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So the transition amplitude for a free Klein-Gordon field for a space-like interval is finite and non-vanishing (decays exponentially). What does one make of this physically, i.e. what is the meaning of a quantum field leaking out of the light cone?

Qmechanic
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Albertus Magnus
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1 Answers1

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The posts linked in the comments give a good explanation of why the microcausality condition $[\phi(x),\phi^\dagger(y)]=0$ for spacelike $x-y$ is sufficient to make sure that only causal signals can propagate. But there are a few ways to think about the physical meaning of the exponential decay in the correlation function $\langle\phi(x)\phi^\dagger(y)\rangle$.

The first way is that even if $x,y$ are spacelike separated, they have a common history. That is, there is another point $z$ in the past of both $x,y$ such that $z-x$ and $z,y$ are timelike. So what happens at $z$ affects what happens at both $x,y$, and so the relevant correlation function is non-zero.

The second way is to realize that in the vacuum state $|0\rangle$, spatial subregions are highly entangled. If you try to split space into two regions $A,B$ and compute the entropy $S(\rho_A) = -\text{Tr}_A[\rho_A \ln \rho_A]$, where $\rho_A = \text{Tr}_B[|0\rangle\langle 0 |]$, you actually find that this is infinite, so the vacuum is extremely entangled. The technical way to phrase this is that observables in continuum QFT form a Type III von Neumann algebra. If spacelike regions were unentangled, then we actually would have that $$\langle0 |\phi(x)\phi^\dagger(y)|0\rangle = \langle 0 |\phi(x) |0\rangle\langle 0|\phi^\dagger(y) | 0 \rangle = 0$$

A third way to think about it is that the Hamiltonian density has a term in it that goes like $\mathcal{H} \sim |\nabla \phi|^2$, so it must be that $\phi(x)$ is correlated at spacelike separation in the vacuum. If this wasn't the case, then there could be sharp jumps in the profile of $\phi(x)$ in the vacuum, and this would produce a large energy. This is very related to the previous point about vacuum entanglement.

Finally, another perspective is that spacelike correlations of a field don't come from one particle, but from multiple identical particles. You might measure a particle at $x$ and your friend might measure another at $y$ to give a non zero answer for $\langle \phi(x)\phi(y)\rangle$, but that doesn't mean they were the same particle. It only says that they are indistinguishable, and come from excitations of the same field.

11zaq
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