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 ...
1
vote
1answer
164 views

Spinor irreducible reps of the Lorentz group and their algebra

Antisymmetric tensor of rank two can be connected with spinor formalism by the formula $$ M_{\mu \nu} = \frac{1}{2}(\sigma_{\mu \nu})^{\alpha \beta}h_{(\alpha \beta )} - \frac{1}{2}(\sigma_{\mu ...
3
votes
1answer
413 views

Commutator of Lorentz boost generators : visual interpretation

I have always struggled to visualize the correctness of the commutation relation for the generators of the boost in the Lorentz group. We have $$[K_i,K_j] = i \epsilon_{ijk} L_k$$ I fail to picture ...
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: ...
12
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 $$\phi(x)\phi^\dagger(y)\ne0.$$ ...