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A trick to solve functional integraltion?

  1. Nov 8, 2005 #1
    we know that the functional integrals are important in quantum field theory,but we have the problem that except for the semiclassical approach,they can not be solved anyway..but if we used the formula:.
    [tex]\int{d[\phi]F[\phi]=\sum_{n=1}^{\infty}(-1)^{n}\phi^n{D^{n}F[\phi]} [/tex]
    where D is the functional derivative [tex] D=\delta/\delta{\phi}[/tex]
    this is the Bernoulli formula for the functional ,we could obtain an approach to the functional integral,wher [tex]\phi=\int{\phi}d^4x[/tex]
     
    Last edited: Nov 8, 2005
  2. jcsd
  3. Nov 22, 2005 #2
    Can you tell me the reference for this expansion formula for functional, because if I take the definition of the first functional derivative i read : [tex] D_\eta[\phi]=\lim_{\epsilon\rightarrow 0}\frac{F[\phi+\epsilon\eta]-F[\phi]}{\epsilon}[/tex]..this is the functional derivative of F in the "direction" [tex]\eta[/tex]..since a functional is a function which has starting elements that are themself functions (functions can build a more than continuus infinite dimensional space)...this leads to the first order approximation : [tex] F[\phi+\epsilon\eta]\approx F[\phi]+\epsilon D_\eta[\phi] [/tex]....the usual derivative is to put [tex]\eta(x)=\delta(x-y)[/tex] and is written D_{\eta(x)=\delta(x-y)}F[\phi]=\frac{\delta F}{\delta\phi}[/tex].
    Hence I get the expansion : [tex] F[\phi+\epsilon\eta]=\sum_{n=0}^\infty \frac{\epsilon^n}{n!}D_\eta F[\phi][/tex]
    In you expansion it is for me not clear what [tex]\phi^n[/tex] means, since phi is a function, there exist different meaning : [tex] \phi(x)^n=\phi(x)\ldots\phi(x)\textrm{ and } \phi^n(x)=\phi(\cdots\phi(x)\cdots)[/tex]
     
    Last edited: Nov 22, 2005
  4. Nov 22, 2005 #3
    LHS is a path integration over [tex]\phi[/tex]. How did you drop that path on RHS i.e. [tex]\int{d[\phi][/tex] ?
     
  5. Dec 5, 2005 #4
    If we consider the functional integration as the inverse of functional derivative let be [tex]G=\delta/\delta{\phi}[/tex] then we have that the solution to the functional equation:

    [tex] F[\phi]+\phi{DF[\phi]}=e^{iS[\phi]/\hbar}[/tex] (1)

    is [tex]F[\phi]=(1/\phi)\int{D[\phi]e^{iS/\hbar}[/tex]

    but the differential equation can be solve by iterations to get the series..

    [tex]F[\phi]=\sum_{n=0}^{\infty}\phi^{n}G^{n}e^{iS/\hbar} [/tex]

    Unless of course is not true that for functionals...

    [tex]G\int{D[\phi]e^{iS/\hbar}=e^{iS/\hbar} [/tex]

    so if we consider the functional integration as the inverse of functional derivative..we can construct a series expansion for the operator D^{-1}, in fact it is supposed that:

    [tex] D[\phi]=Lim{j\rightarrow\infty}\Product{dx1.dx2.dx3.....dxj}[/tex]
     
    Last edited: Dec 5, 2005
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