# Probability waves - instant infinite extent?

1. Nov 9, 2011

### ihatelolcats

Mathematically they are set up so the wave extends instantly to infinity in all directions, right? What would break if you limited the "propagation speed" to c, so if a particle is created at a point in space it would have 0 probability of being detected outside a sphere expanding at c away from that point?

2. Nov 9, 2011

### tommyli

There are two separate issues here.. one is local causality (whether or not information can be transmitted faster than light) and the other is the shape of the wavefunction.

As to the second point, there's a simple answer: it's true that the solution to Schrodinger's equation with no electrostatic potential is a constant wave that extends infinitely; any other function would not be a stationary state and would evolve over time. In reality, particles are created at some point in time with a range of energies, so the wavefunction is initially formed as a wave-packet with a "coherence length". Within this coherence length the particle behaves as a wave and can exhibit interference effects, but outside it behaves as a particle and reproduces classical behaviour. Since this wave-packet is not a stationary solution of the Schrodinger equation it will evolve over time the same way ordinary waves do so in a dispersive medium. Since the coherence length is limited by external "noise" (random fluctuations in the external potential), whether its other particles, or, in a material, defects or impurities, it is impossible to get a single wavefunction that extends infinitely through space.

For the first point, the situation is a little trickier. Speaking purely formally, the wavefunction can exist outside of the light cone, so even if you introduce a finite speed of light built-in to the theory, your particle will be able to break this spead limit with exponential probability. However this is purely an illusion... if you work in the Heisenberg picture, you find that all particle operators commute or anticommute at spacelike separations, and you prevent your theory from violating local causality.

3. Nov 9, 2011

### ihatelolcats

Thanks for the reply, the only thing that doesn't make sense is that you say "it is impossible to get a single wavefunction that extends infinitely through space" since it would take an infinite potential to stop a wavefunction completely.
I first thought about all this when CERN announced the superluminal neutrinos. But if I understand correctly, "time tunneling" of particles is excluded already because of the nature of the operators.

4. Nov 10, 2011

### tommyli

No probs! It's an interesting question. Your point is incorrect, suppose I have a particle with momentum p moving through empty space (there is no local potential). We cannot have a pure state with momentum p since the particle has not existed for infinite time, rather, its a superposition of waves with momenta concentrated within a band of width $$\Delta p$$. Such a wave propagates through space with some finite velocity which is group velocity of the wave (v = dE/dp), but it is localized within a region of space of size $$\Delta x = \frac{\hbar}{\Delta p}$$ (This is the Heisenberg uncertainty relation). The wave absolutely does not extend infinitely through space.. and this is totally consistent with our usual observation of electrons in metals for example.. electrons are delocalized, but even when you get to the superconducting regime, the wavefunction of an electron does not extend beyond some typical length which is much much smaller than the size of the sample.