The eigenfunctions of $H =\frac{p^2}{2m}+ V_0 \log(|x|/a)$ and a curious virial theorem result I'm thinking about applications of the virial theorem to $1d$ quantum systems. The statement of the virial theorem is that all bound states of the Hamiltonian $T+V$ satisfy the constraint
$$\langle 2 T \rangle = \langle x \frac{dV}{dx} \rangle$$
where $T=\frac{p^2}{2m}$ and $V$ are the kinetic and potential energies respectively.
There's something strange that happens for the case of $$H = \frac{p^2}{2m} + V_0 \log(|x|/a)$$ Here, I take $V_0>0$ to ensure that the eigenspectrum is discrete and bounded below.
Plugging in this potential into the virial theorem gives a constant kinetic energy! That is, we have
$$\langle T \rangle = \frac{V_0}{2}$$

I find this result surprising, as it says that all eigenstates of the Hamiltonian, even those varying widely in total energy, have the same kinetic energy. I'm hoping to gain intuition for this result and to understand what the result means for the shape of the eigenfunctions. What do the eigenfunctions of this Hamiltonian look like at low and high energy? In particular, does the constraint with the average kinetic energy being constant in total energy have a fingerprint in the shape of the wavefunctions?
 A: Since $\langle p^2 \rangle$ is constant, one would expect the distance between the nodes of the wave function would not show a strong dependence (if any) on the quantum number, which is shown in figure in the comments.
A: 
There's something strange that happens for the case of $$H = \frac{p^2}{2m} + V_0 \log(|x|/a)$$ Here, I take $V_0>0$
...we have
$$\langle T \rangle = \frac{V_0}{2}$$


What do the eigenfunctions of this Hamiltonian look like at low and high energy?

The link provided by QCD_IS_GOOD in the comments shows what some eigenfunctions look like.

...does the constraint with the average kinetic energy being constant in total energy have a fingerprint in the shape of the wavefunctions?

The average kinetic energy being constant constrains the integral of the absolute value of the derivative of the wave function like this:
$$
\langle T\rangle = \int dx {\left|\frac{d\psi}{dx}\right|}^2 = V_0\;,
$$
where I have chosen units such that $m=\hbar=1$.
This tells you something about the shape of the wavefunction, but not a whole lot.
