# Symmetry Groups

1. Sep 9, 2007

### americanforest

What does it mean to say to say that the electroweak interaction is described by a gauge field theory based on the $$SU(2)_{L}\times$$$$U(1)_Y$$ symmetry group?

I know that SU(2) is a group of unitary matrices and U(1) is the circle group but I don't really see what the sentence means. I haven't taken any quantum mechanics but an doing research on Higgs decays and would like to gain a bit more understanding.

Last edited: Sep 9, 2007
2. Sep 9, 2007

### americanforest

anybody?

3. Sep 15, 2007

### samalkhaiat

4. Sep 15, 2007

### mjsd

ok, if you just wanna understand what this statement means then:
the keyword in this sentence is gauge field theory. The group can be any (Lie) groups really, but gauge invariance means a lot. Suppose you have a group transformation operator, $$U(\vec \alpha) = \exp (i \vec \alpha \cdot \vec g)$$ where
$$\vec g = (g_1, g_2, \ldots, g_n)$$ and the $$g_i$$ is a generator of the group G. here $$\vec \alpha = (\alpha_1, \alpha_2, \ldots, \alpha_n)$$ is just a list of parameters. now if a transformation under $$U(\vec \alpha)$$, say
$$\psi'(x)=U(\vec \alpha) \psi(x)$$ leave your overall Lagrangian (say a Dirac Lagrangian) invariance, then we say the Lagrangian is invariant under a global phase transformation. Now what is this Lagrangian, it is unimportant for someone who do not know Quantum Mechanics. But what is important is that suppose you have some Lagrangian which is made up of some combinations of the $$\psi(x)$$ field and its conjugate put together in such a way that when you apply $$U(\vec \alpha)$$ to them the overall Lagrangian won't change (of course this needs not happen, but let's assume it does).

now we can go further and ask whether each $$\alpha_i$$ are just numbers or can they be dependent on x (the position coordinate) as well? In general, you will find that the Lagrangian under such transformation as $$U(\vec \alpha(x))$$, do change and in order to keep it invariant, you need to introduce a gauge field and the associated gauge transformation of this field. So everytime when apply $$U(\vec \alpha(x))$$ to $$\psi(x)$$, you also apply this gauge tranformation to your gauge field. Now if the overall Lagrangian remains unchanged after these series of tranformations (local gauge transformation), then we say it is gauge invariance. (ok... that's all you really need to know, without seeing the maths)

Now, the obvious by-product of the above process is the introduction of the gauge field which turns out to be, in the SU(2)xU(1) case, corresponding to the electroweak force. So, to answer your question as to what that initial statment means, it is nothing but saying that the gauge field associated with the SU(2)xU(1) group (when you apply this symmetry it leaves the appropriate electroweak lagrangian invariant) turns out to be the electroweak force we observed in reality.

people are motivated to look at this because for quantum electrodynamics (QED), the group is $$U(1)_Q$$ where Q means charge. And it is known for a while that local gauge invariance is exactly how to explain electromagnetic interactions. Electroweak and SU(2)xU(1) is a great triumph for physics and symmetry.

NB: the generators for $$SU(2)_L\times U(1)_Y$$ (for electroweak) are the pauli spin matrices and hypercharge Y (whatever that means.. but the key point is that it is different from Q)

.