Derivative of a complex potential for the $\lambda \Phi^{4}$-model

Question

A charged scalar particle is described by a complex field $\Phi(x) = \phi_{1}(x)+i\phi_{2}(x)$. Consider a Lagrangian of the $\lambda \Phi^{4}$-model whose potential in the Euclidean action is given by:

$V(\Phi)=\frac{m^{2}}{2}|\Phi|^{2}+\frac{\lambda}{4!}|\Phi|^{4} \ \ \ \ \ \ \ \ (1)$

with $|\Phi|^{2}=\Phi^{*}\Phi = \phi_{1}^{2} + \phi_{2}^{2}$

The condition for a minimum is given for $m^{2}<0$

$\frac{\partial V}{\partial \Phi} = m^{2} \Phi +\frac{\lambda}{3!}|\Phi|^{2}\Phi = 0 \ \ \ \ \ \ \ \ (2)$

How did the author get this equation?

The author seems to pass from a potential containing only scalar terms (1) to a derivative containing a vector term (2)?

What would be the second derivative of V?

I tried to express things in terms of real and imaginary parts and take derivatives separately with respect to each of them and combined but It did not seem to work

maybe there is a lack of understanding in my complex analysis.

Answer 1

It would be more accurate to say that the expression (2) is $\partial V / \partial \Phi^*$, not $\partial V / \partial \Phi$. You should think of $\partial / \partial \Phi$ and $\partial / \partial \Phi^*$ as Wirtinger derivatives , which are simple linear combinations of partial derivatives with respect to the real and imaginary parts of $\Phi$. They have the nice property that you can treat $\Phi$ and $\Phi^*$ as independent variables when taking derivatives. But if you prefer, you can think of this just as a shortcut for taking the appropriate partial derivatives with respect to $\phi_1$ and $\phi_2$ and then taking linear combinations.