Does Heisenberg's uncertainty under time evolution always grow? Recently there have been some interesting questions on standard QM and especially on uncertainty principle and I enjoyed reviewing these basic concepts. And I came to realize I have an interesting question of my own. I guess the answer should be known but I wasn't able to resolve the problem myself so I hope it's not entirely trivial.
So, what do we know about the error of simultaneous measurement under time evolution? More precisely, is it always true that for $t \geq 0$
$$\left<x(t)^2\right>\left<p(t)^2\right> \geq \left<x(0)^2\right>\left<p(0)^2\right>$$
(here argument $(t)$ denotes expectation in evolved state $\psi(t)$, or equivalently for operator in Heisenberg picture).
I tried to get general bounds from Schrodinger equation and decomposition into energy eigenstates, etc. but I don't see any way of proving this. I know this statement is true for a free Gaussian wave packet. In this case, we obtain equality, in fact (because the packet stays Gaussian and because it minimizes HUP). I believe this is, in fact, the best we can get and for other distributions, we would obtain strict inequality.
So, to summarize the questions

  
*
  
*Is the statement true?
  
*If so, how does one prove it? And is there an intuitive way to see it is true?
  

 A: No. Here's a simple example where it shrinks:
You have a particle that has a 50% chance of being on the left going right, and a 50% chance of being on the right going left. This has a macroscopic error in both position and momentum. If you wait until it passes half way, it has a 100% chance of being in the middle. This has a microscopic error in position. There will also only be a microscopic change in momentum. (I'm not entirely sure of this as the possibilities hit each other, but if you just look right before that, or make them miss a little, it still works.)
As such, the error in position decreased significantly, but the error in momentum stayed about the same.
A: The question asks about the time dependence of the function
$$f(t) := \langle\psi(t)|(\Delta \hat{x})^2|\psi(t)\rangle
 \langle\psi(t)|(\Delta \hat{p})^2|\psi(t)\rangle,$$
where 
$$\Delta \hat{x} := \hat{x} - \langle\psi(t)|\hat{x}|\psi(t)\rangle, \qquad
\Delta \hat{p} := \hat{p} - \langle\psi(t)|\hat{p}|\psi(t)\rangle, \qquad
\langle\psi(t)|\psi(t)\rangle=1.$$
We will here use the Schrödinger picture where operators are constant in time, while the kets and bras are evolving. 
Edit: Spurred by remarks of Moshe R. and Ted Bunn let us add that (under assumption (1) below) the Schroedinger equation itself is invariant under the time reversal operator $\hat{T}$, which is a conjugated linear operator, so that 
$$\hat{T} t = - t \hat{T}, \qquad \hat{T}\hat{x} = \hat{x}\hat{T}, \qquad \hat{T}\hat{p} = -\hat{p}\hat{T}, \qquad  \hat{T}^2=1.$$
Here we are restricting ourselves to Hamiltonians $\hat{H}$ so that
$$[\hat{T},\hat{H}]=0.\qquad (1)$$
Moreover, if 
$$|\psi(t)\rangle = \sum_n\psi_n(t) |n\rangle$$
is a solution to the Schrödinger equation in a certain basis $|n\rangle$, then 
$$\hat{T}|\psi(t)\rangle := \sum_n\psi^{*}_n(-t) |n\rangle$$
will also be a solution to the Schrödinger equation with a time reflected function $f(-t)$.
Thus if $f(t)$ is non-constant in time, then we may assume (possibly after a time reversal operation) that there exist two times $t_1<t_2$ with $f(t_1)>f(t_2)$. This would contradict the statement in the original question. To finish the argument, we provide below an example of a non-constant function $f(t)$.
Consider a simple harmonic oscillator Hamiltonian with the zero point energy $\frac{1}{2}\hbar\omega$ subtracted for later convenience.
$$\hat{H}:=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^2
-\frac{1}{2}\hbar\omega=\hbar\omega\hat{N},$$
where $\hat{N}:=\hat{a}^{\dagger}\hat{a}$ is the number operator.
Let us put the constants $m=\hbar=\omega=1$ to one for simplicity. Then the annihilation and creation operators are
$$\hat{a}=\frac{1}{\sqrt{2}}(\hat{x} + i \hat{p}), \qquad
\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{x} - i \hat{p}), \qquad
[\hat{a},\hat{a}^{\dagger}]=1,$$
or conversely,
$$\hat{x}=\frac{1}{\sqrt{2}}(\hat{a}^{\dagger}+\hat{a}), \qquad
\hat{p}=\frac{i}{\sqrt{2}}(\hat{a}^{\dagger}-\hat{a}), \qquad 
[\hat{x},\hat{p}]=i,$$
$$\hat{x}^2=\hat{N}+\frac{1}{2}\left(1+\hat{a}^2+(\hat{a}^{\dagger})^2\right), \qquad 
