I've not long started a course on field theories, just got onto spontaneous symmetry breaking but there's something I really can't see. Consider a scalar field with an SU(2) doublet \( \phi \) of complex scalar fields which carry charge 1/2 under a U(1) symmetry. A general \( SU(2) \times U_Y (1) \) transformation can then be written as \( \phi \rightarrow e^{-i \alpha /2} e^{-it^i \tau_i} \phi \approx (I - i\alpha_Y -it^i \tau_i)\phi \) (ok so far) with generators \( \tau_i = \frac{1}{2} \sigma_i \) , \( Y = \frac{1}{2} I \) (still ok so far) The general invariant Lagrangian density is \( \mathfrak{L} = \partial_{\mu}\phi^{\dagger} \partial^{\mu} \phi - V(\phi) \) , \( V = V_0 + m^2 \phi^{\dagger} \phi + \lambda \left(\phi^{\dagger} \phi \right)^2\) Note that the invariance of this Lagrangian density under SU(2) is due to the appearance of the singlet in the Clebsch-Gordon decomposition of \( \bar {\mathbf{2}} \otimes \mathbf{2} = \mathbf{1} + \mathbf{3} \) I really don't get this last sentence. I can see the invariance explicitly follows from V(U \phi) = V(phi) for an SU(2) matrix U, similarly for the first term in the Lagrangian also. What I really don't see is how/why the Clebsch-Gordon decomposition gives the invariance. Why does it, and what significance does the "singlet" have? I can usually see the invariance under certain symmetries for the Lagrangians we've seen simply by plugging in, but maybe there are cases when it won't be so simple and I'll need this Clebsch-Gordon trick?
Take everything I'm about to say with a minor pinch of "Its been 5 years since he did anything to do with the SM" salt but hopefully it'll help. \(\phi^{\dag}\phi\) you can relate to the \(\bar{2}\otimes 2\), since the field transforms as an SU(2) 2 (ie a doublet) and the combination of its conjugate and itself transforms as the tensor product given. If something is 'invariant' under a transformation then it means its a singlet, ie it doesn't mix with anything else or get or lose anything, no matter which transformation within your group you use (ie true for any \(g \in SU(N)\) sort of thing). Now V is basically a polynomial in the \(\bar{2}\otimes 2\) combination of \(\phi^{\dag}\phi\) so if \(\phi^{\dag}\phi\) transforms as a singlet then so will V. But conversely if you want L to transform as a singlet for general coefficients in said polynomial then you need \(\phi^{\dag}\phi\) to transform as a singlet. Thus they are equivalent. If you want a vague analogy which has some kind of mathematical justification then suppose that the \(\bar{2}\otimes 2\) is some kind of matrix which is 4x4 (there's 2 of them so its rank 2 and 2*2 = 4). The whole "This equates to \(1 \oplus 3\)" thing you can view as saying that its possible to block diagonalise said matrix into a 1x1 block and a 3x3 block but you can't do any better. If you had something like \(\bar{3}\otimes 3 = 1 \oplus 8\) (as you'll see in the quark model) then you can view it as a 9 dimensional system splitting into a 1 dimensional one and an 8 dimensional one, as you would the eigenspaces of a matrix. The decomposition means that you can 'pick a basis' (just like a change of basis block diagonalises a matrix) such that you manifestly split the 3 terms which mix together under SU(2) and the single term which doesn't. I've a pair of comments which might be of no use to you. Firstly I'm half remembering this and half looking at it through the lense of the group theoretic decompositions which follow the same \(4 = 1\oplus 3\) principles which lead to the whole string theory obsession with Calabi-Yau manifolds, so my view might be skewed a little by that. Secondly it might be useful to know that in the decomposition of \(\bar{n} \otimes n\) there is always a 1. You've got an example of its use in the SM, it arises in the QCD model as gluons belong to the 8 of the \(\bar{3} \otimes 3 = 1 \oplus 8\) and the dilaton in string theory is the 1 from the \(\bar{8} \otimes 8 = 1 \oplus 28 \oplus 35\).
Wow, thanks for taking the time to reply. That was really helpful. For all your haters I think this shows your self confessed care bear side. I also found a pretty helpful post from BenTheMan in an old thread The thread itself was actually incredible http://www.sciforums.com/showthread.php?p=1476569
You picked a topic close to things I find very interesting and I know you're one of the people who ask honest questions. Ben is definitely the guy to ask, he's done a lot more group theoretic stuff in relation to SUSY and GUTs than I have. The majority of my group theoretic experience is either years old or related to slightly different things, the aforementioned Calabi-Yau stuff which doesn't use tensor products but subgroups to derive singlet things. Its good to see I wasn't utterly wrong in what I said either Please Register or Log in to view the hidden image! Looks interesting. Things like Young Tableux I never had to learn to make use of them but I half tried to teach myself a bit of them so I'd know what people doing GUTs kept referring to. I got the feeling they were neat for the applications a physicist would use but really only came into their own when you were doing more general mathematics. Few physicists ever ask to break SU(NM) into things involving SU(N) and SU(M), since the SU group after SU(5) is less interesting than the exceptional groups, which don't follow such a nice classification (hence the name). Having scrolled down through that thread I'm getting flash backs of sitting in seminars where someone writes down a Lagrangian as long as your arm and then proceeds to examine every term in painful detail. For braver men than I.....
Is it possible the author is actually referring to why the general Lagrangian is allowed to be an SU(2)xSU(2) singlet in the first place, rather than why this particular Lagrangian has such a property? I'm also a bit rusty on Standard Model stuff since it's been about 1.5 years since I took the course, so I apologize if my suggestion has no relevance to the question.
