Quantum Field Theory: Stationary Point of the Effective Action

[latentcorpse]

We have the effective action which obeys $\frac{\delta \Gamma[\varphi]}{\delta \varphi(x)}=J(x)$ where and we are told the stationary point, $\varphi_0$, of this action, $\frac{\delta \Gamma[\varphi_0]}{\delta \varphi(x)}=0$, corresponds to the vacuum expectation value.
(This is out of my notes - there is a discussion around p380 of Peskin & Schroeder on this although it goes a bit more in depth than my notes...)

Anyway, on the next page, he just says that we can write

$\Gamma[\varphi]=i \displaystyle\sum_{n=0}^\infty \frac{1}{n!} \int d^dx_1 \dots \int d^dx_n \varphi(x_1) \dots \varphi(x_n) \Gamma_n (x_1, \dots , x_n)$

Can anybody explain to me where this formula has come from? And what is $\Gamma_n$? He hasn't defined that either.

Thanks.

 

[fzero]

Up to the factor of $$i$$ this is just a power-series expansion for a functional. $$\Gamma_n$$ can be expressed in terms of functional derivatives of $$\Gamma$$ evaluated at $$\phi = 0$$ as written.

 

[latentcorpse]

Up to the factor of $$i$$ this is just a power-series expansion for a functional. $$\Gamma_n$$ can be expressed in terms of functional derivatives of $$\Gamma$$ evaluated at $$\phi = 0$$ as written.

Ok. So I found that $\Gamma_n = - i \frac{\delta^n \Gamma [ \varphi ] }{ \delta \varphi(x_1) \dots \delta \varphi(x_n)}|_{\varphi=0}$

But shouldn't this be evaluated at $\varhpi = \varphi_0$ i.e. the minimum of the effective potential?

 

[fzero]

Ok. So I found that $\Gamma_n = - i \frac{\delta^n \Gamma [ \varphi ] }{ \delta \varphi(x_1) \dots \delta \varphi(x_n)}|_{\varphi=0}$

But shouldn't this be evaluated at $\varhpi = \varphi_0$ i.e. the minimum of the effective potential?

The whole expansion should be made around $$\varphi=\varphi_0$$. Your reference is either being sloppy or you've left out information.