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

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?

• I've transcribed the text of the excerpt for you, but keep in mind for the future that you shouldn't include text in an image. – David Z Nov 4 '13 at 3:06

## 1 Answer

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.

• $\psi = \frac{1}{\sqrt{2\pi} [\int_-\infty^0 \phi (k) e^{-ikx} dk + \int_0^\infty \phi (k) e^{ikx} ]$ – Outrageous Nov 4 '13 at 4:29
• Can't edit the first so I post second yet this is wrong also. Actually I want to write the linear combination of two wavefunction in terms of wave number and then integrate one from negative infinity to zero ,and the others integrate from zero to infinity. Correct to write so? Why is my Latex typing wrong here? Lacking of something? \psi(x)&=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{0}\phi(k)e^{-ikx}dp\+\int\limits_{0}^{infty}\phi(k)e^ikx}dp\\ – Outrageous Nov 4 '13 at 5:03
• @Outrageous : Correct Latex expression : \$\psi(x)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{0}\phi(k)e^{-ikx}dp +\int\limits_{0}^{+\infty}\phi(k)e^{ikx}dp\\ \$ – Trimok Nov 4 '13 at 10:19
• @Outrageous I'm not sure what you are getting at in these comments, but it seems like it might be a separate question. – Dave Nov 4 '13 at 14:31