Parity Eigenstates: X Basis Explanation

[Nitram]

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 eigenfunctions either have even or odd parity?

 

[vanhees71]

I need more context. What's $\omega$ and what is an $\Omega$ basis?

 

[Arne]

I need more context. What's $\omega$ and what is an $\Omega$ basis?

The section in the book is about the parity operator $\Pi$ which acts on the position basis as
\Pi |x\rangle = |-x\rangle
and with that also on the momentum eigenbasis
\Pi |p\rangle = |-p\rangle
The author then points out that the wavefunction in position or momentum space is mirrored under this operator:
\langle x | \psi \rangle = \psi (x) \overset{\Pi}{\longrightarrow} \langle x | \Pi |\psi \rangle = \psi (-x)
and similar the momentum space.

The operator $\Omega$ is just a general hermitian operator with eigenbasis $\{|\omega\rangle\}$. The author then states that the wavefunction in this basis $\psi(\omega) = \langle \omega | \psi \rangle$ does not have to be even or odd, even if $|\psi\rangle$ is a parity eigenstate, and leaves checking this statement to the reader.I'm not completely sure about this, but I thought about the harmonic oscillator (or any 1-particle Hamiltonian with a symmetric potential with bound states) as an example, with Hamiltonian
H = p^2/2m + m\omega^2 x^2.
The eigenstates of the Hamiltonian/energy-operator $|E_n\rangle$ are symmetric or antisymmetric in position space due to the symmetry of the potential. So for example we could look at a superposition of two symmetric eigenstates like $|E_0\rangle$ with energy $E_0 = \hbar \omega / 2$ and $|E_2\rangle$ with $E_2 = 5 \hbar \omega / 2$. The wavefunction in position space will be symmetric and an eigenstate of the parity operator. But trying something similar as above like constructing $|-E_0\rangle$ doesn't even really make sense, since (at least for this Hamiltonian) there aren't even eigenstates with a negative energy. So a state cannot be even or odd in the eigenbasis of the energy operator.

But I was also confused when reading this, because I was not completely sure what the author even meant with even or odd in the eigenbasis of an operator other than in momentum or position space. I personally think that even or odd in the case, at least in the case of the energy eigenbasis of the harmonic oscillator, isn't actually well defined, since I don't think just flipping the sign is a valid operation in this case. And I cannot think of other good possibilities what applying the parity operator might mean in this case, maybe switching some states or something like that.

I hope this helps. Best wishes,

ArneEdit: As a short answer one could maybe say that for example operators like $x^2$ or $p^2$ do not have negative eigenvalues and talking about even or odd wavefunctions or mirroring the wavefunction in their respective eigenbasis doesn't even really make sense, even though the eigenstates will also be eigenstates of the parity operator.