# Complex scalar field, conserved current, expanding functional

1. Jan 11, 2018

### binbagsss

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

Hi

I am looking at this action:

Under the transformation $\phi \to \phi e^{i \epsilon}$

2. Relevant equations

So a conserved current is found by, promoting the parameter describing the transformation- $\epsilon$ say- to depend on $x$ since we know that $\delta S=0$ on-shell trivially. So we find a quantity that has $\delta S \neq 0$ when $\epsilon=\epsilon(x)$, but $\delta S=0$ when $\epsilon$ does not depend on $x$.

I have read that the conserved current is the same as it is without the potential for the addition of a potential $V(\phi \phi*)$ in the Lagrangian.

So it must be that $\delta V(\phi \phi*) =0$ for both $\epsilon$ depending on $x$ or not depending on $x$. Since there are no derivatives this is just that $\delta V(\phi \phi*) =0$, required for the action to be invariant under the transformation anyway.

3. The attempt at a solution

$V(\phi \phi*) \to V(\phi \phi* e^{i \epsilon} e^{-i \epsilon}= V(\phi \phi*)$

So looking at it from this it is obvious.

But, my question / point of my post, if I expand out $V(\phi \phi*)$ how do I show this?

I believe I can write: $V[\phi,\phi*]=V[\phi,\phi*]+\frac{\partial V[\phi,\phi*]}{\partial \phi} \delta\phi + \frac{\partial V[\phi,\phi*]}{\partial \phi*} \delta\phi * [2]$

??

in this case I know that $\delta\phi=-\delta\phi*$ to $O(\epsilon)=O(\delta)$

Therefore looking at [2] I see that it must be that $\frac{\partial V[\phi,\phi*]}{\partial \phi} \delta\phi + \frac{\partial V[\phi,\phi*]}{\partial \phi*} \delta\phi *=0$

$=-\frac{\partial V[\phi,\phi*]}{\partial \phi} \delta\phi* + \frac{\partial V[\phi,\phi*]}{\partial \phi*} \delta\phi *$

so it must be that $\frac{\partial V[\phi,\phi*]}{\partial \phi}=\frac{\partial V[\phi,\phi*]}{\partial \phi*}$

Is this true because rather than $V[\phi,\phi*]$ we have $V[\phi \phi*]$; the potential is symmetric in both $phi$ and $phi*$?

But then I should have a 1 variable taylor expansion rather than 2 variable as I treated it to get [2]?

How would I expand out something like $V[\phi\phi* + \delta\phi* \phi + \delta \phi \phi*]$ treating $\phi\phi*$ as a single variable, rather what I have expanded out is $V[\phi + \delta\phi, \phi* + \delta \phi*]$? (if this is something I've done wrong here, thanks)

Last edited: Jan 11, 2018
2. Jan 16, 2018 at 6:00 PM

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jan 17, 2018 at 9:05 AM

### binbagsss

thanks for the like @Delta² , I don't suppose you may be able to help at all? thanks alot.

4. Jan 17, 2018 at 2:32 PM

### Delta²

Ah, sorry I have my own doubts/questions about this so I wont be much of help if I say anything.

Always nice to see a post with an action integral invariant under some transform, that's why I gave it a like :D.