# Conserved current; Klein-Gordon

1. Apr 12, 2009

### Landau

1. The problem statement, all variables and given/known data

Given the Lagrangian density of a complex relativistic scalar field
$\mathcal L=\frac{1}{2}\partial^\nu\phi^{*}\partial_\nu\phi-\frac{1}{2}m^2\phi^{*}\phi$
where * stands for complex conjugation, compute the conserved current (using Noether's theorem).

2. Relevant equations

I can (should) use $J^k=-\frac{\partial \mathcal L}{\partial (\partial_k \phi_I)}\phi_I$, where summation over I is implied (which just means we have two terms, one for $\phi$ and one for $\phi^{*}$).

3. The attempt at a solution

Well, I computed $\frac{\partial \mathcal L}{\partial (\partial_k \phi)}=\partial^k\phi^{*}$, and similarly $\frac{\partial \mathcal L}{\partial (\partial_k \phi^{*})}=\partial^k\phi$.

Combining these, we simply get $J^k=-\frac{\partial \mathcal L}{\partial (\partial_k \phi_I)}\phi_I=-\frac{\partial \mathcal L}{\partial (\partial_k \phi)}\phi-\frac{\partial \mathcal L}{\partial (\partial_k \phi^*)}\phi^*=-\frac{1}{2}\left(\partial^k\phi^{*}\phi+\partial^k\phi\phi^{*}\right)$

But the correct answer should be $=-\frac{1}{2}\left(\partial^k\phi^{*}\phi-\partial^k\phi\phi^{*}\right)$, differing in a minus sign.

(Just looking at the Lagrangian, $\phi$ and $\phi^*$ are symmetric, right? So the terms in the conserved current should also come in symmetric...but in the correct answer they aren't.)

Apparently $\frac{\partial \mathcal L}{\partial (\partial_k \phi^{*})}=-\partial^k\phi$?

Last edited: Apr 12, 2009
2. Apr 14, 2009

### George Jones

Staff Emeritus
No.

While $\phi$ and $\phi^*$ appear symmetrically in the Lagrangian, they don't come into play symmetrically in the application of Noether's theorem to a Klein-Gordon system.

The Lagrangian is invariant under the continuous transformation $\phi \rightarrow e^{i\alpha} \phi$, which means $\phi^* \rightarrow e^{-i\alpha} \phi^*$.