Quantum Theory: derive EoM of action for a 'general' potential

[binbagsss]

Homework Statement

Action attached:

To find the EoM of $\phi$ / $\phi^*$

Homework Equations

The Attempt at a Solution

[/B]
Without deriving from first principles, using E-L equations I have:

$\partial_{u}\frac{\partial L}{\partial_u \phi} - \frac{\partial L}{\partial \phi} =0$ to get the EoM for $\phi *$

So I get:

$-\partial_u \partial^u \phi* +m^2\phi*+ \frac{\partial V}{\partial \phi*} =0$ (1)

MY QUESTION

Deriving this from first principles I am unsure how to work with the $V(\phi \phi*)$ term to get the $\frac{\partial V}{\partial \phi*}$ in the EoM .

I.e deriving EoM via $S'=S+\delta S+O(\delta^2 S)$ via plugging in $\phi \to \phi + \delta \phi$
I get $V(\phi \phi*) \to V(\phi^*\delta \phi + \phi \delta \phi^* )$

Now to get the EoM I want to factor out $\delta \phi$ and $\delta \phi^*$ from $V(\phi*\delta \phi + \phi \delta \phi^* )$ to get the EoM for $\phi*$ and $\phi$ respectively, arguing that the integrand multiplying the (e.g) $\delta \phi$ must vanish since $\delta \phi$ is arbitrary.

So using the equation (1) above, i.e. sort of cheating and not from first principles I suspect I should be able to show that,

$V(\phi*\delta \phi + \phi \delta \phi* ) = V(\phi*\delta \phi)+V(\phi \delta \phi* )= \frac{\partial V}{\partial \phi*}\delta\phi + \frac{\partial V}{\partial \phi}\delta\phi*$

I am unsure how to show this explicitly. my thoughts are perhaps integration by parts, but i',m confused with this working with the implicit expression of $V$

Any help greatly appreciated. thank you.

 

[thierrykauf]

You need to be given a V(phi) to be able to go further. Often V(phi) = lambda phi^4 is chosen. Your approach is correct, but you need more information to get an explicit answer.

 

[Dazed&Confused]

Why would you not just be able to do $\frac{\partial V}{\partial \phi} = V'(\phi^* \phi) \phi^*$? Also, shouldn't it be $V( \phi \phi^* + \phi \delta \phi^* + \phi^* \delta \phi + \delta \phi \delta \phi^*)-V(\phi \phi^*) =V( \phi \phi^* + \phi \delta \phi^* + \phi^* \delta \phi + \delta \phi \delta \phi^*)-V( \phi \phi^* + \phi \delta \phi^* )+ V(\phi \phi^*+\phi \delta \phi^*)-V(\phi \phi^*)$?

 

[thierrykauf]

All you write is correct, but I don't see how much farther you can go without knowing the exact form of the potential.

 

[binbagsss]

Why would you not just be able to do $\frac{\partial V}{\partial \phi} = V'(\phi^* \phi) \phi^*$? Also, shouldn't it be $V( \phi \phi^* + \phi \delta \phi^* + \phi^* \delta \phi + \delta \phi \delta \phi^*)-V(\phi \phi^*) =V( \phi \phi^* + \phi \delta \phi^* + \phi^* \delta \phi + \delta \phi \delta \phi^*)-V( \phi \phi^* + \phi \delta \phi^* )+ V(\phi \phi^*+\phi \delta \phi^*)-V(\phi \phi^*)$?

ahhh I see, yeh it should.
and what is the purpose of including the two middle terms in the last equality?

 

[Dazed&Confused]

ahhh I see, yeh it should.
and what is the purpose of including the two middle terms in the last equality?

This is the beginning of writing it out as two derivatives, the physicist's way. Simply divide and multiply each by $\delta \phi$ and $\delta \phi^*$ respectively and take the infinitesimal limit.

 

[binbagsss]

This is the beginning of writing it out as two derivatives, the physicist's way. Simply divide and multiply each by $\delta \phi$ and $\delta \phi^*$ respectively and take the infinitesimal limit.

I'm stuck:

$\frac{V(\phi(\phi*+\delta\phi*)}{\delta \phi} \delta \phi - \frac{V(\phi(\phi*+\delta\phi*)}{\delta \phi*} \delta \phi*$
Looking at the limit definition of a derivative I don't see how I get derivatives.

so I need to expand out the potential.

 

[binbagsss]

Why would you not just be able to do $\frac{\partial V}{\partial \phi} = V'(\phi^* \phi) \phi^*$?

sorry what is this, what does the ' denote? is this derivative wrt $\phi*$?

 

[binbagsss]

In order to derive EoM from first principle I need functional expansion right?
Is this given by:

$S[\phi+\delta \phi ] = S[\phi] + \frac{\partial S[\phi]}{\partial \phi} \delta{\phi}$ to the desired order.

But where I've computed $\delta S$ explicitly, where this i given by the prefactor of $\delta{\phi}$ above, except for $V[\phi,\phi*]$
So similarly I have
$V[\phi,\phi*]=V[\phi,\phi*]+\frac{\partial V[\phi,\phi*]}{\partial \phi} \delta\phi + \frac{\partial V[\phi,\phi*]}{\partial \phi*} \delta\phi * [2]$

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I have just seen from the previous question that the Lagrangian given, under the transformation $\phi \to \phi e^{i\epsilon}$ gives the same conserved current as it does without the potential. And in this case I know that $\delta\phi=-\delta\phi*$ to $O(\epsilon)=O(\delta)$ (it's also obvious from $V(\phi \phi*)$ without the expansion and keeping in exponential form where they cancel)..

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 $V[\phi \phi*]$
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?