# Estimating minimum energy with uncertainty principle

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 $\langle|x|\rangle = \sqrt{\langle x^2\rangle} = \Delta x$.

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 $\langle\frac{1}{\lvert x\rvert}\rangle$.
• Yeah, I thought about that, but it's not clear from the question just how accurate of an estimation is expected. Besides, sometimes it makes a huge difference, e.g. it's easy to find a state $\langle 1/\lvert x\rvert\rangle$ is small but $1/\langle\lvert x\rvert\rangle$ is undefined. – David Z Nov 28 '13 at 20:15
• For sure you have to be careful with perhaps ill-defined quantities. But regarding accuracy you make the same type of error when you assume $\Delta p \Delta x = \hbar /2$. It must be bigger (at best equal). In order to know the exact value you need to know the wave function, but then you don't need to estimate anything, as you can simply calculate it. – perplexity Nov 29 '13 at 9:16