Why is the orthogonality of complex functions defined with the integration between its complex conjugate?

1.7k Views Asked by At

As stated in the title, I am not quite understand the orthogonality of complex functions.

For example, for the following function family:
\begin{equation} \phi_k(x)=e^{ikx}, \end{equation} its orthogonality is defined using the integration between
\begin{equation} \phi_l(x)=e^{-ilx}, \end{equation} in [0, 2$\pi$]. This way, the functions used are belonging to two different families, how can it be utilized to define orthogonality of one family of functions?

2

There are 2 best solutions below

3
On BEST ANSWER

It's not different families. It is just their complex conjugates.

This is usual in complex vector spaces. It stems from the very definition of the usual inner or hermitean product $<..,..>:\mathbb{C}^n\times\mathbb{C}^n\rightarrow\mathbb{C} $: $$<(a_1,a_2,...,a_n),(b_1,b_2,...,b_n)>=a_1\overline{b_1}+a_2\overline{b_2}+...+a_n\overline{b_n}$$ The reason lying behind this apparent "peculiarity", is that if we try to "mimic" the usual inner product of $\mathbb{R}^n$ and define $(..,..):\mathbb{C}^n\times\mathbb{C}^n\rightarrow\mathbb{C}$ through $$<(a_1,a_2,...,a_n),(b_1,b_2,...,b_n)>=a_1b_1+a_2b_2+...+a_nb_n$$ (i.e. without using a complex conjugation on the second variable), we end up with a symmetric, non-degenerate, bilinear form on $\mathbb{C}^n$, which however is not an inner (or: hermitean) product in $\mathbb{C}^n$.

What i am trying to say here, is that the conjugation underlies any kind of "measurable" quantification in complex numbers. (measurable here stands for something producing a real value as an indication of magnitude, i.e. a number which can directly be tested to the experiment). Remember that even the magnitude of a complex number $z$ is given by the square root of $z\bar{z}$ and not by the square root of $z^2$.

0
On

Even though you’ve accepted an answer, I’ll offer a slightly different explanation.

Inner products are closely related to norms via a formula you probably already know: $$\|\mathbf v\|^2 = (\mathbf v, \mathbf v).$$ It would be nice if this norm was compatible with an existing norm that the scalar field might have, in particular, we’d like $$\|z\mathbf v\| = |z|\|\mathbf v\|$$ to hold. Now, we like complex numbers to have nonnegative real-valued norms for various reasons, so we have $|z|^2 = z\overline z$. It doesn’t work if you simply square $z$. That conjugate percolates back up through the definitions of complex vector norms and inner products to maintain this consistency.