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If the support of a quantum mechanical position wave function is a bounded interval, and that interval is expanding or contracting, then I think it cannot change in any direction faster than $c$. To clarify, Consider the following example:

If a one dimensional position wave function (which is not acted upon by any internal or external potential) is measured at $t_0$ and the result is position eigenstate $x = 0$ and then, at $t_0 + \Delta t$, we perform a momentum measurement, resulting in a momentum eigenstate $p_{t_0 + \Delta t}$, then at $t \ge (t_0 + \Delta t)$, the position wave function is a plane wave and the length of the interval of its support cannot excede $(|p_{t_0 + \Delta t}|( (t - (t_0+ \Delta t)) + c \Delta t)$, which cannot excede $c( t - t_0)$.  

My question is, if the above is true (which I would think it is), then if the support of a position wave function is a bounded interval, does the Klein-Gordon equation, or Dirac equation, limit the speed of the expansion or contraction of that interval, in any direction, to being no faster than $c$?

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  • $\begingroup$ I probably could have simplified my example: if a one dimensional position wave function (which is not acted on by any internal or external potential) is measured at $t_0$ and the result is position eigenstate $x=0$ then, at $t \ge t_0$, the momentum is completely unknown but the support of the position wave function, for $t \gt t_0$, can be no greater than the interval $[-c(t - t_0), c(t-t_0)]$ and the interval grows in each direction ($+$ and $-$ directions) at a rate that is no greater than $c$. $\endgroup$ – David Aug 4 at 8:23
  • $\begingroup$ For a solution $\psi(x)$ of the Klein-Gordon or Dirac equation, $|\psi(x)|^2$ doesn't represent the probability density for detecting a particle at $x$. The question seems to assume that it does. The appropriate framework for relativistic quantum physics is relativistic quantum field theory (QFT), in which a "solution" of the Klein-Gordon or Dirac equation is a field operator that acts on states, and local observables are constructed from field operators. $\endgroup$ – Chiral Anomaly Aug 4 at 14:49
  • $\begingroup$ In relativistic QFT, a single particle cannot be localized in a strictly bounded region of space. One way to see this is via the Reeh-Schlieder theorem, as I reviewed in my answer to What's the physical meaning of the statement that “photons don't have positions”?. For more quantitative detail, see the answers to The concept of particle in QFT and Getting particles from fields: normalization issue or localization issue? $\endgroup$ – Chiral Anomaly Aug 4 at 14:50

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