# Solving the 1D time-independent Schroedinger's equation with boundary conditions at spatial infinity

In my introductory modern physics class we have examined time-independent solutions to the Schrödinger equation in 1 dimension. We looked at a few cases without finite boundary, e.g., free particles and step potentials with $V_{\mathrm{max}} < E$. In each example, in class and in all introductory material I've found, it was stated without mathematical justification that the reciprocal complex exponential solution should be ignored as "unphysical". Particularly, though a more mathematically complete solution would be of the form $$A\mathrm{e}^{\mathrm{i}\alpha x} + B\mathrm{e}^{-\mathrm{i}\alpha x},$$ I've only seen solutions in the unbounded region where $B=0$ under the interpretation that its term would be the result of some reflection, which can't occur if there is no reachable boundary. The usual statement runs, "the term in $B$ has a negative velocity and is unphysical..." I find this very unsatisfying. I'm not convinced it's proper to use the term "velocity" in this case (do they claim the time-independent wave equation "propagates"?), unless it's with respect to some aspect of probability current (which concept is not in the curriculum). An infinite boundary is an unphysical concept in itself. The solution isn't even square-integrable, as far as I can tell. I can't for the life of me find a mathematical justification, or more than a sentence of explanation (all relying, it seems, on implicit analogy to the physics of a string). It just comes across as arbitrary, or as an excuse to avoid a more complex treatment which I feel would be worthwhile.

Is there a justification for this assumption which can be written mathematically? It just feels like a trick to simplify an underdefined thought experiment.

Here is our interpretation of OP's question: We are essentially talking about the asymptotic form of positive energy scattering states to the time-independent Schrödinger equation (TISE) for the two free regions $x \to \pm \infty$. (As OP notes, scattering states are not normalizable, and therefore do not belong to the Hilbert space. Nevertheless they can be treated mathematically via a rigged Hilbert space formalism.) In one convention$^1$, we have asymptotically

$$\psi(x) ~\sim~ \left\{\begin{array}{rcl}~Ae^{-ikx} + Re^{ikx} & \text{for}& x\to +\infty, \cr\cr Te^{-ikx} & \text{for}& x\to -\infty.\end{array}\right.$$

In other words: We have three waves: an incoming left-mover, a transmitted ($T$) left-mover, and a reflected ($R$) right-mover.

1) What happened to the fourth possibility: an incoming right-mover $Be^{ikx}$ for $x\to -\infty$?

We have kept the fourth possibility zero as a boundary condition to imitate the scattering process of an incoming wave that splits into a reflected and a transmitted wave. (Of course, one could in principle also study scattering of two incoming waves, i.e. all four possibilities. This is e.g. done in this Phys.SE answer.)

2) How can we talk about incoming and outgoing left- and right-movers for the time-independent Schrödinger equation?

That's essentially the question asked in this Phys.SE post.

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$^1$ There is also a left-right mirrored convention.