- #1
binbagsss
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Homework Statement
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Hi
I am looking at this action:
Under the transformation ## \phi \to \phi e^{i \epsilon} ##
Homework Equations
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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)
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