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E.g. For a finite square well that has the following potential: $$ V(x)= \begin{cases} 0, & |x|>a \\ -V_0, &|x|\leq a \end{cases} $$

When one will calculate the wavefunction $\psi(x)$ one can say that the wavefunction should be either even or odd for a bounded state since the potential is even. When I reading Griffiths "Introduction to quantum mechanics" he is not using this symmetrical argument for the scattering state. My question is why it is not possible to use this argument for the scattering state?

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  • $\begingroup$ In this case, solving the SE for $E>0$ yields solutions that are never square integrable; that is, it is always a non-normalizable state. For $E>0$ the spectrum is continuous. $\endgroup$ – Gert Sep 8 at 18:03
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TL;DR: The main point is that boundary conditions (BCs) are important, but they manifest themselves in different ways for bound and scattering states.

Fact: Recall that if we ignore BCs, then for a given arbitrary energy $E\in\mathbb{R}$ (of either sign!), the 1D TISE$^1$ is a 2nd-order homogeneous ODE. Its 2 linearly independent solutions can always be chosen to be both real. If furthermore the potential $V(x)=V(-x)$ is even, then the 2 linearly independent solutions can always be chosen to be even and odd, respectively, cf. e.g. this and this Phys.SE posts.

  1. Bound states $E<0$. If a solution $\psi(x)$ is a bounded function $\mathbb{R}\to \mathbb{C}$, it automatically vanishes at $x=\pm\infty$, so the BCs $\lim_{|x|\to \infty}\psi(x)=0$ are automatically satisfied. Not surprisingly, the above fact continuous to hold: We can choose $\psi$ to be real (and even/odd if the potential $V(x)=V(-x)$ is even).

  2. Scattering states $E>0$. There are generically no reasons why the incoming asymptotic left- & right-moving waves in the 1D scattering problem should display an even/odd symmetry, even if the potential $V(x)=V(-x)$ is even.

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$^1$We will in this answer assume that the potential vanishes asymptotically: $\lim_{|x|\to \infty}V(x)=0$.

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    $\begingroup$ It might be worth saying that the incoming state is asymmetric by assumption: we’re basically asserting the initial state is like a beam-on-target experiment. $\endgroup$ – dmckee Sep 8 at 20:38
  • $\begingroup$ $\uparrow$ Right. $\endgroup$ – Qmechanic Sep 8 at 21:01
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    $\begingroup$ Nevertheless, it's still worth emphasising that the potential does have both even and odd continuum eigenstates at every (suitably positive) energy. We just don't call those scattering states, which have a human requirement to be asymmetric. $\endgroup$ – Emilio Pisanty Sep 8 at 21:06
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Sep 20 at 8:54

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