in this Definite Parity of Solutions to a Schrödinger Equation with even Potential? post in David Z's answer it's stated that the eigenfunctions have parity if the potential has parity/if $[H,P]=0$. In the case of a free particle the potential has parity ($V(x)=V(-x)=0$) and also the commutator vanishes since $$ [H,P]\Psi\propto\frac{\partial^2\Psi(-x)}{\partial x^2}-\frac{\partial^2\Psi(x)}{\partial x^2}\Bigg|_{x\rightarrow -x}=(-1)^2\Psi''(-x)-\Psi''(-x)=0 $$ But the general solution for the free particle $\Psi(x)=e^{ikx}$ with $k^2=\frac{2Em}{\hbar}$ has no parity: $$ \Psi(-x)=cos(-x)+isin(-x)=cos(x)-isin(x)=-\Psi(x)+2cos(x) $$ Why is that ?

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    $\begingroup$ It's not that every solution has definite parity, it's that you can choose a set of solutions which each have definite parity. In this case they're $\cos(kx)$ and $\sin(kx)$, not $e^{\pm ikx}$. $\endgroup$
    – knzhou
    Mar 25, 2022 at 17:42
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/44003/2451 $\endgroup$
    – Qmechanic
    Mar 25, 2022 at 17:50