What are the eigenfunctions of Hamiltonian of a free particle?

Question

From my actual understanding of quantum physics observable are operators, when we measure some observable we will find an eigenvalue of such operator, and the system will collapse in the eigenstate.
The Hamiltonian is the operator related to energy, just like in classical mechanics, and Hamiltonian eigenvalues are, under some assumptions, the energy of a system.
When we have a free particle, the Hamiltonian is:

$$H=-\frac{h^{2}}{2m}\frac{\partial^2 }{\partial x^2}$$

So the eigenfunctions of the Hamiltonian should be the solutions to the second order linear equation:

$$-\frac{h^{2}}{2m}\frac{\partial^2 \psi}{\partial x^2} = E\psi$$

The solutions are a linear combination of $e^{ikx}$ and $e^{-ikx}$, and I'd expect them to be something like

$$\psi(x)=Ae^{ikx}+Be^{-ikx}$$

But different book I saw just give $$\psi(x)=Ae^{\pm ikx}$$ Which looks like mine just half of the solutions I thought. Am I missing something?

Answer 1

$ψ(x)=Ae^{ikx}+Be^{−ikx}$ is the correct general eigenfunction for a given eigenvalue $E$:

But there is the boundary condition that $ψ(x)$ must go to zero at plus and minus infinity. In order to follow it, $E$ can not be a fixed value, but a continuous interval $[E_1,E_2]$.

And $ψ(x)$ must be a Fourier integral.