Why is $\psi = e^{i(kx – \omega t)}$ a valid wavefunction since it isn't finitely integrable on $\Bbb R$?
I'm studying derivations of the Schrödinger Equation, which start with a simple wave function of the form $e^{i(kx – \omega t)}$, and derive the Schrödinger Equation from the time and space derivatives of $\psi$, together with relationships between $k$ and $\omega$ based on $E=\hbar \omega$ and $p=\hbar k$. The calculation is straightforward, but I'm bothered by the philosophy of using $\psi = e^{i(kx – \omega t)}$ as a candidate solution, since
$$
\int_{-\infty}^{+\infty} |\psi(x,t)|^2 dx = \int_{-\infty}^{+\infty} 1 dx = +\infty
$$
and so $\psi$ cannot be normalized. Doesn't that rule it out as a physically meaningful solution to SE? Or does that not matter?
Edit: Or is the point that in reality we're only really interested in wave packets of the form
$$
\int_{-\infty}^{+\infty} A(k) e^{i(kx – \omega t)} dk
$$
which can be made square-integrable? And so looking at $e^{i(kx – \omega t)}$ is just a mathematical convenience to save us from doing the calculation with integrals floating around?
Best Answer
So first of all, the interpretation of $|\Psi(\mathbf{r})|^2$ as a probability density function came after the development of wave mechanics by Schrödinger. The question Schrödinger tried to answer is: What is the wave equation for the particle-wave postulated/developed by de Broglie?
In my first (formal) introduction to quantum mechanics, my Professor put it this way: Wave mechanics is to classical mechanics, what wave optics is to ray optics. Because that is the idea with which Schrödinger got from the Hamilton-Jacobi formalism to his famous Schrödinger equation.
So the first thing to note is that the plane-wave $\mathbf{is}$ indeed a solution to the Schrödinger equation. Question is, how do we interpret this solution? Compare it to, lets say, the lowest energy state of the hydrogen atom. It may seem natural and intuitive to everyone nowadays what the intepretation of a square integrable function describing a bound state is, but that definitely hasn't always been the case.
Only after Born's posutalate, that the square modulus of the wave function should be interpreted as the probability density function of the position of the particle, we can make sense of such a solution to the Schrödinger equation.
To understand what the plane wave solution is in QM, maybe take a step back to wave optics. The plane wave is also a solution to the EM-wave equation. It fills out the whole space at all times. What's the physical interpretation there? Well, we find that due to the linearity of the (em) wave equation, and the fact that the collection of plane waves is a complete orthogonal set of $L^2(\mathbb{R^3})$, we can write every possible solution as a superposition of plane waves (also known as Fourier Transformation). And that is also the case for solutions to the Schrödinger equation.
To save the probability intepretation of $|\Psi(\mathbf{r})|^2$, we can look at it in the following way: Define the density $\rho(\mathbf{r}) = |\Psi(\mathbf{r})|^2$ and the current density $\mathbf{j}(\mathbf{r}) = \frac{\hbar}{2mi}\left(\Psi^*\nabla \Psi - \Psi\nabla\Psi^* \right)$. Then those two quantities fullfill a continuity equation:
$$\frac{\partial \rho}{\partial t} + \nabla \mathbf{j} = 0$$
So you can intepret the plane-wave solution (or superpositions of multiple plane-waves) as a flow of probability. That, in my mind, makes perfect sense for non-bound solutions to the Schrödinger equation, such as the plane-wave. For bound states we have Born (where it of course makes sense to require $\Psi \in L^2(\mathbb{R}^3)$, to have probabilites bound by 0 and 1), and for non-bound solution, we have this intepretation. This whole topic is called "scattering theory", where you describe the current of incoming particles as a plane wave.