Inner product's property in a Hilbert space on $\mathbb{C}$ Recently, I have just learnt about the concept of a Hilbert space.
As far as I can understand, a Hilbert space is a generalized Euclidean space.
When talking about an Euclidean space $E$, indeed there must be a mapping from $E \times E$ to the scalar field, called the "inner product". Back then, when I started to study linear algebra, I studied about the Euclidean space on $\mathbb{R}$. The inner product there has this property
$$\langle x,y \rangle = \langle y,x \rangle $$
But, when expanded to a Hilbert space on $\mathbb{R}$ or $\mathbb{C}$, that property has been changed into
$$\langle x,y \rangle = \overline{\langle y,x \rangle} $$
Indeed, when our scalar field is $\mathbb{R}$, nothing have changed. But what I'm concerning is: Why do we need the inner product of $x$ and $y$ to be the conjugate of the inner product of $y$ and $x$ when the scalar field is $\mathbb{C}$? Why is it neccesary to define that property like that, while we can just simply keep the commutativity like when the scalar field is $\mathbb{R}$?
Sorry if I asked something stupid. Thank you.

It's not a stupid question at all!
One reason is that using conjugate symmetry instead of symmetry allows for a norm to be defined on the space, as with real inner product spaces. We define
$$\|x\| = \sqrt{\langle x, x \rangle},$$
and expect such a thing to be well-defined, real, and non-negative so as to measure distance. Having conjugate symmetry means that,
$$\langle x, x \rangle = \overline{\langle x, x \rangle},$$
making the inner product of $x$ with itself real (of course, further axioms are required to make it non-negative). It better serves a geometric purpose than the slightly more obvious generalisation.
