- #36
latentcorpse
- 1,444
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fzero said:You haven't quite used the chain rule, since you didn't include the term coming from the derivative of
[tex]e^{i S[\phi] + i \int d^d x J(x) \phi(x)},[/tex]
That's actually the most important part.
So, despite having worked all this out for the more copmlicated cases where we have path integrals, I am stumped for the finite dimensional question below
if [itex]Z[V] = \int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} - V( \vec{x} )}[/itex]
where [itex]V(0)=0[/itex] and if [itex]V_{i_1 \dots i_n} = \partial_i_1 \dots \partial_i_n V( \vec{x} )_{ \vec{x}=0}[/itex] with [itex]V_i=V_{ij}=0[/itex], use the result
[itex]G( \frac{\partial}{\partial b}) F(b)= F( \frac{\partial}{\partial u}) G(u) e^{ub}|_{u=0}[/itex]
to show that
[itex]\frac{Z[V]}{Z[0]}= e^{\frac{1}{2} \frac{\partial}{\partial \vec{x}} \cdot A^{-1} \frac{\partial}{\partial \vec{x}}} e^{-V( \vec{x})}|_{\vec{x}=0}[/itex]
I can't figure out what to take as F and what to take as G or why?