Cartesian Coordinates vs Spherical Coordinates vs Spherical Basis

Question

I would like to properly understand spherical coordinates once and for all:

In the years of innocence and youth we are all introduced to cartesian coordinates.

This are pretty simple to understand: we define three fixed, mutually orthogonal, direction in space, represented by three versors $\hat{x},\hat{y},\hat{z}$; and then to represent a point in 3D space¹ we can simply write: $$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z} \ \ \ \ \ \ (1)$$ in fact the instruction to find the point can be simply derived from (1): start from the origin of the coordinate system, take $x$ steps in the $\hat{x}$ direction, then $y$ steps in the $\hat{y}$ direction and the same again for $z$, and then you are done! You have found the point.²

Then soon after we are introduced to spherical coordinates:

Here we fix only two orthogonal direction in space, and we use those to measure the angles $\theta$ and $\phi$, then we can find a point by giving the two angles and the distance from the origin a.k.a the radius $r$. Intuitively this is completely fine, the value of $r$ locks the point into a sphere and then we can use the angles to find the point, but then when we want to explicitly write down the formula for $\vec{r}$ how should we do it? Is pretty common to find the line element written like this: $$d\vec{r}=dr\hat{r}+rd\theta\hat{\theta}+r\sin{\theta}d\phi\hat{\phi}$$ So are we allowed to write $\vec{r}$ as $$\vec{r}=r\hat{r}+r\theta\hat{\theta}+r\sin{\theta}\phi\hat{\phi} \ ? \ \ \ \ \ \ (2)$$ I have my doubts, since $\hat{\theta},\hat{\phi}$ are not constant versors, they kinda rotate around and change direction.³ So how should we write $\vec{r}$ in this coordinate system? Are we forced to write $\vec{r}$ by integrating the line element? Seems pretty inconvenient to me!

And at last, when our innocence is completely gone for sure, we are introduced to the spherical basis.

To be honest when I first heared the term I thought it was a synonymous to spherical coordinates, oh was I wrong! Turns out that the spherical basis is a kind of coordinate system of 3D real space involving complex numbers. On the bright side this coordinate system allows us to write a vector in a simple way reminiscent of the cartesian coordinate system: $$\vec{r}=A_+\hat{e}_++A_-\hat{e}_-+A_0\hat{e}_0 \ \ \ \ \ \ (3)$$ At a first glance this seems just a relabeling of the cartesian basis, no real change whatsoever, but this is not the case! From what I currently understand $\hat{e}_+,\hat{e}_-\hat{e}_0$ are complex vectors and $A_+,A_-,A_0$ are complex numbers! Now I really struggle to understand how this coordinate system works, I mean, I struggle to build an intuition, a picture in my head on how we can represent 3d space using this one. I suppose this is analogous to represent 2D real space using a single complex number, but this supposition seems inadeguate: when we represent 2D real space in this way there are no versors, only one complex number⁴ $$z=re^{i\theta}.$$ Furthermore I would expect something like (3) to represent a 6D real space, not a 3D one. On top of all of this it turns out that this basis is really important to understand fundamental concepts in Quantum Mechanics, such as the Spherical Harmonics, so it is not something that we can brush off as marginal; we really need to build a deep understanding of this one.
So I guess that my question regarding this coordinate system/base is: How does it work? How can it represent 3D real space? How can we build an intuition about this?

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(1): Or to represent a vector in 3D space, since we can think of putting the tip of the vector onto the point and the tail fixed in the origin.
(2): Of course there is the hole question of: How big is your step? Since the versors have module equal of one the answer is one; but from a physical point of view we have to chose an unit of measurement for lenght, and we can do that, but it is beside the point here. So no problem.
(3): On the other hand the fact that $\hat{\theta},\hat{\phi}$ have no constant direction is not a problem when writing the line element, since we are dealing with an infinitesimal vector. In fact in this case we can think of $\hat{\theta},\hat{\phi}$ as constant since we perform the sum in an infinitesimal region of space.
(4):We then take the imaginary part to be $y$ and the real part to be $x$, or alternatively we take $r,\theta$ in polar coordinates.

Answer 1

$\hat{r}$ is defined for $r\ne0$, as $\vec{r}/r$ so $\vec{r}=r\hat{r}$. But $d\vec{r}$ isn't in general parallel to $\hat{r}$, as $\vec{r}$ can evolve in an arbitrary direction, not in general parallel to $\vec{r}$, to $\vec{r}+d\vec{r}$.

Usine Eq. (3B) here, $\vec{r}=(x(\vec{e}_--\vec{e}_+)+iy(\vec{e}_-+\vec{e}_+))/\sqrt{2}+z\hat{z}$, which indeed needs complex coefficients. Note in particular that $r^2$, defined as the unique element of the $1\times1$ matrix $r^\dagger r$, is $\frac{|x+iy|^2+|x-iy|^2}{2}+z^2=x^2+y^2+z^2$.