14

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?

valerio
  • 16,231
JNL
  • 393

1 Answers1

11

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.

knzhou
  • 101,976
  • How would quantum field theory fix this? Does quantum field theory describe collapse of wave function? – JNL Sep 11 '16 at 10:11
  • @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. – Robin Ekman Sep 11 '16 at 11:29
  • 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. – Valter Moretti Sep 11 '16 at 20:36
  • @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. – knzhou Sep 11 '16 at 20:43
  • 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.... – Valter Moretti Sep 11 '16 at 20:44
  • @ValterMoretti Position measurements don't matter for this argument. I'm just trying to show that measurements can't "teleport" particles significantly further than they could move semiclassically. We already know a position measurement can't do this because it can only yield values where $\psi(x)$ is already nonzero. – knzhou Sep 11 '16 at 20:47
  • I think it is a desperate task yours. Schroedinger equation itself is non local as heat equation is. You may have the wave confined in a bounded region (strictly vanishing outside it) at the initial time 0 and, for every arbitrarily small positive time, the wavefunction vanishes nowhere. This without require any mesurement process... – Valter Moretti Sep 11 '16 at 20:53
  • However, I am not able to deal with these, say, experimental physicist's approaches, maybe you are right. Don't worry :) – Valter Moretti Sep 11 '16 at 20:56
  • @ValterMoretti I think my argument explains why that can happen: there are arbitrarily high momentum components in the initial wavefunction, and those components then propagate outwards at velocity $v_g = p/m$. Since $p$ is unbounded, $v_g$ is too. – knzhou Sep 11 '16 at 20:58
  • Yes, I agree with you. – Valter Moretti Sep 11 '16 at 21:08
  • @Robin Ekman It seems to me that it is not one never needs to introduce the collapse of the wavefunction, but one cannot. The $S$-matrix calculate $\langle\psi_f|S|\psi_i\rangle$ is kind of like average of the measurement, but not collapse of the wavefunction. In path integral formalism, one integrates the intermediate momentum up to infinity. This seems to suggest that for each possible path, the speed of light may be violated (and the unbounded momentum gives rise to the divergence in the theory). – JNL Sep 11 '16 at 23:44
  • 2
    How do you know $\Delta t$ is the time it takes to perform the measurement? I have read about the time-energy uncertainty principle and it seems that the interpretation of the principle itself is rather uncertain.. – doublefelix Nov 27 '17 at 16:13
  • "in non-relativistic quantum mechanics, you can have things going faster than the speed of light," This statement is totally wrong. In non-relativistic quantum mechanics everything moves much slower than the speed of light. – my2cts Mar 18 '21 at 22:54