- #1

Aleolomorfo

- 73

- 4

I have a question regarding the causality in QFT.

If I take into consideration a real scalar field and I calculate:

$$[\phi(x),\phi(y)] = 0 \space \space \space \space \space \text{if (x-y)}^2 < 0$$

Thanks to this relation we state that causality in QFT is preserved: a measurement in ##x## cannot talk to another one made in ##y##.

However, this is not the same of saying that the probability of going from ##x## to ##y## is ##0##, as a matter of fact it is ##D(x-y) \neq 0##. So the probability is not zero. I do not understand how we can link these two results coherently. From my perspective they are in contradiction (I know they are not, I'd like to understand why): the latter is what we call strictly causality and so it is broken.

Then usually books make another step forward. If I calculate the probability of going from ##y## to ##x## I found ##D(y-x)##. As long as ##x-y## is spacelike there is a continuos Lorentz transformation between ##x-y## and ##y-x## and since ##D(x)## is Lorentz invariant :

$$P(x \rightarrow y) = P(y \rightarrow y)$$

I understand the single steps, but I do not find the link between them. Thanks in advance!