Thought experiment in relativistic quantum mechanics?

Question

Background

Consider the following thought experiment in the setting of relativistic quantum mechanics (not QFT). I have a particle in superposition of the position basis:

$$ H | \psi \rangle = E | \psi \rangle$$

Now I suddenly turn on an interaction potential $H_{int}$ localized at $r_o = (x_o,y_o,z_o)$ at time $t_o$:

$$ H_{int}(r) = \begin{cases} k & r \leq r_r' \\ 0 & r > r' \end{cases} $$

where $r$ is the radial coordinate and $r'$ is the radius of the interaction of the potential with origin $(x_o,y_o,z_o)$

By the logic of the sudden approximation out state has not had enough time to react. Thus the increase in average energy is:

$$ \langle \Delta E \rangle = 4 \pi k \int_0^{r'} |\psi(r,\theta,\phi)|^2 d r $$

(assuming radial symmetry).

Now, lets say while the potential is turned on at $t_0$ I also perform a measurement of energy at time $t_1$ outside a region of space with a measuring apparatus at some other region $ (x_1,y_1,z_1)$. Using some geometry it can be shown I choose $t_1 > t_0 + r'/c$ such that:

$$ c^2(t_1 - t_0 - r'/c)^2 -(x_1 - x_0)^2 - (y_1 - y_0)^2 - (z_1 - z_0)^2 < 0 $$

Hence, they are space-like separated. This means I could have one observer who first sees me turn on the potential $H_{int}$ and measure a bump in energy $\langle \Delta E \rangle $ but I could also have an observer who sees me first measure energy and then turn on the interaction potential.

Obviously the second observer will observe something different.

Question

How does relativistic quantum mechanics deal with this paradox?

Answer 1

Any realistic observable is local, since any experiment will take place in a finite region of spacetime. One often doesn't need to worry about this in non-relativistic QM, but in relativistic QM (ie, QFT) it is crucial.

Let's denote an observable that takes place at the spacetime point $x$ as $\mathcal{O}(x)$. Then a basic postulate of QFT is that, if $x$ and $y$ are spacelike separated points, for any two observables $\mathcal{O}_1(x)$ and $\mathcal{O}_2(y)$, we have

\begin{equation} [\mathcal{O}_1(x),\mathcal{O}_2(y)]=0, \ \ \{x,y\}\ {\rm spacelike\ separated} \end{equation}

In other words, we can simultaneously diagonalize $\mathcal{O}_1(x)$ and $\mathcal{O}_2(y)$. We can expand the whole state into a product of a superposition of $\mathcal{O}_2(y)$ eigenstates, times a superposition of $\mathcal{O}_1(x)$ eigenstates (times superpositions for other spacetime points). Because of this structure, if the superposition over $\mathcal{O}_2(y)$ eigenstates collapses to a single state, this does not mean there is any collapse of the superposition over $\mathcal{O}_1(x)$ eigenstates.

In your example, a sudden change to the Hamiltonian at $y$, will not cause any changes to observables at $x$. Therefore, no local observers will be able to measure changes in the state due to events happening at spacelike separated intervals.