Why Does the Conserved Current for the Klein-Gordon Equation Differ by a Sign?

[Landau]

Homework Statement

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).

Homework 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^{*}).

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?

 

[George Jones]

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

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^*.