Suppose we have a fermionic field operator $\Sigma_{s,s3}^{\pm}(x)$ with definite transformation properties under parity $\mathcal{P}$. $+$ and $-$ indicates the possible eigenvalues.

The operator $\Sigma^{\pm}_{s,s3}(x)$ will annhilate a particle with a certain parity $\mathcal{S}_1$ and create the antiparticle of its corresponding parity partner antiparticle $\mathcal{S}_\bar{2}$ (since $\Sigma$ has been built in with definite parity, both particles must have the same parity). For example, the proton $\mathcal{S}_1 \equiv P$ and the $\mathcal{S}_\bar{2} \equiv \bar{N^{*}}$, respectively.

Now, let's consider the corresponding two-point function $$C(x,y)=\langle VAC,T| \mathcal{T}\lbrace\Sigma(x)\bar{\Sigma}(y)\rbrace|0,VAC\rangle$$,

which can be expanded as (note that $|VAC\rangle$ has the same quantum numbers of the vacuum but it's not an eigenstate of the Hamiltonian)

$$C(x,y) =\theta(x^0-y^0)\langle VAC,T | \Sigma(x)\bar{\Sigma}(y) |0, VAC \rangle - \theta(y^0-x^0)\langle VAC,T | \bar{\Sigma}(y)\Sigma(x) |0, VAC \rangle. $$

This means that for $x^0 > y^0$, $\mathcal{S}_1$ propagates from $y$ to $x$, while for $x^0 < y^0$, $\mathcal{S}_\bar{2}$ propagates from $x$ to $y$. I was wondering, are there any circumstances (broken symmetries, boundary conditions, ecc) under which the second term $\langle VAC,T | \bar{\Sigma}(y)\Sigma(x) |0, VAC \rangle$ can be zero? i.e. the antiparticle of the parity partner doesn't propagates backward?