Physical meaning of uncertainty principle + question on symmetry A question on the Uncertainty principle.
So we know that it says (for position and momentum) that:
$$ \Delta x \Delta p \ge \hbar/2. $$
Where $\Delta p = \sqrt {\langle p^2\rangle - \langle p\rangle^2 }$  and $\Delta x = \sqrt {\langle x^2\rangle - \langle x\rangle^2} $;
NOW, Potential with a symmetry.
Imagine a square well centred at 0, so extending from -A to A.
The potential is symmetric so (I would expect) the particle is as likely to be in the LHS $(-A<x<0)$ as in the RHS $(0<x<A)$;
This suggests that both $\langle x\rangle$ and $\langle p \rangle$ are 0.
In that case $\Delta x = \sqrt {x^2}$ and $\Delta p = \sqrt{p^2}$;
So effectively the spread is not just a measure of the uncertainty but it is actually telling us the root mean square of both momentum and position.
SO my question is: if I confine a particle in a small volume (assuming a symmetric potential such as a square well so as for the argument above to be applicable), it will have a large momentum and therefore a large kinetic energy, whereas if the square well were larger, its kinetic energy would be smaller.
Where does this extra energy come from?
How can we understand it physically?
(If I were to manually change the width the potential well then I could say that the surplus in energy comes from the work that I've done on the system, but here I'm just considering two separate particles, in different well).
 A: As you mention, if you "manually" change the size of the well the energy difference would be provided by the work done. That's the source.
What you suggest, it's just an idealized situation where you already start with a certain configuration with a definite energy. The origin of the energy is irrelevant, since you don't care about the particular history of the system.
Think of the case when you have in classical mechanics something oscillating. You, in most cases, don't care how it came about to that situation but that it's already oscillating with certain amplitude and frequency.
A: It's not particularly meaningful, in this case, where the energy 'comes from'. It's called the zero-point energy and it's simply there, with not a lot to be done about it.
There are some ways to justify it, of course. Say you have a square potential well, whose bottom has $V=0$, and say you try to put a particle with zero kinetic energy in the centre of the well. Having zero kinetic energy means having zero momentum, and de Broglie's relation then demands that the particle's wavefunction have infinite wavelength. However, having such a large wavelength means that with some probability the particle is in regions with nonzero potential energy, and therefore its total energy is positive. There is just no way to place the particle in a state with zero total energy.

The way to avoid this is, of course, having a wavepacket just small enough that it will fit comfortably inside the well. However, having a limited spatial extent means having a finite wavelength, and therefore (by de Broglie's relation) a nonzero momentum and kinetic energy.
You note rightly that if you were to manually (and adiabatically) decrease the size of the well, you would have to perform work to push the particle in, and this would increase its zero-point energy. If you have two particles in different wells, their difference in energy can best be thought of as a potential to perform work: there exists a physical transformation of the narrower well (i.e. adiabatically allowing it to expand) which takes it to the state of the other particle and at the same time allows it to perform work.
You're no doubt aware of the Casimir effect, in which the effects of the zero-point energy are measurable. This is not simply some mathematical result that's an oddity within the theory; you can indeed perform work using simply the vacuum energy.
