Does quantum mechanics allow faster than light (FTL) travel? Let's suppose I initially have a particle with a nice and narrow wave function[1] (I will leave these unnormed):
$$e^{-\frac{x^2}{a}}$$
where $a$ is some small number (to make it narrow). Let's also suppose that the wave function travels along x axis with constant speed $v$ and it smears out with constant speed $v_s$. So after time $t$ wave function looks like
$$e^{-\frac{(x-vt)^2}{a+v_st}}$$
Now, if speed $v$ is close to $c$ and $v_s$ is big enough too, we could get a situation where the centre of mass moves at subluminal speed (this is the group velocity, I suppose), but the front[2] of the function moves at superluminal speed.
If I measure either the particle's position after some time $T$ or time when it reaches some point $A$ i will mostly and averagely conclude that it has travelled with speed $v$. But in some (less probable) cases it will seem that it has travelled faster, even FTL. Is this a normal thing in quantum mechanics or I understand it all wrong? Maybe there is some constraint not only on $v$, but also on $v+v_s$?
A little background
When talking about superluminal tunneling speeds I usually hear explanations like this:

That speed is apparent. In those cases only some frontal part of the wave function gets through the barrier. Although it's center of mass seems to have travelled FTL, that function would still stay under the initial wave function if it continued it's way without the barrier. The center just shifted because of dropping the rear part.

I have never actually understood why does it solve the problem, because the position of center doesn't change the fact that signal in some cases may arrive FTL. This is where my question comes from.
EDIT:
Sorry, the question was not about what some classic QM models allow, but actual quantum physics. I understand that Schrodinger's equation allows me any speed, I want to know if this situation is allowed in actual quantum physics. I guess the use of term QM was wrong. Sorry, my bad.
Question reformulated
Is it possible to send a particle that's slowly spreads and with some small chance (in some rare cases) measure it arriving FTL? (provided that the mean arrival value still stays under $c$).
I am aware that this is possible when tunneling through a barrier (and the mean value can even move to superluminal by train dropping cars), but is the same possible without any barrier?
Maybe I could equivalently ask, if quantum uncertainties allow occasional exceeding of $c$.
[1] If you say that I had some chance to measure it at any point in the very beginning and that it wasn't entirely localized, we can replace the Gauss' function with a square or triangle one that gets wider along the way.
[2] If you need, we can define the front of the wave, for example first of the points where second derivative is zero. It's position is $vt+\sqrt{\frac{a+v_st}{2}}$.
 A: Non-relativistic quantum mechanical theories allow disturbances to propagate at arbitrarily high velocities.  This has nothing to do with the adjective 'quantum'; the same thing is true of non-relativistic classical mechanical models.  
Relativistic quantum theories impose an additional constraint:   Operators representing observables commute when they describe measurements which occur in space-like separated regions of spacetime.   This constraint prevents information from propagating faster than light.
A: We don't expect the solutions of the Schroedinger equation to furnish representations of the Lorentz group since its invariant only under a the group of Galilean transformations, for example:
\begin{align}
x' &= x - vt\\
y' &= y\\
z' &= z\\
t' &= t.
\end{align}
It's a good -- and very easy -- exercise to check that the Schroedinger equation
\begin{align}
-\frac{\hbar^2}{2m}\nabla^2 \psi(t,\mathbf{r}) + V(\mathbf{r})\psi(t,\mathbf{r}) = i\frac{\partial\psi(t,\mathbf{r})}{\partial t}
\end{align}
for a single particle in an external, time-independent potential is indeed invariant under the Galilean group.
Now, to show that invariance under (only) Galilean group implies that the speed of light isn't a constant irrespective of the motion of the observer (and therefore allows faster-than-light propagation) you would next show that Maxwell's equations are not invariant under Galilean transformation. This, in fact, was one of the tip-offs to Einstein that resulted in the 'light postulate' ($c$ is a constant for all (inertial) observers).
But this isn't the whole story, as previous answers (which are correct) have indicated. What we've done so far, with the Schroedinger equation, is just ask what are the consequences of treating quantum evolution (the Schroedinger equation governs this evolution) with classical, Galilean invariance. Since 1905, the species has been aware that Maxwell's equations support Lorentz invariance, using the same constant, $c$ for the speed of light in any Lorentz frame. The consequences of this observation for quantum evolution of dynamical variables is the subject of quantum field theory.
So the answer to the title question is that QM doesn't make a statement about the propagation of phenomena by faster-than-light signals until one talks about the symmetry group of spacetime -- Galilean or Lorentz. Of course, Lorentz invariance and quantum field theory are forced upon us by experiment. So there's that to consider.
A: Your intuition is correct. You can count it as FTL or alternatively view it as a formation of virtual particle-anti-particle pair from vacuum ahead of the propagating particle with the anti-particle later annihilating with the original one.
As such, you cannot beforehand differentiate such propagating signal from vacuum fluctuation and as such the information cannot be transferred.
