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

1. Jul 18, 2017

### binbagsss

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

Action attached:

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

2. Relevant equations

3. The attempt at a solution

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.

2. Jul 18, 2017

### 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.

3. Jul 18, 2017

### 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 $V( \phi \phi^* + \phi \delta \phi^* + \phi^* \delta \phi + \delta \phi \delta \phi^*)-V(\phi \phi^*) =V( \phi \phi^* + \phi \delta \phi^* + \phi^* \delta \phi + \delta \phi \delta \phi^*)-V( \phi \phi^* + \phi \delta \phi^* )+ V(\phi \phi^*+\phi \delta \phi^*)-V(\phi \phi^*)$?

4. Jul 19, 2017

### 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.

5. Jul 25, 2017

### binbagsss

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

6. Aug 3, 2017

### Dazed&Confused

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.

7. Jan 10, 2018

### binbagsss

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.

8. Jan 10, 2018

### binbagsss

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

Last edited: Jan 10, 2018
9. Jan 10, 2018

### 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?