Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $vt+\sqrt{\frac{a+v_st}{2}}$.

share|improve this question
    
The short answer is yes. You may want to see a discussion of that in Introduction to Quantum Field Theory by Peskin et. al Chapter 2, section 1. –  Jorge Jan 15 '13 at 21:50
1  
I think the thing that you're missing is that QM can be formulated consistent with Galilean invariance or Lorentz invariance. In the former case, FTL is possible. In the latter, FTL is not. The former is not consistent with experiment. –  MarkWayne Jan 17 '13 at 1:00
    
Related: physics.stackexchange.com/q/48025/2451 –  Qmechanic Jan 27 '13 at 18:16
add comment

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
    
Please see the reformulated questions, I want to know about actual physical (experimental) QM, not what classical QM theories state or not. –  Juris Jan 16 '13 at 22:51
add comment

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.

share|improve this answer
    
So the front is actually not allowed to move FTL in relativistic QM? –  Juris Jan 16 '13 at 10:32
    
Sorry, I realized that by QM people understand certain theories, but my aim was to actually ask about actual (experimental?) quantum physics. –  Juris Jan 16 '13 at 22:58
3  
The experimental situation is very simple: superluminal signalling has never been observed despite ample opportunities for it to show up. Recently some people thought neutrinos could go faster than light but that was disproved. –  Michael Brown Jan 17 '13 at 13:33
    
Umm, superluminal signalling. I am not asking for true violation of SRT, just something similar to tunneling but without a barrier. There's no need to violate any theory as long as speed exceeds $c$ :D –  Juris Jan 19 '13 at 10:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.