A Question about derivatives of complex fields

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1. Jul 22, 2017

Chronum

https://arxiv.org/pdf/1705.07188.pdf

Equation 5 in this paper states that
$$\frac{\partial F}{\partial p_i} = 2Re\left\lbrace\frac{\partial F}{\partial x}\frac{\partial x}{\partial p_i}\right\rbrace$$

Here, $p_i$ stands for the $i$'th element of a vector of 'design parameters' $\mathbf{p}$. These design parameters are variables that we directly control.

Just after that equation, the paper states that the derivative of $x$, involves that 2*Re part because $x$ is complex. $x$ is a vector of complex E and H fields.

Now, my question is why is this? Why is the derivative of complex E and H fields, with respect to a certain parameter $p_i$ cause that extra factor of 2 in the front, and why do we only consider the real part?

2. Jul 25, 2017

Daniel Gallimore

There are a couple of things to unpack here. The first is the complex chain rule. If $x(p_i),p_i\in\mathbb{C}$ and $F(x)$ is likewise a complex-valued function, then
$$\frac{\partial F}{\partial p_i}=\frac{\partial F}{\partial x}\frac{\partial x}{\partial p_i}+\frac{\partial F}{\partial x^*}\frac{\partial x^*}{\partial p_i}$$ Next, suppose we have some complex-valued function $z(t)=a(t)+ib(t)$, where $a,b\in\mathbb{R}$. Then
$$\left(\frac{dz}{dt}\right)^*=\left(\frac{da}{dt}+i\frac{db}{dt}\right)^*=\frac{da}{dt}-i\frac{db}{dt}=\frac{dz^*}{dt}$$ and (if $z$ has a well-defined inverse)
$$\frac{dz}{dt^*}=\left(\frac{dt^*}{dz}\right)^{-1}=\left(\left(\frac{dt}{dz}\right)^*\right)^{-1}=\left(\left(\frac{dt}{dz}\right)^{-1}\right)^*=\left(\frac{dz}{dt}\right)^*$$ Furthermore, $z+z^*=(a+ib)+(a-ib)=2a=2\text{Re}(z)$
Putting all of this together,
$$\frac{\partial F}{\partial p_i}=\frac{\partial F}{\partial x}\frac{\partial x}{\partial p_i}+\left(\frac{\partial F}{\partial x}\right)^*\left(\frac{\partial x}{\partial p_i}\right)^*=\frac{\partial F}{\partial x}\frac{\partial x}{\partial p_i}+\left(\frac{\partial F}{\partial x}\frac{\partial x}{\partial p_i}\right)^*=2\text{Re}\left(\frac{\partial F}{\partial x}\frac{\partial x}{\partial p_i}\right)$$ It's not the most rigorous treatment of the problem, but it should be more than enough to give you an idea of why this statement might be true.

3. Jul 28, 2017

Chronum

This is in fact the same rule that the paper author mentioned (I emailed the first author). And then told me to look at Wirtinger derivatives. There are a few assumptions here, but those hold true in this case, and this step has since proceeded to make sense. Many thanks!