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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 $(2)$.

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    $\begingroup$ I think the result is $ (\partial_{\mu} \phi)^* \phi$. The "should obtain" expression is the Noether current. It comes from the variation $\phi \rightarrow e^{i \alpha} \phi$, which gives $\Delta \phi = i \phi $ and $\Delta \phi^* = -i \phi^*$, if the convention is $\phi \rightarrow \phi + \alpha \Delta \phi$. We sum the variations from both $\phi$ and $\phi^*$, gives the current, as Peskin's QFT p18 Eq. (2.16) does. $\endgroup$ – user26143 Sep 14 '13 at 17:22
  • $\begingroup$ More on phase symmetry and Noether's theorem in Klein-Gordon theory: physics.stackexchange.com/q/69891/2451 $\endgroup$ – Qmechanic Sep 14 '13 at 17:54
  • $\begingroup$ @user26143 I have updated my post. $\endgroup$ – xxxxx Sep 14 '13 at 18:12
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    $\begingroup$ The subscript $i$ in Eq. (2) implies summation over $\phi$ and $\phi^*$ (I think). Like the statement under Eq. (2.12) of Peskin, "If the symmetry involves more than one field, the first term of this expression for $j^{\mu}(x)$ should be replaced by a sum of such terms, one for each field", (this expression is similar with Eq. (2) in your post) then under (2.15) "(We treat $\phi$ and $\phi^*$ as independent fields)". $\endgroup$ – user26143 Sep 14 '13 at 18:16
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    $\begingroup$ @Dima: Note that $\phi$ and $\phi^{*}$ are not independent fields, but it turns out to be consistent to treat them as such. See also this Phys.SE post. $\endgroup$ – Qmechanic Sep 18 '13 at 23:39
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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 $\Psi$.

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

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    $\begingroup$ thank you=) My problem was that i didn't understand that we have two fields $\phi$ and $\phi^{*}$. $\endgroup$ – xxxxx Sep 18 '13 at 22:46
  • $\begingroup$ It's important to realize how many degrees of freedom you have. Because $\phi$ is a complex field, it has two. The way how you sum over them is sometimes a matter of taste, for example you could as well have written $\phi=\phi_1+i\phi_2$ and sum over $\phi_1$ and $\phi_2$ instead of $\phi$ and $\phi^*$. As long as you take into consideration all degrees of freedom it should give the same result. $\endgroup$ – Stan Jul 11 '14 at 10:52

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