# Missing parity of free particle [duplicate]

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 ?

• 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 $e^{\pm ikx}$. Mar 25, 2022 at 17:42
• Possible duplicate: physics.stackexchange.com/q/44003/2451 Mar 25, 2022 at 17:50