On page 298 of Shankar's 'Principles of Quantum Mechanics' the author makes the statement :
""In an arbitrary $\Omega$ basis, $\psi(\omega)$ need not be even or odd, even if $| \psi \rangle $ is a parity eigenstate. ""
Can anyone show me how this is the case when in the X basis the parity eigenfunctions $\psi(x)$ can only have even or odd parity?
Assuming ω the momentum, you obtain $\psi(\omega)$ by applying Fourier transformation to $\psi(x)$, thus changing the parity of the eigenfunction. If ω is not the momentum, if you want $\psi(\omega)$ having $\psi(x)$ you do the identity resolution: $\langle \omega \mid \psi \rangle=\psi(\omega)=\langle \omega \mid x \rangle \langle x \mid \psi \rangle=\langle \omega \mid x \rangle \psi(x)$. Now $\langle \omega \mid x \rangle$ is a function that we can call $\phi_x(\omega)$, which can be even, odd or none of the two, thus making the function $\psi(\omega)$ a function not even nor odd.
A good answer is already given by @Pleba, I'm sort of try to give a familiar example.
First In general, $$\Pi |\psi\rangle =\Pi \int |\omega\rangle \langle \omega |\psi\rangle dx$$ As work out by R. Shankar $$\langle \omega|\Pi|\psi\rangle =\psi(-\omega)$$ Let $\psi(-\omega)=\pm\psi(\omega)$ Then for some other basis $$\langle \alpha|\Pi|\psi\rangle =\psi(-\alpha)=\int\langle \alpha|\omega\rangle\langle\omega|\Pi|\psi\rangle d\omega=\int\langle \alpha|\omega\rangle\psi(-\omega) d\omega$$ $$\langle \alpha|\Pi|\psi\rangle=\pm\int \langle \alpha|\omega\rangle\psi(\omega) d\omega \stackrel{?}{=} \pm \phi(\alpha)$$ It depends on the overlap function $ \langle \alpha|\omega\rangle$ what's going to happen.
For example, consider $|\omega\rangle=|x\rangle$ and $|\alpha\rangle=|p\rangle$ with $\langle p|x\rangle \sim e^{-ipx/\hbar}$ which is not an odd or even function. So if $\psi(-x)=\pm \psi(x)$ You can not conclude that $\psi(-p)=\pm \psi(p).$
