Global phase symmetry for complex scalar field theory

Question

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 $(2)$.

Answer 1

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 $$ J^\mu=i(\partial^\mu\phi^*\phi-\partial^\mu\phi\phi^*) $$

Answer 2

You're probably referring to $\mathcal{L}=(\partial_{\mu}\phi)(\partial^{\mu}\phi^*)-m^2\phi^*\phi$. If you rewrite: $$\mathcal{L}=g^{\mu\nu}(\partial_{\mu}\phi)(\partial_{\nu}\phi^*)-m^2\phi^*\phi$$ everything becomes quite simple. You can find two eqns of motion; I will work the out the one resulting from varying with respect to $\phi$

$$\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)=\frac{\partial\mathcal{L}}{\partial\phi}$$

$$g^{\mu\nu}\partial_{\mu}\partial_{\nu}\phi=\square\phi^*=-m^2\phi^* $$

Hence, $$(\square+m^2)\phi^*=0$$

Varying with respect to $\phi^*$ results in $$(\square+m^2)\phi=0$$

As should be expected from the symmetry of the Lagrangian density.

Now, if you're looking for the conserved quantity that looks somewhat like the expression you wrote down, you should consider the change in $\phi$ under a gauge transformation of the first kind ($\phi\rightarrow e^{-i\Lambda} \phi$) in infinitesimal form. This will give you the right quantity to plug into the equation for the conserved current (which follows from Noethers theorem). I've now done all the work for you, except finding $\Psi$.