They said I was daft to start a blog two days before a quantum field theory exam. Well, bah! I say to them. In fact if I post a few equations this might pass as some sort of study. And I want to try out MathJAX.
To illustrate the fact that symmetries lead to conservation laws consider the following simple O(N) invariant theory:
$$ \mathcal{L} = - \frac{1}{2} \partial_{\mu} \phi^{a} \partial^{\mu} \phi^{a}$$
where \( \phi^{a} \) are a set of real scalar fields with the \(a\) index running from 1 to N. To see the effect of the symmetry we vary the action under a symmetry transformation $$\delta \phi^{a} = i \omega_{A} \left( L^{A} \right)^{a}_{b} \phi^{b}$$ where \(\left( L^{A} \right)^{a}_{b}\) is an orthogonal matrix, \(\omega_{A}\) is a set of infinitesimal parameters and \(A\) runs over the adjoint rep \(1,...,\frac{1}{2}N\left(N-1\right)\).
As a trick we let the \(\omega\) coefficients depend on position, although only \( \omega_{A} = \text{constant}\) is actually a symmetry. This lets us pull out the conserved current because, since the action is invariant if \( \omega_{A} = \text{constant}\), \(\omega\) can only enter in to the first order change of the action through its derivatives. Thus $$\delta S = \int \mathrm{d}^{4}x\ j^{A\mu} \partial_\mu \omega_{A}$$
Integrating by parts and using the fact that \(\omega\) is arbitrary gives \(\partial_\mu j^{A\mu}\), the law of current conservation.
Carrying this out we find that the terms not proprtional to \(\partial_\mu \omega_{A}\) vanish by the antisymmetry of \(L^{A}\) and we are left with
$$ \delta S = \int\mathrm{d}^{4}x\ \partial_{\mu}\left(i\partial^{\mu}\phi^{a}\left(L^{\mathrm{A}}\right)_{b}^{a}\phi^{b}\right)\omega_{\mathrm{A}} $$
up to an irrelevant normal ordering constant. So the current is $$ j^{A\mu}=i\partial^{\mu}\phi^{a}\left(L^{\mathrm{A}}\right)_{b}^{a}\phi^{b}$$
and you can check by using the equation of motion \( \square \phi^{a} = 0 \) and antisymmetry of \(L^{A}\) that it is indeed conserved.
Had we added an arbitrary O(N) invariant potential \( V\left(\phi^{a}\phi^{a}\right) \) none of the work would have been affected, but now it would be an interacting theory instead of a noninteracting one. The symmetry of the theory guarantees that no matter how the interactions affect the dynamics of the theory the total charges \( Q^{A} = \int\mathrm{d}^{3}\vec{x}\ \mathrm{N}\left\{ j^{A0}\right\} \) are always conserved in any reaction.
Well, thanks for bearing with that. MathJAX works like a charm. I promise you it won't be this mathy every day. I don't think.
