Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Can someone explain how isospin and hypercharge, can be used to label representations? What is the meaning of the term singlet, doublet etc in this context? In particular how can I use it to label representations of SU(2) embedded in a larger gauge group. I had posted this question, but didn't get any response, so I thought asking it in a more general way would help.

Please feel free to close it down, if you feel it is the exact duplicate.

I have an idea of the isospin from 2 particle systems, and their Clebsch-Gordon coefficients. IN this case, isospin is the eigenvalue of the total angular momentum operator $J$, and the combination of tensor product states with this quantum number, forms a representation, called the doublet, singlet etc depending on its dimension. Is this correct?

share|improve this question
    
Related: physics.stackexchange.com/q/16354/2451 –  Qmechanic Jan 1 '13 at 18:46
4  
ramanujan_dirac, I'm not yet sure if this is an exact duplicate - asking something in a more general way can be enough to distinguish two questions - but your other question has only been up for 8 hours. It's far too early to be saying it hasn't gotten a response; many people probably haven't even seen it. –  David Z Jan 1 '13 at 20:17
add comment

1 Answer

By isospin I assume you mean weak isospin which is an exact gauge symmetry of the standard model. There is another thing called flavour isospin which is an approximate global symmetry of the strong interaction. Also, this is a long reply because I've included background and example that you may or may not need. If I am not explaining things well enough for you might I suggest you try to borrow a copy of Zee's excellent field theory book and reading the sections on grand unification. His examples are fairly similar to mine, though I'm working off memory here. Any mistakes are mine. I'm also not making much of an effort on signs and factors of two for this.

The electroweak gauge group of the standard model is SU(2)xU(1). The SU(2) generators act on isospin indices and the U(1) is the hypercharge. So when someone says that some set of fields $\psi_i$ transform in the $R$ representation they mean that the transformations $$ \delta\psi_i \propto \epsilon_A \left(T^{A(R)}\right)^j_i \psi_j + \epsilon_Y Y \psi_i $$ are a symmetry, where $\epsilon_A$ and $\epsilon_Y$ are the transformation parameters and $T^{A(R)}$ and $Y$ are the generators in the given representation. This is essentially true of any gauge theory. For SU(2) specifically you can use angular momentum theory since the group is the same. The terms singlet, doublet, etc. just refer to the number of fields in the representation. For example a doublet has two fields

$$ \Psi = \left( \begin{array}{c} \psi_1 \\ \psi_2 \end{array} \right) $$

and the SU(2) generators are proportional to the usual Pauli matrices

$$ T^{A(1/2)} = \frac{1}{2} \sigma^A,\ A=1,2,3 $$

These are NOT the angular momentum generators. They have nothing to do with spacetime and the isospin index is not a spacetime index. It just so happens that, because the group is SU(2), the algebra is identical to that of angular momentum. The physical interpretation is completely different.

Now, the main part is your question is best addressed through example. If you embed the standard model gauge group into some larger group (for grand unification, say) you can use the Higgs mechanism to give mass to the new gauge bosons and push them up above the energy scale you are interested in. In the general case you embed SU(2)xU(1) (or whatever) into your larger group, then break the larger group down by giving the extra gauge bosons a heavy mass, then you write down the interaction term for your matter fields and, focusing only on the light gauge bosons, identify the unsuppressed interactions using the explicit forms of the generators and read of the quantum numbers in your unbroken gauge group. Generally an irreducable representation of the larger group will break down into several pieces when only the light bosons are included. This is because the interactions between the pieces are only mediated by heavy bosons and are suppressed.

I'll give a quick example of how this breaking is accomplished. I'll leave you to read up on GUTs and please rely on the standard references to get factors of two and signs right.

Suppose for example you have SU(3) and you break it down to SU(2)xU(1) (the normal GUT strategies are SU(5) and SO(10), but nevertheless). SU(3) has 8 generators while SU(2)xU(1) only has 4. So there are 4 extra gauge bosons we have to get rid of through a judiciously chosen Higgs.

