I read somewhere, that the solution to a real differential equation can always be chosen to be a real-valued function. Is this true? In particular, I am interested in the stationary Schrödinger equation: $$ -\frac{1}{2}\Delta f+Vf-Ef=0 $$ Here, $\Delta$ is the Laplacian, $V$ is a real-valued scalar field and $E$ is a scalar.
In the introduction of quantum mechanics, complex-valued functions often appear in the context of a particle on a ring or the hydrogen atom (i.e. spherical harmonics). Yet, for these systems, there are always multiple complex-valued functions, which have the same eigenvalue $E$. They thus span a vector space, and the basis of this space can also be chosen real without losing any information (cf. real-valued spherical harmonics).
Is there any case, where we cannot get rid of a complex-valued solution by a change of basis in some solution vector space?
If $f$ is a complex-valued solution of a real linear ODE, you can always just take the real part (or the imaginary part) of $f$, and this will also be a solution of the ODE. So the answer is yes. Also note, although you don't say it explicitly, $E$ must be real because the operator $−\frac{1}{2}Δ+V$ is self-adjoint.