It's well known that $$ \langle 0| \phi(\vec x, t) \phi(\vec y, t) |0 \rangle \neq \delta(\vec x - \vec y) . $$ It is then regularly argued that this is not a big problem since the commutator $$ \langle 0| [\phi(\vec x, t), \phi(\vec x, t)]|0 \rangle $$ vanishes for $\vec x \neq \vec y$. (See, for example, page 37 here.) This is motivated by claiming that if for two operators $O_1( \vec x, t)$, $O_2( \vec y,t)$, the commutator $ [O_1(\vec x, t), O_2( \vec y,t)]$ vanishes, "this ensures that a measurement at $\vec x$ cannot affect a measurement at $\vec y$ when $\vec x$ and $\vec y$ are not causally connected."

While this argument certainly makes sense in the context of quantum mechanics, I'm failing to see how it applies to quantum field theory. The field operators are not measurement operators in the usual sense and thus I don't see how the field commutator is related to causality.

Instead, $\langle 0| \phi(\vec x, t), \phi(\vec y, t) |0 \rangle $ is the probability amplitude that we find a field excitation that we prepared at $\vec x$ at the same moment in time at some other location $\vec y$. In other words, "particle-like" field excitations are not completely localized in QFT. This doesn't look like a big deal to me in terms of causality. (At Wikipedia it's argued that this non-localizability is a result of the unavoidable vacuum fluctuations.)

Analogously, $\langle 0| [\phi(\vec x, t), \phi(\vec x, t)]|0 \rangle $ is the probability amplitude that we find a field excitation that we prepared at $\vec x$ at the same moment in time at some other location $\vec y$ minus the probability amplitude that we find a field excitation that we prepared at $\vec y$ at the same moment in time at some other location $\vec x$. In other words, the $\vec x \to \vec y$ amplitude and the $\vec y \to \vec x$ amplitude cancel exactly. (This doesn't seem very surprising if we believe in homogeneity and isotropy of spacetime.) How is this quantity related to causality?


2 Answers 2


I don't know what you mean by the fields not being "measurement operators in the usual sense", but they are the only operators you've got! You can build other operators out of them by addition, multiplication and integration, but QFT doesn't have any other operators. If you want to do measurements you better construct the measured operators out of the fields. And if you do that, then the vanishing commutator makes exactly as much sense as "in the context of quantum mechanics".

In the end, QFT is quantum mechanics, just with more degrees of freedom.


First off, it is incorrect to say that the field operators have nothing to do with measurements (or queries, as I like to call them). The field operators do represent physical parameters whose value can be queried. Namely, The operator

$$\hat{\phi}(t, x, y, z)$$

represents the physical parameter "how strong is the field at space-time coordinate $(t, x, y, z)$", which is very measurable alright: just think about holding an EMF meter up to that point at the given time.

The reason why that the commutator relations relate to causality is related to the fact that there are no passive measurements/queries in quantum theory - all must be active under pain of receipt of zero information received. This is because nontrivial commutation can be straightforwardly interpreted as an informational limit: the Universe only "stores" so much information when it comes to parameters taken together, and hence a gain in information of one when such is requested, must come at the expense of information from the other to avoid blowing that storage limit. And that is physically relevant: when you, say, measure the position of an electron in particle QM well enough, the attendant loss of momentum information necessitates a physical change, which can be seen by considering the statistics of subsequent measurements with enough repeated trials, and this means the measurement must be a physical interaction with real causal effects.

(R)QFT is no different - all the same principles of QM apply, they're just being applied to a different kind of quantum system. When a non-trivial commutator exists between two field parameters $\hat{\phi}(t_1, x_1, y_1, z_1)$ and $\hat{\phi}(t_2, x_2, y_2, z_2)$ at two different spatio-temporal locations, then they act the same way as the position and momentum of the electron in particle QM: suitably accurate measuring on one must, by the same principles, result in a physically relevant change in the other - an actual, physical event. But here, now, because of their denotation, that means actual physical events happening at two different places in space-time with real consequences, as would be seen in repeated trials. Hence, if those two quantities were/are space-like separated, i.e.

$$\Delta t < \frac{1}{c} \Delta s$$

then that would mean a physical, causal interaction between space-like separated points i.e. faster-than-light communication. This is, of course, a no-no for relativistic causality, hence such requires all those commutators be zero or, equivalently, that the joint information in such pairs be unlimited.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.