Variational wavefunctions and "spread" of potential in quantum mechanics A particle in a box has an energy that decreases with the size of the box. In the general case, it is often said that a variational solution for a "narrow and deep" potential is higher in energy than a variational solution for a "wider and shallower" potential, by appealing to the particle in a box as a special case. The former wavefunction is said to be "more confined" in the former than the latter, and the act of being given more space to spread out is said to lower the energy variationally.
Question: are there rigorous mathematical statements of this argument?
 A: The Heisenberg uncertainty principle is probably the best way to understand this phenomenon.
Consider a particle in a confining potential $V(x)$, for instance a box or the quadratic potential of the harmonic oscillator $V(x)=\frac12 k x^2$. Classically, having low energy means that the particle has little potential energy, which means $x \approx 0$, and little kinetic energy, which means $p \approx 0$. But due to the uncertainty principle, these two quantities cannot both be small at the same time, so the particle will find the best trade-off between wiggling around (kinetic energy) and being spread in space (potential energy).
In this light, if the pontial is wide and shallow, then the particle will not need as much energy to spread in space. In contrast, if the potential is deep and narrow, then the particle needs more energy to occupy this region of space.
It is instructive to elaborate this with a formal calculation. Consider the general Hamiltonian
$$ H = \frac{P^2}{2m} + \frac12 V(X) .$$
Introducing hermitian operators $A=\frac1{\sqrt{2m}} P$ and $B=\sqrt{V(X)}$, we have
$$ (A-iB)(A+iB) = A^2 + B^2 + i(AB-BA) = H + i[A,B] .$$
In other words, we can almost express the Hamiltionian as a product of the operator $C=A+iB$ with its hermitian conjugate, except that we cannot get rid of the commutator
$$H = C^\dagger C - i[A,B] .$$
Thus, for any state $|ψ\rangle$, we have
$$ E = \langle ψ|H|ψ\rangle = \langle ψ|C^\dagger C|ψ\rangle - i\langle ψ|[A,B]|ψ\rangle \geq - i\langle ψ|[A,B]|ψ\rangle $$
because the first term is simply the square length of a vector
$$ \langle ψ|C^\dagger C|ψ\rangle = \langle Cψ|Cψ\rangle \geq 0 .$$
For instance, for the harmonic oscillator, we would have
$$ E \geq -i\langle ψ|[A,B]|ψ\rangle = -i\langle ψ|\left[\frac{1}{\sqrt{2m}}P, \sqrt{\frac{k}2} X\right]|ψ\rangle = \frac12 \hbar \sqrt{\frac{k}{m}} = \frac12 \hbar ω$$
which is precisely the zero-point energy of the harmonic oscillator.
As you can see, the non-vanishing commutator between the momentum $P$ and the (square root of the) potential $V(X)$ is responsible for the non-vanishing energy of the lowest energy mode. Also, the more spread out the potential, the smaller the commutator and the uncertainty, hence the smaller this energy.
