0
$\begingroup$

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.

\begin{equation} \frac {\delta F} {\delta \phi(x)}=\phi^{*}(x)\\ \frac {\delta F} {\delta \phi^*(x)}=\phi(x)\tag{2} \end{equation}

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

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

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?

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/89002/2451 $\endgroup$ – Qmechanic Jun 25 '16 at 22:23
  • 3
    $\begingroup$ 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. $\endgroup$ – ACuriousMind Jun 25 '16 at 22:24
  • $\begingroup$ I think this question should be migrated to math.stackexchange.com because it is a purely mathematical question. $\endgroup$ – valerio Jun 26 '16 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.