# Lagrangian invariance under infinitesimal transformations

1. Feb 12, 2016

### ShayanJ

This is my second term in my master's and one of the courses I've taken is QFT1 which is basically only QED.
In the last class, the professor said the Klein-Gordon Lagrangian has a global symmetry under elements of U(1). Then he assumed the transformation parameter is infinitesimal and , under the assumption that the first order transformation leaves the Lagrangian invariant, derived a conserved current.
Now it seems to me that because the Lagrangian is invariant under a finite transformation, and the terms in the Taylor expansion with different orders of parameter are linearly independent, the Lagrangian should also be invariant under each term of the Taylor expansion separately and there should be a conserved current associated to every order.
But when I explained the above line of reasoning to the professor, he said there may be anomalies or something like this. It was really vague actually and I didn't understand what he said. But I figured the above line of reasoning can't go wrong in the case of U(1) so whatever he meant must apply to other symmetry groups. Because for those groups the transformations are done by matrices and the Taylor expansion of matrices is not very straightforward. But still I can't understand what can go wrong.
Can anybody explain further?
Thanks

2. Feb 12, 2016

### samalkhaiat

No, in local field theory, you don’t get any new current. Consider, the K-G field and $U(1)$ transformation $e^{i\epsilon}$. The “currents” associated with odd powers of $\epsilon$ coincide with the usual Noether current associated with first order power, while those associated with even powers of $\epsilon$ are not conserved: To see that look at the transformations to second order:
$$\delta \phi = ( i \epsilon - \frac{1}{2} \epsilon^{2} ) \phi .$$
$$\delta \phi^{\dagger} = (- i \epsilon - \frac{1}{2} \epsilon^{2} ) \phi^{\dagger} .$$
Look at the signs of the second order terms, they are the same, and this prevents you from having conserved “current”. Indeed, if you substitute the above transformations in the $U(1)$ Noether current of the complex KG Lagrangian, you obtain
$$J^{\mu}(\epsilon) = i \epsilon \left( \phi \partial^{\mu} \phi^{\dagger} - \phi^{\dagger} \partial^{\mu} \phi \right) + \frac{1}{2} \epsilon^{2}\left( \phi \partial^{\mu} \phi^{\dagger} + \phi^{\dagger} \partial^{\mu} \phi \right)$$
Now, if you write the current as
$$J^{\mu}(\epsilon) = i \epsilon J_{1}^{\mu} + \frac{1}{2} \epsilon^{2} J_{2}^{\mu}$$
You can identify
$$J_{1}^{\mu}(x) = \phi \partial^{\mu} \phi^{\dagger} - \phi^{\dagger} \partial^{\mu} \phi ,$$
as the conserved Noether current of the $U(1)$ symmetry. But
$$J_{2}^{\mu}(x) = \phi \partial^{\mu} \phi^{\dagger} + \phi^{\dagger} \partial^{\mu} \phi ,$$
is not conserved.