Normal ordering in an interacting field theory

[SheikYerbouti]

I am using Mandl and Shaw and Lahiri to get an introduction to QFT. Something I haven't seen discussed much or at all is whether the commutators or anti-commutators vanish for the normal ordering of a product of different field operators. As a concrete example, let's say that I want to normal order [itex] \bar \psi^- \psi^+ \phi^-[/itex], where [itex]\psi[/itex] is fermionic and [itex]\phi[/itex] is bosonic. Does the term pick up a minus sign or not and why?

 

[eljose79]

Thank you for your post regarding the use of Mandl and Shaw and Lahiri as an introduction to quantum field theory (QFT). I would like to address your question about the commutators or anti-commutators in the normal ordering of a product of different field operators, specifically in the case of \bar \psi^- \psi^+ \phi^- where \psi is fermionic and \phi is bosonic.

Firstly, let me clarify the concept of normal ordering for those who may not be familiar with it. Normal ordering is a mathematical operation in QFT that rearranges the creation and annihilation operators in a product of field operators in such a way that all the creation operators are on the left and all the annihilation operators are on the right. This is important because it allows us to define a vacuum state, which is the state with no particles present, and to calculate the expectation values of physical observables.

Now, to address your specific question, the answer depends on the type of field operators involved. For bosonic field operators, the commutator vanishes, meaning that the normal ordering will not change the sign of the term. However, for fermionic field operators, the anti-commutator does not necessarily vanish and the normal ordering can result in a change of sign.

In the case of \bar \psi^- \psi^+ \phi^-, the term will pick up a minus sign if \psi^- and \psi^+ are fermionic fields. This is because the anti-commutator of two fermionic fields is equal to their product with a negative sign, while the commutator of two bosonic fields is equal to their product with no sign change. Therefore, the normal ordering of this term would result in a minus sign due to the anti-commutation of the fermionic fields.

I hope this helps to clarify your question. It is important to note that the choice of normal ordering is not unique and can vary depending on the specific situation and the desired outcome. I would recommend consulting your textbooks or seeking further guidance from your instructor or peers for a deeper understanding of normal ordering and its implications in QFT.