I have started to study QFT. And I have some difficulties in such classical situation.
Suppose i want to calculate $\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\phi$ for lagrangian density $\mathcal{L}=\partial_\mu \phi^*\partial_\mu \phi-m^2\phi^*\phi$ ($\phi$-complex scalar field). I know I should obtain something like $\left[(\partial_\mu \phi)^*\phi-(\partial_\mu \phi)\phi^*)\right]$ $(1)$ but I don't understand how to get this. It's very new subject for me, so I'll be glad to see any answers.
EDIT
I'm reading Gross D. Lectures on QFT. There are a paragraph called "LOCAL SYMMETRIES". There was proved the following fact:
Consider an internal symmetry transformation
$\phi_{i} \rightarrow \phi_{i}^{'}(x)=\phi_{i}+\Psi_{i\alpha}(x)\omega_{\alpha}(x)$
For this transformation the current is $J^{\alpha\mu}(x)=\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_{i})}\Psi_{i\alpha}(x)$ $(2)$
For described situation the current was written above. So, this equations ($(1)$ and $(2)$) should be equal. But I cant get equation $(1)$ by direct differentiation in $(2)$.
Best Answer
Sometimes statements about the Noether current are somewhat imprecise in literature or in lectures. (I haven't checked your literature though).
The procedure is the following
1. Your Lagrangian is invariant under a continuous symmetry transformation up to a total derivative. I suppose in your case it's $$ \phi\rightarrow e^{i\alpha}\phi, \quad\Rightarrow\quad \phi^*\rightarrow e^{-i\alpha}\phi $$ and being invariant up to a total derivative means (in our case $k_\epsilon^\mu=0$) $$ \mathcal{L}\rightarrow\mathcal{L}+\partial_\mu k_\epsilon^\mu $$ 2. A continuous symmetry always gives rise to a conserved current $J^\mu$. It can be calculated as follows
i. Consider infinitesimal transformations, i.e. replace the transformation parameter $\alpha$ by $\epsilon\ll1$. We get $$ \phi\rightarrow e^{i\epsilon}\phi\approx(1+i\epsilon)\phi=\phi+i\epsilon\phi\equiv\phi+\delta\phi\quad\Rightarrow\delta\phi=\epsilon\cdot i\phi $$ Similarly you can find $\delta\phi^*=-i\epsilon\phi^*$
ii. Use the following formula to calculate the conserved current. Note that we are summing over all fields involved in our Lagrangian, i.e. $\phi$ and $\phi^*$ $$ \epsilon J^\mu=\sum_{\text{fields} X}\frac{\partial\mathcal{L}}{\partial(\partial_\mu X)}\delta_\epsilon X-k_\epsilon^\mu $$ Note that $k^\mu$ is the term that appeared in the transformed $\mathcal{L}$, in our case $k^\mu=0$.
iii. As explained by Danu we have $\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}=\partial^\mu\phi^*$ and $\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi^*)}=\partial^\mu\phi$. Now you just have to insert it in the formula above, the result is $$ \epsilon J^\mu=\partial^\mu\phi^*(\epsilon i\phi)+\partial^\mu\phi(-\epsilon i\phi^*)=\epsilon i(\partial^\mu\phi^*\phi-\partial^\mu\phi\phi^*) $$ Thus $$ J^\mu=i(\partial^\mu\phi^*\phi-\partial^\mu\phi\phi^*) $$