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$u_l(r) = Ar^{l+1}e^{-\kappa r}$ is provided as a solution to the radial wave equation for the Coulomb potential $$-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}u_l(r)+\Bigl[\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} - \frac{e^2}{4\pi\epsilon_0}\frac{1}{r}\Bigr]u_l(r) = E_l(r)$$

I can show this solution satisfies the above differential equation and I get an expression for $E_l(r)$ which depends on $\frac{1}{(l+1)^2}$ and a bunch of constants.

How can I show or argue, on physical grounds, that $u_l(r)$ is the ground state wavefunction for $H_l$, where $H_l$ is $$-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}+\Bigl[\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} - \frac{e^2}{4\pi\epsilon_0}\frac{1}{r}\Bigr].$$

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Rearranging the radial equation, we can write :

$\frac{d^2}{dt^2}u-[{\frac{l(l+1)}{r^2}}-\frac{\alpha}{r}]u -\kappa^2u = 0$ , where $\kappa = \sqrt{\frac{-2mE}{\hbar^2}}$ , and $\alpha = \frac{2me^2}{4\pi\epsilon_0\hbar^2}$

As $r \to \infty$ , The the effect of the term in the $[$ $]$ (which is the "effective Potential" , $V_{eff}$) becomes negligible. In this limit, the above equation reduces to $u" - \kappa^2u = 0$. The physical expectation is that the wavefunction vanishes at $\infty$ (because we are looking for solutions representing bound states). This requires $\kappa^2$ > $0$ (and hence $\kappa$ is real) , or $E$ < $0$ , in which case, $u \to e^{-\kappa r}$ as $r \to \infty$.

In the opposite limit ($r \to 0$), we see that the term that dominates is the first term in the square bracket. In this limit, the radial equation reduces to $\frac{d^2}{dt^2}u - \frac{l(l+1)}{r^2}u = 0$. This equation has two independent solutions : $r^{l+1}$ and $r^{-l}$. We can see that only the first solution makes sense physically, as the second one diverges at $r=0$. So, we have that $u \to r^{l+1}$ as $r \to 0$.

These two statements were derived without any assumptions, so, they must hold true for all eigensolutions of the above radial equation. So, all eigenstates of the radial equation must be of the form $u_n(r) = e^{-\kappa r} r^{l+1} f_n(r) $.

Now, employing the Nodal theorem (as the other answer has alluded to already), we see that the ground state must have zero nodes ; we already know from above that all eigenstate have the form $u_n(r) = e^{-\kappa r} r^{l+1} f_n(r) $, where $n$ can now be understood as identifying the number of nodes in $f_n$. So, for the ground state solution, we need to find $f_0$, which is the "$f$ " with $0$ nodes that satisfies the above radial equation. And as you have found already, $f_0 = 1$.

Also, the nodal theorem can be proved :

"Feynman" version : Can it be argued qualitatively that the ground state of a bound energy spectrum has zero nodes?

Rigorous version : When does the $n$th bound state of a 1-D quantum potential have $n$ maxima/minima?

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  • $\begingroup$ Thank you I was not aware of the nodal theorem, that's very helpful $\endgroup$ – Tapedeck Mar 22 at 20:02
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For one-dimensional problems (as your radial problem) there is the statement that the ground state wave function has no nodes. This does not provide a proof for a given wave function to be a ground state wave function but it clearly is an indication.

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