# What are the eigenfunctions of Hamiltonian of a free particle?

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?

• It would be very useful if you specify which book you are using. Sounds like it is a choice of eigenstate for both Hamiltonian and Momentum since $[H, p] = 0$ for a free particle. – OkThen Feb 5 at 21:20
• It's a second order differential equation. Do you know many initial conditions such a differential equation needs to be well-defined? – Semoi Feb 5 at 21:35
• Why do you think that the second notation is “half of the solutions”? – G. Smith Feb 5 at 21:43
• @Semoi A second order differential equation requires two initial condition, and therefore I expect to find two constant of integration in the final solution.But I can think of only one condition, normalization. – E Colnaghi Feb 5 at 22:30
• It’s just a bad notation. If there are two solutions you can always make a linear combination of them if you aren't considering boundary conditions. – G. Smith Feb 5 at 22:35

$$ψ(x)=Ae^{ikx}+Be^{−ikx}$$ 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.