# Interpretation of the free scalar propagator in terms of field measurement

This is a pretty basic QFT question but I haven't found a discussion of it in depth enough for my taste in any textbook.

We want a relativistic theory to be causal, hence we want the commutator of field operators at different points of spacetime to vanish. This is the case for the free (real) scalar field, i.e. $$[\phi(x), \phi(y)] = 0 .$$ The commutator is often contrasted with the free scalar propagator $$D(x - y) = \langle 0 \mid \phi(x) \phi(y) \mid 0 \rangle$$ which is not zero for space-like intervals -- it appears as the amplitude to 'anihilate at $$y$$ a particle created at $$x$$' is non-zero -- particles can travels faster than time. Usually, this 'paradox' is resolved via the antiparticle interpretation -- in anything physically meaningfull (measurement) it is not possible to single out the case of particle going from $$x$$ to $$y$$ from (anti-)particle going from $$y$$ to $$x$$ (we can Lorentz transform to a frame exchanging the events in time for space-like intervals). Therefore, it is usually argued, in a measurement the amplitude of one of these cancels againts the amplitude of the other one (we can think of this as any possibly faster-than-light particle is anihilated by equally possible (anti-)particle going back in time).

So far, so good. But for a real scalar field, the field operators are themselves hermitian, hence measurable. This suggests that we can also think about the propagator $$D(x - y) = \langle 0 \mid \phi(x) \phi(y) \mid 0 \rangle$$ as measuring the vacuum by the operator $$\phi(x) \phi(y)$$. I see no reason for this to be non-zero in such an interpretation. Doesn't this operator correspond to just measuring the field (in the vacuum state) at the point $$x$$ and $$y$$ (at the same time if we are in a frame in which these are in the same time slice)?

(may be the operator of such interpretation should be $$\phi(x) + \phi(y)$$ but what is then the meaning of the operator $$\phi(x) \phi(y)$$?)

• It is correlation between field in two points. Oct 3, 2022 at 12:32
• Hmm thanks. But intuitively, the correlation between two point of vacuume(-field) still should be zero? Oct 3, 2022 at 12:35
• And I guess that what one can take from the fact that this is a correlation function is that, as 'correlation does not mean causation', the non-vanishing of the propagator is only a relic of existance of some intersection of the past light cones of $x$ and $y$? But it is still somewaht disturbing for the trivially-evolving vacuum? Oct 3, 2022 at 13:08
• There's nothing wrong with having correlations between fields at different points in a particular state. For example, if I yell, then a person to the left of me will hear something, and when they do they'll instantly know that a person to the right of me heard something too. But that's not a violation of relativity, that's just reflecting how I set up the state of the air. Oct 3, 2022 at 17:52
• What we want to avoid is changes in the state propagating faster than light. For example, if I measure at $x$ and get some field value, that's a physical operation at $x$ which changes the state. That had better not affect what somebody measures at $y$ faster than light. Oct 3, 2022 at 17:53

So, what is the vacuum state? In the wave-functional formalism, we can say that the vacuum state gets annihilated by the annihilation operator of a particular mode: \begin{align} a(\mathbf{k}) &= \sqrt{\frac{\sqrt{k^2 + m^2}}{2}} \left(\phi(\mathbf{k}) + \frac{i}{\sqrt{m^2 + k^2}} \pi(\mathbf{k})\right). \end{align} We ignore the normalization concerns needed to make the operator Lorentz invariant, because they aren't relevant here. We hit the wave functional with this operator, and set it equal to zero to find the vacuum state. \begin{align} a\Psi[\phi] &= \sqrt{\frac{\sqrt{k^2 + m^2}}{2}} \left(\phi + \frac{1}{\sqrt{m^2 + k^2}} \frac{\delta}{\delta \phi}\right)\Psi[\phi] \\ &= 0. \end{align} Formally, the above equation is satisfied if the wave functional is \begin{align} \Psi[\phi] &\propto \exp\left(-\int \mathrm{d}^3 k \, \frac{\sqrt{m^2+k^2}}{2} \phi^2(\mathbf{k})\right). \end{align}
I say formally because that integral will tend to diverge. You can use some Bessel function identities to show that the integral is $$\int \mathrm{d}^3x\,\mathrm{d}^3y\, \phi(\mathbf{x}) \phi(\mathbf{y}) (\nabla^2 - m^2) m\frac{K_1(m|\mathbf{x}-\mathbf{y}|)}{8\pi^2|\mathbf{x}-\mathbf{y}|}.$$