Estimating minimum energy with uncertainty principle

Question

I'm currently trying to solve a problem that involves estimating the minimum energy of a particle in the potential:

$$ V(x) = \frac{-V_0a}{|{x}|} $$

I'm quite confused about how to handle the absolute value in the potential. So far I know that the total energy of the particle will be

$$ E = \frac{p^2}{2m} - \frac{V_0a}{|{x}|} $$

The expectation value of the energy is therefore

$$ \langle E\rangle = \frac{\langle p^2\rangle}{2m} - \frac{V_0a}{\langle|{x}|\rangle} $$

Now, can I use the fact that $$\frac{1}{|x|}=\frac{1}{\sqrt{x^2}}$$ and then $\langle|x|\rangle = \sqrt{\langle x^2\rangle} = \Delta x$? Then from the uncertainty principle we know that $\Delta p = \hbar/2\Delta x$ and $\Delta p^2 = \langle p^2\rangle$ ($\langle p\rangle = 0$ and $\langle x\rangle$ = 0 in a minimum energy state)

Plugging this back into the energy equation gives $$ \langle E\rangle = \frac{\hbar^2}{8m\Delta x^2} - \frac{V_0a}{\Delta x} $$

When minimising this I get

$$ E_{min} = \frac{-2mV_0^2a^2}{\hbar^2} $$

Now this doesn't look wildly wrong but I'm not sure if I've used the right approach with $\langle|x|\rangle = \sqrt{\langle x^2\rangle} = \Delta x$.

Answer 1

No, you can't do that. You can write $\lvert x\rvert = \sqrt{x^2}$, and you can then go to $\langle\lvert x\rvert\rangle = \langle\sqrt{x^2}\rangle$, but it's not valid to say $\langle\sqrt{x^2}\rangle = \sqrt{\langle x^2\rangle}$. You can pick almost any wavefunction, $\psi(x) = \pi^{-1/4}e^{-x^2/2}$ for instance, and show that the two are not equal.

In general, $\langle f(x)\rangle \neq f\bigl(\langle x\rangle\bigr)$ unless $f$ is linear. That also means that when you take the expectation value of the energy operator, you can't write $\frac{1}{\langle\lvert x\rvert\rangle}$; you need to leave it as $\langle\frac{1}{\lvert x\rvert}\rangle$.