The generators of SU(3) are the Gell-Mann matrices $g_i$ for $i=1,\cdots,8$. We can identify an SU(2) subgroup generated by $g_1,g_2,g_3$ and a U(1) which commutes with the SU(2) generated by $g_8$. These will become the isospin and hypercharge generators of our unbroken SU(2)xU(1). The other generators have to correspond to gauge bosons which get a mass through the Higgs mechanism.

Let's take an adjoint Higgs (eight components in one to one corresondence with the generators) and go to a gauge where the vacuum expectation value (vev) is in the $g_8$ component:

$$ <\phi> = v g_8 $$

The mass matrix for the gauge bosons becomes

$$ M_{ij} = e^2 \mathrm{Tr}\left( [g_i, <\phi>] [g_j, <\phi>] \right) $$

where $e$ is the (unusual notation here) SU(3) coupling constant and $ [ , ] $ is the matrix commutator. Note I might be missing a sign or numerical factor here. In any case, those generators that commute with the vev correspond to gauge bosons which remain massless, while those which don't commute pick up a mass of the order $ e v $. You'll see if you compute the commutators that this picks out just the bosons we want to remain massless (1,2,3,8) and the ones we want to gain a mass (4,5,6,7). So this Higgs will accomplish the breaking SU(3) -> SU(2)xU(1). We identify the SU(2) generators as $ T^1 = g_1 $ etc. and the U(1) generator as $ Y = g_8 $.

Now we can talk about embedding matter representations in the theory and seeing how they decompose. Take a Dirac field in the fundamental of SU(3). This is a triplet (three components) that transforms like a vector which is acted upon by the Gell-Mann matrices (that's all that "fundamental" means in this context). This is a straightforward generalisation of the SU(2) example from before:

$$ \Psi = \left( \begin{array}{c} \psi_1 \\ \psi_2 \\ \psi_3 \end{array} \right) $$

The interaction term in the Lagrangian will be

$$ L \supset e A^i_\mu \bar{\Psi} g_i \gamma^\mu \Psi $$

where the $A^i_\mu$ are the various gauge bosons. Now start acting on this $\Psi$ with the various generators to see how the different gauge bosons couple. Since the 4,5,6, and 7 bosons are massive they disappear from the theory. These mediate interactions which change $\psi_1$ and $\psi_2$ into $\psi_3$ and vice-versa. These interactions are suppressed by the masses of the heavy bosons and are analogous to very rare processes like proton decay which should happen in GUTs. The remaining massless bosons don't mix the $\psi_1$ and $\psi_2$ with $\psi_3$, so the representation falls into two pieces: a doublet ($\psi_1$ and $\psi_2$) which feels the SU(2) interaction as well as hypercharge, and a singlet $\psi_3$ which only feels the hypercharge interaction. We say that breaking happens in a 3 -> 2 + 1 pattern. You can read off the isospin and hypercharge quantum numbers of the various fields from the interaction Lagrangian.

Another example is a matter field in the same representation as the Higgs, the adjoint (octect). This time the interaction Lagrangian involves the commutator of the generators $g_i$ with the matter field. Again, you can ignore the 4,5,6 and 7 generators and read off the isospin/hypercharge quantum numbers from the Lagrangian. This is a more involved example, but the trick is to write the matter field as a traceless hermitian matrix $\Psi^i_j$ where $i,j=1,2,3$. This breaks up into several pieces: $\Psi^a_b$ where $a,b=1,2$ which breaks down further into a trace (singlet) and a traceless 2x2 matrix (triplet), $\Psi^3_b$ and $\Psi^a_3$ (doublets). $\Psi^3_3$ is actually the singlet trace we already counted. So 8 -> 1 + 3 + 2 + 2 in this case. (I hope this is right, I'm not using any reference and I'm going fairly quickly! Please, somebody correct me if I'm wrong.)

Hope this goes some way towards answering you question.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.