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

  1. Jul 18, 2017 #1
    1. The problem statement, all variables and given/known data

    Action attached:

    everybodywantsapieceoftheaction.png

    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. jcsd
  3. Jul 18, 2017 #2

    thierrykauf

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    Gold Member

    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.
     
  4. Jul 18, 2017 #3
    Why would you not just be able to do [itex] \frac{\partial V}{\partial \phi} = V'(\phi^* \phi) \phi^*[/itex]? Also, shouldn't it be [itex] 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^*) [/itex]?
     
  5. Jul 19, 2017 #4

    thierrykauf

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    Gold Member

    All you write is correct, but I don't see how much farther you can go without knowing the exact form of the potential.
     
  6. Jul 25, 2017 #5
    ahhh I see, yeh it should.
    and what is the purpose of including the two middle terms in the last equality?
     
  7. Aug 3, 2017 #6
    This is the beginning of writing it out as two derivatives, the physicist's way. Simply divide and multiply each by [itex] \delta \phi [/itex] and [itex]\delta \phi^*[/itex] respectively and take the infinitesimal limit.
     
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