In solving the particle in a box, we required $\psi$ to be continuous. We now discuss other requirements the wave function must satisfy.
Since $|\psi|^{2} d \tau$ is a probability, we want to be able to normalize the wave function by choosing a suitable normalization constant $N$ as a multiplier of the wave function. If $\psi$ is unnormalized and $N \psi$ is normalized, the normalization condition (3.59) gives
\(
\begin{align}
1 & =\int|N \psi|^{2} d \tau=|N|^{2} \int|\psi|^{2} d \tau \
|N| & =\left(\int|\psi|^{2} d \tau\right)^{-1 / 2} \tag{3.93}
\end{align}
\)
FIGURE 3.4 Function (a) is continuous, and its first derivative is continuous. Function (b) is continuous, but its first derivative has a discontinuity. Function (c) is discontinuous.
The definite integral $\int|\psi|^{2} d \tau$ will equal zero only if the function $\psi$ is zero everywhere. However, $\psi$ cannot be zero everywhere (this would mean no particles were present), so this integral is never zero. If $\int|\psi|^{2} d \tau$ is infinite, then the magnitude $|N|$ of the normalization constant is zero and $\psi$ cannot be normalized. We can normalize $\psi$ if and only if $\int|\psi|^{2} d \tau$ has a finite, rather than infinite, value. If the integral over all space $\int|\psi|^{2} d \tau$ is finite, $\psi$ is said to be quadratically integrable. Thus we generally demand that $\psi$ be quadratically integrable. The important exception is a particle that is not bound. Thus the wave functions for the unbound states of the particle in a well (Section 2.4) and for a free particle are not quadratically integrable.
Since $\psi^{} \psi$ is the probability density, it must be single-valued. It would be embarrassing if our theory gave two different values for the probability of finding a particle at a certain point. If we demand that $\psi$ be single-valued, then surely $\psi^{} \psi$ will be single-valued. It is possible to have $\psi$ multivalued [for example, $\psi(q)=-1,+1, i$ ] and still have $\psi^{*} \psi$ single-valued. We will, however, demand single-valuedness for $\psi$.
In addition to demanding that $\psi$ be continuous, we usually also require that all the partial derivatives $\partial \psi / \partial x, \partial \psi / \partial y$, and so on, be continuous. (See Fig. 3.4.) Referring back to Section 2.2, however, we note that for the particle in a box, $d \psi / d x$ is discontinuous at the walls of the box; $\psi$ and $d \psi / d x$ are zero everywhere outside the box; but from Eq. (2.23) we see that $d \psi / d x$ does not become zero at the walls. The discontinuity in $\psi^{\prime}$ is due to the infinite jump in potential energy at the walls of the box. For a box with walls of finite height, $\psi^{\prime}$ is continuous at the walls (Section 2.4).
In line with the requirement of quadratic integrability, it is sometimes stated that the wave function must be finite everywhere, including infinity. However, this is usually a much stronger requirement than quadratic integrability. In fact, it turns out that some of the relativistic wave functions for the hydrogen atom are infinite at the origin but are quadratically integrable. Occasionally, one encounters nonrelativistic wave functions that are infinite at the origin [L. D. Landau and E. M. Lifshitz, Quantum Mechanics, 3rd ed. (1977), Section 35]. Thus the fundamental requirement is quadratic integrability, rather than finiteness.
We require that the eigenfunctions of any operator representing a physical quantity meet the above requirements. A function meeting these requirements is said to be well-behaved.