Treat complex conjugate variables as independent while differentiating

Question

In the book "The oxford solid state basics", solutions book, question 6.2.a we can see the following derivation.

The marked statement says that a complex scalar and its conjugate can be treated as independent/separate variables when differentiating one with respect to the other (and thus the result will be 0)

Why is that?
Is there a physical logic behind it? Some approximation?
Is it mathematically correct?
Is more context needed?

There is a somewhat similar question here. The problem is that the question is articulated as a phenomenon of QFT which makes it much harder to understand the question and the answers. I believe that my question here is much more basic and matters as a complex analysis question.

Answer 1

The answer is very simple, although most physics texts don't take the time to explain it. The partial derivative with respect to a complex variable $z = x + i y$ or it s complex conjugate $\bar{z} = x - i y$ is defined by the so-called Wirtinger derivative $$ \frac{\partial}{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right), \quad \frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) $$ If you'd like, you can just treat the derivative symbols on the left as notation, defined by the expressions on the right. But they have the very nice properties (which you should check by hand) $$ \frac{\partial}{\partial z} z = 1, \quad \frac{\partial}{\partial z} \bar{z} = 0, \quad \frac{\partial}{\partial \bar{z}} \bar{z} = 1, \quad \frac{\partial}{\partial \bar{z}} z = 0 $$ and so on. In other words, the Wirtinger derivatives work exactly as if you were differentiating $z$ and $\bar{z}$ as independent variables.

This is especially useful in extremization problems, as in your post. To extremize $E$, you are always welcome to break $\phi_n$ into a real part $x_n$ and an imaginary part $y_n$, and differentiate those. But if $\partial_{x_n} E = 0$, and $\partial_{y_n} E = 0$, then clearly $\partial_{\phi_n} E = 0$ and $\partial_{\bar{\phi}_n} E = 0$ from the definitions of the Wirtinger derivatives. So, you are always allowed to instead use the Wirtinger derivatives to differentiate with respect to $\phi_n$ and its complex conjugate, treating them as independent, and set those derivatives equal to zero.

Answer 2

The complex field $\phi$ can be written in terms of its real and imaginary part: $$ \phi = \xi + \mathrm{i}\eta, $$ where both $\xi, \eta \in \mathbb{R}$ and these would be the two independent variables you should be differentiating over.

But you can uniquely find $\xi$ and $\eta$ from $\phi$ and $\phi^\dagger$: $$ \xi = \frac{\phi + \phi^\dagger}{2} \quad \mathrm{and} \quad \eta = \frac{\phi - \phi^\dagger}{2\mathrm{i}}. $$

So $(\xi, \eta)$ and $(\phi, \phi^\dagger)$ are essentially equivalent, and the latter is more elegant as it allows you to work with the full complex field instead than its sub-components. Also, less notation to keep track.

It's not an approximation, it is indeed mathematically correct, and it is used extensively in Lagrangian mechanics.

Answer 3

This is simply the idea that a complex number contains two real degrees of freedom. To put it more mathematically, say that a complex number $z$ is given by $z=a+bi$, and say we have some arbitrary function of these two degrees of freedom: $f=f(a,b)$. We can always rewrite this function in terms of $z$ and $z^*$ using the following:

$$a =\frac{z+z^*}{2}$$

$$b = \frac{z-z^*}{2i}$$

More concisely, you can think of this process as a simple change of coordinates from $f(a,b) \to f(z,z^*)$.