Complex scalar field, conserved current, expanding functional

[binbagsss]

Homework Statement

[/B]
Hi

I am looking at this action:

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

Homework Equations

[/B]
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.

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)

 

[binbagsss]

Homework Statement

[/B]
Hi

I am looking at this action:

Under the transformation $\phi \to \phi e^{i \epsilon}$View attachment 218290

Homework Equations

[/B]
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.

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)

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

 

[Delta2]

Ah, sorry I have my own doubts/questions about this so I won't 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.