# Variational derivative of $\Phi_a(-\partial^2 - m_0^2 - \Sigma)\Phi_a$

Let me refer to the below link http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-17-phi4on.pdf

In Eq: 18.40, $$\Gamma[\Phi_a, \Sigma]$$ is given as, $$\Gamma[\Phi_a,\Sigma] = NA_{coll}[\Sigma] + \frac{1}{2}\int d^4x \Phi_a(-\partial^2 - m_0^2 - \Sigma)\Phi_a$$

Note: $$A_{coll}[\Sigma]$$ is purely a function of $$\Sigma$$

The in Eq. 18.41, the author states: $$\Gamma_{\Phi_a \Phi_b} = (-\partial^2 - m_0^2 - \Sigma)\delta_{ab}$$

My question is, how is the above possible?

In the first term, since $$A_{coll}[\Sigma]$$ is purely a function of $$\Sigma$$, we have $$\frac{\delta A_{coll}[\Sigma]}{\delta \Phi_a} = 0$$.

So, the first term is not an issue. The issue is with the second term.

In the second term, the factor $$\frac{\delta (\partial^2 \Phi_a)}{\delta \Phi_a} = 0$$.

This means: $$\Gamma_{\Phi_a \Phi_b} = (- m_0^2 - \Sigma)\delta_{ab}$$ and not $$\Gamma_{\Phi_a \Phi_b} = (-\partial^2 - m_0^2 - \Sigma)\delta_{ab}$$

What am I missing?

• Don't you get $\frac{\delta(\partial^2\Phi_a)}{\delta\Phi_b}=\partial^2 \delta_{a,b}$ instead of what you wrote? Commented Nov 20, 2021 at 5:11
• I don't think so. As an example, when one determines the Euler Lagrange equations of motion, one takes $\phi$ and $\partial \phi$ as independent variables. Logically, in functionals, we wish to determine the optimal function, $\Phi$. Hence, one would take $\Phi$ and $\partial \Phi$ to be independent functions, so that one can arrive upon the differential equation, which when solved, gives the optimal $\Phi$. Commented Nov 20, 2021 at 16:13

$$\frac{\delta(\partial^2\Phi_a)}{\delta \Phi_b}=\partial^2\delta_{ab}$$
$$\frac{\delta \mathcal{L}}{\delta\phi}=0$$ or in other words $$\partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}-\frac{\partial \mathcal{L}}{\partial(\phi)}=0$$
One then need to pay attention to the difference between partial derivative and differential. For the equation including $$\partial \mathcal{L}$$ one consider $$\partial_\mu \phi$$ and $$\phi$$ as independant variables, while it is not the case when considering the differential $$\delta \mathcal{L}$$.