Derivative of the complex norm as commonly used in physics

complex numberscomplex-analysisderivatives

On the one hand, I read that the derivative of the complex conjugate $C[z]=\overline{z}$ is not differentiable anywhere (for instance see here). (see 1, below)

On the other hand, I see in physics taking the derivative of a complex scalar field to obtain the equation of motion using the Euler-Lagrange method (for instance see enter link description here (see 2, below)

So which is it, can we or can we not take the derivative?

  1. For case 1, the reference states that a complex function is differentiable if and only if it satisfies the Cauchy-Riemann equations:

$$
f[z]=f[x+iy]=u[x,y]+iv[x,y]
$$

Then f is differentiable if

$$
\frac{\partial u}{\partial x} =\frac{\partial v}{\partial y} \\
\frac{\partial u}{\partial y} =-\frac{\partial v}{\partial x}
$$

Then for the complex conjugate $C[x+iy]=x-iy$ then $\partial u/\partial x =1$ and $\partial v/\partial y=-1$. Consequently $C[z]=\overline{z}$ is not differentiable anywhere in the complex plane.

  1. For case 2, the physics paper defines the Lagrangian of a complex scalar free field as follows:

$$
\mathcal{L}=(\partial \phi^*)(\partial \phi)
$$

Then they claim that

$$
\frac{\partial \mathcal{L}}{\partial (\partial \phi)}=\partial \phi^*\\
\frac{\partial \mathcal{L}}{\partial (\partial \phi^*)}=\partial \phi
$$

To obtain these results I assume they apply the chain rule

$$
\frac{\partial }{\partial (\partial \phi)}(\partial \phi^* \partial \phi)=\partial \phi\frac{\partial }{\partial (\partial \phi)}(\partial \phi^* )+\partial \phi^* \frac{\partial }{\partial (\partial \phi)}(\partial \phi)
$$

Is the following term not an 'illegal' derivative of a complex conjugate function?

$$
\partial \phi\frac{\partial }{\partial (\partial \phi)}(\partial \phi^* )
$$

Why are they allowed to pose it equal to 0?

Best Answer

I've already recommended a discussion here, but it may be worth rewriting to tweak its emphasis. I'll compare three contexts in which one would want to define "derivatives"; in each case, I'll consider a function $f(x)$, rather than switching from the label $x$ to $z$ for the complex case.

  • A function $f:\,\Bbb R\mapsto\Bbb R$ differentiable at $x$ satisfies $f(x+h)\in f(x)+[(Df)(x)]h+o(h)$, with $Df(x)\in\Bbb R$ denoting the derivative of $f$ at $x$.
  • For an integer $n\ge2$, A function $f:\,\Bbb R^n\mapsto\Bbb R^n$ differentiable at $x$ satisfies what looks like the same condition, only now $Df(x)\in\Bbb R^{n\times n}$.
  • But if $S$ is a number system $n$-dimensional over $\Bbb R$, a function $f:\,S\mapsto S$ might satisfy a stronger condition, that we can take $Df(x)\in S$ where, as in the original example, the $O(h)$ term multiplies a number, in this case one in $S$, by another such number, $h$. The Cauchy-Riemann equations characterize which functions $f:\,\Bbb C\mapsto\Bbb C$ succumb to this treatment.

The trick here is respecting the multiplication defined on $\Bbb C$. If we denote the real and imaginary parts of a compex number $w$ as $w_0$ and $w_1$, we cannot in general change$$f_j(x+h)\in f_j(x)+\sum_{k=0}^1[(Df)(x)]_{jk}h_k+o(h)$$to $f_j(x+h)\in f_j(x)+(yh)_j+o(h)$ for some $y\in\Bbb C$. But the general case succumbs to another treatment. Since any $w\in\Bbb C$ satisfies $w_0=(w+w^\ast)/2,\,w_1=(w-w^\ast)/(2i)$,$$\begin{align}f_j(x+h)- f_j(x)&\in[(Df)(x)]_{j0}(h+h^\ast)/2+[(Df)(x)]_{j1}(h-h^\ast)/2i+o(h)\\&=\frac12\left\{[(Df)(x)]_{j0}-i[(Df)(x)]_{j1}\right\}h\\&+\frac12\left\{[(Df)(x)]_{j0}+i[(Df)(x)]_{j1}\right\}h^\ast+o(h).\end{align}$$So now, instead of only having a rate of change against small $h$, we also have a separate one against $h^\ast$. And although $h,\,h^\ast$ are "dependent" in the sense either value determines the other, these two rates of change are uniquely defined. The abbreviation $x:=\partial\phi$ lets us work with the given example from physics:$$\mathcal{L}(x)=x^\ast x\implies\mathcal{L}(x+h)-\mathcal{L}(x)=(x^\ast+h^\ast)(x+h)-x^\ast x\in x^\ast h+xh^\ast+o(h).$$

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