Does collapse of wave function to a momentum eigenstate violate speed of light restriction?

Question

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?

Answer 1

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.

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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.