Going left or going right of the free particle wave function?

Question

In this excerpt from Griffith's quantum mechanics book:

THE FREE PARTICLE

We turn next to what should have been the simplest case of all: the free particle [$V(x) = 0$ everywhere]. As you'll see in a momentum, the free particle is in fact a surprisingly subtle and tricky example. The time-independent Schrödinger equation reads

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi\tag{2.74}$$

or

$$\frac{d^2\psi}{dx^2} = -k^2\psi,\quad\text{where } k \equiv \frac{\sqrt{2mE}}{\hbar}.\tag{2.75}$$

So far, it's the same as inside the infinite square well (Equation 2.17), where the potential is also zero; this time, however, I prefer to write the general solution in exponential form (instead of sines and cosines) for reasons that will appear in due course:

$$\psi(x) = Ae^{ikx} + Be^{-ikx}\tag{2.76}$$

Equation 2.76 is the solution for equation 2.74.

I don't understand why Griffiths says the first term in equation 2.76 represents a wave traveling to the right and the second to the left. Some website says when we apply momentum operator to first term then we get positive value in the positive x direction, but why do we separate two terms? If I apply momentum operator on equation 2.76, I can't get the wavefunction back. Since nobody know the physical meaning of wavefunction, then why do we simply separate it?

Answer 1

2.76 gives the general solution to the differential equation; there are two linearly independent solutions to a second order diff. eq., in this case the two solutions are $e^{ikx},e^{-ikx}$, and the general solution is a linear combination of them.

Note that the two solutions are eigenfunctions of $p$, and the eigenvalues differ by sign. So if we had the case where one of the $A,B$ were zero, we'd be dealing with a system with a specific momentum.

In the general case, both $A,B$ can be non-zero, in this case the state is a mixture of the two different momenta states.