Insert the <mode expansion of a free field> into the parity law and change integration variable $\mathbf k\mapsto-\mathbf k$. Independence of the plane waves gives
$$
\mathcal P a_{\mathbf k}\mathcal P^{-1}=\gamma_p a_{-\mathbf k},
\qquad
\mathcal P a_{\mathbf k}^\dagger\mathcal P^{-1}=\gamma_p a_{-\mathbf k}^\dagger.
$$
Since $\mathcal P|0\rangle=|0\rangle$, an $n$-particle <Fock state> therefore transforms as
$$
\mathcal P|\mathbf k_1,\ldots,\mathbf k_n\rangle
=(\gamma_p)^n|-\mathbf k_1,\ldots,-\mathbf k_n\rangle.
$$

Solved by gpt-5.6-sol high.

Solution

Paper 301 / 2 / b / Solution

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A Fock state has specified occupation numbers. Applying creation operators to the vacuum constructs multiparticle momentum eigenstates.


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