The Free Particle in One Dimension
By a free particle, we mean a particle subject to no forces whatever. For a free particle, integration of (1.12) shows that the potential energy remains constant no matter what the value of $x$ is. Since the choice of the zero level of energy is arbitrary, we may set $V(x)=0$. The Schrödinger equation (1.19) becomes
\( \begin{equation} \frac{d^{2} \psi}{d x^{2}}+\frac{2 m}{\hbar^{2}} E \psi=0 \tag{2.29} \end{equation} \)
Equation (2.29) is the same as Eq. (2.10) (except for the boundary conditions). Therefore, the general solution of $(2.29)$ is (2.13):
\( \begin{equation} \psi=c{1} e^{i(2 m E)^{1 / 2} x / \hbar}+c{2} e^{-i(2 m E)^{1 / 2} x / \hbar} \tag{2.30} \end{equation} \)
What boundary condition might we impose? It seems reasonable to postulate (since $\psi^{*} \psi d x$ represents a probability) that $\psi$ will remain finite as $x$ goes to $\pm \infty$. If the energy $E$ is less than zero, then this boundary condition will be violated, since for $E<0$ we have
\( i(2 m E)^{1 / 2}=i(-2 m|E|)^{1 / 2}=i \cdot i \cdot(2 m|E|)^{1 / 2}=-(2 m|E|)^{1 / 2} \)
and therefore the first term in (2.30) will become infinite as $x$ approaches minus infinity. Similarly, if $E$ is negative, the second term in (2.30) becomes infinite as $x$ approaches plus infinity. Thus the boundary condition requires
\( \begin{equation} E \geq 0 \tag{2.31} \end{equation} \)
for the free particle. The wave function is oscillatory and is a linear combination of a sine and a cosine term [Eq. (2.15)]. For the free particle, the energy is not quantized; all nonnegative energies are allowed. Since we set $V=0$, the energy $E$ is in this case all kinetic energy. If we try to evaluate the arbitrary constants $c{1}$ and $c{2}$ by normalization, we will find that the integral $\int_{-\infty}^{\infty} \psi^{*}(x) \psi(x) d x$ is infinite. In other words, the free-particle wave function is not normalizable in the usual sense. This is to be expected on physical grounds because there is no reason for the probability of finding the free particle to approach zero as $x$ goes to $\pm \infty$.
The free-particle problem is an unreal situation because we could not actually have a particle that had no interaction with any other particle in the universe.