Commutators, Measurement, and Causality in QFT

Question

In Peskin and Schroeder pg. 27-28, they discuss Klein-Gordon theory and causality. For a spacelike separation $(x-y)^2 < 0$, they show that $$\langle 0| \phi(x)\phi(y) |0\rangle \neq 0$$

They go on to say that this doesn't actually break causality. Rather, what one should be looking at isn't whether particles can propagate over space-like intervals, but whether space-like separated measurements can affect one another. Hence, they argue that to understand the measurements of the field $\phi(x)$, one should be trying to understand the the commutator $[\phi(x),\phi(y)]$.

This last statement is opaque to me. In the QM setting, the ordering of the operators can be physically interpreted as one being applied first in time prior to the second. However, this doesn't make sense in the QFT context because one applies the operator at a specific point in space-time; flipping the order is not equivalent to changing the time-order of making the measurements.

1. In the QFT context, what is meant by the "measurement of a field," in analogy to the QM measurement of some operator?
2. Why is the commutator the object of choice when wanting to understand causality? What is the physical interpretation of the commutator here, particularly with respect to measurements of $\phi(x)$?

Answer 1

According to Weinberg in his text, the components of most quantum fields are not really measurable in any obvious way, so it's best not to think in those terms.

However, the fields do have to get commuted past each other when you evaluate the S-matrix, and then the Lorentz invariance of the S-matrix depends crucially on the fields commuting at space like separations.

Lots of aspects of the physical interpretation in QFT are at best subtle, and philosophically weak but plausible-sounding heuristic arguments are not uncommon. (You can already see people disagreeing about something so basic as whether $\phi$ creates a particle in the comments!) I found the early chapters in Peskin hard going for this exact reason- it's much better when you get to phenomenology and the physics is less opaque. If you want a book you can't argue with, try Weinberg- but this does come at the price of taking twice as long to cover the material, unfortunately in a rather idiosyncratic notation that makes it hard to dip in and out of.

Answer 2

This answer is to add some detail to @rwold's answer.

In particle physics experiments we measure the cross section for particle interactions which depends on the S-matrix. The LSZ (Lehmann-Symanzik-Zimmermann reduction formula), which relates the Lorentz invariant S-matrix elements $ \langle f| S|i\rangle$ for $n$ asymptotic momentum eigenstates to an expression involving the quantum fields $\phi(x)$: $$ \langle f|S |i\rangle =\left[i\int d^4x_1 \left(\square+m^2\right) e^{-i p_1 x_1}\right]\cdots \left[i\int d^4x_n\left(\square +m^2\right) e^{+i p_n x_n}\right]\\ \times \langle \Omega |T\left\{\phi\left(x_1 \right) \phi \left(x_2 \right) \phi\left(x_3 \right) \cdots \phi \left(x_n\right)\right\}|\Omega \rangle $$ The $T \{\cdots \}$ refers to the time ordered product and it indicates that all operators should be ordered so that those at later times are always on the left of those at earlier times. E.g. $T \left\{ \phi \left(x_1\right) \phi \left(x_2\right)\right\}=\phi\left(x_2 \right) \phi \left(x_1 \right)$ if $t_2>t_1$ regardless of whether $\phi \left(x_1 \right)$ and $\phi \left(x_2 \right)$ commute or not. However, if $x_1$ and $x_2$ are space like separated then one can change to a different frame which reverses the time ordering. I.e. if we could have $t_2>t_1$ in one frame and $t_1>t_2$ in another. Therefore, for the S-matrix to be Lorentz-invariant (i.e frame independent) we require that $[\phi(x_1),\phi(x_2)]=0$ when $x_1$ and $x_2$ are space like separated.