Spherical quantum oscillator: Is energy smaller than the potential?

Question

A particle with mass $m$ is inside the spherical quantum well $V(r)$: \begin{equation} V(r)= \begin{cases} -V_0, & \text{if}\ r<a \\ 0, & \text{otherwise} \end{cases} \qquad\qquad V_0 \in \mathbb{R^+} \end{equation} Find the ground state by solving the radial equation with $l=0$.

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I do know how to solve this question, however there is one thing bugging me:

I start with the Schrödinger equation: \begin{equation} \hat{H}\psi(r, \vartheta, \varphi)=E \psi(r, \vartheta, \varphi) \end{equation} I then make the seperation ansatz $\psi(r, \vartheta, \varphi) = \phi(r) Y(\vartheta, \varphi)$ which yields (with $l=0$): \begin{equation} \phi''(r)+\frac{2}{r}\phi'(r)+\frac{2m}{\hbar^2}(E-V(r)\phi(r)=0 \end{equation} Now I make the substitution $u(r) = r \phi(r)$ to get: \begin{equation} \bigg(-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial r^2}+V(r) \bigg)u(r) = Eu(r) \end{equation} which I then solve to: \begin{equation} u(r) = A\exp\bigg(\frac{\sqrt{2m(V(r)-E)}}{\hbar}r\bigg)+B\exp\bigg(-\frac{\sqrt{2m(V(r)-E)}}{\hbar}r\bigg) \end{equation}

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I know that this is either a trig or an exponential function depending on the sign of $V(r)-E$. However there seem to be a few possible cases: For the first region we have $V(r)=0$ so there are two cases for the $\text{sgn}(E)$ and for the second there are two more. But when I find (similar) solutions to this problem online, they only write down one of them. Is there a physical argument why we can assume that $-V_0-E<0$ for example?

Answer 1

It is a general fact that for any state $\psi(\vec{r})$, we have $$\langle H \rangle > \langle V \rangle \geq \min V(\vec{r}).$$ The first inequality holds because the kinetic energy operator $\hat{T} = -\frac{\hbar^2}{2m} \nabla^2$ is a positive definite operator on any normalizable state, and $\langle H \rangle = \langle V \rangle + \langle T \rangle$. The second inequality follows from the definition of the expectation value, with equality only holding when $V(\vec{r})$ is a constant.

In your case, for an energy eigenstate, we have $E = \langle H \rangle$ and $\min V(\vec{r}) = - V_0$. The two facts above then imply that $E > - V_0$.