Heisenberg uncertainty and Lorentz contraction Consider a particle in a frame moving with speed $v$ relative to the lab frame. By Lorentz contraction, the width of the wavefunction will be smaller in the lab frame, resulting in smaller $\Delta x$. If $v$ is high enough, then the uncertainty principle $\Delta x \Delta p \ge \hbar/2$ will be violated in the lab frame.
What's wrong here? Does $\Delta p$ increase somehow? This seems unlikely, since simply translating momentum distribution by a constant should not alter the standard deviation. 
 A: You're trying to apply relativity (Lorentz contraction) to a result from nonrelativistic quantum mechanics (Heisenberg uncertainty), so of course you get a contradiction.
In nonrelativistic quantum mechanics, the effect of a boost is given by the Galilean transformation
$$\psi(x) \to \exp((im/\hbar) (vx + v^2t/2))\, \psi(x-vt)$$
as explained in more detail here. You can verify by explicit calculation that this shifts position eigenvalues as
$$x \to x-vt$$
and shifts momentum eigenvalues by
$$p \to p-mv$$
as expected classically. Since both position and momentum are simply shifted, $\Delta x \Delta p$ stays the same, and the uncertainty principle is preserved.
A: With apologies for the many typos (and worse) in the first version of this answer:
Write the wave function as $f(x)$ in the comoving frame.  Then in the lab frame, the wave function is $g(x)=\sqrt{a}f(ax)$ where $a$ is some positive constant.
Write $\hat{f}(x)$ for the Fourier transform of $f$. Then $\hat{g}(x)=\hat{f}(x/a)/\sqrt{a}$.
The change in frame changes the variance of position from $\int x^2 |f(x)|^2$ to $\int x^2 |g(x)|^2$, which means the variance is multiplied by $1/a^2$.  (Check this by substituting $u=a x$ in the second integral.)
The change in frame changes the variance of momentum from $\int x^2|\hat{f}(x)|^2$ to $\int x^2 |\hat{g}(x)|^2$, which means the variance is multiplied by $a^2$. (Check this by substituting $u=x/a$ in the second integral.)
The product of the variances is therefore unchanged.
While I hope the above is enlightening, it's really unnecessary. The key is that $f$ is some arbitrary wave function and $g$ is some other wave function. Some argument must have convinced you that $f$ satisfies the uncertainty principle in the first place.  Whatever that argument is, it applies equally well to $g$.  
