Suppose we have a fermionic field operator Σ±s,s3(x) with definite transformation properties under parity P. + and − indicates the possible eigenvalues.
The operator Σ±s,s3(x) will annhilate a particle with a certain parity S1 and create the antiparticle of its corresponding parity partner antiparticle Sˉ2 (since Σ has been built in with definite parity, both particles must have the same parity). For example, the proton S1≡P and the Sˉ2≡¯N∗, respectively.
Now, let's consider the corresponding two-point function C(x,y)=⟨VAC,T|T{Σ(x)ˉΣ(y)}|0,VAC⟩
,
which can be expanded as (note that
|VAC⟩ has the same quantum numbers of the vacuum but it's not an eigenstate of the Hamiltonian)
C(x,y)=θ(x0−y0)⟨VAC,T|Σ(x)ˉΣ(y)|0,VAC⟩−θ(y0−x0)⟨VAC,T|ˉΣ(y)Σ(x)|0,VAC⟩.
This means that for x0>y0, S1 propagates from y to x, while for x0<y0, Sˉ2 propagates from x to y. I was wondering, are there any circumstances (broken symmetries, boundary conditions, ecc) under which the second term ⟨VAC,T|ˉΣ(y)Σ(x)|0,VAC⟩ can be zero? i.e. the antiparticle of the parity partner doesn't propagates backward?