# Is there a limit on the maximum rate of change of the position expectation value in quantum mechanics to be no greater than the speed of light?

Suppose a particle, constrained to the $$x$$ axis, is measured at $$t_0$$ to be at position eigenstate $$x = 0$$.

Assume for all $$t \gt t_0$$, some external potential acts on the particle.

Nevertheless, at $$t_1 \gt t_0$$, the support of its position wave function must be an interval no longer than $$[-c(t_1 - t_0), c(t_1 - t_0)]$$ in order to satisfy special relativity.
(Here, a position wave function $$\psi$$ is such that $$\psi \psi^*$$ is the position probability density.)

Further, suppose:
At $$t_1$$, the position expectation value is $$-c(t_1 - t_0)\lt x_{e_1} \lt c(t_1 - t_0)$$
At $$t_2 \gt t_1$$ the position expectation value is $$-c(t_1 - t_0) \lt x_{e_2} \lt c(t_1 - t_0)$$ and $$x_{e_2} \ne x_{e_1}$$

Also, assume there is no position measurement, after the one at $$t_0$$, until sometime after $$t_2$$.

Can we have $$|(x_{e_2} - x_{e_1}|/(t_2 - t_1) \gt c$$?

I think that the answer is yes, because the rate of change of the expectation value is not the same as the rate of change of measured values of position, the latter rate of change not being able to exceed $$c$$.

• I think the only consistent relativistic quantum mechanics is relativistic quantum field theory, which doesn’t have a position operator or position wave functions. – G. Smith Aug 3 '19 at 3:43
• thinking more like the Dirac equation and if that has any implications regarding how fast a position expectation value can travel. I know that the Dirac equation is not perfect, but wondering if it implies anything about this topic. – user3536736 Aug 3 '19 at 4:57
• Expectation values of position operators don't travel. Can you be more specific what you mean by that? (What travels is an object or information, not an expectation value.) – Norbert Schuch Aug 3 '19 at 15:04
• Note that the statement you expect is much stronger: If you put a particle at position $x_0=0$ and let it evolve (with a local Hamiltonian, not just "unitary evolution"), then it the probability to find it outside of $|x|\le ct$ is zero. Thus, the expectation value of the particle's position must be in the same range. – Norbert Schuch Aug 3 '19 at 15:15
• It is unclear how you want to prepare a particle in a Gaussian state without acting everywhere. If you only act on a finite spatial region, your particle will be supported in that region. – Norbert Schuch Aug 3 '19 at 19:43

Let the system be at time $$t_1$$ in a state $$\psi(x)$$.
Decompose this state as a superposition of position eigenstates: $$\psi(x,t_1) = \sum_y a_y \delta(y-x)\ .$$ The position expectation value at time $$t_1$$ is thus $$e_1=\sum y |a_y|^2$$.
Further, any of the position eigenstates can at most evolve with the speed of light, this is, if $$\delta(x-x_0)$$ after time $$t_2-t_1$$ evolved into $$\phi_{x_0}(x,t_2) = \sum_y b_{y,x_0} \delta(y-(x-x_0))\ ,$$ we have $$\big|\sum_y y |b_{y,x_0}|^2\big|\le c|t_2-t_1|$$. Note that moreover, $$\sum_y |a_y|^2 = \sum_y y |b_{y,x_0}|^2 = 1$$.
We can now combine the two expressions due to linearity: At time $$t_2$$, we have \begin{align} \psi(x,t_2) &= \sum_z a_z\phi_{z}(x,t_2)\\ & = \sum_{y,z} a_z b_{y,z} \delta(y-(x-z))\ . \end{align} Thus, the position expectation value is \begin{align} |e_2-e_1| &= \Big|\sum_{y,z} (y+z) |a_z|^2 |b_{y,z}|^2 -\sum_z z |a_z|^2 \Big| \\ &= \Big|\sum_{y,z} (y+z) |a_z|^2 |b_{y,z}|^2 -\sum_{y,z} z |a_z|^2 |b_{y,z}|^2\Big| \\ & = \Big|\sum_{y,z} y |a_z|^2 |b_{y,z}|^2 \Big| \\ & \le \sum_{z} |a_z|^2 \Big| \sum_y y |b_{y,z}|^2 \Big| \\ & \le \sum_{z} |a_z|^2 \ c|t_2-t_1| = c|t_2-t_1|\ . \end{align}