Interpretation of the free scalar propagator in terms of field measurement

Question

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 $\phi(x) \phi(y)$?)

Answer 1

"Hmm thanks. But intuitively, the correlation between two point of vacuumesic still should be zero?" This is your mistake. In order to have zero correlation you would have to have a vacuum state that has a much harsher spectrum to it, essentially some kind of white noise. The actual vacuum state, however, suppresses the high-energy modes which leads to a finite correlation length.

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 $$ \int \mathrm{d}^3x\,\mathrm{d}^3y\, \phi(\mathbf{x}) \phi(\mathbf{y}) (\nabla^2 - m^2) m\frac{K_1(m|\mathbf{x}-\mathbf{y}|)}{8\pi^2|\mathbf{x}-\mathbf{y}|}.$$

So, as you can see, the correlations are produced by the fact that the vacuum state suppresses higher energy fluctuations.