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?
Best Answer
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.