For $\psi \in \mathscr S(\Bbb R^3) $ where $ \mathscr S $ is a Schwartz space, the Schrödinger energy functional is given by $$ \mathscr E(\psi) = ||\nabla \psi||_2^2 + \int_\Bbb {R^3} U|\psi|^2.$$ We also have that the ground state energy is given by $$ E_0=inf\{\mathscr E(\psi:\psi \in \mathscr S, ||\psi||=1\}.$$
I understand that $E_0$ is essentially the minimum energy of the system, however I am not clear on how one would compute this for a given potential. For example I have been told that for $U(x) = -{1\over|x|^\alpha} $ where $\alpha >2$, $E_0=-\infty.$
How does one show this?
I started by writing $$ \mathscr E(\psi) =\int|\nabla\psi|^2d^3x-\int_\Bbb {R^3} {1\over|x|^\alpha} |\psi|^2d^3x$$ but am not sure on how to proceed.
Thanks in advance.
I would suggest to try the following family of functions $$\psi_\epsilon(x)= c_\epsilon\frac{e^{-|x|^2}}{(|x|^2+\epsilon^2)^{(3-\alpha)/4}} $$ with $c_\epsilon$ some normalization contant such that $||\psi_\epsilon||=1$. With that you can observe that $\mathscr{E}(\psi_\epsilon) \to -\infty$ for $\epsilon \to 0$. The reason is the potential term. While the kinetic term is some finite positive number. The potential term is diverging in particular, we have that
$$ -\int_{\mathbb R^3} \frac{1}{|x|^\alpha} |\psi_\epsilon|^2 \,d^3 x = - 4\pi c_\epsilon^2 \int_0^\infty r^{2-\alpha} \frac{e^{-2 r^2}}{(r^2+\epsilon^2)^{(3-\alpha)/2} }dr = O(1) - 4\pi c_\epsilon^2 \underbrace{\int_0^1 \frac{r^{2-\alpha}}{(r^2+\epsilon^2)^{(3-\alpha)/2} }dr}_{\sim |\log \epsilon|}$$ such that we find that the infimum is $-\infty$.