# Do wave probabilities violate the speed of light?

1. Aug 11, 2010

### eloheim

I understand that Shrodinger's wave equation describes the probability that a particle will be found at any certain location in space upon its observation. I'm wondering if these 'movements' adhere to the speed limit of light (and if so, why is this okay).

For example, if a particle is observed right 'here', is it possible that it will show up half-way across the universe a split second later? I'm wondering because doesn't the probability wave theoretically extend out infinitely? Could the answer have something to do with the fact that the particle isn't actually moving through space (like how space itself can expand as fast as it wants)?

Thanks for you time.
Jeff.

2. Aug 12, 2010

### Demystifier

This is not possible because the measurement modifies (localizes) the wave function. In older literature it is usually described as wave-function collapse, but in modern physics it is better understood in terms of decoherence. Yet, even decoherence by itself does not provide a complete explanation. Instead, decoherence must be combined with some interpretation of QM, such as many-world or Bohmian interpretation.

3. Aug 12, 2010

### vanhees71

Why do you need some "esoterics" to "interpret" QM? Theoretiscal physics is about the description of objectively observable phenomena in nature, and quantum theory with the Minimal Statistical Interpretation does precisely this, no more no less, with great success. I never understood, why I should accept unobservable elements in the interpretation of the theory that are not necessary to make predictions about observable quantities. I don't need unobservable orbits as in Bohm's theory (which by the way has some troubles with relativistic quantum field theory) or unobservable "parallel universes" as in the many-worlds interpretation.

4. Aug 12, 2010

### eloheim

Yea but an instant later isn't it back to behaving wavelike again? Is it a problem of knowing which is actually 'that' particle or something?

5. Aug 12, 2010

### dreiter

Once you measure an observable (velocity, momentum, etc.) then you are defining a particle to a very specific state, so the probability wave now looks like a delta function, with the peak spiking at your measurement. Now it's true that once that measurement is done, the wave begins to flatten out again, but it doesn't happen instantly. So you cannot measure the location of a particle, then right afterwards measure it to be a million miles away. It takes time for the probability wave to spread...

6. Aug 12, 2010

### vanhees71

A single particle neither behaves like a classical particle nor like a classical wave. The modulus squared, $$|\psi(t,x)|^2$$ of the wave function, $$\psi(t,x)=\langle x|\psi(t) \rangle$$, describes the probability distribution for the position of the particle.

If you have prepared the particle at time, $$t_0$$, to sit at a position, $$x_0$$, with high accuracy, the proability distribution is a sharply peaked function around $$x_0$$ with its momentum around $$p_0$$ also with a certain precision (where the precision of position and momentum of course have to obey the Heisenberg-uncertainty relation $$\Delta x \Delta p \geq \hbar/2$$).

If you let the particle run freely, at a later time, the particle's state is given by a wave packet that runs roughly with velocity $$p_0/m$$. At the same time the width of the particle becomes larger with time (since the dispersion relation E=p^2/(2m) is not linear). This non-relativistic description is of course valid only for small velocities $$p_0/m \ll c$$, and since the probability to find the particle at a far distance from $$x_0+p_0 t/m$$ is very low, there's only a negligible violation of Einstein causality which is due to the nonrelativistic approximation made here.

For a fully relativistic treatment, one has to use QFT. In the most simple cases, one can use linear response theory to prove that observable local quantities run with the retarded propagator defined by the appropriate field operators, and this means automatically that Einstein causality is fulfilled precisely, i.e., the front velocity of the probability wave runs at most with the speed of light in vacuum.

7. Aug 13, 2010

### Demystifier

For example, let me quote Einstein:
"I want to know how God created this world. I am not interested in this or that phenomenon, in the spectrum of this or that element. I want to know His thoughts; the rest are details."

http://xxx.lanl.gov/abs/1007.4946

Last edited: Aug 13, 2010
8. Aug 13, 2010

### Demystifier

Yes, but that new "collapsed" wave function vanishes at large distances from the position of detection.

9. Aug 13, 2010

### Demystifier

If you are a pragmatic physicist, you don't need to accept anything like that. Just like a pragmatic user of TV does not need to know anything about electronics. Of course, a pure pragmatist cannot answer some questions, but from his point of view such questions are irrelevant.

Now you will say: But there is a difference; the existence of electronic elements is proved while that of parallel worlds or Bohmian trajectories isn't. Sure, but if the existence of something is not proved, that is not a reason for not doing research on it. If there are reasonable theoretical reasons that something not yet observed may still exist, it should still be researched. After all, that's why LHC has been built.

In addition, even if various interpretations of QM are not taken seriously, they still may help to have some intuition about otherwise counterintuitive quantum phenomena. And there is no doubt that intuition may have practical merits. For example, Bohmian interpretation helped Bell to discover quantum nonlocality, while many-world interpretation helped Deutch and others in development of the theory of quantum computers.

Finally, let me quote Feynman, who is usually considered to be a practical physicist:
"Physics is like sex: sure, it may give some practical results, but that's not why we do it."

Last edited: Aug 13, 2010
10. Aug 13, 2010

### eloheim

Thanks for taking the time to reply. I'm think I'm at least getting the jist of what you are saying (the answer to my question being, "no"). Later on, after writing the original post, I thought the answer might have something to do with the relationship between quantum mechanics and relativity.

11. Aug 13, 2010

### The_Duck

Note that the Schrodinger equation doesn't respect special relativity, so it's only a low-energy approximation. Relativity + quantum mechanics gives http://en.wikipedia.org/wiki/Quantum_field_theory" [Broken].

Last edited by a moderator: May 4, 2017