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Please help me to understand the following situation.

A fermion field $\Psi(x)$ acting on the vacuum can destroy a particle and create an antiparticle. If however, a chiral field $\Psi_L$ defined as $\Psi_L=\frac{1-\gamma_5}{2}\Psi$ acts on the vacuum, how does one argue that it creates a right-handed antiparticle. Please help me to show this.

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  • $\begingroup$ If you mean by "vacuum" the vacuum state, then it must be destroyed by an annihilation operator. If $\Psi$ is a creation operator, then $\Psi^\dagger$ is a better notation. $\endgroup$
    – Roger V.
    Commented Sep 16, 2020 at 16:58
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    $\begingroup$ @Vadim The question’s notation is standard. Field operators in quantum field theory are a mixture of annihilation operators for particles and creation operators for antiparticles. $\endgroup$
    – Buzz ♦
    Commented Sep 16, 2020 at 17:08
  • $\begingroup$ @Buzz you mean relativistic QFT. In condensed matter QFT operators can be defined differently - depending on how one describes holes. I suppose this is the key to the question: what is meant by "vacuum"? $\endgroup$
    – Roger V.
    Commented Sep 17, 2020 at 7:23
  • $\begingroup$ Here you can find the answer. $\endgroup$
    – Pipe
    Commented Sep 19, 2020 at 6:21

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