\hat{p}^2=\hat{N}+\frac{1}{2}\left(1-\hat{a}^2-(\hat{a}^{\dagger})^2\right).$$
Consider Fock space $|n\rangle := \frac{1}{\sqrt{n!}}(\hat{a}^{\dagger})^n |0\rangle$ 
such that $\hat{a}|0\rangle = 0$. Consider initial state 
$$|\psi(0)\rangle := \frac{1}{\sqrt{2}}\left(|0\rangle+|2\rangle\right), \qquad 
\langle \psi(0)|  = \frac{1}{\sqrt{2}}\left(\langle 0|+\langle 2|\right).$$
Then 
$$|\psi(t)\rangle = e^{-i\hat{H}t}|\psi(0)\rangle
 = \frac{1}{\sqrt{2}}\left(|0\rangle+e^{-2it}|2\rangle\right),$$
$$\langle \psi(t)| = \langle\psi(0)|e^{i\hat{H}t}
 = \frac{1}{\sqrt{2}}\left(\langle 0|+\langle 2|e^{2it}\right),$$
$$\langle\psi(t)|\hat{x}|\psi(t)\rangle=0, \qquad 
\langle\psi(t)|\hat{p}|\psi(t)\rangle=0.$$
Moreover,
$$\langle\psi(t)|\hat{x}^2|\psi(t)\rangle=\frac{3}{2}+\frac{1}{\sqrt{2}}\cos(2t), \qquad 
\langle\psi(t)|\hat{p}^2|\psi(t)\rangle=\frac{3}{2}-\frac{1}{\sqrt{2}}\cos(2t),$$
because $\hat{a}^2|2\rangle=\sqrt{2}|0\rangle$. Therefore,
$$f(t) = \frac{9}{4} - \frac{1}{2}\cos^2(2t),$$
which is non-constant in time, and we are done. Or alternatively, we can complete the counter-example without the use of above time reversal argument by simply performing an appropriate time translation $t\to t-t_0$.
A: Marek,
Think in terms of Harmonic Functions and their Maximum Principle (or Mean Value Theorem).
For simplicity (and, in fact, without loss of generality), let's just think in terms of a free particle, ie, $V(x,y,z) = 0$. When the Potential vanishes, the Schrödinger equation is nothing but a Laplace one (or Poisson equation, if you want to put a source term). And, in this case, you can apply the Mean Value Theorem (or the Maximum Principle) and get a result pertaining your question: in this situation you saturate the equality.
Now, if you have a Potential, you can think in terms of a Laplace-Beltrami operator: all you need to do is 'absorb' the Potential in the Kinetic term via a Jacobi Metric: $\tilde{\mathrm{g}} = 2\, (E - V)\, \mathrm{g}$. (Note this is just a conformal transformation of the original metric in your problem.) And, once this is done, you can just turn the same crank we did above, ie, we reduced the problem to the same one as above. ;-)
I hope this helps a bit.
A: A physical way of seeing this is that the phase space volume of a system is preserved.  Hamiltonian mechanics preserves the volume of a system on its energy surface H = E, which in quantum mechanics corresponds to the Schrodinger equation.  The phase space volume on the energy surface of phase space is composed of units of volume $\hbar^{2n}$ for the momentum and position variables plus the $\hbar$ of the energy $i\hbar\partial\psi/\partial t~=~H\psi$.  This is then preserved.  Any growth in the uncertainty $\Delta p\Delta q~=~\hbar/2$ would then imply the growth in the phase space volume of the system.  This would then mean there is some dissipative process, or the quantum dynamics is replaced by some master equation with a thermal or environmental loss of some form.  For a pure unitary evolution however the phase space volume of the system, or equivalently the $Tr\rho$ and $Tr\rho^2$ are constant.  This means the uncertainty relationship is a Fourier transform between complementary observables which preserve an area $\propto~\hbar$.
A: The Schrodinger equation is time-symmetric. The answer is therefore no.
A: The Heisenberg Uncertainty Principle revolves around the Compton Effect which states that wavelength (w) is inversely proportional to E, p, and f. 
How ever if one gathers up several quantum physics formulas, one can create these 4 formulas and use a form of geometry to solve the Uncertainty Principle but finding the electron's(e-) location and momentum at the same time while being able to approximate where it's presence was, is, and will be. 
[Formulas are of my creation]


*

*((h(c/wi)-h(c/wf))/c^2)+m(e-)=m (e- post photon collision)

*h(c/wi)-h(c/wf)=E(nergy expelled from photon when it collided with the electron)
3.((h(c/wi)-h(c/wf))/c)=p

*(h(c/wi)-h(c/wf)+m(e-)c^2=total e- E


Using a 60/30 rule, one can find the angle that the electron will be launched to and where it was before the photon collided with it by using the original trajectory and the current to find where it intersects. This will allow you to find its location.
If you disagree with this, please tell why so that I can improve it. 
The momentum formula: 
The energy expelled from the photon upon collision: 
