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I know that such events would cancel out in the math, but if an extreme event were to happen in the future (say a black hole forming or something on that par), would a particle in the present react to it? If not, why?

Qmechanic
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Marc DiNino
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1 Answers1

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The path integral is a broad idea, which comes in several different flavors. In non-relativistic quantum mechanics for one particle, you calculate the propagator matrix element $\langle {\bf x}_f| U(t_f,t_i) | {\bf x}_i \rangle$ by summing $\exp(i\, S[{\bf x}(t)]\, /\hbar)$ over paths connecting $(t_i, {\bf x}_i)$ to $(t_f, {\bf x}_f)$ which travel forward in time, so there's clearly no retrocausality. In relativistic quantum field theory, you typically use the LSZ formalism, which involves integrating $\exp(i S[\varphi(x)] / \hbar)$ with $S[\varphi(x)] := \int_{t_i}^{t_f} dt\, \int d^3 {\bf x}\, \mathcal{L}(\varphi(x), \partial_\mu \varphi(x); x)$ in the limit $t_i \to -\infty$ and $t_f \to +\infty$, so that the incoming and outgoing particles are thought of as far-separated and asymptotically noninteracting. (In practice, this limit is carried out through the use of the "$i\epsilon$ trick" in the denominator of the propagator, which sets the boundary conditions.) In neither case are paths that extend later than $t_f$ considered, so there is no retrocausality. (You do find acausal correlations across spacelike separations, but they cannot transmit acausal influences.)

tparker
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    "You do find acausal correlations across spacelike separations, but they cannot transmit acausal influences." Cool! Where could someone read about that? – DanielSank Sep 06 '18 at 21:18
  • @DanielSank Considering scalar field theory for simplicity: the precise statement is that unlike commutators of quantum fields, the Feynman propagator, which is the Green's function for the Klein-Gordon equation with the boundary conditions relevant for calculating correlation functions in relativistic QFT (an implicit $+i \epsilon $ in the denominator with $\epsilon \to 0^+$), does not vanish over spacelike separations, although it turns out to decay exponentially quickly with proper distance. – tparker Sep 06 '18 at 21:26
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    @DanielSank This is why people say that at a given instant, the ground state of a relativistic QFT is spatially entangled: $\langle \varphi(x) \varphi(y) \rangle - \langle \varphi(x) \rangle \langle \varphi(y) \rangle \neq 0$ for spacelike separated $x$ and $y$. But it turns out that it's the expectation value of the commutator of fields, not the simple product, that determines whether causal influences can be transmitted, and this expectation value is zero for spacelike separations. The correlations are the usual EPR/Bell's theorem type correlations that can't be used to communicate. – tparker Sep 06 '18 at 21:28
  • @DanielSank I think is discussed in most QFT books when they derive the Feynman propagator. – tparker Sep 06 '18 at 21:29
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    That is bonkers. Thanks for the clear and concise explanation. – DanielSank Sep 06 '18 at 21:32