Computation of the Wirtinger derivative of a product (continuation) Let us have a real function $f=(A/2)\phi\bar{\phi}$, where $\phi, \bar\phi$ are complex fields. When looking for the stationary state of this function, we can either treat $\phi, \bar\phi$ as independent variables or split them into their real and imaginary parts and then treat those as independent variables. If we adopt the latter, we have $\partial f/\partial \phi^R=A\phi^R$ and $\partial f/\partial \phi^I=A\phi^I$, where $R$ and $I$ refer to the real and imaginary parts, respectively. Now, assume that we want to find the field $\phi$ that minimizes the function $f$ using a simple steepest descent procedure. We thus write $\dot{\phi}^R=- \partial f/\partial \phi^R$ and $\dot{\phi}^I=- \partial f/\partial \phi^I$. But, since $\phi=\phi^R+i\phi^I$, the last two equations can be combined into an evolution equation for the full complex field, $\dot\phi=-(\partial f/\partial\phi^R+i\partial f/\partial\phi^I)=-A\phi$. However, if one calculates $\partial f/\partial \phi$ by the Wirtinger derivative, the evolution equation is different: $\dot\phi=-\partial f/\partial \phi=-(A/2)\bar\phi$. Can someone, please, tell me where is the catch?
 A: 
$\dot\phi=-(\partial f/\partial\phi^R+i\partial f/\partial\phi^I)=-A\phi$

This is the correct version: $\phi$ goes in the direction opposite to the gradient of $f$. 

However, if one calculates $\partial f/\partial \phi$ by the Wirtinger derivative,

... one does not get the gradient of $f$, simply because $\partial f/\partial \phi$ is different from the gradient of $f$. 
Let's use a more traditional notation:  function $F$ of real variables $x,y$, which are also considered as complex variable $z=x+iy$. The gradient of $F$ in the real-variable sense is the vector with components $(F_x,F_y)$. The  Wirtinger derivative $\partial F/\partial   z$ is $\frac{1}{2}(F_x-iF_y)$ which in vector notation is $  \frac12(F_x,-F_y)$. Notice the minus sign here. The other Wirtinger derivative $\partial F/\partial \bar z$ is $\frac{1}{2}(F_x+iF_y)$ which in vector notation is $  \frac12(F_x,F_y)$; this agrees with the real gradient, except for $\frac12$ which does not affect the direction. 
Thus,  the gradient flow can be written as $\dot z=-2\,\partial F/\partial \bar z$ which is easy to compute in your example: $-2\,\partial f/\partial \bar \phi=-A\phi$, consistent with your first formula. 
