I have a hydrogen atom, knowing that its Hamiltonian has been modified turning the standard potential $$ V_{0}(r) = -\frac{Z}{r} $$ into $$ V(r) = -\frac{g}{r^{\frac{3}{2}}} $$ with $g$ a positive constant.
An upper bound to the ground-state energy is to be determined.
I thought of using the variational method, with a trial function $$ \psi(\xi) = A\cdot e^{-\xi r} $$ $A = \frac{\xi^3}{\pi}$ (normalization condition). The idea for such a wave-function comes from having regarded $V(r)$ as a standard hydrogenic potential with a "space-dependent atomic number", $Z(r) = g\cdot r^{-5/2}$.
FIRST QUESTION: Is this a good choice for the trial function? If not, why?
So, just like in a usual hydrogenic atom, $$ \langle T\rangle = \langle\psi(\xi)|-\frac{\nabla^2}{2}|\psi(\xi)\rangle = \frac{\xi^2}{2} $$
I thought of using the virial theorem (being $V(r)$ spherically symmetric and $\propto r^{-3/2}$) to calculate the part $\langle V\rangle$: $$ \langle V\rangle = -\frac{4}{3}\langle T\rangle $$ But I wasn't able to apply it without obtaining something illogical!
Calculating $\langle V\rangle$ explicitly I found
$$
\langle V\rangle = \langle \psi(\xi)|-\frac{g}{r^{-3/2}}|\psi(\xi)\rangle = 4\pi A^2 \int_0^{\infty} \! r^2 e^{-2\xi r}(-\frac{g}{r^{3/2}}) \, \mathrm{d}r =
$$
$$
-4g\xi^3\int_0^{\infty} \! r^{1/2} e^{-2\xi r}\, \mathrm{d}r \xrightarrow[{d}r \rightarrow 2t{d}t]{r \rightarrow t^2} -8g\xi^3\int_0^{\infty} \! t^2 e^{-2\xi t^2}\, \mathrm{d}t =
$$
$$
-8g\xi^3\frac{1}{4{(2\xi)}^{3/2}}\sqrt{\pi} = -\sqrt{\frac{\pi}{2}}g\xi^{3/2}
$$
So, going on applying the variational method, I have to minimize
$$
E(\xi) = \langle\psi(\xi)|H|\psi(\xi)\rangle = \langle\psi(\xi)|-\frac{\nabla^2}{2}-\frac{g}{r^{-3/2}}|\psi(\xi)\rangle =
$$
$$
\langle T\rangle + \langle V\rangle = \frac{\xi^2}{2} -\sqrt{\frac{\pi}{2}}g\xi^{3/2}
$$
calculating
$$
\frac{{d}E(\xi)}{{d}\xi} = \xi - \frac{3}{2}\sqrt{\frac{\pi}{2}}g\xi^{1/2} = 0 \Rightarrow \begin{cases} \xi = 0 \\ \xi = \frac{9\pi}{8}g^2 = \bar{\xi} \end{cases}
$$
The only acceptable solution is the positive one, $\bar{\xi}$, so that an upper bound to the ground-state energy is
$$
E(\bar{\xi}) = \frac{\bar{\xi}^2}{2} -\sqrt{\frac{\pi}{2}}g\bar{\xi}^{3/2}
$$
SECOND QUESTION: How can I properly use the virial theorem?
I asked something like this before and I thought I understood it but - obviously - I have not. Please make this clear to me; thank you in advance!