A parity-odd scalar interaction is rotationally invariant but changes sign under <spatial reflection>. It must contain a <pseudoscalar> contraction, such as a <Levi-Civita symbol> contracted with three spatial gradients.

Parity-odd scalar interaction

The first-order <in-in formalism> formula at observation time $\tau_0$ is
$$
\langle O(\tau_0)\rangle_{\lambda}
=i\int_{-\infty(1-i\epsilon)}^{\tau_0}d\tau\,
\langle[H_{\rm int}(\tau),O_I(\tau_0)]\rangle.
$$
Fourier transforming the three derivatives gives
$$
H_{\rm int}(\tau)
=-i\lambda a^4
\int_{\mathbf p_1\cdots\mathbf p_4}
(2\pi)^3\delta^{(3)}(\mathbf p_1+\cdots+\mathbf p_4)
\,[\mathbf p_2\mathbin\cdot(\mathbf p_3\mathbin\times\mathbf p_4)]
\prod_{a=1}^4\phi_a(\mathbf p_a,\tau),
$$
where $\int_{\mathbf p}=\int d^3p/(2\pi)^3$. Because the four species are distinct, <Wick contraction> pairs each vertex field with the external field of the same species and introduces no permutation factor. If
$$
\mathcal E=\mathbf k_2\mathbin\cdot(\mathbf k_3\mathbin\times\mathbf k_4),
\qquad k_T=k_1+k_2+k_3+k_4,
$$
then
$$
\langle H_{\rm int}(\tau)O(\tau_0)\rangle
=i\lambda a^4(\tau)(2\pi)^3\delta^{(3)}\!\left(\sum_a\mathbf k_a\right)
\mathcal E\prod_{a=1}^4f(k_a,\tau)f^*(k_a,\tau_0).
$$
Combining this with part i gives the requested <time integral>:
$$
\boxed{
\langle O(\tau_0)\rangle_{\lambda}
=2i\lambda(2\pi)^3\delta^{(3)}\!\left(\sum_a\mathbf k_a\right)\mathcal E
\int_{-\infty(1-i\epsilon)}^{\tau_0}d\tau\,a^4(\tau)
\operatorname{Re}\left[i\prod_{a=1}^4f(k_a,\tau)f^*(k_a,\tau_0)\right]}.
$$
The factor $\mathcal E$ is a <pseudoscalar>, so the resulting <primordial trispectrum> is parity odd and purely imaginary in this momentum-space convention.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 312 / 1 / ii / Solution

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A pseudoscalar is invariant under proper rotations but changes sign under a <spatial reflection>. In three dimensions the <scalar triple product> is a pseudoscalar.

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 Incoming links: Pseudoscalar