Heisenberg uncertainty principle and zero point energy On my book is written: for a particle in an infinite square well (supposed 1D and large $a$)
\begin{align}
\Delta x\sim a \to& \text{Heisenberg uncertainty principle} \\
\to& \Delta p _\text{min}\sim\frac{h}{2\pi a} \\
\left( E=\frac{p^2}{2m} \right) \to& E_\text{min}\sim\frac{h^2}{8\pi^2a^2m}\,.
\end{align}
However, I'm not sure that the last passage is legal: how is it possible to consider $\Delta p _\text{min}$ and $p$ the same thing? The first is the standard deviation of the aleatory variable $P$, while the second one is the physical value of momentum, so I'd like to understand if there is another way to determine the zero point energy of a system only using the Heisenberg uncertainty principle.
 A: One way to think of this is in terms of expectation values. When you say $\Delta p$, what you really mean is the standard deviation of $p$.
$$
\Delta p = \sqrt{\langle p^2\rangle-\langle p \rangle^2}
$$
In the case of the ground state, you expect $\langle p \rangle=0$ by symmetry, so you just have $\Delta p = \sqrt{\langle p^2\rangle}$. Then you can consider the expectation value of the energy,
$$
\langle E\rangle = \langle \frac{p^2}{2m}\rangle=\frac{\langle p^2\rangle}{2m}=\frac{(\Delta p)^2}{2m}
$$
So far, everything we've written has been exact. But we want to find the minimum possible value for the energy. A moments thought should tell you that the minimum of $\langle E\rangle$ and the minimum of the energy coincide. So you try to find the smallest possible $\langle E\rangle$, and call that $E_{\min}$. That means you want to find the smallest possible $\Delta p$. But of course you know $\Delta x \lesssim a$, so the smallest $\Delta p$ is $\tilde{}\frac{h}{2\pi a}$. Plugging that in gives you $E_\min$.
They key is realizing that if $\langle p\rangle=0$, then the expectation value of $p^2$ is exactly $(\Delta p)^2$. Of course, everything after that is just approximations, but sometimes they work pretty well!
A: It's not a rigorous derivation, it's an estimation that happens to give the right result. The basic idea is that the minimum possible uncertainty in momentum is going to be of the same order as the minimum possible value of the momentum. This isn't always true, but it's often true enough.
In fact, note that the book conveniently used $\Delta x \Delta p \sim \hbar$ instead of $\Delta x \Delta p \sim \hbar/2$ to get the right result.
As for deriving the energy using only the uncertainty principle, I don't think it's possible. The HUP is just an inequality, the actual uncertainties could be larger than their minimum allowed values. Not to mention the uncertainty in some observable is not necessarily the same as its value.
