# Particle-antiparticle (a)symmetry and two-point function

+ 3 like - 0 dislike
317 views

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?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.