentz invariant (basically because I told you so). What about under SU(3)? Well, under SU(3), a $$\bar{u}$$ transforms like a $$\bar{3}$$ and a $$u$$ transforms like a 3. And I told you earlier that

$$3 \otimes \bar{3} = 8 \oplus 1$$.

So, the mass term can either transform like an 8 (the adjoint of SU(3) ) or a 1, which is a singlet. But this is exactly what we want for it to be invariant under SU(3)---it must transform as a singlet. (Don't worry that it doesn't HAVE to transform as a singlet, just that it CAN.) So, the statement is, one CAN make a mass term for the quarks which is a singlet under SU(3).

But what about under SU(2)? Well, under SU(2), a $$\bar{u}$$ transforms like a $$1$$ and a $$u$$ transforms like a 2. And, one shouldn't need to be convinced that

$$2 \otimes 1 = 2$$.

This is a problem. This says that it is impossible for a quark to have a mass term which is an SU(2) invariant. But we want SU(2) to be a symmetry of the theory, because we know that the standard model gauge group is SU(3)xSU(2)xU(1).

Exercize: How does the mass term transform under U(1)? How SHOULD it transform?

Exercize: Work out the transformations for all of the other SM fields.

Conclusion

Hopefully I have convinced you that there is no way to give the standard model fields masses unless in the usual way. Because we are not willing to abandon the intuition we have gained from the classical theory, we need another way to give the fields masses. So, in this vein, I want to propose the final exercize that everyone who is interested should think about (especially you temur!). The only way you ever understand these things is to work out a few examples for yourself, anyway. If James or Pete is watching, then they can feel free to participate here too.

Final Project: What quantum numbers does the higgs have to have? That is, how does the higgs have to transform under SU(3)xSU(2)xU(1). If you have worked out the two above exercizes, this should be easy.

Hint: The higgs is a lorenz scalar, called $$\phi$$. The interaction between the higgs and the fermions looks like this:

$$\phi \bar{f} f$$.

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I am a little worried that I haven't been as clear as I possibly could have in this, as there haven't been any comments. If you learned something, let me know---if not, then ask some questions!

For me, I understand Lagrangian and all that stuff, but don't understand how you derive properties of real particles just by writing the Lagraingian and looking at how some terms are transformed. As for the exercizes, I can imagine you choose the mass term such that the overall result transforms invariantly under some group, but don't know where to even begin when I want to do it concretely. I feel I am stupid and was thinking I need to read some book (may be about representation theory of Lie algebras/groups) before discussing it here.

but don't understand how you derive properties of real particles just by writing the Lagraingian and looking at how some terms are transformed.

Everything starts from the lagrangian. If you understand how to write the lagrangian down, then you (in principle) know everything about the system. From the lagrangian, one can compute scattering cross-sections and life-times of particles, which are the only two things in high energy physics you can measure.

I can imagine you choose the mass term such that the overall result transforms invariantly under some group, but don't know where to even begin when I want to do it concretely.

Here, I'll walk you through it. Let's take electrons. You are exactly right---the lagrangian should transform as a scalar under the standard model gauge group. Now, here are the givens:
1.You want an interraction term of the form

$$\phi e \bar{e},$$

where $$\phi$$ is the higgs.

2. Under the SM gauge group, the term must be invariant. This means that the hypercharges have to sum to zero. Before, I told you that electrons transform as:

$$e \sim (1,2)_{-1},\\ \bar{e} \sim (1,1)_{2}.$$

Now, so far, the mass term is invariant under SU(3) :$$1\otimes 1 = 1$$. Under SU(2), however, it transforms as a 2 $$1 \otimes 2 \sim 2$$. Under U(1) it transforms as $$-1 + 2 = 1$$.

Well, this isn't too bad. Given what I've already told you, try and figure out the qunaum numbers of the higgs.

I feel I am stupid and was thinking I need to read some book (may be about representation theory of Lie algebras/groups) before discussing it here.

Mathematicians who learn physics have a famously difficult time. We tend to be very willing to throw out rigor at the slightest difficulty, and do things which would make mathematicians sick. The whole subject of field theory is on shaky mathematical footing AT BEST, and there is a million dollar prize for someone who can even write down one mathematically consistent field theory.

So you shouldn't be surprised that it's difficult. Some of the maths you can take at face value---it is the physics that I really care about.

I didn't want to interupt the thread with something trivial, but regarding the top/truth bottom/beauty quark names. I was an active reader of Scientific American in the late 70's and I recall that originally truth and beauty were used. Did a search and found this explaination:

http://education.jlab.org/qa/quark_04.html

Thought it might be of interest.

thanks kevin---I have always been a bit interested in this. I thought it was a European/American thing.

temur: Is it correct that the vectors $$|\eta\rangle$$ and $$|\xi\rangle$$ are in $$\mathbb{C}^n$$?

Ben: Hmm. Yes I believe so.

I don't think so. As Ben notes, $$\left|\eta\right\rangle$$ lives in a Hilbert space. Hilbert spaces are infinite dimensional, whereas $$\mathbb{C}^n$$ is a finite dimensional vector space. So I can't see how $$\left\eta\right\rangle$$ could possibly be an n-tuple of complex numbers.

I think it is more correct to say that $$\left|\eta\right\rangle$$ is an element of a Hilbert space that is defined over the field $$\mathbb{C}$$.

temur: Is it correct that the vectors $$|\eta\rangle$$ and $$|\xi\rangle$$ are in $$\mathbb{C}^n$$?

Ben: Hmm. Yes I believe so.

I don't think so. As Ben notes, $$\left|\eta\right\rangle$$ lives in a Hilbert space. Hilbert spaces are infinite dimensional, whereas $$\mathbb{C}^n$$ is a finite dimensional vector space. So I can't see how $$\left|\eta\right\rangle$$ could possibly be an n-tuple of complex numbers.

I think it is more correct to say that $$\left|\eta\right\rangle$$ is an element of a Hilbert space that is defined over the field $$\mathbb{C}$$.

A Hilbert space can be finite dimensional at least in mathematics. Actually any finite dimensional space can be made into a Hilbert space.