# Functional Differetiation of a complex functional

Suppose I have a simple functional $$F=\int{dx\;\phi^{*}(x)\phi(x)}\tag{1}.$$ Assuming $\phi(x)$ and $\phi^{*}(x)$ are independent and I take a functional differential with respect to $\phi(x)$ and $\phi^{*}(x)$ and I will get the following answers.

$$\frac {\delta F} {\delta \phi(x)}=\phi^{*}(x)\\ \frac {\delta F} {\delta \phi^*(x)}=\phi(x)\tag{2}$$

On the other hand, if I think that functional depends only on real field $\phi$ $$F=\int{dx\;|\phi(x)|^2}. \tag{3}$$ the functional differential will give,

$$\frac {\delta F} {\delta |\phi(x)|}=2|\phi(x)| \\ \frac {\delta F} {\delta |\phi^*(x)|}=0\tag{4}$$

Now I suppose that the field is going to a real field (Imaginary part goes to zero) and we take the limit that $\phi^* \to \phi$ then $|\phi(x)|=\phi(x)$.Then we have a contradiction that $\frac {\delta F} {\delta \phi(x)}=\phi(x)$ but from the second consideration we have $\frac {\delta F} {\delta \phi(x)}=2\phi(x)$. So there is a discrepancy of a factor 2! Can anybody enlighten me?

• – Qmechanic Jun 25 '16 at 22:23
• This has nothing to do with functional differentiation or with physics - your "paradox" also appears for ordinary complex numbers: differentating $z^\ast z$ w.r.t. $z$ gives $z^\ast$, but differentiating $z^2$ w.r.t. $z$ gives $2z$. But I fail to see the problem here - you can't just take the limit of $z$ being real, complex and real differentation are really two different things. – ACuriousMind Jun 25 '16 at 22:24
• I think this question should be migrated to math.stackexchange.com because it is a purely mathematical question. – valerio Jun 26 '16 at 22:26