2
votes
1answer
97 views

In QFT, why do fermions have to anticommute in order to insure causality?

I have seen this question and I believe I understand the answer to it. However, AFAIK, only for bosons the causality condition is a vanishing commutator. For fermions we expect the anticommutator ...
4
votes
1answer
126 views

A question about causality and Quantum Field Theory from improper Lorentz transformation

Related post Causality and Quantum Field Theory In Peskin and Schroeder's QFT p28, the authors tried to show causality is preserved in scalar field theory. Consider commutator $$ [ \phi(x), \phi(y) ...
4
votes
1answer
99 views

Causality in QFT from vanishing commutator and the EPR paradox

The question relates to this post. As shown in Peskin and Schroeder's introduction to quantum field theory p. 28., $$[\phi(x),\phi(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0$$, which implies the ...
1
vote
1answer
111 views

Observables still commute even if fields only anti-commute

In Peskin & Schroeder page 56, after introducing anti commutation relations for the fields instead of commutation relations (in order to fix the negative energy problem as well as to have proper ...
6
votes
2answers
414 views

Causality and Quantum Field Theory

I have a problem with proof of causality in Peskin & Schroeder, An Introduction to QFT, page 28. To avoid confusion I use three vectors notation, rewriting the Eq. (2.53) for $y=0$ as follows: ...
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votes
2answers
865 views

In QFT, why does a vanishing commutator ensure causality?

In relativistic quantum field theories (QFT), $$[\phi(x),\phi^\dagger(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0$$ On the other hand, even for space-like separation $$\phi(x)\phi^\dagger(y)\ne0.$$ ...
7
votes
1answer
266 views

Theories with non-vanishing commutators outside the lightcone

I'm reading Weinberg's new book on Quantum Mechanics, and in Chapter 8.7 "Time-Dependent Perturbation Theory" he derives the usual Dyson series for the $S$ matrix when the interaction Hamiltonian ...