Do stationary states with higher energy necessarily have higher position-momentum uncertainty? For simple potentials like square wells and harmonic oscillators, one can explicitly calculate the product $\Delta x \Delta p$ for stationary states.  When you do this, it turns out that higher energy levels have higher values of 
$\Delta x \Delta p$.  
Is this true for all time-independent potentials?
Certainly, it is possible to find two states $\mid \Psi_1 \rangle$ and $\mid \Psi_2 \rangle$ with $\langle \Psi_1 \mid H \mid \Psi_1 \rangle > \langle \Psi_2 \mid H \mid \Psi_2 \rangle$ and also $\Delta x_1 \Delta p_1 < \Delta x_2 \Delta p_2$.  For example, choose a quadratic potential, let $\mid \Psi_2 \rangle$ be the first state and let $\mid \Psi_1 \rangle$ be a Gaussian coherent state (thus with minimum uncertainty) and fairly high energy.  So I'm asking here just about the stationary states.
As Ron pointed out in the comments, this question is most interesting if we consider potentials with only a single local minimum, and increasing potential to the right of it and decreasing to the left.
 A: The answer is no, and a counterexample is the following plateau potential:
$V(x) = x^2 \ \ \ \ \; \mathrm{for}\ \ \ \  x\ge -A$
$V(x) = A^2 \ \ \ \ \mathrm{for}\ \ \ \  -A-k \le x < -A$
$V(x) = \infty\ \ \; \ \  \mathrm{for}\ \ \ \  x <-A-k$
A is imagined to be a huge constant, and k is a large constant, but not anywhere near as huge as A. The potential has a plateau between -A-k and -A, but is continuous and increasing on either side of the origin. It's loss of uncertainty happens when the energy reaches the Plateau value of $A^2$, and it happens semiclassically, so it happens for large quantum numbers.
Semiclassically, in the Bohr-Sommerfeld (WKB) approximation, the particle has the same eigenfunctions as the harmonic oscillator, until the energy equals $A^2$. At this point, the next eigenfunction oscillates around the minimum, then crawls at a very very slow speed along the plateau, reflects off the wall, and comes back very very slowly to the oscillator.
The time spent on the plateau is much longer than the time spent oscillating (for appropriate choice of A and k) because the classical velocity on the plateau is so close to zero. This means that the position and momentum uncertainty is dominated by the uncertainty on the plateau, and the value of the position uncertainty is much less than the uncertainty for the oscillation if k is much smaller than A, and the value of the momentum uncertainty is nearly zero, because the momentum on the plateau is next to zero.
WKB expectation values are classical orbit averages
This argument uses the WKB expression for the expectation values of functions of x, which, from the WKB wavefunction,
$$\psi(x) = {1\over \sqrt{2T}} {1\over \sqrt{v}} e^{i\int^x p dx}$$,
Where v(x) is the classical velocity a particle would have at position x, and T is just a constant, a perverse way to parametrize the normalization constant of the WKB wavefunction. The expected value of any function of the X operator is equal to
$$\langle f(x)\rangle = \int |\psi(x)|^2 f(x) = {1\over 2 T} \int {1\over v(x)} f(x) dx = {1\over T}\oint f(x(t)) dt$$
Where the last integral is the integral around the full classical orbit. The last expression obviously works for functions of P (it works for any operator using the corresponding classical function on phase space). So the expectation value is just the average value of the quantity along the orbit, the factor of 2 disappears because you go over every x value twice along the orbit, and the strangely named normalization factor "T" is revealed to be the period of the classical orbit, because the average value of the unit operator is 1.
A: I will give a counter-example. Consider the potential below, $V$ is considered to be very large but not infinity. The ground state of the system should be approximate as the ground state of an infinite potential well. The uncertainty is greater than $\hbar/2$.
While when the particle is confined in the parabolic potential well, the local "ground state" corresponding to a Gaussian wave, the uncertainty is just $\hbar/2$.

