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Let's consider a free system where the Hamiltonian is $\hat{p}^2/2m$.

At time $t=0$, we start with a state at position $x$. An instantaneous time $\delta t$ later, where $\delta t\rightarrow 0$, we measure the momentum of the particle, and obtain a value $k$. After the measurement, the wave function collapses to a momentum eigenstate $\psi(x)=e^{ikx}$.

Another interval $\delta t$ later, we measure the position of the particle. Since the particle is in a momentum eigenstate, the measurement can give any value ranging from $-\infty$ to $\infty$.

This seems to suggest that the particle is able to travel through an arbitrarily large distance within a small amount of time $2\delta t$. Does this mean wavefunction collapse violates the speed of light restriction?

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There are two answers to this question. The first is that, yes, in non-relativistic quantum mechanics, you can have things going faster than the speed of light, because relativity is never taken into account. The fix is to learn quantum field theory.


We can also consider this particular situation more closely. Your thought experiment suggests that a momentum measurement can "teleport" a particle infinitely far away in an infinitely short time, which feels unphysical, regardless of relativity.

The solution is that a precise momentum measurement takes a finite amount of time. (And an infinitely precise momentum measurement, as you're suggesting, takes infinitely long.)

To see this, start from the energy-time uncertainty principle $$\Delta E \Delta t \geq \hbar$$ where $\Delta E = \Delta (p^2/2m) = p \Delta p / m$. Then we have the bound $$\frac{p \Delta p \Delta t}{m} \geq \hbar$$ where $p$ is the (average) value of momentum you get, $\Delta p$ is the uncertainty on that momentum, and $\Delta t$ is the time it took to perform the measurement. This tells us that more precise momentum measurements take longer.

Now, the final state after this 'smeared' momentum measurement is a wavepacket centered on the origin with width $\Delta x$, with $$ \Delta x \Delta p \sim \hbar.$$ Finally, combining this with our other result gives $$ \frac{\Delta x}{\Delta t} \leq \frac{p}{m}.$$ That is, the particle is not moving any faster than it would be, semiclassically.

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  • $\begingroup$ How would quantum field theory fix this? Does quantum field theory describe collapse of wave function? $\endgroup$ – JNL Sep 11 '16 at 10:11
  • $\begingroup$ @JNL In a properly relativistic quantum field theory, space-like separated experiments have no influence on each other. One never needs to introduce the "collapse of the wavefunction", whatever that means. The $S$-matrix is sufficient. $\endgroup$ – Robin Ekman Sep 11 '16 at 11:29
  • $\begingroup$ Why the secon time you used Heisenberg principle did you write it with $\sim $ instead of $\geq $ as in the first case? That is the crucial point. I cannot see a physical justification for that. Without that difference you could not have reached a conlusion. $\endgroup$ – Valter Moretti Sep 11 '16 at 20:36
  • $\begingroup$ @ValterMoretti Yeah, this point is tricky. If we assume the momentum measurement error is gaussian, we have equality there. More generally I believe that for physical 'reasonable' error distributions (e.g. is positive on all of [p - \Delta p, p + \Delta p], is continuous, ...) then $\Delta x \Delta p$ can't be that much bigger than $\hbar$ but I don't know how to prove that. $\endgroup$ – knzhou Sep 11 '16 at 20:43
  • $\begingroup$ And what about the second measurement, that concerning position? Is it istantaneous? Frankly speaking, I think this way I mean with "ad hoc hipoteses" one may prove anything and its contrary.... $\endgroup$ – Valter Moretti Sep 11 '16 at 20:44

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