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

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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 S1P 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(x0y0)VAC,T|Σ(x)ˉΣ(y)|0,VACθ(y0x0)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?

asked Feb 21, 2016 in Theoretical Physics by practical matter (50 points) [ revision history ]
reshown Oct 1, 2017 by practical matter

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