@Gupta Oh yes OK, I see that you have $f = a k +b, h = a^2 k + a b + b, H = a^2 F +\frac{a b + b}{1-z}$. In that case, I think this is a complex and interesting problem, and that you may be able to make some headway towards a more general rule using the method I described

@Gupta yes I would expect you to need $G(z) = H(z, F(z), F'(z), \int dz F(z),...)$ more generally, but if you can find some examples with $G(z) = H(z, F(z))$ it would add weight to your conjecture. Currently I'm not convinced it can be extended past the example you gave

I thought the point was that, $Y$ is more spread out than $X$. This means that when we sample from $X$ we know more about what result we're going to get (ie. it will likely be closer to the mean value) than when we sample from $Y$. Hence sampling from $Y$ gives more information, as we knew less about what we'd get prior to sampling

@MichaelEngelhardt what do you not like about the upvoted answer?

Oh yeah thanks that answer gives a good explanation

I wondered if you were making some statement in general for bound states in QM, or if you were making a statement about the hydrogen atom only. I think it does hold including the spin orbit interaction because this is subleading and the statement is only approximate?

Here is a paper that I found doing this based on a quick search. (I remember the one I was looking at a while back was more readable than this one, so could be worth some more searching) researchgate.net/publication/…

Yeah it should be something like so(4) = so(3)×so(3) (at least, at the level of the algebra and maybe with su(2) instead of so(3)). Then one set of $|l,m\rangle$ spherical harmonics is for the rotational motion as in any radial potential, and there's another one for the radial motion in this case due to the augmented symmetry group? Although this should be something like $|n,q\rangle$, I'm not sure what the $q$ quantum number is about. I'm pretty sure I saw a paper doing this a while back

"Wave functions with the same number of nodes have approximately the same energy." is this an observation based on the hydrogen atom and other systems? Or is there some general proof of this? (presumably for systems with a finite number of nodes, are they always bound states?)

@LSpice, an expression for $f_1$ and $f_2$ would be given. Eg. Suppose I have some elementary function which I know has no elementary antiderivative. So, I define its antiderivative to be $f_1$. How can I know whether $f_1$ is expressible in terms of already known special functions or not? Maybe it's equal to $g(\Gamma(x))$, in which case I don't want to define a new special function for it

Starting from $xz'' - Az=0,$ change variables to $z=u/x$ and then $x=e^t$ to reduce the system to $\ddot{u} -2\dot{u} +2u-Au^2,$ with no explicit $t$ dependence. It's now a classical mechanics problem with a damping term $-2\dot{u}$ which obstructs integrating directly. Maybe Mathematica can solve this equation? $v=u-1/A$ also switches the term in $u$ for a constant, although I don't know how much difference that makes.

Thanks for explaining. If I have a point $d\in \mathbb{D}^3$, angular momentum $L$ and linear momentum $p$, do you know the functional form of either $d(p,L)$ or $(p(d),L(d))$?

I understood that your $\mathbb{D}^3$ corresponds to combining together momentum and angular momentum somehow? How is the $\mathbb{D}$ defined?